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en fr Convection-reaction-diffusion systems and interface dynamics Systèmes de convection-réaction-diffusion et dynamique d'interface

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Convection-reaction-diffusion systems and interface
dynamics
Matthieu Alfaro
To cite this version:
Matthieu Alfaro. Convection-reaction-diffusion systems and interface dynamics. Mathematics [math].
Universitж Paris Sud - Paris XI, 2006. English. <tel-00134258>
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N? D?ORDRE : 8392
UNIVERSITE PARIS XI
UFR Scientifique d?Orsay
THESE
pre?sente?e pour obtenir le grade de
Docteur en sciences de l?Universite? Paris XI Orsay, spe?cialite? mathe?matiques
par
Matthieu ALFARO
Sujet :
Syste?mes de convection-re?action-diffusion et dynamique d?interface
Soutenue le 28 septembre 2006 devant la Commission d?examen :
Mme
M.
M.
Mme
Mme
M.
Aline
Franc?ois
Bernard
Danielle
Elisabeth
Hiroshi
BONAMI
HAMEL
HELFFER
HILHORST
LOGAK
MATANO
rapporteur
rapporteur
pre?sident
directrice de the?se
examinateur
examinateur
Remerciements
Je tiens d?abord a? remercier ma directrice de the?se, Danielle Hilhorst. Elle m?a propose? de
nombreux sujets de recherche passionnants dans un domaine mathe?matique tre?s actif. Sa disponibilite?, son e?nergie et son optimisme furent d?un grand secours dans les moments de doute.
Enfin, je lui suis reconnaissant de m?avoir invite? a? de nombreux congre?s, notamment au Japon,
et permis d?entreprendre de stimulantes collaborations scientifiques avec des mathe?maticiens
e?trangers.
Je voudrais e?galement remercier ici le professeur Hiroshi Matano de l?universite? de Tokyo.
Ses ide?es, sa rigueur et sa recherche de la clarte? m?ont beaucoup apporte?. Nos e?changes au
jardin du Luxembourg, a? Cassis ou encore a? Yokohama ont e?te? tre?s fructueux. Son sens de
l?hospitalite? fut e?galement tre?s appre?ciable.
Aline Bonami et Franc?ois Hamel ont accepte? la lourde ta?che d?e?tre les rapporteurs de cette
the?se. J?appre?cie since?rement l?inte?re?t qu?ils portent a? mes travaux ainsi que leurs commentaires
avise?s. Je voudrais aussi exprimer ma gratitude a? Bernard Helffer pour avoir accepte? d?e?tre le
pre?sident du jury et a? Elisabeth Logak pour avoir accepte? d?en e?tre membre.
Re?mi Weidenfeld, dont j?ai en quelque sorte pris la rele?ve, a e?te? un soutien moral efficace ;
d?autre part j?ai plus appris sur les myste?res du TEX en deux heures avec Martin Vohral??k qu?en
plusieurs anne?es de travail autodidacte ! Merci a? eux et a? toute l?e?quipe d?Analyse Nume?rique
et Equations aux De?rive?es Partielles de l?Universite? de Paris-Sud.
Enfin, je voudrais dire a? ma famille et a? mes proches combien j?ai appre?cie? la patience et
l?affection dont ils ont fait preuve durant ces quatre anne?es.
J?aimerais de?dier cette the?se a? Ma??te?, a? mes co?te?s depuis le de?but, et a? Malou, pre?sente
depuis un peu moins longtemps !
Syste?mes de convection-re?action-diffusion et dynamique d?interface
Re?sume?
Le sujet de cette the?se est la limite singulie?re d?e?quations et de syste?mes d?e?quations paraboliques non-line?aires de type bistable, ou? intervient un petit parame?tre ?, avec des conditions
initiales ge?ne?rales. Nous obtenons une estimation nouvelle et optimale de l?e?paisseur et de la
localisation de la zone de transition.
Au Chapitre 1, nous e?tudions une e?quation d?Allen-Cahn ainsi qu?une famille de syste?mes
de re?action-diffusion, notamment le syste?me de FitzHugh-Nagumo et certains syste?mes pre?dateur-proie. Nous conside?rons d?abord l?e?quation d?Allen-Cahn. Nous montrons qu?a? partir
d?une condition initiale arbitraire, la solution devient rapidement proche d?une fonction en
escalier, sauf dans un petit voisinage de l?interface initiale, cre?ant ainsi une zone de transition
abrupte (ge?ne?ration de l?interface). Notre estimation du temps ne?cessaire au de?veloppement
de cette zone est optimale. Dans un deuxie?me temps, l?interface se de?place et la solution reste
proche de la fonction en escalier (de?placement de l?interface). Le de?placement de l?interface,
solution de la limite singulie?re du proble?me parabolique bistable, est induit par sa courbure
moyenne et par un terme de pression. Pour obtenir ces re?sultats, nous construisons deux
paires distinctes de sous- et sur-solutions : l?une pour de?montrer la proprie?te? de ge?ne?ration de
l?interface, et l?autre pour analyser le de?placement de l?interface. En imbriquant ces paires de
sous- et sur-solutions, nous estimons de fac?on optimale, d?une part l?e?paisseur de la zone de
transition, et d?autre part sa localisation. Ensuite, nous e?tendons nos re?sultats a? une classe
assez large de syste?mes de re?action-diffusion : comme nos preuves ne s?appuyent pas sur le
principe de comparaison, nous ne faisons pas d?hypothe?se de monotonie sur les termes de
re?action. L?ide?e est de conside?rer la premie?re e?quation du syste?me comme une perturbation
de l?e?quation d?Allen-Cahn ; les preuves s?appuient sur une le?ge?re modification des re?sultats
pour l?e?quation seule, sur une e?tude de la de?pendance du de?placement de l?interface vis-a?-vis
de diffe?rents parame?tres, et sur de fines estimations a priori.
Le Chapitre 2 est consacre? a? l?e?tude d?un syste?me chimiotactique, dont la premie?re e?quation
est parabolique et non-line?aire, alors que la seconde e?quation est elliptique et line?aire. Il s?agit
d?un mode?le pour une agre?gation d?amibes soumises a? trois effets : la diffusion, la croissance
et le chimiotactisme. Ce dernier phe?nome?ne est une propension de certaines espe?ces a? se
de?placer vers les plus forts gradients de substances chimiques, souvent produites par ces espe?ces
elles-me?mes. En e?tudiant successivement la ge?ne?ration et le de?placement de l?interface, nous
obtenons des estimations optimales de l?e?paisseur de la zone de transition et de sa localisation.
Enfin, au Chapitre 3, nous conside?rons une e?quation quasi-line?aire anisotrope de type AllenCahn, qui intervient en science des mate?riaux et dont le terme de diffusion est inhomoge?ne
et singulier aux points ou? le gradient de la solution s?annule. Nous de?finissons une notion de
solution faible et prouvons un principe de comparaison. Le de?placement de l?interface limite
est induit par une version anisotrope de sa courbure moyenne. Nous effectuons l?analyse en
utilisant la distance associe?e a? une me?trique de Finsler. Nous e?tudions la ge?ne?ration et le
de?placement de l?interface, obtenant une estimation optimale de l?e?paisseur de la zone de
transition.
Mots cle?s : Syste?mes de convection-re?action-diffusion ? Equation d?Allen-Cahn ? Syste?me
de FitzHugh-Nagumo ? Chimiotactisme ? Anisotropie ? Ge?ne?ration d?interface ? Propagation
d?interface ? Epaisseur d?interface.
AMS subject classifications : 35K57, 35K60, 35K50, 35K20, 35R35, 35B20.
Convection-reaction-diffusion systems and interface dynamics
Abstract
This thesis deals with the singular limit of systems of parabolic partial differential equations
involving a small parameter ?, with bistable nonlinear reaction terms and general initial data.
We obtain a new and optimal estimate of the thickness and the location of the transition layer
that develops.
In Chapter 1, we study a perturbed Allen-Cahn equation and a class of reaction-diffusion
systems, which includes the FitzHugh-Nagumo system and some prey-predator systems. We
first consider the case of the single equation. We show that, leaving from arbitrary initial data,
the solution quickly becomes close to a step function, except in a small neighborhood of the
initial interface, creating a steep transition layer (generation of interface). Our estimation of
the time needed to develop such a transition layer is optimal. In the second stage, the interface
starts to move, and the solution remains close to the step function (motion of interface). The
motion of the interface, solution of the singular limit of the original problem, is driven by its
mean curvature and a pressure term. To prove these results, we construct two completely
different pairs of sub- and super-solutions: one for the generation of interface, and the other
for the motion of interface. Fitting these pairs of sub- and super-solutions into each other, we
estimate, in an optimal way, the thickness of the transition layer, and its location. Then, we
extend our results to a large class of reaction-diffusion systems: since our proofs do not rely on
the comparison principle, we do not make any monotony assumptions on the reaction terms.
The idea is to regard the first equation of the system as a perturbed Allen-Cahn equation;
the proofs rely on a slight modification of the results for the single equation, a study of the
dependence of the interface motion on various parameters together with some refined a priori
estimates.
Chapter 2 is devoted to the study of a chemotaxis system, where the first equation is
parabolic and nonlinear, whereas the second equation is elliptic and linear. This is a model
for an aggregation of amoebae which are subjected to three effects: diffusion, growth and
chemotaxis. This last phenomenon is a tendency of some species to move towards higher gradients of chemical substances which they often produce themselves. By successively studying
the generation and the motion of interface, we obtain here as well optimal estimates of the
thickness of the transition layer and of its location.
Finally, in Chapter 3, we consider a quasi-linear anisotropic Allen-Cahn equation, which
arises for instance in material sciences, and whose diffusion term is spatially inhomogeneous
and singular in the points where the gradient of the solution vanishes. We define a notion of
weak solution and prove a comparison principle. The motion of the limit interface is driven by
its anisotropic mean curvature. We perform the analysis using the distance function associated
with a Finsler metric related to the anisotropic diffusion term. We study both the generation
and the motion of interface and obtain an optimal estimate of the thickness of the transition
layer.
Key words: Convection-reaction-diffusion systems ? Allen-Cahn equation ? FitzHugh-Nagumo system? Chemotaxis ? Anisotropy ? Generation of interface ? Motion of interface ? Thickness of interface.
AMS subject classifications: 35K57, 35K60, 35K50, 35K20, 35R35, 35B20.
Table des matie?res
Introduction
9
1 The singular limit of the Allen-Cahn equation and the FitzHugh-Nagumo
system
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Perturbed Allen-Cahn equation . . . . . . . . . . . . . . . . . . . . . . .
1.1.2 Singular limit of reaction-diffusion systems . . . . . . . . . . . . . . . .
1.2 Formal derivation of the interface motion equation . . . . . . . . . . . . . . . .
1.3 Generation of interface: the case g ? ? 0 . . . . . . . . . . . . . . . . . . . . . .
1.3.1 The bistable ordinary differential equation . . . . . . . . . . . . . . . . .
1.3.2 Construction of sub- and super-solutions . . . . . . . . . . . . . . . . . .
1.3.3 Proof of Theorem 1.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.4 Optimality of the generation time . . . . . . . . . . . . . . . . . . . . .
1.4 Generation of interface in the general case . . . . . . . . . . . . . . . . . . . . .
1.4.1 The perturbed bistable ordinary differential equation . . . . . . . . . . .
1.4.2 Construction of sub- and super-solutions . . . . . . . . . . . . . . . . . .
1.4.3 Proof of Theorem 1.3.1 for the general case . . . . . . . . . . . . . . . .
1.5 Motion of interface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.1 A modified signed distance function . . . . . . . . . . . . . . . . . . . .
1.5.2 Construction of sub- and super-solutions . . . . . . . . . . . . . . . . . .
1.5.3 Proof of Lemma 1.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Proof of the main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.1 Proof of Theorem 1.1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.2 Proof of Theorem 1.1.6 . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Application to reaction-diffusion systems . . . . . . . . . . . . . . . . . . . . . .
1.7.1 Preliminaries: global existence . . . . . . . . . . . . . . . . . . . . . . .
1.7.2 Re-examination of the Allen-Cahn equation . . . . . . . . . . . . . . . .
1.7.3 Interface motion under various perturbations . . . . . . . . . . . . . . .
1.7.4 Proof of the main results . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7.5 Proof of Lemmas 1.7.5 and 1.7.6 . . . . . . . . . . . . . . . . . . . . . .
17
18
18
23
26
31
32
36
38
39
40
40
43
44
46
46
47
48
52
52
54
56
56
57
59
62
68
2 The singular limit of a chemotaxis-growth system with general
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Some preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Formal derivation . . . . . . . . . . . . . . . . . . . . . . . .
2.2.2 A comparison principle . . . . . . . . . . . . . . . . . . . .
2.3 Generation of interface . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 The perturbed bistable ordinary differential equation . . . .
71
72
75
75
78
78
79
initial
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
data
. . .
. . .
. . .
. . .
. . .
. . .
8
Table des matie?res
2.4
2.5
2.3.2 Construction of sub and super-solutions
2.3.3 Proof of Theorem 2.3.1 . . . . . . . . .
Motion of interface . . . . . . . . . . . . . . . .
2.4.1 Construction of sub- and super-solutions
2.4.2 Proof of Lemma 2.4.1 . . . . . . . . . .
2.4.3 Proof of Lemma 2.4.2 . . . . . . . . . .
Proof of the main results . . . . . . . . . . . .
2.5.1 Proof of Theorem 2.1.3 . . . . . . . . .
2.5.2 Proof of Theorem 2.1.5 . . . . . . . . .
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3 The singular limit of a spatially inhomogeneous and anisotropic
equation
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Finsler metrics and the anisotropic context . . . . . . . . . . . . .
3.2.1 Finsler metrics . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2 Application to the anisotropic Allen-Cahn equation . . . . .
3.3 Formal derivation of the interface motion equation . . . . . . . . .
3.4 A comparison principle . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Generation of interface . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1 The bistable ordinary differential equation . . . . . . . . . .
3.5.2 Construction of sub- and super-solutions . . . . . . . . . . .
3.5.3 Proof of Theorem 3.5.1 . . . . . . . . . . . . . . . . . . . .
3.6 Motion of interface . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6.1 Construction of sub and super-solutions . . . . . . . . . . .
3.6.2 Proof of Lemma 3.6.1 . . . . . . . . . . . . . . . . . . . . .
3.7 Proof of the main results . . . . . . . . . . . . . . . . . . . . . . . .
Bibliographie
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83
85
87
87
88
93
96
97
98
Allen-Cahn
101
. . . . . . . 102
. . . . . . . 106
. . . . . . . 106
. . . . . . . 107
. . . . . . . 110
. . . . . . . 112
. . . . . . . 115
. . . . . . . 115
. . . . . . . 120
. . . . . . . 121
. . . . . . . 123
. . . . . . . 123
. . . . . . . 124
. . . . . . . 128
131
Introduction
L?objet de cette the?se est l?e?tude d?e?quations aux de?rive?es partielles paraboliques nonline?aires qui interviennent, par exemple, en biologie et en sciences des mate?riaux. Lorsque le
coefficient du terme de diffusion est petit ou lorsque celui du terme de re?action est grand, ces
proble?mes peuvent donner lieu a? des zones de transition abruptes, aussi appele?es interfaces,
entre les diffe?rents e?tats que peut atteindre la solution. Une e?quation mode?le est donne?e par
l?e?quation d?Allen-Cahn :
1
ut = ?u + 2 (u ? u3 ),
?
ou? intervient le petit parame?tre ? > 0. Dans un premier temps, dit de ge?ne?ration de l?interface,
le terme de diffusion ?u peut e?tre ne?glige? devant le terme de re?action ??2 (u?u3 ). Dans l?e?chelle
de temps ? = t/?2 , la solution de l?e?quation d?Allen-Cahn u? se comporte comme la solution
de l?e?quation diffe?rentielle ordinaire u? = f (u) ; ainsi, les valeurs de u? deviennent rapidement
proches de l?un des deux e?quilibres stables 1 ou ?1 et une zone de transition se de?veloppe entre
les deux re?gions {u? ? 1} et {u? ? ?1}. A son voisinage, le terme de diffusion ne peut plus
e?tre ne?glige? et sa combinaison avec le terme de re?action induit, dans un deuxie?me temps, un
de?placement de l?interface. On sait que l?e?paisseur de la zone de transition est lie?e au parame?tre
?.
De nombreux travaux ont porte? sur le comportement asymptotique de l?e?quation d?AllenCahn. En 1979, les physiciens Allen et Cahn [2] obtiennent, par analyse formelle, l?e?quation
du proble?me a? frontie?re libre limite : l?interface se de?place selon sa courbure moyenne. Nous
renvoyons e?galement aux travaux de Kawasaki et Ohta [52], en 1982.
Sous l?hypothe?se que l?interface initiale est une hypersurface, il y a existence et unicite?,
locales en temps, de la solution du proble?me a? frontie?re libre associe?. La convergence vers la
solution classique sur son intervalle de temps d?existence est de?montre?e au de?but des anne?es
90. Citons, par exemple, les re?sultats de Bronsard et Kohn [18], dans le cas de la syme?trie
sphe?rique, de de Mottoni et Schatzman [59, 60] et de X. Chen [20, 21].
En ge?ne?ral, la solution du proble?me limite devient singulie?re en temps fini et il est ne?cessaire
de conside?rer des solutions faibles. Dans les anne?es 90, la notion de solution de viscosite? est
introduite, notamment par Y.G. Chen, Giga et Goto [25]. Puis, la convergence vers la solution
de viscosite? est de?montre?e, notamment par Barles, Soner et Souganidis [6], Evans, Soner et
Souganidis [33], Ilmanen [48] et Barles et Souganidis [7].
Notons que l?inte?grale de la fonction non-line?aire f (u) = u ? u3 entre les deux e?quilibres
stables 1 et ?1 est nulle. Si l?on perturbe cette fonction non-line?aire par un terme d?ordre ?,
de?pendant des variables d?espace, de temps et de la fonction u, alors un terme supple?mentaire
intervient dans l?e?quation du proble?me a? frontie?re libre associe?. Ce re?sultat a e?te? obtenu de
manie?re formelle par Rubinstein, Sternberg et Keller [66] de?s 1989. En 1997, Ei, Iida et Yanagida [30], de?montrent la convergence vers la solution du proble?me a? frontie?re libre limite en
supposant que la condition initiale a une zone de transition de?ja? bien de?veloppe?e dont le profil
10
Introduction
de?pend de ?.
Cette the?se porte sur la limite singulie?re d?e?quations ou syste?mes de convection-re?actiondiffusion qui font intervenir des termes inhomoge?nes et anisotropes, et qui e?tendent l?e?quation
d?Allen-Cahn. Nous de?montrons des proprie?te?s de ge?ne?ration et de de?placement de l?interface ;
l?e?tude de la ge?ne?ration permet de conside?rer une condition initiale tre?s ge?ne?rale. D?autre part,
nous montrons que l?e?paisseur de la zone de transition est d?ordre ?. Nous localisons e?galement
de manie?re optimale l?ensemble des points ou? la solution a pour valeur l?e?quilibre instable
du terme non-line?aire ; plus pre?cise?ment, nous de?montrons que sa distance de Hausdorff a?
l?interface solution du proble?me a? frontie?re libre limite est e?galement d?ordre ?.
Chapitre 1 : Limite singulie?re de l?e?quation d?Allen-Cahn et du
syste?me de FitzHugh-Nagumo
Ce Chapitre fait l?objet d?un article e?crit en collaboration avec D. Hilhorst (Universite?
de Paris-Sud) et H. Matano (Universite? de Tokyo), soumis prochainement pour publication
dans Journal of Differential Equations.
Il comporte deux parties. D?abord nous e?tudions une e?quation d?Allen-Cahn dont le terme
non-line?aire est perturbe?. Ensuite, en exploitant notre e?tude de cette e?quation, ainsi que des
estimations a priori supple?mentaires, nous e?tendons nos re?sultats a? des syste?mes de re?actiondiffusion, notamment au syste?me de FitzHugh-Nagumo et a? certains syste?mes de type pre?dateur-proie.
L?e?quation d?Allen-Cahn perturbe?e
Nous conside?rons le proble?me parabolique non-line?aire
?
1
?
?
ut = ?u + 2 (f (u) ? ?g ? (x, t, u)) dans ? О (0, +?),
?
?
?
?
?u
=0
sur ?? О (0, +?),
?
?
??
?
?
?
dans ?,
u(x, 0) = u0 (x)
(1)
ou? ? est un ouvert borne? et re?gulier de RN , N ? 2. La fonction non-line?aire f admet exactement trois ze?ros ?? < a < ?+ , le caracte?re bistable e?tant assure? par des pentes strictement
ne?gatives aux e?quilibres stables u = ?? , ?+ , et strictement positive a? l?e?quilibre instable u = a ;
d?autre part, nous supposons que l?inte?grale de f entre ?? et ?+ s?annule et que la fonction
g ? est de la forme
g ? (x, t, u) = g(x, t, u) + O(?).
(2)
Le proble?me a? frontie?re libre limite est donne? par
?
Z ?+
?
? Vn = ?(N ? 1)? + c0
g(x, t, r)dr
?
? ? »»
t
t=0
??
sur ?t ,
(3)
= ?0 ,
ou? l?interface initiale ?0 , de?finie par ?0 := {x ? ?, u0 (x) = a}, est une hypersurface sans
bords re?gulie?re ; Vn de?signe la vitesse de de?placement de l?interface le long de la normale, ?
la courbure moyenne de l?interface et c0 une constante lie?e a? la fonction non-line?aire f . Ce
proble?me posse?de une solution classique unique ?t sur un intervalle de temps [0, T ].
Introduction
11
Les re?sultats essentiels de cette partie sont les suivants. Lorsque ? ? 0, la solution classique u? du proble?me (1) converge vers ?? ou ?+ , selon que l?on se trouve a? l?inte?rieur ou
a? l?exte?rieur de l?interface, sur l?intervalle de temps (0, T ]. De plus, l?e?paisseur de la zone
de transition est d?ordre ?, ainsi que la distance de Hausdorff entre l?ensemble des points
??t := {x ? ?, u? (x, t) = a}, et l?interface ?t , solution du proble?me a? frontie?re libre limite (3).
Ces estimations sont optimales.
Concernant la condition initiale, nos hypothe?ses sont peu restrictives. Nous de?montrons
d?abord une proprie?te? de ge?ne?ration d?interface. Dans ce but, nous construisons une paire
de sous- et sur-solutions base?es sur la solution d?une e?quation diffe?rentielle de la forme Y? =
f (Y )+O(?), obtenue en ne?gligeant la diffusion et en travaillant dans l?e?chelle de temps ? = t/?2 .
Par le principe de comparaison, nous de?montrons alors que, apre?s un temps t? d?ordre ?2 | ln ?|,
la solution a de?ja? de?veloppe? une zone de transition tre?s abrupte, autour de l?interface initiale.
Nous montrons e?galement que ce temps de ge?ne?ration t? est optimal, c?est-a?-dire qu?une zone
de transition escarpe?e ne peut s?e?tre de?veloppe?e avant.
Dans un deuxie?me temps, pour de?montrer une proprie?te? de de?placement d?interface, nous
construisons des sous- et sur-solutions base?es, cette fois, sur les deux premiers termes du
de?veloppement asymptotique formel de la solution, qui sont les solutions d?un proble?me stationnaire unidimensionnel associe? et de sa version line?arise?e.
Par imbrication des deux paires de sous- et sur-solutions construites, nous obtenons nos
principaux re?sultats.
Syste?mes de re?action-diffusion
Nous e?tendons ensuite les re?sultats que nous avons obtenus pour l?e?quation seule a? des
syste?mes de re?action-diffusion de la forme
?
? ut = ?u + 1 f ? (u, v) dans ? О (0, +?),
?2
(4)
?
dans ? О (0, +?),
vt = D?v + h(u, v)
ou? le terme de re?action f ? est donne? par
f ? (u, v) = f (u) + ?f1 (u, v) + ?2 f2? (u, v),
avec des conditions aux limites de Neumann homoge?nes et des conditions initiales. Nos preuves
ne s?appuyant pas sur un principe de comparaison, nous ne faisons aucune hypothe?se de monotonie sur les termes de re?action non-line?aires f ? et h. A l?aide de la me?thode des rectangles
invariants, nous de?montrons que la solution (u? , v ? ) existe pour t ? 0.
Ce syste?me contient deux cas particuliers importants, a? savoir le syste?me de FitzHughNagumo qui mode?lise la transmission nerveuse :
?
1
?
? ut = ?u + (f (u) ? ?v),
?2
(5)
?
? v = D?v + ?u ? ?v,
t
et certains syste?mes pre?dateur-proie intervenant en e?cologie :
?
? ut = ?u + 1 А(1 ? u)(u ? 1/2) ? ?v бu,
?2
?
vt = D?v + (?u ? ?v)v.
(6)
12
Introduction
Le proble?me a? frontie?re libre limite est constitue? d?une e?quation de de?placement d?interface
couple?e a? une e?quation parabolique :
Z ?+
(7a)
Vn = ?(N ? 1)? ? c0
f1 (r, v?(x, t))dr sur ?t ,
??
v?t = D?v? + h(u?, v?)
dans ? О (0, T ],
(7b)
ou? la fonction en escalier u?, valant ?? a? l?inte?rieur de l?interface et ?+ a? l?exte?rieur, est comple?tement de?termine?e par l?interface ?t . Ce proble?me admet une solution classique unique (?, v?)
sur un intervalle de temps [0, T ].
Nous de?montrons que nos estimations de l?e?paisseur et de la localisation de la zone de
transition de la solution u? restent vraies pour les syste?mes de re?action-diffusion conside?re?s.
Pour cela, nous conside?rons la premie?re e?quation du syste?me (4) comme une e?quation d?AllenCahn perturbe?e et cherchons a? appliquer les re?sultats obtenus pour l?e?quation seule. Ceci
ne?cessite de prouver l?analogue de (2), c?est-a?-dire l?estimation a priori
v ? (x, t) = v?(x, t) + O(?).
(8)
La difficulte? tient au fait qu?ici, le terme de perturbation de?pend de la fonction v ? , dont le
comportement n?est pas parfaitement connu. De plus, la solution u? convergeant vers la fonction
discontinue u?, la fonction v? pre?sente un de?ficit de re?gularite? face a? v ? ; plus pre?cise?ment, par
les estimations paraboliques, v ? est au moins de classe C 2,1 alors que v? est seulement de classe
1+?
C 1+?, 2 . L?ide?e de la de?monstration est la suivante : on re?examine d?abord l?e?quation pour le
de?placement de l?interface (3) en perturbant le terme non-line?aire g ainsi que l?interface initiale
?0 . Ensuite, tout en accordant une certaine liberte? aux conditions initiales, on construit une
application ? comme suit : a? tout v, on fait correspondre un terme f1 de perturbation du
terme non-line?aire, et donc une interface solution d?une e?quation de la forme (7a) ; a? cette
interface, on associe une fonction en escalier, et donc une solution d?une e?quation de la forme
(7b), note?e ?[v]. Par construction, la solution (?, v?) du syste?me (7a)?(7b) est telle que v? est
un point fixe de ?. En se basant sur des estimations de la solution fondamentale de l?e?quation
de la chaleur vt = D?v, et sur le fait que u? varie peu en dehors d?un voisinage d?ordre ? de
l?interface ?t , on de?montre que ?[v ? ] = v ? + O(?), ou encore que la fonction v ? est presque un
point fixe de ?. Ceci, combine? au fait que ? est contractante, nous permet alors de de?montrer
l?estimation essentielle (8), de laquelle de?coulent les re?sultats pour les syste?mes.
Chapitre 2 : Limite singulie?re d?un syste?me de chimiotactismecroissance avec condition initiale quelconque
Cette partie fait l?objet d?un article soumis pour publication dans Advances in Differential
Equations.
Les Dictyostelides sont des organismes pouvant prendre alternativement une forme unicellulaire (amibe) ou une forme pluricellulaire. On les trouve dans les tapis de feuilles en
de?composition. On a observe? chez eux un cycle de vie assez complexe. Dans un premier temps,
les amibes se dispersent et se nourrissent de bacte?ries. Lorsque ces dernie?res ont toutes e?te?
consomme?es, les amibes e?mettent un attracteur, dit chimiotactique, de fac?on a? attirer les
amibes voisines. Par agre?gation, il se forme un organisme pluricellulaire de centaines de milliers de cellules, sorte de limace de quelques millime?tres de longueur. Cet organisme est compose?
de trois parties, un disque basal, un pied et une masse de spores qui donnent naissance a? de
nouvelles amibes.
Introduction
13
L?e?tude des me?chanismes qui sous-tendent de tels phe?nome?nes ge?ne?rant un organisme pluricellulaire est d?un grand inte?re?t en biologie. En 1970, Keller et Segel [53] ont propose? le
syste?me d?e?quations paraboliques
(
ut = du ?u ? ? и (u??(v)),
(9)
? vt = dv ?v + u ? ?v,
pour la mode?lisation mathe?matique de ce processus d?agre?gation ; la fonction u repre?sente la
concentration d?amibes et v celle de l?attracteur chimiotactique, dont le taux de de?gradation est
donne? par la constante positive ? ; du et dv sont des coefficients de diffusion suppose?s constants ;
? est une constante positive ; la fonction strictement croissante ? exprime l?attraction des
amibes par la substance chimiotactique. Les amibes sont ainsi soumises a? deux phe?nome?nes :
la diffusion et le chimiotactisme, c?est-a?-dire une propension a? se diriger vers la substance
attractrice qu?elles ont elles-me?mes se?cre?te?e. Le proble?me est comple?te? par des conditions
initiales ainsi que des conditions aux limites de Neumann homoge?nes.
De nombreuses analyses mathe?matiques de ce mode?le ont e?te? faites. Il s?ave?re que l?agre?gation, qui se traduit mathe?matiquement par un phe?nome?ne d?explosion en temps fini, n?est pas
syste?matique. Par exemple, elle ne se produit jamais en dimension un d?espace alors qu?en
dimension deux elle ne se produit que si le nombre initial d?amibes est suffisamment e?leve?.
Dans ce Chapitre, nous e?tudions un syste?me d?e?quations propose? par Mimura et Tsujikawa
[57], ou? interviennent un terme de diffusion, un terme de couplage lie? au chimiotactisme ainsi
qu?un terme de croissance. Plus pre?cise?ment, on pose ? = 0 et on e?tudie le proble?me de
Neumann homoge?ne
?
1
?
?
dans ? О (0, +?),
ut = ?u ? ? и (u??(v)) + 2 f? (u)
?
?
?
?
?
?
?0 = ?v + u ? ?v
dans ? О (0, +?),
(10)
?u
?v
?
?
=
=
0
sur
??
О
(0,
+?),
?
?
?
??
??
?
?
?u(x, 0) = u (x)
dans ?,
0
ou? ? > 0 est un petit parame?tre ; le terme de re?action non-line?aire est donne? par
1
f? (u) = u(1 ? u)(u ? ) + ??u(1 ? u),
2
avec ? constante positive.
Le processus d?agre?gation sous-jacent est alors diffe?rent de celui du mode?le sans terme
de croissance. Sous l?hypothe?se d?une condition initiale bien pre?pare?e, c?est-a?-dire pre?sentant
de?ja? une interface, Bonami, Hilhorst, Logak et Mimura [17] ont montre? que, lorsque ? ? 0, la
solution (u? , v ? ) converge vers (u0 , v 0 ), ou? u0 est une fonction en escalier prenant les valeurs 1
et 0. Le proble?me a? frontie?re libre limite est donne? par les e?quations couple?es :
?
??(v 0 ) ?
?
?
+ 2?
V
=
?(N
?
1)?
+
n
?
?
?n
?
?
»
?
??t »
= ?0
sur ?t ,
t=0
?
0 = ?v 0 + u0 ? ?v 0
?
?
?
?
0
?
?
? ?v = 0
??
dans ? О (0, T ],
sur ?? О (0, T ],
(11)
14
Introduction
ou? l?interface initiale ?0 , de?finie par ?0 := {x ? ?, u0 (x) = 1/2}, est une hypersurface
sans bords re?gulie?re ; n de?signe le vecteur normal, unitaire, exte?rieur a? ?t , Vn la vitesse de
de?placement de l?interface le long de la normale et ? la courbure moyenne de l?interface. La
premie?re e?quation traduit le de?placement de la frontie?re libre se?parant les re?gions {u0 = 1} et
{u0 = 0}. Ce proble?me admet une solution classique unique (?, v 0 ) sur un intervalle [0, T ].
Dans ce Chapitre, nous e?tendons les re?sultats de [17] en supposant la condition initiale u0
tre?s ge?ne?rale. Apre?s avoir rappele? le principe de comparaison utilise? dans [17], nous de?montrons
une proprie?te? de ge?ne?ration d?interface. Nous nous appuyons pour cela sur une paire de souset sur-solutions construites a? l?aide de la solution de l?e?quation diffe?rentielle ordinaire ut =
??2 f (u), obtenue en ne?gligeant la diffusion et le chimiotactisme. Nous e?tudions ensuite le
de?placement de l?interface, en nous appuyant sur des sous- et sur-solutions construites a? l?aide
de la solution d?un proble?me stationnaire unidimensionnel associe? ; la de?monstration s?appuie
sur des estimations de la fonction de Green associe?e au proble?me de Neumann homoge?ne sur ?
pour l?ope?rateur ?? + ?. En imbriquant les deux paires de sous- et sur-solutions construites,
on de?montre que l?e?paisseur de la zone de transition de?veloppe?e par la solution u? est d?ordre
?, ainsi que la distance de Hausdorff entre l?ensemble des points ??t := {x ? ?, u? (x, t) = 1/2}
et l?interface ?t , solution du proble?me a? frontie?re libre limite (11). Ces re?sultats sont optimaux.
Chapitre 3 : Limite singulie?re d?une equation d?Allen-Cahn inhomoge?ne et anisotrope
Cette partie de la the?se correspond a? des travaux re?alise?s en collaboration avec H. Garcke
(Universite? de Regensburg), D. Hilhorst (Universite? de Paris-Sud), H. Matano (Universite? de
Tokyo) et R. Scha?tzle (Universite? de Tu?bingen).
Le contexte de cette e?tude est la mode?lisation de mouvements d?interfaces en science des
mate?riaux, ou? la vitesse normale de de?placement de l?interface de?pend de l?angle du vecteur
normal avec une direction fixe. On parle de mouvement anisotrope d?interface.
Nous e?tudions le proble?me de Neumann homoge?ne pour une e?quation d?Allen-Cahn anisotrope et inhomoge?ne :
?
1
?
?
u = ? и ap (x, ?u) + 2 f (u)
dans ? О (0, +?),
?
? t
?
(12)
ap (x, ?u) и ? = 0
sur ?? О (0, +?),
?
?
?
?
dans ?,
u(x, 0) = u0 (x)
ou? ? > 0 est un petit parame?tre. On suppose que le terme non-line?aire f posse?de exactement
trois ze?ros 0 < a < 1, que sa pente est strictement ne?gative aux e?quilibres u = 0 et u = 1,
strictement positive a? l?e?quilibre u = a, ce qui assure son caracte?re bistable ; nous supposons
e?galement que f satisfait la condition inte?grale
Z
1
f (u)du = 0.
0
L?anisotropie intervient dans le terme ? и ap (x, ?u), ou? nous supposons la fonction a(x, p)
strictement positive, strictement convexe et 2 homoge?ne (en la variable p) sur ?ОRN \{0}. Nous
?a
?a
utilisons la notation ap (x, p) pour le vecteur gradient (
,иии ,
)(x, p) et nous supposons
?p1
?pN
3+?
que la fonction a(x, p) est de classe Cloc
seulement sur ? О RN \ {0}, autrement dit que le
Introduction
15
terme de diffusion ? и ap (x, ?u) est singulier aux points ou? le gradient de la solution s?annule.
Il nous faut donc utiliser une notion de solution faible, pour laquelle nous de?montrons un
principe de comparaison.
L?e?quation parabolique dans le proble?me (12) contient, en particulier, l?e?quation inhomoge?ne
1
ut = div(A(x)?u) + 2 f (u),
(13)
?
ou? A(x) est une matrice syme?trique de?finie positive, et l?e?quation anisotrope
А
б
1
ut = div A(?u) + 2 f (u),
?
(14)
ou? les coefficients de la matrice ?p ? A = ?p t A peuvent e?tre singuliers au point p = 0.
Nous commenc?ons par construire une me?trique adapte?e au proble?me anisotrope, en nous
inspirant de re?sultats de Bellettini, Paolini and Venturini sur une me?trique de Finsler, [9] et
[10].
Nous de?montrons que, lorsque ? ? 0, la solution u? converge presque partout vers u?, ou?
u? est une fonction en escalier prenant les valeurs 0 et 1, les re?gions {u? = 0} et {u? = 1} e?tant
se?pare?es par une interface limite qui se de?place. Le proble?me a? frontie?re libre limite est donne?
par :
?
?Vn,? = ?(N ? 1)?? sur ?t ,
»
(15)
??t »»
= ?0 ,
t=0
ou? l?interface initiale ?0 est de?finie par ?0 := {x ? ?, u0 (x) = a}, ou? Vn,? de?signe la vitesse de
de?placement anisotrope de l?interface le long de la normale anisotrope a? ?t et ?? une version
anisotrope de la courbure moyenne de l?interface. L?e?criture du proble?me limite se complique
sensiblement en ge?ome?trie euclidienne :
?
h
i
1
1
?
sur ?t ,
Vn = ?? и p
ap (x, n)
?p
2a(x, n)
2a(x, n)
(16)
»
?
?? »»
=
?
.
t
0
t=0
Si ?0 est assez re?gulie?re, le proble?me limite admet une unique solution classique sur un intervalle
de temps [0, T ].
Avec des hypothe?ses faibles sur le profil de la condition initiale, nous effectuons une analyse
rigoureuse de la ge?ne?ration et du de?placement de l?interface. Nous nous appuyons pour cela sur
deux paires distinctes de sous- et sur-solutions. Pour l?e?tude de la ge?ne?ration de l?interface, on
perturbe la solution de l?e?quation diffe?rentielle ordinaire ut = ??2 f (u), obtenue en ne?gligeant
la diffusion anisotrope. Pour l?e?tude du de?placement de l?interface, on utilise la solution d?un
proble?me stationnaire unidimensionnel associe?. En imbriquant ces deux paires de sous- et sursolutions, on de?montre que l?e?paisseur de la zone de transition de?veloppe?e par la solution est
d?ordre ?, ame?liorant ainsi des re?sultats connus [13], [70].
Chapter 1
The singular limit of the
Allen-Cahn equation and the
FitzHugh-Nagumo system
We consider an Allen-Cahn type equation of the form ut = ?u + ??2 f ? (x, t, u), where ? > 0 is
a small parameter and f ? a bistable nonlinearity associated with a double-well potential whose
well-depths are slightly unbalanced by order ?. Given a rather general initial datum u0 that
is independent of ?, we perform a rigorous analysis of both the generation and the motion of
interface. More precisely we show that the solution develops a steep transition layer within the
time scale of order ?2 | ln ?|, and that the layer obeys the law of motion that coincides with the
formal asymptotic limit within an error margin of order ?. This is an optimal estimate that
has not been known before for solutions with general initial datum, even in the case where
f ? (x, t, u) = f (u).
Next we consider systems of reaction-diffusion equations of the form
(
ut = ?u + ??2 f ? (u, v),
vt = D?v + h(u, v),
which include the FitzHugh-Nagumo system as a special case. Given a rather general initial
datum (u0 , v0 ), we show that the component u develops a steep transition layer and that all
the above-mentioned results remain true for the u-component of these systems.
18
1.1
1.1.1
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
Introduction
Perturbed Allen-Cahn equation
In some classes of nonlinear diffusion equations, solutions often develop sharp internal layers ?
or ?interfaces? ? that separate the spatial domain into different phase regions. This happens,
in particular, when the diffusion coefficient is very small or the reaction term is very large.
The motion of such interfaces is often driven by their curvature. A typical example is the
Allen-Cahn equation ut = ?u + ??2 f (u), where ? > 0 is a small parameter and f (u) is a
bistable nonlinearity, whose meaning is explained below. A usual strategy for studying such
phenomena is to first derive the ?sharp interface limit? as ? ? 0 by a formal analysis, then to
check if this limit gives good approximation of the behavior of actual layers.
In this Chapter we consider a perturbed Allen-Cahn type equation of the form
(P ? )
?
1
?
ut = ?u + 2 (f (u) ? ?g ? (x, t, u)) in ? О (0, +?),
?
?
?
?
?u
=0
on ?? О (0, +?),
?
??
?
?
?
in ?,
u(x, 0) = u0 (x)
and study the behavior of layers near the sharp interface limit as ? ? 0. Here, ? is a smooth
bounded domain in RN (N ? 2) and ? is the Euclidian unit normal vector exterior to ??. The
nonlinearity is given by f (u) := ?W ? (u), where W (u) is a double-well potential with equal
well-depth, taking its global minimum value at u = ?? , ?+ . More precisely we assume that f
is of class C 2 on R and has exactly three zeros ?? < a < ?+ such that
f ? (?▒ ) < 0,
and that
f ? (a) > 0
Z
(bistable nonlinearity),
(1.1)
?+
f (u) du = 0.
(1.2)
??
The condition (1.1) implies that the potential W (u) attains its local minima at u = ?? , ?+ ,
and (1.2) implies that W (?? ) = W (?+ ). In other words, the two stable zeros of f , namely
?? and ?+ , have ?balanced? stability. A typical example is given by the cubic nonlinearity
f (u) = u(1 ? u2 ).
The term ?g ? represents a small perturbation, where g ? (x, t, u) is a function defined on
? О [0, +?) О R. This has the role of breaking the balance of the two stable zeros slightly. In
the special case where g ? ? 0, Problem (P ? ) reduces to the usual Allen-Cahn equation. As we
will explain later, our main results are new even for this special case.
We assume that g ? is C 2 in x and C 1 in t, u, and that, for any T > 0, there exist ? ? (0, 1)
and C > 0 such that, for all (x, t, u) ? ? О [0, T ] О R,
|?x g ? (x, t, u)| ? C??1
and
|gt? (x, t, u)| ? C??1 ,
|gu? (x, t, u)| ? C,
kg ? (и, и, u)k
C 1+?,
1+?
2 (?О[0,T ])
(1.3)
(1.4)
? C.
(1.5)
Moreover, we assume that there exists a function g(x, t, u) and a constant, which we denote
again by C, such that
(1.6)
|g ? (x, t, u) ? g(x, t, u)| ? C?,
1.1 Introduction
19
for all small ? > 0. Note that the estimate (1.5) and the pointwise convergence g ? ? g (as ? ?
0) imply that g satisfies the same estimate as (1.5).
For technical reasons we also assume that
?g ?
=0
??
on ?? О [0, T ] О R,
(1.7)
which, in turn, implies the same Neumann boundary condition for g. Apart from these bounds
and regularity requirements, we do not make any specific assumptions on the perturbation term
g?.
Remark 1.1.1. Since we will consider only bounded solutions in this Chapter, it is sufficient
to assume (1.3)?(1.5) to hold in some bounded interval ?M ? u ? M . Note that if g ? does
not depend on ?, then assumptions (1.3)?(1.5) are automatically satisfied on any bounded
interval ?M ? u ? M .
ц
Remark 1.1.2. The reason why we do not assume more smoothness on g is that we will later
apply our results to systems of equations, including the FitzHugh-Nagumo system, in which
g ? = g ? (x, t) loses C 2,1 -smoothness as ? ? 0.
ц
Remark 1.1.3. The equation in (P ? ) can be expressed in the form
ut = ?u +
1 ?
f (x, t, u),
?2
where f ? is C 2 in x, ? and C 1 in t, u. Conversely, by setting
f ? (x, t, u) ? f (u)
g (x, t, u) = ?
,
?
?
»
»
?f ?
g(x, t, u) = ?
(x, t, u)»»
,
??
?=0
the above equation is reduced to that in (P ? ). The conditions (1.3) and (1.6) then follow
automatically from the above regularity assumptions on f ? . The condition (1.5) holds if we
ц
impose slightly stronger regularity on f ? .
As for the initial datum u0 (x), we assume u0 ? C 2 (?). Throughout the present Chapter
the constant C0 will stand for the following quantity:
C0 := ku0 kC 0 (?) + k?u0 kC 0 (?) + k?u0 kC 0 (?) .
(1.8)
Furthermore we define the ?initial interface? ?0 by
?0 := {x ? ?, u0 (x) = a},
(1.9)
and suppose that ?0 is a C 3+? hypersurface without boundary such that, n being the outward
unit normal vector to ?0 ,
?0 ?? ?
and
u0 > a
?u0 (x) и n(x) 6= 0
in ?+
0,
u0 < a
if x ? ?0 ,
in ??
0,
(1.10)
(1.11)
+
where ??
0 denotes the region enclosed by the hypersurface ?0 and ?0 the region enclosed
between the boundary of the domain ?? and the hypersurface ?0 .
20
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
It is standard that Problem (P ? ) has a unique smooth solution, which we denote by u? . As
? ? 0, a formal asymptotic analysis shows the following: in the very early stage, the diffusion
term ?u is negligible compared with the reaction term ??2 (f (u) ? ?g ? (x, t, u)); it follows
that, in the rescaled time scale ? = t/?2 , the equation is well approximated by the ordinary
differential equation u? = f (u) + O(?). Hence, f being a bistable nonlinearity, the value of
u? quickly becomes close to either ?+ or ?? in most part of ?, creating a steep interface
(transition layer) between the regions {u? ? ?? } and {u? ? ?+ }. Once such an interface
develops, the diffusion term becomes large near the interface, and comes to balance with the
reaction term. As a result, the interface ceases rapid development and starts to propagate in
a much slower time scale.
To study such interfacial behavior, it is useful to consider a formal asymptotic limit of
Problem (P ? ) as ? ? 0. Then the limit solution u?(x, t) will be a step function taking the value
?+ on one side of the interface, and ?? on the other side. This sharp interface, which we will
denote by ?t , obeys a certain law of motion, which is expressed as follows (see Section 1.2 for
details):
?
Z ?+
?
? Vn = ?(N ? 1)? + c0
g(x, t, r)dr on ?t ,
??
(P 0 )
?
? ? »»
t t=0 = ?0 ,
where Vn is the normal velocity of ?t in the exterior direction, ? the mean curvature at each
point of ?t , c0 the constant defined by
h ? Z ?+
i?1
(W (s) ? W (?? ))1/2 ds
c0 =
2
,
(1.12)
??
with W the double-well potential associated with f :
Z s
W (s) = ?
f (r)dr.
a
In the sequel, ? will stand for:
?(x, t) = c0
Z
?+
g(x, t, r)dr.
(1.13)
??
It is well known that Problem (P 0 ) possesses locally in time a unique smooth solution, say
S
1+?
? = 0?t?T (?t О {t}), for some T > 0. More precisely, so as g, the function ? is in C 1+?, 2 ,
which implies, by the standard theory of parabolic equations, that ? is of class C 3+?,
more details, we refer to [23], Lemma 2.1.
Next we set
QT := ? О [0, T ],
3+?
2
. For
+
and for each t ? [0, T ], we denote by ??
t the region enclosed by ?t , and by ?t the region
enclosed between ?? and ?t . We define a step function u?(x, t) by
(
?+ in ?+
t
u?(x, t) =
for t ? [0, T ],
(1.14)
?? in ??
t
which represents the formal asymptotic limit of u? (or the sharp interface limit) as ? ? 0.
The aim of the present Chapter is to make a rigorous and detailed study of the limiting
behavior of the solution u? of Problem (P ? ) as ? ? 0. Our first main result, Theorem 1.1.4,
1.1 Introduction
21
describes the profile of the solution after a very short initial period. It asserts that: given a
virtually arbitrary initial datum u0 , the solution u? quickly becomes close to ?▒ , except in
a small neighborhood of the initial interface ?0 , creating a steep transition layer around ?0
(generation of interface). The time needed to develop such a transition layer, which we will
denote by t? , is of order ?2 | ln ?|. The theorem then states that the solution u? remains close
to the step function u? on the time interval [t? , T ] (motion of interface); in other words, the
motion of the transition layer is well approximated by the limit interface equation (P 0 ).
Theorem 1.1.4 (Generation and motion of interface). Let ? be an arbitrary constant
satisfying 0 < ? < 12 min(a ? ?? , ?+ ? a) and set
х = f ? (a).
Then there exist positive constants ?0 and C such that, for all ? ? (0, ?0 ) and for all t? ? t ? T ,
where t? := х?1 ?2 | ln ?|, we have
?
?
[?? ? ?, ?+ + ?] if x ? NC? (?t )
?
?
?
?
(1.15)
u? (x, t) ? [?? ? ?, ?? + ?] if x ? ??
t \ NC? (?t )
?
?
?
?
? [? ? ?, ? + ?] if x ? ?+ \ N (? ),
+
+
C? t
t
where Nr (?t ) := {x ? ?, dist(x, ?t ) < r} denotes the r-neighborhood of ?t .
Corollary 1.1.5 (Convergence). As ? ? 0, the solution u? converges to u? everywhere in
S
▒
0<t?T (?t О {t}).
The next theorem is concerned with the relation between the actual interface ??t := {x ?
?, u? (x, t) = a} and the formal asymptotic limit ?t , which is given as the solution of Problem
(P 0 ).
Theorem 1.1.6 (Error estimate). There exists C > 0 such that
??t ? NC? (?t )
for 0 ? t ? T.
(1.16)
Corollary 1.1.7 (Convergence of interface). There exists C > 0 such that
dH (??t , ?t ) ? C?
where
for 0 ? t ? T,
(1.17)
Е
ф
dH (A, B) := max sup d(a, B), sup d(b, A)
a?A
b?B
denotes the Hausdorff distance between two compact sets A and B. Consequently, ??t ? ?t as
? ? 0, uniformly in 0 ? t ? T , in the sense of Hausdorff distance.
Note that the estimates (1.16) and (1.17) follow from Theorem 1.1.4 in the range t? ? t ? T ,
but the range 0 ? t ? t? has to be treated by a separate argument. In fact, this is the time
range in which a clear transition layer is formed rapidly from an arbitrarily given initial
datum, therefore the behavior of the solution is quite different from that in the later time
range t? ? t ? T , where things move more slowly.
22
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
The estimate (1.15) in our Theorem 1.1.4 implies that, once a transition layer is formed,
its thickness remains within order ? for the rest of time. Here, by ?thickness of interface? we
mean the smallest r > 0 satisfying
{ x ? ?, u? (x, t) 6? [?? ? ?, ?? + ?] ? [?+ ? ?, ?+ + ?] } ? Nr (??t ).
Naturally this quantity depends on ?, but the estimates (1.15) and (1.17) assert that it is
bounded by 2C? (with the constant C depending on ?), hence it remains within O(?) regardless
of the choice of ? > 0.
Remark 1.1.8 (Optimality of the thickness estimate). The above O(?) estimate is
optimal, i.e., the interface cannot be thinner than this order. In fact, rescaling time and space
as ? := t/?2 , y := x/?, the equation reads as
u? = ?y u + f (u) ? ? g ? .
Thus, by the uniform boundedness of u and by standard parabolic estimates, we have |?y u| ? M
for some constant M > 0, which implies
|?x u(x, t)| ?
M
.
?
From this bound it is clear that the thickness of interface cannot be smaller than M ?1 (?+ ?
?? ) ?, hence, by (1.15), it has to be exactly of order ?. Intuitively, the order ? estimate follows
also from the formal asymptotic expansion (1.24), but the validity of such an expansion is far
from obvious for solutions with arbitrary initial datum.
ц
As far as we know, our O(?) estimate is new, even in the special case where g ? ? 0,
provided that N ? 2. Previously, the best thickness estimate in the literature was of order
?| ln ?| (see [20]), except that Xinfu Chen has recently obtained an order ? estimate for the case
N = 1 by a different argument (private communication). We also refer to a forthcoming article
[51] by Karali, Nakashima, Hilhorst and Matano, in which an order ? estimate is established
for a Lotka-Volterra competition-diffusion system, with large spatial inhomogeneity, whose
nonlinearity is of the balanced bistable type.
Remark 1.1.9 (Optimality of the generation time). The estimate (1.15) also implies
that the generation of interface takes place within the time span of t? . This estimate is optimal.
In other words, a well-developed interface cannot appear much earlier; see Proposition 1.3.10
for details.
ц
The singular limit of Allen-Cahn equation was first studied in the pioneering work of Allen
and Cahn [2] and, slightly later, in Kawasaki and Ohta [52] from the point of view of physicists.
They derived the interface equation by formal asymptotic analysis, thereby revealing that the
interface moves by the mean curvature. Triggered by these early observations, this problem
has become a subject of extensive mathematical studies.
Let us mention for instance the results of Bronsard and Kohn [18] in the case of spherical
symmetry, the articles of de Mottoni and Schatzman [59, 60] and those of Chen [20, 21]. These
results prove convergence to the limit interface equation in a classical framework; that is, under
the assumption that the limit interface ?t is a smooth hypersurface. As for the case where
1.1 Introduction
23
?t is a viscosity or a weak solution of the limit interface equation, we refer to the work of
Barles, Soner and Souganidis [6], Evans, Soner and Souganidis [33], Ilmanen [48] and Barles
and Souganidis [7].
As for Problem (P ? ), whose nonlinearity is slightly unbalanced, the limit interface equation
involves a pressure term as well as the curvature term as indicated in (P 0 ). This fact has been
long known on a formal level; see e.g. Rubinstein, Sternberg and Keller [66]. However, not
much rigorous study has been made. Ei, Iida and Yanagida [30] proved rigorously that the
motion of the layers of Problem (P ? ) is well approximated by the limit interface equation
(P 0 ), on the condition that the initial datum has already a well developed transition layer
whose profile depends on ?. In other words, they studied the motion of interface, but not the
generation of interface.
1.1.2
Singular limit of reaction-diffusion systems
As a matter of fact, our results for the single equation can be extended to a important class of
reaction-diffusion systems. More precisely, we consider systems of parabolic equations of the
form:
?
1
?
?
ut = ?u + 2 f ? (u, v)
in ? О (0, +?),
?
?
?
?
?
?
?
in ? О (0, +?),
vt = D?v + h(u, v)
?
?
?
?u
?v
(RD ? )
=
=0
on ?? О (0, +?),
?
?
??
??
?
?
?
?
in ?,
u(x, 0) = u0 (x)
?
?
?
?
?
v(x, 0) = v0 (x)
in ?,
where D is a positive constant, and f ? , h are C 2 functions such that
(F) there exist C 2 functions f1 (u, v), f2? (u, v) such that
f ? (u, v) = f (u) + ?f1 (u, v) + ?2 f2? (u, v),
(1.18)
where f (u) is a bistable nonlinearity satisfying (1.1), (1.2), and f2? , along with its derivatives in u, v, remain bounded as ? ? 0;
(H) for any constant L, M > 0 there exists a constant M1 ? M such that
h(u, ?M1 ) ? 0 ? h(u, M1 )
for |u| ? L.
(1.19)
The conditions (F) and (H) imply that the system of ordinary differential equations
u? =
1
f? (u, v),
?2
v? = h(u, v),
has a family of invariant rectangles of the form {|u| ? L, |v| ? M }, provided that ? is
sufficiently small. The maximum principle and standard parabolic estimates then guarantee
that the solution (u? , v ? ) of (RD ? ) exists globally for t ? 0 and remains bounded as t ? ?
(see subsection 1.7.1 for details). Apart from (1.19), we do not make any specific assumptions
on the function h.
24
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
Problem (RD ? ) represents a large class of important reaction-diffusion systems including
the FitzHugh-Nagumo system
?
? ut = ?u + 1 (f (u) ? ?v),
?2
(1.20)
?
vt = D?v + ?u ? ?v,
which is a simplified model for nervous transmission, and the following type of prey-predator
system that appears in mathematical ecology:
?
А
б
? ut = ?u + 1 (1 ? u)(u ? 1/2) ? ?v u,
2
?
(1.21)
?
vt = D?v + (?u ? ?v)v.
Remark 1.1.10. In some equations such as the prey-predator system (1.21), only nonnegative
solutions are to be considered. In such a case, we replace the condition (1.19) by
h(u, 0) ? 0 ? h(u, M1 )
for 0 ? u ? L,
and assume f ? (0, v) ? 0. The rest of the argument remains the same.
ц
Now the same formal analysis as is used to derive (P 0 ) in Section 1.2 shows that the
singular limit of (RD ? ), as ? ? 0, is the following moving boundary problem:
?
?
Vn = ?(N ? 1)? ? c0 F1 (v?(x, t))
on ?t ,
?
?
?
?
?
?
v?t = D?v? + h(u?, v?)
in ? О (0, T ],
?
?
?
?
?v?
(RD 0 )
=0
on ?? О (0, T ],
?
??
?
?
»
?
?
?
?t »t=0 = ?0
?
?
?
?
? v?(x, 0) = v (x)
in ?,
0
where u? is the step function defined in (1.14) and
Z ?+
F1 (v) =
f1 (r, v) dr.
??
This is a system consisting of an equation of surface motion and a partial differential equation.
Since u? is determined straightforwardly from ?t , in what follows, by a solution of (RD 0 ) we
mean the pair (?, v?) := (?t , v?(x, t)). In the case of the FitzHugh-Nagumo system (1.20), the
interface equation in (RD 0 ) reduces to
?
? Vn = ?(N ? 1)? + c0 (?+ ? ?? )v?(x, t),
?
v?t = D?v? + ?u? ? ?v?,
while in the prey-predator system (1.21), (RD 0 ) reduces to
?
? Vn = ?(N ? 1)? + c0 v?(x, t)/2,
?
v?t = D?v? + (?u? ? ?v?)u?.
Note that the positive sign in front of the term c0 v?(x, t) in the interface equation implies
an inhibitory effect on u?, since the velocity Vn is measured in the exterior normal direction,
toward which u? decreases.
1.1 Introduction
25
Lemma 1.1.11 (Local existence). Assume that v0 ? C 2 (?) and that ?0 is a C 2+? hypersurface which is the boundary of a domain D0 ?? ?. Then there exists T > 0 such that the
limit free boundary Problem (RD 0 ) has a unique solution (?, v?) in the interval [0, T ]. By the
2+?
1+?
standard theory of parabolic equations, ? is of class C 2+?, 2 and v? is of class C 1+?, 2 .
The existence result was established in [24], Theorem 3.2 and following lemmas. The
uniqueness can be obtained by using Theorem 2 in [21].
Our main results for the system (RD ? ) are the following:
Theorem 1.1.12 (Thickness of interface). Let (1.18) and (1.19) hold (or let the assumptions in Remark 1.1.10 hold). Assume also that u0 satisfies (1.10) and (1.11). Then the same
conclusion as in Theorem 1.1.4 holds for (RD ? ).
Corollary 1.1.13 (Convergence). Under the assumptions of Theorem 1.1.12, the same
conclusion as in Corollary 1.1.5 holds for (RD ? ).
Theorem 1.1.14 (Error estimate). Let the assumptions of Theorem 1.1.12 hold. Then
the same conclusion as in Theorem 1.1.6 holds for (RD ? ). Moreover, there exists a constant
C > 0 such that
kv ? ? v?kL? (?О(0,T )) ? C?.
Corollary 1.1.15 (Convergence of interface). Under the assumptions of Theorem 1.1.12,
the same conclusion as in Corollary 1.1.7 holds for (RD ? ).
The organization of this Chapter is as follows. In Section 1.2, we derive the interface
equation (P 0 ) from (P ? ) by formal asymptotic expansions which involve the so-called signed
distance function. In Sections 1.3 and 1.4, we present basic estimates concerning the generation
of interface for (P ? ). For the clarity of underlying ideas, we first consider the special case
where g ? ? 0 in Section 1.3, and deal with the general case in Section 1.4. In Section 1.5
we prove a preliminary result on the motion of interface (Lemma 1.5.1), which implies that
if the initial datum has already a well-developed transition layer, then the layer remains to
exist for 0 ? t ? T and its motion is well approximated by the interface equation (P 0 ). Our
approach in Sections 1.3 to 1.5 is based on the sub- and super-solutions method, but we use
two completely different sets of sub- and super-solutions. More precisely, the sub- and supersolutions for the motion of interface are constructed by using the first two terms of the formal
asymptotic expansion (1.24), while those for the generation of interface are constructed by
modifying the solution of the equation in the absence of diffusion: ut = ??2 f (u). In Section
1.6, we prove our main results for Problem (P ? ): Theorems 1.1.4, 1.1.6 and their respective
corollaries.
In the final section, we study the reaction-diffusion system (RD ? ) and prove Theorems
1.1.12, 1.1.14 and their corollaries. These results are obtained by applying a slightly modified
version of the results for (P ? ). The strategy is to regard f? (u, v) as a perturbation of f (u).
Indeed, the equation for u in (RD ? ) is identical to that in (P ? ) if we set g ? = ?f1 ? ?f2? .
However, what makes the analysis difficult is the fact that g ? is no longer a given function
but a quantity that depends on the unknown function v ? . In particular, the existence of the
limit g ? ? g (? ? 0) is not a priori guaranteed, and the estimate (1.6) is far from obvious.
As it turns out, the standard Lp or Schauder estimates for v ? would not yield (1.6), because
of the fact that u? converges to a discontinuous function as ? ? 0. In order to overcome this
difficulty, we derive a fine estimate of v ? that is based on estimates of the heat kernel and
the fact that u? remains uniformly smooth outside of an O(?) neighborhood of the smooth
hypersurface ?t .
26
1.2
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
Formal derivation of the interface motion equation
In this section we derive the equation of interface motion corresponding to Problem (P ? ) by
using a formal asymptotic expansion. The resulting interface equation can be regarded as the
singular limit of (P ? ) as ? ? 0. Our argument is basically along the same lines with the formal
derivation given by Nakamura, Matano, Hilhorst and Scha?tzle [63], who studied a similar but
slightly different type of spatially inhomogeneous equations by formal analysis. Let us also
mention some earlier papers [1], [36] and [66] involving the method of matched asymptotic
expansions for problems that are related to ours.
As in [63], the first two terms of the asymptotic expansion determine the interface equation.
Though our analysis in this section is for the most part formal, the observations we make here
will help the rigorous analysis in later sections.
(P ? ). We recall that ??t := {x ? ?,
u? (x, t) = a} is the
Let u? be the solution of Problem
S
S
interface at time t and call ?? := t?0 (??t О {t}) the interface. Let ? = 0?t?T (?t О {t}) be
the solution of the limit geometric motion problem and let de be the signed distance function
to ? defined by:
(
dist(x, ?t ) for x ? ?+
t
e
(1.22)
d(x, t) =
?dist(x, ?t ) for x ? ??
t ,
where dist(x, ?t ) is the distance from x to the hypersurface ?t in ?. We remark that de = 0 on
e = 1 in a neighborhood of ?. We then define
? and that |?d|
Q+
T =
[
0<t?T
(?+
t О {t}),
Q?
T =
[
0<t?T
(??
t О {t}).
We also assume that the solution u? has the expansions
u? (x, t) = ?▒ + ?u1 (x, t) + ?2 u2 (x, t) + и и и
(1.23)
away from the interface ? (the outer expansion) and
u? (x, t) = U0 (x, t, ?) + ?U1 (x, t, ?) + ?2 U2 (x, t, ?) + и и и
(1.24)
near ? (the inner expansion). Here, the functions Uk (x, t, z), k = 0, 1, 2, и и и , are defined for
e t)/?. The stretched space variable ? gives
x ? ?, t ? 0, z ? R and, by definition, ? := d(x,
exactly the right spatial scaling to describe the rapid transition between the regions {u? ? ?? }
and {u? ? ?+ }. We normalize Uk in such a way that
U0 (x, t, 0) = a,
Uk (x, t, 0) = 0,
for all k ? 1 (normalization conditions). To make the inner and outer expansions consistent,
we require that
Uk (x, t, +?) = 0,
U0 (x, t, +?) = ?+ ,
(1.25)
Uk (x, t, ??) = 0,
U0 (x, t, ??) = ?? ,
for all k ? 1 (matching conditions).
In what follows we will substitute the inner expansion (1.24) into the parabolic equation
of Problem (P ? ) and collect the ??2 and ??1 terms. To that purpose we compute the needed
1.2 Formal derivation of the interface motion equation
27
terms and get
det
+ ?U1t + U1z det + и и и
?
?de
+ ??U1 + U1z ?de + и и и
?u? = ?U0 + U0z
?
e2
?de
|?d|
?de
и ?U0z + U0z
+ U0zz 2 + ??U1
?u? = ?U0 + 2
?
?
?
2
e
|?d|
+2?de и ?U1z + U1z ?de + U1zz
+ иии
?
u?t = U0t + U0z
e t)/?).
where the functions Ui (i = 0, 1), as well as their derivatives, are taken at point (x, t, d(x,
Here, ?U0 denotes the derivative with respect to x whenever we regard U0 (x, t, z) as a function
of three variables x, t and z. The symbol ?U0 is defined similarly and this convention applies
to U0z and U1zz as well. We also use the expansions
f (u? ) = f (U0 ) + ?f ? (U0 )U1 + O(?2 ),
А
g ? (x, t, u? ) = g(x, t, u? ) + O(?)
= g(x, t, U0 ) + O(?).
б
? in view of (1.6)
Next, we substitute the expressions above in the partial differential equation in Problem (P ? ).
Collecting the ??2 terms yields
U0zz + f (U0 ) = 0.
In view of the normalization and matching conditions, we can now assert that U0 (x, t, z) =
U0 (z), where U0 (z) is the unique solution of the stationary problem
(
U0 ?? + f (U0 ) = 0,
U0 (??) = ?? ,
U0 (0) = a,
U0 (+?) = ?+ .
(1.26)
This solution represents the first approximation of the profile of a transition layer around
the interface observed in the stretched coordinates.
For example, in the special case where
?
2
f (u) = u(1 ? u ), we have U0 (z) = tanh(z/ 2). In the general case, the following standard
estimates hold.
Lemma 1.2.1. There exist positive constants C and ? such that the following estimates hold.
0 < ?+ ? U0 (z) ? Ce??|z|
0 < U0 (z) ? ?? ? Ce??|z|
for z ? 0,
for z ? 0.
In addition, U0 is a strictly increasing function and, for j = 1, 2,
|Dj U0 (z)| ? Ce??|z|
for z ? R.
Proof. We only give an outline. Rewriting the equation in (1.26) as
u? = v,
v? = ?f (u),
(1.27)
28
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
we see that (U0 (z), U0? (z)) is a heteroclinic orbit of the above system connecting the equilibria (?? , 0) and (?+ , 0). These equilibria are saddle points, with the linearized eigenvalues
{?? , ??? } and {?+ , ??+ }, respectively, where
p
p
?? = ?f ? (?? ),
?+ = ?f ? (?+ ).
Consequently, we have
U0 (z) =
й
?? + C1 e ?? z + o(e ?? z )
?+ + C2 e??+ z + o(e??+ z )
as z ? ??,
as z ? +?
(1.28)
for some constants C1 , C2 . The desired estimates now follow by setting ? = min(?+ , ?? ).
Next we collect the ??1 terms. Since U0 depends only on the variable z, we have ?U0z = 0
? = 1 near ?t , yields
which, combined with the fact that |?d|
e + g(x, t, U0 ).
U1zz + f ? (U0 )U1 = U0 ? (det ? ?d)
(1.29)
This equation can be seen as a linearized problem for (1.26) with an inhomogeneous term. As
is well known (see, for instance, [63]), the solvability condition for the above equation plays
the key role in determining the equation of interface motion. The following lemma is rather
standard, but we give an outline of the proof for the convenience of the reader.
Lemma 1.2.2 (Solvability condition). Let A(z) be a bounded function on ?? < z < ?.
Then the problem
(
z ? R,
?zz + f ? (U0 (z))? = A(z)
(1.30)
?(0) = 0, ? ? L? (R),
has a solution if and only if
Z
A(z)U0 ? (z)dz = 0.
(1.31)
R
Moreover the solution, if it exists, is unique and satisfies, for some constant C > 0,
|?(z)| ? CkAkL? ,
(1.32)
for all z ? R.
Proof. Multiplying the equation by U0 ? and integrating it by parts, we easily see that the
condition (1.31) is necessary. Conversely, suppose that this condition is satisfied. Then, since
U0 ? is a bounded positive solution to the homogeneous equation ?zz + f ? (U0 (z))? = 0, one
can use the method of variation of constants to find the above solution ? explicitly. More
precisely,
Z z│
Z ?
┤
?2
? (?)
?(z) = ?(z)
A(?)?(?) d? d?
0
??
(1.33)
Z ?
Z z│
┤
?2
A(?)?(?) d? d?,
? (?)
= ??(z)
0
?
where ? := U0 ? . The estimate (1.32) now follows from the above expression and (1.28).
From the above lemma, the solvability condition for (1.29) is given by
Z h
i
2
e
U0 ? (z)(det ? ?d)(x,
t) + g(x, t, U0 (z))U0 ? (z) dz = 0
R
1.2 Formal derivation of the interface motion equation
for all (x, t) ? QT . Hence we get
det ? ?de = ?
Z
R
29
g(x, t, U0 (z))U0 ? (z) dz
Z
,
?2
U0 (z) dz
R
which gives
Z
?+
g(x, t, r)dr
??
det = ?de ? Z
?2
.
U0 (z) dz
R
Moreover, multiplying equation (1.26) by U0 ? and integrating it from ?? to z, we obtain
Z z
0 =
(U0 ?? U0 ? + f (U0 )U0 ? )(s)ds
??
1 2
= U0 ? (z) ? W (U0 (z)) + W (?? ),
2
where we have also used the fact that U0 (??) = ?? and U0 ? (??) = 0. This implies that
?
U0 ? (z) = 2(W (U0 (z)) ? W (?? ))1/2 ,
and therefore
Z
?2
U0 (z)dz =
R
Z
R
=
?
? А
б1/2
U0 ? (z) 2 W (U0 (z)) ? W (?? )
dz
Z
2
?+
??
(W (s) ? W (?? ))
1/2
(1.34)
ds.
It then follows, in view of the definition of c0 in (1.12), that
Z ?+
e
e
g(x, t, r)dr.
dt = ?d ? c0
(1.35)
??
e t)) coincides
We are now ready to derive the equation of interface motion. Since ?de (= ?x d(x,
e
with the outward normal unit vector to the hypersurface ?t , we have dt (x, t) = ?Vn , where
Vn is the normal velocity of the interface ?t . It is also known that the mean curvature ? of
e
the interface is equal to ?d/(N
? 1). Thus the equation of interface motion is given by:
Z ?+
g(x, t, r)dr on ?t .
Vn = ?(N ? 1)? + c0
(1.36)
??
Summarizing, under the assumption that the solution u? of Problem (P ? ) satisfies
(
?+
in Q+
T
as ? ? 0,
u? ?
??
in Q?
T
+
we have formally proved that the boundary ?t between ??
t and ?t moves according to the
law (1.36).
30
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
To conclude this section, we give basic estimates for U1 (x, t, z), which we will need in
Section 1.5 to study the motion of interface. Substituting (1.35) into (1.29) gives
(
U1zz + f ? (U0 (z))U1 = g(x, t, U0 (z)) ? ?(x, t)U0 ? (z),
U1 (x, t, и) ? L? (R),
U1 (x, t, 0) = 0,
(1.37)
where ? has been defined in (1.13). Thus U1 (x, t, z) is a solution of (1.30) with
A = A0 (x, t, z) := g(x, t, U0 (z)) ? ?(x, t)U0 ? (z),
(1.38)
where the variables x, t are considered parameters. The problem (1.37) has a unique solution
by virtue of Lemma 1.2.2. Moreover, since A0 (x, t, z) remains bounded as (x, t, z) varies in
? О [0, T ] О R, the estimate (1.32) implies
|U1 (x, t, z)| ? M
for x ? ?, t ? [0, T ], z ? R,
(1.39)
for some constant M > 0. Similarly, since ?U1 is a solution of (1.30) with
│
б┤
А
A = ?x A0 (x, t, z) = ?x g(x, t, U0 (z)) ? ?(x, t)U0 ? (z) ,
and since g is assumed to be C 1 in x, we obtain
|?x U1 (x, t, z)| ? M
for x ? ?, t ? [0, T ], z ? R,
(1.40)
for some constant M > 0.
To obtain estimates as z ? ▒?, we first observe that (1.28) implies
A0 (x, t, z) ? g(x, t, ?▒ ) = O(e??|z| )
as z ? ▒?,
(1.41)
uniformly in x ? ?, t ? [0, T ]. We then apply the following general estimates.
Lemma 1.2.3. Let the assumptions of Lemma 1.2.2 hold, and assume further that A(z)?A▒ =
O(e??|z| ) as z ? ▒? for some constants A+ , A? and ? > 0. Then there exists a constant
? > 0 such that
?(z) ?
A▒
= O(e??|z| ),
f ? (?▒ )
|? ? (z)| + |? ?? (z)| = O(e??|z| ),
(1.42)
as z ? ▒?.
Proof. We only state the outline. To derive the former estimate, we need a slightly
more
А
б
2
?
elaborate version of (1.28). Since f (u) is C , we have f (u) = (u ? ?▒ )f (?▒ ) + O (u ? ?▒ )2 .
Consequently,
й
?? + C1 e ?? z + O(e 2?? z )
as z ? ??,
U0 (z) =
(1.43)
??
z
?2?
z
+
+
?+ + C2 e
+ O(e
)
as z ? +?.
Using the expression (1.33) along with the estimate A(z) ? A▒ = O(e??|z| ) and (1.43), we see
that
б
А
б
А
A▒
as z ? ▒?.
+ O |z|e??▒ |z| + O e? min(?,?▒ )|z|
?(z) = ?
2
(?▒ )
1.3 Generation of interface: the case g ? ? 0
31
This implies the former estimate in (1.42), where ? can be any constant satisfying 0 < ? <
min(?? , ?+ , ?). Substituting this into equation (1.30) gives the estimate for ?zz . Finally, the
estimate for ?z follows by integrating ?zz from ▒? to z.
From the above lemma and (1.41) we obtain the estimate
|U1z (x, t, z)| + |U1zz (x, t, z)| ? Ce??|z| ,
(1.44)
for x ? ?, t ? [0, T ], z ? R. Similarly, since the definition of A0 (1.38) and estimate (1.27)
imply
(?x A0 )(x, t, z) ? (?x g)(x, t, ?▒ ) = O(e??|z| )
as z ? ▒?,
we can apply Lemma 1.2.3 to ? = ?x U1 , to obtain
|?x U1z (x, t, z)| + |?x U1zz (x, t, z)| ? Ce??|z| ,
for x ? ?, t ? [0, T ], z ? R. As a consequence, there is a constant, which we denote again by
M , such that
(1.45)
|?x U1z (x, t, z)| ? M.
Finally we consider the boundary condition. Note that (1.7) implies
i
? h
?
A0 =
g(x, t, U0 (z)) ? ?(x, t)U0 ? (z) = 0
??
??
on ?? О [0, T ] О R.
(1.46)
Consequently, from the expression (1.33), or equivalently the expression
U1 (x, t, z) = U0 ? (z)
Z
z
0
we see that
1.3
│А
?U1
=0
??
б?2
U0 ? (?)
Z
?
??
┤
A0 (x, t, ?)U0 ? (?) d? d?,
on ?? О [0, T ] О R.
(1.47)
Generation of interface: the case g ? ? 0
This section deals with the generation of interface, namely the rapid formation of internal
layers that takes place in a neighborhood of ?0 = {x ? ?, u0 (x) = a} within the time span of
order ?2 | ln ?|. For the time being we focus on the special case where g ? ? 0. We will discuss
the general case in Section 1.4. In the sequel, ?0 will stand for the following quantity:
?0 :=
1
min(a ? ?? , ?+ ? a).
2
Our main result in this section is the following.
Theorem 1.3.1. Let ? ? (0, ?0 ) be arbitrary and define х as the derivative of f (u) at the
unstable equilibrium u = a, that is
х = f ? (a).
(1.48)
Then there exist positive constants ?0 and M0 such that, for all ? ? (0, ?0 ),
? for all x ? ?,
?? ? ? ? u? (x, х?1 ?2 | ln ?|) ? ?+ + ?,
(1.49)
32
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
? for all x ? ? such that |u0 (x) ? a| ? M0 ?, we have that
if u0 (x) ? a + M0 ? then u? (x, х?1 ?2 | ln ?|) ? ?+ ? ?,
if u0 (x) ? a ? M0 ? then u? (x, х?1 ?2 | ln ?|) ? ?? + ?.
(1.50)
(1.51)
The above theorem will be proved by constructing a suitable pair of sub- and supersolutions.
1.3.1
The bistable ordinary differential equation
As mentioned in Section 1.1, the above sub- and super-solutions are constructed by modifying
the solution of the problem without diffusion:
1
f (u?),
?2
u?t =
u?(x, 0) = u0 (x).
This solution is written in the form
u?(x, t) = Y
┤
│t
,
u
(x)
,
0
?2
where Y (?, ?) denotes the solution of the ordinary differential equation
(
Y? (?, ?) = f (Y (?, ?)) for ? > 0,
Y (0, ?)
= ?.
(1.52)
Here ? ranges over the interval (?2C0 , 2C0 ), with C0 being the constant defined in (1.8). We
first study basic properties of Y .
Lemma 1.3.2. We have Y? > 0, for all ? ? (?2C0 , 2C0 ) \ {?? , a, ?+ } and all ? > 0. Furthermore,
f (Y (?, ?))
.
Y? (?, ?) =
f (?)
Proof. First, differentiating equation (1.52) with respect to ?, we obtain
(
Y?? = Y? f ? (Y ),
Y? (0, ?) = 1,
which can be integrated as follows:
Y? (?, ?) = exp
hZ
0
?
i
f ? (Y (s, ?))ds > 0.
We then differentiate equation (1.52) with respect to ? and obtain
(
Y? ? = Y? f ? (Y ),
Y? (0, ?) = f (?),
which in turn implies
Y? (?, ?) = f (?) exp
hZ
?
0
= f (?)Y? (?, ?).
i
f ? (Y (s, ?))ds
(1.53)
1.3 Generation of interface: the case g ? ? 0
33
This last equality, in view of (1.52), completes the proof of Lemma 1.3.2.
We define a function A(?, ?) by
A(?, ?) =
f ? (Y (?, ?)) ? f ? (?)
.
f (?)
(1.54)
Lemma 1.3.3. We have, for all ? ? (?2C0 , 2C0 ) \ {?? , a, ?+ } and all ? > 0,
Z ?
f ?? (Y (s, ?))Y? (s, ?)ds.
A(?, ?) =
0
Proof. Differentiating the equality of Lemma 1.3.2 with respect to ? leads to
Y?? = A(?, ?)Y? ,
(1.55)
whereas differentiating (1.53) with respect to ? yields
Z ?
Y?? = Y?
f ?? (Y (s, ?))Y? (s, ?)ds.
0
These two last results complete the proof of Lemma 1.3.3.
Next we need some estimates on the growth of Y , A and theirs derivatives. We first consider
the case where the initial value ? is far from the stable equilibria, more precisely when it lies
between ?? + ? and ?+ ? ?.
Lemma 1.3.4. Let ? ? (0, ?0 ) be arbitrary. Then there exist positive constants C?1 = C?1 (?),
C?2 = C?2 (?) and C3 = C3 (?) such that, for all ? > 0,
? if ? ? (a, ?+ ??) then, for every ? > 0 such that Y (?, ?) remains in the interval (a, ?+ ??),
we have
(1.56)
C?1 eх? ? Y? (?, ?) ? C?2 eх? ,
and
|A(?, ?)| ? C3 (eх? ? 1);
(1.57)
? if ? ? (?? +?, a) then, for every ? > 0 such that Y (?, ?) remains in the interval (?? +?, a),
(1.56) and (1.57) hold as well,
where х is the constant defined in (1.48).
Proof. We take ? ? (a, ?+ ? ?) and suppose that for s ? (0, ? ), Y (s, ?) remains in the interval
(a, ?+ ? ?). Integrating the equality
Y? (s, ?)
=1
f (Y (s, ?))
from 0 to ? yields
Z
?
Y? (s, ?)
ds
f (Y (s, ?))
0
= ?.
(1.58)
Hence by the change of variable q = Y (s, ?) we get
Z
?
Y (?,?)
dq
f (q)
= ?.
(1.59)
34
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
Moreover, the equality of Lemma 1.3.2 leads to
Z Y (?,?) ?
f (q)
ln Y? (?, ?) =
dq
f (q)
?
Z Y (?,?) ?
Б f (a) f ? (q) ? f ? (a) ц
dq
+
=
f (q)
f (q)
?
Z Y (?,?)
h(q)dq,
= х? +
(1.60)
?
where
h(q) = (f ? (q) ? х)/f (q).
Since
h(q) ? f ?? (a)/f ? (a)
as q ? a,
h is continuous on [a, ?+ ? ?]. Hence we can define
H = H(?) := khkL? (a,?+ ??) .
Since |Y (?, ?) ? ?| takes its values in the interval [0, ?+ ? a ? ?] ? [0, ?+ ? a], it follows from
(1.60) that
х? ? H(?+ ? a) ? ln Y? (?, ?) ? х? + H(?+ ? a),
which, in turn, proves (1.56). Next Lemma 1.3.3 and (1.56) yield
Z ?
??
|A(?, ?)| ? supz?[?? ,?+ ] |f (z)| C?2 eхs ds
0
? C3
(eх?
? 1),
which completes the proof of (1.57). The case where ? and Y (?, ?) are in (?? + ?, a) is similar
and omitted.
Corollary 1.3.5. Let ? ? (0, ?0 ) be arbitrary. Then there exist positive constants C1 = C1 (?)
and C2 = C2 (?) such that, for all ? > 0,
? if ? ? (a, ?+ ??) then, for every ? > 0 such that Y (?, ?) remains in the interval (a, ?+ ??),
we have
C1 eх? (? ? a) ? Y (?, ?) ? a ? C2 eх? (? ? a);
(1.61)
? if ? ? (?? +?, a) then, for every ? > 0 such that Y (?, ?) remains in the interval (?? +?, a),
we have
C2 eх? (? ? a) ? Y (?, ?) ? a ? C1 eх? (? ? a).
(1.62)
Proof. Since
f (q)/(q ? a) ? f ? (a) = х as q ? a,
it is possible to find constants B1 = B1 (?) > 0 and B2 = B2 (?) > 0 such that, for all
q ? (a, ?+ ? ?),
(1.63)
B1 (q ? a) ? f (q) ? B2 (q ? a).
We write this inequality for a < Y (?, ?) < ?+ ? ? to obtain
B1 (Y (?, ?) ? a) ? f (Y (?, ?)) ? B2 (Y (?, ?) ? a).
1.3 Generation of interface: the case g ? ? 0
35
We also write this inequality for a < ? < ?+ ? ? to obtain
B1 (? ? a) ? f (?) ? B2 (? ? a).
Next we use the equality Y? = f (Y )/f (?) of Lemma 1.3.2 to deduce that
B1
B2
(Y (?, ?) ? a) ? (? ? a)Y? (?, ?) ?
(Y (?, ?) ? a),
B2
B1
which, in view of (1.56), implies that
B2
B1
C?1 eх? (? ? a) ? Y (?, ?) ? a ?
C?2 eх? (? ? a).
B2
B1
This proves (1.61). The proof of (1.62) is similar and omitted.
We now present estimates in the case where the initial value ? is smaller than ?? + ? or
larger than ?+ ? ?.
Lemma 1.3.6. Let ? ? (0, ?0 ) and M > 0 be arbitrary. Then there exists a positive constant
C4 = C4 (?, M ) such that
? if ? ? [?+ ??, ?+ +M ], then, for all ? > 0, Y (?, ?) remains in the interval [?+ ??, ?+ +M ]
and
|A(?, ?)| ? C4 ? for ? > 0 ;
(1.64)
? if ? ? [?? ?M, ?? +?], then, for all ? > 0, Y (?, ?) remains in the interval [?? ?M, ?? +?]
and (1.64) holds as well.
Proof. Since the two cases can be treated in the same way, we will only prove the former.
The fact that Y (?, ?), the solution of the ordinary differential equation (1.52), remains in the
interval [?+ ? ?, ?+ + M ] directly follows from the bistable properties of f , or, more precisely,
from the sign conditions f (?+ ? ?) > 0, f (?+ + M ) < 0.
To prove (1.64), suppose first that ? ? [?+ , ?+ +M ]. In view of (1.1), f ? is strictly negative
in an interval of the form [?+ , ?+ + c] and f is negative in [?+ , ?). We denote by ?m < 0
the maximum of f on [?+ + c, ?+ + M ]. Then, as long as Y (?, ?) remains in the interval
[?+ + c, ?+ + M ], the ordinary differential equation (1.52) implies
Y? ? ?m.
By integration, this means that, for any ? ? [?+ , ?+ + M ], we have
Y (?, ?) ? [?+ , ?+ + c]
for ? ? ? :=
M ?c
.
m
In view of this, and considering that f ? (Y ) < 0 for Y ? [?+ , ?+ +c], we see from the expression
(1.53) that
hZ ?
i
hZ ?
i
?
Y? (?, ?) = exp
f (Y (s, ?))ds exp
f ? (Y (s, ?))ds
0
? exp
? exp
hZ
?
?
0
hZ
?
i
f ? (Y (s, ?))ds
sup
0 z?[?? ?M,?+ +M ]
i
|f ? (z)|ds =: C?4 = C?4 (M ),
36
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
for all ? ? ? . It is clear from the same expression (1.53) that Y? ? C?4 holds also for 0 ? ? ? ? .
We can then use Lemma 1.3.3 to deduce that
Z ?
|A(?, ?)| ? C?4
|f ?? (Y (s, ?))|ds
0
┤
? C?4 supz?[?? ?M,?+ +M ] |f ?? (z)| ? =: C4 ?.
│
The case ? ? [?+ ? ?, ?+ ] can be treated in the same way. This completes the proof of the
lemma.
Now we choose the constant M in the above lemma sufficiently large so that [?2C0 , 2C0 ] ?
[?? ? M, ?+ + M ], and fix M hereafter. Then C4 only depends on ?. Using the fact that
? = O(eх? ? 1) for ? > 0, one can easily deduce from (1.57) and (1.64) the following general
estimate.
Lemma 1.3.7. Let ? ? (0, ?0 ) be arbitrary and let C0 be the constant defined in (1.8). Then
there exists a positive constant C5 = C5 (?) such that, for all ? ? (?2C0 , 2C0 ) and all ? > 0,
|A(?, ?)| ? C5 (eх? ? 1).
1.3.2
Construction of sub- and super-solutions
We are now ready to construct the sub- and super-solutions for the study of generation of
interface. For simplicity, we first consider the case where
?u0
=0
??
on ??.
In this case, our sub- and super-solutions are given by
┤
│t
2
w?▒ (x, t) = Y 2 , u0 (x) ▒ ?2 C6 (eхt/? ? 1) .
?
(1.65)
(1.66)
In the general case where (1.65) does not necessarily hold, we have to slightly modify w?▒ (x, t)
near the boundary ??. This will be discussed later.
Lemma 1.3.8. Assume (1.65). Then there exist positive constants ?0 and C6 such that, for
all ? ? (0, ?0 ), (w?? , w?+ ) is a pair of sub- and super-solutions for Problem (P ? ), in the domain
Е
ф
(x, t) ? QT , x ? ?, 0 ? t ? х?1 ?2 | ln ?| ,
satisfying w?? (x, 0) = w?+ (x, 0) = u0 (x). Consequently
w?? (x, t) ? u? (x, t) ? w?+ (x, t)
for x ? ?, 0 ? t ? х?1 ?2 | ln ?|.
Proof. The assumption (1.65) implies
?w?▒
=0
??
on ?? О (0, +?).
Now we define the operator L0 by
L0 u := ut ? ?u ?
1
f (u),
?2
(1.67)
1.3 Generation of interface: the case g ? ? 0
37
and prove that L0 w?+ ? 0 . Straightforward computations yield
L0 w?+ =
1
1
2
Y? + C6 х eхt/? Y? ? ?u0 Y? ? |?u0 |2 Y?? ? 2 f (Y ),
2
?
?
therefore, in view of the ordinary differential equation (1.52),
i
h
Y??
2
L0 w?+ = C6 х eхt/? ? ?u0 ?
|?u0 |2 Y? .
Y?
We note that, in the range 0 ? t ? х?1 ?2 | ln ?|, we have, for ?0 sufficiently small,
2
0 ? ?2 C6 (eхt/? ? 1) ? ?2 C6 (??1 ? 1) ? C0 ,
where C0 is the constant defined in (1.8). Hence
2
? := u0 (x) ▒ C6 (eхt/? ? 1) ? (?2C0 , 2C0 ),
so that we can use the estimate of A = Y?? /Y? in Lemma 1.3.7 and obtain
h
i
2
2
L0 w?+ ? C6 хeхt/? ? |?u0 | ? C5 (eхt/? ? 1)|?u0 |2 Y?
i
h
2
? (C6 х ? C5 |?u0 |2 )eхt/? ? |?u0 | + C5 |?u0 |2 Y? .
Since Y? > 0, this inequality implies that, for C6 large enough,
i
h
L0 w?+ ? C6 х ? C5 C0 2 ? C0 Y? ? 0.
forбProblem (P ? ). Similarly w?? is a sub-solution. Obviously
Hence w?+ is a super-solution
А
w?? (x, 0) = w?+ (x, 0) = Y 0, u0 (x) = u0 (x). Lemma 1.3.8 is proved.
In the more general case where (1.65) is not necessarily valid, one can proceed as follows:
in view of (1.10) and (1.11) there exist positive constants d1 , ? such that u0 (x) ? a + ? if
d(x, ??) ? d1 . Let ? be a smooth cut-off function defined on [0, +?) such that 0 ? ? ? 1,
?(0) = ?? (0) = 0 and ?(z) = 1 for z ? d1 . Then we define
Б
ц
u+
0 (x) = ?(d(x, ??)) u0 (x) + 1 ? ?(d(x, ??)) max u0 (x),
Б
ц x??
(x)
=
?(d(x,
??))
u
(x)
+
1
?
?(d(x,
??))
(a + ?).
u?
0
0
(1.68)
+
+
+
Clearly, u?
0 ? u0 ? u0 , and both u0 and u0 satisfy (1.65). Now we set
w??▒ (x, t) = Y
┤
│t
▒
2
хt/?2
?
1)
.
,
u
(x)
▒
?
C
(e
6
?2 0
Then the same argument as in Lemma 1.3.8 shows that (w??? , w??+ ) is a pair of sub- and super+
solutions for Problem (P ? ). Furthermore, since w??? (x, 0) = u?
0 (x) ? u0 (x) ? u0 (x) =
w??+ (x, 0), the comparison principle asserts that
w??? (x, t) ? u? (x, t) ? w??+ (x, t)
for x ? ?, 0 ? t ? х?1 ?2 | ln ?|.
(1.69)
38
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
1.3.3
Proof of Theorem 1.3.1
In order to prove Theorem 1.3.1 we first present a key estimate on the function Y (?, ?) after
a time interval of order ? ? | ln ?|.
Lemma 1.3.9. Let ? ? (0, ?0 ) be arbitrary; there exist positive constants ?0 and C7 such that,
for all ? ? (0, ?0 ),
? for all ? ? (?2C0 , 2C0 ),
?? ? ? ? Y (х?1 | ln ?|, ?) ? ?+ + ?,
(1.70)
? for all ? ? (?2C0 , 2C0 ) such that |? ? a| ? C7 ?, we have that
if ? ? a + C7 ? then Y (х?1 | ln ?|, ?) ? ?+ ? ?,
if ? ? a ? C7 ? then Y (х?1 | ln ?|, ?) ? ?? + ?.
(1.71)
(1.72)
Proof. We first prove (1.71). For ? ? a + C7 ?, as long as Y (?, ?) has not reached ?+ ? ?, we
can use (1.61) to deduce that
Y (?, ?) ? a + C1 eх? (? ? a)
? a + C1 C7 eх? ?
? ?+ ? ?
provided that ? satisfies
? ? ? ? =: х?1 ln
?+ ? a ? ?
.
C1 C7 ?
Choosing
C7 =
max(a ? ?? , ?+ ? a) ? ?
,
C1
we see that х?1 | ln ?| ? ? ? , which completes the proof of (1.71). Using (1.62), one easily proves
(1.72).
Next we prove (1.70). First, by the bistable assumptions on f , if we leave from a initial
value ? ? [?? ? ?, ?+ + ?] then Y (?, ?) will remain in [?? ? ?, ?+ + ?]. Now suppose that
?+ + ? ? ? ? 2C0 . We check below that Y (х?1 | ln ?|, ?) ? ?+ + ?. First, in view of (1.1), we
can find p > 0 such that
if
if
?+ ? u ? 2C0
? 2C0 ? u ? ??
then
then
f (u) ? p(?+ ? u)
f (u) ? ?p(u ? ?? ).
(1.73)
We then use the ordinary differential equation to obtain, as long as ?+ + ? ? Y ? 2C0 , the
inequality Y? ? p(?+ ? Y ). It follows that
Y?
? ?p.
Y ? ?+
Integrating this inequality from 0 to ? leads to
Y (?, ?) ? ?+ + (? ? ?+ )e?p?
? ?+ + (2C0 ? ?+ )e?p? .
1.3 Generation of interface: the case g ? ? 0
39
?1
Since (2C0 ? ?+ )e?pх | ln ?| ? 0 as ? ? 0, the above inequality proves that, for ? ? (0, ?0 ),
with ?0 = ?0 (?) sufficiently small, Y (х?1 | ln ?|, ?) ? ?+ + ?, which completes the proof of
(1.70).
We are now ready to prove Theorem 1.3.1. By setting t = х?1 ?2 | ln ?| in (1.69), we obtain
┤
│
2
(x)
?
(C
?
?
C
?
)
Y х?1 | ln ?|, u?
6
6
0
│
┤
2
(x)
+
C
?
?
C
?
? u? (x, х?1 ?2 | ln ?|) ? Y х?1 | ln ?|, u+
. (1.74)
6
6
0
Furthermore, by the definition of C0 in (1.8), we have, for ?0 small enough,
2
?2C0 ? u▒
0 (x) ▒ (C6 ? ? C6 ? ) ? 2C0 ,
for x ? ?. Thus the assertion (1.49) of Theorem 1.3.1 is a direct consequence of (1.70) and
(1.74).
Next we prove (1.50). We choose M0 large enough so that M0 ? ? C6 ? + C6 ?2 ? C7 ?. Then,
for any x ? ? such that u?
0 (x) ? a + M0 ?, we have
2
2
u?
0 (x) ? (C6 ? ? C6 ? ) ? a + M0 ? ? C6 ? + C6 ? ? a + C7 ?.
Combining this, (1.74) and (1.71), we see that
u? (x, х?1 ?2 | ln ?|) ? ?+ ? ?,
?
for any x ? ? that satisfies u?
0 (x) ? a + M0 ?. From the definition of u0 in (1.68), it is clear
that
u?
if and only if u0 (x) ? a + M0 ?,
0 (x) ? a + M0 ?
provided that ? is small enough. This proves (1.50). The inequality (1.51) can be shown the
same way. This completes the proof of Theorem 1.3.1.
1.3.4
Optimality of the generation time
To conclude this section we show that the generation time t ? := х?1 ?2 | ln ?| that appears in
Theorem 1.3.1 is optimal. In other words, the interface will not be fully developed much before
t ?.
?
Proposition 1.3.10. Denote by tmin
the smallest time such that (1.15) holds for all t ?
?
[ tmin , T ]. Then there exists a constant b = b(C) such that
?
? х?1 ?2 (| ln ?| ? b)
tmin
for all ? ? (0, ?0 ).
Proof. For simplicity, we deal with the case where (1.65) is valid. In that case, (1.67) holds
for all small ? > 0. For each b > 0, we set
t? (b) := х?1 ?2 (| ln ?| ? b),
and evaluate u? (x, t? (b)) at a point x ? ?+
0 where dist(x, ?0 ) = C?. Since u0 = a on ?t and
since |?u0 | ? C0 by (1.8), we have
u0 (x) ? a + C0 C?.
(1.75)
40
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
It follows from this and (1.61) that
│
┤
w?+ (x, t? (b)) = Y х?1 (| ln ?| ? b), u0 (x) + ?C6 e?b ? ?2 C6
А
? a + C2 e| ln ?|?b u0 (x) + ?C6 e?b ? ?2 C6 ? a)
? a + C2 ??1 e?b (C0 C? + ?C6 e?b )
= a + C2 e?b (C0 C + C6 e?b ).
Now we choose b to be sufficiently large, so that
a + C2 e?b (C0 C + C6 e?b ) < ?+ ? ?.
Then the above estimate and (1.67) yield
u? (x, t? (b)) ? w?+ (x, t? (b)) < ?+ ? ?.
This implies that (1.15) does not hold at t = t? (b), hence t? (b) < t?min . The lemma is proved.
1.4
Generation of interface in the general case
In this section we extend Theorem 1.3.1 to the case where g ? 6? 0. The proof is more technical
than the case g ? ? 0, but the underlying ideas are the same. Hence we will basically follow
the argument of Section 1.3, simply pointing out the main differences.
1.4.1
The perturbed bistable ordinary differential equation
We first consider a slightly perturbed nonlinearity:
f? (u) = f (u) + ?,
where ? is any constant. For |?| small enough, this function is still bistable. More precisely,
we claim that f? has the following properties.
Lemma 1.4.1. Let ?0 be small enough. Then, for all ? ? (??0 , ?0 ),
? f? has exactly three zeros, namely ?? (?) < a(?) < ?+ (?) and there exists a positive
constant C such that
|?? (?) ? ?? | + |a(?) ? a| + |?+ (?) ? ?+ | ? C|?|.
(1.76)
? We have that
? Set
f?
is strictly positive in
f?
is strictly negative in
(??, ?? (?)) ? (a(?), ?+ (?)),
(?? (?), a(?)) ? (?+ (?), +?).
(1.77)
х(?) := f?? (a(?)) = f ? (a(?)),
then there exists a positive constant, which we denote again by C, such that
|х(?) ? х| ? C|?|.
(1.78)
1.4 Generation of interface in the general case
41
Now, for each ? ? (??0 , ?0 ), we define Y (?, ?; ?) as the solution of the following ordinary
differential equation:
(
Y? (?, ?; ?) = f? (Y (?, ?; ?)) for ? > 0,
(1.79)
Y (0, ?; ?) = ?,
where ? varies in (?2C0 , 2C0 ) with C0 being the constant in (1.8).
To prove Theorem 1.3.1, we will construct a pair of sub- and super-solutions for (P ? ) by
simply replacing the function Y (?, ?) in (1.66) by Y (?, ?; ?), with an appropriate choice of ?.
For this strategy to work, we have to check that the basic properties of Y (?, ?) in subsection
1.3.1 carry over to Y (?, ?; ?).
First, it is clear that all the differential and integral identities in subsection 1.3.1 that follow
directly from (1.52) are still valid for (1.79). In particular, Lemmas 1.3.2 and 1.3.3 remain to
hold if we replace Y (?, ?) by Y (?, ?; ?), f by f? and A(?, ?) by A(?, ?; ?), where
A(?, ?, ?) =
f?? (Y (?, ?; ?)) ? f?? (?)
.
f? (?)
Next let us show that the basic estimates which we have established in subsection 1.3.1 are
also valid for the function Y (?, ?; ?). The following lemma, which is an analogue of Lemma
1.3.4, is fundamental.
Lemma 1.4.2. Let ? ? (0, ?0 ) be arbitrary. Then there exist positive constants ?0 = ?0 (?),
e1 (?), C?2 = C?2 (?) and C3 = C3 (?) such that, for all ? ? (??0 , ?0 ), for all ? > 0,
e1 = C
C
? if ? ? (a(?), ?+ ? ?) then, for every ? > 0 such that Y (?, ?; ?) remains in the interval
(a(?), ?+ ? ?), we have
C?1 eх(?)? ? Y? (?, ?; ?) ? C?2 eх(?)? ,
(1.80)
|A(?, ?; ?)| ? C3 (eх(?)? ? 1);
(1.81)
and
? if ? ? (?? + ?, a(?)) then, for every ? > 0 such that Y (?, ?; ?) remains in the interval
(?? + ?, a(?)), (1.80) and (1.81) hold as well.
Proof. In view of (1.76), we can choose a small constant ?0 = ?0 (?) > 0 such that (a(?), ?+ ?
?) ? (a(?), ?+ (?)) for every ? ? [??0 , ?0 ]. Therefore f? (q) does not change sign in the interval
(a(?), ?+ ? ?). Thus, in order to prove the lemma, we just have to write again the proof of
Lemma 1.3.4, simply replacing Y (?, ?) by Y (?, ?; ?). We do not repeat the entire proof here.
Instead, let us explain why C?1 , C?2 and C3 are independent of ?; in view of the proof of Lemma
1.3.4, it is sufficient to estimate, for q ? [a(?), ?+ ? ?], the modulus of the quantity
h? (q) =
f?? (q) ? f?? (a(?))
f? (q)
by a constant depending on ? but not on ? ? [??0 , ?0 ]. Since
h? (q) ?
f??? (a(?))
f ?? (a(?))
=
f?? (a(?))
f ? (a(?))
as q ? a(?),
we see that the function (q, ?) 7? h? (q) is continuous in the compact region { |?| ? ?0 , a(?) ?
q ? ?+ ? ? }. It follows that |h? (q)| is bounded as (q, ?) varies in this region. This completes
the proof of Lemma 1.4.2.
42
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
Corollary 1.4.3. Let ? ? (0, ?0 ) be arbitrary. Then there exist positive constants ?0 = ?0 (?),
C1 = C1 (?) and C2 = C2 (?) such that, for all ? ? (??0 , ?0 ), for all ? > 0,
? if ? ? (a(?), ?+ ? ?) then, for every ? > 0 such that Y (?, ?; ?) remains in the interval
(a(?), ?+ ? ?), we have
C1 eх(?)? (? ? a(?)) ? Y (?, ?; ?) ? a(?) ? C2 eх(?)? (? ? a(?)),
(1.82)
? if ? ? (?? + ?, a(?)) then, for every ? > 0 such that Y (?, ?; ?) remains in the interval
(?? + ?, a(?)), we have
C2 eх(?)? (? ? a(?)) ? Y (?, ?; ?) ? a(?) ? C1 eх(?)? (? ? a(?)).
(1.83)
Proof. We can simply follow the proof of Corollary 1.3.5. In order to prove that C1 and C2 are
independent of ?, all we have to do is to find constants B1 = B1 (?) > 0 and B2 = B2 (?) > 0
such that, for all ? ? [??0 , ?0 ] and all q ? (a(?), ?+ ? ?),
B1 (q ? a(?)) ? f? (q) ? B2 (q ? a(?)).
(1.84)
In view of (1.78), we can choose ?0 > 0 small enough so that, for all ? ? [??0 , ?0 ], we have
х(?) ? х/2 > 0. Since
f? (q)
? х(?) as q ? a(?),
q ? a(?)
it follows that (q, ?) 7? f? (q)/(q ? a(?)) is a strictly positive and continuous function on the
compact region { |?| ? ?0 , a(?) ? q ? ?+ ? ? }, which insures the existence of the constants
B1 and B2 . This completes the proof of the corollary.
Now, it is no trouble to establish an analogue of Lemmas 1.3.6 and 1.3.7 with constants
independent of ?. We claim, without proof, that:
Lemma 1.4.4. Let ? ? (0, ?0 ) and M > 0 be arbitrary. Then there exist positive constants
?0 = ?0 (?, M ) and C4 = C4 (?, M ) such that, for all ? ? (??0 , ?0 ),
? if ? ? [?+ ? ?, ?+ + M ], then, for all ? > 0, Y (?, ?; ?) remains in the interval [?+ ?
?, ?+ + M ] and
|A(?, ?; ?)| ? C4 ?
for ? > 0 ;
(1.85)
? if ? ? [?? ? M, ?? + ?], then, for all ? > 0, Y (?, ?; ?) remains in the interval [?? ?
M, ?? + ?] and (1.85) holds as well.
Lemma 1.4.5. Let ? ? (0, ?0 ) be arbitrary and let C0 be the constant defined in (1.8). Then
there exist positive constants ?0 = ?0 (?), C5 = C5 (?) such that, for all ? ? (??0 , ?0 ), for all
? > 0 and all ? ? (?2C0 , 2C0 ),
|A(?, ?; ?)| ? C5 (eх(?)? ? 1).
1.4 Generation of interface in the general case
1.4.2
43
Construction of sub- and super-solutions
We now use Y (?, ?; ?), the solution of the ordinary differential equation (1.79), to construct a
pair of sub- and super-solutions. The same cut-off argument as in subsection 1.3.2 enables us
to assume (1.65) for simplicity. We set
w?▒ (x, t) = Y
│t
┤
t
2
,
u
(x)
▒
?
r(▒?G,
);
▒?G
,
0
?2
?2
(1.86)
where the function r(?, ? ) is given by
r(?, ? ) = C6 (eх(?)? ? 1),
and the constant G is chosen such that, for all small ? > 0,
|g ? (x, t, u)| ? G
for all (x, t, u) ? ? О [0, T ] О R,
which, in view of (1.5), is clearly possible.
Lemma 1.4.6. There exist positive constants ?0 and C6 such that for all ? ? (0, ?0 ), (w?? , w?+ )
is a pair of sub- and super-solutions for Problem (P ? ), in the domain
Е
ф
(x, t) ? QT , x ? ?, 0 ? t ? х?1 ?2 | ln ?| ,
satisfying w?? (x, 0) = w?+ (x, 0) = u0 (x). Consequently
w?? (x, t) ? u? (x, t) ? w?+ (x, t)
for x ? ?, 0 ? t ? х?1 ?2 | ln ?|.
(1.87)
Proof. First, in view of (1.65), w?▒ satisfy the homogeneous Neumann boundary condition.
We define the operator L by
Lu := ut ? ?u ? ??2 (f (u) ? ?g ? (x, t, u)),
and prove below that Lw?+ ? 0 by slightly modifying the argument which we have used to
prove L0 w?+ ? 0 in subsection 1.3.2. A straightforward calculation yields
Lw?+ =
h
i
Y??
1
1
1
х(?G)t/?2
2
Y
+
Y
х(?G)e
?
|?u
|
?
?u
C
? 2 f (Y ) + g ? (x, t, Y ).
?
6
0
0
?
2
?
Y?
?
?
If ?0 is sufficiently small, we note that ▒?G ? (??0 , ?0 ) and that, in the range 0 ? t ?
х?1 ?2 | ln ?|,
2
|?2 C6 (eх(▒?G)t/? ? 1)| ? ?2 C6 (??х(▒?G)/х ? 1) ? C0 ,
which implies that
u0 (x) ▒ ?2 r(▒?G,
t
) ? (?2C0 , 2C0 ).
?2
These observations allow us to use the results of the previous subsection with ? = t/?2 ,
? = u0 (x) + ?2 r(?G, t/?2 ) and ? = ?G. In particular, the ordinary differential equation (1.79)
yields Y? = f (Y ) + ?G, which implies that
Lw?+ =
h
i
i
Y??
1h
2
G + g ? (x, t, Y ) + Y? C6 х(?G)eх(?G)t/? ? ?u0 ?
|?u0 |2 .
?
Y?
44
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
By the choice of G the first term is positive. Using the estimate of A = Y?? /Y? in Lemma
1.4.5, we obtain, for a constant C5 that is independent of ?,
h
i
2
2
Lw?+ ? Y? C6 х(?G)eх(?G)t/? ? |?u0 | ? C5 (eх(?G)t/? ? 1)|?u0 |2
h
i
2
? Y? (C6 х(?G) ? C5 |?u0 |2 )eх(?G)t/? ? |?u0 | + C5 |?u0 |2 .
In view of (1.78), this inequality implies that, for ? ? (0, ?0 ), with ?0 small enough, and for C6
large enough,
h 1
i
Lw?+ ? Y? C6 х ? C5 C0 2 ? C0 ? 0.
2
+
for Problem
(P ? ). Similarly w?? is a sub-solution. Obviously
Hence w? is a super-solution
А
б
w?+ (x, 0) = w?? (x, 0) = Y 0, u0 (x); ▒?G = u0 (x). Lemma 1.3.8 is proved.
1.4.3
Proof of Theorem 1.3.1 for the general case
As in subsection 1.3.3, we first present a key estimate on the function Y (?, ?; ?) after a time
interval of order ? ? | ln ?|. Roughly speaking, a perturbation ? of order ? does not affect the
result of Lemma 1.3.9.
Lemma 1.4.7. Let ? ? (0, ?0 ) be arbitrary; there exist positive constants ?0 and C7 such that,
for all ? ? (0, ?0 ),
? for all ? ? (?2C0 , 2C0 ),
?? ? ? ? Y (х?1 | ln ?|, ?; ▒?G) ? ?+ + ?,
(1.88)
? for all ? ? (?2C0 , 2C0 ) such that |? ? a| ? C7 ?, we have that
if ? ? a + C7 ? then Y (х?1 | ln ?|, ?; ▒?G) ? ?+ ? ?,
?1
if ? ? a ? C7 ? then Y (х
| ln ?|, ?; ▒?G) ? ?? + ?.
(1.89)
(1.90)
Proof. In view of (1.76), we have, for C7 large enough, a + C7 ? ? a(▒?G) + 12 C7 ?, for all
? ? (0, ?0 ), with ?0 sufficiently small. Hence for ? ? a + C7 ?, as long as Y (?, ?; ▒?G) has not
reached ?+ ? ?, we can use (1.82) to deduce, as done in the proof of Lemma 1.3.9, that
Y (?, ?; ▒?G) ? a(▒?G) + C1 eх(▒?G)? (? ? a(▒?G))
? a ? ?CG + 12 C1 C7 eх(▒?G)? ?
? ?+ ? ?
provided that ? satisfies
??
m0 ? ? + CG?
1
ln
=: х?1 (?)| ln ?|,
1
х(▒?G)
C
C
?
2 1 7
where m0 = max(a ? ?? , ?+ ? a). To complete the proof of (1.89) we must choose C7 so that
х?1 | ln ?| ? х?1 (?)| ln ?| ? 0. A simple computation shows that
х?1 | ln ?| ? х?1 (?)| ln ?| =
х(▒?G) ? х
1
1
m0 ? ? + CG?
+
| ln ?| ?
ln
ln C7 .
1
х(▒?G)х
х(▒?G)
х(▒?G)
2 C1
1.4 Generation of interface in the general case
45
Thanks to (1.78), as ? ? 0, the first term above is of order ?| ln ?| and the second one of
order 1. Hence, for C7 large enough, the upper quantity is positive for all ? ? (0, ?0 ), with ?0
sufficiently small. The proof of (1.90) is similar and omitted.
Next we prove (1.88). First, by taking ?0 sufficiently small, we can assume that the stable
equilibria of f▒?G , namely ?? (▒?G) and ?+ (▒?G), are in [?? ? ?, ?+ + ?]. Hence, f▒?G
being a bistable function, if we leave from a ? ? [?? ? ?, ?+ + ?] then Y (?, ?; ▒?G) will
remain in the interval [?? ? ?, ?+ + ?]. Now suppose that ?+ + ? ? ? ? 2C0 . We check
below that Y (х?1 | ln ?|, ?; ▒?G) ? ?+ + ?. As done in the proof of Lemma 1.3.9, as long as
?+ + ? ? Y ? 2C0 , (1.73) leads to the inequality Y? ? p(?+ ? Y ) + ?G. It follows that
G
Y?
? ?p + ? ,
Y ? ?+
?
which implies, by integration from 0 to ? , that
(?p+? G
)?
?
Y (?, ?; ▒?G) ? ?+ + (2C0 ? ?+ )e
?1
.
?1
Since (2C0 ? ?+ )e(?p+?G? )х | ln ?| ? 0 as ? ? 0, the above inequality proves that, for
? ? (0, ?0 ), with ?0 = ?0 (?) sufficiently small, Y (х?1 | ln ?|, ?; ▒?G) ? ?+ + ?, which completes
the proof of (1.88).
We are now ready to prove Theorem 1.3.1 in the general case. By setting t = х?1 ?2 | ln ?|
in (1.87), we get
│
┤
Y х?1 | ln ?|, u0 (x) ? ?2 r(??G, х?1 | ln ?|); ??G
│
┤
? u? (x, х?1 ?2 | ln ?|) ? Y х?1 | ln ?|, u0 (x) + ?2 r(?G, х?1 | ln ?|); +?G . (1.91)
The point is that, in view of (1.78),
lim
??0
х ? х(▒?G)
ln ? = 0.
х
(1.92)
Hence we have, for ?0 small enough,
1
3
?2 r(▒?G, х?1 | ln ?|) = C6 ?(?(х?х(▒?G))/х ? ?) ? ( C6 ?, C6 ?).
2
2
Hence, as in subsection 1.3.3, the result (1.49) of Theorem 1.3.1 is a direct consequence of
(1.88) and (1.91).
Next we prove (1.50). We take x ? ? such that u0 (x) ? a + M0 ? so that
u0 (x) ? ?2 r(??G, х?1 | ln ?|) ? a + M0 ? ? 32 C6 ?
? a + C7 ?,
if we choose M0 large enough. Using (1.91) and (1.89) we obtain (1.50), which completes the
proof of Theorem 1.3.1.
46
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
1.5
Motion of interface
In Sections 1.3 and 1.4, we proved that the solution u? develops a clear transition layer within
a very short time. The aim of the present section is to show that, once such a clear transition
layer is formed, it persists for the rest of time and that its law of motion is well approximated
by the interface equation (P 0 ).
Let us formulate the above assertion more clearly. By taking the first two terms of the
formal asymptotic expansion (1.24), we get a formal approximation of the solution u? up to
order ?:
│ d(x,
│
e t) ┤
e t) ┤
d(x,
u? (x, t) ? u?? (x, t) := U0
+ ?U1 x, t,
.
(1.93)
?
?
Here U0 , U1 are as defined in (1.26) and (1.37). The right-hand side has a clear transition
layer which lies exactly on ?t . Our goal is to show that this function is a good approximation
of the real solution; more precisely:
If u? becomes rather close to u?? at some time moment t = t0 , then it stays close to
u?? for the rest of time. Consequently, ??t evolves roughly like ?t .
In order to prove such a result, we will construct a pair of sub- and super-solutions u?
? and
for Problem (P ? ) by slightly modifying the above function u?? . It then follows that, if the
solution u? satisfies
?
+
u?
? (x, t0 ) ? u (x, t0 ) ? u? (x, t0 ),
u+
?
for some t0 ? 0, then
?
+
u?
? (x, t) ? u (x, t) ? u? (x, t),
?
?
?
for t0 ? t ? T . As a result, since both u+
? , u? stay close to u? , the solution u also stays close
to u?? for t0 ? t ? T .
The rest of this section is devoted to the construction of these sub- and super-solutions.
We begin with some preparations.
1.5.1
A modified signed distance function
Rather than working with the usual signed distance function de that was introduced in (1.22),
we define a ?cut-off signed distance function? d as follows. First, choose d0 > 0 small enough
e и) is smooth in the tubular neighborhood of ?
so that d(и,
and such that
e t)| < 3d0 },
{(x, t) ? QT , |d(x,
dist(?t , ??) ? 3d0
for all t ? [0, T ].
(1.94)
Next let ?(s) be a smooth increasing function on R such that
?
?
? s
?2d0
?(s) =
?
?
2d0
if |s| ? d0
if s ? ?2d0
if s ? 2d0 .
We then define the cut-off signed distance function d by
А
б
? t) .
d(x, t) = ? d(x,
(1.95)
1.5 Motion of interface
47
e t)| < d0 } and that, in view of (1.94) and
Note that |?d| = 1 in the region {(x, t) ? QT , |d(x,
the above definition, ?d = 0 in a neighborhood of ??. Note also that the equation of motion
(P 0 ), which is equivalent to (1.35), is now written as
dt = ?d ? ?(x, t)
where we recall that
?(x, t) = c0
Z
on ?t ,
(1.96)
?+
g(x, t, r)dr.
(1.97)
??
1.5.2
Construction of sub- and super-solutions
As we stated earlier, we now construct sub- and super-solutions by modifying the function u??
in (1.93). Concerning the second term U1 , which is defined in (1.37), the terms ?U1 and U1t
1+?
do not make sense as we only assume that g(и, и, u) ? C 1+?, 2 . In order to cope with this
lack of smoothness, as g ? (и, и, u) ? C 2,1 , we replace U1 by a more smooth function U1? , which
is defined by
( ?
U1zz + f ? (U0 (z))U1? = g ? (x, t, U0 (z)) ? ? ? (x, t)U0 ? (z),
(1.98)
U1? (x, t, 0) = 0,
U1? (x, t, и) ? L? (R),
where
?
? (x, t) = c0
Z
?+
g ? (x, t, r)dr.
(1.99)
??
Thus U1? (x, t, z) is a solution of (1.30) with
A = A?0 (x, t, z) := g ? (x, t, U0 (z)) ? ? ? (x, t)U0 ? (z),
(1.100)
where the variables x, t, ? are considered parameters. Using (1.5) and the same arguments as
in the end of Section 1.2, we obtain estimates analogous to (1.39) and (1.40), with a constant
M independent of ?:
|U1? (x, t, z)| ? M, |?x U1? (x, t, z)| ? M.
(1.101)
? are solutions of (1.30) with A = ? A?
Moreover, g ? being C 2 in x and C 1 in t, ?x U1? and U1t
x 0
?
and A = A0t , respectively. Thus, in view of (1.3), we obtain
|?x U1? (x, t, z)| ? C/?,
?
|U1t
(x, t, z)| ? C/?,
(1.102)
for a constant C independent of ?. Similarly, (1.4), (1.5) and Lemma 1.2.3 yield estimates
analogous to (1.44) and (1.45) for U1? , for constants C and M independent of ?:
?
?
(x, t, z)| + |U1zz
(x, t, z)| ? Ce??|z| ,
|U1z
(1.103)
?
|?x U1z
(x, t, z)| ? M.
(1.104)
In the rest of this section, C and M will stand for the constants that appear in inequalities
(1.101)?(1.104). Note also that, by the same arguments used to obtain (1.47), we see that
(1.7) implies the homogeneous Neumann boundary condition for U1? :
?U1?
=0
??
on ?? О [0, T ] О R.
(1.105)
48
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
?
We look for a pair of sub- and super-solutions u▒
? for (P ) of the form
u▒
? (x, t) = U0
where
│ d(x, t) ▒ ?p(t) ┤
?
│
d(x, t) ▒ ?p(t) ┤
+ ?U1? x, t,
▒ q(t),
?
(1.106)
2
p(t) = ?e??t/? + eLt + K,
б
А
2
q(t) = ? ?e??t/? + ?2 LeLt .
Note that q = ??2 pt . It is clear from the definition of u▒
? that
(
?+ for all (x, t) ? Q+
T
lim u▒
(x,
t)
=
?
??0
?? for all (x, t) ? Q?
T.
(1.107)
(1.108)
The main result of this section is the following.
Lemma 1.5.1. Choose ?, ? > 0 appropriately. For any K > 1, we can find positive constants
+
?0 and L such that, for any ? ? (0, ?0 ), the functions (u?
? , u? ) are a pair of sub- and super?
+
solutions for Problem (P ) in the range x ? ?, 0 ? t ? T . In other words, u?
? and u? satisfy
the homogeneous Neumann boundary condition and
+
Lu?
? ? 0 ? Lu? ,
in the range x ? ?, 0 ? t ? T , where we recall that the operator L is defined by
Lu := ut ? ?u ? ??2 (f (u) ? ?g ? (x, t, u)).
1.5.3
Proof of Lemma 1.5.1
By virtue of (1.105) and the fact that ?d = 0 near ??, we have
?u▒
?
=0
??
on ?? О [0, T ].
?
In the following we prove inequality Lu+
? ? 0, the inequality Lu? ? 0 following by the same
argument.
Computation of Lu+
?
Straightforward computations yield
dt
?
?
+ pt ) + ?U1t
+ U1z
(dt + ?pt ) + qt ,
?
?
? ?d
+ ??U1? + U1z
?u+
?d,
? = U0
?
?
(u+
? )t = U0 (
2
|?d|2
?
?
? |?d|
?
? ?d
+
??U
+ U1z
+
U
+
2?U
и
?d
+
U
?d,
0
1
1z
1zz
?2
?
?
А
б
at d(x, t) + ?p(t)┤ /?, whereas
where the function U0 , as well as its derivatives, is evaluated
│
А
б
the function U1? , as well as its derivatives, is evaluated at x, t, d(x, t) + ?p(t) /? . Here, ?U1?
denotes the derivative with respect to x whenever we regard U1? (x, t, z) as a function of three
??
?u+
? = U0
1.5 Motion of interface
49
variables x, t and z. The symbol ?U1? is defined similarly. We also expand the reaction terms
as follows.
1
?
2 ??
?
?
f (u+
? ) = f (U0 ) + (?U1 + q)f (U0 ) + (?U1 + q) f (?),
2
?
g(x, t, u+
? ) = g(x, t, U0 ) + (?U1 + q)gu (x, t, ?),
+
where ?(x, t) and ?(x, t) are some functions satisfying U0 < ? < u+
? , U0 < ? < u? . Writing
?
?
g = g + g ? g and combining the above expressions with equations (1.26) and (1.98), we
obtain
Lu+
? = E1 + и и и + E7 ,
where:
1 │ ?
1 ?? ┤
qf (?) + U0 ? pt + qt ,
q
f
(U
)
+
0
?2
2
│ U ?? U ? ┤
0
=
+ 1zz (1 ? |?d|2 ),
?2
?
│U ?
┤
0
?
=
(dt ? ?d + ?),
+ U1z
?
А
б
? p + 1 q g (x, t, ?) ? U ? f ?? (?) ,
= ?U1z
t
u
1
?
1
?
?
= ?? U1z
? (U1? )2 f ?? (?) + U1? gu (x, t, ?) ? 2 ?U1z
и ?d,
2
E1 = ?
E2
E3
E4
E5
? ? ??U ? ,
E6 = ?U1t
1
1
1 ?
1 ?
?
E7 = (g ? ? g)(x, t, u+
? ) ? (g ? g)(x, t, U0 ) + (? ? ?)(x, t)U0 .
?
?
?
Before starting to estimate each of the above terms, let us present some useful inequalities.
First, by the bistability assumption (1.1), there exist positive constants b, m such that
f ? (U0 (z)) ? ?m
if U0 (z) ? [?? , ?? + b] ? [?+ ? b, ?+ ].
(1.109)
On the other hand, since the region {z ? R | U0 (z) ? [?? + b, ?+ ? b] } is compact and since
U0 ? > 0 on R, there exists a constant a1 > 0 such that
U0 ? (z) ? a1
if U0 (z) ? [?? + b, ?+ ? b].
We set
?=
m
,
4
(1.110)
(1.111)
and choose ? that satisfies
0 < ? ? min (?0 , ?1 , ?2 ),
where
?0 :=
a1
,
m + F1
?1 :=
1
,
?+1
?2 :=
(1.112)
4?
,
F2 (? + 1)
with the constant F1 and F2 defined by
F1 :=
max
?? ?u??+
|f ? (u)|,
F2 :=
max
?? ?2?u??+ +2
|f ?? (u)|.
50
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
Combining (1.109) and (1.110), and considering that ? ? ?0 , we obtain
U0 ? (z) ? ?f ? (U0 (z)) ? ?m
for ? ? < z < ?.
(1.113)
Lu+
?
Now let K > 1 be arbitrary. In what follows we will show that
? 0 provided that the
constants ?0 and L are appropriately chosen. We recall that ?? < U0 < ?+ . We go on under
the following assumptions
?0 M ? 1,
?20 LeLT ? 1 .
(1.114)
It follows from (1.101) that, for all ? ? (0, ?0 ), we have ?|U1? (x, t, z)| ? 1. Moreover, ? ? ?1
implies that 0 ? q(t) ? 1, so that, in view of (1.106),
?? ? 2 ? u▒
? (x, t) ? ?+ + 2 .
(1.115)
The term E1
A straightforward computation gives
E1 =
? ??t/?2
(I ? ??) + LeLt (I + ?2 ?L),
e
?2
where
I = U0 ? ? ?f ? (U0 ) ?
? 2 ??
2
f (?)(?e??t/? + ?2 LeLt ).
2
In virtue of (1.113) and (1.115), we have
?2
F2 (? + ?2 LeLT ).
2
Combining this, (1.114) and the inequality ? ? ?2 , we obtain
I ? ?m ?
I ? 2??.
Consequently, we have
E1 ?
?? 2 ??t/?2
e
+ 2??LeLt .
?2
The term E2
First, in the region where |d| < d0 , we have |?d| = 1, hence E2 = 0. Next we consider the
region where |d| ? d0 . We deduce from Lemma 1.2.1 and from (1.103) that:
1
1
+ )(1 + k?dk2? )e??|d+?p|/?
?2
?
2C
? 2 (1 + k?dk2? )e??(d0 /??|p|) .
?
|E2 | ? C(
By the definition of p in (1.107), we have that 0 < K ? 1 ? p ? eLT + K. Consequently, if we
assume
d0
,
(1.116)
eLT + K ?
2?0
d0
d0
then
? |p| ? , so that
?
2?
2C
|E2 | ? 2 (1 + k?dk2? )e??d0 /(2?)
?
32C
? C2 :=
(1 + k?dk2? ).
(e?d0 )2
1.5 Motion of interface
51
The term E3
We recall that
(dt ? ?d + ?)(x, t) = 0
on
?t = {x ? ?, d(x, t) = 0}.
1+?
By equality (1.97) and assumption (1.5), we see that ? is in C 1+?, 2 so that the interface ? is
3+?
of class C 3+?, 2 . Therefore both ?d and dt are Lipschitz continuous near ?t . It follows, from
the mean value theorem applied separately on both sides of ?t , that there exists a constant
N0 > 0 such that:
|(dt ? ?d + ?)(x, t)| ? N0 |d(x, t)|
for all (x, t) ? QT .
Applying Lemma 1.2.1 and estimate (1.103) we deduce that
|E3 | ? 2N0 C
|d(x, t)| ??|d(x,t)/?+p(t)|
e
?
? 2N0 C maxy?R |y|e??|y+p(t)|
А
1б
? 2N0 C max |p(t)|,
?
А
1б
? 2N0 C |p(t)| + .
?
Thus, recalling that |p(t)| ? eLt + K, we obtain
|E3 | ? C3 (eLt + K) + C3 ? ,
where C3 := 2N0 C and C3 ? := 2N0 C/?.
The term E4
? | are bounded by some constant C. Hence,
In view of (1.4) and (1.103), both gu and |U1z
substituting the expression for pt and q, we obtain
|E4 | ? C4
where C4 := C + ?(C + M F2 ).
А 1 ??t/?2
б
?e
+ ?LeLt ,
?
The term E5
In view of (1.97), the term |?| is bounded by c0 (?+ ? ?? )C on ? О [0, T ]. Using successively
(1.103), (1.101), (1.4) and (1.104), we obtain
1
|E5 | ? c0 (?+ ? ?? )CM + M 2 F2 + M C + 2M k?dk2? =: C5 .
2
The term E6
We use (1.102) to deduce that
|E6 | ? 2C =: C6 .
52
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
Finally the term E7
We recall that |g ? ? g| ? C? so that |? ? ? ?| ? c0 (?+ ? ?? )C?. It then follows, in view of
(1.27), that
|E7 | ? 2C + c0 (?+ ? ?? )C 2 =: C7 .
Completion of the proof
Collecting the above estimates of E1 ?E7 gives
Lu+
? ?(
?? 2 C4 ? ??t/?2
)e
+ (2??L ? C3 ? ?C4 L)eLt ? C8 ,
?
?2
?
(1.117)
where C8 := C2 + KC3 + C3 ? + C5 + C6 + C7 . Now we set
L :=
1
d0
ln
,
T
4?0
which, for ?0 small enough, validates assumptions (1.114) and (1.116). If ?0 is chosen sufficiently small (i.e. L large enough), we have, for all 0 < ? < ?0 , that the first term of the
right-hand side of (1.117) is positive, and that
Б
? ??L ? C3 ]eLt ? C8
Lu+
?
? 12 ??L ? C8
? 0.
The proof of Lemma 1.5.1 is now complete, with the choice of the constants ?, ? as in (1.111),
(1.112).
1.6
Proof of the main results
In this section, we prove our main results by fitting the two pairs of sub- and super-solutions,
constructed for the study of the generation and the motion of interface, into each other.
1.6.1
Proof of Theorem 1.1.4
Let ? ? (0, ?0 ) be arbitrary. Choose ? and ? that satisfy (1.111), (1.112) and
?? ?
?
.
3
(1.118)
By the generation of interface Theorem 1.3.1, there exist positive constants ?0 and M0 such
that (1.49), (1.50) and (1.51) hold with the constant ? replaced by ??/2. Since ?u0 и n 6= 0
everywhere on the initial interface ?0 = {x ? ?, u0 (x) = a} and since ?0 is a compact
hypersurface, we can find a positive constant M1 such that
if
if
d0 (x) ?
M1 ?
then
d0 (x) ? ?M1 ?
then
u0 (x) ? a + M0 ?,
u0 (x) ? a ? M0 ?.
(1.119)
1.6 Proof of the main results
53
Here d0 (x) := d(x, 0) denotes the cut-off signed distance function associated with the hypersurface ?0 . Now we define functions H + (x), H ? (x) by
й
if d0 (x) > ?M1 ?
?+ + ??/2
+
H (x) =
?? + ??/2
if d0 (x) ? ?M1 ?,
й
?+ ? ??/2
if d0 (x) ? M1 ?
H ? (x) =
?? ? ??/2
if d0 (x) < M1 ?.
Then from the above observation we see that
H ? (x) ? u? (x, х?1 ?2 | ln ?|) ? H + (x)
for x ? ?.
(1.120)
Next we fix a sufficiently large constant K > 1 such that
U0 (?M1 + K) ? ?+ ?
??
3
and
U0 (M1 ? K) ? ?? +
??
.
3
(1.121)
For this K, we choose ?0 and L as in Lemma 1.5.1. We claim that
?
u?
? (x, 0) ? H (x),
H + (x) ? u+
? (x, 0)
for x ? ?.
(1.122)
We only prove the former inequality, as the proof of the latter is virtually the same. Then it
amounts to showing that
u?
? (x, 0) = U0
А d0 (x)
А
б
б
d0 (x)
? K + ?U1? x, 0,
? K ? ?(? + ?2 L) ? H ? (x).
?
?
(1.123)
By (1.101) we have |U1? | ? M . Therefore, by choosing ?0 small enough so that ?0 M ? ??/6,
we see that
А d0 (x)
б
? K + ?M ? ?(? + ?2 L)
?
б 5
А d0 (x)
? U0
? K ? ??.
?
6
u?
? (x, 0) ? U0
In the range where d0 (x) < M1 ?, the second inequality in (1.121) and the fact that U0 is an
increasing function imply
U0
А d0 (x)
б 5
??
? K ? ?? ? ?? ?
= H ? (x).
?
6
2
On the other hand, in the range where d0 (x) ? M1 ?, we have
U0
б 5
А d0 (x)
5
? K ? ?? ? ?+ ? ?? ? H ? (x).
?
6
6
This proves (1.123), hence (1.122) is established.
Combining (1.120) and (1.122), we obtain
?
?1 2
+
u?
? (x, 0) ? u (x, х ? | ln ?|) ? u? (x, 0).
+
?
Since u?
? and u? are sub- and super-solutions for Problem (P ) thanks to Lemma 1.5.1, the
comparison principle yields
?
?
+
u?
? (x, t) ? u (x, t + t ) ? u? (x, t)
for
0 ? t ? T ? t? ,
(1.124)
54
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
where t? = х?1 ?2 | ln ?|. Note that, in view of (1.108), this is enough to prove Corollary 1.1.5.
Now let C be a positive constant such that
?
2
U0 (C ? eLT ? K) ? ?+ ?
and
?
U0 (?C + eLT + K) ? ?? + .
2
(1.125)
One then easily checks, using successively (1.124), (1.106), (1.125) and (1.118), that, for ?0
small enough, for 0 ? t ? T ? t? , we have
d(x, t) ?
if
if
C?
then
d(x, t) ? ?C?
then
u? (x, t + t? ) ? ?+ ? ?,
u? (x, t + t? ) ? ?? + ?,
(1.126)
and
u? (x, t + t? ) ? [?? ? ?, ?+ + ?],
which completes the proof of Theorem 1.1.4.
1.6.2
Proof of Theorem 1.1.6
In the case where х?1 ?2 | ln ?| ? t ? T , the assertion of the theorem is a direct consequence of
Theorem 1.1.4. All we have to consider is the case where 0 ? t ? х?1 ?2 | ln ?|. We shall use
the sub- and super-solutions constructed for the study of the generation of interface in Section
1.4. To that purpose, we first prove the following lemma concerning Y (?, ?; ?), the solution of
the ordinary differential equation (1.79), in the initial time interval.
Lemma 1.6.1. There exists a constant C8 > 0 such that
if
if
? ? a + C8 ? then
? ? a ? C8 ? then
Y (?, ?; ▒?G) > a
Y (?, ?; ▒?G) < a
for
for
0 ? ? ? х?1 | ln ?|,
0 ? ? ? х?1 | ln ?|.
(1.127)
Proof. We only prove the first inequality. Assume ? ? a + C8 ?. By (1.76), for C8 ? CG, we
have that ? ? a + C8 ? ? a(▒?G). It then follows from (1.82) that
Y (?, ?; ▒?G) ? a(▒?G) + C1 eх(▒?G)? (a + C8 ? ? a(▒?G))
? a ? CG? + C1 (?CG? + C8 ?)
? a + ?(C1 C8 ? CG(C1 + 1))
> a,
provided that C8 is sufficiently large.
Now we turn to the proof of Theorem 1.1.6. We first claim that there exists a positive
constant M2 such that for all t ? [0, х?1 ?2 | ln ?|],
??t ? NM2 ? (?0 ).
(1.128)
To see this, we choose M0 ? large enough, so that M0 ? ? C8 + 2C6 , where C6 is as in Lemma
1.4.6. As is done for (1.119), there is a positive constant M2 such that
if
if
d0 (x) ?
M2 ?
then
d0 (x) ? ?M2 ?
then
u0 (x) ? a + M0 ? ?,
u0 (x) ? a ? M0 ? ?.
(1.129)
1.6 Proof of the main results
55
In view of this last condition, we see that, if ?0 is small enough and if d0 (x) ? M2 ?, then for
0 ? t ? х?1 ?2 | ln ?|,
u0 (x) ? ?2 r(??G,
Б
ц
t
) ? a + M0 ? ? ? ?2 C6 eх(??G)| ln ?|/х ? 1
2
?
Б
ц
? a + ? M0 ? ? C6 ?(х?х(▒?G))/х + ?C6
А
б
? thanks to (1.92)
? a + ?(M0 ? ? 2C6 )
? a + C8 ?.
This inequality and Lemma 1.6.1 imply w?? (x, t) > a, where w?? is the sub-solution defined in
(1.86). Consequently, by (1.87),
u? (x, t) > a
if d0 (x) ? M2 ?.
In the case where d0 (x) ? ?M2 ?, similar arguments lead to u? (x, t) < a. This completes the
proof of (1.128). Note that we have proved that, for all 0 ? t ? х?1 ?2 | ln ?|,
u? (x, t) > a if x ? ?+
0 \ NM2 ? (?0 ),
u? (x, t) < a if x ? ??
0 \ NM2 ? (?0 ).
(1.130)
Next, since ?t depends on t smoothly, there is a constant C? > 0 such that, for all t ?
[0, х?1 ?2 | ln ?|],
(1.131)
?0 ? NC??2 | ln ?| (?t ),
and
+
?+
t \ NC?? (?t ) ? ?0 \ NM2 ? (?0 ),
?
??
t \ NC?? (?t ) ? ?0 \ NM2 ? (?0 ).
(1.132)
As a consequence of (1.128) and (1.131) we get
??t ? NM2 ?+C??2 | ln ?| (?t ) ? NC? (?t ),
which completes the proof of Theorem 1.1.6.
Proof of Corollary 1.1.7. In view of Theorem 1.1.6 and the definition of the Hausdorff
distance, to prove this corollary we only need to show the reverse inclusion, that is
?t ? NC ? ? (??t )
for
0 ? t ? T,
(1.133)
for some constant C ? > 0. To that purpose let C ? be a constant satisfying C ? > max(C?, C),
where C is as in Theorem 1.1.4 and C? as in (1.132). Choose t ? [0, T ], x0 ? ?t arbitrarily and,
n being the Euclidian normal vector exterior to ?t at point x0 , define a pair of points:
x+ := x0 + C ? ?n and x? := x0 ? C ? ?n.
Since C ? > C and since the curvature of ?t is uniformly bounded as t varies over [0, T ], we see
that
?
?
x+ ? ?+
t \ NC? (?t ) and x ? ?t \ NC? (?t ),
if ? is sufficiently small. Therefore, if t ? [х?1 ?2 | ln ?|, T ], then, by Theorem 1.1.4, we have
u? (x? , t) < a < u? (x+ , t).
(1.134)
56
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
On the other hand, if t ? [0, х?1 ?2 | ln ?|], then from (1.130), (1.132) and the fact that C ? > C?,
we again obtain (1.134). Thus (1.134) holds for all t ? [0, T ]. Now, by the mean value theorem,
we see that for each t ? [0, T ] there exists a point x1 such that
x1 ? [x? , x+ ]
and u? (x1 , t) = a.
This implies x1 ? ??t . Furthermore we have |x0 ? x1 | ? C ? ?, since x1 lies on the line segment
[x? , x+ ]. This proves (1.133).
1.7
Application to reaction-diffusion systems
In this section we discuss the singular limit of the reaction-diffusion system (RD ? ) and prove
Theorems 1.1.12, 1.1.14 and their corollaries. Our strategy is to regard the first equation of
(RD ? ) as a perturbed Allen-Cahn equation and apply what we have already proved for this
equation. In order to make this strategy work, we will modify our former arguments slightly.
1.7.1
Preliminaries: global existence
Before studying the singular limit of (RD ? ), we first show that the solution of this system
exists globally for t ? 0, provided that ? is sufficiently small. Recall that the system (RD ? ) is
written in the form
?
б
А
? ut = ?u + 1 f (u) + ? f1 (u, v) + O(?2 ) ,
2
?
?
vt = D?v + h(u, v),
where h(u, v) satisfies the hypothesis (H). The standard parabolic theory guarantees the existence of local solutions for (RD ? ). In order to prove that the solution exists globally for t ? 0,
it suffices to show that the solution remains uniformly bounded. This will be done by using
the method of invariant rectangles.
Given arbitrary u0 , v0 ? C(?), we choose a constant L > 0 such that
f (?L) > 0 > f (L),
?L ? u0 (x) ? L for x ? ?.
(1.135)
Such a constant L exists since f (u) > 0 for u < ?? , and f (u) < 0 for u > ?+ . By the
hypothesis (H), we can choose a constant M1 satisfying
M1 ? kv0 kL? (?) ,
along with the condition (1.19), namely
h(u, ?M1 ) ? 0 ? h(u, M1 )
for |u| ? L.
(1.136)
Now we consider the rectangle
»
R := { (u, v) ? R2 » |u| ? L, |v| ? M1 }.
It follows from (1.135) that, for all sufficiently small ? > 0,
f ? (?L, v) > 0 > f ? (L, v)
for |v| ? M1 .
(1.137)
1.7 Application to reaction-diffusion systems
57
The inequalities (1.136) and (1.137) imply that the rectangle R is a positively invariant region
for the system of ordinary differential equations:
?
? ut = 1 f ? (u, v),
?2
?
vt = h(u, v),
since the vector field (??2 f ? (u, v), h(u, v)) points inwards everywhere on the boundary of R.
The maximum principle then implies that R is also positively invariant for the system (RD ? ).
Consequently, since (u0 (x), v0 (x)) ? R for x ? ?, we have
(u(x, t), v(x, t)) ? R
for x ? ?, t ? 0,
as long as the solution is defined. This uniform bound then implies that the solution exists
globally for t ? 0.
In the case of equations for which only nonnegative solutions are to be considered (see
Remark 1.1.10), we can argue just similarly, by replacing R by the rectangle R+ := {(u, v) | 0 ?
u ? L, 0 ? v ? M1 }. Summarizing, we have proved the following proposition:
Proposition 1.7.1. Let (u0 , v0 ) ? C(?) О C(?). In the case where the conditions of Remark
1.1.10 apply, assume further that u0 , v0 ? 0. Then there exists ?0 > 0 such that for any
? ? (0, ?0 ), the solution (u? , v ? ) of (RD ? ) exists globally for t ? 0 and is uniformly bounded.
Remark 1.7.2. For the details of the method of invariant rectangles, we refer the reader to the
book [68], Chapter 14, Corollary 14.8. See also [28] and [27]. It should be noted that [50] makes
a much earlier study of invariant rectangles for a finite-difference scheme for reaction-diffusion
systems.
1.7.2
Re-examination of the Allen-Cahn equation
Now we turn to the singular limit of (RD ? ). As we have mentioned earlier, our strategy
is to regard the first equation of (RD ? ) as a perturbed Allen-Cahn equation of the form
(P ? ). However, our results for Problem (P ? ), Theorems 1.1.4 and 1.1.6, do not apply to the
system (RD ? ) directly, because of the assumption (1.6), which requires |g ? ? g| to be of order
?. Naturally, such an assumption cannot be made a priori for the system (RD ? ), since the
perturbation term g ? := ?f1 (u, v ? (x, t)) + O(?) depends on the unknown function v ? .
In view of this, we will first re-examine our previous argument for the single equation (P ? )
and see what we can say without the assumption (1.6).
We begin with some notation. Given any function g?(x, t, u) satisfying the conditions (1.3)
and (1.5), and any smooth hypersurface without boundary ??0 such that ??0 ?? ?, we can define
the classical solution of the interface equation (P 0 ) on some time interval 0 ? t < T (g?; ??0 ).
We denote this solution by ?t [ g?; ??0 ] in order to clarify its dependence on g? and ??0 . More
specifically, ?t [ g?; ??0 ] is the solution of the problem
(Pg?,0??0 )
?
Z
?
? Vn = ?(N ? 1)? + c0
?
? ? »»
t
?+
??
t=0
= ??0 .
g?(x, t, r)dr
on ?t ,
58
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
Also, we denote by u? [ g?; u?0 ](x, t) the solution of the problem
?
1
?
?
ut = ?u + 2 (f (u) ? ?g?(x, t, u)) in ? О (0, +?),
?
?
?
?
?
?u
(Pg?,u?
)
0
=0
on ?? О (0, +?),
?
?
??
?
?
?
u(x, 0) = u?0 (x)
in ?,
and define
??t [ g?; u?0 ] := {x ? ?, u? [ g?; u?0 ](x, t) = a}.
+
Once the interface ?t [ g?; ??0 ] is given, we denote by ??
t [ g?; ??0 ], ?t [ g?; ??0 ] the region enclosed
by ?t [ g?; ??0 ] and the one enclosed between ?? and ?t [ g?; ??0 ], respectively. As in (1.14), we
define the step function u?[ g?; ??0 ](x, t) by
(
?+ in ?+
t [ g?; ??0 ]
u?[ g?; ??0 ](x, t) =
for t ? [0, T (g?; ??0 )).
(1.138)
?? in ??
t [ g?; ??0 ]
With these notations, the solution u? , the interfaces ??t , ?t and the step function u? which
we have defined in Section 1.1, can be expressed as follows:
u? = u? [ g ? ; u0 ],
??t = ??t [ g ? ; u0 ],
?t = ?t [ g; ?0 ],
u? = u?[ g; ?0 ].
Henceforth we will fix the initial datum u0 (hence ?0 ) throughout this subsection.
Now let us consider what happens if we do not assume (1.6), i.e. |g ? g ? | = O(?). Sections
1.3 and 1.4, which deal with the generation of interface, remain unchanged since the assumption
(1.6) is not used. The only places where this assumption has been used are the following.
? In the formal derivation of the interface equation in Section 1.2, the assumption (1.6) is
used while collecting the O(??1 ) terms, leading to the conclusion that the second term
U1 of the inner asymptotic expansion is given by a solution of equation (1.29);
? In subsection 1.5.3, the assumption (1.6) is used to show the boundedness of the term
E7 .
Note that the term E7 appears, so to say, as a result of discrepancy between the term U1? and
the distance function d that are used to define u▒
? in (1.106). More precisely, the distance
0 ), while
function d(x, t) is associated with the interface ?t [ g; ?0 ], whose law of motion is (Pg,?
0
? appear
,
both
g
and
g
U1? is associated with g ? via (1.98). Therefore, when we calculate Lu▒
?
without cancelling each other.
On the other hand, if we replace d by the signed distance function d ? associated with the
interface ?t [ g ? ; ?0 ], and define u?▒
? by
│ d ? (x, t) ▒ ?p(t) ┤
│
d ? (x, t) ▒ ?p(t) ┤
?
x,
t,
u?▒
+
?U
▒ q(t),
(1.139)
(x,
t)
=
U
0
?
1
?
?
then the term E7 does not appear in the calculation of Lu?▒
? . Moreover the remaining terms E1
to E6 are virtually the same as those in subsection 1.5.3. Consequently, the new functions u?▒
?
are sub- and super-solutions even without the assumption (1.6). Thus, arguing as in subsection
1.6.1, we obtain the following analogue of (1.124):
?
?
+
u??
? (x, t) ? u (x, t + t ) ? u?? (x, t)
for
0 ? t ? T ? ? t? ,
where t? = х?1 ?2 | ln ?| and T ? is a constant such that 0 < T ? < T (g ? ; ?0 ).
Summarizing, the following proposition holds.
(1.140)
1.7 Application to reaction-diffusion systems
59
Proposition 1.7.3. The conclusions of Theorems 1.1.4 and 1.1.6 hold without assuming (1.6),
provided that ?t [ g; ?0 ] is replaced by ?t [ g ? ; ?0 ]. In particular,
dH (??t [ g ? ; u0 ], ?t [ g ? ; ?0 ]) ? C?
0 ? t ? T ?,
for
(1.141)
where dH denotes the Hausdorff distance.
1.7.3
Interface motion under various perturbations
Next we show that ?t [ g?; ??0 ] depends on g? and ??0 continuously. To this end, we first fix
constants C? > 0, T ? > 0, ? ? (0, 1), and denote by Y the set of functions g?(x, t, u) on
? О [0, T ? ] О R satisfying
sup
u? [?? ,?+ ]
kg?(и, и, u)k
C 1+?,
1+?
2 (?О[0,T ? ])
? C? .
(1.142)
We also fix M a N ? 1 dimensional manifold without boundaries. We denote by Z the set of
C 3+? hypersurfaces without boundary ??0 satisfying ??0 ?? ? and such that
k??0 kC 3+? (M) ? C? ,
(1.143)
where the function ??0 : M 7? [?L, L] is a parametrization of ??0 . For more details we refer to
[23].
Proposition 1.7.4. Let g? ? Y and ??0 ? Z. Let T ? (0, T (g?; ??0 )). Then there exist positive
constants ?, K, M such that, for any g? ? Y and any ??0 ? Z satisfying
kg? ? g?kL? (?О[0,T ]О[?? ,?+ ]) ? ?
dH (??0 , ??0 ) ? ?,
and
there holds that T (g?; ??0 ) > T , where we recall that T (g?; ??0 ) is the maximum time of existence
of a classical solution of Problem (Pg?,0?? ). Furthermore, for each t ? [0, T ],
0
dH ( ?t [g?; ??0 ], ?t [ g?; ??0 ] ) ? K(eM t ? 1) kg? ? g?kL? + eM t dH (??0 , ??0 ) ,
(1.144)
where the L? norm on the right-hand side is taken in ? О [0, t] О [?? , ?+ ].
Proof. By using the local coordinates, one can express ?t [g?, ??0 ], respectively ?t [g?, ??0 ], as a
graph over M and transfer the motion equation (Pg?,0?? ), respectively (Pg?,0?? ), into a quasi-linear
0
0
parabolic equation on the manifold M О [0, T (g?, ??0 )], respectively M О [0, T (g?, ??0 )]. Since g?
and g? satisfy (1.142), and since the embedding
C 1+?,
1+?
2
?
? 1+?
2
?? C 1+? ,
is compact if 0 < ?? < ?, the assumption kg? ? g?kL? ? ? implies
kg?(и, и, u) ? g?(и, и, u)k
? , 1+??
2
C 1+?
? C(?),
where C(?) is a constant satisfying C(?) ? 0, as ? ? 0. Consequently,
Z ?+
Z
░
░ ?+
░
g?(и, и, r)dr ?
g?(и, и, r)dr ░ 1+?? , 1+?? ? (?+ ? ?? )C(?).
??
??
C
2
60
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
By a similar argument, the assumption dH (??0 , ??0 ) ? ? implies
k??0 ? ??0 kC 3+?? (M) ? C ? (?),
where C ? (?) is a constant satisfying C ? (?) ? 0, as ? ? 0. The assertion that T (g?, ??0 ) > T then
follows from the standard local existence theory. For more details, we refer to [3] Theorem 7.1
and its corollary, which are concerned with smooth curvature flows with slightly perturbations
of the velocities and the initial data.
Next we prove the estimate (1.144). This will be done by using the maximum principle.
Let us introduce some notation. For each g? ? Y and each ??0 ? Z, we denote by d(x, t; g?; ??0 )
the signed distance function associated with the interface ?t [ g?; ??0 ]. By ??t ╣ ??t we mean that
??t lies inside of ??t . Clearly we have
?t [ g?; ??0 ] ╣ ?t [ g?; ??0 ] ?? d(x, t; g?; ??0 ) ? d(x, t; g?; ??0 ) for x ? ?.
(1.145)
Now we choose t0 ? [0, T ] arbitrarily and put
?0 := kg? ? g?kL? (?О[0,t0 ]О[?? ,?+ ])
and
?1 := dH (??0 , ??0 ).
Then
g?(x, t, u) ? ?0 ? g?(x, t, u) ? g?(x, t, u) + ?0 ,
for x ? ?, 0 ? t ? t0 , ?? ? u ? ?+ , and
??0 ? ?1 ╣ ??0 ╣ ??0 + ?1 ,
where, by definition
??0 ▒ ?1 := {x ▒ ?1 n; x ? ??0 },
n being the outward unit normal to ??0 . The comparison principle then yields
?t [ g? ? ?0 ; ??0 ? ?1 ] ╣ ?t [ g?; ??0 ] ╣ ?t [ g? + ?0 ; ??0 + ?1 ]
for 0 ? t ? t0 .
Thus, in order to prove (1.144), it suffices to show that there exists constants K, M > 0 such
that, for all small ?0 > 0, ?1 > 0,
(
dH (?t [ g? ? ?0 ; ??0 ? ?1 ], ?t [ g?; ??0 ]) ? K?0 (eM t ? 1) + ?1 eM t ,
(1.146)
dH (?t [ g? + ?0 ; ??0 + ?1 ], ?t [ g?; ??0 ]) ? K?0 (eM t ? 1) + ?1 eM t ,
for 0 ? t ? t0 . We will only show the latter inequality for ?t [ g? + ?0 ; ??0 + ?1 ] since the former
can be shown in the same manner.
Recall that d(x, t; g?; ??0 ) satisfies the equation (1.96), namely
Z ?+
g?(x, t, r)dr on ?t [ g?; ??0 ].
(1.147)
dt = ?d ? c0
??
Choose a constant d0 > 0 such that d(x, t; g?; ??0 ) is smooth ? say, C 3 in x and C 3/2 in t ? in
the neighborhood Nd0 (?t [ g?; ??0 ]), 0 ? t ? T . By equality (1.147) and the mean value theorem
applied separately on both sides of the interface, there exists a constant N0 > 0 such that
Z ?+
»
»
»dt ? ?d + c0
g?(x, t, r)dr» ? N0 |d| in Nd0 /2 (?t [ g?; ??0 ]).
??
1.7 Application to reaction-diffusion systems
Now we put
61
dnew (x, t) := d(x, t; g?; ??0 ) ? K?0 (e2N0 t ? 1) ? ?1 e2N0 t ,
??t := {x ? ? | dnew (x, t) = 0},
where the constant K is to be determined later. If
?0 ? ?0? :=
e?2N0 T d0 ?1
K
4
and
?1 ? ?1? :=
e?2N0 T d0
,
4
then ??t lies within the neighborhood Nd0 /2 (?t [ g?; ??0 ]). Observe that
(dnew )t ? ?dnew = dt ? 2N0 K?0 e2N0 t ? 2N0 ?1 e2N0 t ? ?d
Z ?+
g?(x, t, r)dr + N0 |d| ? 2N0 K?0 e2N0 t ? 2N0 ?1 e2N0 t .
? ?c0
??
Since d = K?0 (e2N0 t ? 1) + ?1 e2N0 t on ??t , we obtain
Z ?+
new
new
? ?c0
(d )t ? ?d
g?(x, t, r)dr ? N0 K?0 e2N0 t ? N0 ?1 e2N0 t
on ??t
??
? ?c0
Z
?+
??
g?(x, t, r)dr ? N0 K?0
on ??t .
Now we set
K = (?+ ? ?? )c0 N0 ?1 .
Then it follows from the above inequality that
Z ?+
(dnew )t ? ?dnew ? c0
g?(x, t, r)dr ? (?+ ? ?? )c0 ?0
on ??t .
??
This inequality and the fact that dnew (x, 0) = d(x, 0 ; g?; ??0 ) ? ?1 imply that ??t satisfies
?
Z ?+
А
б
?
? Vn ? ?(N ? 1)? + c0
g?(x, t, r) + ?0 dr on ??t ,
?
? ?? »»
t
t=0
??
= ??0 + ?1 .
On the other hand, ?t [g? + ?0 ; ??0 + ?1 ] satisfies
?
Z ?+
А
б
?
? Vn = ?(N ? 1)? + c0
g?(x, t, r) + ?0 dr
on ?t [g? + ?0 ; ??0 + ?1 ],
??
?
? ? [g? + ? ; ?? + ? ]»»
t
0
0
1 t=0 = ??0 + ?1 .
By the comparison principle, we obtain
?t [ g?; ??0 ] ╣ ?t [g? + ?0 ; ??0 + ?1 ] ╣ ??t
for 0 ? t ? t0 .
Consequently,
dH (?t [g? + ?0 ; ??0 + ?1 ], ?t [ g?; ??0 ]) ? dH (??t , ?t [ g?; ??0 ]) ? K?0 (e2N0 t ? 1) + ?1 e2N0 t ,
for 0 ? t ? t0 . The proposition is proved.
Before closing this subsection, we remark that our main results for the Allen-Cahn equation
? Theorems 1.1.4 and 1.1.6 ? can also be derived from Propositions 1.7.3 and 1.7.4.
62
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
1.7.4
Proof of the main results
Now we turn to the reaction-diffusion system (RD ? ). In what follows, we fix the initial datum
(u0 , v0 ) and denote the solution of this system by (u? , v ? ). We need some more notation; given
a function v(x, t) on ? О [0, ?), we set
g ? [v](x, t, u) := ?f1 (u, v(x, t)) ? ? f2? (u, v(x, t)),
g[v](x, t, u) := ?f1 (u, v(x, t)),
(1.148)
where f1 , f2? are as in (1.18). The first equation of (RD ? ) is then written in the form
ut = ?u +
б
1А
f (u) ? ? g ? [v ? ](x, t, u) .
2
?
The limit Problem (RD 0 ) is decomposed into two parts:
?
Z ?+
?
? Vn = ?(N ? 1)? + c0
g[v?](x, t, r) dr
?
? ? »»
t
t=0
and
on ?t ,
??
(1.149)
(1.150)
= ?0 ,
?
v?t = D?v? + h(u?, v?)
?
?
?
?
?v?
=0
?
??
?
?
?
v?(x, 0) = v0 (x)
in ? О (0, T ],
on ?? О (0, T ],
(1.151)
in ?,
where u? is the step function associated with the interface ?t . Using the notation given in
subsection 1.7.2, the above interface ?t in (1.150) can be written as ?t [ g[v?]; ?0 ].
In order for Theorems 1.1.4 and 1.1.6 to be applicable to Problem (RD? ), we have to verify
the conditions (1.3) to (1.6). More precisely, we have to show that, for all small ? > 0,
|?x g ? [v ? ](x, t, u)| ? C??1
and
|?t g ? [v ? ](x, t, u)| ? C??1 ,
|?u g ? [v ? ](x, t, u)| ? C,
? C,
kg ? [v ? ](и, и, u)k 1+?, 1+?
2 (?О[0,T ])
C
» ? ?
»
»g [v ](x, t, u) ? g[v?](x, t, u)» ? C?.
Since g[v], g ? [v] are defined by (1.148) and since f1 , f2? are smooth (see assumption (F) in
subsection 1.1.2), it suffices to prove the following estimates for some C > 0 and for all small
? > 0:
|?x v ? (x, t)| ? C??1 and |?t v ? (x, t)| ? C??1 ,
(1.152)
kv ? k
C 1+?,
1+?
2 (?О[0,T ])
? C,
|v ? (x, t) ? v?(x, t)| ? C?.
(1.153)
(1.154)
The first two estimates are elementary. In fact, since v ? satisfies
vt? = D?v ? + h(u? , v ? )
in ? О (0, T ]
(1.155)
along with the homogeneous Neumann boundary condition, it can be expressed as
v ? (x, t) = I1 + I2 ,
(1.156)
1.7 Application to reaction-diffusion systems
where
I1 :=
I2 :=
Z
G(x, y, t)v0 (y)dy,
?
Z tZ
0
63
?
G(x, y, t ? s) h(u? (y, s), v ? (y, s)) dy ds,
with G(x, y, t) being the fundamental solution for equation vt = D?v under the homogeneous
Neumann boundary condition. Since h(u? , v ? ) is uniformly bounded, standard estimates of
G(x, y, t) imply (1.153) for any ? ? (0, 1). In the meanwhile, the same rescaling argument as
in Remark 1.1.8 yields
? C??? .
(1.157)
ku? k ?, ?
C
2
(?О[0,T ])
Indeed, since ?y u? , u?? are bounded, where y = x/?, ? = t/?2 , we have ?x u? = O(1/?),
u?t = O(1/?2 ). Consequently we have
|u? (x, t) ? u? (x? , t? )|
|x ? x? |? + |t ? t? |?/2
?
|u? (x, t) ? u? (x? , t)| |u? (x? , t) ? u? (x? , t? )|
+
|x ? x? |?
|t ? t? |?/2
? |u? (x, t) ? u? (x? , t)|1??
|u? (x, t) ? u? (x? , t)|?
|x ? x? |?
+|u? (x? , t) ? u? (x? , t? )|1??/2
|u? (x? , t) ? u? (x? , t? )|?/2
|t ? t? |?/2
?/2
? (2ku? kL? )1?? k?x u? k?L? + (2ku? kL? )1??/2 ku?t kL?
? C??? .
Combining (1.157) and (1.153), we see that kh(u? , v ? )k
?
C ?, 2 (?О[0,T ])
Schauder estimate,
kI2 k
?
C 2+?,1+ 2 (?О[0,T ])
? C??? , hence, by the
? C??? .
Here the constant C may depend on the choice of ? ? (0, 1). On the other hand, I1 is bounded
in C 2,1 (? О [0, T ]) since v0 ? C 2 (?). Combining these, we obtain that |?x v ? (x, t)| = O(??? ),
hence O(??1 ). By a similar argument, we obtain that |?t v ? (x, t)| = O(??1 ).
It remains to prove (1.154). This requires more elaborate analysis. Let us introduce some
more notation. Given functions u(x, t) and v?0 (x) on ? О [0, ?), we denote by V [u; v?0 ](x, t) the
solution of the problem
?
in ? О (0, T ],
Vt = D?V + h(u(x, t), V )
?
?
?
?
?V
(1.158)
=0
on ?? О (0, T ],
?
??
?
?
?
in ?.
V (x, 0) = v?0 (x)
The solution v ? of (RD ? ) and v? of (RD 0 ) can then be expressed as
v ? = V [u? ; v0 ],
v? = V [u?; v0 ].
Recall also that, with the notation defined in subsection 1.7.2 and in (1.148), the solution u?
of (RD ? ) and the step function u? in (RD 0 ) are expressed as
u? = u? [ g ? [v ? ]; u0 ],
u? = u?[ g[v?]; ?0 ].
64
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
First estimate. Let us compare u? := u? [g ? [v ? ]; u?0 ] with the step function u?? := u?[ g ? [v ? ]; ??0 ].
By (1.140), we have
?
?
+
u??
? (x, t) ? u (x, t + t ) ? u?? (x, t)
for
0 ? t ? T ? t? ,
?
where u?▒
? are as in (1.139), d being the signed distance function associated with the interface
2
?t [g ? [v ? ]; ??0 ]. Since the term e??t/? in q(t), see (1.107), quickly becomes small,
│ d ? (x, t) ? ?p(t) ┤
? ?
+ O(?) in ?+
|u? (x, t) ? u?? (x, t)| ? ?+ ? U0
t [ g [v ]; ??0 ],
?
│ d ? (x, t) + ?p(t) ┤
? ?
|u? (x, t) ? u?? (x, t)| ? U0
? ?? + O(?) in ??
t [ g [v ]; ??0 ],
?
for х1 ?2 | ln ?| ? t ? T , provided that we choose the constant х1 large enough. Consequently,
by Lemma 1.2.1, there exist constants B, C > 0 such that
│
|d ? (x, t)| ┤
+ C?,
|u? (x, t) ? u?? (x, t)| ? B exp ? ?
?
(1.159)
for (x, t) ? ? О [х1 ?2 | ln ?|, T ].
Second estimate. Next we compare v ? := V [u? ; v?0 ] and v? ? := V [u?? ; v?0 ]. Set w := v ? ? v? ? .
Then
А
б
wt = D?w + h(u? , v ? ) ? h(u?? , v? ? ) .
Since
|h(u? , v ? ) ? h(u?? , v? ? )| ? C|w| + C|u? (x, t) ? u?? (x, t)|,
for some constant C > 0, the function w? := e?Ct w satisfies
w?t ? D?w? + Ce?Ct |u? (x, t) ? u?? (x, t)| + C(|w?| ? w?),
hence
w?t ? D?w? + C|u? (x, t) ? u?? (x, t)| + C(|w?| ? w?).
(1.160)
Now let W (x, t) be the solution of the equation
Wt = D?W + C|u? (x, t) ? u?? (x, t)| + C(|W | ? W ),
with initial datum W (x, 0) = |v?0 (x)?v?0 (x)|. Then since (1.160) implies that w? is a sub-solution
of the above equation, and since
w?(x, 0) ? |w?(x, 0)| = |v?0 (x) ? v?0 (x)| = W (x, 0),
we have
for x ? ?, t ? 0.
w?(x, t) ? W (x, t)
(1.161)
Moreover, since W ? 0, the above equation for W can be reduced to
Wt = D?W + C|u? (x, t) ? u?? (x, t)|.
In view of this and W (x, 0) = |v?0 (x) ? v?0 (x)|, we see that
W (x, t) = C
Z tZ
0
?
?
?
G(x, y, t ? s)|u (y, s) ? u? (y, s)| dy ds +
Z
?
G(x, y, t)|v?0 (y) ? v?0 (y)|dy,
1.7 Application to reaction-diffusion systems
65
G(x, y, t) being the fundamental solution that appears in (1.156). This and (1.161) yield
Ct
|w(x, t)| ? Ce
Z tZ
G(x, y, t ? s)|u? (y, s) ? u?? (y, s)| dy ds
?
Z
G(x, y, t)|v?0 (y) ? v?0 (y)|dy. (1.162)
+ eCt
0
?
Combining this and (1.159), we obtain
|w(x, t)| ? BCeCt
Z tZ
0
│
|d ? (y, s)| ┤
dy ds
G(x, y, t ? s) exp ? ?
?
?
+ O(?) + C ? kv?0 ? v?0 kL? (?) , (1.163)
for some constant C ? > 0. In order to estimate the above integral, we need the following
lemma.
Lemma 1.7.5. Let ? be a smooth closed hypersurface in ? and denote by d(x) the signed
distance function associated with ?. Then there exist constants C, r0 > 0 such that for any
function ?(r) ? 0 on R, it holds that
Z
Z r0
C
G(x, y, t) ?(d(y)) dy ? ?
?(r) dr for 0 < t ? T.
(1.164)
t ?r0
|d|?r0
The proof of this lemma will be given in the next subsection. As is easily seen from its proof,
the above estimate remains to hold if ? depends on t smoothly; in other words, the constant C
can be chosen uniformly as ? varies. Applying the above estimate to ?t [ g ? [v ? ]; ??0 ], 0 < t ? T ,
we obtain
Z
│
|d ? (y, s)| ┤
dy
G(x, y, t ? s) exp ? ?
?
?
Z
Z
│
|d ? (y, s)| ┤
+
G(x, y, t ? s) exp ? ?
=
dy
?
|d ? |<r0
|d ? |?r0
А
А ? б
б
+ O e??r0 /?
=O ?
t?s
А ? б
.
=O ?
t?s
It follows from this and (1.163) that
┤
│ Z t
1
?
ds + O(?) + C ? kv?0 ? v?0 kL? (?) ,
|w(x, t)| = O ?
t?s
0
(1.165)
hence
v ? (x, t) ? v? ? (x, t) = O(?) + O(kv?0 ? v?0 kL? (?) ).
(1.166)
Key estimate. Finally we compare the two step functions u?? := u?[ g ? [v ? ]; ??0 ] and u?[ g[v ? ]; ??0 ].
By (1.148), we have kg ? [v ? ] ? g[v ? ]kL? = O(?). In the following we assume dH (??0 , ??0 ) ? ?, so
that, by Proposition 1.7.4, we have
sup dH (?t [ g ? [v ? ]; ??0 ], ?t [ g[v ? ]; ??0 ]) = O(?) + O(dH (??0 , ??0 )).
0?t?T
66
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
This means that u?[ g ? [v ? ]; ??0 ] and u?[ g[v ? ]; ??0 ] differ only in a thin neighborhood of ?t [ g[v ? ]; ??0 ]
of thickness O(?) + O(dH (??0 , ??0 )). Arguing as above and applying Lemma 1.7.5 again, we see
that
Б
ц
Б
ц
kV u?[ g ? [v ? ]; ??0 ]; v?0 ? V u?[ g[v ? ]; ??0 ]; v?0 kL? (?О[0,T ]) = O(?) + O(dH (??0 , ??0 )).
Combining this and (1.166), we obtain
ц
Б
k v ? ? V u?[ g[v ? ]; ??0 ]; v?0 kL? (?О[0,T ]) = O(?) + O(dH (??0 , ??0 )) + O(kv?0 ? v?0 kL? (?) ). (1.167)
Conclusion. In what follows we will show that (1.167) implies our desired estimate (1.154).
Our proof is based on a contraction mapping argument, but this argument applies only to a
certain time interval [0, T1 ] ? [0, T ]. Once we obtain (1.154) for the interval [0, T1 ], we will
repeat the same argument to derive (1.154) on an interval [T1 , 2T1 ], and this ?step by step?
procedure eventually yields (1.154) on the whole interval [0, T ]. To make the above strategy
work, we first introduce some notation. Choose a constant C ? > 0 sufficiently large so that
the estimate (1.153) holds with C = C ? for all small ? > 0, and that kv?k 1+?, 1+? ? C ? . We
2
C
fix such C ? > 0 and define
U
:= { v ? C 1+?,
1+?
2
(? О [0, T ]), kvk
C 1+?,
1+?
2
? C ? },
U? := { v ? U, kv ? v?kL? ? ? }.
Also we choose C? > 0 large enough so that v ? U implies that g[v](x, t, u) satisfies (1.142).
We remark that U? is a closed subset of L? (? О [0, T ]), since kvn k 1+?, 1+? ? C ? and vn ? v
2
C
in L? implies kvk 1+?, 1+? ? C ? . Consequently U? is a complete metric space with respect to
2
C
the L? topology.
Fix v?0 and ??0 . If ? > 0 is chosen small enough, then by Proposition 1.7.4 the classical
solution of (Pg,0 ?? ) with g = g[v] exists on the entire interval [0, T ]; we denote it by ?t [ g[v]; ??0 ] as
0
Б
ц
before. This determines the step function u?[ g[v]; ??0 ], which then determines V u?[ g[v]; ??0 ]; v?0 .
Combining these, we can define a mapping
ц
Б
? : v 7? V u?[ g[v]; ??0 ]; v?0
from U? into L? (? О [0, T ]). Roughly speaking, ? is a contraction mapping on a time interval
[0, T1 ]. More precisely, the following result holds.
Lemma 1.7.6. ? is a Lipschitz continuous map from U? into L? (? О [0, T ]). Moreover, there
exist constants T1 > 0 and ? ? (0, 1) such that
k?[v1 ] ? ?[v2 ]kL? (QT1 ) ? ? kv1 ? v2 kL? (QT1 ) ,
(1.168)
for any v1 , v2 ? U? .
The proof of this lemma will be given in the next subsection. We are now ready to prove the
key estimate (1.154).
? 1st step: We first put u?0 = u0 , v?0 = v?0 = v0 , ??0 = ??0 = ?0 . It follows that
u? := u? [g ? [v ? ]; u?0 ] = u? [g ? [v ? ]; u0 ] = u? ,
v ? := V [u? ; v?0 ] = V [u? ; v0 ] = v ? ,
1.7 Application to reaction-diffusion systems
67
so that (1.167) yields
kv ? ? ?[v ? ]kL? (?О[0,T1 ]) = O(?).
(1.169)
In other words, v ? is almost a fixed point of the mapping ? by an error margin of O(?).
Note that v? is a fixed point of the map ?:
v? = ?[ v? ].
These two observations are sufficient to conclude that v? and v ? are close to each other,
by O(?), until time T1 . Indeed, by the above lemma, we have, working on QT1 ,
k?[v ? ] ? v?kL? = k?[v ? ] ? ?[v?]kL? ? ?kv ? ? v?kL? .
On the other hand,
k?[v ? ] ? v?kL? ? kv ? ? v?kL? ? kv ? ? ?[v ? ]kL? .
Combining these, we obtain
kv ? ? v? kL? ?
1
kv ? ? ?[v ? ] kL? .
1??
(1.170)
In view of (1.169), this proves (1.154) on ? О [0, T1 ]. Hence, conditions (1.152) to (1.154)
are satisfied, at least until time t = T1 ; we can then apply our results for the single
equation to system (RD ? ) and obtain, by Corollary 1.1.7,
dH (?T1 , ??T1 ) = O(?).
(1.171)
? 2nd step: We take T1 as a new initial moment and put u?0 = u? (и, T1 ), v?0 = v ? (и, T1 ),
v?0 = v?(и, T1 ), ??0 = ??T1 := {x ? ?, u? (x, T1 ) = a}, ??0 = ?T1 . It follows that
u? := u? [g ? [v ? ]; u?0 ] = u? [g ? [v ? ]; u? (и, T1 )] = u? ,
v ? := V [u? ; v?0 ] = V [u? ; v ? (и, T1 )] = v ? .
By the result of the first step, we have
kv?0 ? v?0 kL? (?) = kv ? (и, T1 ) ? v?(и, T1 )kL? (?) = O(?).
Moreover, by (1.171),
dH (??0 , ??0 ) = dH (?T1 , ??T1 ) = O(?),
so that (1.167) leads to
kv ? ? ?[v ? ]kL? (?О[T1 ,2T1 ]) = O(?).
Then, using the same arguments as in the first step, we obtain estimate (1.154) on
? О [T1 , 2T1 ], and also an analogue of estimate (1.171) at time t = 2T1 ; repeating this
procedure a finite number of times we obtain estimate (1.154) on ? О [0, T ].
Hence, all the conditions (1.152) to (1.154) are verified so that Theorems 1.1.12 and 1.1.14,
along with their corollaries, follow directly from Theorems 1.1.4, 1.1.6 and their corollaries.
This completes the proof of the main results for (RD ? ).
68
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
1.7.5
Proof of Lemmas 1.7.5 and 1.7.6
Proof of Lemma 1.7.5. We first show that
Z
C
G(x, y, t) dSy ? ?
t
?
for x ? ?, 0 < t ? T.
(1.172)
It suffices to prove this estimate on a small interval [0, t0 ], since the estimate for the remaining
interval [t0 , T ] will follow by simply choosing a large constant C. Note also that, for t sufficiently small, G(x, y, t) is well approximated by the fundamental solution on the entire space
RN :
│ |x ? y|2 ┤
1
.
G0 (x, y, t) :=
exp
?
4Dt
(4?Dt)N/2
In particular, there exists a constant C ? > 0 such that
0 < G(x, y, t) ? C ? G0 (x, y, t)
for x, y ? ?, 0 < t ? t0 ;
for this result, we refer to [35], Chapter I, Section IV.2. Thus it suffices to prove (1.172) for
G0 instead of G.
Given x ? ?, let x0 be the point on ? that is closest to x, and let ?(x0 ) be the outward
normal to ? at x0 . Then x ? x0 = d(x)?(x0 ). Define
Ye := { y ? RN , y и ?(x0 ) = 0 },
Y0 := spanh?(x0 )i,
where и denotes the Euclidean inner product in RN and spanhwi the line spanned by the vector
w. This gives an orthogonal decomposition RN = Ye ?Y0 , and x0 + Ye is the tangent hyperplane
of ? at x0 . Since ? is smooth, it is expressed locally as the graph of a map defined on a subset
of Ye . More precisely, there exist a smooth map
h : Ye ? Y0 ,
and a constant ? > 0 such that h(0) = 0, ?h(0) = 0, and that
S := { x0 + y? + h(y?) , y? ? Ye , |y?| < ? } ? ?,
dist(x0 , ? \ S) ? ?.
Now we decompose the integral (1.172) for G0 as
├Z
!
Z
Z
│ |x ? y|2 ┤
1
dSy .
G0 (x, y, t) dSy =
+
exp ?
4Dt
(4?Dt)N/2
?
S
?\S
Since |x ? y| ? |d(x)| for every y ? ? and since
»
» »
»
|x ? y| ? » |x ? x0 | ? |y ? x0 | » = » |d(x)| ? |y ? x0 | »,
we have
This and (1.173) yield
»
»
|d(x)| + » |d(x)| ? |y ? x0 | »
|y ? x0 |
?
.
|x ? y| ?
2
2
|x ? y| ?
?
2
for y ? ? \ S.
(1.173)
1.7 Application to reaction-diffusion systems
Consequently,
Z
69
│
|x ? y|2 ┤
2
exp ?
dSy ? e?? /16Dt |?|,
4Dt
?\S
(1.174)
where |?| denotes the total area of ?.
On the other hand, for each y ? S, we can express y ? x0 as
y ? x0 = y? + h(y?)
(y? ? Ye , h(y?) ? Y0 ),
and Ye can be identified with RN ?1 . Thus
Z
│ |x ? y|2 ┤
dSy
exp ?
4Dt
S
Z
│ |x ? x ? y? ? h(y?)|2 ┤p
0
=
exp ?
1 + |?h(y?)|2 dy?.
4Dt
|y?|<?
Since h(0) = 0 and ?h(0) = 0, there exists a constant C1 > 0 such that
|?h(y?)| ? C1 |y?|
for |y?| < ?.
(1.175)
Note also that the orthogonality (x ? x0 ? h(y?)) ? y? implies
|x ? x0 ? y? ? h(y?)|2 = |x ? x0 ? h(y?)|2 + |y?|2 ? |y?|2 .
Combining these, we obtain
Z
Z
│ |x ? y|2 ┤
│ |y?|2 ┤q
exp ?
exp ?
1 + C12 |y?|2 dy?
dSy ?
4Dt
4Dt
S
|y?|<?
Z
q
(N ?1)/2
?|z|2 /4D
=t
e
1 + t C12 |z|2 dz,
? ?1
|z|< t
?
where z := y?/ t. Observe that, as t ? 0,
Z
Z
q
?|z|2 /4D
2
2
1 + t C1 |z| dz ?
? ?1 e
|z|< t
?
2 /4D
e?|z|
dz = (4D?)(N ?1)/2 .
RN ?1
?
Consequently,
1
(4?Dt)N/2
Z
│ 1 ┤
│ |x ? y|2 ┤
1
+o ? .
dSy ? ?
exp ?
4Dt
t
4?Dt
S
Combining this and (1.174), we obtain
Z
│ 1 ┤
┤
│ 1 ┤
│ 1
2
G0 (x, y, t) dSy = O ? + O ? N e?? /16Dt = O ? .
t
( t)
t
?
Since ? is a smooth compact hypersurface, its curvatures are bounded. Therefore, the constants
in (1.174), (1.175) can be chosen independent of the choice of x0 ? ?.
? and C1 that appear ?
Hence the above O(1/ t) estimate is uniform with respect to the choice of x ? ?. This proves
the estimate (1.172).
Now, choose a sufficiently small constant r0 > 0 such that the signed distance function
d(x) is smooth in the region { d(x) < 2r0 }. For each r ? [?r0 , r0 ], we define a hypersurface
?(r) by
?(r) := { x ? ?, d(x) = r }.
70
Chapitre 1. The Allen-Cahn equation and the FitzHugh-Nagumo system
Then the curvatures of ?(r) are uniformly bounded as r varies, which implies that there exists
some constant C > 0 such that
Z
C
for 0 < t ? T, r ? [?r0 , r0 ].
G(x, y, t) dSy ? ?
t
?(r)
The estimate (1.164) now follows by integrating in r.
Proof of Lemma 1.7.6. For each t ? [0, T ], we put Qt := ? О [0, t]. Given v1 , v2 ? U? , we
have, in view of (1.148),
kg[v1 ] ? g[v2 ]kL? (Qt О[?? ,?+ ]) ? K1 kv1 ? v2 kL? (Qt ) ,
where
K1 = max |?v f1 (u, v)|,
(u,v)?R
with R being the rectangle defined in subsection 1.7.1. By Proposition 1.7.4,
dH ( ?t [ g[v1 ]; ??0 ], ?t [g[v2 ]; ??0 ] ) ? K(eM t ? 1) kg[v1 ] ? g[v2 ]kL? (Qt О[?? ,?+ ]) .
Combining these, we obtain
dH ( ?t [ g[v1 ]; ??0 ], ?t [g[v2 ]; ??0 ] ) ? KK1 (eM t ? 1) kv1 ? v2 kL? (Qt ) .
(1.176)
Now we define the step functions u?1 := u?[ g[v1 ]; ??0 ] and u?2 := u?[ g[v2 ]; ??0 ]. Since
|u?1 ? u?2 | ? ?+ ? ?? ,
and since the two step functions differ only in the region enclosed between the two surfaces
?t [ g[v1 ]; ??0 ] and ?t [ g[v2 ]; ??0 ], the estimates (1.162) and (1.164) imply that there exists a
constant B1 > 0 such that
Z t
dH ( ?s [ g[v1 ]; ??0 ], ?s [g[v2 ]; ??0 ] )
?
ds.
kV [u?1 ; v?0 ] ? V [u?2 ; v?0 ] kL? (Qt ) ? B1
t?s
0
Combining this and (1.176), we obtain
k ?[v1 ] ? ?[v2 ] kL? (Qt ) ? C1
where C1 = B1 KK1 (eM T ? 1). In particular,
k ?[v1 ] ? ?[v2 ] kL? (QT1 ) ? C1
Z
0
Z
t
0
T1
?
kv1 ? v2 kL? (Qs )
?
ds,
t?s
1
ds kv1 ? v2 kL? (QT1 )
T1 ? s
= ? kv1 ? v2 kL? (QT1 ) ,
?
where ? := 2C1 T1 < 1 for T1 small enough. This proves Lemma 1.7.6.
(1.177)
Chapter 2
The singular limit of a
chemotaxis-growth system with
general initial data
We consider a system of partial differential equations which is a model for an aggregation of
amoebae subjected to three competitive effects: diffusion, growth and chemotaxis, i.e. the
tendency of the specie to move towards higher gradients of a chemical substance. The system
involves a small parameter ? > 0 and a cubic nonlinearity whose stable equilibria are 0 and 1.
We consider rather general initial data u0 that are independent of ?. We denote by (u? , v ? )
the solution. First we prove a generation of interface result namely that, after a time of order
?2 | ln ?| , u? develops a thin transition layer that separates the regions {u? ? 1} and {u? ? 0}.
Then, we make an analysis of the motion of interface: in a much slower time scale, the layer
(0)
starts to propagate. As a consequence, as ? ? 0, the solution u? converges to 0 in ?t and
(1)
(0)
(1)
to 1 in ?t , where ?t and ?t are sub-domains separated by an interface ?t , whose motion
is driven by its mean curvature and a nonlocal drift term. We also show that the thickness of
the transition layer is of order ?.
72
2.1
Chapitre 2. A chemotaxis-growth system with general initial data
Introduction
Let us start by a short description of life-cycles of the cellular slime molds (amoebae). The
cells feed and divide until exhaustion of food supply. Then, the amoebae aggregate to form a
multicellular assembly called a slug. It migrates to a new location, then forms into a fruiting
body, consisting of a stalk formed from dead amoebae and spores on the top (fruiting bodies
that are visible to the naked eye are often referred to as mushrooms). Under suitable conditions
of moisture, temperature, spores release new amoebae. The cycle then repeats itself.
It is known that the aggregation stage is mediated by chemotaxis, i.e. the tendency of
biological individuals to direct their movements according to certain chemicals in their environment. The chemotactant (acrasin) is produced by the amoebae themselves and degraded
by an extracellular enzyme (acrasinase). Moreover, acrasin and acrasinase react to form a
complex whose concentration is assumed to be at a steady state. For more details on the
biological background, we refer to [53], [62] or [38].
So the amoebae have a random motion analogous to diffusion coupled with an oriented
chemotactic motion in the direction of a positive gradient of acrasin. In 1970, Keller and Segel
[53] proposed the following system as a model to describe such movements leading to slime
mold aggregation:
(
ut = ? и (D2 ?u) ? ? и (D1 ?v),
(KS)
vt = Dv ?v + f (v)u ? k(v)v,
inside a closed region ?. Here, u, respectively v, denotes the concentration of amoebae,
respectively of acrasin; f (v) is the production rate of acrasin, and k(v) the degradation rate
of acrasin (due to acrasinase); D2 = D2 (u, v), respectively D1 = D1 (u, v), measures the
vigor of the random motion of the amoebae, respectively the strength of the influence of the
acrasin gradient on the flow of amoebae; Dv is a positive and constant diffusion coefficient.
The problem is completed by initial data u0 and v0 and, assuming that there is now flow of
the amoebae or the acrasin across the boundary ??, by homogeneous Neumann boundary
conditions
?u и ? = ?v и ? = 0 on ?? О (0, +?),
? being the unit outward normal to ??.
An often used simplified model is obtained as follows. By some receptor mechanism, cells
do not measure the gradient of v but of some ?(v), with a sensitive function ? satisfying
?? > 0, so that D1 (u, v) = u?? (v). By taking D2 , f and k as constant functions and using
some rescaling arguments, the system reduces to
(
ut = du ?u ? ? и (u??(v)),
(KS ? )
? vt = dv ?v + u ? ?v,
with du , dv , ? and ? some positive constants.
Many analysis of the Keller-Segel model for the aggregation process were proposed. Chemotaxis having some features of ?negative diffusion?, Nanjundiah [62] suggests that the whole
population concentrates in a single point; we refer to this phenomenon as the chemotactic
collapse. In mathematical terms, this means formation of a Dirac delta-type singularity in
finite time. As a matter of fact, it turns out that the possibility of collapse depends upon the
space dimension. In particular it never happens in the one-dimensional case whereas in two
space dimensions, assuming radially symmetric situations, it only occurs if the total amoebae
number is sufficiently large. The problem of global existence and blow up of solutions has been
intensively studied; we refer in particular to [26], [67], [55], [49], [61], [43], [44].
2.1 Introduction
73
In a different framework, Mimura and Tsujikawa [57], consider aggregating pattern-dynamics arising in the following chemotaxis model with growth:
(
ut = ?2 ?u ? ?? и (u??(v)) + f (u),
?
(M T )
? vt = ?v + u ? ?v,
where ? > 0 Ris a small parameter. The function f is cubic, 0 and 1 being its stable zeros,
1
and satisfies 0 f > 0. In this model, the population is subjected to three competing effects:
diffusion, growth and chemotaxis. The diffusion rate and the chemotactic rate are both very
small compared with the growth rate. They observe that, in a first stage, internal layers ?
which describe the boundaries of aggregating regions ? develop; in a second stage, the motion
of the aggregating regions ? which can be described by that of internal layers ? takes place.
The balance of the three effects (diffusion, growth and chemotaxis) makes the aggregation
mechanism possible. Taking the limit ? ? 0, they formally derive the equation for the motion
of the limit interface and study the stability of radially symmetric equilibrium solutions.
The purpose of this Chapter is to extend some of the results obtained by Bonami, Hilhorst,
Logak and Mimura [17] about the singular limit of a variant of system (M T ? ), where the second
equation is elliptic (? = 0):
?
1
?
?
ut = ?u ? ? и (u??(v)) + 2 f? (u) in ? О (0, +?),
?
?
?
?
?
?
?0 = ?v + u ? ?v
in ? О (0, +?),
(P ? )
?v
?u
?
?
=
=0
on ?? О (0, +?),
?
?
?
??
??
?
?
?u(x, 0) = u (x)
in ?,
0
where ? is a smooth bounded domain in RN (N ? 2), ? is the Euclidian unit normal vector
exterior to ??. We assume that ? is a positive constant and that the nonlinearityf? is given
by
1
f? (u) = u(1 ? u)(u ? ) + ??u(1 ? u)
2
(2.1)
=: f (u) + ?g(u),
with ? > 0. The role of the function g is to slightly break the balance of the two stable zeros.
The sensitive function ? is smooth and satisfies ?? (v) > 0 for v > 0.
We also assume that the initial datum satisfies u0 ? C 2 (?) and u0 ? 0. Throughout the
present Chapter, we fix a constant C0 > 1 that satisfies
ku0 kC 0 (?) + k?u0 kC 0 (?) + k?u0 kC 0 (?) ? C0 .
(2.2)
Furthermore we define the ?initial interface? ?0 by
?0 := {x ? ?, u0 (x) = 1/2}.
We suppose that ?0 is a C 2+? hypersurface without boundary, for a ? ? (0, 1), such that, n
being the Euclidian unit normal vector exterior to ?0 ,
?0 ?? ?
u0 > 1/2
(1)
?0
and
?u0 (x) и n(x) 6= 0
(1)
in ?0 ,
u0 < 1/2
(0)
?0
if x ? ?0 ,
(2.3)
(0)
(2.4)
in ?0 ,
where
denotes the region enclosed by ?0 and
the region enclosed between ?? and ?0 .
The existence of a unique smooth solution to Problem (P ? ) is proved in [17], Lemma 4.2:
74
Chapitre 2. A chemotaxis-growth system with general initial data
Lemma 2.1.1. There exists ?0 > 0 such that, for all ? ? (0, ?0 ), there exists a unique solution
(u? , v ? ) to Problem (P ? ) on ? О [0, +?), with 0 ? u? ? C0 on QT .
To study the interfacial behavior associated with this model, it is useful to consider a
formal asymptotic limit of Problem (P ? ) as ? ? 0. Then the limit solution u0 (x, t) will be a
step function taking the value 1 on one side of the interface, and 0 on the other side. This
sharp interface, which we will denote by ?t , obeys a law of motion, which can be obtained by
formal analysis (see Section 2.2):
?
??(v 0 ) ?
?
?
+ 2? on ?t ,
=
?(N
?
1)?
+
V
?
n
?
?
?n
»
?
?
??t »
= ?0
t=0
(P 0 )
?
??v 0 + ?v 0 = u0
in ? О (0, T ],
?
?
?
?
0
?
?
? ?v = 0
on ?? О (0, T ],
??
where Vn is the normal velocity of ?t in the exterior direction, ? the mean curvature at each
(1)
point of ?t . We set QT := ? О [0, T ] and for each t ? [0, T ], we define ?t as the region
(0)
enclosed by the hypersurface ?t and ?t as the region enclosed between ?? and ?t . The step
function u0 is determined straightforwardly from ?t by
(
(1)
1 in ?t
0
for t ? [0, T ].
(2.5)
u (x, t) =
(0)
0 in ?t
By a contraction fixed-point argument in suitable Ho?lder spaces, the well-posedness, locally in
time, of the free boundary Problem (P 0 ) is proved in [17], Theorem 2.1:
Lemma 2.1.2. There exists a time T > 0 such that (P 0 ) has a unique solution (v 0 , ?) on
[0, T ], with
[
2+?
?=
(?t О {t}) ? C 2+?, 2 ,
0?t?T
and v 0 |? ? C
2+?, 2+?
2
.
Bonami, Hilhorst, Logak and Mimura [17] have proved a motion of interface property;
more precisely, for some prepared initial data, they show that (u? , v ? ) converges to (u0 , v 0 )
as ? ? 0, on the interval (0, T ). So the evolution of ?t determines the aggregating patterns
of the individuals. Here we consider the case of arbitrary initial data. Our first main result,
Theorem 2.1.3, describes the profile of the solution after a very short initial period. It asserts
that, given a virtually arbitrary initial datum u0 , the solution u? quickly becomes close to 1 or
0, except in a small neighborhood of the initial interface ?0 , creating a steep transition layer
around ?0 (generation of interface). The time needed to develop such a transition layer, which
we will denote by t? , is of order ?2 | ln ?|. The theorem then states that the solution u? remains
close to the step function u0 on the time interval [t? , T ] (motion of interface). Moreover, as is
clear from the estimates in the theorem, the ?thickness? of the transition layer is of order ?.
Theorem 2.1.3 (Generation and motion of interface). Let ? ? (0, 1/4) be arbitrary and
set
х = f ? (1/2) = 1/4.
2.2 Some preliminaries
75
Then there exist positive constants ?0 and C such that, for all ? ? (0, ?0 ), for all t? ? t ? T ,
where t? = х?1 ?2 | ln ?|, we have
?
?
[??, 1 + ?]
if x ? NC? (?t )
?
?
?
?
(0)
u? (x, t) ? [??, ?]
(2.6)
if x ? ?t \ NC? (?t )
?
?
?
?
(1)
?[1 ? ?, 1 + ?]
if x ? ?t \ NC? (?t ),
where Nr (?t ) := {x ? ?, dist(x, ?t ) < r} denotes the r-neighborhood of ?t .
Corollary 2.1.4 (Convergence). As ? ? 0, the solution (u? , v ? ) converges to (u0 , v 0 ) evS
(0 or 1)
erywhere in 0<t?T (?t
О {t}).
The next theorem deals with the relation between the set ??t := {x ? ?, u? (x, t) = 1/2}
and the solution ?t of Problem (P 0 ).
Theorem 2.1.5 (Error estimate). There exists C > 0 such that
??t ? NC? (?t )
for 0 ? t ? T.
(2.7)
Corollary 2.1.6 (Convergence of interface). There exists C > 0 such that
dH (??t , ?t ) ? C?
for 0 ? t ? T,
(2.8)
where
dH (A, B) := max{sup d(a, B), sup d(b, A)}
a?A
b?B
denotes the Hausdorff distance between two compact sets A and B. Consequently, ??t ? ?t as
? ? 0, uniformly in 0 ? t ? T , in the sense of Hausdorff distance.
As far as we know, the best thickness estimate in the literature was of order ?| ln ?| (see
[20], [21]). We refer to a forthcoming article [51] in which an order ? estimate is established
for a Lotka-Volterra competition-diffusion system.
The organization of this Chapter is as follows. Section 2.2 is devoted to preliminaries: we
recall the method of asymptotic expansions to derive the equation of the interface motion; we
also recall a relaxed comparison principle used in [17]. In Section 2.3, we prove a generation of
interface property. The corresponding sub- and super-solutions are constructed by modifying
the solution of the ordinary differential equation ut = ??2 f (u), obtained by neglecting diffusion
and chemotaxis. In Section 2.4, in order to study the motion of interface, we construct a pair
of sub- and super-solutions that rely on a related one-dimensional stationary problem. Finally,
in Section 2.5, by fitting the two pairs of sub- and super-solutions into each other, we prove
Theorem 2.1.3, Theorem 2.1.5 and theirs corollaries.
2.2
2.2.1
Some preliminaries
Formal derivation
A formal derivation of the equation of interface motion was given in [16]. Nevertheless we
briefly present it in a slightly different way: we use arguments similar to those in [63] where
76
Chapitre 2. A chemotaxis-growth system with general initial data
the first two terms of the asymptotic expansion determine the interface equation in (P 0 ),
which can be regarded as the singular limit of (P ? ) as ? ? 0. The observations we make here
will help the rigorous analysis in later sections, in particular for the construction of sub- and
super-solutions for the study of the motion of interface in Section 2.4.
(P ? ). We recall that ??t := {x ? ?,
u? (x, t) = 1/2}
Let (u? , v ? ) be the solution of Problem
S
S
is the interface at time t and call ?? := t?0 (??t О {t}) the interface. Let ? = 0?t?T (?t О {t})
be the solution of the limit geometric motion problem and let de be the signed distance function
to ? defined by:
(
(0)
dist(x, ?t ) for x ? ?t
e
d(x, t) =
(2.9)
(1)
? dist(x, ?t ) for x ? ?t ,
where dist(x, ?t ) is the distance from x to the hypersurface ?t in ?. We remark that de = 0 on
e = 1 in a neighborhood of ?. We then define
? and that |?d|
(1)
QT =
[
0<t?T
(1)
(?t О {t}),
(0)
QT =
[
0<t?T
(0)
(?t О {t}).
We assume that the solution u? has the expansions
u? (x, t) = {0 or 1} + ?u1 (x, t) + и и и
(2.10)
away from the interface ? (the outer expansion) and
u? (x, t) = U0 (x, t,
e t)
e t)
d(x,
d(x,
) + ?U1 (x, t,
) + иии
?
?
(2.11)
near ? (the inner expansion). Here, the functions Uk (x, t, z), k = 0, 1, и и и , are defined for
e t)/? gives exactly the right spatial
x ? ?, t ? 0, z ? R. The stretched space variable ? := d(x,
scaling to describe the rapid transition between the regions {u? ? 1} and {u? ? 0}. We use
the normalization conditions
U0 (x, t, 0) = 1/2,
Uk (x, t, 0) = 0,
for all k ? 1. The matching conditions between the outer and the inner expansion are given
by
U0 (x, t, +?) = 0,
Uk (x, t, +?) = 0,
(2.12)
Uk (x, t, ??) = 0,
U0 (x, t, ??) = 1,
for all k ? 1. We also assume that the solution v ? has the expansion
v ? (x, t) = v0 (x, t) + ?v1 (x, t) + и и и
(2.13)
in ? О (0, T ).
We now substitute the inner expansion (2.11) and the expansion (2.13) into the parabolic
? = 1
equation of (P ? ) and collect the ??2 terms. We omit the calculations and, using |?d|
near ?t , the normalization and matching conditions, we deduce that U0 (x, t, z) = U0 (z) is the
unique solution of the stationary problem
(
U0 ?? + f (U0 ) = 0
(2.14)
U0 (??) = 1, U0 (0) = 1/2, U0 (+?) = 0.
2.2 Some preliminaries
77
This solution represents the first approximation of the profile of a transition layer around the
interface observed in the stretched coordinates. Recalling that the nonlinearity is given by
f (u) = u(1 ? u)(u ? 1/2), we have
?
1А
e?z/ 2
z б
? .
U0 (z) = 1 ? tanh ? =
2
2 2
1 + e?z/ 2
(2.15)
We claim that U0 has the following properties.
Lemma 2.2.1. There exist positive constants C and ? such that the following estimates hold.
0 < U0 (z)
? Ce??|z|
0 < 1 ? U0 (z) ?
for z ? 0,
Ce??|z|
for z ? 0.
In addition, U0 is a strictly decreasing function and
|U0 ? (z)| + |U0 ?? (z)| ? Ce??|z|
for z ? R.
Next we collect the ??1 terms. Since U0 depends only on the variable z, we have ?U0z = 0
? = 1 near ?t , yields
which, combined with the fact that |?d|
U1zz + f ? (U0 )U1 = U0 ? (det ? ?de + ?de и ??(v 0 )) ? g(U0 ),
(2.16)
a linearized problem for (2.14). The solvability condition for the above equation, which can
be seen as a variant of the Fredholm alternative, plays the key role for deriving the equation
of interface motion. It is is given by
Z h
i
2
U0 ? (z)(det ? ?de + ?de и ??(v 0 ))(x, t) ? g(U0 (z))U0 ? (z) dz = 0,
R
for all (x, t) ? QT . By the definition of g in (2.1), we compute
Z
?
R
g(U0 (z))U0 (z)dz = ?
Z
0
1
g(u)du = ??/6,
whereas the equality (2.15) yields
Z
R
U0
?2
1
(z)dz = ?
2
Z
0
+?
?
u
du
=
1/6
2.
1 + u4
Combining the above expressions, we obtain
┤
│
?
det ? ?de + ?de и ??(v 0 ) (x, t) = ? 2?.
(2.17)
e t)) coincides with the outward normal unit vector to the hypersurface ?t ,
Since ?de (= ?x d(x,
we have det (x, t) = ?Vn , where Vn is the normal velocity of the interface ?t . It is also known
e
that the mean curvature ? of the interface is equal to ?d/(N
? 1). Thus the above equation
reads as
??(v 0 ) ?
(2.18)
+ 2? on ?t ,
Vn = ?(N ? 1)? +
?n
78
Chapitre 2. A chemotaxis-growth system with general initial data
that is the equation of interface motion in (P 0 ). Summarizing, under the assumption that the
solution u? of Problem (P ? ) satisfies
?
u ?
(
1
0
(1)
in QT
(0)
in QT
as ? ? 0,
(0)
(1)
we have formally showed that the boundary ?t between ?t and ?t moves according to the
law (2.18).
?
One can note that, using the equality (2.15), we clearly have 2?U0 ? + g(U0 ) ? 0 so that,
substituting (2.17) into (2.16) yields U1 ? 0.
2.2.2
A comparison principle
The definition of sub- and super-solutions is the one proposed in [17].
+
?
+
Definition 2.2.2. Let (u?
? , u? ) be two smooth functions with u? ? u? on QT and
?u?
?u+
?
?
?0?
??
??
on ?? О (0, T ).
+
?
By definition, (u?
? , u? ) is a pair of sub- and super-solutions if, for any v which satisfies
?
?
?
?
+
?
on QT ,
?u? ? ??v + ?v ? u?
(2.19)
?
?
? ?v = 0
on ?? О (0, T ),
??
we have
+
Lv? [u?
? ] ? 0 ? Lv ? [u? ],
where the operator Lv? is defined by
Lv? [?] = ?t ? ?? + ? и (???(v ? )) ?
1
f? (?).
?2
As proved in [17], the following comparison principle holds.
Proposition 2.2.3. Let a pair of sub- and super-solutions be given. Assume that, for all
x ? ?,
+
u?
? (x, 0) ? u0 (x) ? u? (x, 0).
Then, if we denote by (u? , v ? ) the solution of Problem (P ? ), the function u? satisfies, for all
(x, t) ? QT ,
?
+
u?
? (x, t) ? u (x, t) ? u? (x, t).
2.3
Generation of interface
In this section we study the rapid formation of internal layers in a neighborhood of ?0 = {x ?
?, u0 (x) = 1/2} within a very short time interval of order ?2 | ln ?|. In the sequel, we shall
always assume that 0 < ? < 1/4. The main result of this section is the following.
2.3 Generation of interface
79
Theorem 2.3.1. Let ? be arbitrary and define х as the derivative of f (u) at the unstable
equilibrium u = 1/2, that is
х = f ? (1/2) = 1/4.
(2.20)
Then there exist positive constants ?0 and M0 such that, for all ? ? (0, ?0 ),
? for all x ? ?,
? ? ? u? (x, х?1 ?2 | ln ?|) ? 1 + ?,
(2.21)
? for all x ? ? such that |u0 (x) ? 12 | ? M0 ?, we have that
if u0 (x) ? 1/2 + M0 ? then u? (x, х?1 ?2 | ln ?|) ? 1 ? ?,
?
?1 2
if u0 (x) ? 1/2 ? M0 ? then u (x, х
? | ln ?|) ? ?.
(2.22)
(2.23)
The above theorem will be proved by constructing a suitable pair of sub and super-solutions.
2.3.1
The perturbed bistable ordinary differential equation
We first consider a slightly perturbed nonlinearity:
f? (u) = f (u) + ?,
where ? is any constant. For |?| small enough, this function is still cubic and bistable; more
precisely, we claim that it has the following properties.
Lemma 2.3.2. Let ?0 > 0 be small enough. Then, for all ? ? (??0 , ?0 ),
? f? has exactly three zeros, namely ?? (?) < a(?) < ?+ (?). More precisely,
f? (u) = (u ? ?? (?))(?+ (?) ? u)(u ? a(?)),
(2.24)
and there exists a positive constant C such that
|?? (?)| + |a(?) ? 1/2| + |?+ (?) ? 1| ? C|?|.
? We have that
? Set
f?
is strictly positive in
f?
is strictly negative in
(??, ?? (?)) ? (a(?), ?+ (?)),
(?? (?), a(?)) ? (?+ (?), +?).
(2.25)
(2.26)
х(?) := f?? (a(?)) = f ? (a(?)),
then there exists a positive constant, which we denote again by C, such that
|х(?) ? х| ? C|?|.
(2.27)
In order to construct a pair of sub and super-solutions for Problem (P ? ) we define Y (?, ?; ?)
as the solution of the ordinary differential equation
(
Y? (?, ?; ?) = f? (Y (?, ?; ?)) for ? > 0,
(2.28)
Y (0, ?; ?) = ?,
for ? ? (??0 , ?0 ) and ? ? (?2C0 , 2C0 ), where C0 has been chosen in (2.2). We present below
basic properties of Y .
80
Chapitre 2. A chemotaxis-growth system with general initial data
Lemma 2.3.3. We have Y? > 0, for all ? ? (?2C0 , 2C0 ) \ {?? (?), a(?), ?+ (?)}, all ? ?
(??0 , ?0 ) and all ? > 0. Furthermore,
Y? (?, ?; ?) =
f? (Y (?, ?; ?))
.
f? (?)
Proof. We differentiate (2.28) with respect to ? to obtain
(
Y?? = Y? f ? (Y ),
Y? (0, ?; ?) = 1,
which is integrated as follows:
Y? (?, ?; ?) = exp
hZ
?
0
i
f ? (Y (s, ?; ?))ds > 0.
(2.29)
Then differentiating (2.28) with respect to ? , we obtain
(
Y? ? = Y? f ? (Y ),
Y? (0, ?; ?) = f? (?),
which in turn implies
Y? (?, ?; ?) = f? (?) exp
hZ
0
?
i
f ? (Y (s, ?; ?))ds ,
which enables to conclude.
We define a function A(?, ?; ?) by
A(?, ?; ?) =
f ? (Y (?, ?; ?)) ? f ? (?)
.
f? (?)
(2.30)
Lemma 2.3.4. We have, for all ? ? (?2C0 , 2C0 ) \ {?? (?), a(?), ?+ (?)}, all ? ? (??0 , ?0 ) and
all ? > 0,
Z ?
f ?? (Y (s, ?; ?))Y? (s, ?; ?)ds.
A(?, ?; ?) =
0
Proof. We differentiate the equality of Lemma 2.3.3 with respect to ? to obtain
Y?? (?, ?; ?) = A(?, ?; ?)Y? (?, ?; ?).
(2.31)
Then differentiating (2.29) with respect to ? yields
Z ?
Y?? = Y?
f ?? (Y (s, ?; ?))Y? (s, ?; ?)ds.
0
These two last results complete the proof of Lemma 2.3.4.
Next we prove estimates on the growth of Y , A and theirs derivatives. We first consider
the case where the initial value ? is far from the stable equilibria, more precisely when it lies
between ? and 1 ? ?.
e1 = C
e1 (?),
Lemma 2.3.5. Let ? be arbitrary. Then there exist positive constants ?0 = ?0 (?), C
C?2 = C?2 (?) and C3 = C3 (?) such that, for all ? ? (??0 , ?0 ), for all ? > 0,
2.3 Generation of interface
81
? if ? ? (a(?), 1 ? ?) then, for every ? > 0 such that Y (?, ?; ?) remains in the interval
(a(?), 1 ? ?), we have
C?1 eх(?)? ? Y? (?, ?; ?) ? C?2 eх(?)? ,
(2.32)
and
|A(?, ?; ?)| ? C3 (eх(?)? ? 1);
(2.33)
? if ? ? (?, a(?)) then, for every ? > 0 such that Y (?, ?; ?) remains in the interval (?, a(?)),
(2.32) and (2.33) hold as well.
Proof. We take ? ? (a(?), 1 ? ?) and suppose that for s ? (0, ? ), Y (s, ?; ?) remains in the
interval (a(?), 1 ? ?). Integrating the equality
Y? (s, ?; ?)
=1
f? (Y (s, ?; ?))
from 0 to ? and using the change of variable q = Y (s, ?; ?) leads to
Z Y (?,?;?)
dq
= ?.
f
? (q)
?
Moreover, the equality in Lemma 2.3.3 enables to write
Z Y (?,?;?) ?
f (q)
ln Y? (?, ?; ?) =
dq
f? (q)
?
Z Y (?,?;?) ?
Б f (a(?)) f ? (q) ? f ? (a(?)) ц
dq
+
=
f? (q)
f? (q)
?
Z Y (?,?;?)
= х(?)? +
h? (q)dq,
(2.34)
(2.35)
?
where
h? (q) =
f ? (q) ? f ? (a(?))
.
f? (q)
In view of (2.27), respectively (2.25), we can choose ?0 = ?0 (?) > 0 small enough so that, for
all ? ? [??0 , ?0 ], we have х(?) ? х/2 > 0, respectively (a(?), 1 ? ?] ? (a(?), ?+ (?)). Since
h? (q) ?
f??? (a(?))
f ?? (a(?))
=
f?? (a(?))
f ? (a(?))
as q ? a(?),
we see that the function (q, ?) 7? h? (q) is continuous in the compact region { |?| ? ?0 , a(?) ?
q ? 1 ? ? }. It follows that |h? (q)| is bounded by a constant H = H(?) as (q, ?) varies in this
region. Since |Y (?, ?; ?) ? ?| takes its values in the interval [0, 1 ? ? ? a(?)] ? [0, 1], it follows
from (2.35) that
х(?)? ? H ? ln Y? (?, ?; ?) ? х(?)? + H,
which, in turn, proves (2.32). Next Lemma 2.3.4 and (2.32) yield
Z ?
??
C?2 eх(?)s ds
|A(?, ?; ?)| ? kf kL? (0,1)
0
?
kf ?? kL? (0,1) C?2 х(?)?
(e
? 1)
х(?)
?
2 ??
kf kL? (0,1) C?2 (eх(?)? ? 1),
х
82
Chapitre 2. A chemotaxis-growth system with general initial data
which completes the proof of (2.33). The case where ? and Y (?, ?; ?) are in (?, a(?)) is similar
and omitted.
Corollary 2.3.6. Let ? be arbitrary. Then there exist positive constants ?0 = ?0 (?), C1 =
C1 (?) and C2 = C2 (?) such that, for all ? ? (??0 , ?0 ), for all ? > 0,
? if ? ? (a(?), 1 ? ?) then, for every ? > 0 such that Y (?, ?; ?) remains in the interval
(a(?), 1 ? ?), we have
C1 eх(?)? (? ? a(?)) ? Y (?, ?; ?) ? a(?) ? C2 eх(?)? (? ? a(?)),
(2.36)
? if ? ? (?, a(?)) then, for every ? > 0 such that Y (?, ?; ?) remains in the interval (?, a(?)),
we have
C2 eх(?)? (? ? a(?)) ? Y (?, ?; ?) ? a(?) ? C1 eх(?)? (? ? a(?)).
(2.37)
Proof. In view of (2.27), respectively (2.25), we can choose ?0 = ?0 (?) > 0 small enough so
that, for all ? ? [??0 , ?0 ], we have х(?) ? х/2 > 0, respectively (a(?), 1 ? ?] ? (a(?), ?+ (?)).
Since
f? (q)
? х(?) as q ? a(?),
q ? a(?)
it follows that (q, ?) 7? f? (q)/(q ? a(?)) is a strictly positive and continuous function in the
compact region { |?| ? ?0 , a(?) ? q ? 1 ? ? }, which insures the existence of constants
B1 = B1 (?) > 0 and B2 = B2 (?) > 0 such that, for all q ? (a(?), 1 ? ?), all ? ? (??0 , ?0 ),
B1 (q ? a(?)) ? f? (q) ? B2 (q ? a(?)).
(2.38)
We write the inequalities (2.38) for q = Y (?, ?; ?) ? (a(?), 1 ? ?) and then for q = ? ?
(a(?), 1 ? ?), which, together with Lemma 2.3.3, implies that
B1
B2
(Y (?, ?; ?) ? a(?)) ? (? ? a(?))Y? (?, ?; ?) ?
(Y (?, ?; ?) ? a(?)).
B2
B1
In view of (2.32), this completes the proof of inequalities (2.36). The proof of (2.37) is similar
and omitted.
We now present estimates in the case that the initial value ? is smaller than ? or larger
than 1 ? ?.
Lemma 2.3.7. Let ? and M > 0 be arbitrary. Then there exist positive constants ?0 =
?0 (?, M ) and C4 = C4 (M ) such that, for all ? ? (??0 , ?0 ),
? if ? ? [1 ? ?, 1 + M ], then, for all ? > 0, Y (?, ?; ?) remains in the interval [1 ? ?, 1 + M ]
and
|A(?, ?; ?)| ? C4 ? for ? > 0 ;
(2.39)
? if ? ? [?M, ?], then, for all ? > 0, Y (?, ?; ?) remains in the interval [?M, ?] and (2.39)
holds as well.
Proof. Since the two statements can be treated in the same way, we will only prove the former.
The fact that Y (?, ?; ?), the solution of the ordinary differential equation (2.28), remains in
the interval [1 ? ?, 1 + M ] directly follows from the bistable properties of f? , or, more precisely,
from the sign conditions f? (1 ? ?) > 0, f? (1 + M ) < 0 valid if ?0 = ?0 (?, M ) is small enough.
2.3 Generation of interface
83
To prove (2.39), suppose first that ? ? [?+ (?), 1 + M ]. By the above arguments, Y (?, ?; ?)
remains in this interval. Moreover f ? is negative in this interval. Hence, it follows from (2.29)
that Y? (?, ?; ?) ? 1. We then use Lemma 2.3.4 to deduce that
|A(?, ?; ?)| ? kf ?? kL? (?M,1+M ) ? =: C4 ?.
The case ? ? [1 ? ?, ?+ (?)] being similar, this completes the proof of the lemma.
Now we choose the constant M in the above lemma sufficiently large so that [?2C0 , 2C0 ] ?
[?M, 1 + M ], and fix M hereafter. Therefore the constant C4 is fixed as well. Using the fact
that ? 7? ? (eх(?)? ? 1)?1 is uniformly bounded for ? ? (??0 , ?0 ), with ?0 small enough (see
(2.27)), and for ? > 0, one can easily deduce from (2.33) and (2.39) the following general
estimate.
Lemma 2.3.8. Let ? be arbitrary and let C0 be the constant defined in (2.2). Then there exist
positive constants ?0 = ?0 (?), C5 = C5 (?) such that, for all ? ? (??0 , ?0 ), all ? ? (?2C0 , 2C0 )
and all ? > 0,
|A(?, ?; ?)| ? C5 (eх(?)? ? 1).
2.3.2
Construction of sub and super-solutions
We now use Y to construct a pair of sub- and super-solutions for the proof of the generation
of interface theorem. We set
│t
┤
t
(2.40)
w?▒ (x, t) = Y 2 , u0 (x) ▒ ?2 r(▒?G, 2 ); ▒?G ,
?
?
where the constant G is defined by
G=
sup
u?[?2C0 ,2C0 ]
|g(u)|,
and the function r(?, ? ) is given by
r(?, ? ) = C6 (eх(?)? ? 1).
For simplicity, we make the following additional assumption:
?u0
=0
??
on ??.
(2.41)
In the general case where (2.41) does not necessary hold, we have to slightly modify w?▒ near
the boundary ??. This will be discussed in the next remark.
Lemma 2.3.9. There exist positive constants ?0 and C6 such that for all ? ? (0, ?0 ), the
functions w?? and w?+ are respectively sub- and super-solutions for Problem (P ? ), in the domain
Е
ф
(x, t) ? QT , x ? ?, 0 ? t ? х?1 ?2 | ln ?| .
Proof. First, (2.41) implies the homogeneous Neumann boundary condition
?w?▒
=0
??
on ?? О (0, +?).
84
Chapitre 2. A chemotaxis-growth system with general initial data
Let v ? be such that
?
?
?
?
+
?
?w? ? ??v + ?v ? w?
(2.42)
?
?
? ?v = 0.
??
According to Definition 2.2.2, what we have to show is
Lv? [w?+ ] := (w?+ )t ? ?w?+ + ? и (w?+ ??(v ? )) ?
1
f? (w?+ ) ? 0.
?2
Let C6 be a positive constant which does not depend on ?. If ?0 is sufficiently small, we note
that ▒?G ? (??0 , ?0 ) and that, in the range 0 ? t ? х?1 ?2 | ln ?|,
2
|?2 C6 (eх(▒?G)t/? ? 1)| ? ?2 C6 (??х(▒?G)/х ? 1) ? C0 ,
which implies that
u0 (x) ▒ ?2 r(▒?G, t/?2 ) ? (?2C0 , 2C0 ).
These observations allow us to use the results of the previous subsection with ? = t/?2 ,
? = u0 (x) + ?2 r(?G, t/?2 ) and ? = ?G. In particular, setting F1 := kf ? kL? (?2C0 ,2C0 ) , this
implies, using (2.29), that
e?F1 T ? Y? ? eF1 T .
Straightforward computations yield
Lv? [w?+ ] =
1
2
Y? + C6 х(?G)eх(?G)t/? Y? ? ?u0 Y? ? |?u0 |2 Y??
2
?
1
1
+ Y? ?u0 и ??(v ? ) + Y ??(v ? ) ? 2 f (Y ) ? g(Y ),
?
?
and then, in view of the ordinary differential equation (2.28), ?G playing the role of ?,
h
ц
1Б
2
Lv? [w?+ ] = G ? g(Y ) + Y? C6 х(?G)eх(?G)t/? ? ?u0
?
i
Y??
Y
?
|?u0 |2 + ?u0 и ??(v ? ) + ??(v ? ) .
Y?
Y?
By the definition of G the first term is positive. Now, using the choice of C0 in (2.2), the fact
that Y?? /Y? = A and Lemma 2.3.8, we obtain, for a C5 independent of ?,
h
2
2
Lv? [w?+ ] ? Y? C6 х(?G)eх(?G)t/? ? C0 ? C5 (eх(?G)t/? ? 1)C02
i
? C0 |??(v ? )| ? 2C0 eF1 T |??(v ? )| .
Moreover, the inequalities in (2.42) can be written as ??v ? +?v ? = h? , with ?2C0 ? h? ? 2C0 ,
so that the standard theory of elliptic equations gives a uniform bound M for |v ? |, |?v ? | and
|?v ? |. Hence, using the smoothness of ?, we have a uniform bound M ? for |??(v ? )| and
|??(v ? )|. It follows that
h
i
2
Lv? [w?+ ] ? Y? (C6 х(?G) ? C5 C02 )eх(?G)t/? ? C0 + C5 C02 ? C0 M ? ? 2C0 eF1 T M ? .
Hence, in view of (2.27), we have, for ?0 small enough (recall that Y? > 0),
h
i
1
Lw?+ ? Y? (C6 х ? C5 C02 ) ? C0 ? C0 M ? ? 2C0 eF1 T M ? ? 0,
2
2.3 Generation of interface
85
for C6 large enough, so that w?+ is a super-solution for Problem (P ? ). We omit the proof that
w?? is a sub-solution.
Now, since w?▒ (x, 0) = Y (0, u0 (x); ▒?G) = u0 (x), the comparison principle set in Proposition 2.2.3 asserts that, for all x ? ?, for all 0 ? t ? х?1 ?2 | ln ?|,
w?? (x, t) ? u? (x, t) ? w?+ (x, t).
(2.43)
Remark 2.3.10. In the more general case where (2.41) is not necessarily valid, one can
proceed in the following way: in view of (2.3) and (2.4) there exist positive constants d1 and
? such that u0 (x) ? 1/2 ? ? if d(x, ??) ? d1 . Let ? be a smooth cut-off function defined on
[0, +?) such that 0 ? ? ? 1, ?(0) = ?? (0) = 0 and ?(z) = 1 for z ? d1 . Then define
u+
0 (x) = ?(d(x, ??))u0 (x) + (1 ? ?(d(x, ??))(1/2 ? ?),
u?
0 (x) = ?(d(x, ??))u0 (x) + (1 ? ?(d(x, ??)) min u0 (x).
x??
Clearly,
u?
0
? u0 ?
u+
0,
and both
u▒
0
satisfy (2.41). Now we set
┤
│t
t
2
w??▒ (x, t) = Y 2 , u▒
(x)
▒
?
r(▒?G,
);
▒?G
.
0
?
?2
Then the same argument as in Lemma 2.3.9 shows that (w??? , w??+ ) is a pair of sub and super+
solutions for Problem (P ? ). Furthermore, since w??? (x, 0) = u?
0 (x) ? u0 (x) ? u0 (x) =
w??+ (x, 0), Proposition 2.2.3 asserts that, for all x ? ?, for all 0 ? t ? х?1 ?2 | ln ?|, we have
ц
w??? (x, t) ? u? (x, t) ? w??+ (x, t).
2.3.3
Proof of Theorem 2.3.1
In order to prove Theorem 2.3.1 we first present a key estimate on the function Y after a time
of order ? ? | ln ?|.
Lemma 2.3.11. Let ? be arbitrary; there exist positive constants ?0 = ?0 (?) and C7 = C7 (?)
such that, for all ? ? (0, ?0 ),
? for all ? ? (?2C0 , 2C0 ),
? ? ? Y (х?1 | ln ?|, ?; ▒?G) ? 1 + ?,
(2.44)
? for all ? ? (?2C0 , 2C0 ) such that |? ? 12 | ? C7 ?, we have that
if ? ? 1/2 + C7 ? then Y (х?1 | ln ?|, ?; ▒?G) ? 1 ? ?,
?1
if ? ? 1/2 ? C7 ? then Y (х
| ln ?|, ?; ▒?G) ? ?.
(2.45)
(2.46)
Proof. We first prove (2.45). In view of (2.25), we have, for C7 large enough, 1/2 + C7 ? ?
a(▒?G) + 12 C7 ?, for all ? ? (0, ?0 ), with ?0 small enough. Hence for ? ? 1/2 + C7 ?, as long as
Y (?, ?; ▒?G) has not reached 1 ? ?, we can use (2.36) to deduce that
Y (?, ?; ▒?G) ? a(▒?G) + C1 eх(▒?G)? (? ? a(▒?G))
? a(▒?G) + 12 C1 C7 ?eх(▒?G)?
?
1
2
? ?CG + 12 C1 C7 ?eх(▒?G)?
?1??
86
Chapitre 2. A chemotaxis-growth system with general initial data
provided that
? ? ? ? :=
1/2 ? ? + CG?
1
ln
.
х(▒?G)
C1 C7 ?/2
To complete the proof of (2.45) we must choose C7 so that х?1 | ln ?| ? ? ? ? 0. A simple
computation shows that
х?1 | ln ?| ? ? ? =
1
1/2 ? ? + CG?
х(▒?G) ? х
| ln ?| ?
ln
х(▒?G)х
х(▒?G)
C1 /2
+
1
ln C7 .
х(▒?G)
Thanks to (2.27), as ? ? 0, the first term above is of order ?| ln ?| and the second one of order
1. Hence, for C7 large enough, the quantity х?1 | ln ?| ? ? ? is positive, for all ? ? (0, ?0 ), with
?0 small enough. The proof of (2.46) is similar and omitted.
Next we prove (2.44). First note that, by taking ?0 small enough, the stable equilibria of
f▒?G , namely ?? (▒?G) and ?+ (▒?G), are in [??, 1+?]. Hence, f▒?G being a bistable function,
if we leave from a ? ? [??, 1 + ?] then Y (?, ?; ▒?G) will remain in the interval [??, 1 + ?]. Now
suppose that 1 + ? ? ? ? 2C0 (note that this work is useless if 2C0 < 1 + ?). We check below
that Y (х?1 | ln ?|, ?; ▒?G) ? 1 + ?. As long as 1 + ? ? Y ? 2C0 , (2.28) leads to the inequality
Y? ? f (1 + ?) + ?G ? 12 f (1 + ?) < 0, for ?0 = ?0 (?) small enough. By integration from 0 to ? ,
it follows that
Y (?, ?; ▒?G) ? ? + 12 f (1 + ?)?
? 2C0 + 12 f (1 + ?)?
? 1 + ?,
provided that
??
2C0 ? 1 ? ?
,
?f (1 + ?)/2
and a fortiori for ? = х?1 | ln ?|, which completes the proof of (2.44).
We are now ready to prove Theorem 2.3.1. By setting t = х?1 ?2 | ln ?| in (2.43), we obtain
│
┤
Y х?1 | ln ?|, u0 (x) ? ?2 r(??G, х?1 | ln ?|); ??G
│
┤
? u? (x, х?1 ?2 | ln ?|) ? Y х?1 | ln ?|, u0 (x) + ?2 r(?G, х?1 | ln ?|); +?G . (2.47)
In view of (2.27),
х ? х(▒?G)
ln ? = 0,
??0
х
so that, for ?0 small enough, we have
lim
(2.48)
1
3
?2 r(▒?G, х?1 | ln ?|) = C6 ?(?(х?х(▒?G))/х ? ?) ? ( C6 ?, C6 ?).
2
2
It follows that u0 (x) ▒ ?2 r(▒?G, х?1 | ln ?|) ? (?2C0 , 2C0 ). Hence the result (2.21) of Theorem
2.3.1 is a direct consequence of (2.44) and (2.47).
Next we prove (2.22). We take x ? ? such that u0 (x) ? 1/2 + M0 ? so that
u0 (x) ? ?2 r(??G, х?1 | ln ?|) ? 1/2 + M0 ? ? 32 C6 ?
? 1/2 + C7 ?,
2.4 Motion of interface
87
if we choose M0 large enough. Using (2.47) and (2.45) we obtain (2.22), which completes the
proof of Theorem 2.3.1.
2.4
Motion of interface
We have seen in Section 2.3 that, after a very short time, the solution u? develops a clear
transition layer. In the present section, we show that it persists and that its law of motion is
well approximated by the interface equation in (P 0 ) obtained by formal asymptotic expansions
in subsection 2.2.1.
More precisely, take the first term of the formal asymptotic expansion (2.11) as a formal
expansion of the solution:
?
?
u (x, t) ? u? (x, t) := U0
│ d(x,
e t) ┤
?
,
(2.49)
where U0 is defined in (2.14). The right-hand side is a function having a well-developed
transition layer, and its interface lies exactly on ?t . We show that this function is a very good
approximation of the solution; more precisely:
If u? becomes very close to u?? at some time moment t = t0 , then it stays close to
u?? for the rest of time. Consequently, ??t evolves roughly like ?t .
+
To that purpose, we will construct a pair of sub- and super-solutions u?
? and u? for Problem
?
?
by slightly modifying u? . It then follows that, if the solution u satisfies
(P ? )
?
+
u?
? (x, t0 ) ? u (x, t0 ) ? u? (x, t0 ),
for some t0 ? 0, then
?
+
u?
? (x, t) ? u (x, t) ? u? (x, t),
?
?
?
for t0 ? t ? T . As a result, since both u+
? , u? stay close to u? , the solution u also stays close
to u?? for t0 ? t ? T .
2.4.1
Construction of sub- and super-solutions
To begin with we present mathematical tools which are essential for the construction of sub
and super-solutions.
A modified signed distance function. Rather than working with the usual signed distance
e defined in (2.9), we define a ?cut-off signed distance function? d as follows. Choose
function d,
e и) is smooth in the tubular neighborhood of ?
d0 > 0 small enough so that d(и,
and such that
e t)| < 3d0 },
{(x, t) ? QT , |d(x,
dist(?t , ??) > 4d0
for all t ? [0, T ].
Next let ?(s) be a smooth increasing function on R such that
?
if |s| ? 2d0
?
? s
?3d0 if s ? ?3d0
?(s) =
?
?
3d0
if s ? 3d0 .
(2.50)
88
Chapitre 2. A chemotaxis-growth system with general initial data
We define the cut-off signed distance function d by
А
б
? t) .
d(x, t) = ? d(x,
(2.51)
Note that |?d| = 1 in the region {(x, t) ? QT , |d(x, t)| < 2d0 } and that, in view of the above
definition, ?d = 0 in a neighborhood of ??. Note also that the equation of motion interface
in (P 0 ), which is equivalent to (2.17), is now written as
dt = ?d ? ?d и ??(v 0 ) ?
?
2?
on ?t .
(2.52)
?
Construction. We look for a pair of sub- and super-solutions u▒
? for Problem (P ) of the
form
│ d(x, t) ? ?p(t) ┤
u▒
▒ q(t),
(2.53)
(x,
t)
=
U
0
?
?
where U0 is the solution of (2.14), and where
2
p(t) = ?e??t/? + eLt + K,
(2.54)
2
q(t) = ?(?e??t/? + ?2 LeLt ).
Note that q = ??2 pt . Let us remark that the construction (2.53) is more precise than the
several procedure of only taking a zeroth order term of the form U0 , since we have shown
in the formal derivation that the first order term U1 vanishes in (2.11). It is clear from the
definition of u▒
? that
(
(1)
1 for all (x, t) ? QT
▒
lim u? (x, t) =
(2.55)
(0)
??0
0 for all (x, t) ? QT .
The main result of this section is the following.
Lemma 2.4.1. There exist positive constants ?, ? with the following properties. For any
K > 1, we can find positive constants ?0 and L such that, for any ? ? (0, ?0 ), the functions
+
?
u?
? and u? are respectively sub- and super-solutions for Problem (P ) in the range x ? ?,
0 ? t ? T.
2.4.2
Proof of Lemma 2.4.1
First, since ?d = 0 in a neighborhood of ??, we have the homogeneous Neumann boundary
condition
?u▒
?
= 0 on ?? О [0, T ].
??
Let v ? be such that (2.19) holds. We have to show that
+
+
+
?
+
?
Lv? [u+
? ] := (u? )t ? ?u? + ?u? и ??(v ) + u? ??(v ) ?
1
f? (u+
? ) ? 0,
?2
the proof of inequality Lv? [u?
? ] ? 0 following by the same arguments.
2.4 Motion of interface
89
Computation of Lv? [u+
?]
By straightforward computations we obtain the following terms:
dt
? pt ) + q t ,
?
? ?d
?u+
,
? = U0
?
?
(u+
? )t = U0 (
??
?u+
? = U0
|?d|2
?d
+ U0 ?
,
?2
?
А
б
where the function U0 , as well as its derivatives, is taken at the point d(x, t) ? ?p(t) /?. We
also use expansions of the reaction terms:
1 2 ??
?
f (u+
? ) = f (U0 ) + qf (U0 ) + q f (?),
2
?
g(u+
? ) = g(U0 ) + qg (?),
+
+
where ?(x, t) and ?(x, t) are some functions satisfying U0 < ? <
? u? , U? 0 < ? < u? . Combining
the above expressions with equation (2.14) and the fact that 2?U0 + g(U0 ) ? 0, we obtain
Lv? [u+
? ] = E1 + и и и + E5 ,
where:
E1 = ?
E2 =
1
1
q[f ? (U0 ) + qf ?? (?)] ? U0 ? pt + qt ,
2
?
2
U0 ??
(1 ? |?d|2 ),
?2
?
U0 ?
(dt ? ?d + ?d и ??(v 0 ) + 2?),
?
1
E4 = ? qg ? (?),
?
E3 =
E5 =
U0 ?
?
?d и ?(?(v ? ) ? ?(v 0 )) + u+
? ??(v ).
?
In order to estimate the terms above, we first present some useful inequalities. As f ? (0) =
= ?1/2, we can find strictly positive constants b and m such that
f ? (1)
if
U0 (z) ? [0, b] ? [1 ? b, 1]
then
f ? (U0 (z)) ? ?m.
(2.56)
On the other hand, since the region {z ? R | U0 (z) ? [b, 1 ? b] } is compact and since U0 ? < 0
on R, there exists a constant a1 > 0 such that
if
U0 (z) ? [b, 1 ? b]
then
U0 ? (z) ? ?a1 .
(2.57)
We then define
F =
sup |f (z)| + |f ? (z)| + |f ?? (z)|,
?1?z?2
?=
m
,
4
(2.58)
(2.59)
90
Chapitre 2. A chemotaxis-growth system with general initial data
and choose ? that satisfies
0 < ? ? min (?0 , ?1 , ?2 ),
where
?0 :=
a1
,
m+F
?1 :=
1
,
?+1
?2 :=
(2.60)
4?
.
F (? + 1)
Hence, combining (2.56) and (2.57), we obtain, using that ? ? ?0 ,
? U0 ? (z) ? ?f ? (U0 (z)) ? 4??
for ? ? < z < ?.
(2.61)
Now let K > 1 be arbitrary. In what follows we will show that Lv? [u+
? ] ? 0 provided that
the constants ?0 and L are appropriately chosen. From now on, we suppose that the following
inequality is satisfied:
?20 LeLT ? 1 .
(2.62)
Then, given any ? ? (0, ?0 ), since ? ? ?1 , we have 0 ? q(t) ? 1, hence, recalling that
0 < U0 < 1,
? 1 ? u▒
(2.63)
? (x, t) ? 2 .
We first estimate the term E1
A direct computation gives
E1 =
? ??t/?2
e
(I ? ??) + LeLt (I + ?2 ?L),
?2
where
I = ?U0 ? ? ?f ? (U0 ) ?
? 2 ??
2
f (?)(?e??t/? + ?2 LeLt ).
2
In virtue of (2.61) and (2.63), we obtain
I ? 4?? ?
?2
F (? + ?2 LeLT ).
2
Then, in view of (2.62), using that ? ? ?2 , we have
I ? 2??.
Consequently, the following inequality holds.
E1 ?
C1
?? 2 ??t/?2
2
e
+ 2??LeLt =: 2 e??t/? + C1 ? LeLt .
2
?
?
As for the term E2
First, in the points where where |d(x, t)| < d0 , we have that |?d| = 1 so that E2 = 0. Next
we consider the points where |d(x, t)| ? d0 . We deduce from Lemma 2.2.1 that:
C
(1 + k?dk2? )e??|d+?p|/?
?2
C
? 2 (1 + k?dk2? )e??(d0 /??|p|) .
?
|E2 | ?
2.4 Motion of interface
91
In view of the definition of p in (2.54), we have that 0 < K ? 1 ? p ? eLT + K, and suppose
from now that the following assumption holds:
eLT + K ?
Then
d0
.
2?0
(2.64)
d0
d0
? |p| ? , so that
?
2?
C
(1 + k?dk2? )e??d0 /(2?)
?2
16C
? C2 :=
(1 + k?dk2? ).
(e?d0 )2
|E2 | ?
Next we consider the term E3
We recall that
dt ? ?d + ?d и ??(v 0 ) +
?
2? = 0
on ?t = {x ? ?, d(x, t) = 0}.
?
? 1+?
2+?
Since v 0 is of class C 1+? , 2 , for any ?? ? (0, 1), and since the interface ?t is of class C 2+?, 2 ,
the functions ?d, ?d, dt and ??(v 0 ) are Lipschitz continuous near ?t . It then follows, from
the mean value theorem applied separately on both sides of ?t , that there exists N0 > 0 such
that:
?
|(dt ? ?d + ?d и ??(v 0 ) + 2?)(x, t)| ? N0 |d(x, t)| for all (x, t) ? QT .
Applying Lemma 2.2.1 we deduce that
|E3 | ? N0 C
|d(x, t)| ??|d(x,t)/?+p(t)|
e
?
? N0 C maxy?R |y|e??|y+p(t)|
А
1б
? N0 C max |p(t)|,
?
А
б
1
? N0 C |p(t)| + .
?
Taking the expression of p into account, we see that |p(t)| ? eLt + K, which implies
|E3 | ? C3 (eLt + K) + C3 ? ,
where C3 := N0 C and C3 ? := N0 C/?.
The term E4
Substituting the expression for q and setting G1 := sup{|g ? (z)|; ?1 ? z ? 2} leads to
|E4 | ? ?G1
│?
2
e??t/? + ?LeLt
?
C4 ??t/?2
e
+ C4 ? ?LeLt .
?
?
┤
92
Chapitre 2. A chemotaxis-growth system with general initial data
We continue with the term E5
This term requires a more delicate analysis. We need a precise estimate of v ? ? v 0 . We recall
that v 0 satisfies ??v 0 + ?v 0 = u0 , with u0 a step function discontinuous when crossing the
interface.
Lemma 2.4.2. There exists a positive constant CG such that, for all (x, t) ? QT ,
│
┤
|v ? | + |?v ? | + |?v ? | (x, t) ? CG ,
│
┤
|v ? ? v 0 | + |?d и ?(v ? ? v 0 )| (x, t) ? CG (?p(t) + q(t)).
(2.65)
(2.66)
We postpone the proof of this lemma and pursue the proof of Lemma 2.4.1. Using the smoothness of ? and (2.65), we obtain a uniform bound CG ? for ??(v ? ). Moreover, we can write
А
б
А
б
?d и ? ?(v ? ) ? ?(v 0 ) = ?? (v ? )?d и ?(v ? ? v 0 ) + ?? (v ? ) ? ?? (v 0 ) ?d и ?v 0 .
?
? 1+?
2
Since v 0 is of class C 1+? ,
again by CG , such that
(2.67)
, for any ?? ? (0, 1), there exists a constant, which we denote
kv 0 kL? (QT ) + k?v 0 kL? (QT ) ? CG ,
which, combined with (2.67), yields
А
б
|?d и ? ?(v ? ) ? ?(v 0 ) | ? k?? k? |?d и ?(v ? ? v 0 )| + |v ? ? v 0 | k??? k? k?dk? CG ,
where the L? -norms of ?? and ??? are considered on the interval (?CG , CG ). It follows from
the above inequality and (2.66) that there exists a constant CG ?? such that, for all (x, t) ? QT ,
А
б
|?d и ? ?(v ? ) ? ?(v 0 ) |(x, t) ? CG ?? (?p(t) + q(t)).
Hence, using (2.63) and the above estimates, we obtain,
|E5 | ?
C
CG ?? (?p(t) + q(t)) + 2CG ? .
?
Then, substituting the expressions for p and q, we easily obtain positive constants C5 , C5 ? and
C5 ?? such that
C5 ? ??t/?2
+ C5 ?? (1 + ?L)eLt .
|E5 | ? C5 +
e
?
Completion of the proof
Collecting the above estimates of E1 ?E5 yields
L
v?
[u+
?]
?
│C
1
?2
┤
C4 + C5 ? ┤ ??t/?2 │
?
?
??
??
e
?
+ L(C1 ? ?C4 ? ?C5 ) ? C3 ? C5 eLt ? C7 ,
?
where C7 := C2 + KC3 + C3 ? + C5 . Now, we set
L :=
d0
1
,
ln
T
4?0
2.4 Motion of interface
93
which, for ?0 small enough, validates assumptions (2.62) and (2.64). If ?0 is chosen sufficiently
small (i.e. L large enough), C1 /?2 ? (C4 + C5 ? )/? is positive, C1 ? ? ?C4 ? ? ?C5 ?? ? 12 C1 ? , and
Б1
?
?? Lt
Lv? [u+
? ] ? 2 LC1 ? C3 ? C5 ]e ? C7
? 14 LC1 ? ? C7
? 0.
The proof of Lemma 2.4.1 is now completed, with the choice of the constants ?, ? as in (2.59),
(2.60).
2.4.3
Proof of Lemma 2.4.2
Lemma 2.4.2 is very inspired by Lemma 4.9 in [17]. We present the proof for the convenience
of the reader. Since our pair of sub- and super-solutions is different from the one in [17], we
need to perform some minor changes. First we give a useful estimate on ?shifted U0 ?.
Lemma 2.4.3. For all a ? R, all z ? R, we have
|U0 (z + a) ? ?]??,0] (z)| ? Ce??|z+a| + ?[?a,a] (z)
Proof. Let us give the proof for a > 0. We distinguish three cases and use the estimates of
Lemma 2.2.1. For z ? ?a, we have |U0 (z + a) ? 1| ? Ce??|z+a| . For ?a < z ? 0, we have
|U0 (z + a) ? 1| ? |U0 (z + a)| + 1 ? Ce??|z+a| + 1. For z > 0, we have |U0 (z + a)| ? Ce??|z+a| .
We proceed in the same way for a < 0.
We turn to the proof of Lemma 2.4.2. First, we recall that v ? is such that (2.19) holds;
hence, in view of (2.63), the estimate (2.65) is a direct consequence of the standard theory of
elliptic equations. Next we prove (2.66). The function w = w? := v ? ? v 0 is solution of
?
?
???w + ?w = h on QT ,
(2.68)
?
? ?w = 0
on ?? О (0, T ),
??
0
?
+
0
0
0
with u?
? ? u ? h = h ? u? ? u , where u is the step function defined by u (x, t) =
?{d(x,t)?0} . The key idea of the proof is the fact that h is exponentially small with respect to
?, except possibly in a thin neighborhood of ?t of width of order ?p(t). More precisely, from
the definitions of u▒
? in (2.53) and from the above lemma for z = d(x, t)/? and a = ▒p(t), we
deduce that
|h(x, t)| ? C(e??|d(x,t)/?+p(t)| + e??|d(x,t)/??p(t)| ) + ?{|d(x,t)|??p(t)} + q(t).
(2.69)
By linearity, we successively consider equation (2.68) with the various terms appearing in the
right-hand side of (2.69). By the standard elliptic estimates, the solution w of (2.68) satisfies
|w(x, t)| + |?w(x, t)| ? C ? sup |h(y, t)|,
(2.70)
y??
which gives the term CG q(t) that appears in the right-hand side of inequality (2.66) for h(y, t) =
q(t). We now suppose that the function h satisfies one of the three following assumptions:
(H1 )
(H2▒ )
|h(y, t)| ? ?{|d(y, t)| ? ?p(t)}
│
┤
d(y, t)
|h(y, t)| ? exp ? ?|
▒ p(t)| ,
?
94
Chapitre 2. A chemotaxis-growth system with general initial data
and write
h(y, t) = h(y, t)?{|d(y,t)|?d0 } + h(y, t)?{|d(y,t)|>d0 } .
We first consider the term h(y, t)?{|d(y,t)|>d0 } . In virtue of (2.64), we have
0 < K ? 1 ? p(t) ? d0 /2?0 .
(2.71)
Under assumption (H1 ), it follows that h is supported in {|d(y, t)| ? d0 /2}, which implies
h(y, t)?{|d(y,t)|>d0 } = 0. Moreover, under assumption (H2▒ ), using again (2.71),
Б
ц
|h(y, t)|?{|d(y,t)|>d0 } ? exp ? ?(d0 /? ? p(t))
? exp(??d0 /2?)
?
2
?
?d0 e
?
1
2
?p(t).
?d0 e K ? 1
Thus, under either of the assumptions (H1 ) or (H2▒ ), the estimate (2.66) ? for the term
h(y, t)?{|d(y,t)|>d0 } ? directly follows from inequality (2.70).
From now on, we assume that h is supported in {|d(y, t)| ? d0 }. We have that
Z
w(x, t) =
G(x, y)h(y, t)dy,
|d(y,t)|?d0
and
?d(x, t) и ?w(x, t) =
Z
|d(y,t)|?d0
(?x G(x, y) и ?d(x, t))h(y, t)dy,
where G is the Green?s function associated to the homogeneous Neumann boundary value
problem on ? for the operator ?? + ?. More precisely, G(x, y) = g? (|x ? y|) + H? (x, y),
where g? (|x ? y|) is the Green?s function associated to the operator ?? + ? on RN and where
H? (x, y) is smooth for x and y far away from ??. It is known that g? is the Bessel function
defined by
Z +?
r2
s ?N +2 ds
,
g? (r) = cN
e? 2s e?? 2 s 2
s
0
with cN > 0 a normalization constant. We use the following estimates (see [17]):
|G(x, y)| ?
|?x G(x, y) и ?d(x, t)| ?
?
?
?
?
?
C
|y ? x|N ?2
for N ? 3
C| ln |y ? x||
for N = 2,
C|d(y, t) ? d(x, t)|
C
+
|y ? x|N
|y ? x|N ?2
This last inequality follows from
|?x G(x, y) и ?d(x, t)| ?
C|?d(x, t) и (y ? x)|
,
|y ? x|N
(2.72)
for N ? 2.
(2.73)
2.4 Motion of interface
95
and from d(y, t)?d(x, t) = ?d(x, t)и(y ?x)+O(|y ?x|2 ). Now, under respectively assumptions
(H1 ), (H2▒ ), we define a function h? = h?? on R О [0, T ], respectively by
?
?
??{|r| ? ?p(t)}
h?(r, t) :=
(2.74)
│
┤
?
?exp ? ?| r ▒ p(t)| .
?
А
б
Note that |h(y, t)| ? h? d(y, t), t . Moreover, recalling (2.71), straightforward computations
show that, under either of the assumptions (H1 ) or (H2▒ ), there exists C? > 0 such that
Z d0
h?(r, t)dr ? C??p(t).
(2.75)
0?
?d0
Finally we have
|w(x, t)| + |?d(x, t) и ?w(x, t)| ? C[A(x, t) + B(x, t)],
with
A(x, t) =
?Z
А
б
2h? d(y, t), t
?
?
?
dy
?
N ?2
?
?
? |d(y,t)|?d0 |y ? x|
?
?
Z
?
?
?
?
?
|d(y,t)|?d0
and
B(x, t) =
Z
if N ? 3
i
А
бh
h? d(y, t), t 1 + | ln |y ? x|| dy
|d(y,t)|?d0
(2.76)
if N = 2,
А
б
|d(y, t) ? d(x, t)|h? d(y, t), t
dy.
|y ? x|N
We now distinguish two cases according to the distance from x to the interface.
? If dist(x, ?t ) ? 2d0 , then |y ? x| ? d0 for all y with |d(y, t)| ? d0 so that
Z
А
б
A(x, t) + B(x, t) ? C
h? d(y, t), t dy,
|d(y,t)|?d0
for a constant C = C(d0 ). Taking d as one of the coordinates in {|d(y, t)| ? d0 }, we get
Z d0
A(x, t) + B(x, t) ? C
h?(r, t)dr,
?d0
and the estimate (2.66) follows from (2.76) and (2.75).
? If dist(x, ?t ) ? 2d0 , then we can make, for any fixed t ? [0, T ], the change of variables
{y ? ?, |d(y, t)| ? 2d0 } ? ?t О [?2d0 , 2d0 ]
y 7? (s, r),
where s = s(y, t) is the projection of y on ?t along the normal of ?t , and r = d(y, t) is the
signed distance function from y to ?t . We write this change of variables as y = X(s, r).
In the case N ? 3, we have
Z d0
Z
ds
A(x, t) ? C
drh?(r, t)
,
N ?2
?d0
?t |X(s, r) ? X(s0 , r0 )|
96
Chapitre 2. A chemotaxis-growth system with general initial data
with x = X(s0 , r0 ). Since X is a smooth diffeomorphism, there exists C > 0 such that
for all (s, r) ? ?t О [?2d0 , 2d0 ], we have
|X(s, r) ? X(s0 , r0 )| ? C(|s ? s0 | + |r ? r0 |) ? C|s ? s0 |.
Hence
A(x, t) ? C
Z
d0
drh?(r, t)
?d0
Z
?t
ds
.
|s ? s0 |N ?2
Since the singularity in 1/|s|N ?2 is integrable on the (N ? 1)-dimensional hypersurface
?t , we have
Z
ds
? C,
N ?2
?t |s ? s0 |
for some C > 0 which does not depend on t ? [0, T ] and s0 ? ?t . Therefore,
A(x, t) ? C
Z
d0
h?(r, t)dr.
(2.77)
?d0
The same estimate is obtained in the case where N = 2, using the fact that ln |s| is
integrable on a finite line.
As for B, we have
B(x, t) ? C
Z
d0
drh?(r, t)
?d0
Z
?t
Б
|r ? r0 |
цN ds.
|s ? s0 | + |r ? r0 |
Making the change of variables s ? s0 = |r ? r0 |s? , we get
Z
?t
|r ? r0 |
ds ?
[|s ? s0 | + |r ? r0 |]N
Z
??t
ds?
? C,
[|s? | + 1]N
where ??t is the image of ?t by the mapping s 7? s? and where C > 0 does not depend
on |r| ? d0 , s0 ? ?t , |r0 | ? 2d0 and t ? [0, T ]. Thus
B(x, t) ? C
Z
d0
h?(r, t)dr.
(2.78)
?d0
Then the estimate (2.66) follows from inequalities (2.76), (2.77), (2.78) and (2.75).
The proof of Lemma 2.4.2 is now complete.
2.5
Proof of the main results
In this section, we prove our main results by fitting the two pairs of sub- and super-solutions,
constructed for the study of the generation and the motion of interface, into each other.
2.5 Proof of the main results
2.5.1
97
Proof of Theorem 2.1.3
Let ? ? (0, 1/4) be arbitrary. Choose ? and ? that satisfy (2.59), (2.60) and
?? ?
?
.
3
(2.79)
By the generation of interface Theorem 2.3.1, there exist positive constants ?0 and M0 such
that (2.21), (2.22) and (2.23) hold with the constant ? replaced by ??/2. Since ?u0 и n 6= 0
everywhere on the initial interface ?0 = {x ? ?, u0 (x) = 1/2} and since ?0 is a compact
hypersurface, we can find a positive constant M1 such that
if
if
d0 (x) ?
M1 ?
then
d0 (x) ? ?M1 ?
then
u0 (x) ? 1/2 ? M0 ?,
u0 (x) ? 1/2 + M0 ?.
(2.80)
Here d0 (x) := d(x, 0) denotes the cut-off signed distance function associated with the hypersurface ?0 . Now we define functions H + (x), H ? (x) by
(
if d0 (x) < M1 ?
1 + 12 ??
+
H (x) =
1
if d0 (x) ? M1 ?,
2 ??
(
if d0 (x) ? ?M1 ?
1 ? 12 ??
?
H (x) =
if d0 (x) > ?M1 ?.
? 12 ??
Then from the above observation we see that
H ? (x) ? u? (x, х?1 ?2 | ln ?|) ? H + (x)
for x ? ?.
(2.81)
??
.
3
(2.82)
for x ? ?.
(2.83)
Next we fix a sufficiently large constant K > 1 such that
U0 (M1 ? K) ? 1 ?
??
3
and
U0 (?M1 + K) ?
For this K, we choose ?0 and L as in Lemma 2.4.1. We claim that
?
u?
? (x, 0) ? H (x),
H + (x) ? u+
? (x, 0)
We only prove the former inequality, as the proof of the latter is virtually the same. Then it
amounts to showing that
u?
? (x, 0) = U0
б
А d0 (x)
+ K ? ?(? + ?2 L) ? H ? (x).
?
(2.84)
In the range where d0 (x) > ?M1 ?, the second inequality in (2.82) and the fact that U0 is a
decreasing function imply
U0
А d0 (x)
б
1
+ K ? ?(? + ?2 L) ? ?? ? ?? ? H ? (x).
?
3
On the other hand, in the range where d0 (x) ? ?M1 ?, we have
U0
А d0 (x)
б
+ K ? ?(? + ?2 L) ? 1 ? ?? ? H ? (x).
?
This proves (2.84), hence (2.83) is established.
98
Chapitre 2. A chemotaxis-growth system with general initial data
Combining (2.81) and (2.83), we obtain
?
?1 2
+
u?
? (x, 0) ? u (x, х ? | ln ?|) ? u? (x, 0).
+
?
Since u?
? and u? are sub- and super-solutions for Problem (P ) thanks to Lemma 2.4.1, the
comparison principle yields
?
?
+
u?
? (x, t) ? u (x, t + t ) ? u? (x, t)
for
0 ? t ? T ? t? ,
(2.85)
where t? = х?1 ?2 | ln ?|. Note that, in view of (2.55), this is enough to prove Corollary 2.1.4.
Now let C be a positive constant such that
U0 (?C + eLT + K) ? 1 ?
?
2
and
U0 (C ? eLT ? K) ?
?
.
2
(2.86)
One then easily checks, using successively (2.85), (2.53), (2.86) and (2.79), that, for ?0 small
enough, for 0 ? t ? T ? t? , we have
if
if
d(x, t) ?
C?
then
d(x, t) ? ?C?
then
u? (x, t + t? ) ? ?,
u? (x, t + t? ) ? 1 ? ?,
(2.87)
and
u? (x, t + t? ) ? [??, 1 + ?],
which completes the proof of Theorem 2.1.3.
2.5.2
Proof of Theorem 2.1.5
In the case where х?1 ?2 | ln ?| ? t ? T , the assertion of the theorem is a direct consequence of
Theorem 2.1.3. All we have to consider is the case where 0 ? t ? х?1 ?2 | ln ?|. We shall use
the sub- and super-solutions constructed for the study of the generation of interface in Section
2.3. To that purpose, we first prove the following lemma concerning Y (?, ?; ?), the solution of
the ordinary differential equation (2.28), in the initial time interval.
Lemma 2.5.1. There exist constants C8 > 0 and ?0 > 0 such that, for all ? ? (0, ?0 ),
if
if
? ? 1/2 + C8 ? then
? ? 1/2 ? C8 ? then
Y (?, ?; ▒?G) > 1/2
Y (?, ?; ▒?G) < 1/2
for
for
0 ? ? ? х?1 | ln ?|,
0 ? ? ? х?1 | ln ?|.
(2.88)
Proof. We only prove the first inequality. Assume ? ? 1/2 + C8 ?. By (2.25), for C8 ? CG,
we have that ? ? 1/2 + C8 ? ? a(▒?G). It then follows from (2.36) that
Y (?, ?; ▒?G) ? a(▒?G) + C1 eх(▒?G)? (1/2 + C8 ? ? a(▒?G))
? 1/2 ? CG? + C1 (?CG? + C8 ?)
? 1/2 + ?(C1 C8 ? CG(C1 + 1))
> 1/2,
provided that C8 is sufficiently large.
Now we turn to the proof of Theorem 2.1.5. We first claim that there exists a positive
constant M2 such that for all t ? [0, х?1 ?2 | ln ?|],
??t ? NM2 ? (?0 ).
(2.89)
2.5 Proof of the main results
99
To see this, we choose M0 ? large enough, so that M0 ? ? C8 + 2C6 , where C6 is as in Lemma
2.3.9. As is done for (2.80), there is a positive constant M2 such that
if
if
d0 (x) ?
M2 ?
then
d0 (x) ? ?M2 ?
then
u0 (x) ? 1/2 ? M0 ? ?,
u0 (x) ? 1/2 + M0 ? ?.
(2.90)
In view of this last condition, we see that, if ?0 is small enough, if d0 (x) ? M2 ?, then for all
0 ? t ? х?1 ?2 | ln ?|,
u0 (x) + ?2 r(?G,
Б
ц
t
?
2 C eх(?G)| ln ?|/х ? 1
)
?
1/2
?
M
?
+
?
0
6
?2
ц
Б
? 1/2 + ? ? M0 ? + C6 ?(х?х(▒?G))/х ? ?C6
? 1/2 + ?(?M0 ? + 2C6 )
? thanks to (2.48)
? 1/2 ? C8 ?.
This inequality and Lemma 2.5.1 imply w?+ (x, t) < 1/2, where w?+ is the sub-solution defined
in (2.40). Consequently, by (2.43),
u? (x, t) < 1/2
if d0 (x) ? M2 ?.
In the case where d0 (x) ? ?M2 ?, similar arguments lead to u? (x, t) > 1/2. This completes
the proof of (2.89). Note that we have proved that, for all 0 ? t ? х?1 ?2 | ln ?|,
u? (x, t) > 1/2
u? (x, t) < 1/2
(1)
if x ? ?0 \ NM2 ? (?0 ),
(0)
if x ? ?0 \ NM2 ? (?0 ).
(2.91)
Next, since ?t depends on t smoothly, there is a constant C? > 0 such that, for all t ?
[0, х?1 ?2 | ln ?|],
?0 ? NC??2 | ln ?| (?t ),
(2.92)
and
(1)
(1)
(0)
(0)
?t \ NC?? (?t ) ? ?0 \ NM2 ? (?0 ),
?t \ NC?? (?t ) ? ?0 \ NM2 ? (?0 ).
(2.93)
As a consequence of (2.89) and (2.92) we get
??t ? NM2 ?+C??2 | ln ?| (?t ) ? NC? (?t ),
which completes the proof of Theorem 2.1.5.
Proof of Corollary 2.1.6. In view of Theorem 2.1.5 and the definition of the Hausdorff
distance, to prove this corollary we only need to show the reverse inclusion, that is
?t ? NC ? ? (??t )
for
0 ? t ? T,
(2.94)
for some constant C ? > 0. To that purpose let C ? be a constant satisfying C ? > max(C?, C),
where C is as in Theorem 2.1.3 and C? as in (2.93). Choose t ? [0, T ], x0 ? ?t arbitrarily and,
n being the Euclidian normal vector exterior to ?t at point x0 , define a pair of points:
x(0) := x0 + C ? ?n and x(1) := x0 ? C ? ?n.
100
Chapitre 2. A chemotaxis-growth system with general initial data
Since C ? > C and since the curvature of ?t is uniformly bounded as t varies over [0, T ], we see
that, if ?0 is sufficiently small,
(0)
(1)
x(0) ? ?t \ NC? (?t ) and x(1) ? ?t \ NC? (?t ).
Therefore, if t ? [х?1 ?2 | ln ?|, T ], then, by Theorem 2.1.3, we have
u? (x(0) , t) < 1/2 < u? (x(1) , t).
(2.95)
On the other hand, if t ? [0, х?1 ?2 | ln ?|], then from (2.91), (2.93) and the fact that C ? > C?,
we again obtain (2.95). Thus (2.95) holds for all t ? [0, T ]. Now, by the mean value theorem,
we see that for each t ? [0, T ] there exists a point x? such that
x? ? [x(0) , x(1) ] and u? (x?, t) = 1/2.
This implies x? ? ??t . Furthermore we have |x0 ? x?| ? C ? ?, since x? lies on the line segment
[x(0) , x(1) ]. This proves (2.94).
Chapter 3
The singular limit of a spatially
inhomogeneous and anisotropic
Allen-Cahn equation
We consider a spatially inhomogeneous and anisotropic reaction-diffusion equation, involving
a small parameter ? > 0 and a bistable nonlinear term whose stable equilibria are 0 and 1,
which arises for instance in material sciences. The diffusion term may be singular in the points
where the gradient of the solution vanishes. We define a notion of weak solution and prove
a comparison principle. We perform the analysis using the distance function associated with
a Finsler metric. We consider rather general initial data u0 that are independent of ?. We
perform a rigorous analysis of both the generation and the motion of interface. More precisely,
we show that, within the time scale of order ?2 | ln ?|, the solution u? develops a steep transition
layer that separates the regions {u? ? 0} and {u? ? 1}. Then, in a much slower time scale,
the layer starts to propagate. As a consequence, as ? ? 0, the solution u? converges almost
+
?
+
everywhere to 0 in ??
t and 1 in ?t , where ?t and ?t are sub-domains of ? separated by an
interface ?t , whose motion is driven by its anisotropic mean curvature. We also prove that
the thickness of the transition layer is of order ?.
102 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
3.1
Introduction
The background of our study is the modelling of anisotropic interface motion, where the normal
velocity of displacement of the interface depends on the angle of the normal vector with a fixed
direction [4], [65]. Related nonlinear reaction-diffusion equations give rise to sharp internal
layers, or interfaces, when the coefficient of the diffusion term is very small or the one of
the reaction term very large. In this Chapter we consider the following anisotropic parabolic
problem involving a spatially inhomogeneous reaction-diffusion equation:
?
1
?
?
u = ? и ap (x, ?u) + 2 f (u) in ? О (0, +?),
?
? t
?
(P ? )
on ?? О (0, +?),
ap (x, ?u) и ? = 0
?
?
?
?
in ?,
u(x, 0) = u0 (x)
where ? is a smooth bounded domain of RN (N ? 2) and ? the Euclidian unit normal vector
exterior to ??.
The parabolic equation in Problem (P ? ) contains, on the one hand, the inhomogeneous
partial differential equation
1
ut = div(A(x)?u) + 2 f (u),
(3.1)
?
where A(x) is a positively definite symmetric matrix depending on the spatial location, and,
on the other hand, the anisotropic equation
А
б
1
ut = div A(?u) + 2 f (u),
?
(3.2)
where the coefficients of the matrix ?p ? A = ?p t A may be singular in the point p = 0.
We suppose that the nonlinear reaction function f is such that f (u) = ?W ? (u), where
W (u) is a double-well potential with equal well-depth, taking its global minimum value at
u = 0 and u = 1. More precisely we assume that f is smooth and has exactly three zeros
0 < a < 1 such that
f ? (0) < 0, f ? (a) > 0, f ? (1) < 0,
(3.3)
and that
Z
1
f (u)du = 0.
(3.4)
0
The assumptions concerning the anisotropic term are the following.
3+?
1. a(x, p) is a real valued function, of class Cloc
(for some 0 < ? < 1) on ? О RN \{0};
2. a(x, p) > 0 for all (x, p) ? ? О RN \{0};
3. a(x, и) is strictly convex for all x ? ?;
4. a(x, p) is 2 homogeneous in the p variable, i.e.
a(x, ?p) = ?2 a(x, p)
for all (x, p) ? ? О RN \{0}, all ? 6= 0.
If p is given by p = (p1 , и и и , pN ), the vector valued function ap is defined by
ap (x, p) =
│ ?a
┤
?a
(x, p), и и и ,
(x, p) ,
?p1
?pN
(3.5)
3.1 Introduction
103
and the matrix valued function app by
app (x, p) =
│ ?2a
┤
(x, p)
,
?pj ?pi
1?i,j?N
for all (x, p) ? ? О (RN \ {0}). Moreover, for a vector p = (p1 , и и и , pN ) and a matrix A =
(aij )1?i,j?N , we use the notations
|p| = max |pi |
i
and
|A| = max |aij |.
i,j
The special case (3.1) is obtained by taking a(x, p) = 12 A(x)p и p, where A(x) is a positively
3+?
on the whole ?ОRN , ap (x, p) = A(x)p
definite symmetric matrix; here, a(x, p) is of class Cloc
and app (x, p) = A(x). Setting A(x) = I leads to the classical Allen-Cahn equation. The special
case (3.2) is obtained by assuming that a(x, p) = a(p) and defining A(p) = ap (p).
Remark 3.1.1. By differentiating (3.5) with respect to p and to ?, we see that, for all (x, p) ?
? О RN \{0}, all ? 6= 0,
ap (x, ?p) = ?ap (x, p),
ap (x, ?p) и p = 2?a(x, p),
app (x, ?p) = app (x, p),
app (x, ?p)p = ap (x, p),
app (x, ?p)p и p = 2a(x, p).
We may define a(x, 0) = 0 and ap (x, 0) = 0, which implies, in view of (3.5) ? i.e. a(x, и) is
2 homogeneous ? and the first equality above ? i.e. ap (x, и) is 1 homogeneous ? that a(x, p)
is of class C 1 on the whole ? О RN .
We also assume that the initial datum u0 ? C 2 (?), and define C0 as
C0 := ku0 kC 0 (?) + k?u0 kC 0 (?) + kD2 u0 kC 0 (?) ,
where D2 u0 (x) =
(3.6)
│ ?2u
┤
0
(x)
. Furthermore we define the ?initial interface? ?0 by
?xj ?xi
1?i,j?N
?0 := {x ? ?, u0 (x) = a},
and suppose that ?0 is a C 4+? hypersurface without boundary such that, n being the Euclidian
unit normal vector exterior to ?0 ,
?0 ?? ?
and
u0 > a
?u0 (x) и ap (x, n(x)) 6= 0
in ?+
0,
u0 < a
if x ? ?0 ,
in ??
0,
(3.7)
(3.8)
+
where ??
0 denotes the region enclosed by ?0 and ?0 the region enclosed between ?? and ?0 .
For T > 0, we set QT = ? О (0, T ). We define below a notion of weak solution for Problem
(P ? ). For this definition, it is sufficient to only suppose that u0 ? H 1 (?) ? L? (?).
Definition 3.1.2. A function u? ? L2 (0, T ; H 1 (?)) ? L? (QT ) is a weak solution of Problem
(P ? ), if
104 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
? u?t ? L2 (QT ),
? ap (x, ?u? (x, t)) ? L? (0, T ; L2 (?)),
? u? (x, 0) = u0 (x) for almost all x ? ?,
? u? satisfies the integral equality
Z tZ h
0
?
u?t ? + ap (x, ?u? ) и ?? ?
i
1
?
f
(u
)?
= 0,
?2
(3.9)
for all nonnegative function ? ? L2 (0, T ; H 1 (?)) ? L? (QT ) and for all t ? [0, T ].
One may prove, using monotonicity and compactness arguments as is done in [12] and [14],
that Problem (P ? ) possesses a unique weak solution which we denote by u? . As ? ? 0, the
qualitative behavior of this solution is the following. In the very early stage, the anisotropic
diffusion term is negligible compared with the reaction term ??2 f (u). Hence, rescaling time by
? = t/?2 , the equation is well approximated by the ordinary differential equation u? = f (u).
Since f is a bistable function, u? quickly approaches the values 0 or 1, the stable equilibria
of f , and an interface is formed between the regions {u? ? 0} and {u? ? 1}. Once such an
interface is developed, the anisotropic diffusion term becomes large near the interface, and
comes to balance with the reaction term so that the interface starts to propagate, in a much
slower time scale.
To understand such interfacial behavior, we have to study the singular limit of (P ? ) as
? ? 0. Then the limit solution u?(x, t) is a step function taking the values 0 and 1 on the sides
of the moving interface ?t . In the case of the usual Allen-Cahn equation, it is well known that
?t evolves according to the mean curvature flow Vn = ?(N ? 1)?. Here we will prove that the
interface evolves according to the law
0
(P )
?
h
i
1
1
?
p
p
?
V
a
=
??
и
(x,
n)
n
p
?
2a(x, n)
2a(x, n)
»
?
?
??t »»
= ?0 .
on ?t ,
t=0
where n is the Euclidian unit normal vector exterior to ?t and Vn the normal velocity of ?t .
We will show below that this equation can be rewritten in the relative geometry associated
with a Finsler metric; it then has the form
?
?
?Vn,? = ?(N ? 1)?? on ?t ,
0
»
(P )
?
??t »»
= ?0 .
t=0
where Vn,? is the anisotropic normal velocity of ?t in the anisotropic exterior direction, and
?? an analogue of the anisotropic mean curvature at each point of ?t .
Almgren, Taylor and Wang [3] have proved that a problem related to the limit Problem
0
(P ) possesses locally in time a unique smooth solution. Here we Swill suppose that there
exists T > 0 such that Problem (P 0 ) has a unique solution ? = 0?t?T (?t О {t}) which
satisfies ? ? C 3+?,(3+?)/2 . For proofs of the local in time existence of solutions of related limit
problems, we also refer to [41].
3.1 Introduction
105
+
For each t ? (0, T ), we define ??
t as the region enclosed by the hypersurface ?t and ?t as
the region enclosed between ?? and ?t . Further we define a function u?(x, t) by
(
1 in ?+
t
for t ? (0, T ).
(3.10)
u?(x, t) =
0 in ??
t
The aim of this Chapter is to study the limiting behavior of the solution u? of Problem (P ? )
as ? ? 0. We extend previous studies [13], [70] about a related anisotropic parabolic equation.
It is convenient to present our main result, Theorem 3.1.3, in the form of a convergence
theorem, mixing generation and propagation of interface. It describes the profile of the solution
after a very short initial period. It asserts that: given a virtually arbitrary initial datum u0 ,
the solution u? quickly becomes close to 0 or 1, except in a small neighborhood of the initial
interface ?0 , creating a steep transition layer around ?0 (generation of interface). The time t?
for the generation of interface is of order ?2 | ln ?|. The theorem then states that the solution
u? remains close to the step function u? on the time interval (t? , T ) (motion of interface).
Moreover, as is clear from the estimates in the theorem, the thickness of the transition layer
is of order ?.
Theorem 3.1.3 (Generation and motion of interface). Let ? be an arbitrary constant
satisfying 0 < ? < 12 min(a, 1 ? a) and set
х = f ? (a).
Then there exist positive constants ?0 and C such that, for all ? ? (0, ?0 ), for almost all (x, t)
such that t? ? t ? T , where
t? := х?1 ?2 | ln ?|,
(3.11)
we have,
?
?
?
? [??, 1 + ?]
?
?
u? (x, t) ? [??, ?]
?
?
?
?
? [1 ? ?, 1 + ?]
if
x ? NC? (?t )
if
x ? ??
t \ NC? (?t )
if
x ? ?+
t \ NC? (?t ),
(3.12)
where Nr (?t ) := {x ? ?, dist? (x, ?t ) < r} denotes the r-neighborhood of ?t ; by dist? (x, ?t ),
we mean the ?? distance to the set ?t , where ?? is the distance associated to a Finsler metric,
whose definition is given in Section 3.2.
Corollary
3.1.4 (Convergence). As ? ? 0, the solution u? converges to u? almost everywhere
S
in 0<t?T (?▒
t О {t}).
The organization of this Chapter is as follows. In Section 3.2, we recall notations and
results concerning the Finsler metrics that turn out to be very efficient in the anisotropic
context. In Section 3.3, we perform asymptotic expansions in order to derive the equation of
the interface motion. In Section 3.4 we prove a weak comparison principle for Problem (P ? ).
In Section 3.5, we prove a generation of interface property. The sub- and super-solutions for
this time range are constructed by modifying the solution of the ordinary differential equation
ut = ??2 f (u). In Section 3.6, we construct a pair of sub- and super-solutions for the study
of the motion of interface, by using a related one-dimensional stationary problem. In Section
3.7, by fitting these two pairs of sub- and super-solutions into each other, we prove our main
results for (P ? ): Theorem 3.1.3 and its corollary.
106 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
For the proof of a propagation of interface property in the case of a related problem we
refer to Bellettini, Colli Franzone and Paolini [8], who also give a precise approximation of
the moving interface. We also refer to articles about the convergence to classical and viscosity
solutions, in the case of well prepared initial data, by Elliott and Scha?tzle [31], [32], for a
homogeneous function a = a(p).
3.2
Finsler metrics and the anisotropic context
For the convenience of the reader, we first recall properties stated by Bellettini, Paolini and
Venturini, [9] and [10]. For more details and proofs, see these references where the idea is to
endow RN with the distance obtained by integrating the Finsler metric and to work in relative
geometry.
3.2.1
Finsler metrics
Suppose that ? : ? О RN ? [0, +?[ is a continuous function satisfying the properties
?(x, ??) = |?|?(x, ?)
?0 |?| ? ?(x, ?) ? ?0 |?|
for all (x, ?) ? ? О RN and all ? ? R,
N
for all (x, ?) ? ? О R ,
(3.13)
(3.14)
for two suitable constants 0 < ?0 ? ?0 < +?. We say that ? is strictly convex if, for any
x ? ?, the map ? 7? ?2 (x, ?) is strictly convex on RN . We shall indicate by
B? (x) = {? ? RN , ?(x, ?) ? 1}
the unit sphere of ? at x ? ?.
The dual function ?0 : ? О RN ? [0, +?[ of ? is defined by
Е
ф
?0 (x, ? ? ) = sup ? ? и ?, ? ? B? (x)
(3.15)
for any (x, ?) ? ? О RN . One can prove that ?0 is continuous, convex, satisfies properties
(3.13) and (3.14), and that ?00 , the dual function of ?0 , coincides with the convex envelope of
? with respect to ?.
We say that ? is a (strictly convex smooth) Finsler metric, and we shall write ? ? M(?)
if, in addition to properties (3.13) and (3.14), ? and ?0 are strictly convex and of class C 2 on
? О (RN \ {0}). In particular ?00 = ?.
We denote by ?? the integrated distance associated to ? ? M(?), that is, for any (x, y) ? ?,
we set
o
nZ 1
А
б
?? (x, y) = inf
(3.16)
?(?(t), ??(t))dt , ? ? W 1,1 [0, 1]; ? , ?(0) = x , ?(1) = y .
0
In the special case of the Euclidian metric, the function ? is given by ?(x, p) = ?(p) =
(p1 2 + и и и + pN 2 )1/2 , so that ?? reduces to the usual distance.
Given ? ? M(?) and x ? ?, let T 0 (x, и) : RN ? RN be the map defined by
(
?0 (x, ? ? )?0p (x, ? ? ) if ? ? ? RN \ {0}
0
?
(3.17)
T (x, ? ) =
0
if ? ? = 0.
Here, ?0p denotes the gradient with respect to p whenever we regard ?0 (x, p) as a function
of two variables x and p. In the following, for better readability, some x and t dependencies
3.2 Finsler metrics and the anisotropic context
107
are omitted. If u : ? ? R is a smooth function with non vanishing gradient, we define the
anisotropic gradient by
?? u = T 0 (x, ?u) = ?0 (x, ?u)?0p (x, ?u).
(3.18)
In a similar way as in the isotropic case, if ?t is a smooth hypersurface of ? at time t, and
n the outer normal vector to ?t (in the Euclidian sense), we define n? the ?-normal vector to
?t and ?? , an analogue of the ?-mean curvature of ?t ? which differs from ?? defined in [9],
[10] ? by
1
n? = ?0p (x, n), ?? =
div n? .
(3.19)
N ?1
Furthermore, we have the following formulas: if ? is a smooth function with non vanishing
gradient such that ?t = {x ? ?, ?(x, t) = 0}, and ? is positive outside ?t and negative inside,
then
??
,
|??|
??
1
div
,
?=
N ?1
|??|
n? = ?0p (x, ??),
n=
?? =
(3.20)
1
div ?0p (x, ??),
N ?1
(3.21)
on ?t . We also define the normal velocity of ?t and the ?-normal velocity of ?t by
Vn = ?
?t
,
|??|
Vn,? = ?
?t
.
0
? (x, ??)
(3.22)
To conclude these preliminaries, we quote a theorem proved in [10].
Theorem 3.2.1. Let ? be connected, and let ? ? M(?). Let ?? be the integrated distance
associated to ?. Let C ? ? be a closed set, and let dist? (x, C) be the ?? distance to the set C
defined by
Е
ф
dist? (x, C) = inf ?? (x, y) , y ? C .
(3.23)
Then
?0 (x, ?dist? (x, C)) = 1,
(3.24)
at each point x ? ? \ C where dist? (и, C) is differentiable.
In the special case of the Euclidian metric, (3.24) reduces to the property that |?d| = 1.
3.2.2
Application to the anisotropic Allen-Cahn equation
We set, for all (x, p) ? ? О RN ,
?0 (x, p) =
p
2a(x, p).
(3.25)
First, since a(x, и) is 2 homogeneous, ?0 satisfies assumptions (3.13) and (3.14) with the constants
?0 = [2
min
x??,|p|=1
a(x, p)]1/2 > 0
and
?0 = [2
max
a(x, p)]1/2 > 0,
(3.26)
x??,|p|=1
also using that a is continuous and strictly positive on the compact set ? О S N ?1 , with S N ?1
the unit sphere of RN . By the hypotheses on a(x, p), we see that ?0 is strictly convex and of
108 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
class C 2 on ? О (RN \ {0}); moreover, by Remark 3.1.1, ?0 is continuous on ? О RN . It follows
that ? is a Finsler metric and the theory of the above subsection applies. We have
(
ap (x, p) if p ? RN \ {0}
0
T (x, p) =
(3.27)
0
if p = 0.
S
Let ? = 0?t?T (?t О {t}) be the unique solution of the limit geometric motion Problem (P 0 )
and let de be the signed distance function to ? defined by
(
dist(x, ?t ) for x ? ?+
t ,
e t) =
d(x,
(3.28)
? dist(x, ?t ) for x ? ??
t ,
where dist(x, ?t ) is the distance from x to the hypersurface ?t in ?. Let de? be the anisotropic
signed distance function to ? defined by
(
dist? (x, ?t ) for x ? ?+
t ,
(3.29)
de? (x, t) =
? dist? (x, ?t ) for x ? ??
t ,
where dist? (x, ?t ) denotes the ?? distance to the set ?t defined in (3.23). By Theorem 3.2.1,
the following equality holds
2a(x, ?de? (x, t)) = 1
in a neighborhood of ?t .
(3.30)
We then write the second equalities in (3.20), (3.21), (3.22), once with ? = de and once with
? = de? to obtain two formulas to express the ?-normal vector n? , the analogue of the ?-mean
curvature ?? and the ?-normal velocity Vn,? :
1
e = ap (x, ?de? ),
n? = q
ap (x, ?d)
e
2a(x, ?d)
"
#
h
i
1
1
e = 1 div ap (x, ?de? ) ,
div q
?? =
ap (x, ?d)
N ?1
N ?1
e
2a(x, ?d)
1
Vn,? = ? q
det = ?(de? )t .
e
2a(x, ?d)
(3.31)
(3.32)
(3.33)
The end of this section is devoted to the operator
div ?? u = div T 0 (x, ?u) = ? и ap (x, ?u),
(3.34)
which differs from the anisotropic Laplacian ?? u defined in [9], [10]. In the case of the Finsler
metric, it turns out that the term div ?? u may be less regular than ?u. Nevertheless, we
show below a boundedness property.
Lemma 3.2.2. There exists a positive constant CL such that, for all u ? C 2,1 (? О [0, T ]), the
following inequality holds.
»
»
»? и ap (x, ?u(x, t))» ? CL (|?u(x, t)| + |D2 u(x, t)|)
for all (x, t) ? QT .
(3.35)
3.2 Finsler metrics and the anisotropic context
109
Proof. We can, with no loss of generality, ignore the dependence in time. First, assume that x
is such that ?u(x) 6= 0. Regarding a(x, p) as a function of two variables x and p = (p1 , и и и , pn ),
we obtain, by a straightforward calculation, that
X ?2a
X ?2a
?2u
? и ap (x, ?u(x)) =
(x, ?u(x)) +
(x, ?u(x))
(x).
(3.36)
?xj ?pj
?pi ?pj
?xi ?xj
j
i,j
2
a
(x, и) is 1 homogeneous as well,
We recall that ap (x, и) is 1 homogeneous, and therefore ?x?j ?p
j
and that app (x, и) is 0 homogeneous. It follows that
X »» ? 2 a
?u(x) »»
|? и ap (x, ?u(x))| ? |?u(x)|
)»
(x,
»
?xj ?pj
|?u(x)|
j
X »» ? 2 a
?u(x) »»
)»
(x,
+ |D2 u(x)|
»
?pi ?pj
|?u(x)|
i,j
» ?2a
»
X
»
»
? |?u(x)|
max »
(y, p)»
?x
?p
y??,|p|=1
j
j
j
» ?2a
»
X
»
»
max »
(y, p)»,
+ |D2 u(x)|
?p
?p
y??,|p|=1
i
j
i,j
б
А
А
б
where we have used that a ? C 3+? ? О RN \ {0} ? C 2 ? О S N ?1 . This proves (3.35) under
the assumption ?u(x) 6= 0.
Now assume that x is such that ?u(x) = 0. We have to proceed in a slightly different way
since app (x, 0) does not make sense. The operator ap (x, и) is 1 homogeneous so that, for any
direction ?,
t?1 (ap (x, t?) ? ap (x, 0)) = ap (x, ?).
We denote by (e1 , и и и , eN ) the Euclidian basis of RN . It follows from the above equality that
ap (x, и) admits at the point 0 partial derivatives in any direction ei and
which, in turn, implies that
?ap (x, и)
(0) = ap (x, ei ),
?pi
(3.37)
?a
? ?a
(x, 0) =
(x, ei ).
?pi ?pj
?pj
(3.38)
2
a
(x, и) is 1 homogeneous as well,
Note that, since ap (x, и) is 1 homogeneous, and therefore ?x?j ?p
j
the first term in (3.36) vanishes at point (x, 0). It follows, from (3.36) and (3.38) that, in the
case where ?u(x) = 0,
» X ?a
»
?2u
»
»
|? и ap (x, ?u(x))| = »
(x, ei )
(x)»
?pj
?xi ?xj
i,j
» ?a
»
X
»
»
? |D2 u(x)|
max »
(y, p)»,
y??,|p|=1 ?pj
i,j
which gives (3.35) in this case as well.
Remark 3.2.3. By similar arguments, one can obtain a positive constant CT such that, for
all u ? C 2,1 (? О [0, T ]),
»?Б
»
»
ц»»
»
ap (x, ?u(x, t)) » ? CT »Dx1 Dt1 u(x, t)»
for all (x, t) ? QT .
(3.39)
»
?t
110 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
3.3
Formal derivation of the interface motion equation
In this section we derive the equation of interface motion corresponding to Problem (P ? ) by
using a formal asymptotic expansion. The resulting interface equation can be regarded as the
singular limit of (P ? ) as ? ? 0. Our argument goes basically along the same lines with the
formal derivation given by Nakamura, Matano, Hilhorst and Scha?tzle [63]: the first two terms
of the asymptotic expansion determine the interface equation. Though our analysis in this
section is for the most part formal, the results we obtain will help the rigorous analysis in later
sections.
S
Let u? be the solution of (P ? ). Let ? = 0?t?T (?t О {t}) be the solution of the limit
geometric motion problem and let de? be the anisotropic signed distance function to ? defined
in (3.29). We then define
[
[
+
?
=
(?
О
{t}),
Q
=
(??
Q+
t
t О {t}).
T
T
0<t?T
0<t?T
We also assume that the solution u? has the expansions
u? (x, t) = 0 or 1 + ?u1 (x, t) + и и и ,
(3.40)
away from the interface ? (the outer expansion) and
u? (x, t) = U0 (x, t, ?) + ?U1 (x, t, ?) + ?2 U2 (x, t, ?) + и и и ,
(3.41)
near ? (the inner expansion), where Uj (x, t, z), j = 0, 1, 2, и и и , are defined for x ? ?, t ? 0,
z ? R and ? := de? (x, t)/?. The stretched space variable ? gives exactly the right spatial scaling
to describe the sharp transition between the regions {u? ? 0} and {u? ? 1}. We normalize Uk
in such a way that
U0 (x, t, 0) = a, Uk (x, t, 0) = 0,
for all k ? 1 (normalization conditions). To make the inner and outer expansions consistent,
we require that
Uk (x, t, +?) = 0,
U0 (x, t, +?) = 1,
(3.42)
Uk (x, t, ??) = 0,
U0 (x, t, ??) = 0,
for all k ? 1 (matching conditions).
In what follows we will substitute the inner expansion (3.41) into the parabolic equation of
Problem (P ? ) and collect the ??2 and ??1 terms. To that purpose, note that if V = V (x, t, z)
and v(x, t) = V (x, t, ?) are real valued functions then we have ?v = 1? Vz ?de? + ?x V and
vt = 1? (de? )t Vz + Vt ; if v and V are vector valued functions we obtain ? и v = 1? ?de? и Vz + ?x и V .
A straightforward computation yields
1
u?t = (de? )t U0z + U0t + (de? )t U1z + ?U1t + и и и
?
1
?u? = U0z ?de? + ?x U0 + U1z ?de? + ??x U1 + и и и
?
1
?
ap (x, ?u ) = ap (x, U0z ?de? + ??x U0 + ?U1z ?de? + ?2 ?x U1 + и и и )
?
1
= ap (x, U0z ?de? ) + app (x, U0z ?de? )(?x U0 + U1z ?de? ) + и и и
?
1
= U0z ap (x, ?de? ) + app (x, ?de? )(?x U0 + U1z ?de? ) + и и и ,
?
3.3 Formal derivation of the interface motion equation
111
where we have used the various homogeneity properties of a and its derivatives. It follows that
1
? и ap (x, ?u? ) = ?de? и ?z (ap (x, ?u? )) + ?x и (ap (x, ?u? ))
?
i
1 e h1
= ?d? и U0zz ap (x, ?de? ) + app (x, ?de? )(?x U0z + U1zz ?de? )
?
?
ц
1Б
+ ?x U0z и ap (x, ?de? ) + U0z ? и ap (x, ?de? ) + и и и
?
1h
1
= 2 U0zz 2a(x, ?de? ) + 2a(x, ?de? )U1zz + 2?x U0z и ap (x, ?de? )
?
? i
+ U0z ? и ap (x, ?de? ) + и и и ,
where we have used Remark 3.1.1 and where the functions Ui (i = 0, 1), as well as their
de? (x, t)
). Hence, in view of (3.30), we obtain, in a
derivatives, are taken at the point (x, t,
?
neighborhood of ?t ,
? и ap (x, ?u? ) =
i
1
1h
e
e
U
+
+
2?
U
и
a
(x,
?
d
)
+
U
?
и
a
(x,
?
d
)
+ иии
U
0zz
1zz
x 0z
p
0z
p
?
?
?2
?
We also use the expansion
f (u? ) = f (U0 ) + ?U1 f ? (U0 ) + и и и .
Next, we substitute the expressions above in the partial differential equation in Problem (P ? ).
Collecting the ??2 terms yields
(3.43)
U0zz + f (U0 ) = 0.
In view of the normalization and matching conditions, we can now assert that U0 (x, t, z) =
U0 (z), where U0 is the unique solution of the one-dimensional stationary problem
(
U0 ?? + f (U0 ) = 0,
U0 (??) = 0,
U0 (0) = a,
U0 (+?) = 1.
(3.44)
This solution represents the first approximation of the profile of a transition layer around
the interface observed in the stretched coordinates. We recall standard estimates on U0 , see
Chapter 1 for more details.
Lemma 3.3.1. There exist positive constants C and ? such that the following estimates hold.
0 < 1 ? U0 (z) ? Ce??|z|
0 < U0 (z)
? Ce??|z|
for z ? 0,
for z ? 0.
In addition to this U0 ? > 0 and, for all j = 1, 2,
|Dj U0 (z)| ? Ce??|z|
for z ? R.
Since U0 depends only on the variable z, we have ?x U0 ? = 0. Then, by collecting the ??1
terms, we obtain
U1zz + f ? (U0 )U1 = (de? )t U0 ? ? ? и ap (x, ?de? )U0 ? ,
(3.45)
112 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
which can be seen as a linearized problem for (3.43). The solvability condition for the above
equation plays the key role in deriving the equation of interface motion. By Lemma 1.2.2 in
Chapter 1, which is a variant of the Fredholm alternative, it is given by
Z h
i
(de? )t (x, t)U0 ? (z) ? ? и ap (x, ?de? (x, t))U0 ? (z) U0 ? (z)dz = 0,
R
for all (x, t) ? QT . Since
R
R (U0
? 2
)
> 0, it follows that
(de? )t = ? и ap (x, ?de? ).
(3.46)
In virtue of the expressions of ?? and Vn,? in (3.32) and (3.33), the above equation, written in
relative geometry, reads as
(3.47)
Vn,? = ?(N ? 1)?? on ?t ,
that is the interface motion equation (P 0 ). Using again the formulas (3.32) and (3.33), one
can come back to the Euclidian geometry and obtain the equivalent interface motion equation
h
i
1
1
p
on ?t .
(3.48)
Vn = ?? и p
ap (x, n)
2a(x, n)
2a(x, n)
Summarizing, under the assumption that the solution u? of Problem (P ? ) satisfies
(
1
in Q+
?
T
as ? ? 0, almost everywhere,
u ?
0
in Q?
T
+
we have formally proved that the boundary ?t between ??
t and ?t moves according to the
law (3.47) or (3.48).
Remark 3.3.2. To conclude this section, note that combining (3.46) with (3.45) yields U1 = 0.
In fact, the second term of the asymptotic expansion vanishes because the two stable zeros
Rof1 the nonlinearity f have ?balanced? stability, or more precisely because of the assumption
0 f (u)du = 0. If we perturb the non linearity by order ?, say f (u) ?? f (u) ? ?g(u), the
equation of the free boundary problem contains an additional term and U1 no longer vanishes.
3.4
A comparison principle
In this section, we prove a comparison principle for Problem (P ? ). To begin with, we define
a notion of sub- and super-solution for Problem (P ? ). To that purpose, we suppose that
u0 ? H 1 (?) ? L? (?).
2
1
?
Definition 3.4.1. A function u+
? ? L (0, T ; H (?)) ? L (QT ) is a weak super-solution for
Problem (P ? ), if
2
? (u+
? )t ? L (QT ),
+
?
2
? ?? u +
? (x, t) = ap (x, ?u? (x, t)) ? L (0, T ; L (?)),
? u+
? (x, 0) ? u0 (x) for almost all x ? ?,
? u? satisfies the integral inequality
Z tZ h
i
1
+
+
)
?
+
a
(x,
?u
)
и
??
?
f
(u
)?
? 0,
(u+
p
?
? t
?
?2
0
?
for all nonnegative function ? ? L2 (0, T ; H 1 (?)) ? L? (QT ) and for all t ? [0, T ].
(3.49)
3.4 A comparison principle
113
We define a weak sub-solution u?
? in a similar way, by changing ? in (3.49) by ?, and
with the condition u?
(x,
0)
?
u
(x),
for almost all x ? ?.
0
?
The following remark will reveal efficient when constructing our smooth sub- and supersolutions in later sections.
Remark 3.4.2. Note that, by Lemma 3.2.2, if u ? C 2,1 (QT ), then the function
L0 u := ut ? ? и ap (x, ?u) ?
1
f (u),
?2
is well-defined in QT . Also, using Lemma 3.2.2, we deduce that ? и ap (x, ?u) ? L? (QT ).
2,1 (Q ) satisfies L u+ ? 0 almost everywhere,
Integrating by parts, we deduce that if u+
0 ?
T
? ?C
)и?
=
0
on
??О(0, T ), and u+
the anisotropic Neumann boundary condition ap (x, ?u+
?
? (x, 0) ?
+
?
u0 (x) for almost all x ? ?, then u? is a super-solution for Problem (P ); an analogous remark
2,1 (Q ).
stands for a sub-solution u?
T
? ?C
We prove below an inequality which expresses the strong monotonicity of the function
= ap (x, p).
T 0 (x, p)
Lemma 3.4.3. There exists a constant ? > 0 such that, for all x ? ?, for all p1 , p2 ? RN ,
(ap (x, p2 ) ? ap (x, p1 )) и (p2 ? p1 ) ? ?|p2 ? p1 |2 .
(3.50)
Proof. First we consider the case that sp1 + (1 ? s)p2 6= 0 for all s ? [0, 1]. Then, the function
s 7? a(x, sp1 + (1 ? s)p2 ) is of class C 2 on [0, 1] and there exist s1 , s2 such that
1
a(x, p2 ) ? a(x, p1 ) = ap (x, p1 ) и (p2 ? p1 ) + (p2 ? p1 ) и app (x, s1 p1 + (1 ? s1 )p2 )(p2 ? p1 ),
2
and
1
a(x, p1 ) ? a(x, p2 ) = ap (x, p2 ) и (p1 ? p2 ) + (p1 ? p2 ) и app (x, s2 p1 + (1 ? s2 )p2 )(p1 ? p2 ).
2
We claim that there exist 0 < ?2 ? ?2 such that, for all x ? ?, all p ? RN \ {0}, all p? ? RN ,
?2 |p?|2 ? app (x, p)p? и p? ? ?2 |p?|2 .
(3.51)
Indeed, it follows from the strict convexity of a(x, и) that app (x, p) is a positively definite
symmetric matrix. Hence the function (x, p, p?) 7? app (x, p)p? и p? is strictly positive and continuous on the compact set ? О S N ?1 О S N ?1 which, combined with the fact that app (x, и) is 0
homogeneous, proves (3.51). It then follows that
?2
|p2 ? p1 |2 ,
2
?2
a(x, p1 ) ? a(x, p2 ) ? ap (x, p2 ) и (p1 ? p2 ) + |p2 ? p1 |2 .
2
a(x, p2 ) ? a(x, p1 ) ? ap (x, p1 ) и (p2 ? p1 ) +
(3.52)
(3.53)
Adding up inequalities (3.52) and (3.53) yields the desired inequality, with the constant ? = ?2 .
114 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
In the case that sp1 + (1 ? s)p2 = 0 for some s ? [0, 1], p1 and p2 are colinear and we may
suppose that there exists l ? R such that p2 = lp1 . We can assume l 6= 0, l 6= 1 and p1 6= 0.
By using the different homogeneity properties in Remark 3.1.1, we obtain that
(ap (x, p2 ) ? ap (x, p1 )) и (p2 ? p1 ) = (l ? 1)2 ap (x, p1 ) и p1
= 2(l ? 1)2 a(x, p1 )
= 2a(x, (l ? 1)p1 )
? ?0 2 |(l ? 1)p1 |2 = ?0 2 |p2 ? p1 |2 ,
where ?0 has been defined in (3.26). The proof is now completed.
We are now ready to prove the following comparison principle.
?
Lemma 3.4.4. Suppose that u+
? , respectively u? , is a super-solution, respectively a sub?
solution, for Problem (P ); we have that
?
+
u?
? ? u ? u?
almost everywhere in QT .
Proof. By subtracting equality (3.9) for the solution u? and inequality (3.49) for the super2
1
?
solution u+
? , we obtain that, for all ? ? L (0, T ; H (?)) ? L (QT ) such that ? ? 0, and for
all t ? [0, T ],
Z tZ h
i
?
+
(u? ? u+
? )t ? + (ap (x, ?u ) ? ap (x, ?u? )) и ??
0
?
Z tZ
┤
1│
?
+
)
?
f
(u
)
?
f
(u
?
?
2
0
? ?
Z tZ
|u? ? u+
(3.54)
? C1
? |?,
0
?
where C1 is the positive constant defined by
C1 = ??2 kf ? k
А
б.
+
?
L? ?max(ku? k? ,ku+
? k? ),max(ku k? ,ku? k? )
+
2
1
?
Next we set ? = (u? ? u+
? ) , which belongs to L (0, T ; H (?)) ? L (QT ); it follows from
(3.50) that
Z tZ
(ap (x, ?u? ) ? ap (x, ?u+
? )) и ??
0
?
Z tZ
?
+
(ap (x, ?u? ) ? ap (x, ?u+
=
? )) и (?u ? ?u? )
{u? ?u+
? ?0}
0
??
Z tZ
0
{u? ?u+
? ?0}
2
|?u? ? ?u+
?| .
Then, substituting this inequality into (3.54) yields
Z
Z │
Z tZ
Z tZ
┤2
1 t d
+
?
+ 2
2
)
+
?
|?u
?
?u
|
?
C
(u? ? u+
(u? ? u+
1
?
?
?) ,
+
+
2 0 dt ?
?
?
{u ?u? ?0}
{u ?u? ?0}
0
0
and therefore
Z │
?
?
(u ?
+
u+
?)
┤2
Z tZ │
┤2 Z │
┤2
?
+ +
+
(u ? u? )
(u? ? u+
(t) ? 2C1
+
)
(0).
?
0
?
?
3.5 Generation of interface
115
Using Gronwall?s Lemma, we find that
Z │
Z │
┤2
┤2
?
+ +
2C1 t
+
(t) ? e
)
(0),
(u ? u? )
(u? ? u+
?
?
?
Since u? (x, 0) ? u+
? (x, 0) for almost all x ? ?, it follows that
u? ? u+
?
a.e. in QT .
?
Similarly one can show that u?
? ? u a.e. in QT .
Lemma 3.4.5. Let u? be a solution of Problem (P ? ). Then
?ku0 kL? (?) ? u? ? max(1, ku0 kL? (?) )
a.e. in QT .
Proof. We remark that ?ku0 kL? (?) and that max(1, ku0 kL? (?) ) are sub- and super-solutions
for Problem (P ? ).
3.5
Generation of interface
This section deals with the generation of interface, namely the rapid formation of internal
layers that takes place in a neighborhood of ?0 = {x ? ?, u0 (x) = a} within the time span of
order ?2 | ln ?|. In the sequel, ?0 will stand for the quantity
?0 :=
1
min(a, 1 ? a).
2
Our main result in this section is the following.
Theorem 3.5.1. Let ? ? (0, ?0 ) be arbitrary and define х as the derivative of f (u) at the
unstable equilibrium u = a, that is
х = f ? (a).
(3.55)
Then there exist positive constants ?0 and M0 such that, for all ? ? (0, ?0 ),
? for almost all x ? ?,
? ? ? u? (x, х?1 ?2 | ln ?|) ? 1 + ?,
(3.56)
? for almost all x ? ? such that |u0 (x) ? a| ? M0 ?, we have that
if u0 (x) ? a + M0 ? then u? (x, х?1 ?2 | ln ?|) ? 1 ? ?,
?
?1 2
if u0 (x) ? a ? M0 ? then u (x, х
? | ln ?|) ? ?.
(3.57)
(3.58)
We will prove this result by constructing a suitable pair of sub and super-solutions.
3.5.1
The bistable ordinary differential equation
The sub- and super-solutions mentioned above will be constructed by modifying the solution
of the problem without diffusion:
u?t =
1
f (u?),
?2
u?(x, 0) = u0 (x).
116 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
This solution is written in the form
┤
│t
,
u
(x)
,
0
?2
u?(x, t) = Y
where Y (?, ?) denotes the solution of the ordinary differential equation
(
Y? (?, ?) = f (Y (?, ?))
Y (0, ?)
for ? > 0,
= ?.
(3.59)
Here ? ranges over the interval (?2C0 , 2C0 ), with C0 being the constant defined in (3.6). We
first collect basic properties of Y .
Lemma 3.5.2. We have Y? > 0, for all ? ? (?2C0 , 2C0 )\{0, a, 1} and all ? > 0. Furthermore,
Y? (?, ?) =
f (Y (?, ?))
.
f (?)
Proof. First, differentiating equation (3.59) with respect to ?, we obtain
(
Y?? = Y? f ? (Y ),
Y? (0, ?) = 1,
(3.60)
which can be integrated as follows:
Y? (?, ?) = exp
hZ
0
?
i
f ? (Y (s, ?))ds > 0.
(3.61)
We then differentiate equation (3.59) with respect to ? and obtain
(
Y? ? = Y? f ? (Y ),
Y? (0, ?) = f (?),
which in turn implies
Y? (?, ?) = f (?) exp
hZ
?
0
= f (?)Y? (?, ?).
i
f ? (Y (s, ?))ds
(3.62)
(3.63)
This last equality, in view of (3.59), completes the proof of Lemma 3.5.2.
We define a function A(?, ?) by
A(?, ?) =
f ? (Y (?, ?)) ? f ? (?)
.
f (?)
Lemma 3.5.3. We have, for all ? ? (?2C0 , 2C0 ) \ {0, a, 1} and all ? > 0,
A(?, ?) =
Z
0
?
f ?? (Y (s, ?))Y? (s, ?)ds.
(3.64)
3.5 Generation of interface
117
Proof. Differentiating the equality of Lemma 3.5.2 with respect to ? leads to
Y?? = A(?, ?)Y? ,
(3.65)
whereas differentiating (3.61) with respect to ? yields
Z ?
Y?? = Y?
f ?? (Y (s, ?))Y? (s, ?)ds.
0
These two last results complete the proof of Lemma 3.5.3.
Next we need some estimates on Y and its derivatives. First, we perform some estimates
when the initial value ? lies between ? and 1 ? ?.
Lemma 3.5.4. Let ? ? (0, ?0 ) be arbitrary. Then there exist positive constants C?1 = C?1 (?),
C?2 = C?2 (?) and C3 = C3 (?) such that, for all ? > 0,
? if ? ? (a, 1 ? ?) then, for every ? > 0 such that Y (?, ?) remains in the interval (a, 1 ? ?),
we have
C?1 eх? ? Y? (?, ?) ? C?2 eх? ,
(3.66)
and
|A(?, ?)| ? C3 (eх? ? 1);
(3.67)
? if ? ? (?, a) then, for every ? > 0 such that Y (?, ?) remains in the interval (?, a), (3.66)
and (3.67) hold as well,
where х is the constant defined in (3.55).
Proof. We take ? ? (a, 1 ? ?) and suppose that for s ? (0, ? ), Y (s, ?) remains in the interval
(a, 1 ? ?). Integrating the equality
Y? (s, ?)
=1
f (Y (s, ?))
from 0 to ? yields
Z
0
?
Y? (s, ?)
ds
f (Y (s, ?))
= ?.
Hence by the change of variable q = Y (s, ?) we get
Z Y (?,?)
dq
= ?.
f (q)
?
Moreover, the equality of Lemma 3.5.2 leads to
Z Y (?,?) ?
f (q)
dq
ln Y? (?, ?) =
f (q)
?
Z Y (?,?) ?
Б f (a) f ? (q) ? f ? (a) ц
=
dq
+
f (q)
f (q)
?
Z Y (?,?)
h(q)dq,
= х? +
?
where
h(q) = (f ? (q) ? х)/f (q).
(3.68)
(3.69)
(3.70)
118 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
Since
f ?? (a)
as q ? a,
f ? (a)
the function h is continuous on [a, 1 ? ?]. Hence we can define
h(q) ?
H = H(?) := khkL? (a,1??) .
Since |Y (?, ?) ? ?| takes its values in the interval [0, 1 ? a ? ?] ? [0, 1 ? a], it follows from (3.70)
that
х? ? H(1 ? a) ? ln Y? (?, ?) ? х? + H(1 ? a),
which, in turn, proves (3.66). Lemma 3.5.3 and (3.66) yield
Z ?
??
|A(?, ?)| ? supz?[0,1] |f (z)| C?2 eхs ds
0
? C3
(eх?
? 1),
which completes the proof of (3.67). The case where ? and Y (?, ?) are in (?, a) is similar and
omitted.
Corollary 3.5.5. Let ? ? (0, ?0 ) be arbitrary. Then there exist positive constants C1 = C1 (?)
and C2 = C2 (?) such that, for all ? > 0,
? if ? ? (a, 1 ? ?) then, for every ? > 0 such that Y (?, ?) remains in the interval (a, 1 ? ?),
we have
C1 eх? (? ? a) ? Y (?, ?) ? a ? C2 eх? (? ? a);
(3.71)
? if ? ? (?, a) then, for every ? > 0 such that Y (?, ?) remains in the interval (?, a), we
have
(3.72)
C2 eх? (? ? a) ? Y (?, ?) ? a ? C1 eх? (? ? a).
Proof. Since
f (q)/(q ? a) ? f ? (a) = х
as q ? a,
it is possible to find B1 = B1 (?) > 0 and B2 = B2 (?) > 0 such that, for all q ? (a, 1 ? ?),
B1 (q ? a) ? f (q) ? B2 (q ? a).
(3.73)
We write this inequality for a < Y (?, ?) < 1 ? ? to obtain
B1 (Y (?, ?) ? a) ? f (Y (?, ?)) ? B2 (Y (?, ?) ? a).
We also write this inequality for a < ? < 1 ? ? to obtain
B1 (? ? a) ? f (?) ? B2 (? ? a).
Next we use the equality Y? = f (Y )/f (?) of Lemma 3.5.2 to deduce that
B1
B2
(Y (?, ?) ? a) ? (? ? a)Y? (?, ?) ?
(Y (?, ?) ? a),
B2
B1
which, in view of (3.66), implies that
B2
B1
C?1 eх? (? ? a) ? Y (?, ?) ? a ?
C?2 eх? (? ? a).
B2
B1
This proves (3.71). The proof of (3.72) is similar and omitted.
Next we present estimates in the case where the initial value ? is smaller than ? or larger
than 1 ? ?.
3.5 Generation of interface
119
Lemma 3.5.6. Let ? ? (0, ?0 ) and M > 0 be arbitrary. Then there exists a positive constant
C4 = C4 (?, M ) such that
? if ? ? [1 ? ?, 1 + M ], then, for all ? > 0, Y (?, ?) remains in the interval [1 ? ?, 1 + M ]
and
|A(?, ?)| ? C4 ? for ? > 0 ;
(3.74)
? if ? ? [?M, ?], then, for all ? > 0, Y (?, ?) remains in the interval [?M, ?] and (3.74)
holds as well.
Proof. Since the two statements can be treated in the same way, we will only prove the former.
The fact that Y (?, ?), the solution of the ordinary differential equation (3.59), remains in the
interval [1 ? ?, 1 + M ] directly follows from the bistable properties of f , or, more precisely,
from the sign conditions f (1 ? ?) > 0, f (1 + M ) < 0.
To prove (3.74), suppose first that ? ? [1, 1 + M ]. In view of (3.3), f ? is strictly negative
in an interval of the form [1, 1 + c] and f is negative in [1, ?). We denote by ?m < 0 the
maximum of f on [1 + c, 1 + M ]. Then, as long as Y (?, ?) remains in the interval [1 + c, 1 + M ],
the ordinary differential equation (3.59) implies
Y? ? ?m.
By integration, this means that, for any ? ? [1, 1 + M ], we have
M ?c
.
m
In view of this, and considering that f ? (Y ) < 0 for Y ? [1, 1 + c], we see from the expression
(3.61) that
hZ ?
i
hZ ?
i
?
f (Y (s, ?))ds exp
f ? (Y (s, ?))ds
Y? (?, ?) = exp
Y (?, ?) ? [1, 1 + c]
for ? ? ? :=
0
? exp
? exp
hZ
?
?
0
hZ
?
i
f ? (Y (s, ?))ds
sup
0 z?[?M,1+M ]
i
|f ? (z)|ds =: C?4 = C?4 (M ),
for all ? ? ? . It is clear from the same expression (3.61) that Y? ? C?4 holds also for 0 ? ? ? ? .
We can then use Lemma 3.5.3 to deduce that
Z ?
|A(?, ?)| ? C?4
|f ?? (Y (s, ?))|ds
0
┤
? C?4 supz?[?M,1+M ] |f ?? (z)| ? =: C4 ?.
│
The case ? ? [1 ? ?, 1] can be treated in the same way. This completes the proof of the
lemma.
Now we choose the constant M in the above lemma sufficiently large so that [?2C0 , 2C0 ] ?
[?M, 1 + M ], and fix M hereafter. Then C4 only depends on ?. Using the fact that ? =
O(eх? ? 1) for ? > 0, one can easily deduce from (3.67) and (3.74) the following general
estimate.
Lemma 3.5.7. Let ? ? (0, ?0 ) be arbitrary and let C0 be the constant defined in (3.6). Then
there exists a positive constant C5 = C5 (?) such that, for all ? ? (?2C0 , 2C0 ) and all ? > 0,
|A(?, ?)| ? C5 (eх? ? 1).
120 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
3.5.2
Construction of sub- and super-solutions
We are now ready to construct the sub- and super-solutions for the study of generation of
interface. To make the proof less technical, we make the additional assumption
ap (x, ?u0 (x)) и ? = 0
on ??.
(3.75)
In this case, our sub- and super-solutions are given by
│t
┤
2
w?▒ (x, t) = Y 2 , u0 (x) ▒ ?2 C6 (eхt/? ? 1) .
?
(3.76)
In the general case where (3.75) does not necessarily hold, we have to slightly modify w?▒ (x, t)
near the boundary ??. This can be done by using some cut-off initial data u▒
0 (see Chapter
1, Section 1.3).
Lemma 3.5.8. There exist positive constants ?0 and C6 such that, for all ? ? (0, ?0 ), (w?? , w?+ )
is a pair of sub- and super-solutions for Problem (P ? ), in the domain
ф
Е
(x, t) ? QT , x ? ?, 0 ? t ? х?1 ?2 | ln ?| .
│t
┤
Proof. First, we remark that w▒ (x, 0) = Y 2 , u0 (x) = u0 (x). Next we define the operator
?
L0 by
1
(3.77)
L0 u := ut ? ? и ap (x, ?u) ? 2 f (u),
?
and prove that L0 w?+ ? 0. Straightforward calculations yield
1
2
Y? + хC6 eхt/? Y? ,
2
?
?w?+ = ?u0 (x)Y? .
(w? + )t =
First, using (3.75) and the fact that ap (x, и) is 1 homogeneous, we see that w?▒ satisfy the
anisotropic Neumann boundary condition
ap (x, ?w?▒ ) и ? = 0
on ?? О (0, +?).
In view of the ordinary differential equation (3.59), we obtain
2
L0 w?+ = хC6 eхt/? Y? ? ? и ap (x, ?w?+ ).
By the estimate in Lemma 3.2.2, it follows that
2
L0 w?+ ? хC6 eхt/? Y? ? CL (|?w?+ (x, t)| + |D2 w?+ (x, t)|),
(3.78)
where we recall that
|D2 w?+ (x, t)| = max |?i ?j w?+ (x, t)|.
i,j
A straightforward calculation yields
?i ?j w?+ (x, t) = (?i ?j u0 )Y? + (?i u0 ?j u0 )Y?? .
Recalling that Y? > 0, we now combine the expression of ?w?+ , the above expression and
inequality (3.78) to obtain
2
L0 w?+ /Y? ? хC6 eхt/? ? CL C0 ? C0 ? C0 2
|Y?? |
,
Y?
(3.79)
3.5 Generation of interface
121
where C0 is the constant defined in (3.6). We note that, in the range 0 ? t ? х?1 ?2 | ln ?|, we
have
2
0 ? ?2 C6 (eхt/? ? 1) ? ?2 C6 (??1 ? 1) ? C0 ,
if ?0 is sufficiently small. Hence
2
? := u0 (x) + C6 (eхt/? ? 1) ? (?2C0 , 2C0 ),
so that, by the results of the previous subsection, Y remains in (?2C0 , 2C0 ). In view of (3.65),
Y?? /Y? is equal to A so that, combining the estimate of A in Lemma 3.5.7 and (3.79) yield
2
L0 w?+ /Y? ? (хC6 ? C0 2 C5 )eхt/? ? CL C0 ? C0 .
Now, choosing
C6 ?
б
А
2
max C0 2 C5 , C0 (CL + 1)
х
proves L0 w?+ /Y? ? 0. Since Y? > 0, it follows that L0 w?+ ? 0. Hence, by Remark 3.4.2, w?+ is
a super-solution for Problem (P ? ). Similarly w?? is a sub-solution. Lemma 3.5.8 is proved.
Consequently, by the comparison principle proved in Lemma 3.4.4,
w?? (x, t) ? u? (x, t) ? w?+ (x, t),
(3.80)
for almost all (x, t) ? QT that satisfies 0 ? t ? х?1 ?2 | ln ?|.
3.5.3
Proof of Theorem 3.5.1
In order to prove Theorem 3.5.1 we first present a key estimate on the function Y after a time
interval of order ? ? | ln ?|.
Lemma 3.5.9. Let ? ? (0, ?0 ) be arbitrary; there exist positive constants ?0 and C7 such that,
for all ? ? (0, ?0 ),
? for all ? ? (?2C0 , 2C0 ),
? ? ? Y (х?1 | ln ?|, ?) ? 1 + ?,
(3.81)
? for all ? ? (?2C0 , 2C0 ) such that |? ? a| ? C7 ?, we have that
if ? ? a + C7 ? then Y (х?1 | ln ?|, ?) ? 1 ? ?,
if ? ? a ? C7 ? then Y (х?1 | ln ?|, ?) ? ?.
(3.82)
(3.83)
Proof. We first prove (3.82). For ? ? a + C7 ?, as long as Y (?, ?) has not reached 1 ? ?, we
can use (3.71) to deduce that
Y (?, ?) ? a + C1 eх? (? ? a)
? a + C1 C7 eх? ?
? 1 ? ?,
provided that ? satisfies
? ? х?1 ln
1?a??
=: ? ? .
C1 C7 ?
122 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
Choosing
C7 =
max(a, 1 ? a) ? ?
,
C1
we see that х?1 | ln ?| ? ? ? , which completes the proof of (3.82). Using (3.72), one easily proves
(3.83).
Next we prove (3.81). First, by the bistable assumptions on f , if we leave from a ? ?
[??, 1 + ?] then Y (?, ?) will remain in [??, 1 + ?]. Now suppose that 1 + ? ? ? ? 2C0 . We
check below that Y (х?1 | ln ?|, ?) ? 1 + ?. First, in view of (3.3), we can find p > 0 such that
if
if
1 ? u ? 2C0
? 2C0 ? u ? 0
then
then
f (u) ? p(1 ? u)
f (u) ? ?pu.
(3.84)
We then use the ordinary differential equation (3.59) to obtain, as long as 1 + ? ? Y ? 2C0 ,
the inequality Y? ? p(1 ? Y ). It follows that
Y?
? ?p.
Y ?1
Integrating this inequality from 0 to ? leads to
Y (?, ?) ? 1 + (? ? 1)e?p?
? 1 + (2C0 ? 1)e?p? .
?1
Since (2C0 ? 1)e?pх | ln ?| ? 0 as ? ? 0, the above inequality proves that, for ? ? (0, ?0 ), with
?0 = ?0 (?) sufficiently small, Y (х?1 | ln ?|, ?) ? 1 + ?, which completes the proof of (3.81).
We are now ready to prove Theorem 3.5.1. By setting t = х?1 ?2 | ln ?| in (3.80), we obtain,
for almost all x ? ?,
│
┤
Y х?1 | ln ?|, u0 (x) ? (C6 ? ? C6 ?2 )
│
┤
(3.85)
? u? (x, х?1 ?2 | ln ?|) ? Y х?1 | ln ?|, u0 (x) + C6 ? ? C6 ?2 .
Furthermore, by the definition of C0 in (3.6), we have, for ?0 small enough,
?2C0 ? u0 (x) ▒ (C6 ? ? C6 ?2 ) ? 2C0 ,
for x ? ?. Thus the assertion (3.56) of Theorem 3.5.1 is a direct consequence of (3.81) and
(3.85).
Next we prove (3.57). We choose M0 large enough so that M0 ? ? C6 ? + C6 ?2 ? C7 ?. Then,
for any x ? ? such that u0 (x) ? a + M0 ?, we have
u0 (x) ? (C6 ? ? C6 ?2 ) ? a + M0 ? ? C6 ? + C6 ?2 ? a + C7 ?.
Combining this, (3.85) and (3.82), we see that
u? (x, х?1 ?2 | ln ?|) ? 1 ? ?,
for almost all x ? ? that satisfies u0 (x) ? a + M0 ?. This proves (3.57). The inequality (3.58)
can be shown the same way. This completes the proof of Theorem 3.5.1.
3.6 Motion of interface
3.6
123
Motion of interface
We have seen in Section 3.5 that, after a very short time, the solution u? develops a clear
transition layer. In the present section, we show that it persists and that its law of motion is
well approximated by the interface equation (P 0 ).
More precisely, take the first term of the formal asymptotic expansion (3.41) as a formal
expansion of the solution:
u? (x, t) ? u?? (x, t) := U0
│ de (x, t) ┤
?
.
?
(3.86)
The right-hand side of (3.86) is a function having a well-developed transition layer, and its
interface lies exactly on ?t . We show that this function is a very good approximation of the
solution; therefore the following holds:
If u? becomes rather close to u?? at some time moment, then it stays close to u?? for
the rest of time.
+
To that purpose, we will construct a pair of sub- and super-solutions u?
? and u? for Problem
?
?
by slightly modifying u? . It then follows that, if the solution u satisfies
(P ? )
?
+
u?
? (x, t0 ) ? u (x, t0 ) ? u? (x, t0 ),
for some t0 ? 0 and for almost all x ? ?, then
?
+
u?
? (x, t) ? u (x, t) ? u? (x, t),
?
for almost (x, t) ? QT that satisfies t0 ? t ? T . As a result, since both u+
? , u? stay close to
?
?
?
u? , the solution u also stays close to u? for t0 ? t ? T .
3.6.1
Construction of sub and super-solutions
To begin with we present mathematical tools which are essential for the construction of sub
and super-solutions.
A modified anisotropic signed distance function. Rather than working with the anisotropic signed distance function de? , defined in (3.29), we define a ?cut-off anisotropic signed
distance function? d? as follows. Choose d0 > 0 small enough so that de? (и, и) is smooth in the
tubular neighborhood of ?
{(x, t) ? QT , |de? (x, t)| < 3d0 },
and that
dist? (?t , ??) > 3d0
for all t ? [0, T ].
(3.87)
Next let ?(s) be a smooth increasing function on R such that
?
if |s| ? d0
?
? s
?2d0 if s ? ?2d0
?(s) =
?
?
2d0
if s ? 2d0 .
We define the cut-off anisotropic signed distance function d? by
А
б
d? (x, t) = ? de? (x, t) .
(3.88)
124 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
Note that, in view of (3.30),
2a(x, ?d? (x, t)) = 1
in a neighborhood of ?t ,
(3.89)
more precisely in the region {(x, t) ? QT , |d? (x, t)| < d0 }. Moreover, in view of (3.87), we
have
far away from ?t ,
(3.90)
2a(x, ?d? (x, t)) = 0
i.e. in the region {(x, t) ? QT , |d? (x, t)| ? 2d0 }. Furthermore, we recall, see (3.46), that an
equation for ? is given by
(3.91)
(d? )t = ? и ap (x, ?d? ) on ?t .
?
Construction. We look for a pair of sub- and super-solutions u▒
? for (P ) of the form
│ d (x, t) ▒ ?p(t) ┤
?
u▒
▒ q(t),
(3.92)
(x,
t)
=
U
0
?
?
where U0 is the solution of (3.43), and where
2
p(t) = ?e??t/? + eLt + K,
2
q(t) = ?(?e??t/? + ?2 LeLt ).
Note that q = ??2 pt . It is clear from the definition of u▒
? that
(
1 for all (x, t) ? Q+
T
(x,
t)
=
lim u▒
?
??0
0 for all (x, t) ? Q?
T.
(3.93)
(3.94)
The main result of this section is the following.
Lemma 3.6.1. There exist positive constants ?, ? with the following properties. For any
K > 1, we can find positive constants ?0 and L such that, for any ? ? (0, ?0 ), the functions u?
?
and u+
? satisfy the homogeneous anisotropic Neumann boundary condition and
+
L0 u?
? ? 0 ? L0 u?
in ? О [0, T ],
where the operator L0 has been defined in (3.77).
3.6.2
Proof of Lemma 3.6.1
+
+
We show below that L0 u+
? := (u? )t ? ? и ap (x, ?u? ) ?
L0 u?
? ? 0 following by the same arguments.
1
f (u+
? ) ? 0, the proof of inequality
?2
Computation of L0 u+
?
In the sequel, the function U0 and its derivatives are taken at the point (d? (x, t) + ?p(t))/?.
Straightforward computations yield
1
?
(u+
? )t = ( (d? )t + pt )U0 + qt ,
?
1 ?
?u+
? = U0 ?d? ,
?
1 ??
1 ?
? и ap (x, ?u+
? ) = 2 U0 ?d? и ap (x, ?d? ) + U0 ? и ap (x, ?d? )
?
?
1 ?
1 ??
= 2 U0 2a(x, ?d? ) + U0 ? и ap (x, ?d? ),
?
?
3.6 Motion of interface
125
where we have used properties from Remark 3.1.1. Note that, d? being constant in a neigh+
borhood of ??, we have that ?u+
? = 0 on ?? О (0, T ) and u? satisfies the homogeneous
anisotropic Neumann boundary condition
ap (x, ?u+
? )и? =0
on ?? О (0, T ).
Further, we use the expansion
1 2 ??
?
f (u+
? ) = f (U0 ) + qf (U0 ) + q f (?),
2
+
for some function ?(x, t) satisfying U0 < ? < u? .
Combining the above expressions with equation (3.43), we obtain
L0 u+
? = E1 + E2 + E3 ,
where:
б
1 А ?
1
q f (U0 ) + qf ?? (?) + U0 ? pt + qt ,
2
?
2
┤
?? │
U0
E2 = 2 1 ? 2a(x, ?d? ) ,
?
┤
U0 ? │
(d? )t ? ? и ap (x, ?d? ) .
E3 =
?
E1 = ?
In order to estimate the above terms, we first present some useful inequalities. As f ? (0)
and f ? (1) are strictly negative, we can find strictly positive constants b and m such that
if
U0 (z) ? [0, b] ? [1 ? b, 1]
f ? (U0 (z)) ? ?m.
then
(3.95)
On the other hand, since the region {(x, z) ? ? О R | U0 (z) ? [b, 1 ? b] } is compact and since
U0 ? > 0 on R, there exists a constant a1 > 0 such that
if
U0 (z) ? [b, 1 ? b]
then
U0 ? (z) ? a1 .
(3.96)
We define
F =
sup |f (z)| + |f ? (z)| + |f ?? (z)|,
?1?z?2
?=
(3.97)
m
,
4
(3.98)
and choose ? that satisfies
0 < ? ? min(?0 , ?1 , ?2 ),
where
(3.99)
a1
1
4?
, ?1 :=
, ?2 :=
.
4? + F
?+1
F (? + 1)
Hence, combining (3.95) and (3.96), we obtain, using that ? ? ?0 ,
?0 :=
U0 ? (z) ? ?f ? (U0 (z)) ? 4??
for z ? R.
(3.100)
L0 u+
?
? 0 provided that
Now let K > 1 be arbitrary. In what follows we will show that
the constants ?0 and L are appropriately chosen. From now on, we suppose that the following
inequality is satisfied:
?20 LeLT ? 1 .
(3.101)
Then, given any ? ? (0, ?0 ), since ? ? ?1 , we have 0 ? q(t) ? 1, hence, recalling that
0 < U0 < 1,
? 1 ? u▒
(3.102)
? (x, t) ? 2 .
126 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
We first estimate the term E1
A direct computation gives
E1 =
? ??t/?2
e
(I ? ??) + LeLt (I + ?2 ?L),
?2
where
I = U0 ? ? ?f ? (U0 ) ?
? 2 ??
2
f (?)(?e??t/? + ?2 LeLt ).
2
In virtue of (3.100) and (3.102), we obtain
I ? 4?? ?
?2
F (? + ?2 LeLT ).
2
Then, in view of (3.101), using that ? ? ?2 , we have
I ? 2??.
Consequently, we have
E1 ?
?? 2 ??t/?2
C1
2
+ 2??LeLt =: 2 e??t/? + C1 ? LeLt .
e
?2
?
As for the term E2
First, in the points where |d? | < d0 , by (3.89), we have E2 = 0. Next we consider the points
where |d? | ? d0 . We deduce from the definition of ?0 in (3.26) that
0 ? 2a(x, ?d? (x, t)) ? (?0 )2 |?d? (x, t)|2
? (?0 )2 k?d? k2? := D < ?.
Applying Lemma 3.3.1 yields
C
(1 + D)e??|d? +?p|/?
?2
C
? 2 (1 + D)e??(d0 /??|p|) .
?
|E2 | ?
By the definition of p in (3.93) we have that 0 < K ? 1 ? p ? eLT + K; we suppose from now
that the following assumption holds:
eLT + K ?
Then
d0
.
2?0
d0
d0
? |p| ?
so that, defining C ? := C(1 + D),
?
2?
C ? ??d0 /(2?)
e
?2
16C ?
? C2 :=
.
(e?d0 )2
|E2 | ?
(3.103)
3.6 Motion of interface
127
Next we consider the term E3
We set
F(x, t) = (d? )t (x, t) ? ? и ap (x, ?d? (x, t)).
We recall that d? ? C 3+?,(3+?)/2 in a neighborhood V of ?, say
V = {(x, t) ? QT , |d? (x, t)| < d0 }.
Combining the fact that
with the definition of
?0
2a(x, ?d? (x, t)) = 1
in V,
in (3.26), we see that
|?d? | ?
1
?0
in V.
(3.104)
We also recall that (x, p) 7? a(x, p) is of class C 3+? on ? О RN \ {0}. Since by (3.104) |?d? |
is bounded away from zero, it follows that x 7? ? и ap (x, ?d? (x, t)) is in C 1+? (Vt ), where
Vt := {x ? ?, (x, t) ? V}.
Moreover the function x 7? (d? )t (x, t) is in C 1+? (Vt ). Therefore the function x 7? F(x, t) is
Lipschitz continuous on Vt . By equation (3.91), we have that
F(x, t) = 0
on ?t = {x ? ?, d? (x, t) = 0},
and it follows from the mean value theorem applied separately on both sides of ?t that there
exists a constant N1 such that
|F(x, t)| ? N1 |d? (x, t)|
for all (x, t) ? V.
(3.105)
Next, using Lemma 3.2.2, we remark that F is bounded on ? О [0, T ]\V so that there exists
a constant N2 such that
sup |F(x, t)| ? N2 .
(3.106)
?О[0,T ]\V
By the inequalities (3.105) and (3.106), we deduce that
|F(x, t)| = |(d? )t (x, t) ? ? и ap (x, ?d? (x, t))| ? N0 |d? (x, t)|
with N0 := max(N1 , N2 /d0 ). Applying Lemma 3.3.1 we deduce that
|E3 | ? N0 C
|d? (x, t)| ??|d? (x,t)/?+p(t)|
e
?
? N0 C maxy?R |y|e??|y+p(t)|
А
1б
? N0 C max |p(t)|,
?
А
1б
? N0 C |p(t)| + .
?
Thus, recalling that |p(t)| ? eLt + K, we obtain
|E3 | ? C3 (eLt + K) + C3 ? ,
where C3 := N0 C and C3 ? := N0 C/?.
in QT ,
128 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
Completion of the proof
Collecting the above estimates of E1 , E2 and E3 yields
L0 u+
? ?
C1 ??t/?2
+ (LC1 ? ? C3 )eLt ? C4 ,
e
?2
where C4 := C2 + KC3 + C3 ? . Now, we set
L :=
d0
1
ln
,
T
4?0
which, for ?0 small enough, validates assumptions (3.101) and (3.103). If ?0 is chosen sufficiently small (i.e. L sufficiently large), we obtain, for all ? ? (0, ?0 ),
? (LC1 ? ? C3 )eLt ? C4
L0 u+
?
? 12 LC1 ? ? C4
? 0.
The proof of Lemma 3.6.1 is now completed.
3.7
Proof of the main results
In this section, we prove Theorem 3.1.3 and Corollary 3.1.4 by fitting the two pairs of suband super-solutions, constructed for the study of the generation and the motion of interface,
into each other.
Let ? ? (0, ?0 ) be arbitrary. Choose ? and ? that satisfy (3.98), (3.99) and
?? ?
?
.
3
(3.107)
By the generation of interface Theorem 3.5.1, there exist positive constants ?0 and M0 such
that (3.56), (3.57) and (3.58) hold with the constant ? replaced by ??/2. Since, by the
hypothesis (3.7) and the equality (3.31), ?u0 (x) и n? (x) 6= 0 everywhere on the initial interface
?0 = {x ? ?, u0 (x) = a} and since ?0 is a compact hypersurface, we can find a positive
constant M1 such that
if
if
d? (x, 0) ?
M1 ?
then
d? (x, 0) ? ?M1 ?
then
Now we define functions H + (x), H ? (x) by
й
1 + ??/2
H + (x) =
??/2
й
1 ? ??/2
H ? (x) =
???/2
u0 (x) ? a + M0 ?,
u0 (x) ? a ? M0 ?.
(3.108)
if d? (x, 0) > ?M1 ?
if d? (x, 0) ? ?M1 ?,
if d? (x, 0) ? M1 ?
if d? (x, 0) < M1 ?.
Then from the above observation we see that
H ? (x) ? u? (x, х?1 ?2 | ln ?|) ? H + (x),
for almost all x ? ?.
(3.109)
3.7 Proof of the main results
129
Next we fix a sufficiently large constant K > 1 such that
U0 (?M1 + K) ? 1 ?
??
3
and
U0 (M1 ? K) ?
??
.
3
(3.110)
For this K, we choose ?0 and L as in Lemma 3.6.1. We claim that
?
u?
? (x, 0) ? H (x),
H + (x) ? u+
? (x, 0),
(3.111)
for all x ? ?. We only prove the former inequality, as the proof of the latter is virtually the
same. Then it amounts to showing that
u?
? (x, 0) = U0
б
А d? (x, 0)
? K ? ?(? + ?2 L) ? H ? (x).
?
(3.112)
In the range where d? (x, 0) < M1 ?, the second inequality in (3.110) and the fact that U0 is an
increasing function imply
U0
б
А d? (x, 0)
? K ? ?(? + ?2 L) ? U0 (M1 ? K) ? ?? ? ??2 L
?
??
? ??
?
3
? H ? (x).
On the other hand, in the range where d? (x, 0) ? M1 ?, we have
U0
А d? (x, 0)
б
? K ? ?(? + ?2 L) ? 1 ? ??
?
? H ? (x).
This proves (3.112), so that (3.111) is established.
Combining (3.109) and (3.111), we obtain
?
?1 2
+
u?
? (x, 0) ? u (x, х ? | ln ?|) ? u? (x, 0),
+
for almost all x ? ?. Since, by Lemma 3.6.1, u?
? and u? are sub- and super-solutions for
Problem (P ? ), the comparison principle yields
?
?
+
u?
? (x, t) ? u (x, t + t ) ? u? (x, t),
(3.113)
for almost all (x, t) ? QT that satisfies 0 ? t ? T ? t? , where we recall that t? = х?1 ?2 | ln ?|.
Note that, in view of (3.94), this is sufficient to prove Corollary 3.1.4. Now let C be a positive
constant such that
?
?
U0 (C ? eLT ? K) ? 1 ?
and U0 (?C + eLT + K) ? .
(3.114)
2
2
One then easily checks, using (3.113), (3.92) and (3.107), that, for ?0 small enough, for almost
all (x, t) ? QT with 0 ? t ? T ? t? , we have
if
if
d? (x, t) ?
C?
then
d? (x, t) ? ?C?
then
u? (x, t + t? ) ? 1 ? ?
u? (x, t + t? ) ? ?,
and
u? (x, t + t? ) ? [??, 1 + ?],
which completes the proof of Theorem 3.1.3.
(3.115)
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Syste?mes de convection-re?action-diffusion et dynamique d?interface
Cette the?se porte sur la limite singulie?re d?e?quations et de syste?mes d?e?quations paraboliques non-line?aires de type bistable, avec des conditions initiales ge?ne?rales. Nous prouvons des
proprie?te?s de ge?ne?ration d?interface et analysons le de?placement d?interface. Nous obtenons
une estimation nouvelle et optimale de l?e?paisseur et de la localisation de la zone de transition,
ame?liorant ainsi des re?sultats connus pour diffe?rents proble?mes mode?les.
Au Chapitre 1, nous conside?rons d?abord une e?quation d?Allen-Cahn. Le de?placement de
l?interface limite est induit par sa courbure moyenne et par un terme de pression. Nous e?tendons
ensuite nos re?sultats a? une classe assez large de syste?mes de re?action-diffusion. Pour cela, nous
conside?rons la premie?re e?quation du syste?me comme une perturbation de l?e?quation d?AllenCahn, e?tudions la de?pendance du de?placement de l?interface vis-a?-vis de diffe?rents parame?tres,
et prouvons de fines estimations a priori. Le Chapitre 2 est consacre? a? l?e?tude d?un syste?me qui
mode?lise une agre?gation d?amibes soumises a? la diffusion, a? la croissance et au chimiotactisme.
Ce dernier phe?nome?ne est une propension de certaines espe?ces a? se de?placer vers les plus forts
gradients de substances chimiques, souvent produites par ces espe?ces elles-me?mes. Enfin, au
Chapitre 3, nous conside?rons une e?quation anisotrope, qui intervient en science des mate?riaux
et dont le terme de diffusion est inhomoge?ne et singulier aux points ou? le gradient de la
solution s?annule. Nous de?finissons une notion de solution faible et prouvons un principe de
comparaison. Le de?placement de l?interface limite est induit par une version anisotrope de sa
courbure moyenne. Nous utilisons la distance associe?e a? une me?trique de Finsler.
Mots cle?s : Syste?mes de convection-re?action-diffusion ? Equation d?Allen-Cahn ? Syste?me
de FitzHugh-Nagumo ? Chimiotactisme ? Anisotropie ? Ge?ne?ration d?interface ? Propagation
d?interface ? Epaisseur d?interface.
Convection-reaction-diffusion systems and interface dynamics
This thesis deals with the singular limit of systems of parabolic partial differential equations,
with bistable nonlinear reaction terms and general initial data. We prove some generation of
interface properties and study the motion of interface. We revisit a variety of model problems
and obtain a new and optimal estimate of the thickness and the location of the transition layer
that develops.
In Chapter 1, we first consider a perturbed Allen-Cahn equation. The motion of the limit
interface is driven by its mean curvature and a pressure term. Then, we extend our results to a
large class of reaction-diffusion systems. The idea is to regard the first equation of the system
as a perturbed Allen-Cahn equation ; the proofs are based upon a study of the dependence
of the interface motion on various parameters together with some refined a priori estimates.
Chapter 2 is devoted to the study of a chemotaxis system. This is a model for the aggregation
of amoebae in the presence of diffusion, growth and chemotaxis. This last phenomenon is a
tendency of some species to move towards higher gradients of chemical substances which they
often produce themselves. Finally, in Chapter 3, we consider an anisotropic equation, which
arises for instance in material sciences, and whose diffusion term is spatially inhomogeneous
and singular in the points where the gradient of the solution vanishes. We define a notion of
weak solution and prove a comparison principle. The motion of the limit interface is driven by
its anisotropic mean curvature. We use the distance function associated with a Finsler metric.
Key words : Convection-reaction-diffusion systems ? Allen-Cahn equation ? FitzHughNagumo system? Chemotaxis ? Anisotropy ? Generation of interface ? Motion of interface ?
Thickness of interface.
AMS subject classifications : 35K57, 35K60, 35K50, 35K20, 35R35, 35B20.
e solution of the ordinary differential equation
ut = ??2 f (u). In Section 3.6, we construct a pair of sub- and super-solutions for the study
of the motion of interface, by using a related one-dimensional stationary problem. In Section
3.7, by fitting these two pairs of sub- and super-solutions into each other, we prove our main
results for (P ? ): Theorem 3.1.3 and its corollary.
106 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
For the proof of a propagation of interface property in the case of a related problem we
refer to Bellettini, Colli Franzone and Paolini [8], who also give a precise approximation of
the moving interface. We also refer to articles about the convergence to classical and viscosity
solutions, in the case of well prepared initial data, by Elliott and Scha?tzle [31], [32], for a
homogeneous function a = a(p).
3.2
Finsler metrics and the anisotropic context
For the convenience of the reader, we first recall properties stated by Bellettini, Paolini and
Venturini, [9] and [10]. For more details and proofs, see these references where the idea is to
endow RN with the distance obtained by integrating the Finsler metric and to work in relative
geometry.
3.2.1
Finsler metrics
Suppose that ? : ? О RN ? [0, +?[ is a continuous function satisfying the properties
?(x, ??) = |?|?(x, ?)
?0 |?| ? ?(x, ?) ? ?0 |?|
for all (x, ?) ? ? О RN and all ? ? R,
N
for all (x, ?) ? ? О R ,
(3.13)
(3.14)
for two suitable constants 0 < ?0 ? ?0 < +?. We say that ? is strictly convex if, for any
x ? ?, the map ? 7? ?2 (x, ?) is strictly convex on RN . We shall indicate by
B? (x) = {? ? RN , ?(x, ?) ? 1}
the unit sphere of ? at x ? ?.
The dual function ?0 : ? О RN ? [0, +?[ of ? is defined by
Е
ф
?0 (x, ? ? ) = sup ? ? и ?, ? ? B? (x)
(3.15)
for any (x, ?) ? ? О RN . One can prove that ?0 is continuous, convex, satisfies properties
(3.13) and (3.14), and that ?00 , the dual function of ?0 , coincides with the convex envelope of
? with respect to ?.
We say that ? is a (strictly convex smooth) Finsler metric, and we shall write ? ? M(?)
if, in addition to properties (3.13) and (3.14), ? and ?0 are strictly convex and of class C 2 on
? О (RN \ {0}). In particular ?00 = ?.
We denote by ?? the integrated distance associated to ? ? M(?), that is, for any (x, y) ? ?,
we set
o
nZ 1
А
б
?? (x, y) = inf
(3.16)
?(?(t), ??(t))dt , ? ? W 1,1 [0, 1]; ? , ?(0) = x , ?(1) = y .
0
In the special case of the Euclidian metric, the function ? is given by ?(x, p) = ?(p) =
(p1 2 + и и и + pN 2 )1/2 , so that ?? reduces to the usual distance.
Given ? ? M(?) and x ? ?, let T 0 (x, и) : RN ? RN be the map defined by
(
?0 (x, ? ? )?0p (x, ? ? ) if ? ? ? RN \ {0}
0
?
(3.17)
T (x, ? ) =
0
if ? ? = 0.
Here, ?0p denotes the gradient with respect to p whenever we regard ?0 (x, p) as a function
of two variables x and p. In the following, for better readability, some x and t dependencies
3.2 Finsler metrics and the anisotropic context
107
are omitted. If u : ? ? R is a smooth function with non vanishing gradient, we define the
anisotropic gradient by
?? u = T 0 (x, ?u) = ?0 (x, ?u)?0p (x, ?u).
(3.18)
In a similar way as in the isotropic case, if ?t is a smooth hypersurface of ? at time t, and
n the outer normal vector to ?t (in the Euclidian sense), we define n? the ?-normal vector to
?t and ?? , an analogue of the ?-mean curvature of ?t ? which differs from ?? defined in [9],
[10] ? by
1
n? = ?0p (x, n), ?? =
div n? .
(3.19)
N ?1
Furthermore, we have the following formulas: if ? is a smooth function with non vanishing
gradient such that ?t = {x ? ?, ?(x, t) = 0}, and ? is positive outside ?t and negative inside,
then
??
,
|??|
??
1
div
,
?=
N ?1
|??|
n? = ?0p (x, ??),
n=
?? =
(3.20)
1
div ?0p (x, ??),
N ?1
(3.21)
on ?t . We also define the normal velocity of ?t and the ?-normal velocity of ?t by
Vn = ?
?t
,
|??|
Vn,? = ?
?t
.
0
? (x, ??)
(3.22)
To conclude these preliminaries, we quote a theorem proved in [10].
Theorem 3.2.1. Let ? be connected, and let ? ? M(?). Let ?? be the integrated distance
associated to ?. Let C ? ? be a closed set, and let dist? (x, C) be the ?? distance to the set C
defined by
Е
ф
dist? (x, C) = inf ?? (x, y) , y ? C .
(3.23)
Then
?0 (x, ?dist? (x, C)) = 1,
(3.24)
at each point x ? ? \ C where dist? (и, C) is differentiable.
In the special case of the Euclidian metric, (3.24) reduces to the property that |?d| = 1.
3.2.2
Application to the anisotropic Allen-Cahn equation
We set, for all (x, p) ? ? О RN ,
?0 (x, p) =
p
2a(x, p).
(3.25)
First, since a(x, и) is 2 homogeneous, ?0 satisfies assumptions (3.13) and (3.14) with the constants
?0 = [2
min
x??,|p|=1
a(x, p)]1/2 > 0
and
?0 = [2
max
a(x, p)]1/2 > 0,
(3.26)
x??,|p|=1
also using that a is continuous and strictly positive on the compact set ? О S N ?1 , with S N ?1
the unit sphere of RN . By the hypotheses on a(x, p), we see that ?0 is strictly convex and of
108 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
class C 2 on ? О (RN \ {0}); moreover, by Remark 3.1.1, ?0 is continuous on ? О RN . It follows
that ? is a Finsler metric and the theory of the above subsection applies. We have
(
ap (x, p) if p ? RN \ {0}
0
T (x, p) =
(3.27)
0
if p = 0.
S
Let ? = 0?t?T (?t О {t}) be the unique solution of the limit geometric motion Problem (P 0 )
and let de be the signed distance function to ? defined by
(
dist(x, ?t ) for x ? ?+
t ,
e t) =
d(x,
(3.28)
? dist(x, ?t ) for x ? ??
t ,
where dist(x, ?t ) is the distance from x to the hypersurface ?t in ?. Let de? be the anisotropic
signed distance function to ? defined by
(
dist? (x, ?t ) for x ? ?+
t ,
(3.29)
de? (x, t) =
? dist? (x, ?t ) for x ? ??
t ,
where dist? (x, ?t ) denotes the ?? distance to the set ?t defined in (3.23). By Theorem 3.2.1,
the following equality holds
2a(x, ?de? (x, t)) = 1
in a neighborhood of ?t .
(3.30)
We then write the second equalities in (3.20), (3.21), (3.22), once with ? = de and once with
? = de? to obtain two formulas to express the ?-normal vector n? , the analogue of the ?-mean
curvature ?? and the ?-normal velocity Vn,? :
1
e = ap (x, ?de? ),
n? = q
ap (x, ?d)
e
2a(x, ?d)
"
#
h
i
1
1
e = 1 div ap (x, ?de? ) ,
div q
?? =
ap (x, ?d)
N ?1
N ?1
e
2a(x, ?d)
1
Vn,? = ? q
det = ?(de? )t .
e
2a(x, ?d)
(3.31)
(3.32)
(3.33)
The end of this section is devoted to the operator
div ?? u = div T 0 (x, ?u) = ? и ap (x, ?u),
(3.34)
which differs from the anisotropic Laplacian ?? u defined in [9], [10]. In the case of the Finsler
metric, it turns out that the term div ?? u may be less regular than ?u. Nevertheless, we
show below a boundedness property.
Lemma 3.2.2. There exists a positive constant CL such that, for all u ? C 2,1 (? О [0, T ]), the
following inequality holds.
»
»
»? и ap (x, ?u(x, t))» ? CL (|?u(x, t)| + |D2 u(x, t)|)
for all (x, t) ? QT .
(3.35)
3.2 Finsler metrics and the anisotropic context
109
Proof. We can, with no loss of generality, ignore the dependence in time. First, assume that x
is such that ?u(x) 6= 0. Regarding a(x, p) as a function of two variables x and p = (p1 , и и и , pn ),
we obtain, by a straightforward calculation, that
X ?2a
X ?2a
?2u
? и ap (x, ?u(x)) =
(x, ?u(x)) +
(x, ?u(x))
(x).
(3.36)
?xj ?pj
?pi ?pj
?xi ?xj
j
i,j
2
a
(x, и) is 1 homogeneous as well,
We recall that ap (x, и) is 1 homogeneous, and therefore ?x?j ?p
j
and that app (x, и) is 0 homogeneous. It follows that
X »» ? 2 a
?u(x) »»
|? и ap (x, ?u(x))| ? |?u(x)|
)»
(x,
»
?xj ?pj
|?u(x)|
j
X »» ? 2 a
?u(x) »»
)»
(x,
+ |D2 u(x)|
»
?pi ?pj
|?u(x)|
i,j
» ?2a
»
X
»
»
? |?u(x)|
max »
(y, p)»
?x
?p
y??,|p|=1
j
j
j
» ?2a
»
X
»
»
max »
(y, p)»,
+ |D2 u(x)|
?p
?p
y??,|p|=1
i
j
i,j
б
А
А
б
where we have used that a ? C 3+? ? О RN \ {0} ? C 2 ? О S N ?1 . This proves (3.35) under
the assumption ?u(x) 6= 0.
Now assume that x is such that ?u(x) = 0. We have to proceed in a slightly different way
since app (x, 0) does not make sense. The operator ap (x, и) is 1 homogeneous so that, for any
direction ?,
t?1 (ap (x, t?) ? ap (x, 0)) = ap (x, ?).
We denote by (e1 , и и и , eN ) the Euclidian basis of RN . It follows from the above equality that
ap (x, и) admits at the point 0 partial derivatives in any direction ei and
which, in turn, implies that
?ap (x, и)
(0) = ap (x, ei ),
?pi
(3.37)
?a
? ?a
(x, 0) =
(x, ei ).
?pi ?pj
?pj
(3.38)
2
a
(x, и) is 1 homogeneous as well,
Note that, since ap (x, и) is 1 homogeneous, and therefore ?x?j ?p
j
the first term in (3.36) vanishes at point (x, 0). It follows, from (3.36) and (3.38) that, in the
case where ?u(x) = 0,
» X ?a
»
?2u
»
»
|? и ap (x, ?u(x))| = »
(x, ei )
(x)»
?pj
?xi ?xj
i,j
» ?a
»
X
»
»
? |D2 u(x)|
max »
(y, p)»,
y??,|p|=1 ?pj
i,j
which gives (3.35) in this case as well.
Remark 3.2.3. By similar arguments, one can obtain a positive constant CT such that, for
all u ? C 2,1 (? О [0, T ]),
»?Б
»
»
ц»»
»
ap (x, ?u(x, t)) » ? CT »Dx1 Dt1 u(x, t)»
for all (x, t) ? QT .
(3.39)
»
?t
110 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
3.3
Formal derivation of the interface motion equation
In this section we derive the equation of interface motion corresponding to Problem (P ? ) by
using a formal asymptotic expansion. The resulting interface equation can be regarded as the
singular limit of (P ? ) as ? ? 0. Our argument goes basically along the same lines with the
formal derivation given by Nakamura, Matano, Hilhorst and Scha?tzle [63]: the first two terms
of the asymptotic expansion determine the interface equation. Though our analysis in this
section is for the most part formal, the results we obtain will help the rigorous analysis in later
sections.
S
Let u? be the solution of (P ? ). Let ? = 0?t?T (?t О {t}) be the solution of the limit
geometric motion problem and let de? be the anisotropic signed distance function to ? defined
in (3.29). We then define
[
[
+
?
=
(?
О
{t}),
Q
=
(??
Q+
t
t О {t}).
T
T
0<t?T
0<t?T
We also assume that the solution u? has the expansions
u? (x, t) = 0 or 1 + ?u1 (x, t) + и и и ,
(3.40)
away from the interface ? (the outer expansion) and
u? (x, t) = U0 (x, t, ?) + ?U1 (x, t, ?) + ?2 U2 (x, t, ?) + и и и ,
(3.41)
near ? (the inner expansion), where Uj (x, t, z), j = 0, 1, 2, и и и , are defined for x ? ?, t ? 0,
z ? R and ? := de? (x, t)/?. The stretched space variable ? gives exactly the right spatial scaling
to describe the sharp transition between the regions {u? ? 0} and {u? ? 1}. We normalize Uk
in such a way that
U0 (x, t, 0) = a, Uk (x, t, 0) = 0,
for all k ? 1 (normalization conditions). To make the inner and outer expansions consistent,
we require that
Uk (x, t, +?) = 0,
U0 (x, t, +?) = 1,
(3.42)
Uk (x, t, ??) = 0,
U0 (x, t, ??) = 0,
for all k ? 1 (matching conditions).
In what follows we will substitute the inner expansion (3.41) into the parabolic equation of
Problem (P ? ) and collect the ??2 and ??1 terms. To that purpose, note that if V = V (x, t, z)
and v(x, t) = V (x, t, ?) are real valued functions then we have ?v = 1? Vz ?de? + ?x V and
vt = 1? (de? )t Vz + Vt ; if v and V are vector valued functions we obtain ? и v = 1? ?de? и Vz + ?x и V .
A straightforward computation yields
1
u?t = (de? )t U0z + U0t + (de? )t U1z + ?U1t + и и и
?
1
?u? = U0z ?de? + ?x U0 + U1z ?de? + ??x U1 + и и и
?
1
?
ap (x, ?u ) = ap (x, U0z ?de? + ??x U0 + ?U1z ?de? + ?2 ?x U1 + и и и )
?
1
= ap (x, U0z ?de? ) + app (x, U0z ?de? )(?x U0 + U1z ?de? ) + и и и
?
1
= U0z ap (x, ?de? ) + app (x, ?de? )(?x U0 + U1z ?de? ) + и и и ,
?
3.3 Formal derivation of the interface motion equation
111
where we have used the various homogeneity properties of a and its derivatives. It follows that
1
? и ap (x, ?u? ) = ?de? и ?z (ap (x, ?u? )) + ?x и (ap (x, ?u? ))
?
i
1 e h1
= ?d? и U0zz ap (x, ?de? ) + app (x, ?de? )(?x U0z + U1zz ?de? )
?
?
ц
1Б
+ ?x U0z и ap (x, ?de? ) + U0z ? и ap (x, ?de? ) + и и и
?
1h
1
= 2 U0zz 2a(x, ?de? ) + 2a(x, ?de? )U1zz + 2?x U0z и ap (x, ?de? )
?
? i
+ U0z ? и ap (x, ?de? ) + и и и ,
where we have used Remark 3.1.1 and where the functions Ui (i = 0, 1), as well as their
de? (x, t)
). Hence, in view of (3.30), we obtain, in a
derivatives, are taken at the point (x, t,
?
neighborhood of ?t ,
? и ap (x, ?u? ) =
i
1
1h
e
e
U
+
+
2?
U
и
a
(x,
?
d
)
+
U
?
и
a
(x,
?
d
)
+ иии
U
0zz
1zz
x 0z
p
0z
p
?
?
?2
?
We also use the expansion
f (u? ) = f (U0 ) + ?U1 f ? (U0 ) + и и и .
Next, we substitute the expressions above in the partial differential equation in Problem (P ? ).
Collecting the ??2 terms yields
(3.43)
U0zz + f (U0 ) = 0.
In view of the normalization and matching conditions, we can now assert that U0 (x, t, z) =
U0 (z), where U0 is the unique solution of the one-dimensional stationary problem
(
U0 ?? + f (U0 ) = 0,
U0 (??) = 0,
U0 (0) = a,
U0 (+?) = 1.
(3.44)
This solution represents the first approximation of the profile of a transition layer around
the interface observed in the stretched coordinates. We recall standard estimates on U0 , see
Chapter 1 for more details.
Lemma 3.3.1. There exist positive constants C and ? such that the following estimates hold.
0 < 1 ? U0 (z) ? Ce??|z|
0 < U0 (z)
? Ce??|z|
for z ? 0,
for z ? 0.
In addition to this U0 ? > 0 and, for all j = 1, 2,
|Dj U0 (z)| ? Ce??|z|
for z ? R.
Since U0 depends only on the variable z, we have ?x U0 ? = 0. Then, by collecting the ??1
terms, we obtain
U1zz + f ? (U0 )U1 = (de? )t U0 ? ? ? и ap (x, ?de? )U0 ? ,
(3.45)
112 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
which can be seen as a linearized problem for (3.43). The solvability condition for the above
equation plays the key role in deriving the equation of interface motion. By Lemma 1.2.2 in
Chapter 1, which is a variant of the Fredholm alternative, it is given by
Z h
i
(de? )t (x, t)U0 ? (z) ? ? и ap (x, ?de? (x, t))U0 ? (z) U0 ? (z)dz = 0,
R
for all (x, t) ? QT . Since
R
R (U0
? 2
)
> 0, it follows that
(de? )t = ? и ap (x, ?de? ).
(3.46)
In virtue of the expressions of ?? and Vn,? in (3.32) and (3.33), the above equation, written in
relative geometry, reads as
(3.47)
Vn,? = ?(N ? 1)?? on ?t ,
that is the interface motion equation (P 0 ). Using again the formulas (3.32) and (3.33), one
can come back to the Euclidian geometry and obtain the equivalent interface motion equation
h
i
1
1
p
on ?t .
(3.48)
Vn = ?? и p
ap (x, n)
2a(x, n)
2a(x, n)
Summarizing, under the assumption that the solution u? of Problem (P ? ) satisfies
(
1
in Q+
?
T
as ? ? 0, almost everywhere,
u ?
0
in Q?
T
+
we have formally proved that the boundary ?t between ??
t and ?t moves according to the
law (3.47) or (3.48).
Remark 3.3.2. To conclude this section, note that combining (3.46) with (3.45) yields U1 = 0.
In fact, the second term of the asymptotic expansion vanishes because the two stable zeros
Rof1 the nonlinearity f have ?balanced? stability, or more precisely because of the assumption
0 f (u)du = 0. If we perturb the non linearity by order ?, say f (u) ?? f (u) ? ?g(u), the
equation of the free boundary problem contains an additional term and U1 no longer vanishes.
3.4
A comparison principle
In this section, we prove a comparison principle for Problem (P ? ). To begin with, we define
a notion of sub- and super-solution for Problem (P ? ). To that purpose, we suppose that
u0 ? H 1 (?) ? L? (?).
2
1
?
Definition 3.4.1. A function u+
? ? L (0, T ; H (?)) ? L (QT ) is a weak super-solution for
Problem (P ? ), if
2
? (u+
? )t ? L (QT ),
+
?
2
? ?? u +
? (x, t) = ap (x, ?u? (x, t)) ? L (0, T ; L (?)),
? u+
? (x, 0) ? u0 (x) for almost all x ? ?,
? u? satisfies the integral inequality
Z tZ h
i
1
+
+
)
?
+
a
(x,
?u
)
и
??
?
f
(u
)?
? 0,
(u+
p
?
? t
?
?2
0
?
for all nonnegative function ? ? L2 (0, T ; H 1 (?)) ? L? (QT ) and for all t ? [0, T ].
(3.49)
3.4 A comparison principle
113
We define a weak sub-solution u?
? in a similar way, by changing ? in (3.49) by ?, and
with the condition u?
(x,
0)
?
u
(x),
for almost all x ? ?.
0
?
The following remark will reveal efficient when constructing our smooth sub- and supersolutions in later sections.
Remark 3.4.2. Note that, by Lemma 3.2.2, if u ? C 2,1 (QT ), then the function
L0 u := ut ? ? и ap (x, ?u) ?
1
f (u),
?2
is well-defined in QT . Also, using Lemma 3.2.2, we deduce that ? и ap (x, ?u) ? L? (QT ).
2,1 (Q ) satisfies L u+ ? 0 almost everywhere,
Integrating by parts, we deduce that if u+
0 ?
T
? ?C
)и?
=
0
on
??О(0, T ), and u+
the anisotropic Neumann boundary condition ap (x, ?u+
?
? (x, 0) ?
+
?
u0 (x) for almost all x ? ?, then u? is a super-solution for Problem (P ); an analogous remark
2,1 (Q ).
stands for a sub-solution u?
T
? ?C
We prove below an inequality which expresses the strong monotonicity of the function
= ap (x, p).
T 0 (x, p)
Lemma 3.4.3. There exists a constant ? > 0 such that, for all x ? ?, for all p1 , p2 ? RN ,
(ap (x, p2 ) ? ap (x, p1 )) и (p2 ? p1 ) ? ?|p2 ? p1 |2 .
(3.50)
Proof. First we consider the case that sp1 + (1 ? s)p2 6= 0 for all s ? [0, 1]. Then, the function
s 7? a(x, sp1 + (1 ? s)p2 ) is of class C 2 on [0, 1] and there exist s1 , s2 such that
1
a(x, p2 ) ? a(x, p1 ) = ap (x, p1 ) и (p2 ? p1 ) + (p2 ? p1 ) и app (x, s1 p1 + (1 ? s1 )p2 )(p2 ? p1 ),
2
and
1
a(x, p1 ) ? a(x, p2 ) = ap (x, p2 ) и (p1 ? p2 ) + (p1 ? p2 ) и app (x, s2 p1 + (1 ? s2 )p2 )(p1 ? p2 ).
2
We claim that there exist 0 < ?2 ? ?2 such that, for all x ? ?, all p ? RN \ {0}, all p? ? RN ,
?2 |p?|2 ? app (x, p)p? и p? ? ?2 |p?|2 .
(3.51)
Indeed, it follows from the strict convexity of a(x, и) that app (x, p) is a positively definite
symmetric matrix. Hence the function (x, p, p?) 7? app (x, p)p? и p? is strictly positive and continuous on the compact set ? О S N ?1 О S N ?1 which, combined with the fact that app (x, и) is 0
homogeneous, proves (3.51). It then follows that
?2
|p2 ? p1 |2 ,
2
?2
a(x, p1 ) ? a(x, p2 ) ? ap (x, p2 ) и (p1 ? p2 ) + |p2 ? p1 |2 .
2
a(x, p2 ) ? a(x, p1 ) ? ap (x, p1 ) и (p2 ? p1 ) +
(3.52)
(3.53)
Adding up inequalities (3.52) and (3.53) yields the desired inequality, with the constant ? = ?2 .
114 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
In the case that sp1 + (1 ? s)p2 = 0 for some s ? [0, 1], p1 and p2 are colinear and we may
suppose that there exists l ? R such that p2 = lp1 . We can assume l 6= 0, l 6= 1 and p1 6= 0.
By using the different homogeneity properties in Remark 3.1.1, we obtain that
(ap (x, p2 ) ? ap (x, p1 )) и (p2 ? p1 ) = (l ? 1)2 ap (x, p1 ) и p1
= 2(l ? 1)2 a(x, p1 )
= 2a(x, (l ? 1)p1 )
? ?0 2 |(l ? 1)p1 |2 = ?0 2 |p2 ? p1 |2 ,
where ?0 has been defined in (3.26). The proof is now completed.
We are now ready to prove the following comparison principle.
?
Lemma 3.4.4. Suppose that u+
? , respectively u? , is a super-solution, respectively a sub?
solution, for Problem (P ); we have that
?
+
u?
? ? u ? u?
almost everywhere in QT .
Proof. By subtracting equality (3.9) for the solution u? and inequality (3.49) for the super2
1
?
solution u+
? , we obtain that, for all ? ? L (0, T ; H (?)) ? L (QT ) such that ? ? 0, and for
all t ? [0, T ],
Z tZ h
i
?
+
(u? ? u+
? )t ? + (ap (x, ?u ) ? ap (x, ?u? )) и ??
0
?
Z tZ
┤
1│
?
+
)
?
f
(u
)
?
f
(u
?
?
2
0
? ?
Z tZ
|u? ? u+
(3.54)
? C1
? |?,
0
?
where C1 is the positive constant defined by
C1 = ??2 kf ? k
А
б.
+
?
L? ?max(ku? k? ,ku+
? k? ),max(ku k? ,ku? k? )
+
2
1
?
Next we set ? = (u? ? u+
? ) , which belongs to L (0, T ; H (?)) ? L (QT ); it follows from
(3.50) that
Z tZ
(ap (x, ?u? ) ? ap (x, ?u+
? )) и ??
0
?
Z tZ
?
+
(ap (x, ?u? ) ? ap (x, ?u+
=
? )) и (?u ? ?u? )
{u? ?u+
? ?0}
0
??
Z tZ
0
{u? ?u+
? ?0}
2
|?u? ? ?u+
?| .
Then, substituting this inequality into (3.54) yields
Z
Z │
Z tZ
Z tZ
┤2
1 t d
+
?
+ 2
2
)
+
?
|?u
?
?u
|
?
C
(u? ? u+
(u? ? u+
1
?
?
?) ,
+
+
2 0 dt ?
?
?
{u ?u? ?0}
{u ?u? ?0}
0
0
and therefore
Z │
?
?
(u ?
+
u+
?)
┤2
Z tZ │
┤2 Z │
┤2
?
+ +
+
(u ? u? )
(u? ? u+
(t) ? 2C1
+
)
(0).
?
0
?
?
3.5 Generation of interface
115
Using Gronwall?s Lemma, we find that
Z │
Z │
┤2
┤2
?
+ +
2C1 t
+
(t) ? e
)
(0),
(u ? u? )
(u? ? u+
?
?
?
Since u? (x, 0) ? u+
? (x, 0) for almost all x ? ?, it follows that
u? ? u+
?
a.e. in QT .
?
Similarly one can show that u?
? ? u a.e. in QT .
Lemma 3.4.5. Let u? be a solution of Problem (P ? ). Then
?ku0 kL? (?) ? u? ? max(1, ku0 kL? (?) )
a.e. in QT .
Proof. We remark that ?ku0 kL? (?) and that max(1, ku0 kL? (?) ) are sub- and super-solutions
for Problem (P ? ).
3.5
Generation of interface
This section deals with the generation of interface, namely the rapid formation of internal
layers that takes place in a neighborhood of ?0 = {x ? ?, u0 (x) = a} within the time span of
order ?2 | ln ?|. In the sequel, ?0 will stand for the quantity
?0 :=
1
min(a, 1 ? a).
2
Our main result in this section is the following.
Theorem 3.5.1. Let ? ? (0, ?0 ) be arbitrary and define х as the derivative of f (u) at the
unstable equilibrium u = a, that is
х = f ? (a).
(3.55)
Then there exist positive constants ?0 and M0 such that, for all ? ? (0, ?0 ),
? for almost all x ? ?,
? ? ? u? (x, х?1 ?2 | ln ?|) ? 1 + ?,
(3.56)
? for almost all x ? ? such that |u0 (x) ? a| ? M0 ?, we have that
if u0 (x) ? a + M0 ? then u? (x, х?1 ?2 | ln ?|) ? 1 ? ?,
?
?1 2
if u0 (x) ? a ? M0 ? then u (x, х
? | ln ?|) ? ?.
(3.57)
(3.58)
We will prove this result by constructing a suitable pair of sub and super-solutions.
3.5.1
The bistable ordinary differential equation
The sub- and super-solutions mentioned above will be constructed by modifying the solution
of the problem without diffusion:
u?t =
1
f (u?),
?2
u?(x, 0) = u0 (x).
116 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
This solution is written in the form
┤
│t
,
u
(x)
,
0
?2
u?(x, t) = Y
where Y (?, ?) denotes the solution of the ordinary differential equation
(
Y? (?, ?) = f (Y (?, ?))
Y (0, ?)
for ? > 0,
= ?.
(3.59)
Here ? ranges over the interval (?2C0 , 2C0 ), with C0 being the constant defined in (3.6). We
first collect basic properties of Y .
Lemma 3.5.2. We have Y? > 0, for all ? ? (?2C0 , 2C0 )\{0, a, 1} and all ? > 0. Furthermore,
Y? (?, ?) =
f (Y (?, ?))
.
f (?)
Proof. First, differentiating equation (3.59) with respect to ?, we obtain
(
Y?? = Y? f ? (Y ),
Y? (0, ?) = 1,
(3.60)
which can be integrated as follows:
Y? (?, ?) = exp
hZ
0
?
i
f ? (Y (s, ?))ds > 0.
(3.61)
We then differentiate equation (3.59) with respect to ? and obtain
(
Y? ? = Y? f ? (Y ),
Y? (0, ?) = f (?),
which in turn implies
Y? (?, ?) = f (?) exp
hZ
?
0
= f (?)Y? (?, ?).
i
f ? (Y (s, ?))ds
(3.62)
(3.63)
This last equality, in view of (3.59), completes the proof of Lemma 3.5.2.
We define a function A(?, ?) by
A(?, ?) =
f ? (Y (?, ?)) ? f ? (?)
.
f (?)
Lemma 3.5.3. We have, for all ? ? (?2C0 , 2C0 ) \ {0, a, 1} and all ? > 0,
A(?, ?) =
Z
0
?
f ?? (Y (s, ?))Y? (s, ?)ds.
(3.64)
3.5 Generation of interface
117
Proof. Differentiating the equality of Lemma 3.5.2 with respect to ? leads to
Y?? = A(?, ?)Y? ,
(3.65)
whereas differentiating (3.61) with respect to ? yields
Z ?
Y?? = Y?
f ?? (Y (s, ?))Y? (s, ?)ds.
0
These two last results complete the proof of Lemma 3.5.3.
Next we need some estimates on Y and its derivatives. First, we perform some estimates
when the initial value ? lies between ? and 1 ? ?.
Lemma 3.5.4. Let ? ? (0, ?0 ) be arbitrary. Then there exist positive constants C?1 = C?1 (?),
C?2 = C?2 (?) and C3 = C3 (?) such that, for all ? > 0,
? if ? ? (a, 1 ? ?) then, for every ? > 0 such that Y (?, ?) remains in the interval (a, 1 ? ?),
we have
C?1 eх? ? Y? (?, ?) ? C?2 eх? ,
(3.66)
and
|A(?, ?)| ? C3 (eх? ? 1);
(3.67)
? if ? ? (?, a) then, for every ? > 0 such that Y (?, ?) remains in the interval (?, a), (3.66)
and (3.67) hold as well,
where х is the constant defined in (3.55).
Proof. We take ? ? (a, 1 ? ?) and suppose that for s ? (0, ? ), Y (s, ?) remains in the interval
(a, 1 ? ?). Integrating the equality
Y? (s, ?)
=1
f (Y (s, ?))
from 0 to ? yields
Z
0
?
Y? (s, ?)
ds
f (Y (s, ?))
= ?.
Hence by the change of variable q = Y (s, ?) we get
Z Y (?,?)
dq
= ?.
f (q)
?
Moreover, the equality of Lemma 3.5.2 leads to
Z Y (?,?) ?
f (q)
dq
ln Y? (?, ?) =
f (q)
?
Z Y (?,?) ?
Б f (a) f ? (q) ? f ? (a) ц
=
dq
+
f (q)
f (q)
?
Z Y (?,?)
h(q)dq,
= х? +
?
where
h(q) = (f ? (q) ? х)/f (q).
(3.68)
(3.69)
(3.70)
118 Chapitre 3. A spatially inhomogeneous and anisotropic Allen-Cahn equation
Since
f ?? (a)
as q ? a,
f ? (a)
the function h is continuous on [a, 1 ? ?]. Hence we can define
h(q) ?
H = H(?) := khkL? (a,1??) .
Since |Y (?, ?) ? ?| takes its values in the interval [0, 1 ? a ? ?] ? [0, 1 ? a], it follows from (3.70)
that
х? ? H(1 ? a) ? ln Y? (?, ?) ? х? + H(1 
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