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Nonlocal Cahn-Hilliard Equation

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2/05 EEIRC/Dateoueindd-pjs

NONLOCAL CAHN-HILLIARD EQUATION
By

J ianlong Han

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

2005

ABSTRACT

NONLOCAL CAHN-HILLIARD EQUATION
By

Jianlong Han

The thesis includes three parts. In the first part, we study a nonlocal Oahu-Hilliard
equation with no flux boundary condition and prove the existence, uniqueness and
continuous dependence on initial data of the solution to this equation. We also apply
a nonlinear Poincare inequality to show the existence of an absorbing set in each
constant mass affine space. In the second part, we study the existence, uniqueness
and continuous dependence on initial data of the solution to a nonlocal Cahn-Hilliard
equation with homogeneous Dirichlet boundary conditions on a bounded domain.
Under a nondegeneracy assumption the solutions are classical but when this is re-
laxed, the equation is satisfied in a weak sense. Also we prove that there exists a
global attractor in some metric space. In the third part, we establish the existence,
uniqueness and continuous dependence on initial values for classical solutions to the
Cauchy problem of a nonlocal Oahu-Hilliard equation. We also prove that under cer-
tain conditions, there exists a discontinuous steady state solution for this equation in

a bounded domain.

To my mother

iii

ACKNOWLEDGMENTS

I would like to thank Professor Peter W. Bates, my dissertation advisor, for his
invaluable guidance, expert advice, constant encouragement and support during my
graduate study at Michigan State University and Brigham Young University.

I would also like to thank Professor Kening Lu and Professor Tiancheng Ouyang
for their advice when I studied at Brigham Young University.

I wish to thank Professor Yumin Huang and Professor Yiming Long for their
support during my study at Nankai University.

Many thanks to my dissertation committee members Professor Chichia Chiu, Pro-
fessor Keith Promislow, Professor Xiaodong Yan, and Professor Zhengfang Zhou for

their time and advice.

iv

TABLE OF CONTENTS

1 Introduction 1

2 The Neumann boundary problem for a nonlocal Cahn-Hilliard equa-

tion 8
2.1 Existence and uniqueness .......................... 8
2.2 Long term behavior in the LP norm ..................... 22
2.3 Long term behavior in the H 1 norm ..................... 28
2.4 Applications to other nonlocal problems .................. 37
3 The Dirichlet boundary problem for a nonlocal Oahu-Hilliard equa-

tion 44
3.1 Existence, uniqueness and continuous dependence on initial data for clas-

sical solutions ............................... 44
3.2 Existence, uniqueness and continuous dependence on initial data for gen-

eralized solutions ............................. 49
3.3 Long term behavior in the H1 norm ..................... 57
3.4 Existence of a global attractor ........................ 69
4 The Cauchy problem and steady state solutions for a nonlocal Cahn-

Hilliard equation 71
4.1 The Cauchy problem for a nonlocal Cahn-Hilliard equation ....... 71
4.2 Steady state solutions for a nonlocal Cahn-Hilliard equation ....... 76
BIBLIOGRAPHY 83

CHAPTER 1

Introduction

An interesting phenomenon is observed when a molten binary alloy is rapidly cooled to
a lower temperature. We find that the sample becomes inhomogeneous very quickly,
decomposing into a very fine-grained structure - two concentration phases, one rich in
one component and one rich in the other component. As time passes, the fine-grained
structure becomes more coarse with larger particles growing and smaller particles
tending to dissolve. The sudden appearance of a fine grained structure is called
Spinodal decomposition. The coarsening process is called Ostwald Ripening.
In 1953, materials scientists John Cahn and John Hilliard derived the following

equation:
U, = —A(52Au — F’(u)) for x E Q C IR" and t > 0, (1.1)

with the natural boundary conditions

Bu 6 , _
Bh—O and a—n(Au—/\F(u))—O on 69.

They conjectured that (sometimes called the Principle of Spinodal Decomposition),
“Most solutions to the Cahn-Hilliard equation that start with initial data near a fixed
constant in the Spinodal region exhibit fine-grained decomposition.” Since the conjec-

ture agrees with the outcome of physical experiments, the Cahn-Hilliard equation has

been accepted as a meaningful model of the dynamics of phase separation in binary
alloys.

To derive the Cahn-Hilliard equation as a model for the evolution of a concentra-
tion field in a binary alloy, we take the point of view that microstructure changes in
such a way as to decrease the total free energy of the sample, consistent with the sec-
ond law of thermodynamics. Not only should the free energy decrease but it should
do so as quickly as possible.

The Helmholtz free energy of a state is
E = H — TS,

where H is the total interaction energy, T is the absolute temperature, and S is
the total entropy of mixing. Using accepted definitions of H and S, for a scaled

concentration field it at fixed subcritical temperature, one can derive the expression

E(u)=gffm-yxm)—u(y>)2dxdy+/F(u(x))dx. (1.2)

where C is a positive constant, F is a double well function, having local minima at
:l:1, and the interaction kernel J is assumed to be integrable with positive integral
and with J (-a:) = J (x) by the symmetry of interaction between sites.

The goal is to find dynamics for the field it which decreases E(u) optimally, con-

sistent with thermodynamic principles. This suggests the evolution law.
8
78—: = —gradE(u), (1.3)

where 'y > O is called the relaxation coefficient since it determines the rate at which
u approaches equilibrium.

In the evolution equation, for each t, u(t) is a function of position, that is, u(t)
lies in some space of functions X defined on a spatial domain, and so 92 6 X for

at
all t > 0. On the other hand, E: X —-> IR is a nonlinear functional and grad E(u) is

therefore the linear functional on X defined by

< gradE(u),v >= 5%E(u + hv)|h=o, (1.4)

where <, > is the duality pairing. In order to have (1.3) make sense, X must be a

Hilbert space. First we consider the case X = L2(IR"). In this case, we get
gradL2E(u) = C[(/ J(z)dz)u - J in u] + F'(u),

where =l= is convolution. On a bounded domain (2, this is given by J a: u E fn J (a: —

y)u(y)dy. Taking f J = 1, C = 1 and denoting f = F’, equation (1.3) becomes
7%1—‘=J*u-u—f(u). (1.5)
We call this the Nonlocal Allen—Cahn equation. In (1.2), if we make the approximation
WI) - My) 1’ V“(15) - (Iv - y),

and assume J is isotropic, equation (1.3) leads to

an

the appropriate boundary conditions being
Bu
fl — 0.

This is called the Allen-Cahn equation.

Note that both versions of the Allen-Cahn equation do not preserve the average
value of u. This violates conservation of species if we are modeling phase change in
binary alloys. One way to correct this is to select a new metric space with respect to
which we take the gradient of the energy. So following Fife [23], we consider a new
Hilbert space H51, where H “1 is the dual of the Sobolev space H1 and the subscript
zero refers to mean value zero. Recall that, if f E L2 and f f = 0, there is a unique

(1) such that

—A(/>=f, $20, and/czo.

3

We use the notation ab = (—Ao)‘1 f. Since (—A(,)‘1 is a positive self adjoint operator,
fractional powers of it are well defined. The space H0"1 is the completion of the space

of smooth functions of mean value zero in the norm
1

Hanna-1 = H(—Ao)_2UHL2.

The inner product is given by
1 1
< u,v >,,0.1=<(—A0) 2n, (—Ao) 2v >L2

for u, v belonging to Ho’l. If u 6 Ho‘1 and v 6 L2, then
<u,v >H0—1=<(—Ao)‘1u,v>.L2

This means that the representative of grad E(u) in H51 is (—A)(gradL2E(u)) and
instead of the Nonlocal Allen-Cahn equation, we have the following Nonlocal Cahn-

Hilliard equation

2% = A((./n J(m — y)dyu(m) - [9'1” — y)u(y)dy + f(u)) (1'7)

with natural boundary condition

(fnJ( y)dyu w()- L, J($ — y)U(y)dy + f (u))
an

Integrating the equation over 9, using the Divergence Theorem and the boundary

d
BEAU—0,

so species are conserved. Also, to see that energy decreases along trajectories, note

=0.

 

condition we have

that

LZE-“l =2 f f we -— yxum — auntie) — u.(y))dzdy + / F'<u>u.dx

=2//J(my)—:( mm)(ut m—)dmdy 2//J(ym—)):(m u(y)dmdy—
2J(/j m—y y)(mut m)dmdy+2//( m—y y)yut( )(dmdy+/f) )utdm.

(1.8)

For the case of a bounded domain (I, using the symmetry of J, if we write a(m) =

In J(m — y)dy, J * a(m): fnJ( m - y)u( y)dy, and [C(11) = a(m)u — J =l= u(m) + f(u),

then we have

 

dE(U) = /(a(m)u(m) _ J * a(m) + f(u))utdm

Again using a first order approximation for u(m) — u(y), the local equation corre-

sponding to (1.7) is

65% = —A(dAu — f(u))) for x E Q C IR” and _t > 0, (1.10)

with the natural boundary conditions
6

a—:=O and B—an (Au—f(u)))=0 on 39.

This is the Oahu-Hilliard equation (1. 1).

Equations (1.5), (1.6), (1.7), and (1.10) are important in the study of materi-
als science in modeling certain phenomena such as Spinodal decomposition, Ostwald
ripening, and grain boundary motion. Equations (1.7) and (1.10) share some common
features, for example, the mass is conserved and the energy is decreased. There is a
lot of work on equation (1.10), see for example [2], [3], [4], [5], [8], [9], [10], [11], [18],
[19], [21], [35], [39] and references contained in those articles.

However, for equation (1.7), there are very few results. To the best of our knowl-
edge, the only results related to equation (1.7) were given in [26] by H. Gajewski and

K. Zacharias and in [29] by G.Giacomin and J. Lebowitz .

In [26], H. Gajewski and K. Zacharias considered the equation

 

gtfflw — w» — v - ow) -—- 0.
“OWN,” = 0, (1.11)
n

71(2), 0) = u0(x),
where w (=m) fn" (Im — y])( )(1- 2u(y))dy, v— - f’(u )+ w, and f(u) = ulogu + (1 —

u) log(1 — u).

The mobility it has the form ,u = a(m, V”)

f”(U) ’

(a(m, sl)sl - a(m, 32)32)(51 — 32) Z aolsl — $2[2, 81, 32 E IR+,

where a satisfies:

a
(a(m, 31)31 — a(m, 32)32)(sl — 32) S -—1|sl — 32].
3

They proved the existence and uniqueness of a solution to equation ( 1.11).

In [29], G. Giacomin and J. Lebowitz considered the equation

 

5F
am: v w p>V(——- °( ))1
on Td, the torus IR“ mod Z“, where 61??) is the L2 gradient of F0. Here, a is a

function from [0, 1] taking nonnegative values and such that 0(0) = 0(1) 2 0, and

= (a f.(p(r))dr + i f [W J<r — r')((p<r> — WWW.

. . 1 . . .
The function fc has a double well structure, symmetric about 2’ With the minimum

at values p+ and p“ < p+.

Denote g(p) = fc(p)+-J—@(p—%)2, where J( (0): deJ( r.)dr In [29], it is assumed
that:

2

(1) There exists a constant c > 0 such that

S 0(0) E a(p)g"(p) S c

OIH

for all p 6 (0,1).

(2) Both 9 and 0 are symmetric with respect to g.

(3) J 6 C2(T"), J 2 0 and J(r) depends only on [r].

With the above assumptions, G. Giacomin and J. Lebowitz indicated how one
might prove the existence and uniqueness of solutions.

In chapter 2, we study equation (1.7) with no flux boundary condition and prove
the existence, uniqueness and continuous dependence on initial data of the solution to
this equation. We also apply a nonlinear Poincaré inequality to show the existence
of an absorbing set in each constant mass affine space. In Chapter 3, we study the
existence, uniqueness and continuous dependence on initial data of the solution to
equation (1.7) with Dirichlet boundary conditions on a bounded domain. Under
a nondegeneracy assumption the solutions are classical but when this is relaxed,
the equation is satisfied in a weak sense. Also we prove that there exists a global
attractor in some metric space. In chapter 4, we establish the existence, uniqueness
and continuous dependence on initial values for classical solutions to the Cauchy

problem of equation (1.7). We also prove that under certain conditions, there exists

a discontinuous steady state solution for equation (1.7).

CHAPTER 2

The Neumann boundary problem
for a nonlocal Cahn-Hilliard

equation

2.1 Existence and uniqueness

Consider the integro-differential boundary value problem

a—u=A(/J(m—y)dyu()--‘/{;J(113—y)u(tlu)dZ¢/‘i'f())m Qit>0,

(fnJ( y)dyu m()— fnJ( m — y)u( (y)dy + f(u))

6n

= O on 852, t > 0, (2'1)

 

u(m, 0) = uo(m).

In order to prove the existence of a classical solution to (2.1), we need the initial

2 + B
data to satisfy the boundary condition. So we assume uo(m) E 02 +fl 2 (Q) for

some 6 > 0, and uo(m) satisfies the compatibility condition:

W y)-—dyu0(m) fQJ( y)uo (y)dy+ f (210))

6111320 on ('39. (2.2)

 

Rewrite (2.1) as

 

1;“ = a(m,u)Au + b(m,u, Vu) in Q, t > 0,

Bu 6a(m) BJ (m — y)

_ .— —— = 2.
a(m,u)6n + 6n a(m) [n 6n u(y)dy 0 on 09, t > 0, ( 3)

u(m, O) = uo(m),

where
a(flI, U) = a(95) + NH).
a(m) = [0 Jo — gm.
b(m, u, Vu) = 2Va - Vu + f"(u)|Vu|2 + uAa — (AJ) at 11.
We assume the following conditions:
(A1) a(m) E Cz+5(S-2), f 6 CQ+5(R).
(A2) There exist c1 > 0, c2 > 0, and r > 0 such that

a(x, U) = a(:13) + f’(U) 2 61+ C2IUI2’-

(A3) 60 is of class 0'2”.
Note that (A2) implies

F(u) = A“ f(s)ds 2 c3|u|2"+2 — c4 (2.4)

for some positive constants c3 and c4.

For any T > 0, denote QT = Q x (0, T). We first establish an a priori bound for

solutions of (2.1).
Theorem 2.1.1 If u(m,t) E C2'1(QT) is a solution of equation (2.1), then

”(1281K Mm, t)| S C(uo) (2-5)

for some constant C(uo).

In order to prove the theorem, we need the following lemma.

Lemma 2.1.2 If a(m, t) E C2’I(QT) is a solution of equation (2.1), then there is a
constant C(uo) such that

SUP IIU('at)llq S C(Uo) (2-5)

0957‘

for any q 3 2r + 2.
PROOF. Let
E(u) = i//J(m — y)(u(m) — u(y))2dmdy + /F(u(m))dm. (2.7)

It follows from (1.9) that

dE (u)
dt

 

_<_ 0.
Therefore E(u) g E(uo), i.e.,
:1,- f f J<x — m(ue) - a(ywxdy + / F<u<x>>dx
s i / / Jo — mace) — warmly + f mamas.

From condition (A1), (2.4), and Young’s inequality, we obtain

/ |u|2r+2dm g C(uo).
9

Since this is true for any t > 0, we have

sup [lulQszm S C(uo),
n

0937‘
where C (uo) does not depend on T.

Since 0 is bounded, it follows that

Slip “1th S C(no)
03:37“

for any q S 21‘ +2.

10

We will prove the theorem with an iteration argument.

PROOF. For p > 1, multiply equation (2.1) by 1414”"1 and integrate over £2, to

 

obtain
fululp"1utdm = —/a(m, u)Vu-V(u|u]”“1(m))dm
- f f we - y)u(w)V(ulul”‘1(x))dydx
+//VJ(m—y)u(y)V(u|u[”’1(m))dydm.
Since
fa(m,u)Vu-V(u]u|p“l)dm =p/a(m,u)|u]”'1|Vu|2dm
n n
and
1&1 2
WM 2 l2 = QZ—WuIP-‘lwr.
with condition (A2), we have
19 +1
[a(m, u)Vu - V(u|u|”‘l)dm_>( 4PC11)2/ |V|u| 2 [2dm
a
p + 2m + 1

 

41”? / IVIu] 2 (2012.

(p + 21' + 1)2
This yields

4 p+1
p+1dt./|u Ipldx 19+i()2 final-)1]: 2 [26113

+ Sf / We — y)u(y)V(quI"-‘(x))dydw-

From Cauchy-Schwartz and Young’s inequalities, together with

V(U|Ul”’1) = PIUIP‘IVu,

11

(2.10)

(2.11)

(2.12)

(2.13)

we have
“ I I We - y)u(2)V(ululp"‘(2))dyd2

s Mlp/ lluIPVu(2)|
Q
12—1 p+1 (2.14)
5M1p / Iul 2 1w )Ilul 2 dz:

p_+_1

0119 2 | 1 1:
—' <(p I 1)2 «/S:2 I I I I 1} 2p (II I

for some positive constant M2 which does not depend on p, and M1 = sup f IVJ (m —

 

y)Idy. Also we have
/ [We — y)||u(y)IV(UIUI”"1(m))dydx

= p] f 117112: — y)“2(2)Ilu(2)l’”“|Vu(2)ld2dy

_-_ 11:1
312/1142): 2 qu(2)||u(2r)| 2 flVJ(2—y)llu(y)ldydx
<15P/Iu(517)Ip1IVU()I2dm+[I’UIIIIIPII/IVJQL'—y)llu()Idyl2

s .p/ a(m))r‘lvmxnwx
— 1 p + 1 2

p— .—
+ c<e>p1 f 1n<x>12+1d212 + 1([1/ we — 2111212112112 2 dx>2 + 1

.<_ 610/ went-wwww
L11 -2—
+ a(e1p1/ lu(2)|”“dxlp +11] 121(2):?“de + 1M?

5 [ep |u(2)|”‘1|Vu(2)|2d2 + 40pr f |u(2:)|”+1d2
p+1
Cl]? 2 +1
1)2‘/‘IVIuI—— 2 Idm—l-M3p/Iul” dm
Q

 

(2.15)

for some constant M3 which does not depend on p. Inequalities (2.12)-(2.15) imply

p__+1
di/f‘zlulp+ld$(p + 21:31 1)/IVIU I 2 _I2d$SC'(p+1)2/IUIp+ld-T- (2.16)
O

 

12

Now we need the following Gagliardo—Nirenberg inequality,

 

 

 

 

 

 

 

 

 

 

HDj'UllLs S Clllevll‘irllvllF + Czllvlqu, (2-17)
where
j 1 j 1 m 1
—< <1 —=— ——— 1— —. 2.1
m—“— ,8 n+a(, n)+( a)q ( 8)
In (2.17), set 3 = 2, j = 0, r = 2, m =1, to get
ll’vllg S CIIIDvII§“IIvIl3“'“’ + Czll’vlli- (2-19)
“I: +1
2 _ 1
Letv=|u| 2 ,pk=2",q=—£flil-+—),and
#1: +1
n(2 — q) n
= = . 2.20
a n(2—q)+2q n+2+22“‘ ( )
Using Young’s inequality this yields
in: +1 a #1: +1
[lull‘k +1dx g 6/ |V|u| 2 |2dx + cc 1— “(f |u|”k*1+1dx)“k-l+1.
n n n
(2.21)
If we set p = pk in (2.16) and plug (2.21) into (2.16), we obtain
d 2 “k +1
— u“’°+1dx+—£l—#i/Vu 2 2dx
dtntI uk+ln||| I
#2 +1 1121
s cm. +1)2<ef IVIuI 2 lzdx + cc 1- a(j Iul“’°—1 + Ida/1H +1).
0 , a
(2.22)
Choosing e = 1 ' 61M]: , we have
C(11); +1)2 [1,]; +1
Mk +1 Mk +1
d we +1 2 2 #k—l +1 [1 +1
2; .2 II dx + 01(2) 9 IVIuI I dz s cm n In! dx) ,
(2.23)

13

1 a 2

= 01/11: = 1—a.. CW" —1—a. l—a
whereC’lUc) uk+1,Cg(k) C c (#k+1) (uk+1) .

Choosing e = 1 in (2.21), this and (2.23) also imply

 

 

 

 

lite-1'1
%/|u|”k+1dx+01(k)/ |u|“k+1dxg04(k)(/luluk-1+1dx)l1k—1+1
n n n

 

where (321(k): 02(k ) + c.

By Gronwall’s inequality, we have

 

 

 

 

Mk +1
Alulflk+ldx< [a luo |1uk+1dx +——C4()k)(sup/n |u|“k1+1d1:)”k-1+1
0109) t>0 (2 24)
11;. +1 ’
6(2) max(MI‘* + llfll, (sup f lul“‘°‘1 “(12212—1 +1},
:20 n
2 . . .
where 6(k) = c(1 + pk)“, a = i———a, and M0 = sup luol. ThlS implles
— 2:60
Mk +1
f lul“" + 1am 3 6(2) max(Mo“'° + llfll, (sup / lul“’°‘1 + Mam-1+1}
Q >0 (I
”I: +1 ”I: +1
< H< ((lfllék — 2))111- -1+ 1 max{M”’° +1 (sup /n lul2d2) 2 }.
i=0
(2.25)
Since 111 < 2‘, we have
#k—i +1

Mic-1'1 Hie-1’1 pk+1
5(k)<5(k _1)uk—1 + 150c _ 2)]1‘k—2 + 1 ...5(1) 2

501+2+~-2"“.(ga)k+(k—1)2+---+(k—z')2"+---+2"-1 (2.26)

 

k k+1
< 02 — 1(2OI—k+2 — 2

and

MIC-+1 file-1‘1

—— k
IQI-IQI“’=-1+1~-IQI 2 gm? +1. (2'27)

14

Estimates (2.25)-(2.27) and Lemma 2.1.2 imply

1
(/ lull/4k +1dx)flk +1 S CIQI220 max{Mo,sup(/ IuI2d$)§} S C(UO) (2.28)
n t_>_0 n

H

where C(uo) does not depend on k. Since this is true for any k, letting k —> oo in
(2.28), we have
IIUHoo S C(Uo),

and therefore,

sup IIuIIoo g C(uo). (2.29)
0937‘

Since it E C (QT), it follows that

max Mat)! S C(Uo)

QT

Remark 2.1.3 In (2.29), since C(uo) does not depend on T, we also obtain a global

bound for u whenever there is global existence of a classical solution.

Since maxQT IuI _<_ M, after a slight modification of the proof of Theorem 7.2 in

Chapter V in [30], using the equivalent form (2.3) we have

Theorem 2.1.4 For any solution it E Cum—QT) of equation (2.1) having maxQT IuI g

C, one has the estimates
1%“ IVuI < K1, lul‘H” < K2, (2.30)

where constants K1, K2, and 6 depend only on C, IIUOIIC2(Q) and Q, I I8”) is the

H older norm given in [30].

In (2.3), setting v(:c, t) = a(x, t) — 220(1), we obtain the equivalent form

git): 51(27, 1) ,u0)Av+b(:c, 1), Vi) ,uo) in Q, t > O,
é(:1:,v, u0)5—+ +iL(:1:,v,u0) = 0, on 80, t > 0, (2-31)
v(:r,0) = ,

15

where

(1(5):, '1), U0) = 0(2), 'U + U0),

5(1):, 'U, V'U, U0) = a(m) 'U + U0)AUO + b($, 'U + UO, V('U + 210)):
and

(v(:1:,t) +uo(:1: )) +a( .‘L‘, U ,uo)%——::(

—/——— 3" x ' (1.2+ Uo(y))dy

Since (2.2) implies 2/3(x, 0, 21.0) = 0, the compatibility condition for (2.31) is also sat-
isfied.

_an

Denote

Lv = _ a(x, v, m)m — E(x, 22, W. (to).

9|?

and

[101)- — 'a—v' — CIA’U,

at

where c1 is the constant in condition (A2).

Consider the following family of problems:
ALv + (1 — A)Lov = 0 in QT:
.. 012 ~ 82)
A(a(x,u,uo)5; + 1/J(:r,v,uo)) + (1 — A)(c1(a—n)) = 0 on 852 x [0, T], (2.32)
v(2:, 0) = 0.

Lemma 2.1.5 If v(x,t, A) E C2’1(QT) is a solution of (2.32), then
max |v($,t,/\)| _<_ K, (233)
QT

where K does not depend on x\.

16

PROOF. Since A&(x,v,uo) + (1 — A)c1 2 A61 + (1 — A)c1 = cl > O, the terms in
(2.32) also satisfy (A1) — (A2) and so (2.33) follows from Theorem 2.1.1.

Consequently one may also conclude from Lemma 2.1.5 and Theorem 2.1.4 that:
Lemma 2.1.6 If v(:r, t, A) 6 02’1(QT) is a solution of equation (2.32), then
rgaxIVv(:c,t,/\)I 5 K1, Iv(:c,t,A)I8,;'6) 3 K2, (2.34)
where constants K1, K2, and 6 do not depend on A.

We will use the following abstract result(see [30]):
Theorem 2.1.7 (Leray-Schauder Fixed Point Theorem) Consider a transformation
y = T($, A)

where 2:,y belong to a Banach space X and 0 g A g 1.

Assume:
(a) For any fixed A, T(-,A) is continuous on X.
(b) For .2: in bounded sets of X, T(a:, A) is uniformly continuous in A on [0,1].

(6) For any fixed A, T (-,A) is a compact transformation, i.e., it maps bounded

subsets of X into precompact subsets of X.

((1) There exists a constant K such that every possible solution :1: of a: - T(a:, A) = 0
with A 6 [0,1] satisfies : IIxII s K

(e) The equation a: - T(2:,0) = 0 has a unique solution in X.

Then there exists a solution of the equation :1: — T(;1:, 1) : 0.

Define a Banach space

1:9
X = {v(a:,t) E C1 + fl, 2 (QT) : v(2:,0) = 0}

17

with the usual Holder norm.
For any function w e X satisfying conditions maxQleI S M and maxQTIVwI 3

M1, we consider the following linear problem

vt —— (A&(x,w,uo) + (1 — A)cl)Av + Ab(:c, w, Vw, uo) = O in QT,
~ 3v ~ 6v
A(a(:z:, w,uo)57; + (Mm, w, uo))+(1— Mela—n — O on (99 x [0,T],
’U(.’E, 0) = 0.
(2.35)
2 + fl M
It is clear that there exists a unique solution v(:2:, t, A) E C ’ 2 (QT) of (2.35).

Define T(w, A) by
T(w, A) = v(2:,t, A).
Lemma 2.1.8 For w being in a bounded set of X, T(w, A) is uniformly continuous
in A.

PROOF. Let w E X with IIwIIx S M and let v1 = T(w,A1), v2 = T(w,A2), and
v 2 v1 — v2. We have:

vt — (A1&(a:,w, uo) + (1 — A1)c1)Av = (A1 — A2)h(x, w, v2),

- 0v
(Ala(x)w)u0) + (1 — 206113; = (A1 — A2)g(:c,w,v2), (2-36)
v(a:,0) = 0,
where
h(a:, w, v2) = (a(m, w, uo) — c1)Av2 — b(x,w, Vw, no),
and
(xwv)-c§2—a(xwu)%
9,,2—18n IIOan-

Since lex g M and A2&(x,w,u0) + (1 — A2)c1 2 c1 > 0, from (2.35) we have

”02(37, 1, A2)IIC2,1(QT) S N for some constant N independent of A2. Therefore

max Ih(z,w,vg)l 3 N1, max Ig(2:,w,v2)| 3 N2

18

for constants N1 and N2 that do not depend on A2. Note also that Aa(x, w, uo) + (1 —
A)c1 _>_ c1 > 0 for all A 6 [0,1]. It then follows from linear parabolic theory that the
solution of equation (2.36) will approach zero in X as I/\1 — A2| —> 0.

Similarly, one can see that for any fixed A, 1T(x, A) is continuous in X. Fur-

2 + B
. 2 + 5, — 1+ @—
thermore, smce C 2 (QT) ‘——> C 2 (QT) is compact, we see that

T (w, A) is a compact transformation.
These observations, Lemma 2.1.5—Lemma 2.1.8 and the Leray-Schauder Theorem

imply the existence of a solution v(x, t) of (2.31), and therefore:

Theorem 2.1.9 Let ,8 > 0. For uo E C2+5(O) satisfying the boundary condition
2 + B
+ ,— -
(2.2), there exists a solution u to (2.1) with u E C B 2 (QT).

We complete our goal of establishing well-posedness with the following:

Theorem 2.1.10 ( Uniqueness and continuous dependence on initial data)
If u1(x, t) and u2(x, t) are two solutions corresponding initial data u10(x) and

u20(x) of equation (2.1), then
sup / Iul — ugldx g C/ Iulo — UgoIdIB, (2.37)
0931‘ n n

where C only depends on T.

PROOF. For any r E (0,T), 0 E C2'1(Q,) with 3—0— = 0 on 60 x (0,T), we have

Lu,(x,r)0(x,r)dx 2]” u,-(x, 0)0(x, O)dx+f /n(u,0t + B(x, u,)A6)dxdt

+/ AHAJak u ,dxdt+/0 [a 0— * uidxdt,
on

where B(x,u) = a(x)u + f(u). Hence,
Al“ — u2)6(x, r)dx — [00110 — u20)0(x, 0)dx
+/; [)(m — u2)( )(0, + HA9) dxdt +/ L0AJ*(u1 — u2)dxdt (2.39)

(2.38)

+/OTa 905-7; * (U1 - 11.2)dflidt,

19

where

B(IL',’U.1)— B($,U2) for 1115‘ 1‘2

H ,t = “1 ‘ “2
(5L. ) BB(x,u1)
T fOI' 71.1 = 11.2.
Let 0 be the solution to the final value problem
60_ .
Bt— -—,H(x t)A0+fil9 1n 9,05tST,
92 = 0 on 69, O S t S 7', (2-40)
an
B(x, 7') = h(x),

where h(x) E C8°(Q), 0 S h S 1 and B > 0 is a constant.

By the comparison theorem, we have
0 S 0 S 65“”).
Therefore, from (2.39) we have
[(11.1 — 11.2)hd113
n
=/0(u10 — u20)0(x, 0) dx +/ /( ul —- ufiflddxdt (2.41)
+/ f0AJ*(u1 — u2)dxdt+/ [8 0— * (ul — u2)dxdt.
an
Hence,

/(u1 — u2)hdx

n

S / Iulo - u20Ie’fide +/ flul — uglfiefilt—Tldxdt (2.42)
n o n

+C1/ flul —u2IeB(‘—T)dxdt+C2/ flul —u2IeB("’)dxdt.
o n o n

Letting B —) 0 and h —> sign (ul — u2)+ in (2.42), we have

[(ul — U2)+dl‘ S / I’ulo — U20Id$ + 03/ / I’ltl - U2Id$dt. (2.43)
D Q 0 Q

Interchanging ul and U2 gives

[I111 — UgIdl‘ S / I’ulo — U20Id$ + C3/ [I114 — UgIdIL‘dt. (2.44)
Q Q 0 Q

20

By Gronwall’s inequality, (2.44) yields

/ I111 - U2ld$ S C(Tlf lulo - U20Id$- (2-45)
(2 (2

Remark 2.1.11 If uo(x) E L°°(Q), we can consider weak solutions as follows:

Define
X={f(x)€C°°(Q)| x)=/J(y)f(x— y)dy,—|ao=0}
and let

B 2 Closure of X in the L2 norm.

Definition 2.1.12 A weak solution of (2.1) is a function u E C([0,T],L2(Q)) fl
L°°(QT) fl L2([0,T],H1(Q)), with at E L2([O,T],H‘1(Q)) and Vh(x,u) E
L2((0, T), L2(Q)) such that

< ut(x,t),i,b(x) > 4.1/S18) Vh(x,u) - th(x)dx
(2.46)

—‘/Q(VJ =I< u( )Vi,b(x)dx =0
for all w E H1(Q) and ac. time 0 S t g T, where h(x,u) = a(x)u + f(u), a(x) =
fa J($ " y)dy, and

WE 0) — 210(2) (247)

Theorem 2.1.13 If (A1) — (A3) are satisfied and no 6 L°°(Q) (I B, then there exists

a unique weak solution u of (2.1)

Essentials of the proof: Since no 6 L°°(Q) D B, there exists a sequence uIIk) E X

such that
(k)
””0 " “OIIL2 "’ 0,

2.48
“225.11”... < c, ( )

21

where C does not depend on 1:. Consider equation (2.1) with initial data use).
There exists a unique classical solution um. By the energy estimate and other a
priori bounds, one can find a subsequence and a weak limit u such that
um —1 u in L2((0,T),H1(S2)),
2111—) u in L2((O,T),L2(Q)),
IIUIloo S C.
(2.49)
h(x,u(kl) —\ h(x, u) in L2((0,T),H1(Q)),
h(x,u(kl) —-) h(x, u) in L2((0,T),L2(Q)),
ugh) —1ut in L2((o,T),H-1(o)),

and u satisfies equation (2.46).

2.2 Long term behavior in the LP norm

First, we establish a nonlinear version of the Poincaré inequality.

Proposition 2.2.1 Let Q C R" be smooth and bounded. For p > 1, there is a

constant C (O, p) such that for all u E W1'2P(fl) with fa u = 0,

/ Iulzpdx S C(Q,p)/ IVIuIPIzdx. (2.50)

o o
PROOF. If (2.50) is not true, there exists a sequence {uk} C mem) such that
fuk = 0, flukl2pd$ > k/ IVIukIpIQdIII. (2.51)

n n n
11101: = ——u—k——, then it follows that
llukll2p
1
/ wk = 0, / Iwk|2pdx = 1, / IVIwIIPIQda: < 75' (2.52)
n n (2

Therefore, there exists a subsequence (still denoted by {IwkIP}) and w E H1(Q) such

that

Iwklp —1 w in H1 and IwkIp —> w in L2. (2.53)

22

Since In IVIwkIP|2dx S 12’ for any (p E C8°(Q), we have

 

 

3Iwk|p
d 2.
a 02:.- (p x—>0 ( 54)
for i=1,- - - ,n. Therefore,
6w
d = 2.
Leap x 0 ( 55)

for i=1,- - - ,n and (p E C8°(Q). So Vw = 0 a.e in Q, and w is constant in 9.

By taking a subsequence, (2.52) and (2.53) yield
1 1

w - (Isl-l); and IwkIp -—> (Ell—Ifi a.e in Q. (2.56)
So, we have
1
IwkI —> (—1—)E’ a.e in Q. (2.57)

IQI

Since f w, = 0, there exists a unique solution (pk to

—A(,o=w,c in 0,
By) _

B—n _. 0 on an, (2.58)
[mix = 0.
(I
From (2.58), we obtain
] (wit = / w. s llwkll L2|l<pkll 1.2- (2.59)

Since f9 (pkdx = 0, by Poincaré’s inequality, IIkaIILz S cIIVwkIIL'z, therefore (2.58)
and (2.59) imply

IIVSOkllL2 S CllwkllL2 (2-60)

and
/V(IwkIp'lwk)V(pkdx_—./IwkIp+1dx. (2.61)
n n

23

Since

V(Iwklp—l’wk) = lekIp—lvwk,

we have

|V(|wk|p—lwk)l = leklp"llvwk| = IVIwklpl.

Hence, from (2.61), we have

fIwkIPde=/V(Iwk|p-1wk)V(pkdx
Q Q
s f IVIwkIPHVs/Jkldx
Q

S IlVlwilpllellvcpklle
—> 0

as k -—) 00, by (2.52) and (2.60). Hence, along a subsequence,
Iwk|p+1 —-) 0 a.e in Q,
i.e,
|wk| —> 0 a.e in $2.

This contradicts (2.57).

(2.62)

(2.63)

(2.64)

(2.65)

Remark 2.2.2 In [6], the same result was established independently by Alikakos and

Rostamian, which was brought to my attention by Professor Alikakos.

The following lemma may be found in [38]

Lemma 2.2.3 ( Uniform Gronwall inequality) Let y be a positive absolutely continu-

ous function on (0, 00) which satisfies
y' + W” S 5

with p >1,1/ > 0,6 2 0. Then, fort Z 0, we have
1 —1

56) s (923 + (u(p —1)t)P- 1.

24

(2.66)

We use this to prove the following:

1
O I c — '
Proposrtlon 2.2.4 Let an < (-C—1-)27‘, where 01 and c2 are the constants in assump-
2

tion (A2), and let 210 = 1511—] In uodx. If u(x) is a solution of (2.1), and I110] 3 do, then

for any q > 1, we have

q+1

02” )" 2r (2.67)

q+1

 

flu—uoIQ+1dx<C1+(
n
where C1 depends on 020 and q, and C2 depends on q.

PROOF. Let u = v + 17.0 in equation (2.1). Since

222:
6t-8t'
Vv=Vu,

and
Va(x)v — (VJ) * v = Va(x)u - (VJ) =1: u

(2. 1) becomes

% = V - (a(x, u)Vv + vVa(x) — (VJ) * v), (2.68)

with the boundary condition
(a(x, u)Vv + vVa(x) - (VJ) * v) - n = 0 on 852.

Multiplying equation (2.68) by Iqu'lv, integrating by parts, and using Hblder’s and

Young’s inequalities gives

Edi/n Iqu+1dx:q/Q a(x,u)|qu— 1|VvI2dx
5+1 (2.69)

Se(q)/IVIvI— 2 |2dx+M(e( q/DIL’IQ+1dx.

25

Since IvI2’" S 22""1(IuI2’ + Iuolzr), with condition (A2) we have

 

 

 

 

a(x, u) > cl -— CgI’ltoIzr + 26-52;IvI2'. (2.70)
It follows from (2.69) and (2.70) that
_ q + 1
4+1, 44(c -02IUo|2')/ v ~3— 2d
q+1dt,/n|vI + (q+1)2 “I M l x
4 q + 2r +1
(102 2 (2.71)
2 d
q__ +1
< e(q)/ IVIvI 2 _Izdx + M(e (q))] Iqu +1dx.
n
i 4 21')
Choosing 00 < (2)2r, ((q) = (J(: _: 33° in (2.71), for Iaol 3 do, we obtain
1 d 4 q + 2r +1
——-—-—- ‘1 +12 + ‘10” f v 2 2d
q+1dt./I;|v| x 22’(q(+21‘+1)2 I IUI I x (2.72)
< M((: (1),...) / Iv|q+1dx.
Since f vdx :2 0, Proposition 2.2.1 implies
q + 2r + 1
/ IUIQ+2r+1d$_ < 0/ IVIvI 2 I2d$ (2.73)
It follows from (2.72)-(2.73) and Holders and Young’s inequalities that
q + 2r + 1
d _—
d—, —fn lvlq+1d2 + 03(4 )(f |v|"+‘d2) 2 +1 3 0.6.4.) 12-74)
for constants C3(q) and C4(q, do). By Lemma 2.2.3, we have
C __‘1_+_1__ C t2 (1 +1
fIqude < (fly + 2r +1 + (__3(_Q)__T) 2r , (2.75)
n 4((1100) (I +1

From Proposition 2.2.4, we have

Theorem 2.2.5 Assume do > 0 is given in Proposition 2.2.4, q > 1 , and

1 1

C __ __
H> lef—chaW—l-2T+l +00IQIQ+1-

26

Then for any solution of (2.1) with —| f uodx|= luol g do, there exists a time

I‘ll
to(ao,q) 2 0 such that

||u||q+1 < u, for all t > to(ao,q). (2.76)

Remark 2.2.6 Applying Proposition 2.2.1 to the standard Cahn—Hilliard equation,
we can prove that there exists an absorbing set in each constant mass affine L2 space
directly as follows:

Let u satisfy

((9):: A(— —dAu+f(u)) in Q, t> 0
i“ :0, 59—42 :0 on an, t> 0, (2.77)
an an

u(x, O) = uo(x),

where

2p—l
=ZaJ-u 3, a2p_1 >0, pEN, p22.
For simplicity, assume IQI— — 1, let f uo(x x)dx — uo, and u— — u — uo.

Multiply equation (2.77) by v and integrate over 9 to get
d 2d
—f——— IdvtI x +d/ IAvl2 + C/ |V|v|P|2_ < Cl flu]2 + h(uo),

where h(uo) depends only on 210.

Since I v = 0, by Proposition 2.2.1, we have

waSC/WMW.

BY H older’s and Young’s inequalities we have

W + cg] |v|2)p s 19(0).

SO by Gronwall’s inequality, there exists an absorbing set in the affine space Hao

1 _
{ue L2 why—:1“);

27

2.3 Long term behavior in the H1 norm

In Section 2.2, we considered the long term behavior of the solution in the LP norm
for any given 19 Z 1. In particular, there exists a “local absorbing set” in the sense
that if I f uol is not too large, the solution enters a fixed bounded set in the affine
space 210 + L” in finite time (note that 210 = K12]- fn uo is conserved by the evolution).
In this section we consider the long term behavior of the solution in the H1 norm. In
this case, we do not need any restriction on | f uol.

Note that (A2) implies

f (u)u Z c5|u|""'+2 — c6 for some constants c5 and c6.
We make additional assumptions on the nonlinearity,
(A4) |f(u)l S C7|ul2r+1+ 63,

(A5) F(u) = I: f(8)d3 S CglU|2r+2 + C10, and 65 > C9.
Remark 2.3.1 (A2), (A4), and (A5) mm for f(u) = clulzru+ lower terms.

Denote 213 = ia- fa 4de and write (,0 = 2/2 - it.
For (p E L2(Q), satisfying (,5 = O, we consider the following equation:

—A0=cp

66
a—nlan = 0 (2.78)

[6:0
52

The equation (2.78) has a unique solution 6 2: (—A0)‘1(<p). Denote HWH—i =
1

(In <P(—Ao)’1(<p)dx)2. This is a continuous norm on L2(Q).

Since a = no is constant, we may write the equation as

B(u — a) _

28

where K (u In J (x y)dyu(x) -—J(fn x— y )u(y)dy+ f ( ). Applying the operator

(—A(,)‘1 to both sides of equation (2.79), we obtain

d(-Ao)—l(u “ '3) + K(u) = 0. (2.80)

 

Taking the scalar product with u — a in L262), we have

1 d 2
gallu—uH- +(K(u) u—u)=0 (2-81)

From condition (A2)—(A5), we have

(K(u)7u—fi)
— —f(fJ(x— y(-)dyux) [J(x— Hmong/”(m)( < )—a)dx

=f /( J( x -— x)dydx — f /( J( x — y) )u(y)u(x)dydx
+/fW x):—u/fW
f/an )(-u (yu))2dxdy+/uf( )dx—u/f(u)d$ (2.82)
_ >2 -/ f.“ a: — — u(y))uxdy + [(Cslul2'” — 66W“?
— |u| [((c7|u|2'+1+ c3)dx
2 g f / J(x — yxum — u(y))2dxdy + c. f lul2’+2dx
— e / |u|2’+2dx — c(u, e)
for any 6 > 0. Choosing e = c5 — c9, we have

(K (,u) u—u)

__ >-21—//J( x-y — u( (y))2dxdy+c9/|u|2'+2dx — c(u )

gill/[fl x(-—y —u(yu))2dxdy+/F( )dx—c(u) (2-83)
> E( (u) — 0(3)
= E(u) -— c(uo).

29

Also from (2.83), we have
(K(u), u — 11) _>_ c/ |u|2r+2dx — c(flo) (2.84)

for some positive constants c and c(fio).

Since ||.||_1 is a continuous norm on L2(Q), we have

llu - fill—1 S Cllu - fillz. (2-85)
Therefore,

”U — gull—1 S CU“ — “Mb
5 CH“ — fioll2r+2 (2-86)
S Cllull2r+2 + C(30)

for some positive constants C and C(fio). From (2.81), (2.84), and (2.86), it follows

that
d — 2 — 2r+2 —
all“ - “OH—1 + CH“ — “OH—1 S C(uo)- (237)

By Lemma 2.2.3, we obtain

 

C (110)

C )r+l +(C(r)t) r . (2.88)

H“ — floll2—1 S(

 

Thus, we have proved:

 

Theorem 2.3.2 There exists M (fig) such that for any p > M (110) 27' + 2 , there exists

a time to such that
“u — floll—l S P, V t Z '50- (2'89)

me (2.81) and (2.82), we also obtain

1 d
mnu — 220112.. + E(u) s cm). (2.90)

30

Integrating from t to t + 1, then (2.89) implies
t+l p
/ E(U(S))d3 S 6*(120) E C(flo) + 3 (2.91)
t
for t 2 to. Since E (u(t)) is decreasing, (2.91) implies
E(U(t)) S c*(a0) (2-92)

fort2t0+1.

Since, from (2.4),

EM» 2 i f / J(x - m(ua) — u(y))2dxdy + / Fonda:

(2.93)
2 c3/lu12’"+2 — C4,
inequalities (2.92) and (2.93) yield
/ |n|2r+2 g c4210) (2.94)

for t _>_ to.

1
Corollary 2.3.3 There exists c.(a0) > M (ao)2r + 2 such that for any p > 6.070),

 

there exists a time t3 such that
flul"r1 S c.(u0) for t 2 t5. (2.95)

Next we estimate ||Vu||2.

Denote h(x, u) = a(x)u + f (u) Multiplying (2.1) by h(x, u) and integrating over

a, we have

/ h(x,u)ut + / |Vh(x,u)|2 = / VJ*u-Vh(x,u). (2.96)
Since

h(x, u)... : (a(x)u + f(u))ut = % %a(x)u2 + F(u)], (2-97)

31

and
/ W =1 u - was, u) s cuuni + gnvzza. a)”; (2.98)
equation (2.96) yields

51, fig-aw + F<u>1+§ f NW. 2012 s cllullfi. (299)

Integrate (2.99) from t to t + 1, and use assumption (A2) and Corollary 2.3.3, to

obtain

/t‘t+1/|Vh(x,u)|2 S c (2.100)

for some constant c and all t 2 t5.

Multiply (2.1) by h(x, u), and integrate on Q to obtain

[h(x,u)tu¢+/Vh(x,u)'Vh(x,u)t =/VJ*u-Vh(x,u)t. (2.101)
Since

h(x, u)¢ut = a(x )uf + f'(u )uf 2 c1213,

/Vh()x,u Vh(x,u) =—2-d—-t-/|Vh(,)|xu2,

(2.102)
and
fVJ*’u.° Vh(x,)u =5—t-/VJ*u Vh(x,u)— /VJ*ut Vh(x,,u)
we have
CI/I’utlz +-—/|Vh(,)|xu2
2‘“ (2.103)
S —t/VJ*u-Vh(x,u) — /VJ*u,-Vh(x,u).
Estimate (2.103) with the Cauchy-Schwartz, and Young’s inequalities imply
d 2 d 2
d—t |Vh(x,u)| g a? 2VJ =1: u . Vh(x, u) + 7 IVh(x,u)| (2.104)

for some constant 7 > 0.

32

For t < s < t+ 1, multiplying (2.104) by e“t 3), we have

%l67(t-s)/|Vh(x,U)l2l S 87("3)%/2VJ*U ° Vh($,u)- (2'105)

Integrating (2.105) between 3 and t + 1, we obtain

_7/ IVh(1;,u(x,t+1))I2—87(t_8)/|Vh($,u($,8))|2
9 @mm

t+l
g/ e7(‘"‘)%/ 2VJ* u(-,u) - Vh(x,u(x,u))dxdu.
s 9

Write
t+l d
f e"“‘”)— dflfn 2VJ =1: u( )-V,h(x u(x, u))dxdu

= aw) [d 2VJ =1 u(-. u) - Vh(x, u(x, u))d$|§“
o (2.107)

1+1
—/ (—ry)e7(“")/2VJ*u(-,u) - Vh(x,u(x,u))dxdu
s n

=n+a

Also,
1, = 879-“) / 2VJ * u(-,u) - Vh(x, u(x,;1))dx|§+1
n

= -7/ 2VJ * u(-,t+ 1) - Vh(x,U($,t+ 1))
n

— e7(t“)/ 2VJ * u(o, 8) ~ Vh(x, u(x, s))dx.
:1

am&

Using the Cauchy-Schwartz and Young’s inequalities, this is bounded above by

—7
e IVh(x,u(x,))t+1 )|)2+C'/n|u(,xt+1)|2

Tn
+‘/Q|Vh(x,u(x,s))|2+C'/lu(515s3)l2

for some constant C. Furthermore,

HI
[2: / 767(t‘”)L2VJ*u(-, 11-) Vh(x, u(x, 11))dxdu
(2.109)

t+l
<C/ [fanh(,(,))«l—xuxpl2 fnlu(,xu)|.2]du

33

Estimate (2.106) becomes

67/” |Vh(x,u(x,t+1))|2 — refit-3)] |Vh(x,u(x, 3))l2 g

8—7

_ |Vh(:z:,,u(a:t+1))|2+C’/IU(SIJt+1)+|2 fthquxsmz (2.110)

t+l
+C/|u(xs)|2+C/ {/IVh(xu(xu)) ))+]2 +L|u(xp)| |)2]d,u

Therefore,

“'7
8—- |Vh(x, u(x, t + 1)))2
2 n

S6'7“")1‘;IVh(x,u(x,s))|2+0!”|u(x,t+1)|2+‘/‘;|Vh(x,u(x,s))|2 (2.111)

+ C fn IU(I,S)|2 + C [HI/n IVh(x,u(x,u))12 + [Q IU(-'r,u)|2]du-

Integrating (2.111) from t to t + 1 with respect to s, we have

-7
6? IVh(x, u(x, t + 1))|2dx
n

t+l 1+1
3] e7<t‘5)/IVh(x,u(x,s))|2dxds+C/ f|u(x,t+l)|2dxds
tH-l 9 HI t n
+/ fth(x,u(x,s))|2dxds+C/ /|u(x,s)|2dxds
t t+1n t+1 t {2
+0] f If IVh(x,U(x,u))|2dx+ / IU(w,u)|2dx]duds (2.112)
t-l-l ’ a ‘2
Sf [IVh(x,u(x,s))|2dxds+C/|u(x,t+1)|2dx
3+1 ‘7 ”1+1
+/ fth(x,u(x,s))|2dxds+C/ /|u(x,s)|2dxds
t n t n

t+l
+ 0 f. (), — mfg IVh(x, «1(me1de + [a |u(:r, u)l2dxld#.

By (2.95) and (2.100), estimate (2.112) yields
/ IVh(x, u(x, t + 1))I2dx 3 C(00) (2.113)
n

for t 2 to(1‘10) and some C(flo) > 0.

Since
Vh(x.u(x,t+1)) = (a(x) + f’(u(t + 1)))Vu(x, t + 1) — u(x,t+1)Va(x), (2.114)

34

we have
/|Vh( (x ,u,(x t+1) )|2> é/‘;I(a( x)+f (u( (t+ 1)))|2|Vu(x,t+1)|2
— / |u(x,t+1)Va(x)| (2.115)

2 f2 lc1|vu(:1~, 1+ 1)) —D(u0)
()2
for t 2 to(ao) and some constant D010).

Estimates (2.113) and (2.115) imply
[a |'§7u(x,t+1)|2 3 C(20), (2.116)
for t Z t3(i‘10) and C(20) > 0. Thus, we have
Theorem 2.3.4 There exists a time t5 (110) such that
||u||H1 g c(uo) for t 2 tab-10). (2.117)
Remark 2.3.5 [38] gives a similar result for the Cahn—Hilliard equation.
Also we have the following theorem

Theorem 2. 3. 6 If u is a solution of (2.1), and Q(u =(fnJ( y)—dy)u(x)

u(x) + f (u(x)), then there exist a sequence {tk} and u“ such that

u(tk) ——> u‘ weakly in H1,

 

(2.118)
Q(u(tk)) —-) Q(u') weakly in H1,
and Q(u‘) is a constant, i.e. u" is a steady state solution of (2.1).
PROOF. If u is a solution of (2.1), from (1.9), we have
dE
11M: —/ |VQ(u |.)2dx (2.119)
This implies
T
/ / |VQ(u)|2dxdt = E(u(0)) — E(u(T)). (2.120)
0 n

35

Recall that

E(u) = :1,- f / J(x — y)(u<x) - u(y))2dxdy + f F<u<x>>dx.
Using (2.4), we have
—E(u(T)) S C, (2.121)

where C does not depend on T.
(2.120)-(2.121) imply

[00/ |VQ(u)|2dxdt S C', (2.122)
0 n

for some positive number C'. So there exists a sequence {tk} with tk 6 [k, k + 1], such

that

/ |VQ(u(tk))|2dx —> 0. (2.123)
a
From (2.29), Remark 2.1.3, and (2.117), we have

l|u(tk)lloo S 01.
(2.124)
llu(tk)llH1 S 02.

Observations (2.123) and (2.124) imply that there exists a subsequence of {tk} (still

denoted by {tk}) such that

u(tk) —> u“ weakly in H1,

u(tk) —+ u' strongly in L2,

(2.125)
Q(u(tk)) ——> u weakly in H1,
Vv = 0 a.e in Q.
Since ||u(tk)lloo S Cl and Hu‘tlloo S 01, we have
||f(u(tk)) — f(u*)||z.2 s C||u(t.) — u'llbz. (2.126)

36

'x'l

for some constant C.
(2.125) and (2.126) imply
v = Q(u”) a.e in Q,

12 = constant, (2.127)

fu‘dxz/uodx.
n Q

So u“ is a steady state solution of (2.1).

2.4 Applications to other nonlocal problems

The method for the nonlocal Oahu-Hilliard equation can also be applied to other
nonlocal problems. For example, we consider the following integrodifferential equation
that may be related to interacting particle systems with Kawasaki dynamics (see [18],

[31], [32], [33], [34])=

% = A(u — tanh(flJ =1 11)) in 9, t > 0,
a(u ‘ ””th * 21)) = 0 on an, t > 0, (2-128)

 

8n
u(x, 0) = u()(x),

where fl is a constant and J is a smooth function.

Note that the average of u, a is constant in time.
2 + a

2 (QT) is a solution of (2.128), multiplying equation

a,

2+
If u(x,t) E C

(2.128) by u and integrating by parts, we have

.1..d__f|u|2+/|Vu |2= /V( t(anh([3.]*u))Vu (2-129)

Since

45V] * u

V(tanh(5J * 11)) = (e—fiJ at u + eflJ * U)2

 

37

 

and

(e—BJ at u + 86.] >1: u)2 Z 4,

we have
/ |\7(tanh(fiJ*u))|2 g / |fi|2|VJ=ku|2 3 C(0, J, 3) / |u|2. (2.130)
The Cauchy-Schwartz inequality and (2.129)-(2.130) imply
ldflul2 1 2 2
—— - < . .
2 at + 2/IVuI _C(Q,J,B)/|u| (2131)

By Gronwall’s lemma, we obtain

flu|2 S c(T,uo). (2.132)
A similar argument to that in the proof of Theorem 2.1.1 yields

sup |u| S C(u0,T). (2.133)

T

The analogues of Theorems 2.1.4-2.1.10 yield

Theorem 2.4.1 If a(x) = fJ(x — y)dy E C2+°(O), 00 is of class 02+“ for some

a > 0, and u()(x) E C2+°(Q) satisfies the compatibility condition, then there exists a
2 + a

_ _ 2 + a, —— -
unique solution u(x,t) E C 2 (QT) to (2.128).

For the long term behavior of the solution, we consider

% = AK(u), (2.134)

Where K(u) = u — tanh([3J * 11).
Apply the operator (—A())‘1 to both sides of (2.134), where (-A())‘1 is defined
in (2.78). We obtain

 

+ K(u) = 0. (2.135)

38

Taking the scalar product with u — 110 in L2(Q), we have

1 d
2d—tllu-fioll ,+(K(..) u—uo)= 0

Note that
(K(u), u — 110) = (u — tanh(flJ * u), u — 1‘10)

=(u — uo + no, u — uo) —(tanh(BJ :1: u), u — uo)

>/Iu-—uoI —IuoI/Iu—uoI- flu-nol
_>."2"/lu—fiol2 "B(fio)

for some constant B (110).

Continuity of the embedding gives
H“ - {toll—1 S CH“ — fioll2-
Equation (2.136) yields
d — 2 — 2 —
all“ " “OH—1 + CH“ — “OH—1 S 2B(uo).

Gronwall’s inequality implies

 

||u — {tonal _<_ m + Ke'Ct.
C
So, there exists ()(m) > ZBéUO) and to := to(a0) such that for t Z to

llu - fiollil S C(fio),

and so there exists an absorbing set in the Ho’1 norm.

For t > to, integrating (2.136) from t to t + 1 gives

HUM + 1) - fio||31-||U(-.t)- flollz—l

3+1
+/ / [u(x, s) — u0|2dxds g B(ao).
t n

39

(2.136)

(2.137)

(2.133)

(2.139)

(2.140)

(2.141)

(2.142)

Inequalities (2.141)-(2.142) imply
1+1
/ / |u(x, s) — uo|2dxds 3 ()(m). (2.143)
1 n
for some constant C (110). This yields
1+1 _
f / |u(x, s)|2dxds g C(ao). (2.144)
1 9
So there is a t1 6 [t,t + 1], where L, |u(x,t1)|2dx 3 C(20). From (2.131) we have

1+1
f|u(x,t+1)|2dx S / |u(x,t1)|2dx+C(Q, J, 5)] [|u(x,s)|2dxds
n n 1 o

(2.145)
S 01010)
for t > to.
By (2.132) and (2.145), we have
sup/ |u(x,t)|2dx 3 C(50). (2.146)
2‘0 0

By (2.131) and (2.146), using a similar argument to that in the proof of Theorem

2.1.1, we have
SUP Hallo.) S C(uo)- (2-147)
120

Next we estimate ||Vu||2

Integrating (2.131) from t to t + 1 with t Z to, we have

1+1 t+1
||u(-.It+1)ll2 — ||u(-.t)||2 + qulzdx S C ||u(-.s)l|§- (2-148)
2 2 1 n 1

Inequalities (2.145) and (2.148) yield

1+1
/ / |Vu|2dx g ()(m). (2.149)
1 n

Multiplying (2.128) by u, and integrating over (2, we obtain
[(1102 +/Vu-Vut=/V(tanh(,8.1 :1: ”(1))-Vat. (2.150)

40

Note that
d
/V(tanh(flJ =1: u)) - V11; 2 a /V(tanh(BJ * u)) - Vu

-— {:(tanthakth - Va,

and

|V(tanh(flJ * u))tl =

(45VJ* u,)(e-W * u + efiJ * U) — 8321 =1 m(eBJ * u - e-W * “)(VJ) * 11

(2.151)

(2.152)

 

I (6—3.7 * u + efiJ * u)3

S Cllutllz-

It follows from (2.150)-(2.153) that
__ 2
[(1102 +2ddt/lvul
= dit / V(tanh([3J* u)) - Vu — / V(tanh(flJ 441)), - Vu
S %/V(tanh(flJ :1: u)) - Vu+/C||ut||2|Vu|
g gE/VUanth at u)) - Vu + éllutllg + C/ |Vu|2.

Therefore,

2%]qu g gZ/ZVUanth =1: 21)) ° Vu + 0/ IV’UI2,

where C depends on 110, J and 9.

For t < s < t+ 1, multiplying (2.155) by eat—3), we have

d —s
&;[/|Vu(x,s)lzec(t )dx]

3 ecu—gig] 2V(tanh(flJ * u(-, s))) - Vu(x, s)dx].

41

(2.153)

(2.154)

(2.155)

(2.156)

Integrating (2.156) between 3 and t+ 1, we obtain

/|Vu(x,t+1)|2e’Cdx—/|Vu(x,s)|2ec(t"‘)dx
t+1 d
_<_ / eatefiIf wean/203.1 * 24.10)) - v.33, 1042:1414

= 11.

We compute
I1 = lea“) / 2V(tanh(flJ * 14-110))'V’U($.u)alflvl|§+1
1+1
+/ 0600-11) /2V(tanh(flJ * u(-,u))) - Vu(x, 11)dxda
= e-0 / 2V(tanh(,8J 1 u(-,t+ 1))) - V1101, 1 + 1)dx

— eat-3) /2VtanhfiJ * u(-, s) - Vu(x, s)dx

1+1
+/ CeC(““l /2V(tanh(6J =1: u(,u))) - Vu(x, ,u)dxd;1

E P1 + P2 + P3.
First,

P1 = e‘C/2V(tanh()8J =1: u(-, t + 1))) 1 Vu(x, t + 1)dx

—C
S 6? |Vu(x, t + 1)|2dx + C/|u(x,t+1)|2dx

for some constant C. Also since t < s < t+ 1, eat”) 3 1, and we have

P2 3 /|Vu(x,s)|2dx+C/|u(x,s)|2dx,

and

1+1
P3 3 f [/ |Vu(x,p)|2dx + C/ Iu(x,p)|2dx]dp.
Estimate (2.157) becomes

/|Vu(x,t+1)|2e"cdx — / |Vu(x,s)|2e0(“3)dx

—C

(2.157)

(2.158)

(2.159)

(2.160)

(2.161)

g 32— |Vu(x,t+l)|2dx+C/|U(I,t+1)|2d$+/IVU(1‘13)I2‘117 (2162)

+0 / |u(x, s)|2dx+ / 1+1[ / |Vu(x, 11)|2dx+C / |u(x,,u)|2dx]d/1.

42

This yields

fqu(x,t+1)|2dx
g C[/ |Vu(x, s)|2dx +/lu(x,t+1)|2dx + / |u(x, s)|2dx (2.163)

1+1 1+1
+/ /|Vu(x,p)|2dxdu+/ /|U($1#)l2d$d#l-
1 1

Integrating (2.163) from t to t + 1 with respect to s, we obtain

1+1
/|Vu(x, t + 1)|2dx S C[/ fqu(x, s)|2dxds
1

1+1 1+1
+ f / |u(x.t+ 112414“ f / |u(x,s)|2dxds
t t

1+1 1+1
+/ /|Vu(x,u)|2dxdu+/ flu(x,p)|2dxdu] (2.164)
t 1+1 t
= C[/ [IVu(x, s)|2dxds +/|u(x,t+1)|2dxds
1

+£t+l/|u(x,s)|2dxds].

Therefore, estimates (2.144), (2.145), (2.149), and (2.164) imply
/ |Vu(x,t+1)|2dx g C(00) (2.165)

for t 2 to(1'10) and for some constant C (1'10).
This means that there exists an “absorbing set” in the affine space Hfio relative

to the H1 norm.

43

CHAPTER 3

The Dirichlet boundary problem

for a nonlocal Cahn—Hilliard

equation

3.1 Existence, uniqueness and continuous depen-
dence on initial data for classical solutions

In this part, we study the following integrodifferential equation

%=A(/. J(x—y>dyu(x)— / J(x—y)u(y)dy+ f(u)) 1.. QT,

u = 0 on ST, (3'1)

u(x, 0) = u()(x),
Where as before QT = Q x (0,T), ST = 652 x (0, T), Q C IR" is a bounded domain,
J(—x) = J (x) for x 6 IR", and f is bistable. We do not assume that J is nonnegative
but its integral is assumed to be positive.

Rewrite (3.1) as
”“1 = (19(x) + f’(u))Au + 2Vp(x) - Vu + f"(u)Vu - Vu + 11131) — (AJ) * u, (3.2)

44

where
p(x) 211.1(1) — y)dy, and (AJ) at u E anJ(x — y)u(y)dy.
We make the following assumptions
(Bl) J 6 C2+7(R"), f E C2+7(R) for some 7 > 0,
(Bg) There exists c1 > 0 such that a(x, u) E p(x) + f’(u) 2 cl,
(3;) 69 is of class C2”.
We first establish an a priori bound for the solution of (3.2).

Remark 3.1.1 Note that bistability of f is not important for our results. However,
the nonlinearity cannot have a slope that is too negative, whereas there is no such
restriction for the local Cahn-Hilliard equation. This is not just a technicality since

the equation has no solution with p(x) + f’ (u) < 0.

Proposition 3.1.2 Assume (B1) — (B3). If u(x,t) E C(QT) flC2'1(QT) is a solution
of (3.2), then

max |u| g C(Q,T,uo) (3.3)
QT
for some positive constant C (Q, T, uo).

PROOF. Set u(x,t) 2 ve‘", where a is to be determined. Then Vu = eU‘Vv,

Au 2 eU‘Av, and (3.2) becomes

eatvt + veata = (p(x) + f'(u))e"‘Av + 2Vp(x) - Vve‘“

(3.4)
+ f"(u)Vv - Vvem + Apve‘" — (AJ) 21 ve‘".
Multiplying (3.4) by v and using vAv = éAv2 — |Vv|2, we obtain
1 2 2 1 I 2 I 2 2
5(1) )1 + v 0 = 50906) + f (u))Av -* (PU?) + f (u))lel + V1413) ' V” (3 5)
+ éf"(u)Vv - sze‘" + Apv2 — v(AJ) =1: v.

45

If there exists (Po,to) 6 QT with to > 0 such that v2(Po,to) = max v2, then

Av2(Po,t0) S 0, V’Uz(Po,to) = 0, (’02)¢(Po,to) Z 0, and (3.5) yields

(a — Ap)v2(Po,to) S — jg; AJ(P0 — y)v(y,to)dyv(P0,to). (3.6)

Choose a large enough such that a — max(Ap) > 6 > 0, we have

M
Inaxl'vl S 75- ] |v(y,to)|dy S Mie""‘° / IU(y.to)ldy S Mze‘”‘°IIUIl2 (3-7)
0 n

for some positive constants M, M1 and M2.
On the other hand, multiplying (3.1) by u and integrating over (2, it follows from

H61der’s and Young’s inequalities and condition (82) that

1d

__ 2 < 2
2dt nudx_K‘/nudx, (3.8)

where K depends only on J and 9. This yields

fu2dx S C(T)/u3dx. (3.9)
n n
It follows from (3.7) and (3.9) that
lv(Po1t0)| S Cllluollz- (3-10)
Therefore,
maxlvl S max{C1||u0||2,max|u0|}. (3.11)

Since max |u| S e"rT max Iv], (3.3) follows from (3.11).
If u(x, t) is a solution of (3.1), maxQT In] S C, after a slight modification of

Theorem 7.2 in Chapter V in [30], we have

. 2+m1+3_ .
Theorem 3.1.3 For any solution u(x,t) 6 C 2 (QT) of equation (3.1) hav-
ing maxQT |u| S C one has the estimates
maXIVuI g K1, |u|?” _<_ K2, (3.12)
QT T

where constants K1, K2 and 'y depend only on C, J, C), ac, and the boundary of Q
(l ° I8?) is a Hélder norm defined in [30]).

46

In order to prove the existence of a solution, we use Schaefer’s fixed point theorem

from [22] or [25].

Theorem 3.1.4 (Schaefer’s Fixed point Theorem). Suppose X is a Banach space,

and
A : X —-) X
is a continuous and compact mapping. Assume further that the set
{u 6 Xlu = uA[u] for some 0 S p S 1}
is bounded. Then A has a fixed point.

The a priori bounds established above will be used in conjunction with this to

prove

Theorem 3.1.5 Suppose conditions (Bl) — (B3) hold, uo(x) E C2+7(Q) and Uolan =

. . 2 + 2,1 + 1 - .
0. Then there exists a solution u(x, t) E C 2 (QT) of equation (3.1).

PROOF. Let v = u — uo, then (3.2) becomes

vt = h(x, v)Av + b(x, v, Vv),

'Ulan : —Uo, (3.13)
v(0, x) = 0,
where
a(x, v) = a(x, v + U0),
b(x, v, Vv) = a(x, v + uo)Au0 + b(x, ’0 + U0, V(v + 710)),
and

b(x, u, Va) 2 2Vp - Vu + f"(u)|Vu|2 + uAp — (AJ) at u.

47

1+7

Define X = {w E C1 + 7’ 2 (Qfll w(0,x) = 0}.

For any w E X, consider the following linear equation

vt : a(‘r, 1U)A’U + B(x1w1Vw))

”Ian 2 —u0(x), (3-14)
v(0, x) = 0.
7
Since w 6 X, a(x, w) and b(x,w,Vw) belong to C7, E(QT), and so there exists a
unique solution v E C2 + 7’ 1 + %(QT) of equation (3.14).

Define an operator A : X —+ X such that
v = A[w].

Claim 1: A is continuous from X to X.

In fact, if v1 = A[wl], v2 = A[Wg], and q = v1 — v2, then q satisfies

qt 2 h(x, w1)Aq + k(x,w1,w2),
(Ilan = 0, (3.15)
«1(0, x) = 0,

where

k(x,w1, 102) = (a(x,w1) — 11(mesz + b(x,w1, le)

~

—b(x, 1.02, ng).
Fixing w, as w] approaches w2 in X, we have
,7 1
((1031101) — (“$1702”sz —> 0 in C , 2(QT)1
7

b(x,w1,Vw1) — b(x, wg, ng) —> 0 in C7, E(QT).

7
Therefore, k(x,w1,w2) -> 0 in 07’ 2 (QT)

48

This implies q -+ 0 in X as w) —> Mg, so A is continuous in X.

Claim 2: If w = ,uA[w], there exists a uniform bound C such that
llwllx S C-

1 1
In fact, if ,u = 0, then w = 0. If 0 < u S 1, since A[w] = fiw, ;w is a solution of
(3.14), we have
wt 2 a(x, w)Aw + pb(x, w, Vw),
wlao = —Hu0(~’13)1 (3~16)

w(0, x) = 0.
Pr0position 3.1.2 and Theorem 3.1.3 imply

Ilwll + 1+, so,
C ’71 2
where C does not depend on 11.
Finally, the compactness of A follows from the fact that C ’ 2 (Ch) 9-)
1 +
1 + 71 —l - . .
C 2 (QT) 18 compact. ThlS completes the proof.

We will prove the uniqueness and continuous dependence on initial values of the

solution in the next section.

3.2 Existence, uniqueness and continuous depen-
dence on initial data for generalized solutions

In section 3.1, under the assumption (B2), equation (3.1) is a nondegenerate parabolic
equation. In this section, we consider the degenerate case. Consider the following

equation with u() E L°°(Q)

‘33—: = mum) — [a u(y)AJ(:c — 1041 in 0.,
u = 0 on ST, (3.17)

49

where

h(x, u) = P($)u($) + f (U)

with p(a:) defined in section 2. Instead of nondegeneracy condition (B2), we assume:

8h(:r:, u)
Bu

(Bé) For every fixed 2:, h(x, 0) = O, and 2 d1 |u|r1 for some positive constants

T1 and d1.

Definition 3.2.1 A generalized solution of (3.17) is a function
u E C([O,T] : L1(Q)) fl L°°(QT) such that

Au(z,t)z/)(a:,t)dx—//tu($,t)z/)8(a:,s)d$ds=f/‘h(x,u)Az/J(x,s)da:ds

(3.18)
— ff (AJ * u(-, s))w(:r, s)d:1:ds + / u(x,0)w(x, O)da:
t n
for all 1,!) E C2'1(QT) such that 1,0(sc, t) = 0 for x E 652 and 0 S t g T, and
u(x,0) = uo(:z:). (3.19)

We first prove the uniqueness.

Proposition 3.2.2 Let u1,u2 be two solutions of equation (3.17) with initial data

um, ugo E L°°(Q), then

”141(7) - “2(TlllL1m) S C(Tlllulo — U2o||L1(n)

for each 7‘ E (0, T), and some constant C(T).

PROOF. For any 1' E (O, T), and I/J E C2’1(QT) with «plan 2 0 for 0 < t < 7', after

multiplying (3.17) by w and integrating over 0 x (0, T), we have

[,u,-(x, T)w(:z:, T)dx =/u,-(:1:,0)1/2(:r,0)d3:+/0 fn( 21,112; + h(x, u )Aw)d:rdt

+fo [2(AJ11‘ ui) ).7,bd:1:dt

50

(3.20)

Setting z 2 ul — U2 and 20 = um — U20, equation (3.20) gives

[a z(a:,r)z/J(:1:,r)dx= f9 zo(a:)i,b(a:,0)dx

 

1 1 (3-21)
+/ fzwt +b(:z:, t)Ai/1)dxdt+/ [(AJ*z)¢rdxdt,
o n o o
where
h(x, ul) —— h(IL',’U2) for m 7‘5 ”2’
b(x, t) = "'1 — “2
hu($,u1) for u1 = u2.
Following the idea in [7], we consider the problem:
(gt/jz—bA¢+1/w in Q, 0<t<r,
w = 0 on 652, 0 < t < 1', (3-22)
$0117.) 2 9(33):

where g(a:) E C8°(Q) , 0 S g S 1, and V > 0 is constant.
Since b just belongs to L°°(QT) and may be equal to zero, we perturb to get a
1
nondegenerate equation, by setting bn = pn 2:: b+ ;, where pn is a mollifier in IR", and

f0? fn(Pn * b — b)2da:dt _<_ 52-. Consider

%=_bnAzp+uz/z in Q, 0<t<r.

on 89, O < t < r, (323)

1%
||

«b(x, T) = 51(30)-
1
Since b7, 2 a, the equation is a nondegenerate parabolic equation, and so there
exists a solution 1,0,. 6 C“(Q1).
Lemma 3.2.3 The solution of (3.23) has the following properties

(2') )0<¢n <eU(t- 7’),

(ii) / [bu IA( #4,) )_lzdxdt < C,

(iii) sup flen|2deC,

0<t_<_r

where the constant C depends only on g.

51

PROOF. Since 1 2 g 2 0, by the comparison principle, em”) 2 11),, 2 0, this

proves (i).

For (ii) and (iii), multiplying equation (3.23) by Aw” and integrating over 9 x

(t, r), we have

% / IVibn(a:, t)|2dx+ [1 / anAI/Jnl2dxds
o

+u/ fle/Jnlzdxds— - -‘/Q [Vi/1,.(x, r)|2dx.
Since ¢n(a:,r) = 9(2), V’l/Jn($,7') = Vg, we have

sup [IV‘t/Jnl2d23 g C(g)

Ogtgr Q
and
f f bnlAwanxdt s 0(9).
t (1
Therefore,
f/bnlAwn|2dxdth.
o o

Replacing ’l/) by 1b,, in( (3 21) ,and using (3. 23) we obtain

/2( z$(,'r)g ()dx—[O/z (b bn )Aibndxdt

= [92(a:,0)z/2n(0)da: + //.(AJ * z + V2)¢ndxdt.

Since

[1] 2(3— b ,,)Az/Jndxdt
0
1
32/0“ (b b"”’ )dx)2dt l(f0f3.4113,,(laxatifi

<———+0,

J5
equation (3.26) implies

[14:17, T)g(:1:)d:1:

g / |z(2:, O)|e"(t"’)d;1: +f |AJ at z + uzle”(t‘7)d:1:dt.
:1 Q.

 

52

(3.24)

(3.25)

(3.26)

(3.27)

Letting 1/ ——> 0 and 9(1) —> sign z+(:1:, r) in (3.27), we have

f(ul—u2)+dx§ / |u10—ugolda:+ // |AJ*z|d:1:dt. (3.23)
9 Q Q,-

Interchanging ul and ug yields
/ l'ltg — ulldx S / lugo — ulolda: + C/ |u2 — ulldxdt. (3.29)
o a Q.
(3.29) and Gronwall’s inequality imply the conclusion.

Remark 3.2.4 Since every classical solution is also a weak solution, this also proves

the uniqueness and continuous dependence on initial values for classical solutions.

To prove the existence of a solution to (3.18), we consider the regularized problem

and take uo E C2+7(S_2) for some 7 > 0, with Uglan = 0.

g;- = A(h€(:1:,u)) — LAJUB — y)u(y)dy in QT,
u = 0 I on ST, (3'30)

where

h€(:1:,u) = p(:1:)u(:1:) + f(u) + eu.

2 + '7
2 + ,— _
By Theorem 3.1.4, there exists a classical solution u€(x, t) E C 7 2 (QT)
These solutions are uniformly bounded:
Lemma 3.2.5 There exists a constant C, independent of e, such that
max Ine($,t)| S C (331)

QT

forallO<e§1.

53

PROOF. Multiplying equation (3.30) by u and integrating over (2, we have

zen/u +‘/'V“'2+ [(PC’L‘H f’(u))IVu|2= é/uQApCr) -/uAJ*u.

(3.32)

Using Holder’s and Young’s inequalities, we obtain

max u2 3 C1,
0951"

where Cl does not depend on e. A similar argument to that in the proof of Proposition
3.1.2 yields (3.31).
Now we can prove the existence of a generalized solution.

We need the following lemma.

Lemma 3.2.6 Ifr > O is a constant, u, v 6 IR, then we have

1
|(IUI'u - I'Ul’v)| Z -2-,|u - UI’H- (3-33)

6h‘(:1:, u)
at

[611653) +ddt/IVhe(x,u) W/i—hfa: u)(_ AJ*u) (3.34)

Bh‘E (11:, u)
at

[WaAhm g C/(c+p(a:)+f’(U))|utl

</\/(e+p )+f'u( ))((|’ut|\/€+P$) )+f( (u))

<-;-/(e+p(:v’U)+f( ))lul2+C/((6+P) )+f'(u))

Multiplying equation (3.30) by and integrating over 52, we have

Since u is uniformly bounded, and =(e + 19(3): ) + f’(u))ut, we have

 

 

(3.35)

where C does not depend on 6.

Equation (3.34) and inequality (3.35) imply

é/Ot/QI—lE—étfflut+/|Vh€($,u)|2 S fth€(;r,uo)|2+C. (3.36)

54

6
Note that the first term is positive from the expression for —. Therefore,

at

sup f |Vh€(:v.u)|"’ s c,

05th

and

T ahé ’
‘/0 /—é—_:iflut S Ci

where C does not depend on 6.

Since u is uniformly bounded, we also have

[OT/’flligfllzzflT/(€+p(xl+ f’(u))2|Ui|2
sci/L3

S Cl:

where Cl does not depend on e.
We show the dependence on e by writing 71.5. We have

6h6 (11:, ug)
llTllL2((0,T),L2(fl)) S C,

6
<
0132;); |th (x.Ue)IIL2m) _ C)

max|h€(x,u€)| S C,
QT

where C does not depend on 6.
Also from condition (Bl) and (3.40), we obtain

3(luclr‘uc)
llTllL2((0,T),L2(Q)) S C,

7‘1 <
OglngIIVUUeI “Elllmm _ C,

max luglr1+1 S C.
QT

(3.37)

(3.38)

(3.39)

(3.40)

(3.41)

It follows from (3.40) that h‘(a:,u€) is equicontinuous from [0, T] into L2(Q) with

values in a bounded subset of H162). Since H1(Q) '—-> L2(Q) is compact, by

55

Arzela-Ascoli’s lemma, there is a sequence 6,, —> 0 such that h‘n(:c,u€n) —+ v in
C([0,T],L2(Q))-

Also from (3.41), we know that |ue|rlug is equicontinuous from [0, T] into L262)
with values in a bounded subset of H1(Q). By Arzela—Ascoli’s lemma, there exists
6,. —> 0 such that lunlrlun —-> v1 in C([0, T], L2(Q)), where un 2 Men.

By Lemma 3.2.6, we have

/ Iun — uml"+1d:c S 2'] ||un|’1un — lumlrlumldx. (3.42)
n (2

Therefore, {an} is a Cauchy sequence in C ([0, T], Lrl+1(Q)) and there exists u such
that 11,, -) u in C([0, T], L'1+1(Q)).

By Lemma 3.2.5, we can also conclude that u 6 L°°(QT). Since f is differentiable,
we have h€"(:1:,u€n) —> h(x, u) in C([0, T], L’1+1(Q)).

Letting 6,, —> 0, we see that u satisfies equation (3.18), and u is a generalized
solution with initial data no 6 C§+7(Q).

For no 6 L°°(Q), choose v.0" E C§+7(s'1) such that
||u0n — U0||L1(Q) —) 0 as n —) oo. (3.43)
By Proposition 3.2.2, we have

sup Hum“) - “u(tlllLWfl) S CHUOm - uOnllL‘M): (3°44)
0937‘

where C does not depend on m, 71. Furthermore, there is a constant C1, depending
only on ||uo||Loo, such that HujllL°° S C1.

By (3.43) and (3.44), there exists u E C([0,T],L1(Q)) such that um(t) —> u in
C ([0, T], L1(Q)), clearly u is a generalized solution.

We have proved:

Theorem 3.2.7 For any T > 0 and uo E L°°(Q), if conditions (BI), (35), and (83)
are satisfied, then there exists a unique function u E C ([0, T], L1(Q)) flL°°(QT) which

satisfies equation (3.18)

56

3.3 Long term behavior in the H1 norm

In this section, we prove that there exists a continuous semigroup associated with

equation (3.1). Then we consider the boundedness in time of the solution.

Definition 3.3.1 A weak solution of (3.1) is a function
U G C([0,T], 112(9)) 0 [C([0,T], L°°(Q)) r) L2(l0)Tl,H3(Q)), U: E L2(l0,T], H“(Q)),
h(x, u) E L2((0, T), H1(Q)) such that

< u¢($,t),’l/J(.’L') > +/th(2:, u) - V¢(x)d:1:

(3.45)
= [)(m * u(-, s))z,b(a:)d:1:

for all i!) 6 H3 (9) and a.e. time 0 S t S T, where h(x, u) = p(x)u + f(u), and
u(x, 0) = uo(a:). (3.46)

Theorem 3.3.2 If (Bl) — (B3) are satisfied and no 6 L°°(Q), then there exists a

unique solution u of (3.45).

PROOF. Since uo E L°°(Q), there exists a sequence ugk) E C2+7(Q) with 7 > 0

such that

k
llué ’ — 21on —> o,

(3.47)
))uék’n... < C,
where C does not depend on k.
We consider the following problem
Bu .
at— = A(h(:1:,u) — J a): u) 1n QT,
u = 0 on ST, (3-48)

2+
2+7,—Z

By Theorem 3.1.4, there exists a classical solution 11““) E C 2 (QT), and

max Iu‘kla, t)| _<. 0, (3.49)

QT
where C does not depend on k.
Multiplying equation (3.48) by um and integrating over (2, we have

d In |u(k)|2d:1:
dt

Since Vh(:c, u(kl) - VuU‘) 2 c1|Vu(")|2 + qu - Vu‘k), where Cl is defined in condition

 

+f Vh(:1:,u(k)) -Vu(k)d:1: 2 [(AJ * u(k))u(k)d:c. (3.50)
n n

(32), from equation (3.50), we also have

sup ||u(k)||L2 S C1(T), (3.51)
0951‘
T
f / |vu<k>|2dxdtgcz(r), (3.52)
0 (I

where Cl(T), C2(T) do not depend on k.

Since by (3.49), u(k) is uniformly bounded, from inequality (3.52), we have

T T
f / |Vh(x,u('°))|2da:dt =/ / |u(k)Vp+ (p(:c) + f’(u(")))Vu(k)|2d:1:dt
o o o o

(3.53)
S C3(T)
for some positive constant C3 (T) which does not depend on k.
From equality (3.53) and equation (3.48), we also have
))ufik’nwofl 11—132)) 3 CAT). (354)

where C4(T) does not depend on k.
Inequalities (3.51)-(3.54) imply that there exist subsequence of {71"} (still denoted

by {uk}) and v, u, 9 such that
h(mwm) —‘ 'U in L2((0,T),H1(Q)),
u(k) —\ u in L2((0,T),H1(Q)), (3-55)

ué’“) —-1 g in Bums-1(a)).

58

Since “u(klllLoo S C and 21"“) —> u in L2((0,T), L2(Q)), we have

llullL0° S C,

h(x,u(kl) —> h(x,u) in L2((0,T),L2(Q)),
(3.56)

.9 : ”t:
v = h(x,u).

This implies that u is a weak solution of (3.45). Uniqueness follows from Proposition

3.2.2.

Corollary 3.3.3 If u, 6 113(5)) n L°°(Q), and if u e C([0,T],L2(o)) n
L°°([O,T],L°°(Q)) n L2([0,T],Hg(o)), with u, e L2([0,T],H-1(o)), satisfies equa—
tion (3.45), then u e C([0,T],L2(Q)) r) L°°([0,T],L°°(Q)) r1 L°°([0,T],H3(Q)), and
u, e L2([0, T], L2(o)). Furthermore, ifdimfl = 1, we also have u e L2([0,T], 112(9)),
and u e C([0,T],H1(Q)).

PROOF. Since uo E H362) fl L°°(Q), we may assume ||u3k)||H1 S C for some

constant C which does not depend on k in (3.47). Multiplying equation (3.48) by

65 (k)
M and integrating over (2, we have

(,,))u (k)
/6__,_h( x, u +2d—t _/ WW? ”2 = /@%E‘J(_AJWU°)). (3.57)

A similar argument to that in the proof of (3.34)-(3.38) in section 3.2 shows

sup /|Vh(:1:,u(k))|2 S C, (3.58)
OStST
and
T (k)
/ [(14:55:53) 3 C, (3.59)
o

where C does not depend on k.

59

Condition (B;) and (3.58)-(3.59) imply

sup / |vu<k>|2 3 Cl, (3.60)

ogth

T
/ fluik)l2 _<_ 01, (3.61)
0

Therefore, by passing to limits as a subsequence of k —+ 00, we deduce
u E L°°([0,T],H6(Q)), at E L2([0.Tl. 142(9)). and
h(x, u) E L°°([0,T],H(](Q)).

and

where 01 does not depend on k.

If dim!) = 1, multiplying (3.48) by —Au(") and integrating over 9, we have
2 dt —/ IVu(")|2d:1:+ / Ah( (3: ,—u(k))Au(k)d:1: — / AJ*u(k)Au(")dx. (3.62)
Since
Ah($, u(kl) = u(klAp + 2Vp - Vum + f"(u('°))|Vu(k)I2 + (p + f'(u(k)))Au(k), (3.63)
by Hélder’s and Young’s inequalities, and using (3.49) and (3.60), we have
/ AJ * u(klAu(k)d:1: g e / |Au(’°)|2d:c + 0(6), (3.64)

and
[Ah(x,u(k))Au(k)dx Z f(p+ f'(u(k)))|Au(k)|2d:r — C(6)/]Apu(k)|2dx

—6/|Au(k)|2d$—C(6)/|Vp~Vu(k)|2dx

—6/|Au(k)|2dx—C(6)/|f”(u(k))|2|Vu(kll4dx (3.65)
—6/|Au(k)|2d$
> (c1 —36 [)lAu(k| 2dr— C/(6)|.Vu(k)|4d:1:

60

In order to estimate f |Vu‘kll4dzr, we need the following Gagliardo—Nirenberg inequal-

ity.
llevllLs S Clllevll‘irllvllfi“ + Czll’Ulqu, (3-66)
where

j 1 j 1 m 1

—< <1_=_ _—— 1— —. .
m—“— ,3 ”+a(, n)+( a)q (367)

In (3.66),sets=4,j=0,r=2,m=1,n=1,r=2,a=Z,q=2toget
l E
((410 s CIHDvuguvng + (721)402- (3623)
Let v = Vum, then (3.68) and (3.60) give
1 § 1
IIVu"°’|I4 s 0.0444(2)); ((5743113) + 02074302 5 CHAN)”; + C. (3.69)
This and Young’s inequality imply
IIVu‘kllli s eIIIAu‘k’Hé + 0(6). (3.70)
Inequalities (3.65) and (3.70) imply that
[Ah(a:,u(k))Au(k)dx 2 (c1 — 36 — C61) / IAumlzdx — C(6, 61), (3.71)
where constant C (6, 61) does not depend on 1:.
Equation (3.62), inequalities (3.64) and (3.71) imply
1 d (k) 2 (k) 2
2d—t IVu | dx +(c1— 46 — 061) |Au I dz S C(€,€1). (3.72)

Choose 6 and 61 small enough such that Cl — 46 — C61 2 6—21, and integrate over (0,T)

to obtain
1 (k) 2 1 (k) 2 01 T (k) 2
-2- IVu (T)| (1113—5 IVu (0)] dx+ 2 IAu | d1: S C.
o

61

(3.73)

Therefore, there exists a subsequence such that uk -\ u in L2((0, T), H 2((2)). Since
u E L2((0,T),H2(f2)) and ut E L2((0,T),L2(Q)), we also have u E C([0,T], H1(Q)).
This completes the proof.

In order to prove the existence of an absorbing set, instead of (B2), we assume
(32) There exist positive constants c1, c2 and r such that a(x, u) 2 c2|u|" + Cl.
Also, we assume
(B4) There exist positive constants c3 and c4 such that a(x, u) S c3|u|' + c4.

First we study long term behavior in the L” norm.

We need the following version of Gronwall’s lemma (see Temam [38]):

Lemma 3.3.4 (Uniform Gronwall inequality) Let y be a positive absolutely continu-

ous function on (0,00) which satisfies
3' + 113/” S 6

with p >1,1/ > 0,6 _>_ 0. Then, fort 2 0, we have

y(t) s (—)P + (402 — 1)t)P — 1. (3.74)
With this we can establish

Proposition 3.3.5 If u is a solution of (3.1), then for p Z 1, we have
P+1 p+1
d (P) (1209)” ‘
uP+1dx< —‘—P+"+1+ —— r 3.75
[fill (0,200)) (pH) ( )

where d1 (p) and d2(p) are constants which do not depend on the initial data.

62

PROOF. Multiplying equation (3.1) by quI‘"‘1 and integrating over O, we obtain

fululp‘lutd1:= —/a(:r, u)Vu-V(u|u|p'1)d1:

f/AJ(1:—y)u1:( ))ulp- ldyda: (3.76)

—//AJ(1:—y)u(1:)|u|p ldydx

+ f f we — y) - Vu(z)u(z)lu|”“dydx-

 

 

Since
1
11-1 _ p+l
Lulu] utdx— p+——_1d—dt/nlu| d115, (3.77)
fa($,u)Vu-V(quIp_l)dx=p/a(x,u)Iu|p"1IVuI2d1:, (3.78)
n :1
2:1 1
WM 2 )2: (1” ’ ——Iu (HIV/u)? (3.79)
and
p+r+1 2
WM 2 (2: (“2“) |u|”+"‘|Vul2. (3.80)
from (1.32) we have
(4 +1
La(1:,u)Vu~ V(u|u|")1) pcly/IVIuIE—f |2d1:
4 p+r+1 (3'81)
[902 __ 2
—— V 2 d .
>(p(p+r+1)2/..' (u) I 4

Equations (3.76)-(3.80) and inequality (3.811) yield

4 4 p+r+1
.4. pa. 2 4 m f —2— 2
p-I-lj—t/Iul dz IIIp-l-( +1)2/IVIUI_ lldxjuif— +1. +1)2 IVIuI Ida;
[ISfAJ(1:—y)|u|"+1dyal1:-/fAJ(1:—y)u(:1:)IuIID ldyda:
+ f / VJ” - y) ' VU($)U($)Iu|p_ld3/d$-
n o
(3.82)

63

By Holder’s and Young’s inequalities, we have

f/u(x)'|u|”1VJ(z —y)-Vu(:c)dydx
__ Bil
3114/90qu 2 IVu(x)l)lu| 2 dx

 

 

 

(3.83)
S _46/ lulp‘IIVu(x)|2dx+C(e)/ Iulp+1dx
9 o
p + 1
(12+ —-1—4)2 ——2/ Mn! 2 mm + C(e) [a |u|P+1dx.
Also,
f f we — mnuomuwwydx
1’ 1
S (/ I'ttll“"’ci:v)1 +P(f(/ |AJ(:c —- y)l|u(y)|dy)l+”dx)1 + (3-84)
3 Cf |u|P+1dx,
o
and
f f we — y)|lu(r)l”+‘dyd:c s c f lulwx. (3.85)
n n n
Inequality (3.82) and estimates (3.83)-(3.85) imply
1 4 p + 1 4 p + r + 1
2+1 “1 ‘2— 2 P62 / ‘T— 22
p:——1dc:4/nlul d3: +(p—+—1)2/IVIUI 'dx+(p+r+1)2 anlul | :1:
p__ + 1
———2/0 IVIUI 2 lzdz + 0(2) / |u|”“dx.
19:61)2 a
(3.86)
Let t'" — p—gin n.(3 86), then we obtain
(2 I? + 1 4 P + 7‘ + 1
p+1d P01 2 2 p62 / __2__ 2
P+1dt/| | )2/‘IVIUI— |d$+(p+r+1)2 Qlvlul ldl‘
SC/ |u|p+ldz.
n
(3.87)

64

Therefore,

d p+r+l
— / lulp+1d2 + mp) / IVIuI 2 122x 5 .22.) / |u|”“d2 “-82
dt (2 a a
for some constants d1 (p) and d2 (p).
+r+1
19— 2(p + 1)
Set 2) = |u| 2 and 7 = —, so that (3.88) becomes
p+r+1
i / |v|7dx+d1(p) / |Vv|2dx s 212(2)) / wax. (3.89)
dt :2 n o

By Poincaré’s inequality, fa |v|2dx S C In IVvlzdx, we have

22, / |v|7dx + 21(2) / lvlidx _<. 22(2) / |v|7d22 (320)
o 9 9

where d1 (p) and d2(p) have been redefined.

Since '7 < 2, it follows from H older’s and Young’s inequalities that
2

g; [2 12m +d2(p)(/n WW2); s 22(2). (291)

where d1 (p) and d2(p) have been redefined.
The conclusion follows from Lemma 3.3.4.
Using a similar argument to that in the proof of Theorem 2.1.1 in Chapter 2, we

obtain
Proposition 3.3.6 If no 6 L°°(SZ), then
sup ||u||oo g C(uo). (3.92)
:20
Next we need to estimate ||Vu||2.

Theorem 3.3.7 Assume that u is a solution of (3.1) and conditions

(Bl), (32), (33) and (B4) are satisfied. There exists to > 0 such that ift 2 to then
sup ||Vu|l2 < C, (3.93)
tZto

where constant C does not depend on initial data.

65

PROOF. Multiplying (3.1) by a(x, mg; and integrating over (2, we obtain

[naming-Was 2/nVo(a(2:,u)Vu)a(:c,u)%t7-‘-dx+

a_u

L/QVJO: - y) - Vu(:1:)a(a:, u) at dxdy+

Bu (3.94)
LL(AJ)($ — y)u(:c)a(x,u)5t-dxdy—
Bu
[0 [n(AJxx — y)u(y)a(:z:, madxdy.
Since
Bu Bu
fnv - (a(x,u)Vu)a(x,u)-a—tdx = — fna(m,u)Vu . V(a(a:,u)-5?)dx
6n
= — /n(a(a:,u)Vp(:L‘) ' Vu—at-dm— (3-95)
,, 20a 2 6Vu
a($,u)f (u)|Vu| 5t— — a (x,u)VuF)dx,
and

if. 2 2 _/ n 292 2 6V1].
2dt na ($2u)|Vu| dx— n(a(ac,'u)f (u)|Vu| 6t +a (2:,u)Vu———at )dx, (3.96)

this yields

/ V ~ (a(x,u)Vu)a(:L',u)%%dx = — Edi a2(x, u)|Vu|2dx
” 2 t 9 (3.97)

—L(a(m,u)Vp(m)-Vuaat—udx.

It follows from (3.94) and (3.97) that

/ a(x,u)|a—u|2dx = — 11/ a2($,u)|Vu|2dx
a at a

2 dt
Bu
— (ammo) . V2522
Bu
+ [2 [2 VJ(2 — y) - v2(2)2(2, ogdxdy (328)

Bu
+/‘;/Q(AJ)(:C ——y)u(a:)a(a:, 21)-6761111?!

{332

— f9/fl(AJ)($ —- y)u(y)a(:c,u) 6t dzdy.

66

Note that

fa(a:,u)Vp(a:) - Vu@dx S 6/a(:1:,u) £92sz +C(e)/ a(x,u)|Vu|2dx, (3.99)
Q at Q at Q

Bu Bu 2 2
fnanJb: — y) - vu(x)a(:c, waded?! S ‘ [a “(x’ ""50" dx + 0“) f9 “(33’ ”MW 33,

(3.100)
021 an 2
AL(AJ)($ - y)U($)a($,U)b—tdxdy S 6110(2), u)|-a—t|2dx + C(c)/‘;a(x,u)u dz,
(3.101)

and
0a Bu 2 x
[a [wuss — y)“(y)a($:u)adxdy s e [a a(m)-a? d

+ 0(a) f «m)(/“(A00 — y)u(y)dy)2dx.
(3.102)

Choosing e = i, it follows from (3.98)-(3.102) that

$5:- n(12(Iv,u)|Vu|2d:z: g (J(];2 “(3,U)|Vu|2dx
+/{;G($,u)u dx+/‘;a($,u)(/‘; AJ(a:—y)u(y)dy) dz).
(3.103)

From condition (B4), we have

fa(x,u)|Vu|2d$ _<_ f(c3lul' +c4)|Vu|2dx
o n
r + 2 (3-104)

2 2 —
= — v 2 2d / v 2d,
[n(r'l'?) c3| |u| | x+ Qc4| ul 2:

[a(x,u)u2da: g f(cglulr +c4)u2d:c
‘1 9 (3.105)

=/03|u|'+2dz+/c4u2dz,
n n

67

and

[0005: I‘M/“(A005 — y)u(y)dy)2d$ _<_ [n(Calulr + C4)(/Q(AJ)($ — y)U(y)d3/)2d$
g C [a u(y)2dy [n (c3114r +04)da:.

(3.106)

Note that Proposition 3.3.5 implies

fc3|u(:c,t)|’+2da:+/c4u(:1:,t)2d:1: S C (3.107)
n n

for t Z to and for some constant C which does not depend on initial data and

fu(y)2dy/(c3|ul' + c4)dx 5 C (3.108)
n n

for t 2 to and for some constant C which does not depend on initial data.

Also, inequality (3.87) and Proposition 3.3.5 yield
t+1 2 2 :13 t+l
/ / (—) |V|u| 2 |de + f / c4|Vu|2dx g C (3.109)
t n 7' + 2 t n
for t 2 to and for some constant C which does not depend on initial data,

and

t+l t+l
/ [(1203, u)|Vu|2dxds Sf f(c3|u|'+c4)2|Vu|2dxds
t n z n
t+l
g f f c5(|V|u|'+1|2+|Vu|2)dxds (3110)
t o

S C
for t 2 to and for some constant C which does not depend on initial data.

It follows from (3.103)—(3.108) that

r + 2
- 02(x,U)qul2da: 5 Cl(/ |V|u| 2 |2da: +/ 1%de + 02) (3.111)
dt 0 n n
for some constants C1 and C2 which do not depend on initial data and for t 2 to.

For to < t < s < t+ 1, integrating inequality (3.111) between 5 and t+ 1, we

obtain

68

/ a2(:1:,u(x,t + 1))IVu(2:, t + 1)|2dx — / a2(a:, u(x, s))IVu(a:, s)|2dx
n 51

HI 1' + 2 (3-112)
3 00/ [f9 0710000 2 Izdx + /n (Vuwwczx + 02100.

Integrating (3.112) from t to t + 1 with respect to s, we have

/ “2(5’3 “(17 t+1))|Vu(a: t+1)l2dx_</t+1/{;a2(,1: u( 2:, s))qu(a:, s)|2dxds

r+2

t+1 t+l
+C1( (/ f [L |V|u( a: ,u)| 2 —|2dm+/ |Vu($,u)|2dx+C2]dpds).
o
(3.113)
By (3.109), (3.110), (3.112) and Fubini Theorem, we obtain
/a2(a:,u(:z:,t+1))IVu(:1:,t+1)|2d:I: S C (3.114)
n
for some constant C and for t Z to.
Condition (B3) and (3.114) yield
/ |Vu($,t+1)|2d:c S C (3.115)
n

for t 2 to and for some constant C which does not depend on initial data.

3.4 Existence of a global attractor

In this section, we prove that there exists a global attractor for weak solutions in
some metric space for n = 1.
Let H be a metric space, S (t) (t 2 0) be a family of operators, which map H into

itself and enjoy the usual semigroup properties

S(t+s)=S(t)-S(s) Vs,t20,
(3.116)

5(0) = I (Identity in H),

69

S (t) is a continuous operator from H into itself for all t Z 0. (3.117)

The following lemma may be found in [38].

Lemma 3.4.1 Assume that H is a metric space and that the operators S (t) satisfy
(3.116) and (3.117) and the following condition:

For every bounded set 8 there exists to which may depend on B such that
UtZ¢OS(t)B is relatively compact in H.

Assume that there exists an open set U and a bounded subset B of u such that
B is absorbing in u. Then the w-limit set of B, A = w(B), is a compact attractor
which attracts the bounded sets of u ( for the inclusion relation). Furthermore, if H

is a Banach space and U is convex and connected, then A is connected too.

Let X = L°°(Q) with the metric from L1(Q). From Theorem 3.3.2 and Corollary
3.3.3, there exists a semigroup S (t) associated with equation (3.1).

From Proposition 3.3.6, we have that S(t) maps L°°(Q) to L°°(Q).

Since for n = 1, H162) 9-) C"(Q) is compact, from Theorem 3.3.7, we see that
There exists an absorbing set in H 1 n X.

Therefore, by Lemma 3.4.1, we have the following theorem.

Theorem 3.4.2 For n = 1, if conditions (B1), (32), (83), and (B4) are satisfied,
then the semigroup associated with (3.1) possesses an attractor A C H1(Sl) flX which

is maximal and compact.

70

CHAPTER 4

The Cauchy problem and steady
state solutions for a nonlocal

Cahn-Hilliard equation

4.1 The Cauchy problem for a nonlocal Cahn-
Hilliard equation

We consider the following equation:

at (4.1)

{Q = A(¢(u) -— J *u) in Rn X (0,T))
u(x,0) = u0($):

where <p(u) = u + f (u), f is bistable (e.g. f (u) = u(u2 — 1)), * is convolution.

For T > 0, let QT = IR" x (0, T). We make the following assumptions:
(Cl) f E C2+5(1R) and (5(a) Z c for some positive constants c and 6,
(C2) J E C2+B(R"), AJ 6 L1(R")fl L°°(IR"), and fan J = 1.

First, we prove the uniqueness and continuous dependence of solutions on initial

data. We have

71

Proposition 4.1.1 Let u.- (i = 1,2) be two solutions of equation (4.1) with initial
data um (i = 1, 2). If conditions (Cl) and (C2) are satisfied, if u.- E C ([0, T], L1(lR"))fl
L°°(QT), and if U30 6 L1(lR") fl L°°(lR") (i = 1, 2), then

sup / |u1 — u2|dx S C(T)/ lulo — UgoldIE (4.2)

ogth

for some constant C (T)

PROOF. For any 1' E (0, T), and w E C2'1(Q,), with w = O for |x| large enough,

after multiplying (4.1) by w, and integrating over [0, r] x R", we have

[a u,(x,r)w(x,r)dx = [n u,-(x,0)w(x,0)dx

+ [7] (“Mt + cp(u,-)A1/1)dxdt — [7 wAJ * uidxdt.
0 n 0 R"

(4.3)

Set 2 2 ul — U2, 20 = um — U20, then (4.3) gives

[n z(x,r)i,b(x, r)dx = f" zo(x)1/2(x, 0)dx
+ [T]. Z(.’L‘,t)(I/)t + b(x,t)A1,b)dxdt — [T wAJ * z(x, t)dxdt,
0 " 0 Rn
(4.4)

where
(p('U.1) _ (p012) for U1 ¢ "’2,

b(x, t) = “I - “2 (4.5)
<p'(u1) for u1 2 U2.

 

Let g(x) E C8°(R") have compact support, 0 _<_ g(x) S 1, and take A > 0. We will
choose w, above, to satisfy certain conditions. First, consider the following final value

problem on a large ball 83(0)

96% = —b(x,t)Az/) + Ad for |x| < R, 0 < t < r,
1)):0 on |x|=R,0<t<r, (4'6)
1/1($,T) = g(x) lirl S R-

72

There exists a unique solution of (4.6) w E C2'1(BR(0) x (0,T)) which satisfies the

following properties:

0 s w s eW-T), (4.7)
f / b(x, t)|Aw|2dxdt g C, (4.8)
0 33(0)
sup / |V¢|2dx S C, (4.9)
03%" 33(0)

where the constant C only depends on g.
In order to extend w to be zero outside of 83(0), we define {R E C§°(IR") such

that
O S 6R S 1,

£R===1 if kE|<:}2-Iq

§R===0 if|afl >*}2-‘%,

(4.10)

IV€R($)Ia |A§R($)l S C'

for some constant C which does not depend on R.
Let 7 = (Rib, where w satisfies (4.6) in BR(0) and is zero outside. Using 7 instead
of w in (4.4), we have

[Rn z(x, T)g€Rdx — L {a(x)zo(x)2/J(x, 0)dx + f/fAJ at z — Az)§m/;dxdt
=/ 144,020, t)(2V€R-Vz/2+vA§R)dxdt (4,11)
0.

E C (z, B).
Since ul and U2 belong to L°°(QT), and since b is positive, from estimates (4.7)-(4.9)

and (4.10), we have
1002.101 _<. f / (blul-U2|((2|V€RIIV¢|+libllAézzl»
0 BR\BR-1

30/] b(lull+|ug|)(|V7,/;|+1)dxdt (4.12)
0 BR\BR—l

g c/ / (|u1|+ (0204444.
0 BR\BR-l

73

Since ul and U2 belong to L1(QT), letting R —> 00 we have C(z, R) —> 0.

This implies

/ 2(x, r)g(x)dx g |z0(x)|e_’\’dx +/ / (|AJ * z — Adam—”dxdt. (4.13)
n o 11

IR"

Letting A —+ 0 and g(x) —> sign 2+ (x, r), we obtain

/ (ul — u2)+dx S / lulo — ugoldx + C/ |u1 — u2|dxdt. (4.14)
1; IR“ 0 R"

Interchanging ul and 212 yields
T
|u1 — ugldx S / IU10 — UgoldIE + C/ lul — UgldIL'dt. (4.15)
Rn n o Rn

Inequality (4.2) follows from (4.15) and Gronwall’s inequality.

Next we prove the existence of a solution to equation (4.1).

Theorem 4.1.2 For any T > 0, if uo(x) E C3+B(R"), and if (0 and J satisfy as-

sumptions (C1) and (C2), then there exists a unique solution of (4.1) which belongs

2+3 2——
toC ’ 2 (QTlnL1(QT)nL°°(QT)'

PROOF. Since uo(x) = 0 for |x| large enough, we consider

all: Acp( (u )— J*u) in BR(0) x (0,T),

u(x,t) = 0 on 333(0) x (0,T), (4.16)
“(it 0) — 140(3).
From Theorem 3.1.5 in Chapter 3, there exists a unique solution of (4.16) u(x, t) 6
2+ 3— 2 + B
C 2 (33(0) X (0,T))-

Let u(x, t) = veIt in (4.16), then we have

etvt + vet = <p'(u)e‘Av + <p"(u)|Vv|262t — e‘AJ * v. (4-17l

74

1
Multiplying (4.17) by v and using uAv = 5A1? - |Vv|2, we obtain

$022); + v2 = é—cp’(u)Av2 + égo"(u)Vv - sze‘ — <,o'(u)|V'U|2 — vAU * 1)). (4-18)

If there exists (Po,to) 6 83(0) x (0, T] such that u2(Po,to) = maxvz, then

Av2(Po,to) S O, Vu2(Po,to) = 0, Vv(Po,to) = 0, vf(Po,t0) 2 0, and (4.18) yields

U2(P0, to) S — AJ(P0 - y)'U(y, t0)dy'U(Po, to). (4.19)
BR

This yields

max M S M lv(y, to)ldy (4-20)
BR

for some constant M which does not depend on R.
Since u = 0 is also a solution of (4.16) with initial data uo = 0, by Proposition

3.2.2 in Chapter 3, we have

/ |u(x, t) — Oldx g C(T) lug — Oldx (4.21)
BR

33
for some constant C' (T) which does not depend on R.
Inequalities (4.20) and (4.21) imply
max |u| g C(T) / luoldx. (4.22)
BR
Since no 6 L1(1R"), we have

max lo] 3 B(T) (4.23)

for some constant B(T) which does not depend on R.

This yields

max |u| S B(T)eT (4.24)

75

for some constant B (T) which does not depend on R.

Now we have proved the solution of (4.16) is uniformly bounded, i.e.,

max |u(x, t)| < C
BRX[0,T]

for any R > 0, where C does not depend on R.

A similar argument to that in the proof of Theorem 3.1.3 in Chapter 3 yields
HURHMB s can T) (4.25)

for any R > K E constant, where ml is a solution of (4.16) in BR x (0, T) and C(K, T)
is a constant which does not depend on R (H - ”2+5 is a Holder norm defined in [30]).
By employing the usual diagonal process, we can choose a sequence {R,-} such
that “R“ Dug“ and Dzug‘. converge to u, Du, and Dzu pointwise, and u satisfies
equation (4.1). From (4.21) and (4.24), we also have u E L1(QT) fl L°°(QT).

Uniqueness follows from Proposition 4.1.1.

4.2 Steady state solutions for a nonlocal Cahn-
Hilliard equation

In this section, we consider the following equation:

/:(:—y )dyu( )—-/QJ(y)u(zr— y)u=dy+f() CinQ,
(4.26)

where Q is a bounded domain, C is a constant. The case when 9 2 IR or R" has been

treated by others (see [10], [12], [16], [17] and references therein).

Proposition 4.2.1 Suppose Q C R" is a closed and bounded set, J (:12) Z 0 and is
continuous on R", supp I) B5(O) for some positive constant 6, and f is nondecreasing.

Then the only continuous solution of equation (4.26) is zero.

76

PROOF. Without loss of generality, we assume that f (0) = 0. If f (0) 75 0, we
may use f (u) — f (0) instead of f (u) in (4.26).

Case 1: C s O in equation (4.26).

If the conclusion is not true, since f uda: = O, and u is continuous on (2, there
exists Po 6 0 such that u(Po) = max u(x) > 0.

Let A = {y 6 Q|u(y) = maxu(a:)}.

We claim: There exist Po 6 6A and r > 0 such that K := (Q \ A) D B,(Po) has
positive measure. If this is not true, we have meas(Q\A) = O. This and u(x) = max u
on A imply fn u = A u > O. This contradicts fn u = 0.

Since SuppJ 3 85(0) implies SuppJ(Po — -) D 85(Po), choosing 7" 1 = min{6,r}

gives
meas(K 0 BT, (Po)) > O, (4.27)
J(Po — y) > 0 on K H B,,(Po), (4.28)
and
u(Po) — u(y) > 0 on K D B,1 (P0). (4.29)

Inequalities (4.27)-(4.29) imply

[a J (Po - y)(U(Po) - n(y))dy _>. [K J (Po - y)(U(Po) - u(s))aly > 0- (4-30)

This and f (u(Po)) _>_ 0 imply

[J(Po — y)U(Po)dy - / J(Po — y)U(y)dy + f(U(P0)) > 0. (4-31)
9 Q

contradicting (4.26).

Case 2: C > 0 in (4.26).

In this case, taking P0 such that u(Po) = minu < 0 leads to a contradiction in a
similar way.

If f’ (u) changes sign, we make the following assumptions:

77

(E1) 52 =(—1,1)1rdimo = 1, o = (—1, 1) x o' ifdimQ >1.

(E2) J(III) = J(IJJI), J(III) Z 0. and

MZsup/J(x—y)dy2inf/J(z—y)dy2m>0
Q 1160 {2

2:60

for positive constants M and m.

(E3) f E 01(R), f is odd, f(1) = 0, there exist 6 > 0 and a 6 (0,1) such that
f’(:42) Z 6 on [a,oo), and f(-—a) Z (1 +a)M.

(E4) 0 = o in (4.26).

Remark 4.2.2 Condition (E3) implies that f(—1) = 0, f’(u) 2 6 on (—00, —a], and
—f(a) 2 (1+ a)M.

Let j(:r) = fa J(a: — y)dy. From (E2), we have
m < j(:z:) < M. (4.32)

Dividing equation (4.26) by j (2:), we consider

__1_ ._ .. f_<u_<4>_>___
u(x) ,(x) f.“ y) <y)dy+ 0,

fl”) (4.33)
/ u(x)d:c = 0.
9

Theorem 4.2.3 If assumptions (E1) — (E4) are satisfied, then there exists a solution

of equation (4.33) such that

Zafora: E M15(0,1)X (2',
u(x) (4.34)
S —afor:z: 6 M2 E (—1,0) x 52’.

Moreover, we have

—1 S u(m) _<_ 1. (4.35)

78

PROOF. Following the idea in [10], let
B = {u E L°°(Q)| u(—:121,:r') = —u(:1:1,:1:'), u(x) E [a, 1] for :12 6 M1}.

The definition of B implies that u(x) E [—1, —a] for x 6 M2.
Define

Tu<z>=u<x>+h1——. 14/,“ (ya:— My)dy-U($)-j(—)(iw-())l

We want to show T : B ——> B is a contraction map if h is small enough.
In fact, since j (=2:) In J (:12 — y)dy, with assumption (E2), we have j(—:1:1,:c’) =

j(:1:1,:c’). And if u(x) E B, we have:

I I h 1 I I I I
T(u(—:1:1,:c))— 14—501,”? ) + m] n' “—331 — 211,33 — y )U(y1,y)dy1dy
h

j(_$lvxl)

h l
= _ I _ __ I_ I Id d I
u(xl,:1:) j——($1,$’)/ 11"“ $1+zl,x y)u(zl,y) zl y

— hu(- 111,33 33)+ f(u(—311$’))

+ hu(:1:1,:1: )— “M17113,”

1&1 {1)
= ”4331,17 ‘77—“, 1,4) [1111' J(III —Zl,$' — y')u(21,y')dzldy'
+ hu($1,x') — )m f(u(:z:1,x' ))

 

= —(u(:1:1,a:’ fl:1$h$')/11./J( —21,:1:'— y')u(zl,y')dzldy'

— hu(a21,x') + “7431117)”

Jth’)
.—_—_ —-T(u(:c1,x')).

usm

Choose h small enough such that

heij0<1—h msn
A”

for u E [—1,—a] U [a, 1] and x E Q.

79

This implies that u — h[u + 3(1—1'5f(u)] is increasing in u on [a, 1]. Since u(y) Z a

for y 6 M1, and u(y )> —1 for y 6 M2, we have for a: 6 M1

MET/J(x—yu)( y)——dy+u—h[u+ j(1:1:) f(u)]

W_h:(—x)/‘;J(z—y y)u(y)dy+a—ha—hmf(a)
1

zhm MlJ(:1:—y yuy)()dy+h—,(1—)- M2J(:r—y m)()dy+a-ha—hj(—x)f(a)
I l
>haj(—$) MlJ(a2—y)dy—hm M2J(x-y)dy+a—ha—hj—(:I—)f(a)
l 1
—a—haJ—(—5 M2J(z—y)dy—hm M2J(:c— y)dy—j(—)f(a)
>a-.—(h-)[(1+a) M J(x- y)dy+f(a)l
2a
(4.38)
by (E3)
Also
Ta:()= h——)/nJ(a:—yy)u( (y)dy+u—h[u+j—1(15f(u)]
g hTé’S/n“ x—y)u(y)dy+1—h—hJ—,(—5f(1) (439)
$1
forxeMl.

Estimates (4.36)-(4.39) imply that T maps B to B.

80

1
For u, v E B, choosing h small enough so that 0 < 1 — h(l + 6M) < 1, we have

IITu — Tun... =11<u — v) + 32—) A J(4 — y)(uo) — u(y))dy

- («u(x) — u(x)) - 7350(4) — f(v))||..

hf’(0u+(1—0)v) _v _h_ m_ u _v
1(4) )(u )+ 1(4) / J( y)( (y) (y)u(/H...

5(1— 141 +6-1[7))llu- vlloo+h||(u-v)||..

so - hafimu — v11...

 

=||(1 - h -

(4.40)

where 0(3) 6 (0, 1) for all x E Q. Here we used (E3) and the fact that for any :1: E Q
either u(x), u(x) Z a or u(x), u(x) g —a.
Therefore, T is a contraction map from B to B and so there exists a unique point

u E B such that Tu = u. Estimates (4.35) follows from the definition of B.

Remark 4.2.4 If we just consider the solution to

/ J(z — y)u(:c)dy — / J(x — y)u(y)dy + f(u) = 0 in Q (4.41)
a 11

without the condition In udx = 0, then the conditions that f is odd and J (:12) = J ([22])
are not necessary. In this case, we can use a similar method to that in [10] to prove the
existence of a discontinuous solution under conditions (E2), (E3)’, and (E4), where

(E3)’: f 6 Cl(R), f(—1) = f(c) = f(1) = 0 for c 6 (—1,1), there exist 6 > 0,
a 6 (0,1), b E (—l,0) such that f’(:z) Z 6 on [a,oo) U (—oo,b), f(a) _<_ -(1+ a)M,
and f(b) 2 (1+ b)M, where M is defined in (E2).

81

BIBLIOGRAPHY

82

[1]

l2]

[3]

[4]

l5]

[6]

[7]

l8]

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