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

Abstract

This paper is concerned with the nonlinear equation involving the fractional Laplacian: ( ) s v(x)=b(x)f(v(x)), xR, where s(0,1), b:RR 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 s1 are also obtained.

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

1 Introduction

In this paper we study the fractional Laplacian

( ) s v(x)=b(x)f ( v ( x ) ) ,xR,
(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 dy.

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=12s(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(xz)dzfor 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 ) ) ,xR

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 dxdy+ 0 Ω b(x)G ( u ( x , 0 ) ) dx.
(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:RR 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(0,12s)) 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 |1v|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 s1 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 ) } dt.
(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 y0;

  1. (3)
    u L ± L ( B R + ( x , 0 ) ) 0as x±;
    (2.4)
  2. (4)
    x u L ( B R + ( x , 0 ) ) + y a u y L ( B R + ( x , 0 ) ) 0as 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 nZ. 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 xR.

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 ̲ εR
(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 (xl,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 xR, 0 y a | u | 2 dy<. In addition, the integral can be differentiated with respect to xR under the integral sign. We have

lim M + M y a | u | 2 dy=0
(2.13)

uniformly in xR. If u is a layer solution of (1.2),

lim | x | + 0 y a | u | 2 dy=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 xR, 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 ) } dx=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 dz
(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(πs)Γ(2s)>0
(2.20)

and, as a consequence,

lim x ± | x | 2 s | ( v s t ) ( x ) 1 | =t 2 π sin(πs)Γ(2s)>0.
(2.21)

3 Existence and asymptotic estimates

To prove the existence of layer solutions, we introduce a Liouville result where a0 is required. This is the reason why we restrict ourselves to the case s 1 2 in Theorem 1.1.

Proposition 3.1 Let a0. 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 uC a.e. in R + 2 .

Proof Since a0, 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 uC 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)0f(1),

the energy functional E(u,Ω) admits a minimizer u C w 0 , a ={w H 1 ( y a ,Ω),1w1 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 dxdy 0 Ω b(x) f (u) ξ 2 dx0,
(3.3)

for every ξ H 1 (Ω, y a ) such that ξ0 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 dxdy+ 0 Ω b(x) G ˜ ( u ( x , 0 ) ) dx.
(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 1p< 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 dxdy 0 Ω b(x)f(1) ( u 1 ) + dx=0.

Since f(1)0, Ω y a | ( u 1 ) + | 2 dxdy0. Thus ( u 1 ) + 0 a.e. in Ω, i.e., u1 a.e. in Ω. Similarly we also get u1 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 y0}.

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

  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 y0,

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 dxdy R b(x) f ( u ( x , 0 ) ) ξ 2 dx0
(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 y0.

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 x0. Thus u R 0 for x>0 and y0. 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 y0 and u R +ξ[1,1]. Since 1< u R <1, u R +(1ϵ)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+(1ϵ)ξ[1,1]. Our claim is proved.

If w(x,y)=w(x,y),

E ( w , B s + ) =2E ( 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,y0}. Therefore u is also a local minimizer of in R + + n + 1 ={(x,y) R + 2 ,x>0,y0} 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)|<1ϵ for some 0<ϵ<1 and xR. Hence 0u(x,y)<1ϵ for all x>0 and y0 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 (|(xl,y)|). Let w=1 φ R +(1 φ R )u H 1 ( y a , B R + (l,0)), wu 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 dxdy+2 b ̲ min [ 0 , 1 ϵ ] GR.

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, u1 as x. u1 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 (1v)b(x) f ( ξ 1 )(1v)=0in 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 ) =0in 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 )>1v in (, x 0 ], which can be done since v s t , v1 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 )(1v)}<0, it is achieved at some point x ̲ ( x 0 ,+). Since d>0 in ( x 0 ,+), ( x x ) s {C(1 v s t )(1v)}( 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 (1v)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 )<1v 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 {(1v)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 {(1v)c(1 v s t )}( x ¯ )0 from the first inequality of (3.17), which contradicts the fact that ( x x ) s {(1v)c(1 v s t )}( x ¯ )<0. Thus c(1 v s t )(1v) for some 0<c<C given from above.

Therefore,

c x 2 s |1v|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 s1

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 =12 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 ξdxdy C a k R b(x)f ( u a k ( x , 0 ) ) ξdx=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 dx R b(x)f ( u 1 ( x , 0 ) ) ξ 1 dxas  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)|dyϵ 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 dy= 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 dy=C 1 + a k a k δ a k 0as  a k 1.
(4.9)

By (4.6)-(4.9),

(1+ a k ) 0 y a k ξ 2 (y) x u a k dy u 1

and

(1+ a k ) R + 2 y a k ξ 1 (x) ξ 2 (y) x u a k dxdy R ξ 1 (x) u 1 dx,
(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)dx= R b(x)f ( u 1 ( x ) ) ξ 1 (x)dx.
(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 ) } dxfor 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 dxdy I ( u 1 ) 2 2 dx,
(4.16)
(1+ a k ) I × [ 0 , R ] y a k x u a k (1ϵ) ξ 1 (x) ξ 2 (y)dxdy I u 1 (x)(1ϵ) ξ 1 (x)dx,
(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 f0 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 ) } dx2 b ̲ Rϵ

for some ϵ>0.

Let ξ C 0 1 (lR,l+R), ξ=1 if |xl|<(1η)R and ξ=0 if |xl|>R where η will be determined later, | ξ | 1 η R . Define w=1ξ+(1ξ) 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. □

References

  1. Cabré X, Solà-Morales J: Layer solutions in a half-space for boundary reactions. Commun. Pure Appl. Math. 2005, 58(12):1678-1732. 10.1002/cpa.20093

    Article  Google Scholar 

  2. Cabré X, Sire Y: Nonlinear equations for fractional Laplacians I: regularity, maximum principles, and Hamiltonian estimates. Ann. Inst. Henri Poincaré, Anal. Non Linéaire 2013. 10.1016/j.anihpc.2013.02.001

    Google Scholar 

  3. Cabré, X, Sire, Y: Nonlinear equations for fractional Laplacian II: existence, uniqueness, and qualitative properties of solutions. arXiv:1111.0796 (2011)

    Google Scholar 

  4. Caffarelli L, Silvestre L: An extension problem related to the fractional Laplacian. Commun. Partial Differ. Equ. 2007, 32(8):1245-1260. 10.1080/03605300600987306

    Article  MathSciNet  Google Scholar 

  5. Fabes EB, Kenig CE, Serapioni RP: The local regularity of solutions of degenerate elliptic equations. Commun. Stat., Theory Methods 1982, 7(1):77-116.

    MathSciNet  Google Scholar 

  6. Palatucci G, Savin O, Valdinoci E: Local and global minimizers for a variational energy involving a fractional norm. Ann. Mat. Pura Appl. 2013, 192(4):673-718. 10.1007/s10231-011-0243-9

    Article  MathSciNet  Google Scholar 

  7. Nekvinda A: Characterization of traces of the weighted Sobolev space W 1 , p (Ω, d M ε ) on M . Czechoslov. Math. J. 1993, 43(4):695-711.

    MathSciNet  Google Scholar 

  8. Alama S, Bronsard L, Gui C:Stationary layered solutions in R 2 for an Allen-Cahn systems with multiple well potential. Calc. Var. Partial Differ. Equ. 1997, 5(4):359-390. 10.1007/s005260050071

    Article  MathSciNet  Google Scholar 

  9. Ghoussoub N, Gui C: On a conjecture of De Giorgi and some related problems. Math. Ann. 1998, 311: 481-491. 10.1007/s002080050196

    Article  MathSciNet  Google Scholar 

  10. Ghoussoub N, Gui C: On De Giorgi’s conjecture in dimensions 4 and 5. Ann. Math. 2003, 157(1):313-334. 10.4007/annals.2003.157.313

    Article  MathSciNet  Google Scholar 

  11. Adams RA Pure and Applied Mathematics 65. In Sobolev Spaces. Academic Press, New York; 1975.

    Google Scholar 

  12. Gilbarg D, Trudinger NS: Elliptic Partial Differential Equations of Second Order. Springer, Berlin; 1997.

    Google Scholar 

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Acknowledgements

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

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Hu, Y. Layer solutions for a class of semilinear elliptic equations involving fractional Laplacians. Bound Value Probl 2014, 41 (2014). https://doi.org/10.1186/1687-2770-2014-41

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