# Existence Results for a Perturbed Problem Involving Fractional Laplacians

Authors: Yan Hu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We extend the results of Cabre and Sire (2011) to show the existence of layer solutions of fractional Laplacians with perturbed nonlinearity(-Δ)su=b(x)f(u) in ℝ with s∈(0,1). Here b is a positive periodic perturbation for f, and -f is the derivative of a balanced well potential G. That is, G∈C2,γ satisfies G(1)=G(-1)<G(τ)∀τ∈(-1,1),G'(1)=G'(-1)=0. First, for odd nonlinearity f and for every s∈(0,1), we prove that there exists a layer solution via the monotone iteration method. Besides, existence results are obtained by variational methods for s∈(1/2,1) and for more general nonlinearities. While the case s≤1/2 remains open.


---

## Body

## 1. Introduction

We consider the classical fractional Laplacian(1)(-Δ)su=b(x)f(u)inℝ,
where s∈(0,1) and
(2)(-Δ)su=csP.V.∫ℝu(x)-u(y)|x-y|1+2sdy.
Here P.V. stands for the Cauchy principle value and cs is a positive constant depending only on s.In the whole paper we suppose thatb∈(C1,γ∩L∞)(ℝ)(γ>max⁡{0,1-2s}) is a positive function and b(x+1)=b(x) for every x∈ℝ.It is well known that the fractional Laplacian is a nonlocal operator which can be localized by adding a variable(3)div⁡(ya∇u(x,y))=0inℝ+2,-dslim⁡y↓0+ya∂u∂y=b(x)f(u)on⁡∂ℝ+2=ℝ,
where a=1-2s∈(-1,1), u(x,y)|∂ℝ+2=u(x), and
(4)u(x,y)=ps∫ℝ‍y2s(y2+z2)(1+2s)/2u(x-z,0)dz
which is called the s-extension of u(x,0). ps is a positive constant multiplier depending only on s.Obviously, (1) and (3) have their own variational structures. For nonlocal equation (1), its corresponding nonlocal energy functional on any open interval I⊂ℝ is defined by
(5)ℰnon(u,I)=12∬I|u(x)-u(y)|2|x-y|1+2sdydx+∫I‍∫ℝ∖I‍|u(x)-u(y)|2|x-y|1+2sdydx+∫I‍b(x)G(u)dx
with prescribed boundary data outside of I.As for elliptic equation (3), its local energy functional is given by
(6)ℰloc⁡(u,Ω)=ds2∫Ω‍ya|∇u|2dxdy+∫∂Ω∩∂ℝ+2‍b(x)G(u)dx
with Ω⊂ℝ+2 an open domain. Here G is the primitive function of -f. The details for the fractional Laplacian can be seen in [1–6].In view of the De Giorgi conjecture [7–9], Cabré and Sire [1–3] consider layer solutions of the autonomous equation
(7)(-Δ)su=f(u)inℝ.
The layer solution of (7) is a strictly increasing solution with limits ±1 at ±∞. In [2, 3] the necessary and sufficient conditions for existence of layer solutions were presented by a Hamiltonian equality and a Modica-type estimate via variational methods, that is, G′(1)=G′(-1)=0 and G(1)=G(-1)<G(τ) for every τ∈(-1,1). In other words, (7) is an Allen-Cahn-type equation.Equation (7) does not depend on the variable x explicitly; layer solutions were proved to be strictly increasing and are the unique local minimizer by the sliding method. Furthermore, if f is odd and f′(±1)<0, the layer solution is proved odd symmetric about some point where it equals zero. Many properties about the fractional Laplacian were established such as the regularity, Hopf lemma, harnack, and maximum principle; for details see [2, 3].We give the definition of layer solutions of (1).Definition 1.
A functionu∈(Cloc⁡β∩L∞)(ℝ)(0<β<1) is said to be a layer solution of (1) if it satisfies
(8)(-Δ)su=b(x)f(u)inℝ,u⟶±1asx⟶±∞.Different from (7), (1) depends on the variable x explicitly; the sliding method cannot be used anymore and layer solutions of (1) have no monotonicity. Therefore, methods in [2, 3] for existence of layer solutions cannot be used here; our problem becomes complicated. In addition, the fractional Laplacian is a nonlocal case. All these cause some difficulties for existence results of layer solutions of (1).In our previous work [10], assuming that f was odd and b was even, we proved the existence of layer solutions of (1) by variational methods and a Liouville theorem for s∈[1/2,1). The restriction on s was due to no Liouville-type conclusion for 0<s<1/2. We also obtained a Hamiltonian equality and asymptotic estimates as x→±∞ for layer solutions. Furthermore, the asymptotic behavior of layer solutions as s↑1 is also investigated.In this paper, we still study existence results of layer solutions of (1). At first we assume that f is odd and b is even. For this case, we mainly consider the s-extension of u; layer solutions are constructed by the subsupersolution technique for all s∈(0,1).Theorem 2.
Suppose thatf∈C1,γ(ℝ)(γ>max⁡{0,1-2s}) and(1)
f(-τ)=-f(τ) for every τ∈[-1,1];(2)
f(1)=f(-1)=f(0)=0, f≥0 in [0,1] and f≤0 in [-1,0].
Obviously, ifG(u)=∫u-1‍f(s)ds,
(9)G(1)=G(-1)<G(τ)forτ∈(-1,1);G′(±1)=0.
Ifb is even, then there exists a layer solution u∈C2,α(ℝ)(0<α<1) of (1) which is also odd,
(10)(-∂xx)su=b(x)f(u)inℝ,u⟶±1asx⟶±∞.For more general nonlinearitiesf and s∈(1/2,1), existence results are proved by variational methods and a careful energy comparison. This part needs some tricks.Theorem 3.
Lets∈(1/2,1). Let G∈C2,γ(ℝ)(γ>max⁡{0,1-2s}) be the primitive function of -f and(a)
G(1)=G(-1)<G(τ) for τ∈(-1,1), G′(1)=G′(-1)=0;(b)
G′′(1)>0 and G′′(-1)>0.
Then, there exists a layer solutionu∈C2,α(ℝ) of (1) for some 0<α<1,
(11)(-∂xx)su=b(x)f(u)inℝ,u⟶±1asx⟶±∞.Remark 4.
From the assumption above, there exists a0<δ<1/2 such that G is strictly increasing in [-1,-1+δ] and decreasing in [1-δ,1], G(s)>min⁡{G(1-δ),G(-1+δ)} for s∈(-1+δ,1-δ).
Fors≤1/2, the global energy may be infinite. The desired asymptotic behavior of layer solutions at infinity, that is, u→±1 as x→±∞, cannot be obtained. This case is open.
The paper is organized as follows. In Section2, we prove existence results of layer solutions of (1) under odd assumptions of f. Section 3 is devoted to the Proof of Theorem 3. Finally, we give some results about regularity and gradient estimates and a Hopf lemma about (1) and (3) in the appendix.
For convenience, we give some notations.
One has(12)b¯=max⁡ℝb(x),b_=min⁡ℝb(x).
Let u∈Hs(I),
(13)∥u∥Hs(I)=∥u∥H˙s(I)+(∫I‍u2dx)1/2,
where
(14)∥u∥H˙s(I)=12∬I|u(x)-u(y)|2|x-y|1+2sdydx+∫I‍∫ℝ∖I‍|u(x)-u(y)|2|x-y|1+2sdydx
is a seminorm.


## 2. Odd Nonlinearities

The main aim of this section is to solve existence problem for odd nonlinearities. We assume thatf satisfies assumptions in Theorem 2. For example, f(u)=u-u3, the nonlinearity of Allen-Cahn equation satisfies this condition. Here we construct a layer solution of (1) by the subsupersolution method. For this purpose, let us recall results in [3].Lemma 5.
Letf∈C1,γ(ℝ)(γ>max⁡{0,1-2s}), where s∈(0,1), such that f=-G′. Then there exists a solution u of
(15)(-∂xx)su=f(u)inℝ
such that u′>0 in R and lim⁡x→±∞u=±1 if and only if
(16)G′(1)=G′(-1)=0,G>G(1)=G(-1)in(-1,1).
If in additionf′(-1)<0 and f′(1)<0, then this solution is unique up to translations.
As a consequence, iff is odd and f′(±1)<0, then the solution is odd with respect to some point. That is, u(x+b)=-u(-x+b) for some b∈ℝ.Remark 6.
Denoteu* to be the odd solution of (15) such that u*(0)=0. Define u¯(x)=u*(b¯1/2sx) and u_(x)=u*(b_1/2sx); they are a supersolution and a subsolution of (1) in (0,∞). Indeed, by simple calculations,
(17)(-Δ)su¯=(-Δ)su*(b¯1/2sx)=b¯f(u*(b¯1/2sx))≥b(x)f(u¯)in(0,∞),(-Δ)su_=(-Δ)su*(b_1/2sx)=b_f(u*(b_1/2sx))≤b(x)f(u_)in(0,∞).
Furthermore, u¯>u_ in (0,∞) by monotonicity of u* and the fact that b¯>b_.In order to prove Theorem2, we give the following two lemmas.Lemma 7.
Letu¯(·,·) and u_(·,·) be the s-extensions of u¯(·,0) and u_(·,0), respectively, in ℝ+2; then u¯>u_ for x>0 and y≥0. Furthermore, one has
(18)div⁡(ya∇u¯)=0inℝ+2∩{x>0},-dslim⁡y↓0+ya∂u¯∂y≥b(x)f(u¯)on∂ℝ+2∩{x>0},div⁡(ya∇u_)=0inℝ+2∩{x>0},-dslim⁡y↓0+ya∂u_∂y≤b(x)f(u_)on∂ℝ+2∩{x>0}.Proof.
For everyu(·,0)∈(C∩L∞)(ℝ), u(x,y)(y>0), the s-extension of u(x,0) can be expressed by
(19)u(x,y)=ps∫ℝ‍y2s(z2+y2)(1+2s)/2u(x-z,0)dz=ps∫ℝ‍1{z2+1}(1+2s)/2u(x-yz,0)dz.
If v(x,0)=u(ax,0), for some a∈ℝ,
(20)v(x,y)=ps∫ℝ1(z2+1)(1+2s)/2v(x-yz,0)dz=ps∫ℝ1(z2+1)(1+2s)/2u(ax-ayz,0)dz=u(ax,ay).
Thus, u¯(x,y)=u*(b¯1/2sx,b¯1/2sy) and u_(x,y)=u*(b_1/2sx,b_1/2sy).
In addition, ifu(-x)=-u(x),
(21)u(x,y)=ps∫ℝ‍1(z2+1)(1+2s)/2u(x-yz,0)dz=ps∫-∞x/y‍1(z2+1)(1+2s)/2u(x-yz,0)dz+ps∫x/y+∞‍1(z2+1)(1+2s)/2u(x-yz,0)dz=ps∫-∞0‍1{(z+(x/y))2+1}(1+2s)/2u(-yz,0)dz+ps∫0+∞‍1{(z+(x/y))2+1}(1+2s)/2u(-yz,0)dz=ps∫0+∞‍{1[((x/y)-z)2+1](1+2s)/2-1[((x/y)+z)2+1](1+2s)/2}×u(yz,0)dz.
Thus, for x>0 and y>0,
(22)u¯(x,y)-u_(x,y)=ps∫0∞‍{1[((x/y)-z)2+1](1+2s)/2-1[((x/y)+z)2+1](1+2s)/2}×{u*(b¯1/2syz,0)-u*(b_1/2syz,0)}dz>0
since u* is increasing.
Equation (18) is the consequence of direct calculations; the proof of this lemma is complete.Now letΩ⊂ℝ+2 be a Lipschitz bounded domain; define the energy functional
(23)ℰ~loc⁡(w,Ω)=ds2∫Ωya|∇w|2dxdy+A2∫∂Ω∩∂ℝ+2w2dx-∫∂Ω∩∂ℝ+2g(x)wdx,
where A>0 is a constant and g∈C1,γ(ℝ)∩L∞(ℝ)(γ>max⁡{0,1-2s}) is a nonnegative function.Lemma 8.
Givenw0∈C2,α(ℝ+2) for some 0<α<1. Let Γ={w∈H1(ya,Ω), w=w0 on ∂Ω∩ℝ+2¯}. Then there exists a minimizer w∈Γ such that w∈C0,β(Ω¯) for some 0<β<1 and satisfies the Euler-Lagrange equation:
(24)div⁡(ya∇w)=0inΩ,-dslim⁡y↓0+ya∂w∂y+Aw=g(x)on∂Ω∩∂ℝ+2,w=w0on∂Ω∩ℝ+2¯.Proof.
One has(25)ℰ~loc⁡(w,Ω)=12∫Ω‍ya|∇w|2+A2∫∂Ω∩∂ℝ+2‍w2-∫∂Ω∩∂ℝ+2‍g(x)w≥12∫Ω‍ya|∇w|2+A2∫∂Ω∩∂ℝ+2‍w2-A4∫∂Ω∩∂ℝ+2‍w2-1A∥g∥L∞2·|∂Ω∩∂ℝ+2|=12∫Ω‍ya|∇w|2+A4∫∂Ω∩∂R+2‍w2-1A∥g∥L∞2·|∂Ω∩∂ℝ+2|.
Obviously,w-w0∈H1(ya,Ω); we now extend it by zeroes outside of Ω in ℝ+2, w-w0∈H1(ya,ℝ+2) and by the trace theorem and a compact imbedding theorem [11–13], w-w0∈Hs(ℝ)↪↪Lp(ℝ) for every 1≤p<∞. Therefore, E~ is well defined, bounded below, and coercive in Γ; there is a minimizer denoted again by w. By variational calculations,
(26)div⁡(ya∇w)=0inΩ,-dslim⁡y↓0+ya∂w∂y+Aw=g(x)on⁡∂Ω∩∂ℝ+2,w=w0on⁡∂Ω∩ℝ+2¯,
in weak sense. Furthermore, w∈C0,β(Ω¯) for some 0<β<1 by regularity discussion.Now we come to prove Theorem2.Proof of Theorem2.
By Lemma7, u¯=u*(b¯1/2sx,b¯1/2sy) and u_=u*(b_1/2sx,b_1/2sy) are a supersolution and a subsolution of (3) in ℝ+2∩{x>0}, respectively. Let ΩR=(0,R)×(0,R) and u0=u¯. Consider the mixed-boundary problem
(27)div⁡(ya∇u1)=0inΩR,-dslim⁡y↓0+ya∂u1∂y+Au1=b(x)f(u0)+Au0iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiion⁡∂ΩR∩∂ℝ+2,u1=u¯on⁡∂ΩR∩ℝ+2¯,
where A>max⁡{0,b¯max⁡[0,1](-f′)}. By Lemma 8, there is a solution u1∈C0,β(Ω¯R) for some 0<β<1, and further u1>0 in ΩR by the strong maximum principle and Hopf lemma.
Denotew=u1-u0,
(28)div⁡(ya∇w)=0inΩR,-dslim⁡y↓0+ya∂w∂y+Aw≤0on⁡∂ΩR∩∂ℝ+2,w=0on⁡∂ΩR∩ℝ+2¯.
Again, by the strong maximum principle and Hopf lemma, u1<u0 in ΩR.
Now we start the iteration procedure. Givenun and un-1 such that 0≤un≤un-1≤1 in ΩR, the problem
(29)div⁡(ya∇un+1)=0inΩR,-dslim⁡y↓0+ya∂un+1∂y+Aun+1=b(x)f(un)+Auniiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiion⁡∂ΩR∩∂ℝ+2,un+1=u¯on⁡∂ΩR∩ℝ+2¯
has a solution un+1∈C0,β(Ω¯R) for some 0<β<1 by Lemma 8 and un+1>0 in ΩR by the strong maximum principle and Hopf lemma.
Letw=un+1-un,
(30)div⁡(ya∇w)=0inΩR,-dslim⁡y↓0+ya∂w∂y+Aw={b(x)f(un)-f(un-1)un-un-1+A}(un-un-1)iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiion⁡∂ΩR∩∂ℝ+2,w=0on⁡∂ΩR∩ℝ+2¯.
Again by the strong maximum principle and Hopf lemma, un+1<un in ΩR.
Assume thatun≥u_ in ΩR for some n∈Z+. Let w=un+1-u_,
(31)div⁡(ya∇w)=0inΩR,-dslim⁡y↓0+ya∂w∂y+Awiiiiii={b(x)f(un)-f(u_)un-u_+A}(un-u_)iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiion⁡∂ΩR∩∂ℝ+2,w=u¯-u_≥0on⁡∂ΩR∩ℝ+2¯;un+1>u_ in ΩR by maximum principle and Hopf lemma.
Thus there is a sequence{un}, u¯=u0>u1>u2>⋯>un>⋯>u_, un→uR by monotone convergence theorem and
(32)div⁡(ya∇uR)=0inΩR,-dslim⁡y↓0+ya∂uR∂y=b(x)f(uR)on⁡∂ΩR∩∂ℝ+2,uR=u¯on⁡∂ΩR∩ℝ+2¯;u_≤uR≤u¯ in ΩR and ∥uR∥C0,β(Ω¯R)≤C where C does not depend on R by regularity discussion.
Up to a subsequenceuR→u in Cloc⁡0(ℝ+2∩{x>0}), ya(uR)y→yauy in Cloc⁡0(ℝ+2¯∩{x>0}) as R→∞, and
(33)div⁡(ya∇u)=0inℝ+2∩{x>0},-dslim⁡y↓0+ya∂u∂y=b(x)f(u)on⁡∂ℝ+2∩{x>0},u(0,y)=0fory≥0;u_<u<u¯ in ℝ+2¯∩{x>0} by the strong maximum principle and Hopf lemma. Therefore u→1 as x→∞ from asymptotic behavior of u¯ and u_ as x→∞.
We defineu(x,y)=-u(-x,y) for x<0 and y≥0; then
(34)div⁡(ya∇u)=0inℝ+2,-dslim⁡y↓0+ya∂u∂y=b(x)f(u)on⁡∂ℝ+2,
and u→±1 as x→±∞. The desired solution is achieved. We complete the proof.


## 3. Nonsymmetric Nonlinearities

In this section, we deal with more general situations; that is,G∈C2,γ(ℝ)(γ>max⁡{0,1-2s}):(1)
G′(1)=G′(-1)=0, G(1)=G(-1)<G(τ) for τ∈(-1,1);(2)
G′′(1)>0, G′′(-1)>0.For simplicity we can assume thatG(±1)=0 by adding a constant.Consider the nonlocal energy functional defined on an open intervalI⊂ℝ,
(35)ℰnon(w,I)=12∬I|w(x)-w(y)|2|x-y|1+2sdydx+∫I‍∫ℝ∖I‍|w(x)-w(y)|2|x-y|1+2sdydx+∫I‍b(x)G(w)dx,
where w∈Hs(I) with prescribed boundary data outside of I.Define a nondecreasing competitor(36)h(x)={1if|x|≥1,xif|x|<1.
If |I|>4, by direct calculations,
(37)ℰnon(h,I)≤{Cs(1+|I|1-2s)ifs∈(0,12),Cs(1+log⁡|I|)ifs=12,Csifs∈(12,1),
where Cs is a positive constant depending only on s (see [6]). In order to prove Theorem 3, we first give several lemmas.Lemma 9.
LetI=(a,b) with b-a>4. Let ΓI={w∈Hs(I), w=-1 for x≤a and w=1 for x≥b, |w|≤1}. Then, for every s∈(0,1), there is a minimizer u(a,b)∈ΓI of ℰnon,
(38)(-Δ)su(a,b)=b(x)f(u(a,b))in(a,b)
and u(a,b)∈C2,α(a,b) for some 0<α<1.Proof.
We can change the values ofG for |x|>1 such that G has the linear growth there which is denoted by G~. Given the admissible set
(39)Γ~I={w∈Hs(I),w=-1forx≤a,w=1forx≥b}.
Clearly ΓI⊂Γ~I and h(x-((a+b)/2)) is an element of them.
Consider the energy functional(40)ℰ~non(w,I)=12∬I|w(x)-w(y)|2|x-y|1+2sdydx+∫I‍∫ℝ∖I‍|w(x)-w(y)|2|x-y|1+2sdydx+∫I‍b(x)G~(w)dx.
If the minimizer w of ℰ~non in Γ~I satisfies |w|≤1, then w∈ΓI and ℰ~non(w,I)=ℰnon(w,I); that is, w is also a minimizer of ℰnon in ΓI.
Obviously, the energy is nonincreasing by cuttingw at ±1 and
(41)ℰ~non(w,I)=12∥w∥H˙s(I)2+∫I‍b(x)G~(w)dx
is well defined, bounded below, and coercive in Γ~I. Thus, there is a minimizer u(a,b)∈ΓI,
(42)(-Δ)su(a,b)=b(x)f(u(a,b))in(a,b),ℰnon(u(a,b),(a,b))≤{Cs(1+|b-a|1-2s)ifs∈(0,12),Cs(1+log⁡⁡|b-a|)ifs=12,Csifs∈(12,1)
by (37).Lemma 10.
Lets∈(1/2,1). Let u(0,R) be the minimizer of ℰnon(·,(0,R)) with R>4; then
(43)ℰnon(u(0,R),(-l,l))≤ℰnon(u(0,R)+ϕ,(-l,l))+αl(R)
for any ϕ∈C0∞(-l,l) and l∈Z+, where αl→0 as l→∞.Proof.
DenoteZR(x)=u(-l,R+l)(((R+2l)/R)x-l). Clearly, ZR=1 as x≥R and ZR=-1 as x≤0.
One has(44)ℰnon(ZR,(0,R))=12∬0R|u(-l,R+l)(R+2lRx-l)-u(-l,R+l)(R+2lRy-l)|2×(|x-y|1+2s)-1+∫0R‍∫(-∞,0)∪(R,+∞)‍|u(-l,R+l)(R+2lRx-l)-u(-l,R+l)(R+2lRy-l)|2×(|x-y|1+2s)-1+∫0R‍b(x)G(u(-l,R+l)(R+2lRx-l))=(RR+2l)1-2s12∬-lR+l|u(-l,R+l)(x)-u(-l,R+l)(y)|2|x-y|1+2s+(RR+2l)1-2s×∫-lR+l‍∫(-∞,-l)∪(R+l,+∞)‍|u(-l,R+l)(x)-u(-l,R+l)(y)|2|x-y|1+2s+RR+2l∫-lR+l‍b(RR+2l(x+l))G(u(-l,R+l))≤(RR+2l)1-2sEnon(u(-l,R+l),(-l,R+l))+(RR+2l)1-2s×∫-lR+l‍{b(RR+2l(x+l))-b(x)}G(u(-l,R+l)).
Denote
(45)βl(R)=(RR+2l)1-2s×∫-lR+l‍{b(RR+2l(x+l))-b(x)}G(u(-l,R+l))≤(RR+2l)1-2sCssup⁡ℝ|b((RR+2l)(x+l))-b(x)|⟶0;
here we use the periodic of b;
(46)γl(R)={(RR+2l)1-2s-1}ℰnon(u(-l,R+l),(-l,R+l))≤Cs{(RR+2l)1-2s-1}⟶0
as R→∞.
Therefore,(47)ℰnon(ZR,(0,R))≤ℰnon(u(-l,R+l),(-l,R+l))+γl(R)+βl(R).
Since u(0,R) is the minimizer of ℰnon(·,(0,R)),
(48)ℰnon(u(0,R),(0,R))≤ℰnon(u(-l,R+l),(-l,R+l))+γl(R)+βl(R).
Let
(49)ηl(R)=ℰnon(u(0,R),(-l,R+l))-ℰnon(u(0,R),(0,R))=∫-l0‍∫R∞‍|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s+∫RR+l‍∫-∞-l‍|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s≤4∫-l0‍∫R∞‍|x-y|-1-2sdydx+4∫RR+l‍∫-∞-l‍|x-y|-1-2sdydx=2s(2s-1){R1-2s-(R+2l)1-2s}⟶0
as R→∞.
By (48)-(49) and the fact that u(-l,R+l) is the minimizer in (-l,R+l),
(50)ℰnon(u(0,R),(-l,R+l))≤ℰnon(u(-l,R+l),(-l,R+l))+γl(R)+βl(R)+ηl(R)≤ℰnon(u(0,R)+ϕ,(-l,R+l))+γl(R)+βl(R)+ηl(R);ℰnon(u(0,R)+ϕ,(-l,R+l))-ℰnon(u(0,R),(-l,R+l))=12∬-ll|(u(0,R)+ϕ)(x)-(u(0,R)+ϕ)(y)|2|x-y|1+2s-12∬-ll|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s+∫-ll‍∫l+∞‍|(u(0,R)+ϕ)(x)-(u(0,R)+ϕ)(y)|2|x-y|1+2s-∫-ll‍∫l+∞‍|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s+∫-ll‍∫-∞-l‍|(u(0,R)+ϕ)(x)-(u(0,R)+ϕ)(y)|2|x-y|1+2s-∫-ll‍∫-∞-l‍|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s+∫-ll‍b(x){G(u(0,R)+ϕ)-G(u(0,R))}=ℰnon(u(0,R)+ϕ,(-l,l))-ℰnon(u(0,R),(-l,l)).
Therefore,
(51)ℰnon(u(0,R),(-l,l))≤ℰnon(u(0,R)+ϕ,(-l,l))+αl(R),
where αl(R)=γl(R)+βl(R)+ηl(R)→0 as R→∞. The proof of Lemma 10 is completed.It is noticed that∥u(0,R)∥C2,α≤C for some C independent of R; then by the canonical diagonal process, up to a subsequence, u(0,R)→u∞ in Cloc⁡2(ℝ) as R→∞ and
(52)(-∂xx)su∞=b(x)f(u∞)in(0,∞),u∞=-1in(-∞,0].Foru∞, we have the following.Lemma 11.
u
∞ is a local minimizer of the nonlocal energy; that is, for every l>0 and for any ϕ∈C0∞(-l,l),
(53)ℰnon(u∞,(-l,l))≤ℰnon(u∞+ϕ,(-l,l)).Proof.
Sinceu(0,R)→u∞ in Cloc⁡2(ℝ),
(54)12∬-ll|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s⟶12∬-ll|u∞(x)-u∞(y)|2|x-y|1+2s,∫-ll‍b(x)G(u(0,R))-∫-ll‍b(x)G(u∞).
For |y|≥2l and |x|≤l, |x-y|≥|y|-|x|≥|y|/2,
(55)∫-ll‍∫ℝ∖(-2l,2l)‍|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s≤4∫-ll‍∫ℝ∖(-2l,2l)‍|x-y|-1-2s≤4∫-ll‍dx∫ℝ∖(-2l,2l)‍21+2s|y|1+2sdy≤16sl1-2s.
Thus, by the Dominated convergence theorem and locally uniformly convergence,
(56)∫-ll‍∫ℝ∖(-l,l)‍|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s=∫-ll‍∫ℝ∖(-2l,2l)‍|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s+∫-ll‍∫(-2l,2l)∖(-l,l)‍|u(0,R)(x)-u(0,R)(y)|2|x-y|1+2s⟶∫-ll‍∫ℝ∖(-l,l)‍|u∞(x)-u∞(y)|2|x-y|1+2s.
By (54) and the above,
(57)ℰnon(u(0,R),(-l,l))⟶ℰnon(u∞,(-l,l))
as R→∞. The same discussion leads to the fact that
(58)ℰnon(u(0,R)+ϕ,(-l,l))⟶ℰnon(u∞+ϕ,(-l,l))
as R→∞ and for any ϕ∈C0∞(-l,l).
Therefore, (53) follows from Lemma 10, and
(59)(-∂xx)su∞=b(x)f(u∞),-1≤u∞≤1inℝ,u∞=-1in(-∞,0].
Furthermore u∞≡-1 by the strong maximum principle.Letu(-R,R) be the minimizer of ℰnon(·,(-R,R)) in Γ(-R,R). By continuity, there is a PR=sup⁡{x∈(-R,R),u(-R,R)≤0} such that u(-R,R)(PR)=0 and u(-R,R)>0 for x>PR.Lemma 12.
One hasR-|PR|→∞ as R→∞.Proof.
To check this, we suppose by contradiction that|R+PR|<+∞ for some sequence R→∞. Up to a subsequence, R+PR→k as R→∞ for some constant k.
One has(60)-1⟵u(0,2R)(x)=u(-R,R)(x-R)>0
for x>k and R large enough. This contradiction verifies our conclusion.Now we start to prove our conclusion.Proof of Theorem3.
LetuR(x)=u(-R,R)(x+[PR]) for x∈(-R-[PR],R-[PR]), where [PR] is denoted by the integer part of PR. And uR(x) satisfies
(61)(-Δ)suR=b(x)f(uR)in(-R-[PR],R-[PR]).
Up to a subsequence which is denoted again by uR, uR→u in Cloc⁡2(ℝ) as R→∞ and
(62)(-Δ)su=b(x)f(u)in(-∞,∞).
Since uR(PR-[PR])=u(-R,R)(PR)=0 and 0≤PR-[PR]<1, for the above subsequence, there is a sub-subsequence such that PR-[PR]→P∈[0,1], u(P)=0, and u≥0 for x≥P.
One has(63)ℰnon(uR,(-R-[PR],R-[PR]))=ℰnon(u(-R,R),(-R,R))≤Cs.
By Fatou’s lemma and Lemma 12, ℰnon(u,ℝ)≤Cs and ∫-∞+∞‍b(x)G(u)<+∞.
We claim that|u|→±1 as |x|→∞. Indeed, if there exists a sequence xn→∞ such that liminf⁡x→∞|u|=lim⁡xn→∞|u(xn)|≠1, by the C2 regularity of u, there must be a ball Bρ(xn) such that G(u)>ϵ in Bρ(xn) for some ϵ>0,
(64)∞>∫-∞+∞‍b(x)G(u)≥∑n=1∞∫Bρ(zn)‍b(x)G(u(x))≥b_·ϵ∑n=1∞2ρ=∞.
Therefore, u→1 or -1 as x→∞. The proof as x→-∞ is similar. Since u≥0 for x>P∈[0,1], u→1 as x→∞.
In the following we exclude the case thatu→1 as x→-∞.
Assume thatu→1 as x→-∞ by contradiction. Denote C0=min⁡Ru=u(x0) for some x0∈R. C0>-1 by the strong maximum principle.
One has(65)∫ℝ‍u(x0)-u(y)|x-y|1+2sdy=(-Δ)su(x0)=b(x0)f(u(x0))<0,f(u(x0))=f(C0)<0. Therefore C0∉[1-δ,1] by Remark 4.
Choose a small positive constantδ*<min⁡{δ,1-((|C0|+1)/2)} such that G(1-δ*)<min⁡{G(1-δ),G(-1+δ),G(-(|C0|+1)/2)}. Denote S=inf⁡{x∈ℝ,u<1-δ*} and O=sup⁡{x∈R,u<1-δ*}; S and O are well defined since u→±1 as x→±∞.
Denote(66)TR=inf⁡{x∈(-R,R),u(-R,R)≥1-δ*},SR=inf⁡{x∈(TR,R),u(-R,R)<1-δ*},OR=sup⁡{x∈(-R,R),u(-R,R)<1-δ*},C0,R=min⁡(SR,OR)u(-R,R)=u(-R,R)(xR),QR=inf⁡{x>PR,u(-R,R)=|C0|+12}.
If R is large enough, TR, SR, OR, C0,R, xR, and QR are all well defined. By local uniform convergence, |C0,R|∈(|C0|/2,(|C0|+1)/2) and |OR-SR|<|O-S|+1=M:
(67)|C0|+12=|u(-R,R)(PR)-u(-R,R)(QR)|≤∥u(-R,R)∥C2,β(R)|PR-QR|≤C|PR-QR|
for some positive constant C independent of R, |PR-QR|≥L>0.
For simplicity, denoteI1=[-R,TR], I2=[TR,SR], I3=[SR,OR], and I4=[OR,R]; then [-R,R]=∪i=14Ii.
Define(68)uR*={1-δ*ifx∈I3∩{u(-R,R)<1-δ*}=I3*,u(-R,R)ifelsewhere.
We compare the energy of uR* and of u(-R,R):
(69)ℰnon(u(-R,R),(-R,R))-ℰnon(uR*,(-R,R))=12∫I1∪I2∪I3∪I4‍∫I3*‍{|u(-R,R)(x)-u(-R,R)(y)|2|x-y|1+2s-|uR*(x)-uR*(y)|2|x-y|1+2s}+∫I3*‍∫(-∞,-R)∪(R,+∞)‍{|u(-R,R)(x)-u(-R,R)(y)|2|x-y|1+2s-|uR*(x)-uR*(y)|2|x-y|1+2s}+∫I3*‍b(x){G(u(-R,R))-G(uR*)}.
Since u→1 as x→±∞, |OR-TR|>|SR-TR|→∞ as R→∞.
One has(70)12∫I1∫I3*|u(-R,R)(x)-u(-R,R)(y)|2|x-y|1+2s+∫I3*∫-∞-R|u(-R,R)(x)-u(-R,R)(y)|2|x-y|1+2s-12∫I1∫I3*|uR*(x)-uR*(y)|2|x-y|1+2s-∫I3*∫-∞-R|uR*(x)-uR*(y)|2|x-y|1+2s≥-4∫-∞TR∫I31|x-y|1+2sdy=-2s(2s-1)(|SR-TR|1-2s-|OR-TR|1-2s)⟶0
as R→∞.
Sinceu(-R,R)≥1-δ* in I2∪I4∪(I3∖I3*) and u(-R,R)<1-δ* in I3*,
(71)12∫I2∪I4∪(I3∖I3*)‍∫I3*‍(|u(-R,R)(x)-u(-R,R)(y)|2|x-y|1+2s-|u(-R,R)(x)-uR*(y)|2|x-y|1+2s)+∫I3*‍∫R∞‍|u(-R,R)(x)-1|2|x-y|1+2s-∫I3*‍∫R∞‍|uR*(y)-1|2|x-y|1+2s≥0,12∬I3*|u(-R,R)(x)-u(-R,R)(y)|2|x-y|1+2s-12∬I3*|uR*(x)-uR*(y)|2|x-y|1+2s>0,∫I3*‍b(x){G(u(-R,R))-G(uR*)}≥b_∫PRQR‍{G(u(-R,R))-G(uR*)}≥b_L{min⁡(G(1-δ),G(-1+δ),G(-1+|C0|2))-G(1-δ*)(G(1-δ),G(-1+δ),G(-1+|C0|2))}>0.
Therefore, ℰnon(u(-R,R),(-R,R))>ℰnon(uR*,(-R,R)) for R large enough. This contradiction deduces the fact that u→-1 as x→-∞. u is the desired solution.



---

*Source: 548301-2014-03-27.xml*