Open Access

Layer solutions for a class of semilinear elliptic equations involving fractional Laplacians

Boundary Value Problems20142014:41

https://doi.org/10.1186/1687-2770-2014-41

Received: 4 November 2013

Accepted: 4 February 2014

Published: 17 February 2014

Abstract

This paper is concerned with the nonlinear equation involving the fractional Laplacian: ( ) s v ( x ) = b ( x ) f ( v ( x ) ) , x R , where s ( 0 , 1 ) , b : R R is a periodic, positive, even function and −f is the derivative of a double-well potential G. That is, G C 2 , γ ( 0 < γ < 1 ), G ( 1 ) = G ( 1 ) < G ( τ ) τ ( 1 , 1 ) , G ( 1 ) = G ( 1 ) = 0 . We show the existence of layer solutions of the equation for s 1 2 and for some odd nonlinearities by variational methods, which is a bounded solution having the limits ±1 at ±∞. Asymptotic estimates for layer solutions as | x | + and the asymptotic behavior of them as s 1 are also obtained.

MSC:35B20, 35B40, 49J45, 82B26.

Keywords

fractional Laplacianlayer solutionsexistencelocal minimizers

1 Introduction

In this paper we study the fractional Laplacian
( ) s v ( x ) = b ( x ) f ( v ( x ) ) , x R ,
(1.1)
where s ( 0 , 1 ) , and ( ) s is the fractional Laplacian defined by
( ) s v = C s  P.V.  R v ( x ) v ( y ) | x y | 1 + 2 s d y .

Here P.V. stands for the Cauchy principle value and C s is a positive constant multiplier depending only on s.

The fractional Laplacian is a nonlocal operator which can be localized as
{ div ( y a u ) = 0 in  R + 2 , lim y 0 + y a u y = 1 d s b ( x ) f ( u ) on  R + 2 ,
(1.2)
where a = 1 2 s ( 1 , 1 ) , d s = 2 2 s 1 Γ ( s ) Γ ( 1 s ) and u ( x , 0 ) = v ( x ) . Moreover u ( , ) can be expressed by a Poisson kernel,
u ( x , y ) = P s ( , y ) v = p s R y 2 s ( | z | 2 + y 2 ) 1 + 2 s 2 v ( x z ) d z for every  y > 0 ,

which is called the s-extension of v. p s is a positive constant depending only on s. For more details as regards the fractional Laplacian, readers can refer to [17] and the references therein.

In view of the celebrated De Giorgi conjecture (see [810]), Cabré and Sire [2, 3] considered layer solutions of the nonlocal equation
( ) s v = f ( v ) in  R .
(1.3)
The necessary and sufficient conditions for the existence of one-dimensional layer solutions were given as
G ( 1 ) = G ( 1 ) < G ( s ) s ( 1 , 1 ) , G ( 1 ) = G ( 1 ) = 0 ,

where G = f . All these were obtained by a Hamiltonian equality and a Modica-type estimate for layer solutions. By the sliding method, the layer solution of (1.3) was proved to be the unique local minimizer which increases in x with values varying from −1 to 1. The regularity, Hopf principle, maximum principle as well as a Harnack inequality for (1.3) or for its extension equation (1.2) (in this case b = 1 ) were given. Some of them will be used in our paper.

If b is not a constant and is periodic, the perturbed equation (1.1) becomes complicated. The aim of this paper is to study the layer solution of (1.1) with periodic perturbed nonlinearity.

Definition 1.1 A function v ( L C β ) ( R ) ( 0 < β < 1 ) is said to be a layer solution of (1.1), if v solves (1.1),
( ) s v ( x ) = b ( x ) f ( v ( x ) ) , x R
and
lim x ± v ( x ) = ± 1 .
Definition 1.2 A function u L ( R + 2 ) C β ( R + 2 ¯ ) is said to be a layer solution of (1.2), if u solves (1.2),
{ div ( y a u ) = 0 in  R + 2 , lim y 0 + y a u y = 1 d s b ( x ) f ( u ) on  R + 2
and
lim x ± u ( x , 0 ) = ± 1 .

Namely, u ( x , 0 ) is the corresponding layer solution of (1.1).

Different from the unperturbed case (1.3), the inhomogeneous term b ( x ) f ( u ) depends explicitly on x in (1.1) and (1.2); the sliding method cannot be used and layer solutions of them have no monotonicity in the direction of x. The method for obtaining layer solutions in [2] and [3] cannot be used in our case directly; some difficulties need to be solved.

In the paper, we consider the extension problem (1.2). Obviously, (1.2) has a variational structure.

Denote
Ω R + 2 , a bounded Lipschitz domain , B R ( x , y ) R 2 , a ball centered at  ( x , y ) R 2  with radius  R , B ϵ + ( x , 0 ) = B ϵ ( x , 0 ) R + 2 , 0 Ω = { ( x , 0 ) Ω R + 2 | ϵ > 0 , B ϵ + ( x , 0 ) Ω } , + Ω = Ω R + 2 ¯ .
For u H 1 ( y a , Ω ) , the norm is
u H 1 ( y a , Ω ) = ( Ω y a | u | 2 d x d y ) 1 2 + ( Ω y a | u | 2 d x d y ) 1 2 .
The energy functional of u on Ω is given by
E ( u , Ω ) = d s Ω y a 2 | u | 2 d x d y + 0 Ω b ( x ) G ( u ( x , 0 ) ) d x .
(1.4)

We state our main results in the following.

We show, via a Liouville result, the existence of layer solutions of (1.1) for s 1 2 and for some odd nonlinearities.

Theorem 1.1 Let s 1 2 . Assume that b , f C 1 , γ ( R ) ( 0 < γ < 1 ):
  1. (1)

    b : R R is 1-periodic, even, not constant and positive; denote b ¯ = max R b and b ̲ = min R b ;

     
  2. (2)

    f ( τ ) = f ( τ ) for any τ [ 1 , 1 ] , f ( 1 ) = f ( 1 ) = f ( 0 ) = 0 , f > 0 in ( 0 , 1 ) and f < 0 in ( 1 , 0 ) .

     
Obviously, if G = f ,
G ( 1 ) = G ( 1 ) < G ( τ ) for  τ ( 1 , 1 ) , G ( 1 ) = G ( 1 ) = 0 .
There exists a layer solution v C 2 , β ( R ) (for some 0 < β < 1 ) of (1.1):
{ ( x x ) s v ( x ) = b ( x ) f ( v ( x ) ) in  R , v ± 1 as  x ± .
(1.5)

In addition, v is odd.

Furthermore we obtain asymptotic estimates of the layer solutions of (1.1) by comparing with a layer solution of the unperturbed equation (1.3).

Theorem 1.2 Let b C 1 , γ L is positive. Let f C 1 , γ ( R ) ( γ > max ( 0 , 1 2 s ) ) satisfy
  1. (i)

    G ( 1 ) = G ( 1 ) < G ( τ ) for τ ( 1 , 1 ) , G ( 1 ) = G ( 1 ) = 0 ;

     
  2. (ii)

    G ( 1 ) > 0 , G ( 1 ) > 0 .

     
If v is a layer solution of (1.1), then the following asymptotic estimates hold:
c x 2 s | 1 v | C x 2 s for  x > 1 ,
(1.6)
c | x | 2 s | 1 + v | C | x | 2 s for  x < 1
(1.7)

for some constants 0 < c < C .

Finally we investigate the asymptotic behavior of v s as s 1 and obtain a local elliptic equation, which is stated as follows.

Theorem 1.3 Let s [ 1 2 , 1 ) . Let { v s k } be a sequence of layer solutions of (1.1) in Theorem  1.1. Then, there exists a subsequence denoted again by { v s k } converging locally uniformly to a function v 1 C 2 ( R ) as s k 1 , which is also a layer solution of the local elliptic equation
{ v x x 1 ( x ) = b ( x ) f ( v 1 ( x ) ) in  R , lim x ± v 1 ( x ) = ± 1 .
(1.8)
In addition,
1 2 ( v x 1 ) 2 = b ( x ) { G ( v 1 ( x ) ) G ( 1 ) } + x + b ( t ) { G ( v 1 ( t ) ) G ( 1 ) } d t .
(1.9)

For convenience of the presentation we will use C for a general positive constant; such a C is usually different in different contexts.

2 Some preliminaries and properties

In this paper, we mainly study the extension equation (1.2). To make our problems clear, we present several properties of layer solutions.

Lemma 2.1 Let u be a bounded solution of (1.2),
{ div ( y a u ) = 0 in  R + 2 , lim y 0 + y a u y = 1 d s b ( x ) f ( u ) on  R + 2
and
lim x ± u ( x , 0 ) = L ±
(2.1)
with two constants L ± . Then,
  1. (1)
    f ( L + ) = f ( L ) = 0 ;
    (2.2)
     
  2. (2)
    lim x ± u ( x , y ) = L ±
    (2.3)
     
for every y 0 ;
  1. (3)
    u L ± L ( B R + ( x , 0 ) ) 0 as  x ± ;
    (2.4)
     
  2. (4)
    x u L ( B R + ( x , 0 ) ) + y a u y L ( B R + ( x , 0 ) ) 0 as  x ± .
    (2.5)
     

Proof Our proof uses the invariance of the problem under periodic translations in x and a compactness argument.

Denote u n ( x , y ) = u ( x + n , y ) for n Z . Since b is 1-periodic, u n still satisfies the equations
{ div ( y a u n ) = 0 in  R + 2 , lim y 0 + y a u n y = 1 d s b ( x ) f ( u n ( x , 0 ) ) on  R + 2 .
(2.6)
By regularity results in [2] and [5], we see that up to a subsequence,
u n u ± in  C loc 0 ( R + 2 ¯ ) , x u n x u ± in  C loc 0 ( R + 2 ¯ ) , y a u n y y a u y in  C loc 0 ( R + 2 ¯ )
as n ± . Then u ± solves the equations
{ div ( y a u ± ) = 0 in  R + 2 , lim y 0 + y a u ± y = 1 d s b ( x ) f ( u ± ) on  R + 2 ,
(2.7)

and it follows that u ± ( x , 0 ) L ± for every x R .

Consider the Dirichlet problem
{ div ( y a u ± ) = 0 in  R + 2 , u ± ( x , 0 ) L ± on  R + 2 .
(2.8)

u ± L ± is the unique solution of (2.8) by Corollary 3.5 in [2]. As a consequence, (2.2) and (2.5) are obvious. □

The following lemma is a necessary condition for a local minimizer of the energy functional .

Lemma 2.2 Let u be a local minimizer of the energy functional under perturbations in [ 1 , 1 ] . That is, for any bounded Lipschitz domain Ω R + 2 and for any ξ H 1 ( y a , Ω ) having compact support in Ω 0 Ω such that u + ξ [ 1 , 1 ] ,
E ( u , Ω ) E ( u + ξ , Ω ) .
Let
lim x ± u ( x , 0 ) = ± 1 .
(2.9)
Then
G ( 1 ) = G ( 1 ) G ( τ ) for all  τ ( 1 , 1 ) .
(2.10)

Proof To show (2.10), it is sufficient to prove that G ( 1 ) G ( τ ) and G ( 1 ) G ( τ ) for all τ [ 1 , 1 ] . Suppose G ( τ 0 ) < G ( 1 ) for some point τ 0 [ 1 , 1 ] by contradiction. For simplicity, assume that G ( τ 0 ) = 0 by adding a constant.

By (2.9),
lim inf l + E ( u , B R + ( l , 0 ) ) lim inf l + 0 B R + ( l , 0 ) b ( x ) G ( u ( x , 0 ) ) 2 b ̲ ε R
(2.11)

for some ε > 0 .

Let ξ R be a cut-off function with values in [ 0 , 1 ] ,
ξ R = { 1 in  B ( 1 η ) R , 0 in  R n + 1 B R ,

where η ( 0 , 1 ) will be specified later, and | ξ R | 1 η R .

Define ξ R , l ( x , y ) = ξ R ( x l , y ) . Let w = τ 0 ξ R , l + ( 1 ξ R , l ) u , then w = u on + B R + ( l , 0 ) and w τ 0 in B ( 1 η ) R + ( l , 0 ) . We have
lim sup l + E B R + ( l , 0 ) ( w ) = lim sup l + { d s B R + ( l , 0 ) y a 2 | ( 1 ξ R , l ) u + ( τ 0 u ) ξ R , l | 2 + 0 B R + ( l , 0 ) b ( x 1 ) G ( w ) } 2 d s B R + y a | ξ R , l | 2 + 2 b ¯ max [ 1 , 1 ] G η R C d s R a η 2 + 2 b ¯ max [ 1 , 1 ] G η R .
(2.12)

We use (2.5) in the first inequality above.

Having chosen η = b ̲ ε 2 b ¯ max [ 1 , 1 ] G , by (2.11) and (2.12),
lim sup l + E ( w , B R + ( l , 0 ) ) < lim inf l + E ( u , B R + ( l , 0 ) )

for large R > 1 . This contradiction leads to G ( 1 ) G ( τ ) for all τ [ 1 , 1 ] . By the same discussion, G ( 1 ) G ( τ ) for all τ [ 1 , 1 ] . Thus we complete the proof. □

As in [2], we construct a Hamiltonian equality which will be used in the proof of Theorem 1.3. For this purpose a lemma is in order, for whose proof see Lemma 5.1 in [2].

Lemma 2.3 Let u L ( R + 2 ) be a solution of (1.2). Then for every x R , 0 y a | u | 2 d y < . In addition, the integral can be differentiated with respect to x R under the integral sign. We have
lim M + M y a | u | 2 d y = 0
(2.13)
uniformly in x R . If u is a layer solution of (1.2),
lim | x | + 0 y a | u | 2 d y = 0 .
(2.14)

Proposition 2.1 (Hamiltonian equality)

Let u be a layer solution of (1.2) for a ( 1 , 1 2 ) , i.e.,
{ div ( y a u ) = 0 in  R + 2 , lim y 0 + y a u y = 1 d s b ( x ) f ( u ) on  R + 2 , u ( x , 0 ) ± 1 as  x .
For every x R , the Hamiltonian equality holds:
d s 0 y a 2 { u x 2 u y 2 } d y = b ( x ) { G ( u ( x , 0 ) ) G ( 1 ) } + x b ( t ) { G ( u ( t , 0 ) ) G ( 1 ) } d t .
(2.15)
As a consequence,
+ b ( x ) { G ( u ( x , 0 ) ) G ( 1 ) } d x = 0 .
(2.16)
Proof We note that the integral in (2.16) is well defined since s = 1 a 2 ( 1 4 , 1 ) and
G ( u ( x , 0 ) ) G ( 1 ) = G ( t ) 2 ( u ( x , 0 ) 1 ) 2 = O ( | x | 4 s ) as  | x | ,

where t is some point between u ( x , 0 ) and 1.

By Lemma 2.3, the left integral in (2.15) can be differentiated with respect to x,
d d x d s 0 y a 2 { u x 2 u y 2 } d y = d s 0 y a { u x u x x u y u y x } d y = d s 0 y a { u x x + u y y + a y u y } u x d y + d s lim y 0 + y a u y u x = b ( x ) f ( u ( x , 0 ) ) u x ( x , 0 ) .
In the second equality above we use the fact that lim y y a u y u x = 0 (see [2]). We have
d d x { b ( x ) ( G ( u ( x , 0 ) ) G ( 1 ) ) + x + b ( t ) ( G ( u ( t , 0 ) ) G ( 1 ) ) d x } = b ( x ) f ( u ( x , 0 ) ) u x ( x , 0 ) .
Thus,
d s 0 y a 2 { u x 2 u y 2 } d y b ( x ) { G ( u ( x , 0 ) ) G ( 1 ) } + x + b ( t ) { G ( u ( t , 0 ) ) G ( 1 ) } d t + C .
(2.17)

Let x + , the left of (2.17) converging to zero by (2.14); thus C = 0 and (2.15) is proved. Letting x , (2.16) is also obtained. □

To study asymptotic estimates of layer solutions of (1.1), we recall an explicit layer solution of the unperturbed problem (1.3).

Lemma 2.4 ([3], Theorem 3.1)

Let s ( 0 , 1 ) . For every t > 0 , the C function
v s t ( x ) = sign ( x ) 2 π 0 sin ( z ) z e t ( z | x | ) 2 s d z
(2.18)
is the layer solution to the fractional equation
( x x ) s v s t = f s t ( v s t ) in  R ,
(2.19)
for a nonlinearity f s t C 1 ( [ 1 , 1 ] ) which is odd and twice differentiable in [ 1 , 1 ] and which satisfies
f s t ( 0 ) = f s t ( 1 ) = 0 , f s t > 0  in  ( 0 , 1 ) , ( f s t ) ( ± 1 ) = 1 t .
In addition, the following limits exist:
lim | x | | x | 1 + 2 s ( x v s t ) ( x ) = t 4 s π sin ( π s ) Γ ( 2 s ) > 0
(2.20)
and, as a consequence,
lim x ± | x | 2 s | ( v s t ) ( x ) 1 | = t 2 π sin ( π s ) Γ ( 2 s ) > 0 .
(2.21)

3 Existence and asymptotic estimates

To prove the existence of layer solutions, we introduce a Liouville result where a 0 is required. This is the reason why we restrict ourselves to the case s 1 2 in Theorem 1.1.

Proposition 3.1 Let a 0 . Suppose u is a bounded nonnegative function which satisfies weakly the problem
{ div ( y a u ) 0 in  R + 2 , lim y 0 + y a u y 0 on  R + 2 .
(3.1)

Then u C a.e. in R + 2 .

Proof Since a 0 , R a 1 for R > 1 . Let ξ be a smooth function with values in [ 0 , 1 ] , ξ = 1 in B R and ξ = 0 outside of B 2 R , | ξ | C R 1 . Multiplying (3.1) with u ξ 2 and integrating by parts, we have that
B R + y a | u | 2 B 2 R + y a | u | 2 ξ 2 2 R + 2 y a ξ u | u | | ξ | 2 { B 2 R + B R + y a | u | 2 ξ 2 } 1 2 { B 2 R + B R + y a | ξ | 2 u 2 } 1 2 C { B 2 R + B R + y a | u | 2 ξ 2 } 1 2 ( R R 1 + a R 2 ) 1 2 C { B 2 R + B R + y a | u | 2 ξ 2 } 1 2 .

Thus R + 2 y a | u | 2 C for some constant C independent of R. Let R , B 2 R + B R + y a | u | 2 ξ 2 0 . We deduce that R + 2 y a | u | 2 = 0 and u C a.e. in R + 2 . □

Next we prove an existence result about the local minimizer of .

Lemma 3.1 Let Ω R + 2 be a bounded Lipschitz domain. Let w 0 C 0 ( Ω ¯ ) H 1 ( y a , Ω ) be a given function with | w 0 | 1 ; b is a bounded positive function.

Suppose that
f ( 1 ) 0 f ( 1 ) ,
the energy functional E ( u , Ω ) admits a minimizer u C w 0 , a = { w H 1 ( y a , Ω ) , 1 w 1  a.e. in  Ω , w = w 0  on  + Ω  in the weak sense } , which solves weakly
{ div ( y a u ) = 0 in  Ω , lim y 0 + y a u y = 1 d s b ( x ) f ( u ( x , 0 ) ) on  0 Ω , u = w 0 on  + Ω .
(3.2)
Moreover, u is a stable solution of (3.2), i.e.,
d s Ω y a | ξ | 2 d x d y 0 Ω b ( x ) f ( u ) ξ 2 d x 0 ,
(3.3)

for every ξ H 1 ( Ω , y a ) such that ξ 0 on + Ω in the weak sense.

Proof Consider the set H w 0 , a ( Ω ) = { w H 1 ( y a , Ω ) , w w 0  on  + Ω  in the weak sense } C w 0 , a , H w 0 , a ( Ω ) since w 0 H w 0 , a ( Ω ) . Denote
f ˜ = { f ( 1 ) if  t 1 , f if 1 < t < 1 , f ( 1 ) if  t 1 ,

and G ˜ = 0 u f ˜ . Up to an additive constant, G ˜ = G in [ 1 , 1 ] .

Consider the energy functional
E ˜ ( u , Ω ) = d s Ω y a 2 | u | 2 d x d y + 0 Ω b ( x ) G ˜ ( u ( x , 0 ) ) d x .
(3.4)

If E ˜ has an absolute minimizer u in C w 0 , a ( Ω ) , the statement of Lemma 3.1 is proved.

For every function w H w 0 , a ( Ω ) , w w 0 H 1 ( y a , Ω ) and vanishes on + Ω in the weak sense. We can extend w w 0 in R + 2 by zeroes outside of Ω and w w 0 H 1 ( y a , R + 2 ) . By the trace theorem and the Sobolev imbedding theorem (see [7, 11, 12]),
H 1 ( y a , R + 2 ) L p ( R )

for p = 2 1 2 s if s < 1 2 or for any 1 p < if s 1 2 . Moreover, H 1 ( y a , R + 2 ) L 2 ( 0 Ω ) .

Since G ˜ has linear growth at infinity, E ˜ is well defined, bounded below and coercive in H w 0 , a . There exists an absolute minimizer u H w 0 , a . By the first order variation, we have
{ div ( y a u ) = 0 in  Ω , lim y 0 + y a u y = 1 d s b ( x 1 ) f ˜ ( u ( x , 0 ) ) on  0 Ω .
(3.5)
Multiply ( u 1 ) + with (3.5) and integrate in Ω,
d s Ω y a | ( u 1 ) + | 2 d x d y 0 Ω b ( x ) f ( 1 ) ( u 1 ) + d x = 0 .

Since f ( 1 ) 0 , Ω y a | ( u 1 ) + | 2 d x d y 0 . Thus ( u 1 ) + 0 a.e. in Ω, i.e., u 1 a.e. in Ω. Similarly we also get u 1 a.e. in Ω. Hence u C w 0 , a ( Ω ) . (3.2) follows from (3.5), and (3.3) comes from the second order variation of . □

Remark 3.1 Suppose that b is an even function, f and w 0 are odd with respect to x, with a slight modification we can also show that there is an odd minimizer in the admissible set { w C w 0 , a | w ( x , y ) = w ( x , y )  for every  y 0 } .

Now we start to show the existence of layer solutions of (1.2).

Theorem 3.1 Let s 1 2 . Let b ( C 1 , γ L ) ( R ) and f C 1 , γ ( R ) ( 0 < γ < 1 ):
  1. (a)

    b : R R is an even positive function, b ( x + 1 ) = b ( x ) x R ,

     
  2. (b)

    f ( τ ) = f ( τ ) for any τ [ 1 , 1 ] , f ( 1 ) = f ( 1 ) = f ( 0 ) = 0 , f > 0 in ( 0 , 1 ) and f < 0 in ( 1 , 0 ) .

     
Then there exists a layer solution u of (1.2) in R + 2 :
{ div ( y a u ) = 0 in  R + 2 , lim y 0 + y a u y = 1 d s b ( x ) f ( u ( x , 0 ) ) on  R + 2 ,
(3.6)
which is odd with respect to x, i.e., u ( x , y ) = u ( x , y ) , and, for every y 0 ,
lim x ± u ( x , y ) = ± 1 .
(3.7)
Furthermore, u is a local minimizer of the energy functional under odd perturbations in [ 1 , 1 ] , and it is stable in the sense that
d s R + 2 y a | ξ | 2 d x d y R b ( x ) f ( u ( x , 0 ) ) ξ 2 d x 0
(3.8)

for every function ξ C 1 ( R + 2 ¯ ) with compact support in R + 2 ¯ , ξ ( x , y ) = ξ ( x , y ) and u + ξ [ 1 , 1 ] .

Proof The proof is divided into three parts. For simplicity, we make G ( 1 ) = G ( 1 ) = 0 by adding a constant.

Step 1. We show that there exists a solution with values in [ 1 , 1 ] of (3.6) which is odd with respect to the variable x for every y 0 .

Let Q R = [ R , R ] × [ 0 , R ] and w 0 = arctan x arctan R . Define the admissible set
C w 0 , a , o = { w C w 0 , a ( Q R ) , y 0 , w ( x , y ) = w ( x , y ) } .
By Remark 3.1, there is a minimizer u R in C w 0 , a , o ,
{ div ( y a u R ) = 0 in  Q R , lim y 0 + y a u R y = 1 d s b ( x ) f ( u R ( x , 0 ) ) on  0 Q R , u R = w 0 on  + Q R .
(3.9)
Define
u R : = { u R ( x , y ) if  u R ( x , y ) < 0  and  x > 0 , u R ( x , y ) if  u R ( x , y ) 0  and  x > 0

and u R ( x , y ) : = u R ( x , y ) for x 0 . Thus u R 0 for x > 0 and y 0 . Obviously u R is still a minimizer of E ( , Q R ) .

By the regularity results in [2], u R , x u R , y a u R y C β ( Q R ) for some 0 < β < 1 and the continuous module is uniform bounded. Up to a subsequence, u R u , ( u R ) x u x and y a u R y y a u y in C 0 ( B s + ¯ ) as R for all R > s + 2 . By the canonical diagonal procedure, u solves
{ div ( y a u ) = 0 in  R + 2 , lim y 0 + y a u y = 1 d s b ( x ) f ( u ( x , 0 ) ) on  R + 2 , u ( x , y ) = u ( x , y ) in  R + 2 ¯ ,
(3.10)

and by the Hopf maximum principle 1 < u < 1 .

Step 2. We show that there exists at least a subsequence x n such that u ( x n , 0 ) 1 .

First we claim that u is a local minimizer under odd perturbations in [ 1 , 1 ] . That is,
E ( u , Ω ) E ( w , Ω )

for any Ω R + 2 and for any odd function w H 1 ( y a , Ω ) with | w | 1 and w = u on + Ω in the weak sense.

Let ξ C c 1 ( B s + 0 B s + ) is odd with respect to x for every y 0 and u R + ξ [ 1 , 1 ] . Since 1 < u R < 1 , u R + ( 1 ϵ ) ξ ( 1 , 1 ) for ϵ ( 0 , 1 ) . We have
E ( u R , B s + ) E ( u R + ( 1 ϵ ) ξ , B s + ) for  R > s + 2 .
Let R , and
E ( u , B s + ) E ( u + ( 1 ϵ ) ξ , B s + )

for every s > 0 and u + ( 1 ϵ ) ξ [ 1 , 1 ] . Our claim is proved.

If w ( x , y ) = w ( x , y ) ,
E ( w , B s + ) = 2 E ( w , B s + + ) = 2 { d s B s + + y a 2 | w | 2 d x d y + 0 B s + + b ( x ) G ( w ) d x } ,
where B s + + = { ( x , y ) B s + , x > 0 , y 0 } . Therefore u is also a local minimizer of in R + + n + 1 = { ( x , y ) R + 2 , x > 0 , y 0 } with perturbations in [ 1 , 1 ] , i.e.,
E ( u , Ω ) E ( w , Ω )

for any Ω R + + 2 and for any w H 1 ( y a , Ω ) with | w | 1 and w = u on + Ω in weak sense.

Suppose u ( x n , 0 ) 1 for any sequence x n by contradiction. | u ( x , 0 ) | < 1 ϵ for some 0 < ϵ < 1 and x R . Hence 0 u ( x , y ) < 1 ϵ for all x > 0 and y 0 by the fact that u ( , y ) = P s ( , y ) u ( , 0 ) .

Let R > 1 . Let φ R be a cut-off function with values 1 in B ( 1 η ) R + and zeroes outside of B R + , | φ R | C η R for some 0 < η < 1 determined later.

Denote φ R = φ R ( | ( x l , y ) | ) . Let w = 1 φ R + ( 1 φ R ) u H 1 ( y a , B R + ( l , 0 ) ) , w u on + B R ( l , 0 ) . For l > R ,
E ( w , B R + ( l , 0 ) ) = d s B R + ( l , 0 ) y a 2 | ( 1 φ R ) u + ( 1 u ) φ R | 2 d x d y + 0 B R + ( l , 0 ) b ( x ) G ( w ) d x d s B R + ( l , 0 ) y a 2 | u | 2 d x d y + d s B R + ( l , 0 ) y a 2 | φ R | 2 d x d y + d s { B R + ( l , 0 ) y a | u | 2 d x d y } 1 2 { B R + y a | φ R | 2 d x d y } 1 2 + 0 ( B R + B ( 1 η ) R + ) b ( x ) G ( w ) d x d s B R + ( l , 0 ) y a 2 | u | 2 d x d y + ( C η 2 R 2 R R 1 + a ) + { C R [ 1 R y a y 2 d y + 0 1 ( y a + y a ) d y ] } 1 2 ( C η 2 R a ) 1 2 + 2 b ¯ max [ 0 , 1 ] G η R d s B R + ( l , 0 ) y a 2 | u | 2 d x d y + C η 2 R a + C η 1 R 1 + a 2 + 2 b ¯ max [ 0 , 1 ] G η R .

Here the constant C does not depend on R, we use the gradient estimates (see [2]) in the second line from the bottom.

On the other hand,
E ( u , B R + ( l , 0 ) ) d s B R + ( l , 0 ) y a 2 | u | 2 d x d y + 2 b ̲ min [ 0 , 1 ϵ ] G R .

Choose η = b ̲ min [ 0 , 1 ϵ ] G 2 b ¯ max [ 0 , 1 ] G , E ( u , B R + ( l , 0 ) ) > E ( w , B R + ( l , 0 ) ) for large R. This contradiction leads to the result that there exists at least a sequence x n such that u ( x n , 0 ) 1 .

Step 3. We show that u is the layer solution, i.e., lim x ± u ( x , 0 ) = ± 1 .

Let u n ( x , y ) = u ( x + n , y ) and n Z + . By the regularity results [2], up to a subsequence,
u n u in  C loc 0 ( R + 2 ¯ ) , u x n u x in  C loc 0 ( R + 2 ¯ ) , y a u n y y a u y in  C loc 0 ( R + 2 ¯ )
as n .
{ div ( y a u ) = 0 in  R + 2 , lim y 0 + y a u y = 1 d s b ( x ) f ( u ( x , 0 ) ) on  R + 2 , 0 u 1 in  R + 2 .
(3.11)
Define u ˜ = 1 u , we have
{ div ( y a u ˜ ) = 0 in  R + 2 , lim y 0 + y a u ˜ y = 1 d s b ( x ) f ( u ( x , 0 ) ) 0 on  R + 2 , 0 u ˜ 1 in  R + 2 .
(3.12)

u ˜ C by Proposition 3.1, f ( u ( x , 0 ) ) = f ( C ) 0 and u 0 or 1. Thus u 1 by step 2. That is, u 1 as x . u 1 as x is achieved by odd symmetry.

u is the desired layer solution. □

Proof of Theorem 1.1 It follows from Theorem 3.1; for the regularity of v see [2]. □

Lastly we give asymptotic estimates for layer solutions of (1.1) as | x | .

Proof of Theorem 1.2 Let v be a layer solution of (1.1),
{ ( x x ) s v ( x ) = b ( x ) f ( v ( x ) ) in  R , lim x ± v = ± 1 .
(3.13)
Then
( x x ) s ( 1 v ) b ( x ) f ( ξ 1 ) ( 1 v ) = 0 in  R ,
(3.14)

where ξ 1 is some point between v ( x ) and 1.

Consider the layer solution v s t of the unperturbed problem in Lemma 2.4,
( x x ) s ( 1 v s t ) ( f s t ) ( ξ 2 ) ( 1 v s t ) = 0 in  R
(3.15)

with ξ 2 is some point between v s t ( x ) and 1.

Since ( f s t ) ( 1 ) = 1 t , choose t large enough such that 2 t < b ̲ f ( 1 ) and choose x 0 R such that ( f s t ) ( ξ 2 ) < 2 t < b ̲ f ( ξ 1 ) for all x > x 0 .

Choose C > 0 such that C ( 1 v s t ) > 1 v in ( , x 0 ] , which can be done since v s t , v 1 as x .

Define
d ( x ) = { 2 t in  ( x 0 , + ) , C f s t ( v s t ) b ( x ) f ( v ) C ( 1 v s t ) ( 1 v ) in  ( , x 0 ] ,
d ( x ) L . We have
{ ( x x ) s { C ( 1 v s t ) ( 1 v ) } + d ( x ) { C ( 1 v s t ) ( 1 v ) } 0 in  R , C ( 1 v s t ) ( 1 v ) > 0 in  ( , x 0 ] .
(3.16)
Obviously, if inf R { C ( 1 v s t ) ( 1 v ) } < 0 , it is achieved at some point x ̲ ( x 0 , + ) . Since d > 0 in ( x 0 , + ) , ( x x ) s { C ( 1 v s t ) ( 1 v ) } ( x ̲ ) 0 from the first inequality of (3.16), which contradicts with the fact that
( x x ) s { C ( 1 v s t ) ( 1 v ) } ( x ̲ ) = R { C ( 1 v s t ) ( 1 v ) } ( x ̲ ) { C ( 1 v s t ) ( 1 v ) } ( y ) | x ̲ y | 1 + 2 s d y < 0 .

Therefore ( 1 v ) C ( 1 v s t ) for C > 0 given from above.

On the other hand, choose small t > 0 such that b ¯ f ( 1 ) < 1 2 t and choose x 0 R such that b ¯ f ( ξ 1 ) < 1 2 t < ( f s t ) ( ξ 2 ) for all x > x 0 . Choose c > 0 such that c ( 1 v s t ) < 1 v in ( , x 0 ] .

Define
d ˜ ( x ) = { 1 2 t in  ( x 0 , + ) , b ( x ) f ( v ) c f s t ( v s t ) ( 1 v ) c ( 1 v s t ) in  ( , x 0 ]

and obviously d ˜ ( x ) L .

Then,
{ ( x x ) s { ( 1 v ) c ( 1 v s t ) } + d ˜ ( x ) { ( 1 v ) c ( 1 v s t ) } 0 in  R , ( 1 v ) c ( 1 v s t ) > 0 in  ( , x 0 ] .
(3.17)

If inf R { ( 1 v ) c ( 1 v s t ) } < 0 , it is only achieved at some point x ¯ ( x 0 , + ) . Since d ˜ > 0 in ( x 0 , + ) , ( x x ) s { ( 1 v ) c ( 1 v s t ) } ( x ¯ ) 0 from the first inequality of (3.17), which contradicts the fact that ( x x ) s { ( 1 v ) c ( 1 v s t ) } ( x ¯ ) < 0 . Thus c ( 1 v s t ) ( 1 v ) for some 0 < c < C given from above.

Therefore,
c x 2 s | 1 v | C x 2 s for  x > 1
by Lemma 2.4. Similarly,
c | x | 2 s | 1 + v | C | x | 2 s for  x < 1 .

Here c and C maybe different from above. □

4 Asymptotic as s 1

In this section we prove Theorem 1.3, which consists of two lemmas.

Lemma 4.1 Let { v s k } be a sequence of layer solutions of (1.1) in Theorem 1.1. Then there exists a subsequence denoted again by { v s k } , converging locally uniformly to v 1 which solves the local elliptic equation
v x x 1 ( x ) = b ( x ) f ( v 1 ) in  R .
(4.1)
Proof Consider u a k , the s-extension of v s k , which solves
{ div ( y a k u a k ) = 0 in  R + 2 , ( 1 + a k ) lim y 0 + y a k y u a k = C a k b ( x ) f ( u a k ( x , 0 ) ) on  R + 2 ,
(4.2)

where a k = 1 2 s k and C a k = 1 + a k d s k = 2 ( 1 s k ) d s k . Obviously a k 1 as s k 1 .

Let ξ C c 1 ( R + 2 ¯ ) . Multiplying (4.2) with ξ and integrating in R + 2 ,
( 1 + a k ) R + 2 y a k u a k ξ d x d y C a k R b ( x ) f ( u a k ( x , 0 ) ) ξ d x = 0 .
(4.3)
Choose ξ ( x , y ) = ξ 1 ( x ) ξ 2 ( y ) , ξ 1 C c 1 ( R ) and ξ 2 is a cut-off function which equals 1 in [ 0 , 1 ] and 0 in [ 2 , ) , | ξ 2 | C for some constant C > 0 . Thus (4.3) can be rewritten as
( 1 + a k ) R + 2 y a k { ξ 1 ( x ) ξ 2 ( y ) x u a k + ξ 1 ( x ) ξ 2 ( y ) y u a k } d x d y = C a k R b ( x ) f ( u a k ( x , 0 ) ) ξ 1 ( x ) d x .
(4.4)
By the regularity results in [2], the continuous module does not depend on s for s > s 0 > 1 2 . Up to a subsequence,
u a k u 1 in  C loc 0 ( R + 2 ¯ ) , ( u a k ) x ( u 1 ) x in  C loc 0 ( R + 2 ¯ )
and
C a k = 2 ( 1 s k ) d s k = 2 ( 1 s k ) 2 2 s k 1 Γ ( s k ) Γ ( 1 s k ) 1
as s k 1 (or equivalently a k 1 ). Then
C a k R b ( x ) f ( u a k ( x , 0 ) ) ξ 1 d x R b ( x ) f ( u 1 ( x , 0 ) ) ξ 1 d x as  a k 1 .
(4.5)
For the first integral in (4.4), we consider
( 1 + a k ) 0 y a k ξ 2 ( y ) x u a k d y = ( 1 + a k ) 0 δ y a k ξ 2 ( y ) { x u a k u 1 ( x ) } d y + ( 1 + a k ) 0 δ y a k ξ 2 ( y ) u 1 ( x ) d y + ( 1 + a k ) δ y a k ξ 2 ( y ) x u a k d y = I 1 + I 2 + I 3 ,
(4.6)
| I 1 | ( 1 + a k ) 0 δ y a k ξ 2 ( y ) | x u a k u 1 ( x ) | d y ϵ δ 1 + a k
(4.7)
for 0 < δ < 1 and small ϵ > 0 . Here we use the fact that x u a k u 1 ( x ) locally uniformly in R + 2 ¯ . We have
I 2 = u 1 ( x ) ( 1 + a k ) 0 δ y a k d y = δ 1 + a k u 1 u 1 as  a k 1 .
(4.8)
Since | u a k | C y for y > 0 and C independent of a k (see [2]),
| I 3 | C ( 1 + a k ) δ y a k 1 d y = C 1 + a k a k δ a k 0 as  a k 1 .
(4.9)
By (4.6)-(4.9),
( 1 + a k ) 0 y a k ξ 2 ( y ) x u a k d y u 1
and
( 1 + a k ) R + 2 y a k ξ 1 ( x ) ξ 2 ( y ) x u a k d x d y R ξ 1 ( x ) u 1 d x ,
(4.10)
| ( 1 + a k ) R + 2 y a k ξ 1 ( x ) ξ 2 ( y ) y u a k d x d y | R | ξ 1 ( x ) | d x ( 1 + a k ) 1 2 y a k | ξ 2 ( y ) | | y u a k | d y C ( 1 + a k ) 1 2 y a k 1 d y = C 1 + a k a k ( 2 a k 1 ) 0
(4.11)

as a k 1 .

Therefore, by (4.4), (4.5), (4.10), and (4.11),
R u 1 ( x ) ξ 1 ( x ) d x = R b ( x ) f ( u 1 ( x ) ) ξ 1 ( x ) d x .
(4.12)
That is,
v x x 1 = b ( x ) f ( v 1 )
(4.13)

in the weak sense ( u 1 = v 1 ). By the regularity theory of elliptic equations, v 1 is also a classical solution of (4.13). □

Lemma 4.2 v 1 is also a layer solution of (4.1), i.e., v 1 ± 1 as x ± .

Proof Claim 1. v 1 is a local minimizer in ( 0 , ) under perturbations in [ 1 , 1 ] . That is,
F ( v 1 , I ) F ( v 1 + ξ 1 , I )
(4.14)
for any bounded open interval I ( 0 , ) and for any ξ 1 C 0 1 ( I ) such that | v 1 + ξ 1 | 1 , where
F ( w , I ) : = I { | w x | 2 2 + b ( x ) G ( w ) } d x for every  w H 1 ( I ) .
Indeed, for the test function ξ in Lemma 4.1 with the additional property that | u a k + ξ | 1 , we have
0 E ( u a k + ( 1 ϵ ) ξ , I × [ 0 , R ] ) E ( u a k , I × [ 0 , R ] ) = 1 + a k 2 I × [ 0 , R ] y a k | ( u a k + ( 1 ϵ ) ξ ) | 2 d x d y + C a k I b ( x ) G ( u a k + ( 1 ϵ ) ξ ) d x 1 + a k 2 I × [ 0 , R ] y a k | u a k | 2 d x d y C a k I b ( x ) G ( u a k ) d x = 1 + a k 2 I × [ 0 , R ] y a k | x u a k + ( 1 ϵ ) ξ 1 ( x ) ξ 2 ( y ) | 2 d x d y 1 + a k 2 I × [ 0 , R ] y a k | x u a k | 2 d x d y + ( 1 + a k ) I × [ 0 , R ] y a k y u a k ( 1 ϵ ) ξ 1 ( x ) ξ 2 ( y ) d x d y + 1 + a k 2 I × [ 0 , R ] y a k ( ( 1 ϵ ) ξ 1 ( x ) ξ 2 ( y ) ) 2 d x d y + C a k I b ( x ) G ( u a k + ( 1 ϵ ) ξ 1 ( x ) ) d x C a k I b ( x ) G ( u a k ) d x .
(4.15)
As in the discussions in Lemma 4.1, let a k 1 , and we have
1 + a k 2 I × [ 0 , R ] y a k ( x u a k ) 2 d x d y I ( u 1 ) 2 2 d x ,
(4.16)
( 1 + a k ) I × [ 0 , R ] y a k x u a k ( 1 ϵ ) ξ 1 ( x ) ξ 2 ( y ) d x d y I u 1 ( x ) ( 1 ϵ ) ξ 1 ( x ) d x ,
(4.17)
1 + a k 2 I × [ 0 , R ] y a k ( ( 1 ϵ ) ξ 1 ( x ) ξ 2 ( y ) ) 2 d x d y = 1 2 I ( 1 ϵ ) 2 ( ξ 1 ( x ) ) 2 d x { 0 1 ( 1 + a k ) y a k d y + 1 R ( 1 + a k ) y a k ( ξ 2 ( y ) ) 2 d y } 1 2 I ( ( 1 ϵ ) ξ 1 ( x ) ) 2 d x .
(4.18)
By (4.16)-(4.18),
1 + a k 2 I × [ 0 , R ] y a k ( x u a k + ( 1 ϵ ) ξ 1 ( x ) ξ 2 ( y ) ) 2 d x d y 1 + a k 2 I × [ 0 , R ] y a k ( x u a k ) 2 d x d y 1 2 I ( u 1 ( x ) + ( 1 ϵ ) ξ 1 ( x ) ) 2 d x 1 2 I ( u 1 ( x ) ) 2 d x ,
(4.19)
( 1 + a k ) I × [ 0 , R ] y a k y u a k ξ 1 ( x ) ξ 2 ( y ) d x d y = ( 1 + a k ) I × [ 1 , 2 ] y a k y u a k ξ 1 ( x ) ξ 2 ( y ) d x d y C ( 1 + a k ) 1 2 y a k 1 d y = C ( 1 + a k ) a k { 2 a k 1 } 0 as  a k 1 ,
(4.20)
( 1 + a k ) 2 I 0 R y a k ( ξ 1 ( x ) ξ 2 ( y ) ) 2 d x d y = ( 1 + a k ) 2 I 1 2 y a k ( ξ 1 ( x ) ξ 2 ( y ) ) 2 d x d y C ( 1 + a k ) 1 2 y a k d y = C ( 2 a k + 1 1 ) 0 as  a k 1 ,
(4.21)
C a k I b ( x ) G ( u a k + ( 1 ϵ ) ξ 1 ( x ) ) d x C a k I b ( x ) G ( u a k ) d x I b ( x ) G ( u 1 + ( 1 ϵ ) ξ 1 ( x ) ) d x I b ( x ) G ( u 1 ) d x .
(4.22)

By (4.15), (4.19)-(4.22), our claim is proved.

Claim 2. v 1 1 as x .

Define v 1 , n ( x ) = v 1 ( x + n ) for n Z + , up to a subsequence, v 1 , n v 1 , in C loc 2 as n ,
{ v x x 1 , ( x ) = b ( x ) f ( v 1 , ( x ) ) , x R , 0 v 1 , 1 .
(4.23)

Since f 0 and b > 0 , v x x 1 , 0 in and v 1 , 0 or 1.

We show that v 1 0 or 1 as x . Indeed, if there are two sequences { x n } and { y n } such that v 1 ( x n ) 0 and v 1 ( y n ) 1 as n , there must exist z n ( x n , y n ) such that v 1 ( z n ) = 1 2 .

Denote v ˜ n 1 ( x ) = v 1 ( x + [ z n ] ) where [ z n ] is the integer part of z n . v ˜ n 1 ( z n [ z n ] ) = v 1 ( z n ) = 1 2 and up to a subsequence v ˜ n 1 v ˜ 1 in C loc 2 , v ˜ 1 solves equation (4.23). Therefore v ˜ 1 0 or 1. For the above subsequence, there is a subsubsequence such that z n [ z n ] z [ 0 , 1 ] as n and v ˜ 1 ( z ) = 1 2 . This contradiction verifies v 1 0 or 1 as x .

To check v 1 1 as x , suppose that v 1 0 as x by contradiction. Then,
lim inf l + F ( v 1 , ( l R , l + R ) ) = lim inf l + l R l + R { | v 1 | 2 2 + b ( x ) G ( v 1 ) } d x 2 b ̲ R ϵ

for some ϵ > 0 .

Let ξ C 0 1 ( l R , l + R ) , ξ = 1 if | x l | < ( 1 η ) R and ξ = 0 if | x l | > R where η will be determined later, | ξ | 1 η R . Define w = 1 ξ + ( 1 ξ ) v 1 , then w ( l ± R ) = v 1 ( l ± R ) . We have
lim sup l + F ( w , ( l R , l + R ) ) = lim sup l + l R l + R ( 1 2 | ( 1 ξ ) v x 1 + ( 1 v 1 ) ξ x | 2 + b ( x ) G ( 1 ξ + ( 1 ξ ) v 1 ) ) d x l R l + R ξ x 2 d x + b ¯ max [ 1 , 1 ] G 2 η R 1 η 2 R + b ¯ max [ 1 , 1 ] G 2 η R .
Choose η = ϵ b ̲ 2 b ¯ max [ 1 , 1 ] G ,
lim sup l + F ( w , ( l R , l + R ) ) < lim inf l + F ( v 1 , ( l R , l + R ) )

for R > 1 large enough. Therefore v 1 1 as x , by odd symmetry, v 1 1 as x , i.e., v 1 is a layer solution of the local elliptic equation (4.13).

By the Hamiltonian equality (2.15),
b ( x ) { G ( v 1 ( x ) ) G ( 1 ) } + x + b ( t ) { G ( v 1 ( t ) ) G ( 1 ) } d t = 1 2 ( v x 1 ) 2 = lim a k 1 ( 1 + a k ) 0 y a k 2 ( x u a k ) 2 = lim a k 1 ( 1 + a k ) 0 y a k 2 ( y u a k ) 2 + lim a k 1 C a k b ( x ) { G ( u a k ( x , 0 ) ) G ( 1 ) } + lim a k 1 C a k x b ( t ) { G ( u a k ( t , 0 ) ) G ( 1 ) } d t .
Therefore,
lim a 1 ( 1 + a ) 0 y a 2 ( y u a ) 2 = x + b ( t ) { G ( v 1 ( t ) ) G ( 1 ) } d t lim a k 1 C a k x b ( t ) { G ( u a k ( t , 0 ) ) G ( 1 ) } d t .

 □

Proof of Theorem 1.3 It follows from Lemmas 4.1 and 4.2. □

Declarations

Acknowledgements

This research has been supported by National Natural Science Foundation of China (Grant No. 11371128).

Authors’ Affiliations

(1)
College of Mathematics and Econometrics, Hunan University

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© Hu; licensee Springer. 2014

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