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Nonlinear AnalysisR e c e n t A d v a n c e s i n
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Editors
Michel ChipotUniversity of Zurich, Switzerland
Chang-Shou LinNational Taiwan University, Taiwan
Dong-Ho TsaiNational Tsing Hua University,Taiwan
Hsinchu, Taiwan 2025 November 2006
P r o c eed ing s o f t he
Nonlinear AnalysisR e c e n t A d v a n c e s i n
N E W J E R S E Y L O N D O N S I N G A P O R E BEIJ ING S H A N G H A I H O N G K O N G T A I P E I C H E N N A I
International Conference on Nonlinear Analysis
world cientific
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British Library Cataloguing-in-Publication Data
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For photocopying of material in this volume, please pay a copying fee through the Copyright
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ISBN-13 978-981-270-924-0
ISBN-10 981-270-924-X
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Copyright 2008 by World Scientific Publishing Co. Pte. Ltd.
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Printed in Singapore.
RECENT ADVANCES IN NONLINEAR ANALYSIS
Proceedings of the International Conference on Nonlinear Analysis
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PREFACE
This book collects different articles of research on nonlinear analysis. These
results were presented in the framework of the International Conference on
Nonlinear Analysis which took place in the National Center for Theoreti-
cal Sciences at the National Tsing Hua University in Hsinchu, Taiwan in
November 2006.
This conference which was a very successful event has covered a large variety
of topics in partial differential equations that the reader will discover by
looking shortly at the content.
We would like to thank the National Center for Theoretical Sciences for
its support in organizing this meeting. Our appreciation goes also to Mrs
Schacher who arranged the final version of the whole book and to Ms Zhang
and World Scientific for their editing work.
Zurich, November 2007
Michel Chipot
Chang-Shou Lin
Dong-Ho Tsai
v
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CONTENTS
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
Toyohiko Aiki and Jana Kopfova
A mathematical model for bacterial growth
described by a hysteresis operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Nelly Andre and Itai Shafrir
On a vector-valued singular perturbation problem
on the sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
H. Brezis, M. Chipot and Y. Xie
On Liouville type theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Takesi Fukao
Free boundary problems of the nonlinear heat equationscoupled with the Navier-Stokes equations . . . . . . . . . . . . . . . . . . . . . . . 67
Yoshikazu Giga, Yukihiro Seki and Noriaki Umeda
Blow-up at space infinity for nonlinear heat equations . . . . . . . . . . . 77
Stuart Hastings, David Kinderlehrer and J. Bryce McLeod
Diffusion mediated transport with a look at motor proteins . . . . . 95
Kota Ikeda and Masayasu Mimura
Mathematical study of smoldering combustionunder micro-gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
Tetsuya Ishiwata
Motion of non-convex polygons by crystalline curvature
and almost convexity phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
Yoshitsugu Kabeya and Hirokazu Ninomiya
Fundamental properties of solutions
to a scalar-field type equation on the unit sphere . . . . . . . . . . . . . . . 135
vii
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viii
Risei Kano, Nobuyuki Kenmochi and Yusuke Murase
Existence theorems for elliptic quasi-variational inequalities
in Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
Kazuhiro Ishige and Tatsuki Kawakami
Asymptotic behavior of solutions
of some semilinear heat equations in RN . . . . . . . . . . . . . . . . . . . . . . . 171
Ken Shirakawa, Masahiro Kubo and Noriaki Yamazaki
Well-posedness and periodic stability for quasilinear parabolic
variational inequalities with time-dependent constraints . . . . . . . . 181
Marcello Lucia and Mythily Ramaswamy
Global bifurcation for semilinear elliptic problems . . . . . . . . . . . . . . 197
Makoto Narita
Global existence and asymptotic behavior
of Gowdy symmeric spacetimes with nonlinear scalar field . . . . . . 217
J. Harterich and Kuni Sakamoto
Interfaces driven by reaction, diffusion and convection . . . . . . . . . 225
Shota SatoLife span of solutions for a superlinear heat equation . . . . . . . . . . . 237
Noriaki Yamazaki
Optimal control problems
of quasilinear elliptic-parabolic equation . . . . . . . . . . . . . . . . . . . . . . . 245
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A MATHEMATICAL MODEL FOR BACTERIAL GROWTHDESCRIBED BY A HYSTERESIS OPERATOR
TOYOHIKO AIKI
Department of Mathematics, Faculty of Education, Gifu University, Gifu,
501-1193, Japan
JANA KOPFOVAMathematical Institute, Silesian University in Opava, 746 01 Opava,
Czech Republic
Abstract: We consider a mathematical model for a bacterial growth in a Petri
dish. The model consists of partial differential equations and a hysteresis operator
describing the relationship between some variables of the equations. We consider
the hysteresis relation to be represented by a completed relay operator. We prove
the existence of a solution to the system by using the standard approximation
method.
1. Introduction
In this paper we consider the mathematical model for a bacterial growth
in a Petri dish. The model was proposed by Hoppensteadt-Jager [2],
Hoppensteadt-Jager-Poppe [3]. Let h be the histidine (amino acid) con-
centration in a Petri dish,b be the size of the bacterial population, g be the
concentration of the growth mediums buffer and v be the metabolic activ-
ity of bacteria. In this model the metabolic activity of bacteriav plays a
very important and interesting role, and is decided by a hysteresis operator
with input functions h andg. We denote by RN, N2, the bounded
domain with the smooth boundary and putQ(T) = (0, T) , 0< T
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bacteria. Then, it holds that
h
t =Dhh vb,
g
t =Dgg vb in Q(T), (1.2)
where Dh and Dg are diffusion coefficients of the histidine and the buffer,
respectively, and and are positive constants.
The problem was also investigated by Visintin in [7] and Chapter 11
in [8]. Here, we discuss the mathematical description of the metabolic
activity along Visintins idea (Chapters 6 and 11 in [8]). Let = (h, g) be
a continuous function with 0 1 on [0, )[0, ), andM0={(h, g)R2 : (h, g) = 0}, M = {(h, g) R2 : 0 < (h, g) < 1}, M1 = {(h, g)
R2 : (h, g) = 1} (see Figure 1). Next, we denote by r the corresponding
relay operator. For any u C([0, T]), 0 < T
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is, (h, g) M0, then the bacteria can not grow (v= 0). Otherwise ((h, g)M), the metabolic activity is decided by the historical data. Hence, we
have obtained the system consisting of (1.1), (1.2), (1.3), the homogeneous
Neumann boundary condition (1.4) and initial condition (1.5):h
=
g
= 0 on (0, T) , (1.4)
where is the outward normal vector to the boundary;
b(0, x) =b0(x), h(0, x) =h0(x), g(0, x) =g0(x), (1.5)
where b0, h0 and g0 are given functions.
The mathematical treatment of the relay operator is difficult, because
this operator is closed, we therefore follow the idea of A. Visintin and replace
it with a completed relay operator.
considered the system with the completed relay operator instead of the
relay operator in [7]. Now, we introduce the completed relay operator,
which is denoted by k. For any = (1, 2) R2 with 1 < 2, u
C([0, T]) and [1, 1], w k(u, ) if and only if w is measurable in
(0, T),
ifu(t)=1, 2, then w is constant in a neighbourhood oft,ifu(t) =1, then w is non-increasing in a neighbourhood oft,
ifu(t) =2, then w is nondecreasing in a neighbourhood oft,
(1.6)
w(0)
{1} ifu(0)< 1,
[1, ] ifu(0) =1,
{} if1 < u(0)< 2,
[, 1] if u(0) =2,
{1} ifu(0)> 2,
(1.7)
w(t)
{1} ifu(0)< 1,
[1, 1] if 1u(0) 2,
{1} ifu(0)> 2.
It is clear that w BV(0, T). Thus the multi-valued operator k :C([0, T]) BV([0, T]) is well-defined.
Moreover, we assume that the bacteria diffuse:b
t =Dbb+cvb in Q(T),
b
= 0 on (0, T) , (1.8)
where Db is a positive constant.
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Here, we give a brief list of previous works related to the above systems.
In [2, 3] numerical results for the system (1.1) (1.5) were obtained. Also,
Visintin proved the existence of a solution to the following system in [7]:
For i= 1, 2
uit Diui+cis= 0 in Q(T),
s= w+12 on Q(T),
w k((u1, u2), 2s0 1) in Q(T),ui = 0 on (0, T),
ui(0) =u0i,
bt =cs in Q(T),
b(0) =b0 in Q(T),
where 1 = 0, 1 = 1, c, Di and ci are positive constants, i = 1, 2, s0 and
b0
are initial values ofs and b, respectively. Although there are not direct
relations with bacterial growth, Visintin studied the following problems:
ut u+w= f in Q(T),
w k(u, w0) in Q(T),
u= 0 on (0, T) ,
u(0) =u0,
ut u+w=f in Q(T),
w r(u, w0) in Q(T),
u= 0 on (0, T) ,
u(0) =u0,
(1.9)
and
(u+w)t u= g in Q(T),
w k(u, w0) in Q(T),
u= 0 on (0, T) ,
u(0) =u0,
(1.10)
where f and g are given functions on Q(T), w0 is the initial function. For
(1.9) the existence theorem was established and for (1.10) the existence and
uniqueness were proved in [7, 8].
In the present paper we have slightly modified Visintins result. Pre-
cisely, our main result of this paper is to prove the existence of a solutionto the system P :={ (1.8), (1.2), (1.11), (1.4), (1.5)}:
v= w+ 1
2 , and w k((h, g), w0) on Q(T). (1.11)
In the next section we show the assumptions for data and precise state-
ment of the main result, after we provide the definition and basic properties
on the completed relay operator due to [8]. At the end of this paper we
prove the existence theorem.
2. Statement of the result
First, we provide a weak and space structured systems of completed
relay operator due to [8]. To do so we introduce a function space
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and the following two continuous functions and . We denote by
L2w(; BV(0, T)) the set of all functions w : BV(0, T) satisfying
x [0,T] f dw(x) +T
0 w(x)dt is measurable on for any (, f)
R C0([0, T]) and w(x)BV(0,T) L2(), where[0,T] f dw(x) is the
Lebesgue-Stieltjes integral, C0([0, T]) ={f C([0, T]); f(0) = f(T) = 0},vBV(0,T) =
T0
|v|dx+ Var(v) and Var(v) stands the total variation ofv on [0, T]. Clearly, L2w(; BV(0, T)) is the Banach space with norm
([0,T]
w(x)2BV(0,T)dx)1/2. For any = (1, 2) R2 with 1 < 2 we
set () = ( 2)+ ( 1), () = () for R.
Definition 2.1. For any u L2(; C([0, T])) L2w(; BV(0, T)) and any
L
(;[1, 1]) w k(u, ) if and only ifw L2w(; BV(0, T)), (1.7)
holds a.e. in , and w L2w(; BV(0, T)),Q(T)
w((u) )dxdt
Q(T)
(|(u)| ||)dxdt for L1(Q(T)),
dx
[0,T]
((u) (z))dw 0 for z L1(; C([0, T])).
Next, for simplicity we can rewrite P as follows:
bt b= cbv on Q(T), (2.1)
uit iui+cibv= 0 on Q(T), i= 1, 2, (2.2)
v = w+ 1
2 , w k((u1, u2), w0) on Q(T), (2.3)
b
= 0,
ui
= 0, i= 1, 2, on (0, T) , (2.4)
b(0, x) =b0(x), ui(0, x) =u0i(x), i= 1, 2, (2.5)
where , c, i, ci, i = 1, 2, are positive constants, v0, b0, u0i, i= 1, 2, are
initial functions and w0 = 2v0 1.
We define a solution of the system (2.1) (2.5) in the following way.
Definition 2.2. We call that the quadruplet {b, u1, u2, v} of functions, b,u1, u2 and v on Q(T), T >0, is a solution of P on Q(T) if and only if the
conditions (C1), (C2), (C3) hold:
(C1) b S(T) := W1,2(0, T; L2()) L(0, T; H1()), ui S(T),i= 1, 2, v L(Q) L2w(; BV(0, T)), and (2.3) holds.
(C2) (2.1), (2.2) and (2.4) hold in the usual weak sense.
(C3) (2.5) holds for a.e. x .
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In order to give a main result of this paper as Theorem 2.1 we prepare
the following notations.
f
(r) = 1 for r 1,
0 for r < 1, f(r) = 1 for r > 2,
0 for r 2.
Theorem 2.1. Let T > 0, = (1, 2) R2, , 1 and 2 be positive
numbers, c, c1 and c2 be real numbers, and b0 H1(), u0i H1(),
i = 1, 2, v0 L(). Also, assume that C(R2) W1,(R2), and
the compatibility condition for the initial data, f((u01, u02)) w0 f ((u01, u02)) on , where w0 = 2v0 1. Then there exists at least one
solution{b, u1, u2, v} of P on Q(T).
3. Completed relay operator
In the proof of Theorem 2.1 we shall approximate the completed relay
operator k by play operators. The aim of this section is to introduce the
approximation ofk and to show a useful property for the approximation.
First, we show a boundedness of the operator k.
Lemma 3.1. (cf. [8] ) Let= (1, 2) with 1 < 2 and L(0, 1). If
L2(; BV(0, T)) andw k(, ), then
(
w(x)2BV(0,T)dx)1/2
2
2 1(
(x)2BV(0,T)dx)1/2+(T+2)||1/2.
Next, for any >0 we put
f(r) =
1 for r 1,1 (r 1) + 1 for1 2 < r < 1,
1 for r 1 2,
f(r) =
1 for r 2+ 2,1 (r 2) 1 for 2 < r < 2+ 2,
1 for r 2,
and define by I(; ) : L2() (, +] the indicator function of the
interval [f(), f
()] for R, that is, for z L2()
I(; z) =
dx, (x)
0 ifz(x) [f(), f
()], otherwise,
for a.e. x .
Here, we consider the following ordinary differential equation,
wt+I(; w) 0 on [0, T], w(0) =w0, (3.1)
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where I(; w) is its subdifferential, is a continuous function on [0, T]
andw0is the initial value ofw. If W1,2(0, T; L2()), then there exists a
unique solution w W1,2(0, T; L2()) of (3.1). Also, it is well-known that
in this case the mapping fromtowcorresponds to the play operator. Thisfact was already pointed out in [8] and used for the analysis for the system
including play and stop operators in [8, 6, 4, 1, 5]. Moreover, we can obtain
the next lemma concerned with the relationship between the completed
operator and the play operator. We note that we can find the assertion of
the lemma as a comment in [9] .
Lemma 3.2. LetT >0, = (1, 2) R2 with1 < 2, >0, = (1
, 2+ )R
2
, W1,2
(0, T; L2
()), andw0 L
() withf((0))w0 f ((0)) a.e. on . Then, w k( w, w0) if and only if
w W1,2(0, T; L2()) is a solution of
wt(, x) +I((, x); w(, x)) 0 on [0, T] for a.e. x ,
w(0) =w0 a.e. on.
4. Proof of Theorem 2.1
Throughout this section we use same notations used in the previous sec-
tions.
First, we consider the following approximate problem P for 0 < 0 :=
214 :
bt b=cbv on Q(T), (4.1)
uit iui+cibv = 0 on Q(T), i= 1, 2, (4.2)
v =
w+ 1
2 , wt+I((u1, u2); w) 0 on [0, T], (4.3)w(0) =w0 on , (4.4)
b
= 0,
ui
= 0, i= 1, 2, on (0, T) , (4.5)
b(0, x) =b0(x), ui(0, x) =u0i(x), i= 1, 2. (4.6)
The first lemma is concerned with the well-posedness for P.
Lemma 4.1. Let T > 0, 0 < 0. If all assumptions of Theo-rem 2.1 hold, then there exists one and only one solution{b, u1, u2, w}
of P in the following sense: b S(T), ui S(T), i = 1, 2, w W1,2(0, T; L2()); (4.1) (4.6) hold in the usual sense.
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This lemma can be proved in the similar way that of Section 5 in [4] so
that we omit its proof. Here, we give the proof of Theorem 2.1.
Proof. For 0 < 0 let {b, u1, u2, w} be a solution of P. Here, forsimplicity we put = (u1, u2). Then, since w [f(), f()]
a.e. on Q(T), we have |w| 1 a.e. on Q(T) so that |v| 1 a.e.
on Q(T). Hence, by the standard argument for parabolic equations we
observe that {b}, {ui}, i = 1, 2, are bounded in L(0, T; H1()) andW1,2(0, T; L2()).
Next, Lemmas 3.1 and 3.2 imply wk( w, w0) and
( w
2
BV(0,T)dx)
1/2
2
2 1(
w2BV(0,T)dx)
1/2 + (T+ 2)||1/2
2
2 1(
2BV(0,T)dx)
1/2 +
2 1(
w2BV(0,T)dx)
1/2
+(T+ 2)||1/2.
On account of the assumption 0< 0 we have
12
(
w2BV(0,T)dx)
1/2
2
2 1(
2BV(0,T)dx)
1/2 + (T+ 2)||1/2
2
2 1(
(
T0
(|| + |
t|)dt)
2dx)1/2 + (T+ 2)||1/2
2T1/2
2 1(
T
0
(||2L2()+ |
t
|2L2())dt)
1/2 + (T+ 2)||1/2.
Hence, by the assumption forthe set {w} is bounded inL2(; BV(0, T)).
Then by applying Proposition 2.3 mentioned in [8] we can take a subse-
quence {j} {} such that wj w weakly* in L2w(; BV(0, T)) and
weakly* in L(Q(T)), bj b, uij ui weakly* in L(0, T; H1()),
weakly in W1,2(0, T; L2()) and in C([0, T]; H) as j fori= 1, 2.
Here, we show that w k((u1, u2), w0). In fact, it is clear that w L2w(; BV(0, T)) and (1.7) holds a.e. on because of the compatibility
condition for the initial data. Accordingly, it is sufficient to proveQ(T)
w((0) )dxdt
Q(T)
(|(0)| ||)dxdt for L1(Q(T)),
(4.7)
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dx
[0,T]
((0)) (z))dw 0 for zL1(; C([0, T])), (4.8)
where 0
= (u1
, u2
). For 0< 0
we set
= (1
, 2
+ ). Then,
we see that for each j and L1(Q(T))Q(T)
wj (j (j jwj ) )dxdt
Q
(|j (j jwj )| ||)dxdt.
(4.9)
By letting j in (4.9) we can get (4.7) since wj w weakly* inL(Q(T)) and (u1j , u2j ) (u1, u2) in C([0, T]; L
2()) as j .
Next, we prove (4.8). Let z L1(; C([0, T])). Then for eachj it holds
thatdx
[0,T](j (j jwj ) j (z))dwj 0. Here, for simplicity
we put
dx
[0,T]
(j (j jwj ) j (z))dwj
[0,T]
((0) (z))dw
=
dx[0,T]
j (j jwj )dwj
dx[0,T]
(0)dwj
+
dx
[0,T]
(0)dwj
dx
[0,T]
(0)dw
dx
[0,T]
(j (0) (0))dwj
(
dx[0,T]
(0)dwj
dx[0,T]
(0)dw)
=:4
i=1
Iij .
By using some properties of Lebesgue-Stieltjes integral we observe that
I1j
wjBV(0,T)|j (j jwj ) (0)|C([0,T])dx
(
wj2BV(0,T)dx)
1/2((
|j 0|2C([0,T])dx)
1/2
+j(
|wj |2C([0,T])dx)
1/2 + (
|j (0) (0)|2C([0,T])dx)
1/2).
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We note that
|(x) 0(x)|2C([0,T]) 2
T
0
|(x) 0(x)||t(x) 0t(x)|dx
for a.e. x , and for each j
|j (x) 0(x)|2C([0,T])dx 2|jt 0t|L2(Q(T))|j 0|L2(Q(T)).
By elementary calculations we obtain
|j (0) (0)| j for j and (t, x) Q(T).
Accordingly, we haveI1j 0 as j . Also, it is obvious that Iij 0 as
j for i = 2, 3, 4. Hence, (4.8) holds. Moreover, it is easy to see that(C2) and (C3) hold.
References
[1] T. Aiki, E. Minchev and T. Okazaki, A prey - predator model with hysteresiseffect, SIAM J. Math. Anal., 36 (2005), 2020-2032.
[2] F. C. Hoppensteadt, W. Jager, Pattern formation by bacteria, Lecture Notesin Biomathematics 38, 68-81, 1980.
[3] F. C. Hoppensteadt, W. Jger, C. Poppe, A hysteresis model for bacterialgrowth patterns, Modelling of patterns in space and time (Heidelberg, 1983),123134, Lecture Notes in Biomath., 55, Springer, Berlin, 1984.
[4] N. Kenmochi, E. Minchev and T. Okazaki, On a system of nonlinear PDEswith diffusion and hysteresis effects, Adv. Math. Sci. Appl., 14 (2004), 633-664.
[5] E. Minchev, A diffusion-convection prey-predator model with hysteresis,Math. J. Toyama Univ., 27(2004), 51-69.
[6] E. Minchev, T. Okazaki and N. Kenmochi, Ordinary differential systems
describing hysteresis effects and numerical simulations, Abstr. Appl. Anal.,7(2002), 563-583.[7] A. Visintin, Evolution problems with hysteresis in the source term, SIAM J.
Math. Anal., 17(1986), 11131138.[8] A. Visintin, Differential Models of Hysteresis, Appl. Math. Sci., Vol. 111,
Springer-Verlag, Berlin, 1993.[9] A. Visintin, Quasilinear first-order PDEs with hysteresis, J. Math. Anal.
Appl., 312(2005), 401-419.
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ON A VECTOR-VALUED SINGULAR PERTURBATIONPROBLEM ON THE SPHERE
NELLY ANDRE AND ITAI SHAFRIR
1. Introduction
Let 1 and 2 be two disjoint smooth, simple closed curves in R2 of lengths
l(1) and l(2), respectively, such that 1 lies inside 2 and the origin 0
lies inside 1. Let W : R2 [0, ) be a smooth function (i.e., at least ofclass C4) satisfying
W >0 on R2 \ (1 2) and W= 0 on 1 2 . (H1)
SinceWattains its minimal value zero on 1
2 we have clearlyWn = 0
on j , j = 1, 2, where Wn denotes the derivative in the direction of the
exterior normal to j . We assume then that we are in the generic case, i.e.,
that
Wnn >0 on 1 2 . (H2)
Finally, we add the following coercivity assumption on the behavior ofW
at infinity: there exist constantsR0 >0 and C0 >0 such that
W(x)C0|x|for|x| R0 . (H3)
Let G be either a bounded smooth domain in RN, or a smooth N-
dimensional manifold. For each >0 consider the energy functional
E(u) =
G
|u|2 + W(u)2
(1.1)
for u H1
(G,R2
). Let Rc be a positive number such that the circleSRc ={|x| = Rc} separates the two curves 1 and 2. We shall assume
departement de mathematiques, universite de tours, 37200 tours, francedepartment of mathematics, technion israel institute of technology, 32000 haifa, israel
11
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w.l.o.g. that 1 lies inside SRc which lies inside 2. The number Rc rep-
resents the constraint in the following minimization problem that we shall
study:
min{E(u) : uH1(G,R2), G
|u|= Rc} , (P)
where
G
|u| := 1(G)
G
|u|. Denoting by u a minimizer in (P), we areinterested in the asymptotic behavior of the minimizers{u} and theirenergies E(u), as goes to 0.
A first study of this problem was carried out by Sternberg [7]. He proved
a -convergence result, which has the following consequences:
(i) If a subsequence{un} converges to a limit u0, then u0 belongs to theset
S :={uBV(G, 1 2) : G
|u|= Rc}, (1.2)
and
PerG{u01}= min{PerG{u1} : u S} . (1.3)
(ii) The asymptotic expansion for the energy E(u), as 0, isE(u) = 2D min
uSPerG{u1} + o(1) . (1.4)
We refer to the books [5, 1] for the definition of the perimeter and other
notions from the theory of BV-functions, that we shall use in the sequel.
The constant D which appears in (1.4) is a certaindistance between
the two curves 1 and 2 which we shall now define. First, for any pair of
points x, y
R2 we set
dW(x, y) = inf Lip([0,1],R2),(0)=0,(1)=y
L() , (1.5)
where
L() =
10
W((t))
1/2 |(t)| dt . (1.6)Then, we define
D:= inf Lip([0,1],R2),(0)1,(1)2
L() . (1.7)
It was proved by Sternberg (see [7] ) that there exists a geodesic realiz-
ing the infimum D in (1.7). There may be of course more than one such
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geodesic; their number may be infinite (as is the case of 1, 2 which are
concentric circles andWradially symmetric). We denote the set of all these
geodesics by
G ={(i) :i I} , (1.8)whereIis some set of indices. For eachi Iwe denote by (i)1 =(i)(0)1 and
(i)2 =
(i)(1)2 the endpoints of the geodesic (i), and then setZj ={(i)j }iI, for j = 1, 2.
The results of Sternberg left some important questions unresolved:
(1) Existence of a converging subsequence is not known, i.e., a compactness
result is missing.
(2) Even if we assume convergence of a subsequence towards a limit u0,which is a map inS satisfying (1.3), we cannot say whereon 1 2, u0takes its values.
In [3] we made some progress on this problem in the case where G is
a domain in RN. In particular, we demonstrated the major role played by
the geometry ofG in determining the asymptotic behavior of{u}. In fact,when G is convex, and under some additional technical assumptions, the
limit u0 takes only two values, one in 1 and the other one in 2. On the
other hand, when G is nonconvex, the limit u0 may be more complicated
(i.e., the restriction of u0 to{x G : u0(x) j}, j = 1, 2, is notnecessarily identically constant). However, there is still no complete answer
to the above questions in general.
In the present article we shall concentrate on the special case where G
is the sphere S2. Thanks to the symmetry properties of the minimizers
{u
}in this case, we are able to give a quite complete analysis for both the
asymptotic behavior of the minimizers and their energies. We believe that
the case of the sphere will give some indication on the expected behavior
of the minimizers in the case of a convex domain G. First, notice that from
the symmetrization method of O. Lopes, see Theorem IV.3 in [6], it follows
that for each, the minimizer u is axially symmetric with respect to some
axis. We shall assume in the sequel, without loss of generality, a common
axis of symmetry for{u}, i.e.,
eachu is symmetric w.r.t. the e3 axis. (1.9)In view of (1.9) we can view each u as a function of a single variable ,
i.e.,
u=u(), [/2, /2] . (1.10)
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Next we introduce some notation needed in order to state our results.
Set
mj = minxj |
x|
and Mj = maxxj |
x|(j = 1, 2) , (1.11)
and
Mj ={xj :|x|= mj} , j = 1, 2 .We shall also assume the following:
bothM1 andM2 consist of a finite number of points . (H4)
Consider anyv Sand let Gj ={xS2
: v(x)j}, j = 1, 2. Denotingby the standard measure on the sphere, we have
(G1)m1+ (G2)m2(S2)Rc =S2
|v| (G1)M1 + (G2)M2 , (1.12)
Clearly, (1.12) and (1.11) imply that
m2 Rcm2 m1 (G1)
M2 RcM2 M1 . (1.13)
Recall now the well-known solution to the isoperimetric problem on thesphere. For a given value t (0, 4), the domain on S2 of area surface twith minimal perimeter is a disk on the sphere with surface t. The perimeter
of this disk is given by the function
I(t) =
4t t2 , (1.14)which is a concave function on (0, 4), symmetric about the middle point
2. Therefore, from (1.13) we deduce that in order to obtain G1 with
minimal perimeter we must have:
either (i) (G1)/(S2) =
m2 Rcm2 m1
or (ii) (G1)/(S2) =
M2 RcM2 M1 .
(1.15)
In order to know which of the two possibilities in (1.15) is preferable we
should check which possibility realizes the minimum in min m2Rcm2m1
, 1
M2RcM2M1 . Without loss of generality, we shall assume in the sequel thatpossibility (i) holds in (1.15), i.e., that
:= m2 Rcm2 m1
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For as defined in (1.16), we denote by 0(0, /2) the angle satisfying
2 /2
0
cos d/4= . (1.17)In other words, the area of the disc on the sphere given, in spherical coor-
dinates, by{(, ) : [0, /2]}, is 4. Moreover, thanks to (1.16) wehave:
(1 sin 0)m1+ (1 + sin 0)m2 = 2Rc . (1.18)Finally, on each of the curves j ,j = 1, 2 we introduce a kind of a distance
function to the setMj by setting for eachxj ,
j(x, Mj) = inf
0
|w|2+|w|mj dt : wH1([0, ), j), w(0) =x .(1.19)
Note that the minimization problem in (1.19) is actually scalar, since w
takes its values on a curve. In the next lemma we give some of the properties
of the solution to (1.19).
Lemma 1.1. For anyxj there exists a minimizerwj that realizes theminimum in (1.19). Moreover, there exists a point y = y(x) Mj suchthat limt wj(t) =y and
j(x, Mj)2
=
0
|wj |2 dt= 0
(|wj | mj) dt . (1.20)
Proof. The analysis of problem (1.19) is identical to that of scalar problem
with a one-well potential. Therefore, all the the assertion of the lemma are
known and standard.
Now we are in position to state our main result.
Theorem 1.1. Assume thatWsatisfies hypotheses(H1)(H4),(1.16)andthat (1.9) holds, i.e., each u is of the form (1.10). Then, up to replacing
each u by u R, whereR is the reflection w.r.t. the (e1, e2)-plane, wehave:
(i) For a subsequence
un(/2,0)x(2) + (0,/2)x(1) , (1.21)uniformly on (2 +, 0 )(0 +, 2 ), for every > 0, withx(j) Mj, j = 1, 2.
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(ii) The asymptotic expansion of the energy is given by
E(u)
2 = cos 0
2D
+
2D
K
1/2 + o(
12 ), (1.22)
where
K= min{1((i)1 , M1) + 2((i)2 , M2) : i I} , (1.23)and:= tan0m2m1 .
Note that Theorem 1.1 provides us with a criterionfor identifying the limit
in (i). Indeed, the points x(1), x(2) that realize the limit form the pair of
points, one inM1 and the other one inM2, which are the closest (in anappropriate sense) to a geodesic inG.Our results can be generalized without difficulty to the problem on SNfor anyN3. Indeed, the axial symmetry of the minimizers is guaranteedby Lopes method, so that the one dimensional formulation involves the
weight function (cos )N1. Furthermore, although we have not examined
in detail such cases, it is very likely that most of the results of this paper
may be extended, by the same techniques, to more general potentialsW, for
example, with zero set consisting of two compact surfaces in R3, separated
by a sphere of radius Rc.
2. A first upper-bound
We shall first introduce some more notation that will be needed in the
sequel. Using dW we define the corresponding distance functions to the
curves 1, 2 by
j() =dW(, j) := infxj
dW(, x) , j = 1, 2 . (2.1)
We also set = min(1, 2) . (2.2)It is well known (c.f. [7, 4]) that for j = 1, 2, j Lip(R2) is a solution ofthe eikonal-type equation
|j()|2 =W() a.e. on R2 . (2.3)It was further shown in [2] that j is regular in a neighborhood of j , i.e.,
d0>0 s.t. j is of class C2 in{x : j(x)< d0} , (2.4)for j = 1, 2. Moreover, we have
j(x)W(x)dist2(x, j) on{x : j(x)< d0} . (2.5)
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Clearly,dW(x1, x2) =D for everyx11andx22. In order to identifythe end points of geodesics from Gwe shall use yet another distance functionbetween points from 1 and 2. We denote by the domain lying between
1 and 2.Recall that in [2] it was shown that for each x0 j , j = 1, 2, there is
a curveG(j)x0 parametrized on (, t(x0)] which satisfies the equationG(j)x0 =(G(j)x0),G(j)x0() =x0 ,
(2.6)
such that the union of these curves (over all x0j) covers without inter-sections
{j
d0
}+, which is the part of
{j
d0
}lying between 1 and
2 (see (2.4)). Similarly, the remaining parts{jd0}, j = 1, 2, can becovered by an analogous family of curves. Using these curves we can now
define a projection map sj from{j d0} to j which associates to eachx {j d0} the unique point x0 = sj(x)j for which the curveG(j)x0passes by x.
For any small >0 set
={x :(x)> } . (2.7)For any x11 and x22 we lety1, y2 be the points determinedbydW(yj , xj) =, j = 1, 2. We then define
d()W(x1, x2) = 2+ inf
Lip([0,1],),(0)=y1,(1)=y2
L() , (2.8)
where L() is defined in (1.6). Note that for each > 0 we have
d()W(x1, x2)D for every x11 andx22 with equality if and only if
x1 and x2 are the end points of a geodesic inG
.
The main result of this section is a simple upper-bound for the energy.
It is not optimal, but it gives the exact first term in the energy expansion
(of the order 1 ), and the order1/2 of the next term.
Proposition 2.1. There exists a constantC1 >0 such that
E(u)2 cos 02D
+
C11/2
(2.9)
Proof. We shall construct a function v = v() which satisfies the con-straint in (P), i.e., /2
/2
|v| cos d= 2Rc , (2.10)
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as well as the bound (2.9). Fix two pointsxj Mj, j = 1, 2 and a geodesic :=(i0) G (see (1.8)) of lengthL = len(). We shall denote for short by
p1 and p2 the endpoints (i0)1 and
(i0)2 of, respectively, so that choosing
the arclength parametrization forwe have,(0) =p1and(L) =p2. Thefollowing functionz(s) will be used in a choice of a certain parametrization
of the curve. It is defined as the solution of the ODE:
dz
ds =
W((z(s))) , z(0) =L/2 . (2.11)
It is easy to see that z is defined on the whole real line and satisfies
lims
z(s) = 0 and lims
z(s) =L .
Furthermore, sinceW((z(s)))d((z(s)), 1)z(s) as s
and W((z(s)))d((z(s)), 2)L z(s) as s ,
where d stands for the euclidean distance, we have:
0z(s)C2ec1s as s , (2.12)0L z(s)C2ec2s as s , (2.13)
for some positive constants c1, c2 and C1, C2. From (2.11) we deduce that
for any function v() defined on an interval [1, 2] by
v() =(z(c
)) ,
we have 21
|v|2 + W(v)
2
cos d=
2
2
21
W(v)cos d
= 2
21
W(v)|v()| cos d .
(2.14)
Set = 0 1/2,with= to be determined later. We define v()on [
/2, /2] as follows:
(i) On [ + 1/2, /2] we set vx(1).(ii) On [,+1/2] we follow the curve 1 in a constant velocity from
p1(=(i0)1 ) to x
(1) (in the shortest way between the two possibilities).
(iii) On [ , ] we let v be the linear function which equals to
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(z( ln 1/c1 )) at = and to p1 at = .(iv) On [ ( 1c1 + 1c2 ) ln 1 , ] we set
v() =z ln 1/c1 .(v) On [ 2 ( 1c1 + 1c2 ) ln 1 , ( 1c1 + 1c2 ) ln 1 ] we let v to be thelinear function which equals to p2(=
(i0)2 ) at =
2 ( 1c1 + 1c2 ) ln 1and to (z( ln 1/c2 )) at=
( 1c1 + 1c2 ) ln 1 .(vi) On [ 2 ( 1c1 + 1c2 ) ln 1 1/2, 2 ( 1c1 + 1c2 ) ln 1 ] we followthe curve 2 in a constant velocity from x
(2) top2 (using the shortest way
among the two possibilities).(vii) On [/2, 2 ( 1c1 + 1c2 ) ln 1 1/2] we set vx(2).We shall next compute the contribution to the integral of|v| from
each of the intervals (i)(vii) and show that a bounded solution for is
determined by the constraint (2.10). Note first that /2+1/2
cos |x(1)|
= 1 sin(0 ( 1)1/2)|x(1)|=
1 sin 0+ ( 1)1/2 cos 0|x(1)| + O() .(2.15)
Here and in the rest of the proof we denote by O(f()) a function g(t, )
satisfying|g(t, )| C(t)f() withC(t) bounded for boundedt. Next, it iseasy to verify that +1/2
|v| cos = 1/21(x(1), p1)cos 0+ O() , (2.16)
with
1(x(1), p1) =
J1(x(1),p1)
|z|d1(x
(1), p1) ,
where J1(x(1), p1) denotes the (shortest) segment of 1 between x
(1) and
p1, whose length is denoted then byd1(x(1), p1). Clearly
|v| cos = O() . (2.17)
Similarly, for the intervals in (iv) and (v) we find, respectively, ( 1c1
+ 1c2) ln 1
|v| cos = O( ln1
) , (2.18)
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and
( 1c1
+ 1c2) ln 1
2( 1
c1
+ 1
c2
) ln 1
|v| cos = O() . (2.19)
As in (2.16) we find 2( 1c1+ 1c2 ) ln 12( 1c1
+ 1c2) ln 1
1/2
|v| cos = 1/22(x(2), p2)cos 0+ O() , (2.20)
whereJ2 and2 are defined analogously toJ1 and1. Finally, as in (2.15),
we have
2( 1c1+ 1c2 ) ln 11/2/2
cos |x(2)|=
1 + sin 0 (+ 1)1/2 cos 0|x(2)| + O( ln1
) .
(2.21)
Summing up the integrals in (2.15)(2.21) and comparing to (1.18) and
(2.10) yields
1/2 cos 0m1(1) + 1(x(1), p1) + 2(x(2), p2)m2(+ 1)= O( ln1
) ,
from which we can solve for = that satisfies
= 1(x
(1), p1) + 2(x(2), p2) m1 m2
m2 m1 + O(1/2 ln
1
) .
Clearly remains bounded as goes to zero.
Next we compute the contribution to the energyE(v) from each of the
segments (i) to (vii). We denote these contributions byI1I7, respectively.Clearly we have
I1 =I7 = 0 , (2.22)
and
I2+ I6 =O(1/2) . (2.23)
From (2.12)(2.13) we infer that
|p1 (z( ln 1/c1 ))|= O() and |p2 (z( ln 1/c2 ))|= O() ,which implies that
I3+ I5 =O(1) . (2.24)
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Therefore, choosing an appropriate C yields the existence of 0 = 0
(0, C1/2) satisfying
(u(0))1/2 . (3.3)We shall assume in the sequel that (3.3) holds with(u(0)) = 2(u(0))(this implies, of course, that u(0) is close to 2).
Next, define, if it exists,
+1 = inf{0 : 1(u())1/2} ,
and then
+
1 = sup{+1 : 2(u())1/2} .In general, for an even j let
+j = inf{+j1 : 2(u())1/2} ,+j = sup{+j : 1(u())1/2} ,
while for an odd j:
+j = inf{+j1 : 1(u())1/2} ,+
j = sup{+j : 2(u())1/2} .We stop atj =k+ 1 ifj =k+ is the first index for which either +j doesnot exist, or+j >
2 1/2. For eachj = 1, . . . , k+ 1 we have (see (2.14))
1
2E
(u
, A+
j ,+j
)
2
cos +
j +j
+j W(u)|u| 2
(D 21/2)cos +j .
(3.4)
In particular, we deduce
1
2E(u, A+j ,
+j
) 2
(D 21/2)sin 1/2 C1/2
, (3.5)
which together with the upper bound (2.9) implies that the process of
selection of pairs (+j , +j ) must terminate, with the bound k
+ C1/2
.
Similarly, set, if it exists,
1 = sup{0 : 1(u())1/2} ,
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and then
1 = inf{1 : 2(u())1/2} .
In general, for an even j let
j = sup{j1 : 2(u())1/2} ,
j = inf{j : 1(u())1/2} ,while for an odd j:
j = sup{j1 : 1(u())1/2} ,j = inf{j : 2(u())1/2} .
We stop at j =k 1 ifj =k is the first index for which either j doesnot exist, orj
1/2 on each A+2j ,+2j+1
andA2j+1,
2j, which implies that
{1(u)
1/2
} A1/2
2 ,
21/2
V . (3.9)
On the other hand, since 2(u)> 1/2 on each A+2j1,
+2j
and A2j ,
2j1,
we have
V {1(u)1/2} {(u)1/2} . (3.10)
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Note that from (2.26) it follows that there exists (0, 1) such that|({1(u)1/2}) 4| C1/2
and |({2(u)1/2
}) 4(1 )| C1/2
.
(3.11)
Note that on the sets{j(u)< d0} we have|usj(u)|2 W(u) (see(2.5)), hence
{j(u)1/2}
|u| {j(u)1/2}
|sj(u)|
{j(u)1/2}
|u sj(u)|
C {j(u)1/2}
W(u)1/2 C1/2 .(3.12)
By (H3), (2.5) and (2.26) we also have{(u)>1/2}
|u| R0{(u)> 1/2}{ |u| R0}
+ 1
C0
{(u)>1/2}{|u|>R0}
W(u)C 1/2 .(3.13)
Set
m(j) = 1
({j(u)1/2}){j(u)1/2}
|sj(u)| , forj = 1, 2 . (3.14)
Using (3.12)(3.14) and the constraint we obtain
m(1) ({1(u)1/2}) + m(2) ({2(u)1/2})=
{1(u)1/2}|s1(u)| +
{2(u)1/2}|s2(u)|
={1(u)1/2}
|u| +{2(u)1/2}
|u| + O(1/2)
= 4Rc+ O(1/2) .
(3.15)
Form (3.15) and (3.12) it follows that there exists a number K, which is
uniformly bounded, such that
m(1) (+ K1/2) + m(2) (1 K1/2) =Rc . (3.16)
Since m(j) [mj , Mj ], j = 1, 2, it follows from (3.16), in particular, that is bounded away from 0 and 1. Therefore, by the definitions of andI
(see (1.14)),
I(4(+ K1/2))I(4) = 2 cos 0. (3.17)
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By (3.9)(3.10),(3.17) and the Lipschitz property ofIwe conclude that
22[ k
+
2 ]
j=1 cos +j +2[ k
2 ]
j=1 cos j = Per VI((V))2 cos 0 C1/2 .
(3.18)
On the other hand, summing-up the inequalities in (3.4) and (3.6) yields
1
2E
u,k+1j=1
A+j ,
+j
k1j=1
Aj ,
j
2D
k+1j=1
cos +j +
k1j=1
cos j C1/2 .
(3.19)
Combining (3.19) with the upper-bound (2.9) we obtain
k+j=1
cos +j +
kj=1
cos j cos 0+ C1/2 . (3.20)
Note that we used the fact that cos +k+ = cos k = sin
1/2 = O(1/2).
Therefore, combining (3.18) with (3.20) we get
k+j=1
cos +i +kj=1
cos i cos 0C 1/2 . (3.21)
Our next objective is to show that V consists of essentially one an-
nulus.
Claim: We have:
either(VA
+
j ,
+
j+1
) =O(),
or (VAj+1 ,
j) =O(), for some j.
(3.22)
Proof of the Claim: If V consists of only one annulus, then either V =
A+1 ,+2
orV=A2 ,
1(see (3.8)) and the claim holds. It remains to treat
the case where V consists of more than one annulus. Note that since I is
concave andI(0) = 0, we have
I(a + b)I(a) + I(b) , (3.23)for all admissible values ofa and b. By assumption, we may write V as adisjointunionV =B C with(V)/2(B)< (V). From (1.14) itis clear that there exists 0>0 such that
I()2I((V) ) , 0 . (3.24)
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We distinguish two cases:
(i) (C)> 0.
(ii) (C)
0.
In case (i) we have, for some constant >0, a stronger form of (3.23),
namely
I((V)) + I((B)) + I((C)) . (3.25)From (3.25) and (3.21) it follows that
I((V)) + I((B)) + I((C))Per(B) + Per(C) = Per(V)2 cos 0+ C1/2 ,
which clearly contradicts (3.18), for small enough. Therefore, only case
(ii) should be considered.
Notice the following simple consequence of the concavity ofI:
(C)I((B))I((V)) I((B)) . (3.26)
By (3)(3.26) and (3.24) we get
2 cos 0+ C1/2
I((B)) + I((C))I((V)) (C)I
((B)) + I((C))
I((V)) +12
I((C))2 cos 0 C1/2 +12
I((C)) ,
which yields, I((C))C 1/2, i.e.,(C)C . (3.27)
To conclude the proof of the claim, we shall show that one of the components
ofV has measure larger than (V)
C. By the above computation it is
enough to consider the case where V consists of r 3 components (i.e.,annuli) that we shall now denote by A1, A2, . . . , Ar with
(A1)(A2) (Ar) .Furthermore, if(A1) (V)2 then the conclusion follows from the argu-ment that led to (3.27). Thus assume that
(A1)< (V)
2 . (3.28)
We shall see that this is impossible. Indeed, (3.28) implies the existence of
j01 which is the largestindex for which
(Aj0+1) + (Aj0+2) + + (Ar) (V)
2 .
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Setting
B =Aj0+1 Aj0+2 Ar and C =A1 A2 Aj0,
we obtain from the argument that led to (3.27) that (C)C . But fromthe definition ofj0 it is easy to see that we must have then(B) 34(V),i.e., (V) =(B) + (C) 34(V) + C, which is a contradiction (for small enough). The proof of the claim is then complete.
Consider now the fat component A1, and assume without loss of gen-
erality that it lies in the upper hemisphere, i.e., A1 = A+j ,+j+1
for some
j. Under this assumption we shall show that (3.1) holds, for 1 =+j (the
other possibility leads to (3.2) by an identical argument). By (3.21) we
deduce that
| cos +j + cos +j+1 cos 0| C 1/2 , (3.29)while by (3.18) we have
|I((V)) 2 cos 0| C 1/2 . (3.30)Using the Lipschitzity of the (local) inverse of the function I we deduce
from (3.30) and (3.22) that
sin +j+1 sin +j = 1 sin 0+ O(1/2) . (3.31)It is easy to conclude from (3.29) and (3.31) that
|+j 0| C 1/2 and |+j+1
2| C 1/2 .
Finally, we must havej = 1, since otherwise we will get a contradiction to
(3.21). We can therefore set 1
=+1
and (3.1) follows.
In the sequel we shall assume, without loss of generality, that (3.1) holds.
We then set
2 =2() = sup{1 : 2(u()) =1/2} . (3.32)Note that by (3.1) we have
E(u, A2,1) 2cos 1
(D 21/2
)2D cos 0
C1/2
. (3.33)
The next lemma provides pointwise estimates that roughly speaking show
thatu() is close to 2 for below 2, while u() is close to 1 for
above1.
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Lemma 3.1. There existsc3 >0 such that
1(u()) cos c31/2 , 1 , (3.34)2
(u
()) cos
c3
1/2 ,
2
. (3.35)
Proof. By (2.9) and (3.33), for each 1 we haveC
1/2 E(u, A1,)
2
1
cos d
d
1(u())
d
2
1(u())cos 1(u(1))cos 1+
1
sin 1(u()) d
2 cos 1(u()) cos 1 1/2 ,
and (3.34) follows.
Next, for each 2 we have by the same computation as aboveC
1/2 E(u, A,2)
2
2
cos d
d2(u())
d
2 cos 22(u(2)) + cos 2(u())
2
sin 2(u()) d
.
(3.36)
Denoting += max(, 0), we have 2
sin 2(u()) d
= +
sin 2(u()) d+ 2+
sin 2(u()) d
2+
sin 2(u()) d .
(3.37)
Since for [0, 2] we have sin Ccos and since by (3.33) and (2.9),2
2+
cos 2(u()) d C1/2
,
we obtain from (3.36)(3.37) that
C
1/2 2
cos 2(u()) C
1/2 ,
and (3.35) follows as well.
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4. Proof of the lower-bound
In this section we prove the lower-bound energy estimate of Theorem 1.1.
The next lemma provides a crucial estimate for the distance between 1and2.
Lemma 4.1. We have1 2 ln 1 .
Proof. Put
2 = sup{(2, 1) : 2(u()) = d02} , (4.1)
1 = inf{(2, 1) : 1(u()) = d0
2} . (4.2)
Note that(u) d02 on [2,1] (see (2.2)). Therefore, alsoW(u)c >0on [2, 1] for some constant c, by (2.5). From the simple upper-bound
1
2
A2,1
W(u) C
we then easily conclude that
1 2C . (4.3)
It is therefore sufficient to prove the following:
2 2 ln1
and 1 1 ln1
. (4.4)
Clearly it is enough to prove the first estimate in (4.4), as the proof of the
second one is identical.We define the function 2() by the equation
u() =Gs2(u())(2()) , [2, 1] . (4.5)
We can writeu() as a sum of two orthogonal components, the first, (u)
in the direction :=2 and the second, (u) in the direction ofs2.We therefore have
|u|2 =|(u)|2 + |(u) |2 . (4.6)It is easy to verify that the first component is given by
(u)=2(u())2() . (4.7)
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Therefore,
2
2 |u|2 +
W(u)
2 cos
22
|(u)|2 +
W(u)
2
cos
=
22
|2(u)|2(2)2 +
W(u)
2
cos
=
22
W(u)
(2)
2 + 1
2
cos
= 22
W(u)(2 1 )2 +2 dd2(u) cos .
(4.8)
We also have 12
|u|2 +
W(u)
2
cos
12
2
d
d
2(u)
cos . (4.9)
Note that by (3.1) we have
1
2
d
d2(u) cos = (D 1/2)cos 1 1/2 cos 2
+
12
2(u)sin D cos 0 C1/2 .
(4.10)
Combining (4.8)(4.10) with (3.33) yields
2
2
W(u)(2
1
)2
C 1/2 . (4.11)
Since W(u)C 1/2 on [1, 2] we obtain from (4.11) that 22
(21
)2 C
.
Applying the last estimate together with the Cauchy-Schwarz inequality
yields
|2(2) 2(2)
2
2
| 2
2 |
21
| C2 2 1/2 . (4.12)From (2.6) we obtain that
2(2) =O(1) and 2(2) ln1
. (4.13)
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Plugging (4.13) in (4.12) yields the first estimate in (4.4).
The next lemma provides an estimate for the distance between 2 (or
1) and 0 in terms of the two averages
m1, =
/21
|u| cos 1 sin 1 and m2, =
2/2
|u| cos 1 + sin 2
. (4.14)
Lemma 4.2. There exist two constants1, 2 >0 (independent of) such
that
cos 2 cos 0 =1(m1, m1) + 2(m2, m2) + O( ln1
) . (4.15)
Proof. Because of the constraint and Lemma 4.1 we have 0/2
cos
m2+ /2
0
cos
m1
= 2
/2
cos
m2,+ /2
2
cos
m1,+ O( ln1
) ,
which can be rewritten as
(sin 0 sin 2)(m2 m1)= (sin 2+ 1)(m2, m2)
+ (1 sin 2)(m1, m1) + O( ln1
) .
(4.16)
Since sin 2 sin 0 = O(1/2) (see (3.1) and Lemma 4.1), it follows from(4.16) that also
m2, m2=O(1/2
) andm1, m1 =O(1/2
) . (4.17)
Therefore,
sin 2 sin 0 = 1m2 m1 (sin 0+ 1)(m2, m2)
1m2 m1 (1 sin 0)(m1, m1) + O( ln
1
) .
(4.18)
Finally, a simple computation using Taylor formula leads from (4.18) to
(4.15) with
1= tan 0(1 sin 0)
m2 m1 and 2 =tan 0(1 + sin 0)
m2 m1 . (4.19)
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Next we shall introduce some more notation. For s for which j(u())d0 we define vj =vj,() = sj(u()), j = 1, 2. In particular we set
p1
=v1
(1
) = s1
(u
(1
)) and p2
=v2
(2
) = s2
(u
(2
)) . (4.20)
Recall the distance function d()W that was defined in (2.8). We shall next
use it for =1/2.
Lemma 4.3. We have
1
2E(u)
2/2+1/3
|v2|2 +2d(1/2)
W
(p
1
, p
2
)
(|v2()| m2) cos +
/21/31
|v1|2 +
2d(1/2)W (p
1, p
2)
(|v1()| m1)
cos
+2d
(1/2)W (p
1, p
2)cos 0
C5/12 ,
(4.21)
where= tan0m2m1 .
Proof. Note first that for [/2 + 1/3, 2] we have cos C 1/3, soby Lemma 3.1
2(u())C1/6 , (4.22)hencev2 is well-defined. Similarly,v1is well-defined for[1, /21/3].By the definition of1 and2, (4.15) and (4.17) we have
12
E(u, A2,1)2 cos 1 (d(1/2)W (p
1, p
2) 21/2)
2
cos 0+
2j=1
j(mj, mj) + O( ln1
)
(d(1/2)W (p
1, p
2) 21/2)
2d(1/2)W (p
1, p
2)
cos 0+
2j=1
j(mj, mj) 4cos 0
1/2 + O(| ln |) .
(4.23)
From (4.23) and the upper bound (2.9) we deduce that
E(u, S2 \ A1,2)
C
1/2. (4.24)
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By (4.24)
2
0
2(u)cos
C
2
0
W(u)cos
C 2E(u, S
2
\A1,2)
C 21/2 =C 3/2 ,
so there exists 2(0, 2) such that
2(u(2))C 3/2 . (4.25)
From (4.25) we get by the same computation as in (3.36) the lower bound
for the normalenergy (see (4.7)(4.8)), i.e.,
1
2E (u, A2,2)
:=
22
|(u)|2 +
W(u)
2
cos
2
cos 2
1/2 cos 12(u(2)) + 21
sin 2(u)
2cos 2
1/2 C1/2
2cos 0
1/2 C .
(4.26)
Similarly, there exists 1 (1, 1 + 1/2) such that 1(u(1)) C.Hence,
1
2E (u, A1,1)
2
1
1
(cos 0 C1/2)d
d1(u) 2
(cos 0 C1/2)(1/2 C) 2cos 0
1/2 C .
(4.27)
Next we take into account also the contribution from the tangential en-
ergy, i.e., of (u). As in [2] we have
(u) =u
s2(u)
|s2(u)
|
=
dd (s2(u))
|s2(u)
|
= v2
|s2(u)
|
, (4.28)
at points where 2(u)< d0. A simple modification of the argument of the
proof of [2] shows that (4.28) implies that
|(u) |2 |v2|2(1 cd(v2, 2)) |v2|2(1 1/22 (u)) , (4.29)
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for some constant >0. Combining (4.26) with (4.29) in conjunction with
(4.22) and (4.24) leads to
1
2 E(u, A/2+1/3
,2)
2cos 01/2
+
2/2+1/3
cos (1 1/22 (u))|v2|2 C
2cos 01/2
+
2/2+1/3
|v2|2 cos C5/12 .
(4.30)
Similarly,
1
2
E(u, A1,/21/3)
2cos 01/2
+
/21/31
|v1|2 cos C5/12 .(4.31)
Adding together (4.23) with (4.30) and (4.31) leads to
1
2E(u) 2d
(12)
W (p1, p
2)cos 0
+
2d(
12)
W (p1, p
2)
2j=1
j(mj, mj)
+ 22+
13
|v2|2 cos + /21
3
1
|v1|2 cos C 512 . (4.32)
Next we estimate the second term on the r.h.s. of (4.32). First, write
m1,(1 sin 1) =
2
1
|u| cos
= 2
2 13 |
u|
cos + 2
13
1 |v1
|cos +
2
13
1
(|u
| |v1
|)cos . (4.33)
Since (H3) implies that|u| C(1 + W(u)) we get, using also (4.24), that /2/21/3
|u| cos C(1 cos(1/3)) + C /2/21/3
W(u)cos
C(2/3 + 3/2) .(4.34)
Using (2.5), (4.24) and the Cauchy-Schwarz inequality we obtain
/21/31
(|u| |v1|)cos C
/21/31
1/21 (u)cos C 3/4 .
(4.35)
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From (4.33)(4.35) we get
m1,(1
sin 1) =
/21/3
1 |v1
|cos + O(2/3) ,
from which we obtain that
m1, m1 = 11 sin 1
/21/31
(|v1| m1)cos + O(2/3) . (4.36)
A similar argument yields
m2,
m2 = 1
1 + sin 2 2
/2+1/3(|v2
| m2)cos + O(
2/3) . (4.37)
Since|20|, |10| C1/2 (see (3.1) and Lemma 4.1), we finallydeduce (4.21) from (4.32) and (4.36)(4.37).
The following lemma is an immediate consequence of Lemma 4.3.
Lemma 4.4. There exists a constantC, independent of, such that
d(1/2)W (p1, p2)D + C1/2 . (4.38)
Proof. It suffices to combine the upper bound (2.9) with (4.21) (taking
into account only the third term on the r.h.s.).
The next lemma provides a lower bound for the two integrals on the right
hand side of (4.21).
Lemma 4.5. We have /21/31
|v1|2 +
2d(1/2)W (p
1, p
2)
(|v1()| m1)
cos
cos 01/2
2d
(1/2)W (p
1, p
2) 1(p
1, M1) + o(1) .
(4.39)
and
2/2+1/3
|v2|2 +2d(1/2)W (p1, p2) (|v2()| m2) cos cos 0
1/2
2d
(1/2)W (p
1, p
2) 2(p
2M2) + o(1) .
(4.40)
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Proof. Clearly it suffices to prove (4.39). First, note that
J1() :=12
2
13
1 |v1|2 +2d
(12)
W (p1, p
2)
(
|v1()
| m1) cos
C ,(4.41)
by the upper-bound (2.9) and Lemma 4.3. We introduce a new variable t
by= 1+ t1/2 and then define a new function by
w(t) =v1(1+ t1/2) , 0t /2
1/3 11/2
. (4.42)
We have
J1() 12 1+ 141
|v1|2 +
2d(
12)
W (p1, p
2)
(|v1()| m1)
cos d
12 (cos 1 c 14 ) 1+ 141
|v1|2 +
2d(
12)
W (p1, p
2)
(|v1()| m1)
d
= (cos 1
c14 )
14
0 |w|2 + 2d(12)
W (p
1
, p
2)(
|w(t)
| m1) dt . (4.43)
By (4.41) we have 1/40
|w|2 + 2d(1/2)W (p1, p2)(|w(t)| m1)
dtC .
Hence, there exists t(0, 1/4/2) such thaty:=w(t) satisfies
|y
| m1
C 1/4 . (4.44)
From (4.44) it follows that there exists z 1 M1 such that|z1 y|= o(1).It follows then that there exists a function () =o(1) such that
|z| m1() ,zJ1(y, z1) , (4.45)see after (2.16) for the definition ofJ1. We shall next define a new function
won [0, ). First, on the interval [0, t] we set
w= w. Then, on the interval
(t, t+(())1/2] we letwgo fromw(t) toz1along 1in constant velocity.Finally, we setw(t)z1 on (t+ (())1/2, ). It is clear from the above
construction thatw satisfies t
| w|2 + 2d(1/2)W (p1, p2)(| w(t)| m1) dt= o(1) . (4.46)
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From (4.46) and (1.19) it follows that
1/4
0 |w|2 + 2d(1/2)W (p
1, p
2)(
|w(t)
| m1) dt
2d(1/2)W (p
1, p
2) 1(p
1, M1) + o(1) .
(4.47)
Combining (4.47) with (4.43) we are led to (4.39).
In order to complete the proof of the lower-bound part of (1.22) we need to
identify the limit of the points p1 and p2. This is the purpose of the next
lemma.
Lemma 4.6. Suppose that for a subsequence, limpn1 = p1 and limpn2 =
p2. Thenp1 andp2 are the end points of some geodesic fromG.
Proof. Consider a constant unit velocity geodesic n realizing
d(1/2n )W (p
n1 , p
n2 ), which passes between the points y
n1 = un(1) and
yn2 =un(2) (see (4.20)). For each >0 we letp1,n be the last point on
that geodesic where 1(p
1,n) = . Similarly, let p
2,nbe the first point on
that geodesic where 2(p2,n) =. Denote by
nthe part ofnbetweenp
1,n
andp2,n. Passing to a subsequence if necessary we obtain the convergence
of n towards a limiting geodesic with end points p1 and p
2 which must
satisfy dW(p1, p2) = D2 (because of (4.38)). Repeating this process
with a sequencem0 and passing to a diagonal subsequence, we deduce(for a further subsequence) that n converges to a curve joining a point
q1 1 to a point q2 1 with the following property: for each small
enough > 0 we have dW(p
1, p
2) = D2, where p
j , j = 1, 2, are thepoints on that geodesic satisfying j(p1) =, respectively. It follows that
G . The result of the lemma would follow once we show thatq1 = p1andq2 =p2.
Looking for a contradiction, assume for example that q1= p1. Let be a small positive number that we shall fix later. We may choose n large
enough such that |p1,np1|< . We now denote by na re-parametrizationon the interval [0, 1] of the part ofn between y
n1 and p
1,n such that
W(n) |(n)| const on [0, 1] .
Using the same notations as in the proof of Lemma 4.1 we can decompose
(n) as a sum of two orthogonal components: (n)
in the direction of
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1, and (n) in the direction ofs1. We have
1
0 W(n) |(n)|
2
= 1
0
W(n)|(n)|2 + |(n) |2 . (4.48)Next, note that 1
0
W(n) |(n)|2 1
0
W(n) |(n)|
2= 1
0
|1(n)| |(n)|2
( 1/2n )2 . (4.49)
Using (2.5) we find
10
W(n) |(n)|2 c11/2n 10
|(n)|2 c21/2n , (4.50)for some positive constants c1 and c2 which do not depend on n or on
(forsmall enough). Plugging (4.49)(4.50) in (4.48) yields (provided is
chosen small enough), 10
W(n) |(n)|
( 1/2n )2 + c21/2n
1/2
2 + c31/2n 1/2 + c4 1/2n .(4.51)
From (4.51) we deduce that
d(1/2n )W (p
n1 , p
n2 )D+ 1/2n + c4
1/2n
,
which contradicts (4.38), provided is chosen small enough.
The next proposition provides the lower-bound estimate needed for asser-tion (ii) of Theorem 1.1.
Proposition 4.1. We have
1
2E(u)cos 0
2D
+
2D K
1/2
+ o(12 ) . (4.52)
Proof. Plugging the estimates (4.39)(4.40) in (4.21) and using the obvi-
ous estimate d(1/2)
W (p
1, p
2)D yields1
2E(u) 2D cos 0
+
cos 01/2
2D
1(p
1, M1)
+ 2(p2, M2)
+ o(1/2) .
(4.53)
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Therefore, (4.52) is a direct consequence of (4.53), Lemma 4.6 and the
definition (1.23) ofK.
5. A refined upper-bound and the convergence result
Next we prove the upper-bound part of the energy expansion (1.22).
Proposition 5.1. We have
1
2E(u)cos 0
2D
+
2D K
1/2
+ o(12 ) . (5.1)
Proof. We are going to refine the construction used in the proof of Propo-
sition 2.1, using the insight we got from the lower-bound estimates weestablished so far. We first fix x(1) M1, x(2) M2 and(i0) G that re-alizes the minimum for Kin (1.23) and denote by p1 =
(i0)1 andp2=
(i0)2
the end points of (i0). For these values of x(1), x(2), p1 and p2 we shall
slightly modify the construction of a test functionv, that was described in
the proof of Proposition 2.1 on 7 intervals, from (i) to (vii). Actually, we
shall modify the construction only on the intervals (i)(ii) and (vi)(vii).
From Lemma 1.1 it follows that there exists a minimizer w1 that realizes1(p1, M1) with limt w1(t) =x(1). We then definew1(t) =w1(2Dt) , t[0, ) ,so that
0
| w1|2 + 2D(| w1| m1) dt= 2D 1(p1, M1) .
Let be a number whose distance to 0 is of the order O(1/2), that will
be determined later. We first set
v() =w1( 1/2
) for [,0+ /22
] . (5.2)
Then, on the interval [0+/22 , /2] define v in such a way that it follows
the curve 1 from the point v(0+/2
2 ) to the point x(1) in a constant
velocity. From (1.20) and (5.2) it easily follows that
1
2 E(v, A,/2) = cosD/2 1(p1, M1)1/2 + o(1/2) . (5.3)The above construction replaces the construction on the intervals (i) and
(ii) in the proof of Proposition 2.1. From here we follow exactly that con-
struction on the intervals (iii)(v). Finally, the construction on the intervals
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(vi)(vii) is modified in a similar manner to the above, and yields the anal-
ogous estimate to (5.3), namely
1
2 E(v, A/2,2( 1c1+ 1c2 ) ln 1 )
= cos
D/2 2(p2, M2)1/2 + o(1/2) .(5.4)
Combining (5.3) and (5.4) with the estimates from the proof of Proposi-
tion 2.1 yields
1
2E(v)cos
2D
+
2D
1(p1, M1) + 2(p2, M2)
1/2
+ o(
1/2
) .
(5.5)
It remains to fix the value ofin such a way that the constraint (2.10)
is satisfied. Put
1 =
/2
|v| cos d1 sin and 2 =
/2
|v| cos d1 + sin
.
By our construction ofv and (1.20) we have
(1 m1)(1 sin) = /2
(|v| m1)cos
= 1(p1, M1)
2
2D (cos)1/2 + o(1/2)
(5.6)
and
(2
m2)(1 + sin) =
/2
(
|v
| m2)cos
=2(p2, M2)
2
2D (cos)1/2 + o(1/2) .
(5.7)
We have also a third equation coming from the constraint,
1(1 sin) + 2(1 + sin) =m1(1 sin 0) + m2(1 + sin 0) . (5.8)
Writing = 0
1/2 and using first order Taylor expansion yields from
(5.6)(5.8) a perturbed linear system of 3 equations in the 3 unknowns1, 2 and which has a solution
=
8D
1(p1, M1) + 2(p2, M2)
cot 0+ o(1) . (5.9)
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For the value ofgiven by (5.9) we find by the Taylor expansion for the
cosine function that
cos= cos 0+ 1/2 sin 0+ o(
1/2)
= cos 0
1 +
8D(1+ 2)
1/2
+ o(1/2) .(5.10)
Finally, plugging (5.10) in (5.5) gives (5.1).
Proof of Theorem 1.1 completed. Since the energy estimate (1.22) follows
from Proposition 4.1 and Proposition 5.1, it remains to prove the conver-
gence result, i.e., assertion (i). Passing to a subsequence, we may deduce
from Lemma 4.6 that limp
n
1 =p1=
(i0)
1 and limp
n
2 =p2=
(i0)
2 for somegeodesic (i0) G. Moreover, from (4.53) and (5.1) we deduce that there isa pair of corresponding points, x(j) Mj , j = 1, 2,that together with(i0),realize the minimumK in (1.23). Next we shall show that unx(1) uni-formly on [0+ ,
2] for any >0 (the uniform convergenceunx(2)
on [2 + , 0 ] is proved in the same manner). A direct consequence of(3.34) is that
d(u, 1)
0 uniformly on [0+ ,
2] . (5.11)
Therefore, whenever a sequence{n} [0+ , 2 ] satisfies un(n)x1, then necessarilyx11. Combining (4.21) with the upper-bound (5.1)(or (2.9)) we get /21/3
1
|v1,|2 +
2d(1/2)W (p
1, p
2)
(|v1,()| m1)
cos
C 1/2 ,(5.12)
where v1, = s1(u). A direct consequence of (5.12) is that for each >0
we have
meas{[0+ ,
2 ]} :|v1,()| m1 >
C()1/2 . (5.13)
Since (5.13) implies that lim0 |v1,()| = m1 in measure, for a subse-quence we have limn0 |v1,n()| = m1 a.e. on (0, 2 ). Consider then apoint [0 +, 2] such that (possibly for a further subsequence)lim
n0u
n
() = l i mn0 v1,
n
() = x1 M
1. From the proof of
Lemma 4.5 it follows that x1 must coincide with x(1); otherwise we would
get a contradiction to the upper bound (5.1). In the last conclusion we
used hypothesis (H4), which implies that the distance according to the
expression in (1.19) between two distinct points of 1 is positive. The
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above argument implies that if limn0un(n) = limn0 v1,n(n) = x,
with {n} [0+, 2 ], thenx = x(1). The claimed uniform convergenceon [0+ ,
2 ] follows.
References
[1] L. Ambrosio, N. Fusco and D. Pallara, Functions of Bounded Variation andFree Discontinuity Problems, Oxford Mathematical Monographs. Oxford Uni-versity Press, New York, 2000.
[2] N. Andre and I. Shafrir,On a singular perturbation problem involving acircular-well potential, Trans. Amer. Math. Soc. 359(2007), 47294756.
[3] N. Andre and I. Shafrir, On a minimization problem with a mass constraint
involving a potential vanishing on two curves, preprint.[4] I. Fonseca and L. Tartar,The gradient theory of phase transitions for systemswith two potential wells, Proc. Roy. Soc. Edinburgh Sect. A 111 (1989), 89102.
[5] E. Giusti, Minimal Surfaces and Functions of Bounded Variation, Monographsin Mathematics, 80, Birkhauser Verlag, Basel, 1984.
[6] O. Lopes, Radial and nonradial minimizers for some radially symmetric func-tionals, Electron. J. Differential Equations 3 (1996) (electronic).
[7] P. Sternberg, The effect of a singular perturbation on nonconvex variationalproblems, Arch. Rational Mech. Anal. 101(1988), 209260.
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SOME REMARKS ON LIOUVILLE TYPE THEOREMS
H. BREZIS
Laboratoire Jacques-Louis Lions, universite Pierre et Marie Curie, 175, rue du
Chevaleret, 75013 Paris, France and Rutgers University, Department of
Mathematics, Hill Center, Busch Campus, 110, Frelinghuysen Road,
Piscataway, NJ 08854, USA, e-mail: [email protected]
M. CHIPOT, Y. XIE
Institut fur Mathematik, Abt. Angewandte Mathematik, Universitat Zurich,
Winterthurerstrasse 190, CH8057 Zurich, Switzerland, e-mail:
[email protected], [email protected]
Abstract: The goal of this note is to present elementary proofs of statements
related to the Liouville theorem.
1. Introduction
We denote by A(x) = (aij(x)) a (k k)-matrix where the functions aij ,
i, j = 1, . . . , k are bounded measurable functions defined on Rk and which
satisfy, for some , > 0, the usual uniform ellipticity condition:
||2 (A(x) ) ||2 a.e. x Rk, Rk. (1.1)
We address here the issue of existence of solutions to the equation:
(A(x)u(x)) +a(x)u(x) = 0 in D(Rk), (1.2)
where aLloc(Rk) and a 0. When a= 0 and (A) = , the usual
Laplace operator, the above equation is the so called stationary Schrodinger
equation for which a vast literature is available (see [16], [20]). When a = 0
it is well known that every bounded solution to (1.2) has to be constant
(see e.g. [5], [11], [12] and also [4], [19] for some nonlinear versions). The
case where a = 0, and k 3 is very different and in this case non trivial
bounded solutions might exist.
Many of the results in this paper are known in one form or another
(see for instance [1], [2], [3], [10], [9], [14], [16], [17]) but we have tried to
develop here simple self-contained pde techniques which do not make use of
43
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probabilities, semigroups or potential theory as is sometimes the case (see
e.g. [2], [3], [8], [17], [18]). One should note that some of our proofs extend
also to elliptic systems.
This note is divided as follows. In the next section we introduce anelementary estimate which is used later. In Section 3 we present some
Liouville type results, i.e., we show that under some conditions on a, (1.2)
does not admit nontrivial bounded solutions. Finally in the last section we
give an almost sharp criterion for the existence of nontrivial solutions.
2. A preliminary estimate
Let us denote by a bounded open subset ofR
k
with Lipschitz boundaryand starshaped with respect to the origin. For any r R we set
r =r. (2.1)
Let us denote by a smooth function such that
0 1, = 1 on 1/2, = 0 outside , (2.2)
|| c, (2.3)
wherec denotes some positive constant.Lemma 2.1. Suppose thatu H1loc(R
k)satisfies(1.2)withA(x)satisfying
(1.1). Then there exists a constantC independent ofr such thatr
{|u|2 +au2}2x
r
dx
C
r
r\r/2
|u|22
x
r dx
12
r\r/2
u2 dx
12
,
(2.4)
where | | denotes the usual euclidean norm inRk.
Proof. By (1.2) we have for every v H10 (r)r
Au v+auv dx= 0. (2.5)
Taking
v=u
2xr
= u
2
(2.6)yields
r
Au {u2} +au22 dx= 0. (2.7)
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Since
2 = 2
r
x
r we obtain
r
{Au u}2 +au22 dx= 1
r
r\r/2
Au 2u dx
C1
r
r\r/2
|u||u| dx,
whereC1 is a constant depending on aij and c only. Using the ellipticity
condition (1.1) it follows easily that
min(1, )
r
{|u|2 +au2}2 dx C1
r
r\r/2
|u||u| dx.
By the CauchySchwarz inequality we haver
{|u|2 +au2}2 dx
C1
r min(1, )
r\r/2
|u|22 dx1/2
r\r/2
u2 dx1/2
.
This completes the proof of the lemma.
3. Some Liouville type results
3.1. The case where the growth of u is controled
In this case we have
Theorem 3.1. Under the asumptions of Lemma 2.1, let u be solution to
(1.2)such that forr large,
1
r2
r\r/2
u2 dx C (3.1)
whereC is a constant independent ofr, thenu= constant and ifa 0 or
k 3 one hasu= 0.
Proof. From (2.4) we derive thatr
|u|22 dx C
r
r
|u|22 dx
1/2r\r/2
u2 dx
1/2
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and thus
r/2|u|2 dx
r|u|22 dx
C
r2 r\r/2u2 dx CC.
It follows that the nondecreasing function
r
r
|u|2 dx
is bounded and has a limit when r +. Going back to (2.4) we have
r/2
|u|2 dx c
r
r\r/2
|u|2 dx1
2
r
for some constant c. This implies
r/2
|u|2 dx c
r
|u|2 dx
r/2
|u|2 dx
12
0
as r + and the result follows.
Remark 3.1. Whenk 2 condition (3.1) is satisfied ifuis bounded, andin this case the only bounded solution of (1.2) is u = 0. Therefore we
will assume throughout the rest of this paper that k 3.
We denote byrthe first eigenvalue of the Neumann problem associated
to the operator A +a in r \ r/2, i.e., we set
r = Infr\r/2
Au u+au2 dx: u H1(r \ r/2),r\r/2
u2 dx= 1
.
(3.2)
One remarks easily that ifuis a minimizer of (3.2) so is |u|. One can show
then that the first eigenvalue is simple. Moreover we have
Theorem 3.2. Under the assumptions of Lemma 2.1, suppose that for
some constantsC0 >0,
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Proof. From the definition ofr we haver\r/2
u2 dx
1r
r\r/2
Au u+au2 dx
u H1(r \ r/2).
(3.4)
Going back to (2.4) we findr
(|u|2 +au2)2x
r
dx
C
r r
(|u|2 +au2)2x
r dx1/2
r\r/2
u2 dx1/2
,
which leads tor
(|u|2 +au2)2x
r
dx
C
r2
r\r/2
u2 dx
for some constant C independent ofr. Using in particular (2.2) we obtainr/2
|u|2 +au2 dx C
r2
r\r/2
u2 dx, r >0. (3.5)
From (3.3) and (3.4) we derive that, for some constant C,r/2
|u|2 +au2 dx C
r2
r\r/2
|u|2 +au2 dx
C
r2
r
|u|2 +au2 dx r >0. (3.6)
Iteratingp-times this formula leads to
r/2
|u|2
+au2
dx
Cp
r(2)p2p1r
|u|2
+au2
dx.
By (3.5) it follows that it holdsr/2
|u|2 +au2 dx Cp
r(2)p+2
2pr
u2 dx,
for some constantCp independent ofr. If nowu is supposed to be bounded
byMwe get
r/2
|u|2 +au2 dx Cpr(2)p+2
M2|2pr| = Cp||M2
(2p
r)k
r(2)p+2 .
(|2pr| denotes the Lebesgue measure of the set 2pr). Choosing (2)p +
2> k the result follows by letting r +.
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Remark 3.2. Under the assumption of Theorem 3.2 we have obtained in
fact that (1.2) can not admit a nontrivial solution with polynomial growth.
Of course this result is optimal since Re(ez) =ex1 cos x2 is harmonic in Rk
for any k 2. One should note that Theorem 3.2 applies also to systemssatisfying the Legendre condition when auv is replaced by a nonnegative
bilinear forma(u, v) (see [6], [7]).
We now discuss some conditions on a which imply (3.3). We have
Theorem 3.3. Suppose that for |x| large enough
a(x) c
|x|, 0.
We have
r\r/2
Ar v+arv dx= rr\r/2
rv dx v H1(r \ r/2).
Taking v= 1 yieldsr\r/2
a(x)rdx= r
r\r/2
rdx.
Using (3.7) we derive, for some constant C,
C
rr\r/2 rdx r
r\r/2 rdx
and the result follows.
We now consider other cases where (3.3) holds, in particular when no
decay is imposed to a. We are interested for instance in the case where at
infinity a has enough mass locally. We start with the following lemma:
Lemma 3.1. Let (for instance) Q = (0, 1)k be the unit cube in Rk. For
any > 0 and > 0 there exists = (, ) such that if the function asatisfies
0 a a.e. x Q,
Q
a dx , (3.8)
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then
Q
v2 dx
Q
|v|2 +av2 dx v H1(Q). (3.9)
Proof. If not, there exists , and a sequence of functions an, vn such
thatan satisfies (3.8) and vn H1(Q) is such that
1
n
Q
v2ndx
Q
|vn|2 +anv
2ndx. (3.10)
Dividing by |vn|2theL2-norm ofvnwe can assume without loss of generality
that Q
v2ndx= 1. (3.11)
By (3.10), (3.11) we have thenQ
|vn|2 dx
1
n,
Q
v2ndx= 1 (3.12)
andvn is uniformly bounded in H1(Q). Therefore
vn 1 in H1(Q). (3.13)
From (3.10) we have Q
anv2ndx
1
n. (3.14)
Thus
Q
andx= Q
anv2ndx+
Q
an(1 v2n) dx
1
n+|1 vn|2|1 +vn|2 0
whenn +. Impossible. This completes the proof of the lemma.
With the notation of Section 2 we set
= (1, 1)k. (3.15)
We consider the lattice generated by Q= (0, 1)k i.e., the cubes
Qi=Qzi =zi+Q zi Zk.
Then we have
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Theorem 3.4. Suppose that forn large enough,
Qia(x) dx Qi R
k \ n, (3.16)
then
2n 1
1
n (3.17)
whereis defined in Lemma 3.1 and denotes the maximum of two num-
bers.
Proof. Indeed by Lemma 3.1 after a simple translation fromQi intoQ we
have
Qi
u2 dx
Qi
|u|2 +au2 dx Qi Rk \ n u in H
1(Qi).
This leads clearly to
2n\n
u2 dx
2n\n
|u|2 +au2 dx
2n\n
1
Au u+au2 dx
1
1
2n\n
Au u+au2 dx u H1(2n \ n).
The result follows then from (3.2).
Remark 3.3. Combining Theorems 3.2 and 3.4 it follows that (1.2) cannothave a nontrivial bounded solution (or of polynomial growth) when (3.16)
holds. This is the case when at infinity
a a0 >0
or more generally
a ap (3.18)
whereap is a periodic function with period Q.
In the case when (3.3) holds with = 2 the technique of Theorem 3.2
cannot be applied. However, we will show that the non existence of nontriv-
ial solutions can be established in this case too i.e., condition (3.3) is not
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sharp if we impose certain growth condition on {aij(x)}. Before turning
to this let us prove some general comparison result. For simplicity we will
denote also by A the operator Au= xi(aijxj ).
Proposition 3.1. Suppose thatO is a bounded open subset ofRk. Leta1,
a2 be two bounded functions satisfying
a1 a2 0 a.e. inO. (3.19)
Letu1, u2 H1(O) be such thatAu2+a2u2 Au1+a1u1 0 inO,
u2 (u1 0) onO,(3.20)
then
u2 (u1 0) inO. (3.21)
Proof. The inequality
Au+au 0 inO
means O
aijxjuxiv+auv dx 0 v H10 (O), v 0.
Consideringv =u2 and Au2+a1u2 0 leads tou2 0. Next considering
v= (u1 u2)+ H10 (O) and (3.20) we obtain
O
aijxju1xi(u1 u2)+ +a1u1(u1 u2)
+ dx
O aijxju2xi(u1 u2)
+
+a2u2(u1 u2)+
dx.
HenceO
aijxj (u1 u2)xi(u1 u2)+ + (a1u1 a2u2)(u1 u2)
+ dx 0.
Now on u1 u2 one has a1u1 a1u2 a2u2 and it follows that (u1
u2)+ = 0.
Next we prove
Theorem 3.5. Assume that there existsR, large enough, such that
a(x) c0r2
|x| R >0
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wherec0 is a positive constant. In addition to (1.1), suppose thataij(x)
C1(Rk\B(0, R)) satisfies for some positiveD:
xi(aij(x))xj D |x| > R.
(In the above inequality we make the summation convention of repeated
indices). Then the equation
xi(aij(x)xj u) +a(x)u= 0 (3.22)
cannot have nontrivial bounded solution.
Proof. Letun be the solution toxi(aij(x)xj un) +a(x)un = 0 in B(0, n),un = 1 on B(0, n),
(3.23)
whereB(0, n) denotes the ball of center 0 and radius n. From Propositi