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BY 4.0 license Open Access Published by De Gruyter February 25, 2022

Thresholds for the existence of solutions to inhomogeneous elliptic equations with general exponential nonlinearity

  • Kazuhiro Ishige , Shinya Okabe EMAIL logo and Tokushi Sato

Abstract

In this paper we study the existence and the nonexistence of solutions to an inhomogeneous non-linear elliptic problem

(P) Δu+u=F(u)+κμinRN,u>0inRN,u(x)0as|x|,

where F = F(t) grows up (at least) exponentially as t ∞. Here N ≥ 2, κ > 0, and μLc1(RN)\{0} is nonnegative. Then, under a suitable integrability condition on μ, there exists a threshold parameter κ* > 0 such that problem (P) possesses a solution if 0 < κ < κ* and it does not possess no solutions if κ > κ*. Furthermore, in the case of 2 ≤ N ≤ 9, problem (P) possesses a unique solution if κ = κ*.

MSC 2010: 35B09; 35J61; 35B32

1 Introduction

This paper is concerned with the existence and the nonexistence of solutions to an inhomogeneous nonlinear elliptic problem

(P) {Δu+u=F(u)+κμinRN,u>0inRN,u(x)0as|x|,

where N ≥ 2, FC1([0, ∞)) ∩ C2((0, ∞)), κ > 0, and μLc1(RN)\{0} is nonnegative. Throughout this paper we assume the following conditions on the nonlinear term F:

  1. (Behavior of F as t → 0+)

    F(0) = F′(0) = 0 and F″(t) = O(tα−1) as t → +0 for some α ∈ (0, 1);

  2. (Exponential nonlinearity of F)

    F is a convex function in (0, ∞) such that

limtF(t)=,liminftF(t)F(t)F(t)2=1.

Conditions (F1) and (F2) are satisfied in the following cases:

  • F(t) = tp exp(tq) with p > 1 and q > 0;

  • F(t) = tp exp(exp(tq)) with p > 1 and q > 0.

In this paper, under a suitable integrability condition on μ, we prove the existence of the threshold parameter κ* on the solvability of problem (P). Furthermore, in the case of 2 ≤ N ≤ 9, we show that problem (P) possesses a unique solution if κ = κ*, and give a complete classification of the solvability of problem (P).

Let us recall some results on an inhomogeneous elliptic problem with power nonlinearity

(P’) {Δu+u=up+κμinRN,u>0inRN,u(x)0as|x|,

where N ≥ 2, κ > 0, and μ is a nonnegative Radon measure in RN with compact support. The structure of solutions to problem (P’) has been studied in many papers. See e.g., [1,2,3,4,5,6,11,12,13,14,19,20,21,22,27,28,30,32,33,34,35], and references therein. Among others, under the assumption that

μH1(RN)and|x|N2μL(RN),

Deng and Li [11,12] proved the following assertions:

  1. There exists κ* > 0 such that problem (P’) possesses a solution if 0 < κ < κ* and it possesses no solutions if κ > κ*;

  2. Let κ = κ*. Then problem (P’) possesses a unique solution if 1 < ppS;

  3. Assume either 1 < p < pS or p = pS with 3 ≤ N ≤ 5. Then problem (P’) possesses at least two solutions in H1(RN) if 0 < κ < κ*;

  4. Let p = pS and N ≥ 6. Under a suitable symmetric condition on μ, problem (P’) possesses a unique solution in H1(RN) for small enough κ > 0.

Here

pS=ifN=1,2,pS:=N+2N2ifN3.

Subsequently, under more general conditions on μ, in [21], the authors of this paper obtained assertion (A) for problem (P’). Furthermore, we proved that assertion (B) holds in the Joseph-Lundgren subcritical case, that is,

1<p<pJL:={ifN10,(N2)24N+8N1(N2)(N10)ifN11.

In this paper, motivated by 21], we study the existence and the nonexistence of solutions to inhomogeneous elliptic problem (P) with exponential nonlinearity and obtain similar results to assertion (A). Furthermore, we prove that assertion (B) holds in the case of 2 ≤ N ≤ 9.

The structure of solutions to elliptic problems with exponential nonlinearity has also been studied in many papers. See e.g. [7,8,15,22,24,25,26] and references therein.

We introduce some notations and formulate a definition of solutions to problem (P). For xRN and R > 0, let B(x, R) := {yRN : |xy| < R}. Define

C0(RN):={fC(RN):lim|x|f(x)=0},Lcr(RN):={fLr(RN):fhasacompactsupportinRN},

where 1 ≤ r ≤ ∞. Similarly, we define Wc1,r(RN) , where 1 ≤ r ≤ ∞. We denote by G the fundamental solution of −Δv + v = 0 in RN, that is,

(1.1) G(x):=1(2π)N/2|x|(N2)/2K(N2)/2(|x|),xRN\{0},

where K(N−2)/2 is the modified Bessel function of order (N − 2)/2.

Definition 1.1

Let μ be a nonnegative measurable function in RN and κ > 0.

  1. We say that u is a solution to problem (P) if uC0(RN)+Lc(RN) and u satisfies

    (1.2) u(x)=[G*F(u)](x)+κ[G*μ](x)>0foralmostallxRN.

  2. We say that u is a supersolution to problem (P) if uC0(RN)+Lc(RN) and u satisfies

    u(x)[G*F(u)](x)+κ[G*μ](x)>0foralmostallxRN.

  3. Let u be a solution to problem (P). We say that u is a minimal solution to problem (P) if, for any solution v to problem (P), u(x) ≤ v(x) holds for almost all xRN.

Now we are ready to state our results of this paper. Theorem 1.1 corresponds to assertion (A) and shows the existence of threshold parameter κ* for the existence of solutions to problem (P).

Theorem 1.1

Let N ≥ 2 and FC1([0, ∞)) ∩ C2((0, ∞)). Assume conditions (F1) and (F2). Let μLc1(RN)\{0} be nonnegative in RN and

(1.3) G*μL(RN).

Then there exists κ* ∈ (0, ∞) with the following properties:

  1. If 1 < κ < κ*, then problem (P) possesses a minimal solution uκ. Furthermore, uκ(x) = O(G(x)) as |x| → ∞;

  2. If κ > κ*, then problem (P) possesses no solutions.

Theorem 1.2 corresponds to assertion (B) and shows the existence and the uniqueness of solutions to problem (P) with κ = κ*. See also Theorem 6.1.

Theorem 1.2

Let N ≥ 2 and FC1([0, ∞)) ∩ C2((0, ∞)). Assume conditions (F1) and (F2). Let μLc1(RN)\{0} be nonnegative in RN and 2 ≤ N ≤ 9. Then problem (P) with κ = κ* possesses a unique solution if either

  1. limsuptF(t)F(t)< and G * μBC(RN);

  2. There exists μ¯Wc1,r(RN) with r = 4N/(N + 2) such that μμ¯ in RN.

We remark that G * μBC(RN) if μLcr for some r > N/2. (See (G2) in Section 2.)

Theorem 1.1 is proved by a modification of the proof of [21, Theorem 1.1]. For the proof of Theorem 1.2, we obtain uniform local L-estimates of {uκ} with respect to 0 < κ < κ*, where uκ and κ* are as in Theorem 1.1. In this step, the arguments in [21] are not available since some key inequalities in [21] are not useful for exponential nonlinearity. Furthermore, it is not easy to introduce a general framework treating exponential nonlinearity. We prepare some inequalities peculiar to exponential nonlinearity under conditions (F1) and (F2), and obtain energy estimates of solutions to problem (P) to show

sup0<κ<κ*supzRN(1+F(uκ))1νH1(B(z,1))<

with a suitable choice ν ∈ (1/2, 1) (see (6.1)). Then we apply the Sobolev imbedding theorem to obtain

sup0<κ<κ*supzRNF(uκ)LpS+1ν(B(z,1))<.

On the other hand, it follows from 2 ≤ N ≤ 9 that

limν1/2pS+1ν=4NN2>N2.

Then we apply elliptic regularity theorems with a suitable choice of ν to obtain uniform local boundedness and equi-continuity of {F(uκ)} with respect to κ ∈ (0, κ*). See Section 6.1. These imply that the limit function u* := limκκ*−0 uκ is bounded in RN and it solves integral equation (1.2). Here limκκ *−0 uκ is the limit of uκ as κκ* with κ ∈ (0, κ*). Next we prove that u*(x) → 0 as |x| → ∞ by using the comparison between u* and its mirror reflection with respect to hyperplanes (see Section 6.2). Finally we modify the arguments in [21, Section 6] to complete the proof of Theorem 1.2 (see Section 6.3).

The rest of this paper is organized as follows. In Section 2 we recall some properties of the fundamental solution and an eigenvalue problem to the elliptic operator −Δ+1. In Sections 3 and 4 we modify the arguments in [21] to obtain some estimates of approximate solutions and prove Theorem 1.1. In Section 5 we prove some inequalities related to the nonlinearity F, which play an important role in Section 6.1. In Section 6 we prove Theorem 1.2, and give a complete classification of the solvability of problem (P) in the case of 2 ≤ N ≤ 9.

2 Preliminaries

In this section we recall some properties on the fundamental solution G = G(x). In what follows, for any nonnegative functions f and g in RN, we say that f(x) ≍ g(x) as x → 0 if there exists c > 0 such that c−1g(x) ≤ f(x) ≤ cg(x) in a neighborhood of 0. Similarly, we say that f(x) ≍ g(x) as x → ∞ if there exists c > 0 such that c−1g(x) ≤ f(x) ≤ cg(x) in a neighborhood of the space infinity. The letter C denotes generic positive constants and it may have different values also within the same line.

2.1 Fundamental solution G

We collect some properties of the fundamental solution G. It follows from (1.1) that G is radially symmetric in RN and monotone decreasing with respect to |x|. Furthermore,

(2.1) G(x){|x|(N2)ifN3,log|x|ifN=2,as|x|0,G(x)|x|N12e|x|as|x|.

We recall the following properties on the fundamental solution G. These follow from the Hölder inequality, the Hardy–Littlewood–Sobolev inequality, the Sobolev inequality, and elliptic regularity theorems (see e.g., [21, Section 2] and [27, Appendix]).

  1. Let r > N/2. Then

    G*vC0(RN)andG*vL(RN)CvLr(RN)

    for vLr(RN).

  2. For vL1(RN) ∩ Lr(RN) with some r > N/2, then

    G*vC0(RN)L1(RN)H1(RN).

  3. Let vLcr(RN) for some r > 1. Then

    [G*v]LcNrNr(RN)if1<r<Nand[G*v]BC(RN)ifr>N.

Set g := G * χB(0,1), where χB(0,1) is the characteristic function of the ball B(0, 1). Then

(2.2) gC,g(x)>0inRN,g(x)G(x)as|x|,|g|gBC(RN).

Furthermore, for any σ > 1, there exists C > 0 such that

(2.3) 0<[G*gσ](x)Cg(x)forxRN.

We define

gBC(RN):={f:g1fBC(RN)},

which is a Banach space with the norm |||f||| := supxRN |g(x)−1f(x)|.

2.2 Eigenvalue problem

Consider the eigenvalue problem

(2.4) Δϕ+ϕ=λa(x)ϕinRN,ϕH1(RN),

where aLN/2(RN)∩Lr(RN)\{0} for some r > N/2 and a(x) ≥ 0 for almost all xRN. By the same arguments in the proofs of [27, Lemmas B2 and B3] we have the following lemma.

Lemma 2.1

Then problem (2.4) has the first eigenvalue λ1 > 0 and the corresponding eigenfunction φ1 with φ1 > 0 in RN. Furthermore,

λ1=inf{ψH1(RN)2/RNaψ2dx:ψH1(RN),RNaψ2dx0}.

3 Approximate solutions

Let μLc1(RN)\{0} be nonnegative in RN and satisfy (1.3). Then

μ0(x):=[G*μ](x)>0foralmostallxRN,μ0C(RN\B(0,R)).

For any κ > 0, we define {Ujκ}j=0 and {Vjκ}j=0 by

(3.1) U0κ:=κμ0,Ujκ:=G*F(Uj1κ)+κμ0,j=1,2,,V0κ:=U0κ,Vjκ:=UjκUj1κ,j=1,2,.

On the other hand, by (F1), for any M > 0, we find CM > 0 such that

(3.2) 0F(t)CMtα+1,0F(t)CMtα,0F(t)CMtα1,

for t ∈ (0, M]. Notice that α ∈ (0, 1).

Let u be a solution to problem (P). Let K > 0 and j ∈ {0, 1, . . . }. Then, by the same arguments as in [21, Section 3] we obtain the following properties:

(3.3) 0<C1g(x)Ujκ(x)Uj+1κ(x)u(x),0Ujκ+ϵ(x)Ujκ(x)CϵUjκ(x),0Vjκ+ϵ(x)Vjκ(x)Cϵ,g1UjκL(RN)andg1UjκL(RN)Cκ,g1Vj+1κBC(RN)andg1Vj+1κBC(RN)Cκα(j+1)+1,

for xRN, 0 < κK, and ∈ (0, 1]. Furthermore, we have:

Lemma 3.1

Assume the same conditions as in Theorem 1.1. Let κ > 0. Then, for any j ∈ {0, 1, . . . }, there exists C > 0 such that

(3.4) 0Vjκ+ϵ(x)Vjκ(x)Cϵαg(x),xRN,

for 0 < ≤ 1.

Proof

Let 0 < ≤ 1. Since V0κ+ϵV0κ=κ1ϵU0κ , by (3.3) we have (3.4). Furthermore, by (F2), (3.1), (3.2), and (3.3) we see that

V1κ+ϵV1κ=G*[F(U0κ+ϵ)F(U0κ)]G*[F(U0κ+ϵ)(U0κ+ϵU0κ)]CϵG*[(U0κ)α+1]Cϵ[G*gα+1].

This together with (2.3) implies (3.4) with j = 1.

Assume that

(3.5) g1[Vjκ+ϵVjκ]L(RN)Cϵα,0<ϵ1,

for some j ∈ {1, 2, . . . }. It follows from (3.1) that

(3.6) Vj+1κ+ϵ(x)Vj+1κ(x)=G*[F(Ujκ+ϵ)F(Uj1κ+ϵ)]G*[F(Ujκ)F(Uj1κ)]=G*[h(x,1)h(x,0)],

where

h(x,t):=F(α(x,t))F(β(x,t)),α(x,t):=tUjκ+ϵ(x)+(1t)Uj1κ+ϵ(x),β(x,t):=tUjκ(x)+(1t)Uj1κ(x).

On the other hand, we have

(3.7) h(x,t)=F(α(x,t))Vjκ+ϵF(β(x,t))Vjκ=F(α(x,t))(Vjκ+ϵVjκ)+(F(α(x,t))F(β(x,t)))Vjκ

for 0 ≤ t ≤ 1. Furthermore, by (3.3) we see that

(3.8) 0α(x,t)β(x,t)=t(Ujκ+ϵ(x)Ujκ(x))+(1t)(Uj1κ+ϵ(x)Uj1κ(x))Cϵ(Ujκ(x)+Uj1κ(x))Cϵg(x),max{α(x,t),β(x,t)}=α(x,t)Ujκ+ϵ(x)CUjκ(x)Cg(x),min{α(x,t),β(x,t)}=β(x,t)Uj1κ(x)C1g(x),

for xRN, 0 ≤ t ≤ 1, and 0 < ≤ 1. Then, by (3.2), (3.5), and (3.8) we have

(3.9) 0F(α(x,t))(Vjκ+ϵ(x)Vjκ(x))Cϵαg(x)α(Vjκ+ϵ(x)Vjκ(x))Cϵαg(x)α+1,

(3.10) (F(α(x,t))F(β(x,t)))Vjκ(x)Cβ(x,t)α1(α(x,t)β(x,t))g(x)Cϵg(x)α+1,

for xRN, 0 ≤ t ≤ 1, and 0 < ≤ 1. Combining (3.6), (3.7), (3.9), and (3.10), by (2.3) we obtain

0Vj+1κ+ϵ(x)Vj+1κ(x)Cϵα[G*gα+1](x)Cϵαg(x)

for xRN and 0 < ≤ 1. Therefore we see that (3.5) holds with j replaced by j + 1. Then, by induction we obtain the desired conclusion. Thus Lemma 3.1 follows.

4 Proof of Theorem 1.1

Let κ > 0. Assume the same conditions as in Theorem 1.1. By conditions (F1) and (F2) we apply the same arguments in [21, Section 4] to obtain the following properties.

  1. The following two conditions are equivalent:

    1. u=w+U1κ is a solution to problem (P);

    2. wC0(RN) is positive in RN and w satisfies

      (4.1) w=G*[F(w+U1κ)F(U0κ)]inRN.

  2. Let w satisfy (W1)-(b). Then 0 ≤ w(x) ≤ Cg(x) in RN. Furthermore, wH1(RN) and w is a weak solution of

    (4.2) Δw+w=F+(w+U1κ)F(U0κ)inRN,

    that is,

    (4.3) RN[wψ+wψ]dx=RN[F+(w+U1κ)F(U0κ)]ψdx

    for ψH1(RN). Here F+(t) := F(max{t, 0}) for tR.

  3. Let v be a supersolution to problem (P). Then Ujκ(x)v(x) in RN for j ∈ {0, 1, . . . }. In addition, the limit function

    Uκ(x):=limjUjκ(x)

    exists in RN and it is a minimal solution to problem (P).

Furthermore, similarly to the proof of [21, Lemma 4.4], we have:

Lemma 4.1

Assume the same conditions as in Theorem 1.1. Set

t*:=sup{t0:F(t)=0}.

Let κ* ≥ 0 be such that t* = κ*G * μL(RN). For any κ > 0, set

Sκ:={u:uisasolutiontoproblem(P)}.

Then the following properties hold.

  1. There exists > 0 such that Sκ ≠ ∅ for κ ∈ (0, κ* + ).

  2. If Sκ ≠ ∅ for some κ > 0, then problem (P) possesses a minimal solution uκ.

Notice that t* < ∞ and κ* < ∞ under the assumptions of Lemma 4.1.

Proof

We prove assertion (i). If 0 < κκ*, then F(U0κ)=F(κG*μ)=0 in RN. This together with (3.3) implies that uκ=U0κ in RN. So it suffices to consider the case of κ > κ*.

Let κ* < κ < κ* + 1, and define

Tκ[v]:=G*[F+(v+U1κ)F(U0κ)]forvgBC(RN).

We prove that Tκ with κ* < κ < κ* + is a contraction mapping on

BgBC(0,δ)¯:={fgBC(RN):|||f|||δ}

for some δ > 0 and ∈ (0, 1). It follows from (F1) and (F2) that

t*=sup{t0:F(t)=0}=sup{t0:F(t)=0}.

Then, for any L > 0, by (F1) we see that

(4.4) 0F(t)=0tF(τ)dτ=stF(τ)dτCstτα1dτCst(τs)α1dτC(ts)α

for 0 ≤ s < t < t* + L with t*s. Since 0U1κ*=U0κ*t* and V1κ*=0 in RN, by Lemma 3.1, (3.3), and (4.4) we have

|F+(v+U1κ)F(U0κ)|F(|v|+U1κ)(|v|+V1κ)=F(|v|+U1κ)(|v|+V1κV1κ*)C(|v|+U1κU1κ*)α(|v|+|V1κV1κ*|)C(|||v|||+|||U1κU1κ*|||)α(|||v|||+|||V1κV1κ*|||)gα+1C(|||v|||+κκ*)α(|||v|||+(κκ*)α)gα+1C(|||v|||+(κκ*)α)α+1gα+1

for vgBC(RN) with |||v||| ≤ 1. This together with (2.3) implies that

(4.5) |||Tκ[v]|||C(|||v|||+(κκ*)α)α+1

for vgBC(RN) with |||v||| ≤ 1. Similarly, by (3.3) and (4.4) we have

|F+(v+U1κ)F+(v˜+U1κ)||F(|v|+|v˜|+U1κ)||vv˜|C(|v|+|v˜|+U1κU1κ*)α|vv˜|C(|||v|||+|||v˜|||+|||U1κU1κ*|||)α|||vv˜|||gα+1C(|||v|||+|||v˜|||+κκ*)α|||vv˜|||gα+1

for vgBC(RN) with |||v||| ≤ 1. This together with (2.3) implies that

(4.6) |||Tκ[v]Tκ[v˜]|||C(|||v|||+|||v˜|||+κκ*)α|||vv˜|||

for v, v˜gBC(RN) with |||v|||, |||v˜|||1 .

Let δ ∈ (0, 1) and ∈ (0, 1) be small enough. By (4.5) and (4.6) we have

BgBC(0,δ)¯vTκ[v]BgBC(0,δ)¯,|||Tκ[v]Tκ[v˜]|||12|||vv˜|||forv,v˜BgBC(0,δ)¯,

for κ* < κ < κ* + . This means that Tκ is a contraction mapping on BgBC(0,δ)¯ .

Define a sequence {wk} ⊂ gBC(RN) by

w0:=0,wk:=Tκ[wk1](k=1,2,).

Due to (3.3), by induction we see that 0 < w1 < w2 < · · · < wk < · · · in RN. Applying the fixed point theorem, we can find a positive function wgBC(RN) such that

limk|||wwk|||=0,w=Tκ[w].

Furthermore, we see that w satisfies (4.1). Then we deduce that u=U1κ+w is a solution to problem (P), and obtain assertion (i). Assertion (ii) follows from property (W3). Thus the proof of Lemma 4.1 is complete.

Let κ > κ* and uκ be a minimal solution to problem (P). Then uκ > uκ* and F′(uκ) ≢ 0 in RN. Consider the linearized eigenvalue problem to problem (P) at uκ

(Eκ) Δϕ+ϕ=λF(uκ)ϕinRN,ϕH1(RN).

By Lemma 2.1 we see that problem (Eκ) has the first eigenvalue λ1κ>0 and

(4.7) λ1κ=inf{ψH1(RN)2/RNF(uκ)ψ2dx:ψH1(RN),RNF(uκ)ψ2dx0}.

Let ϕ1κ be the first eigenfunction to problem (Eκ) such that ϕ1κ>0 in RN. By the same arguments as in the proof of [21, Lemma 4.5] we see that

(4.8) ϕ1κ(x)=λ1κRNG(xy)F(uκ(y))ϕ1κ(y)dy,

(4.9) c1g(x)ϕ1κ(x)cg(x),

for xRN. Furthermore, we have:

Lemma 4.2

Assume the same conditions as in Theorem 1.1. Define

κ*:=sup{κ>0:Sκ}.

For any κ* < κ < κ*, let λ1κ be the first eigenvalue to problem (Eκ). Then λ1κ>1 and

(4.10) RN[|ψ|2+ψ2]dxλ1κRNF(uκ)ψ2dx>RNF(uκ)ψ2dx

for ψH1(RN) if

RNF(uκ)ψ2dx0.

Furthermore,

(4.11) 1<λ1κλ1κifκ*<κκ<κ*.

Proof

Let κ* < κ < κ′ < κ*. Set wκ:=uκU1κ , wκ:=uκU1κ , and ψ := wκ′wκ. It follows from (3.3) and property (W3) that

(4.12) wκ=limj[UjκU1κ]=limj[Vjκ+Vj1κ++V2κ]>limj[Vjκ+Vj1κ++V2κ]=limj[UjκU1κ]=wκ.

Since the function [t*, ∞) ∋ sF(t + s) − F(s) is strictly monotone increasing for any fixed t > 0, by (3.3) we see that

(4.13) F(wκ+U1κ)F(U0κ)=F(wκ+V1κ+U0κ)F(U0κ)>F(wκ+V1κ+U0κ)F(U0κ)F(wκ+V1κ+U0κ)F(U0κ)=F(wκ+U1κ)F(U0κ)0

for xE:={xRN:U0κ>t*} . It follows from κ > κ* that |E| > 0. Similarly, since the function [0, ∞) ∋ sF(t + s) − F(s) is monotone increasing for any fixed t > 0, by (3.3) we have

(4.14) F(wκ+U1κ)F(U0κ)=F(wκ+V1κ+U0κ)F(U0κ)F(wκ+V1κ+U0κ)F(U0κ)F(wκ+V1κ+U0κ)F(U0κ)=F(wκ+U1κ)F(U0κ)

for xRN \ E. Since ϕ1κ is the eigenfunction of (Eκ) and positive, by (4.12) we have

(4.15) λ1κRNF(uκ)ψϕ1κdx>0.

Since |E| > 0, by (4.3), (4.13), and (4.14) we obtain

(4.16) λ1κRNF(uκ)ψϕ1κdx=RN[ψϕ1κ+ψϕ1κ]dx=RN[F(wκ+U1κ)F(U0κ)F(wκ+U1κ)+F(U0κ)]ϕ1κdx>RN[F(wκ+U1κ)F(wκ+U1κ)]ϕ1κdxRNF(uκ)ψϕ1κdx.

Combining (4.15) with (4.16), we see that λ1κ>1 . This together with (4.7) implies (4.10) and (4.11). Thus Lemma 4.2 follows.

Now we are ready to complete the proof of Theorem 1.1.

Proof of Theorem 1.1

We show that κ* < ∞. Consider the eigenvalue problem

(4.17) Δψ+ψ=λψinB(0,1),ψ=0onB(0,1),ψH01(B(0,1)).

Let λB and ψB be the first eigenvalue and the first eigenfunction to problem (4.17), respectively, such that ψB > 0 in B(0, 1). Set ψB = 0 outside B(0, 1). Then ψBH1(RN) and

(4.18) RN[|ψB|2+ψB2]dx=B(0,1)[|ψB|2+ψB2]dx=λBB(0,1)ψB2dx>0.

On the other hand, it follows from Lemma 4.2, (F2), and (3.3) that

(4.19) RN[|ψB|2+ψB2]dx>RNF(uκ)ψB2dxRNF(U0κ)ψB2dxRNF(C1κg(x))ψB2dxC1F(C1κminxB¯(0,1)g(x))

for κ* < κ < κ*. Since F′(t) → ∞ as t → ∞, by (4.18) and (4.19) we see that κ* < ∞. This together with Lemma 4.1 implies assertions (i) and (ii) of Theorem 1.1. Thus Theorem 1.1 follows.

5 Some inequalities related to (F1) and (F2)

In this section we obtain some inequalities related to (F1) and (F2), which are used in Section 6.1. The following two lemmas concern with the growing up rate of F(t) and F′(t) as t → ∞.

Lemma 5.1

Assume the same conditions as in Theorem 1.1. Then

limtF(t)tq=limtF(t)tq=forq>0.

Proof

Let q > 0. Let > 0 be such that 1/ > q + 1. By (F2) we find T > 0 such that

(logF(t))(1ϵ)(logF(t)),tTϵ.

This implies that

logF(t)(1ϵ)logF(t)C,tTϵ,

that is,

F(t)eCF(t)1ϵ,tTϵ.

Then

F(t)[F(Tϵ)ϵ+ϵeC(tTϵ)]1ϵ,tTϵ.

On the other hand, by (F1) and (F2) we have

F(t)=F(t)F(0)tF(t),t0.

Therefore we deduce that

limtF(t)tqlimtF(t)tq+1=,limtF(t)tq=limtF(t)tq+1tF(t)F(t)=.

The proof is complete.

Lemma 5.2

Assume the same conditions as in Theorem 1.1. Set

(5.1) γ(t):=F(t)1+F(t)fort0.

Then F(t)γ(t) ≤ F′(t) for t ≥ 0. Furthermore,

(5.2) liminftγ(t)γ(t)2=0,limtγ(t)t=.

Proof

It easily follows from (5.1) that F(t)γ(t) ≤ F′(t) for t ≥ 0. Furthermore, by (F2) we have

liminftγ(t)γ(t)2=liminftF(t)(1+F(t))F(t)2(1+F(t))2(1+F(t))2F(t)2=liminft[F(t)(1+F(t))F(t)21]=0.

We prove that limt→∞ γ(t)t = ∞. Set

η(t):=γ(t)t=tF(t)1+F(t).

It follows that

(5.3) η(t)=(F(t)+tF(t))(1+F(t))tF(t)2(1+F(t))2=F(t)1+F(t)+tF(t)(1+F(t))F(t)2(1+F(t))2=t1η(t)+t1η(t)2[F(t)(1+F(t))F(t)21],t>0.

Then we have

(5.4) limsuptη(t)=.

Indeed, if not, then L := supt>0 η(t) < ∞. Let 0 < < 1/2L. Then, by (F2) and (5.3) we find T > 0 such that

(5.5) η(t)t1η(t)ϵt1η(t)212t1η(t),tT.

This implies that

logη(t)logη(T)12logtT,tT.

Then we see that η(t) → ∞ as t → ∞, which is a contradiction. Thus (5.4) holds.

Let L′ > 0. Assume that lim inft→∞ η(t) < L′. Then, by (5.4) we find a sequence {tj} ⊂ (0, ∞) such that

(5.6) limjtj=,η(tj)L,η(tj)0.

Similarly to (5.5), by (5.3) we obtain

η(tj)tj1η(tj)ϵtj1η(tj)212tj1η(tj)>0withϵ=12L.

This contradicts (5.6). So we see that lim inft→∞ η(t) ≥ L′. Since L′ is arbitrary, we obtain limt→∞ η(t) = ∞. Thus (5.2) holds, and Lemma 5.2 follows.

Let t* ≥ 0 be as in Lemma 4.1. Then it follows from (F1) and (F2) that

t*=sup{t0:F(t)=0}=sup{t0:γ(t)=0}.

For ν ∈ (0, 1), set

(5.7) Hν(t):=2ν0t(1+F(τ))2νγ(τ)2dτ,hν(t):={Hν(t)γ(t)fort>t*,0for0tt*.

By (F1) and (F2) we find δ > 0 such that

Hν(t)=2νt*t(1+F(τ))2νγ(τ)2dτCt*tF(τ)2dτCF(t)2(tt*)Cγ(t)2(tt*)

for t ∈ [t*, t* + δ). Then we see that hν is continuous in [0, ∞) and Hν(t) = γ(t)hν(t) for t ≥ 0. In the following lemmas we obtain some inequalities on Hν and hν.

Lemma 5.3

Assume the same conditions as in Theorem 1.1. Let 0 < ν < 1 and 0 < δ ≤ 1. Then there exists C > 0 such that

(1+F(t))2νδHν(t)t+Cfort0.

Proof

Let 0 < ν < 1. It follows from (F2) that

(1+F(t))2νγ(t)2=(1+F(t))2ν2F(t)2ast,

which implies that limt→∞ Hν(t) = ∞. On the other hand, by Lemma 5.1 we see that limt→∞ F(t)2/ν/t = ∞. Since

[t1(1+F(t))2ν]=t1(1+F(t))2ν[t1+2ν(1+F(t))1F(t)]=t1(1+F(t))2νγ(t)[1γ(t)t+2ν],

it follows from L’Hospital's rule and (5.2) that

limtHν(t)t(1+F(t))2ν=limt2ν(1+F(t))2νγ(t)2/[t1(1+F(t))2ν]=limt2νγ(t)t/[1γ(t)t+2ν]=.

Then we obtain the desired conclusion, and the proof is complete.

Lemma 5.4

Assume the same conditions as in Theorem 1.1. Let 0 < ν < 1 and 0 < δ ≤ 1. Then there exists C > 0 such that

(5.8) hν(t)(1+δ)(1+F(t))2ν+Cfort0.

Furthermore,

  1. hν(t)22ν(1+F(t))2νfort0 ,

  2. (2νδ)(1+F(t))2ννhν(t)0fort0if0<δ2ν(1ν)2ν .

Proof

Let ∈ (0, 1]. By (5.2) we find t ≥ 0 such that

(5.9) γ(t)ϵγ(t)2forttϵ.

Then, combining (5.9) with the definition of γ, we have

2νtϵt(1+F(τ))2νγ(τ)2dτ=tϵt[ddτ(1+F(τ))2ν]γ(τ)dτ=[(1+F(τ))2νγ(τ)]τ=tϵτ=ttϵt(1+F(τ))2νγ(τ)dτ[(1+F(τ))2νγ(τ)]τ=tϵτ=t+ϵtϵt(1+F(τ))2νγ(τ)2dτ

for tt. This together with (5.7) implies that

(5.10) (1νϵ2)(Hν(t)Hν(tϵ))(1+F(t))2νγ(t)(1+F(tϵ))2νγ(tϵ)

for tt. Then we deduce from Lemma 5.1, (5.2), and (5.7) that

(1νϵ2)limsupthν(t)(1+F(t))2ν1.

Since is arbitrary, we deduce that

limsupthν(t)(1+F(t))2ν1,

which implies (5.8). On the other hand, by the definition of γ and (F2) we have

γ(t)2+γ(t)=F(t)1+F(t)0.

This means that inequality (5.9) holds with = 1 and t = 0. Therefore, by (5.10) with = 1 and t = 0 we see that

(1ν2)Hν(t)(1+F(t))2νγ(t)

for t ≥ 0. This implies assertion (1). Furthermore, it follows from assertion (1) that

(2νϵ)(1+F(t))2ννhν(t)(2νϵ2ν2ν)(1+F(t))2ν0

for t ≥ 0 if

2νϵ2ν2ν0.

Setting δ = , we obtain assertion (2). Thus Lemma 5.4 follows.

Lemma 5.5

Assume the same conditions as in Theorem 1.1. Assume that

(5.11) limsuptF(t)F(t)<.

Then, for any ν ∈ (0, 1), δ ∈ (0, 1), and M > 0, there exists C > 0 such that

(5.12) Hν(t)γ(s+t)[(1+δ)(1+F(t))2ν+C]

for s ∈ [0, M] and t ≥ 0.

Proof

Let ∈ (0, 1] and t ≥ 0 be as in (5.9). It follows from (5.11) that

γ¯:=supt0γ(t)<.

Then, by (5.9) we have

ddt[log(1+F(τ))]τ=tτ=t+ϵ=γ(t+ϵ)γ(t)=tt+ϵγ(τ)dτϵtt+ϵγ(τ)2dτϵγ¯tt+ϵγ(τ)dτ=ϵγ¯[log(1+F(τ))]τ=tτ=t+ϵ

for tt and ∊′ > 0. This implies that the function

[tϵ,)teϵγ¯t[log(1+F(t+ϵ))log(1+F(t))]

is monotone increasing. Then we obtain

log(1+F(t+ϵ))log(1+F(t))ϵeϵγ¯slog(1+F(t+s+ϵ))log(1+F(t+s))ϵ

for tt, s ≥ 0, and ∊′ > 0. Letting ∊′ → +0, we see that

(5.13) γ(t)eϵγ¯sγ(t+s)

for tt, and s ≥ 0.

Let M > 0 and 0 < δ′ < δ. By (5.8) and (5.13), taking small enough ∈ (0, 1), we have

(5.14) Hν(t)eϵγ¯Mγ(t+s)[(1+δ)(1+F(t))2ν+C]γ(t+s)[(1+δ)(1+F(t))2ν+C]

for tt and s ∈ [0, M]. On the other hand, by the convexity of F and (5.8) we see that

(5.15) Hν(t)F(t)1+F(t+s)1+F(t+s)1+F(t)[(1+δ)(1+F(t))2ν+C]Cγ(t+s)[(1+δ)(1+F(t))2ν+C]

for 0 ≤ tt and s ∈ [0, M]. We combine (5.14) and (5.15) to obtain (5.12). Thus Lemma 5.5 follows.

6 Proof of Theorem 1.2

For 0 < κ < κ*, let uκ be the minimal solution to problem (P). Set

wκ:=uκU1κ.

In Section 6.1 we obtain uniform L-local estimates of {wκ} with respect to 0 < κ < κ* by using inequalities given in Section 5. Then we see that u* := limκκ*−0 uκ is bounded and it solves integral equation (1.2). In Section 6.2 we prove that u*(x) → 0 as |x| → ∞. In Section 6.3 we complete the proof of Theorem 1.2.

6.1 Uniform estimates of wκ

The main purpose of this subsection is to prove the following proposition.

Proposition 6.1

Let 2 ≤ N ≤ 9, and assume the same conditions as in Theorem 1.1. Further assume either

  1. limsuptF(t)F(t)<or

  2. There exists μ¯Wc1,r(RN) with r = 4N/(N + 2) such that μμ¯ in RN.

Then {wκ} are uniformly bounded in RN and equi-continuous for compact sets in RN with respect to 0 < κ < κ*. Here κ* is as in Theorem 1.1.

For this aim we prove the following lemma by the use of Lemma 5.4.

Lemma 6.1

Let ν ∈ (1/2, 1) be such that

(6.1) 2ν12ν(1ν)2ν.

Then, for any δ ∈ (0, 2ν − 1], there exists C > 0 such that

(2νδ)[(1+F(v)gz]1νH1(RN)2(v,Hν(v)gz2ν)H1(RN)+C

for nonnegative functions vH1(RN) and zRN. Here gz(x) := g(xz) for x, zRN.

Proof

Let ν ∈ (1/2, 1) be as in (6.1) and zRN. Let vH1(RN) be nonnegative and set

Ψ(x):=[(1+F(v(x)))gz(x)]1ν,xRN.

Since

Ψ=1ν[(1+F(v))gz]1ν1[F(v)gzv+(1+F(v))gz]=1ν[(1+F(v))gz]1ν[γ(v)v+gzgz],

we have

(6.2) |Ψ|2=1ν2[(1+F(v))gz]2ν[γ(v)2|v|2+2γ(v)vgzgz+|gz|2gz2].

On the other hand, it follows that

(6.3) [Hν(v)gz2ν]=2νgz(x)2ν[(1+F(v))2γ(v)2v+Hν(v)gzgz].

By (6.2) and (6.3) we have

(6.4) (2νδ)[(1+F(v))gz]1νH1(RN)2(v,Hν(v)gz2ν)H1(RN)=1ν2RN[δ(1+F(v))2νγ(v)2|v|2+ν2Hν(v)v]gz2νdx+2ν2RN[(2νδ)(1+F(v))2νγ(v)νHν(v)]vgzgzgz2νdx+2νδν2RN(1+F(v))2ν[ν2+|gz|2gz2]gz2νdx.

Let 0 < δ ≤ 2ν − 1. Then it follows from (6.1) that

0<δ<2ν12ν(1ν)2ν.

By Lemma 5.4 we see that

(2νδ)(1+F(v))2ννhν(v)0.

This together with (2.2) and (6.4) implies that

(2νδ)[(1+F(v)gz]1νH1(RN)2(v,Hν(v)gz2ν)H1(RN)1ν2RN[δ(1+F(v))2νγ(v)2|v|2+ν2Hν(v)v]gz2νdx+2Cν2RN(2νδ)(1+F(v))2νγ(v)|v|gz2νdx+CRN(1+F(v))2νgz2νdxRNHν(v)vgz2νdx+CRN(1+F(v))2νgz2νdx.

Then, by Lemma 5.3, (2.1), and (2.2) we obtain

(2νδ)[(1+F(v))gz]1νH1(RN)2(v,Hν(v)gz2ν)H1(RN)C.

Then Lemma 6.1 follows.

We prove Proposition 6.1 under condition (i).

Proof of Proposition 6.1 under condition (i)

Since wκ = 0 if 0 < κκ*, it suffices to consider the case of κ* < κ < κ*.

Assume condition (i). Let ν ∈ (1/2, 1) be as in (6.1) and 0 < δ < (2ν − 1)/2. It follows from (4.2) that

(wκ,Hν(wκ)gz2ν)H1(RN)=RN[F(wκ+U1κ)F(U0κ)]Hν(wκ)gz2νdxRNF(wκ+U1κ)Hν(wκ)gz2νdx

for zRN and κ* < κ < κ*. Then, by Lemma 5.5 with M=U1κ* and (5.1) we obtain

(wκ,Hν(wκ)gz2ν)H1(RN)RNF(uκ)γ(uκ)[(1+δ)(1+F(wκ))2ν+C]gz2νdx(1+δ)RNF(uκ)[(1+F(wκ))gz]2νdx+CRNF(uκ)gz2νdx.

Applying Lemmas 4.2 and 6.1, we obtain

(2νδ)[(1+F(wκ))gz]1νH1(RN)2C(wκ,Hν(wκ)gz2ν)H1(RN)(1+δ)[(1+F(wκ))gz]1νH1(RN)2+Cgz1νH1(RN)2

for zRN and κ* < κ < κ*. Since 2νδ > 1 + δ, we deduce that

(6.5) supκ*<κ<κ*supzRN[(1+F(wκ))gz]1νH1(RN)2<.

Consider the case when 3 ≤ N ≤ 9. Since

pS:=N+2N2>32,

we can assume, without loss of generality, that

12<ν<pS1=4N2,

which implies that

(6.6) pS+1ν=1ν2NN2>N2.

Applying the Sobolev inequality, by (6.5) we have

sup0<κ<κ*supzRN[(1+F(wκ))gz]1νLpS+1(RN)<,

which implies that

(6.7) sup0<κ<κ*supzRNF(wκ)LpS+1ν(B(z,1))<.

On the other hand, it follows from condition (i) that F(s + t) ≤ eCsF(t) + C for t ∈ [0, ∞) and s ≥ 0. This implies that

(6.8) F(uκ)F(U1κ*+wκ)CF(wκ)+CinRN

for κ* < κ < κ*. Combining (6.7) and (6.8), we obtain

(6.9) sup0<κ<κ*supzRNF(uκ)LpS+1ν(B(z,1))<.

Recalling uκ=wκ+U1κ and U0κL(RN) , by (6.6) we apply the elliptic regularity theorems (see e.g. [31, Theorems 1 and 7]) to elliptic equation (4.2). Then we see that {wκ}κ*<κ<κ* are uniformly bounded in RN and equi-continuous for compact sets in RN. Thus Proposition 6.1 follows in the case of 3 ≤ N ≤ 9 under condition (i).

Similarly, in the case of N = 2, by the Sobolev inequality and (6.5) we see that

supκ*<κ<κ*supzRNF(uκ)Lr(B(z,1))<

for r ∈ [1, ∞), instead of (6.9). Then, applying the elliptic regularity theorems again, we see that {wκ}κ*<κ<κ* are uniformly bounded in RN and equi-continuous for compact sets in RN. Thus Proposition 6.1 follows in the case of N = 2 under condition (i). The proof of Proposition 6.1 is complete under condition (i).

We prepare the following lemma for the proof of Proposition 6.1 under condition (ii).

Lemma 6.2

Let 2 ≤ N ≤ 9 and μ be a nonnegative measurable function in RN with compact support. Let μ¯Wc1,r(RN) with r = 4N/(N + 2) such that 0μ(x)μ¯(x) for almost all xRN. Then

G*μH1(RN)C0(RN),μ¯1νHc1(RN)for12<ν<1.

Proof

Let 2 ≤ N ≤ 9 and μ¯Wc1,r(RN) with r = 4N/(N + 2). Then r = 2 if N = 2 and r < N if N ≥ 3. Let 1/2 < ν < 1. The Sobolev imbedding theorem implies that

(6.10) μ¯1νLc2(RN)ifN=2,μ¯LcNrNr(RN)ifN3.

Since

2ν<4<NrNrwithr=4NN+2,

we have μ¯1/νLc2(RN) if N ≥ 3. So we see that μ¯1/νLc2(RN) .

On the other hand, it follows from r = 4N/(N + 2) and 1/2 < ν < 1 that

NNr=2r2>1r2(2ν2).

This implies that

(6.11) NrNr/(2ν2)>rr2=(r2).

Here (r/2) is the Hölder conjugate of r/2 ≥ 1. Since |μ¯|2Lcr/2(RN) and μ¯LcNrNr(RN) , by (6.11) we have

|μ¯1ν|2=1ν2μ¯2ν2|μ¯|2Lc1(RN).

Combining μ¯1/νLc2(RN) , we see that μ¯1/νHc1(RN) .

On the other hand, it follows from N ≤ 9 and r = 4N/(N + 2) that

NrNr>N2ifN3.

Then it follows from (G2), (6.10), and 0μμ¯ in RN that G * μC0(RN) ∩ H1(RN). Thus Lemma 6.2 follows.

Proof of Proposition 6.1 under condition (ii)

Similarly to the case of condition (i), it suffices to consider the case when κ* < κ < κ*. Assume condition (ii). Let ν ∈ (1/2, 1) be as in (6.1) and 0 < δ < (2ν − 1)/2. It follows from (P), (5.1), and (5.7) that

(uκ,Hν(uκ)gz2ν)H1(RN)=RN(F(uκ)+κμ)Hν(uκ)gz2νdxRNF(uκ)hν(uκ)gz2νdx+κRNμ¯γ(uκ)hν(uκ)gz2νdx

for zRN and κ* < κ < κ*. This together with Lemmas 5.4 and 6.1 implies that

(2νδ)[(1+F(uκ))gz]1νH1(RN)2C(uκ,Hν(uκ)gz2ν)H1(RN)RNF(uκ)hν(uκ)gz2νdx+2κ2νRNμ¯γ(uκ)(1+F(uκ))2νgz2νdxRNF(uκ)[(1+δ)(1+F(uκ))2ν+C]gz2νdx+2κ2νRNμ¯F(uκ)(1+F(uκ))2ν1gz2νdx

for zRN and κ* < κ < κ*. By Lemma 4.2 and the Hölder inequality we have

(2νδ)[(1+F(v))gz]1νH1(RN)2C(1+δ)[(1+F(v))gz]1νH1(RN)2+Cgz1νH1(RN)2+2κ2ν(RNμ¯(x)2νF(uκ)gz2νdx)ν2(RNF(uκ)(1+F(uκ))2νgz2νdx)2ν2(1+δ)[(1+F(v))gz]1νH1(RN)2+C+2κ2νμ¯1νgz1νH1(RN)ν[(1+F(v))gz]1νH1(RN)2ν

for zRN and κ* < κ < κ*. Since 2νδ > 1 + δ, we deduce that

supκ*<κ<κ*supzRN[(1+F(uκ))gz]1νH1(RN)2<.

Then, similarly to the proof under condition (i), the Sobolev inequality implies that

supκ*<κ<κ*supzRNF(uκ)Lr(B(z,1))<forsomer>N/2.

Applying the elliptic regularity theorems to elliptic equation (4.2), by (1.3) we see that {wκ}κ*<κ<κ* are uniformly bounded in RN and equi-continuous for compact sets in RN. Thus Proposition 6.1 follows under condition (ii). The proof of Proposition 6.1 is complete.

As a corollary of Proposition 6.1, we have:

Proposition 6.2

Assume the same conditions as in Proposition 6.1. Then there exists the limit function

(6.12) u*(x):=limκκ*0uκ(x)inRN

such that u*BC(RN)+Lc(RN) and it satisfies (1.2) in RN.

Proof

Let 0 < κ < κ′ < κ*. Since uκ′ and uκ are a supersolution and the minimal solution to problem (P), respectively, we have

(6.13) uκuκ=wκ+U1κwκ+U1κ*inRN.

Since U1κ*L(RN) , by Proposition 6.1 we see that {uκ} is uniformly bounded in RN and monotone increasing with respect to κ ∈ (0, κ*). This implies that the limit function u* can be defined by (6.12) and it is bounded in RN. Furthermore, we see that u* satisfies

u*(x)=[G*F(u*)](x)+κ[G*μ](x)>0inRN.

On the other hand, it follows that

w*(x):=u*(x)U1κ*(x)=limκκ*0[uκ(x)U1κ(x)]=limκκ*0wκ(x)inRN.

By Proposition 6.1 we apply the Arzelá–Ascoli theorem to see that w*BC(RN), that is, u*BC(RN)+Lc(RN) . Thus Proposition 6.2 follows.

6.2 Decay of w*

The aim of this subsection is to prove the following proposition.

Proposition 6.3

Assume the same conditions as in Proposition 6.1. Then u* is the minimal solution to problem (P) with κ = κ*, that is, u* = uκ*.

Thanks to Proposition 6.2 and (6.13), it suffices to prove that u*(x) → 0 as |x| → ∞. In the proof, it is crucial to compare u* with its mirror reflection with respect to hyperplanes.

For any θSN−1 and > 0, we define a half-space Πθ of RN by

Πθ:={xRN:xθ}.

Notice that 0Πθ and dist(0,Πθ)= . For xRN, set

xθ:=x+2(θx)θ,

which is the mirror reflection point of x with respect to the hyperplane Πθ .

Definition 6.1

Let f be a measurable function in RN. For R > 0, we say that f satisfies property (DR) if

f(xθ)f(x)inΠθ

for θSN−1 and ℓR.

The following lemma concerns with property (DR) and the convolution of functions.

Lemma 6.3

Let f be a measurable function such that (G * f)(x) is defined for almost all RN. If f satisfies property (DR) for some R > 0, then G * f also satisfies property (DR).

Proof

Assume that f satisfies property (DR) for some R > 0. Since G is radially symmetric in RN and monotone decreasing with respect to |x|, we have

(G*f)(x)(G*f)(xθ)=Πθ(G(xy)G(xθy))f(y)dyΠθ(G(xyθ)G(xθyθ))f(yθ)dy=Πθ(G(xy)G(xθy))(f(y)f(yθ))dy0

for xΠθ , θSN−1, and R. Here we used the relations

|xy|=|xθyθ|,x,yRN,|xy||xθy|,x,yΠθ.

Thus Lemma 6.3 follows.

We prove Proposition 6.3 by the use of property (DR).

Proof of Proposition 6.3

Let R > 0 be such that supp μB(0, R). We easily see that μ satisfies property (DR). This together with Lemma 6.3 implies that U0κ satisfies property (DR). Since F is monotone increasing, we see that F(U0κ) also satisfies property (DR). Similarly, thanks to Lemma 6.3, we see that G*F(U0κ) satisfies property (DR), which implies that U1κ satisfies property (DR). Repeating this arguments, we deduce that Ujκ satisfies property (DR) for j ∈ {0, 1, . . . }. Therefore, by property (W3) we see that uκ satisfies property (DR) for 0 < κ < κ*, which together with (6.12) implies that u* satisfies property (DR). Then the following properties hold.

  • For any θSN−1, the function [R, ∞) ∋ ru*() is monotone decreasing.

  • For any xRN, there exists L > 0 such that u*(x) ≥ u*(y) for yRN \ B(0, L).

Thanks to these properties, for any x0RN it follows that

limsup|x|u*(x)u*(x0).

Since x0 is arbitrary, we see that

limsup|x|u*(x)liminf|x|u*(x).

These imply that

(6.14) u*(x)u*:=lim|x|u*(x)inRN\B(0,R).

Then it follows from Proposition 6.2 that

RNG(xy)[F(u*(y))u*]dy=u*(x)u*U0κ*(x)0,

which implies that u*=F(u*) .

Assume that u*>0 . Then, by (F1) and (F2) we see that F(u*)>1 . On the other hand, by (4.10) and (6.12) we have

(6.15) RN\B(0,R)[|ψ|2+ψ2]dxRN\B(0,R)F(u*)ψ2dx

for ψH01(RN\B(0,R)) . Then, by (F2) and (6.14) we obtain

RN\B(0,R)|ψ|2dx=RN\B(0,R)[F(u*)1]ψ2dx[F(u*)1]RN\B(0,R)ψ2dxC1RN\B(0,R)ψ2dx

for ψH01(RN\B(0,R)) . This means that the Poincaré inequality holds in RN \B(0, R). This is a contradiction, and we see that u*=0 . Therefore, combining Proposition 6.2, we deduce that u* is a solution to problem (P) with κ = κ*. Thus Proposition 6.3 follows.

6.3 Proof of Theorem 1.2

We complete the proof of Theorem 1.2. It remains to prove the uniqueness of solutions to problem (P) with κ = κ*. For this aim, we prove the following lemma.

Lemma 6.4

Assume the same conditions as in Proposition 6.1. Then λ1κ*=1 .

Proof

It follows from (F2) and (6.15) that

RN[|ψ|2+ψ2]dxRNF(uκ*)ψ2dx

for ψH1(RN) \ {0}. This implies that λ1κ*1 .

Assume that λ1κ*>1 . For ∈ (0, 1] and a ∈ (0, 1], set

u¯:=U1κ*+ϵ+wκ*+aϕ1κ*C0(RN)+Lc(RN),

where wκ*:=uκ*U1κ* . Then

(6.16) u¯G*F(u¯)(κ*+ϵ)G*μ=(wκ*+aϕ1κ*)G*F(u¯)+G*F(U0κ*+ϵ)=G*[F(wκ*+U1κ*)F(U0κ*)]G*F(u¯)+G*F(U0κ*+ϵ)+aϕ1κ*=G*[H(1)H(0)]+aϕ1κ*,

where

H(x,t):=F(P(x,t))F(Q(x,t)),P(x,t):=tU1κ*+ϵ(x)+(1t)U1κ*(x)+wκ*(x)+atϕ1κ*(x),Q(x,t):=tU0κ*+ϵ(x)+(1t)U0κ*(x).

Furthermore, we have

(6.17) H(t)=F(P(x,t))(U1κ*+ϵU1κ*+aϕ1κ*)F(Q(x,t))(U0κ*+ϵU0κ*)=F(uκ*)aϕ1κ*+{F(P(x,t))F(uκ*)}aϕ1κ*+F(P(x,t))(V1κ*+ϵV1κ*)+{F(P(x,t))F(Q(x,t))}(U0κ*+ϵU0κ*)

for xRN and t ∈ (0, 1).

On the other hand, it follows from property (W2) that 0 < wκ* (x) ≤ Cg(x) in RN. Then, by (3.3) and (4.8) we have

(6.18) 0P(x,t)Q(x,t)=tV1κ*+ϵ(x)+(1t)V1κ*(x)+wκ*(x)+atϕ1κ*(x)V1κ*+1(x)+wκ*(x)+ϕ1κ*(x)Cg(x),max{P(x,t),Q(x,t)}U1κ*+ϵ(x)+wκ*(x)+atϕ1κ*(x)CU1κ*(x)+Cg(x)=C[U0κ*(x)+V1κ*(x)]+Cg(x)Cg(x),min{P(x,t),Q(x,t)}C1g(x).

These together with (3.2) and (3.3) imply that

(6.19) 0{F(P(x,t))F(Q(x,t))}(U0κ*+ϵ(x)U0κ*(x))Cϵg(x)α+1

for xRN and t ∈ (0, 1). Furthermore, by Lemma 3.1 we have

(6.20) F(P(x,t))(V1κ*+ϵV1κ*)Cϵαg(x)α+1

for xRN and t ∈ (0, 1). Similarly to (6.18), we have

0P(x,t)[U1κ*(x)+wκ*(x)]=t[U1κ*+ϵ(x)U1κ*(x)]+atϕ1κ*(x)CϵU1κ*(x)+Cag(x)C(ϵ+a)g(x),max{P(x,t),U1κ*(x)+wκ*(x)}U1κ*+ϵ(x)+wκ*(x)+aϕ1κ*(x)Cg(x),min{P(x,t),U1κ*(x)+wκ*(x)}C1g(x).

These imply that

(6.21) 0F(P(x,t))F(uκ*)C(ϵ+a)g(x)α

for xRN and t ∈ (0, 1). Combining (6.17), (6.19), (6.20), and (6.21), we obtain

H(x,t)aF(uκ*(x))ϕ1κ*(x)+Ca(ϵ+a)g(x)α+1+C(ϵα+ϵ)g(x)α+1

for xRN and t ∈ (0, 1). Therefore we deduce from (2.3), (4.8), (4.9), and (6.16) that

u¯(x)[G*F(u¯)](x)(κ*+ϵ)[G*μ](x)a[G*(F(uκ*)ϕ1κ*)](x)C[a(ϵ+a)+ϵα+ϵ][G*gα+1](x)+aϕ1κ*(x)a(11λ1κ*)ϕ1κ*(x)C[a(ϵ+a)+ϵα+ϵ]g(x)a(11λ1κ*)ϕ1κ*(x)C[a(ϵ+a)+ϵα+ϵ]ϕ1κ*(x)

in RN. Then, taking suitable small enough a > 0 and > 0, we observe that

u¯(x)[G*F(u¯)](x)(κ*+ϵ)[G*μ](x)>0inRN.

This means that ū is a supersolution to problem (P) with κ = κ* + . Then, by property (W3) we find a solution to problem (P) with κ = κ* + . This contradicts the definition of κ*. Therefore we deduce that λ1κ*=1 , and the proof is complete.

Now we are ready to prove Theorem 1.2.

Proof of Theorem 1.2

Thanks to Proposition 6.3, it suffices to prove the uniqueness of solutions to problem (P) with κ = κ*.

Set wκ*=uκ*U1κ* . Let ũ be a solution to problem (P) with κ = κ* such that ũuκ*. Since uκ* is the minimal solution, we have ũuκ* and ũuκ* in RN. Then it follows from (1.2) and the elliptic regularity theorems (see e.g. [31, Theorems 1 and 7]) that ũuκ*C(RN) and

(6.22) u˜(x)uκ*(x)>0inRN.

Set w˜=u˜U1κ* . Then wκ* and w˜ are solutions to (4.2) with κ = κ* and the function z:=w˜wκ* satisfies

(6.23) z>0inRN,Δz+z=F(w˜+U1κ*)F(wκ*+U1κ*)inRN.

Let ϕ1κ* be the eigenfunction to problem (Eκ*) corresponding to λ1κ* such that ϕ1κ*>0 in RN. By Lemma 6.4 we have

(6.24) Δϕ1κ*+ϕ1κ*=F(wκ*+U1κ*)ϕ1κ*inRN.

Multiplying (6.24) by z and integrating it on RN, we obtain

RN[zϕ1κ*+zϕ1κ*]dx=RNF(wκ*+Uj*κ*)ϕ1κ*(w˜wκ*)dx.

On the other hand, by (6.23) we see that

RN[zϕ1κ*+zϕ1κ*]dx=RN[F(w˜+U1κ*)F(wκ*+U1κ*)]ϕ1κ*dx.

These imply that

RN[F(w˜+U1κ*)F(wκ*+U1κ*)F(wκ*+U1κ*)(w˜wκ*)]ϕ1κ*dx=0,

that is,

(6.25) RN[F(u˜)F(uκ*)F(uκ*)(u˜uκ*)]ϕ1κ*dx=0.

On the other hand, it follows from condition (F2) that F(t) ≥ F(s) + F′(s)(ts) for ts. Then, by (6.22) and (6.25) we see that

(6.26) F(u˜)F(uκ*)F(uκ*)(u˜uκ*)=0inRN.

Consider the case when t* > 0, where t* is as in Lemma 4.1. By assumption (i) of Theorem 1.2 with (3.3) and Lemma 6.2 we have G * μC0(RN). Recalling that uκ*=U1κ*+wκ* , we see that uκ*C0(RN). On the other hand, it follows from Lemma 4.1 (i) that κ* > κ*, which implies that

(6.27) uκ*L(RN)U1κ*L(RN)U0κ*L(RN)=κ*G*μL(RN)>κ*G*μL(RN)=t*.

Then we find x*RN such that uκ* (x*) = t*, which implies that

F(u˜(x*))=F(u˜(x*))F(uκ*(x*))F(uκ*(x*))(u˜(x*)uκ*(x*))=0.

Then we have ũ(x*) ≤ t*, which contradicts (6.22). Thus uκ* is the unique solution to problem (P) with κ = κ* in the case when t* > 0.

Consider the case when t* = 0. Since F(0) = 0 and F(t) > 0 if t > 0, by the convexity of F we find a sequence {tj} ⊂ (0, ∞) such that limj→∞ tj = 0 and F″(tj) > 0. This together with uκ*C0(RN) implies that

(6.28) F(uκ*(x))>0forsomexRN.

Then, by (F2), (6.22), and (6.28) we have

F(u˜(x))F(uκ*(x))F(uκ*(x))(u˜(x)uκ*(x))=01(1τ)d2dτ2F(τu˜(x)+(1τ)uκ*(x))dτ=(u˜(x)uκ*(x))201(1τ)F(τu˜(x)+(1τ)uκ*(x))dτ>0.

This contradicts (6.26), and we see that uκ* is the unique solution to problem (P) with κ = κ* in the case when t* = 0. Thus Theorem 1.2 follows.

At the end of this paper we give other sufficient conditions for the uniqueness of solutions to problem (P) with κ = κ* in the case when lim supt→∞ F′(t)/F(t) < ∞. We remark that G * μ is not necessarily continuous in RN under the assumptions of Theorem 6.1.

Theorem 6.1

Assume conditions (F1) and (F2). Let μLc1(RN)\{0} be nonnegative in RN, and assume (1.3). Let 2 ≤ N ≤ 9 and lim supt→∞ F′(t)/F(t) < ∞. Then problem (P) with κ = κ* possesses a unique solution if either t* = 0 or F is strictly convex in (t*, ∞).

Proof

Similarly to Theorem 1.2, it suffices to prove the uniqueness of solutions to problem (P) with κ = κ*. Consider the case when t* = 0. Since μ has a compact support, we see that G * μ is continuous in RN \ B(0, R) for some R > 0. Then we apply the same argument as in the proof of Theorem 1.2 in the case when t* = 0 to see that identity (6.26) does not hold. On the other hand, if F is strictly convex in (t*, ∞), then, by (6.27) we see that identity (6.26) does not hold. Then, similarly to the proof of Theorem 1.2, we complete the proof.

Acknowledgements

The first and the second authors were supported in part by JSPS KAKENHI Grant Number JP19H05599. The second author was supported in part by JSPS KAKENHI Grant Number 20KK0057 and 21H00990.

  1. Conflict of interest statement:

    The authors state no conflict of interest.

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Received: 2021-09-13
Accepted: 2021-12-13
Published Online: 2022-02-25

© 2022 Kazuhiro Ishige et al., published by De Gruyter

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