Skip to main content

Advertisement

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

Article metrics

  • 610 Accesses

  • 3 Citations

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. 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

  2. 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

  3. 3.

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

  4. 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

  5. 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.

  6. 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

  7. 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.

  8. 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

  9. 9.

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

  10. 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

  11. 11.

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

  12. 12.

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

Download references

Acknowledgements

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

Author information

Correspondence to Yan Hu.

Additional information

Competing interests

The author declares that she has no competing interests.

Author’s contributions

The author read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and Permissions

About this article

Keywords

  • fractional Laplacian
  • layer solutions
  • existence
  • local minimizers