# Existence Results for Quasilinear Elliptic Equations with Indefinite Weight

Authors: Mieko Tanaka
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We provide the existence of a solution for quasilinear elliptic equation-div(a∞(x)|∇u|p-2∇u+ã(x,|∇u|)∇u)=λm(x)|u|p-2u+f(x,u)+h(x) in Ω under the Neumann boundary condition. Here, we consider the condition that ã(x,t)=o(tp-2) as t→+∞ and f(x,u)=o(|u|p-1) as |u|→∞. As a special case, our result implies that the following p-Laplace equation has at least one solution: -Δpu=λm(x)|u|p-2u+μ|u|r-2u+h(x) in Ω,∂u/∂ν=0 on ∂Ω for every 1<r<p<∞, λ∈ℝ, μ≠0 and m,h∈L∞(Ω) with ∫Ωmdx≠0. Moreover, in the nonresonant case, that is, λ is not an eigenvalue of the p-Laplacian with weight m, we present the existence of a solution of the above p-Laplace equation for every 1<r<p<∞, μ∈ℝ and m,h∈L∞(Ω).


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## Body

## 1. Introduction

In this paper, we consider the existence of a solution for the following quasilinear elliptic equation:(P;λ,m,h)-div⁡A(x,∇u)=λm(x)|u|p-2u+f(x,u)+h(x)inΩ,∂u∂ν=0on⁡∂Ω,whereΩ⊂ℝN is a bounded domain with C2 boundary ∂Ω, ν denotes the outward unit normal vector on ∂Ω, λ∈ℝ, 1<p<∞ and m,h∈L∞(Ω). We assume that f is a Carathéodory function on Ω×ℝ satisfying
(1.1)lim⁡|t|→∞f(x,t)|t|p-2t=0uniformlyinx∈Ω,
and that f(x,t) is bounded on a bounded set (admitting f≡0 in the nonresonant case). Here, A:Ω-×ℝN→ℝN is a map which is strictly monotone in the second variable and satisfies certain regularity conditions (see the following assumption (A)). The equation (P;λ,m,h) contains the corresponding p-Laplacian problem as a special case. Although the operator A is nonhomogeneous in the second variable in general, we assume that A(x,y) is asymptotically (p-1)-homogeneous at infinity in the following sense (AH).Throughout this paper, we assume that the mapA satisfies the following assumptions (AH) and (A):(AH)
there exist a positive functiona∞∈C1(Ω-,ℝ) and a continuous function a~(x,t) on Ω-×ℝ such that
(1.2)A(x,y)=a∞(x)|y|p-2y+a~(x,|y|)yforeveryx∈Ω,y∈RN,lim⁡t→+∞a~(x,t)tp-2=0uniformlyinx∈Ω-.(A)
A(x,y)=a(x,|y|)y, where a(x,t)>0 for all (x,t)∈Ω-×(0,+∞) and
(i)
A∈C0(Ω-×ℝN,ℝN)∩C1(Ω-×(ℝN∖{0}),ℝN);(ii)
there existsC1>0 such that
(1.3)|DyA(x,y)|≤C1|y|p-2foreveryx∈Ω-,y∈RN∖{0};(iii)
there existsC0>0 such that
(1.4)DyA(x,y)ξ⋅ξ≥C0|y|p-2|ξ|2foreveryx∈Ω-,y∈RN∖{0},ξ∈RN;(iv)
there existsC2>0 such that
(1.5)|DxA(x,y)|≤C2(1+|y|p-1)foreveryx∈Ω-,y∈RN∖{0}.A similar hypothesis to (A) is considered in the study of quasilinear elliptic problems (cf. [1, Example  2.2], [2–6]). It is easily seen that many examples as in the above references satisfy the condition (AH). Also, the following example satisfies our hypotheses:
(1.6)div⁡((|∇u|p-2+|∇u|q-2)(1+|∇u|q)(p-q)/q∇u)for1<p≤q<∞.
In particular, for A(x,y)=|y|p-2y, that is, divA(x,∇u) stands for the usual p-Laplacian Δpu, we can take C0=C1=p-1 in (A). Conversely, in the case where C0=C1=p-1 holds in (A), by the inequalities in Remark 1.4 (ii) and (iii), we see a(x,t)=|t|p-2 whence A(x,y)=|y|p-2y.Concerning the weightm,  throughout this paper, we assume that
(1.7)|{m>0}|:=|{x∈Ω;m(x)>0}|>0
holds, where |X| denotes the Lebesgue measure of a measurable set X.BecauseA(x,y) is asymptotically (p-1)-homogeneous at infinity, the solvability of our equation is related to the following homogeneous equation (see Theorem 1.1):
(EV;m)-div⁡(a∞(x)|∇u|p-2∇u)=λm(x)|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω,
where a∞ is the positive function as in (AH). We say that λ∈ℝ is an eigenvalue of (EV;m) if the equation (EV;m) has a nontrivial solution.There are few existence results of a solution to our equation (and also thep-Laplace equation). For example, if λ<0 and m≡1 hold, then the standard argument guarantees the existence of a solution. For the p-Laplacian as a special case of our problem, it is shown in [7] that the equation
(1.8)-Δpu=λm|u|p-2u+hinΩ∂u∂ν=0on⁡∂Ω
has a unique positive solution provided 0<λ<λ*(m), ∫Ω‍mdx<0 and 0≢h∈L∞(Ω)+, where λ*(m) is the principal eigenvalue defined in Section 2.1 with a∞≡1. In [8], although the resonant case where λ=λ1(m) or λ=λ2(m) is considered under the assumptions to f(x,u)=f(u), its result does not cover the case of f(u)=|u|r-2u with 1<r<p, where λi(m) (i=1, 2) is ith eigenvalue of the p-Laplacian with weight m. For the Laplace problem under the Neumann boundary condition, we can refer to [9, 10]. Under the Dirichlet boundary condition, the existence results for the Laplace problem are well known when m≡1 and λ is not an eigenvalue of the Laplacian (cf. [11]). Moreover, under the Dirichlet (or blow-up) boundary condition, many authors study various equations involving the p-Laplace (Laplace) operator with (indefinite) weight. For example, we refer to [12] for boundary blow-up problems with Laplacian, [13] for periodic reaction-diffusion problems and [14, 15] for singular quasilinear elliptic problems.Recently, the present author shows the existence of a solution for our problem in the case whereλ is between the principal eigenvalue and the second eigenvalue in [6] (for f≡0). In addition, a similar situation is treated in [5]. However, existence results are not seen in the case when λ is greater than the second eigenvalue for our problem. Therefore, the first purpose of this paper is to present an existence result of a solution in the nonresonant case where λ is not an eigenvalue of (EV;m). Then, it studied the existence of at least one solution in the resonant case under assumptions that cover the case f(u)=μ|u|r-2u with 1<r<p and μ≠0.For the proof of our result, it is necessary to study the weighted eigenvalue problem(EV;m). Thus, in Section 2, we introduce two sequences {λn(m)}n and {μn(m)}n of an eigenvalue of (EV;m) defined by Ljusternik-Schnirelman theory or Drábek-Robinson's method (cf. [16]), respectively. Then, we show several properties of above eigenvalues. In Section 3, we give the proof in the nonresonant case by using {μn(m)}n. In Sections 4 and 5, we handle the resonant case.


### 1.1. Statements of Our Existence Results

First, we state the existence result of a solution in the nonresonant case.Theorem 1.1.
Assume thatλ∈ℝ is not an eigenvalue of (EV;m). Then, (P;λ,m,h) has at least one solution.To state our existence result in the resonant case, we introduce some conditions. Set(1.9)F(x,u):=∫0uf(x,s)ds,G~(x,y):=∫0|y|a~(x,t)tdt,
where a~ is the function as in (AH).  (H+)
there exist0≤q≤p-1 and H0>0 such that
(1.10)lim⁡|y|→∞pG~(x,y)-a~(x,|y|)|y|2|y|1+q=+∞uniformlyina.e.x∈Ω,f(x,t)t-pF(x,t)≥-H0(1+|t|1+q)fora.e.x∈Ω,everyt∈R;(H-)
there exist0≤q≤p-1 and H0>0 such that
(1.11)lim⁡|y|→∞pG~(x,y)-a~(x,|y|)|y|2|y|1+q=-∞uniformlyina.e.x∈Ω,f(x,t)t-pF(x,t)≤H0(|t|1+q+1)fora.e.x∈Ω,everyt∈R;(HF+)
there exist0≤q≤p-1 and H0>0 such that
(1.12)pG~(x,y)-a~(x,|y|)|y|2≥-H0(1+|y|1+q)foreveryx∈Ω,y∈RN,lim⁡|t|→∞f(x,t)t-pF(x,t)|t|1+q=+∞uniformlyina.e.x∈Ω;(HF-)
there exist0≤q≤p-1 and H0>0 such that
(1.13)pG~(x,y)-a~(x,|y|)|y|2≤H0(1+|y|1+q)foreveryx∈Ω,y∈RN,lim⁡|t|→∞f(x,t)t-pF(x,t)|t|1+q=-∞uniformlyina.e.x∈Ω.Theorem 1.2.
Assume one of the following conditions:(i)
λ=0 and (HF+) or (HF-) hold;(ii)
λ≠0, ∫Ω‍mdx≠0 and one of (H+), (H-), (HF+) and (HF-) hold;(iii)
λ≠0, ∫Ω‍mdx=0 and (H+) or (HF+) hold; Then, (P;λ,m,h) has at least one solution.In the special case wherea~(x,t)≡0 and f(x,u)=μ|u|r-2u for 1<r<p, we easily see that (HF+) or (HF-) holds with 0≤q<r-1 provided μ<0 or μ>0, respectively. Therefore, the following result is proved according to Theorem 1.2.Corollary 1.3.
Let1<r<p<∞, μ≠0 and ∫Ω‍mdx≠0. Then, the following equation has at least one solution:
(1.14)-div⁡(a∞(x)|∇u|p-2∇u)=λm(x)|u|p-2u+μ|u|r-2u+h(x)inΩ,∂u∂ν=0on⁡∂Ω.


### 1.2. Properties of the MapA

In what follows, the norm onW1,p(Ω) is given by ∥u∥p:=∥∇u∥pp+∥u∥pp, where ∥u∥q denotes the norm of Lq(Ω) for u∈Lq(Ω) (1≤q≤∞). Setting G(x,y):=∫0|y|‍a(x,t)tdt, then we can easily see that
(1.15)∇yG(x,y)=A(x,y),G(x,0)=0
for every x∈Ω-.Remark 1.4.
It is easily seen that the following assertions hold under condition (A):  (i)
for allx∈Ω-, A(x,y) is maximal monotone and strictly monotone in y;(ii)
|A(x,y)|≤(C1/(p-1))|y|p-1 for every (x,y)∈Ω-×ℝN;(iii)
A(x,y)y≥(C0/(p-1))|y|p for every (x,y)∈Ω-×ℝN;(iv)
G(x,y) is convex in y for all x and satisfies the following inequalities:
(1.16)A(x,y)y≥G(x,y)≥C0p(p-1)|y|p,G(x,y)≤C1p(p-1)|y|p,
for every (x,y)∈Ω-×ℝN, where C0 and C1 are the positive constants in (A).The following result is proved in [3]. It plays an important role for our poof.Proposition 1.5 (see [3, Proposition 1]).
LetA:W1,p(Ω)→W1,p(Ω)* be the map defined by
(1.17)〈A(u),v〉=∫Ω‍A(x,∇u)∇vdx,
for u, v∈W1,p(Ω). Then, A has the (S)+ property, that is, any sequence {un} weakly convergent to u with limsupn→∞〈A(un),un-u〉≤0 strongly converges to u.


## 1.1. Statements of Our Existence Results

First, we state the existence result of a solution in the nonresonant case.Theorem 1.1.
Assume thatλ∈ℝ is not an eigenvalue of (EV;m). Then, (P;λ,m,h) has at least one solution.To state our existence result in the resonant case, we introduce some conditions. Set(1.9)F(x,u):=∫0uf(x,s)ds,G~(x,y):=∫0|y|a~(x,t)tdt,
where a~ is the function as in (AH).  (H+)
there exist0≤q≤p-1 and H0>0 such that
(1.10)lim⁡|y|→∞pG~(x,y)-a~(x,|y|)|y|2|y|1+q=+∞uniformlyina.e.x∈Ω,f(x,t)t-pF(x,t)≥-H0(1+|t|1+q)fora.e.x∈Ω,everyt∈R;(H-)
there exist0≤q≤p-1 and H0>0 such that
(1.11)lim⁡|y|→∞pG~(x,y)-a~(x,|y|)|y|2|y|1+q=-∞uniformlyina.e.x∈Ω,f(x,t)t-pF(x,t)≤H0(|t|1+q+1)fora.e.x∈Ω,everyt∈R;(HF+)
there exist0≤q≤p-1 and H0>0 such that
(1.12)pG~(x,y)-a~(x,|y|)|y|2≥-H0(1+|y|1+q)foreveryx∈Ω,y∈RN,lim⁡|t|→∞f(x,t)t-pF(x,t)|t|1+q=+∞uniformlyina.e.x∈Ω;(HF-)
there exist0≤q≤p-1 and H0>0 such that
(1.13)pG~(x,y)-a~(x,|y|)|y|2≤H0(1+|y|1+q)foreveryx∈Ω,y∈RN,lim⁡|t|→∞f(x,t)t-pF(x,t)|t|1+q=-∞uniformlyina.e.x∈Ω.Theorem 1.2.
Assume one of the following conditions:(i)
λ=0 and (HF+) or (HF-) hold;(ii)
λ≠0, ∫Ω‍mdx≠0 and one of (H+), (H-), (HF+) and (HF-) hold;(iii)
λ≠0, ∫Ω‍mdx=0 and (H+) or (HF+) hold; Then, (P;λ,m,h) has at least one solution.In the special case wherea~(x,t)≡0 and f(x,u)=μ|u|r-2u for 1<r<p, we easily see that (HF+) or (HF-) holds with 0≤q<r-1 provided μ<0 or μ>0, respectively. Therefore, the following result is proved according to Theorem 1.2.Corollary 1.3.
Let1<r<p<∞, μ≠0 and ∫Ω‍mdx≠0. Then, the following equation has at least one solution:
(1.14)-div⁡(a∞(x)|∇u|p-2∇u)=λm(x)|u|p-2u+μ|u|r-2u+h(x)inΩ,∂u∂ν=0on⁡∂Ω.


## 1.2. Properties of the MapA

In what follows, the norm onW1,p(Ω) is given by ∥u∥p:=∥∇u∥pp+∥u∥pp, where ∥u∥q denotes the norm of Lq(Ω) for u∈Lq(Ω) (1≤q≤∞). Setting G(x,y):=∫0|y|‍a(x,t)tdt, then we can easily see that
(1.15)∇yG(x,y)=A(x,y),G(x,0)=0
for every x∈Ω-.Remark 1.4.
It is easily seen that the following assertions hold under condition (A):  (i)
for allx∈Ω-, A(x,y) is maximal monotone and strictly monotone in y;(ii)
|A(x,y)|≤(C1/(p-1))|y|p-1 for every (x,y)∈Ω-×ℝN;(iii)
A(x,y)y≥(C0/(p-1))|y|p for every (x,y)∈Ω-×ℝN;(iv)
G(x,y) is convex in y for all x and satisfies the following inequalities:
(1.16)A(x,y)y≥G(x,y)≥C0p(p-1)|y|p,G(x,y)≤C1p(p-1)|y|p,
for every (x,y)∈Ω-×ℝN, where C0 and C1 are the positive constants in (A).The following result is proved in [3]. It plays an important role for our poof.Proposition 1.5 (see [3, Proposition 1]).
LetA:W1,p(Ω)→W1,p(Ω)* be the map defined by
(1.17)〈A(u),v〉=∫Ω‍A(x,∇u)∇vdx,
for u, v∈W1,p(Ω). Then, A has the (S)+ property, that is, any sequence {un} weakly convergent to u with limsupn→∞〈A(un),un-u〉≤0 strongly converges to u.


## 2. The Weighted Eigenvalue Problems

### 2.1. Preliminaries

The following lemmas can be easily shown by way of contradiction because∫Ω‍a∞|∇u|pdx is equivalent to ∥∇u∥pp (note that a∞ is positive). Here, we omit the proofs (refer to [7]).Lemma 2.1.
Assume∫Ω‍mdx<0. Then, there exists a constant c>0 such that ∫Ω‍a∞|∇u|pdx≥c∥u∥pp for every u∈W1,p(Ω) with ∫Ω‍m|u|pdx>0.Lemma 2.2.
Assume that∫Ω‍mdx≠0 and ξ>0. Then, there exists a constant b(m,ξ)>0 such that
(2.1)∫Ω‍a∞|∇u|pdx-ξ∫Ω‍m|u|pdx≥b(m,ξ)∫Ω‍|u|pdx
for every u∈B(m):={u∈W1,p(Ω);∫Ω‍m|u|pdx≤0}.Lemma 2.3.
Assume thatm≥0 in Ω. Then, for every ξ>0 there existed d(m,ξ)>0 such that
(2.2)∫Ω‍a∞|∇u|pdx-ξ∫Ω‍m|u|pdx≥d(m,ξ)∫Ω‍|u|pdx
for every u∈W1,p(Ω).First, we recall the following principle eigenvalueλ*(m):
(2.3)λ*(m)≔inf⁡⁡{∫Ω‍a∞|∇u|pdx;u∈W1,p(Ω),∫Ω‍m|u|pdx=1}.
Because of ∞>supx∈Ωa∞(x)≥infx∈Ωa∞(x)>0, we have the following result as the same argument as in the case of the p-Laplacian.Proposition 2.4 (see [7, Proposition  2.2]).
The following assertions hold:(i)
If∫Ω‍mdx≥0 holds, then λ*(m)=0;(ii)
If∫Ω‍mdx<0 holds, then λ*(m)>0 is a simple eigenvalue and it admits a positive eigenfunction. In addition, the open interval (0,λ*(m)) contains no eigenvalues of (EV;m).Lemma 2.5.
Assume∫Ω‍mdx<0. Then, one has λ*(m+ɛ)<λ*(m)<λ*(m-ɛ′) for every ɛ>0 and ɛ′>0 with |{m-ɛ′>0}|>0.Proof.
We choose a minimizeru for λ*(m) because Proposition 2.4 guarantees the existence of it. Then, for every ɛ>0, we have
(2.4)λ*(m+ɛ)≤∫Ω‍a∞|∇u|pdx∫Ω‍(m+ɛ)|u|pdx<∫Ω‍a∞|∇u|pdx∫Ω‍m|u|pdx=∫Ω‍a∞|∇u|pdx=λ*(m)
by the definition of λ*(m+ɛ). By applying the same argument to a minimizer for λ*(m-ɛ), we obtain λ*(m)<λ*(m-ɛ′) for ɛ′>0 with |{m-ɛ′>0}|>0.


### 2.2. Other Eigenvalues

Here, we introduce two unbounded sequences{λn(m)}n and {μn(m)}n as follows:
(2.5)J(u):=∫Ω‍a∞|∇u|pdxforu∈W1,p(Ω),J~:=J∣S(m),S(m):={u∈W1,p(Ω);∫Ω‍m|u|pdx=1},Sn(m):={X⊂S(m);compact,symmetricandγ(X)≥n},Fn(m):={g∈C(Sn-1,S(m));gisodd},λn(m):=inf⁡X∈Sn(m)max⁡u∈XJ~(u),μn(m):=inf⁡g∈Fn(m)max⁡z∈Sn-1J~(g(z)),
where γ(X) denotes the Krasnoselskii genus of X (see [17, Definition 5.1] for the definition) and Sn-1 denotes the usual unit sphere in ℝn. We see that λn(m) is defined by Ljusternik-Schnirelman theory and it is known that the definition of μn(m) is introduced by Drábek and Robinson ([16]) under the p-Laplace Dirichlet problem with m≡1.Remark 2.6.
The following assertions can be shown easily:(i)
λ1(m)=μ1(m)=λ*(m);(ii)
𝒮n(m)≠∅ and ℱn(m)≠∅ for every n∈ℕ;(iii)
g(Sn-1)⊂𝒮n(m) for every g∈ℱn(m);(iv)
μn(m)≥λn(m) for every n∈ℕ;(v)
λn+1(m)≥λn(m) and μn+1(m)≥μn(m) for every n∈ℕ,see [18] for the proof of  (ii).Define aC1 functional Φm on W1,p(Ω) by Φm(u):=∫Ω‍m|u|pdx for u∈W1,p(Ω). Because 1∈ℝ is a regular value of Φm, it is well known that the norm of the derivative at u∈S(m) of the restriction of J to S(m) is defined as follows:
(2.6)‖J~′(u)‖*:=min⁡⁡{‖J′(u)-tΦm′(u)‖W1,p(Ω)*;t∈R}=sup⁡⁡{〈J′(u),v〉;v∈Tu(S(m)),‖v‖=1},
where Tu(S(m)) denotes the tangent space of S(m) at u, that is, Tu(S(m))={v∈W1,p(Ω);∫Ω‍m|u|p-2uvdx=0}. Here, we recall the definition of the Palais-Smale condition for J~.Definition 2.7.
J
~ is said to satisfy the  bounded Palais-Smale condition if any bounded sequence un∈S(m) such that ∥J~′(un)∥*→0 has a convergent subsequence. Moreover, we say that J~ satisfies the Palais-Smale condition at level c∈ℝ if any sequence un∈S(m) such that J~(un)→c and ∥J~′(un)∥*→0 as n→∞ has a convergent subsequence. In addition, we say that J~ satisfies the Palais-Smale condition if J~ satisfies the Palais-Smale condition for every c∈ℝ.The following result can be proved by the same argument as in [19, Proposition 3.3]  (which treats the case of the p-Laplacian, i.e., a∞≡1) because of ∞>supx∈Ωa∞(x)≥infx∈Ωa∞(x)>0. Here, we omit the proof.Lemma 2.8.
The following assertions hold:(i)
J~ satisfies the bounded Palais-Smale condition;(ii)
J~ satisfies the Palais-Smale condition provided ∫Ω‍mdx≠0.Proposition 2.9.
λ
n
(
m
) and μn(m) are eigenvalues of (EV;m) such that
(2.7)lim⁡n→∞λn(m)=lim⁡n→∞μn(m)=+∞.Proof.
In the case of∫Ω‍mdx≠0, since J~ satisfies the Palais-Smale condition, we can apply the first deformation lemma on C1 manifold (refer to [20]). Thus, by the standard argument, we can prove that λn(m) and μn(m) are critical values of J~. This means that λn(m) and μn(m) are eigenvalues of (EV;m) by the Lagrange multiplier rule. In addition, we can easily show limn→∞λn(m)=+∞ by the standard argument via the first deformation lemma on C1 manifold (refer to [21, Proposition  3.14.7], [22] or [17] in the case of a Banach space). Hence, limn→∞μn(m)=+∞ holds because of μn(m)≥λn(m) for every n∈ℕ.
In the case of∫Ω‍mdx=0, by the same argument as in [18], our conclusion can be proved. For readers' convenience, we give a sketch of the proof. For ɛ>0, we define Jɛ(u):=J(u)+ɛ∥u∥pp and J~ɛ:=Jɛ∣S(m). Moreover, we set minimax values λnɛ(m) and μnɛ(m) of J~ɛ by
(2.8)λnɛ(m):=inf⁡X∈Sn(m)max⁡u∈XJ~ɛ(u),μnɛ(m):=inf⁡g∈Fn(m)max⁡z∈Sn-1J~ɛ(g(z)).
Because any Palais-Smale sequence of J~ɛ is bounded, it is easily shown that J~ɛ satisfies the Palais-Smale condition (refer to [19, Proposition  3.3]) Hence, it can be proved that λnɛ(m) and μnɛ(m) are critical values of J~ɛ. Furthermore, it follows from the argument as in [18, Lemma  3.5] that λnɛ(m)→λn(m) and μnɛ(m)→μn(m) as ɛ→0+. Therefore, by noting that Jɛ is p-homogeneous, we can obtain a solution uɛ with ∥uɛ∥=1 for -div(a∞|∇u|p-2∇u)=cɛm|u|p-2u in Ω, ∂u/∂ν=0 on ∂Ω, where cɛ=λnɛ(m) or μnɛ(m). Because of ∥uɛ∥=1, it follows from the standard argument that uɛ has a subsequence strongly convergent to a solution u for
(2.9)-div⁡(a∞|∇u|p-2∇u)=cm|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω,
where c=limɛ→0+cɛ. Thus, λn(m) and μn(m) are eigenvalues of (EV;m). To prove limn→∞λn(m)=+∞, by considering a function mδ(x):=max{m(x),δ}  for δ>0, we have λn(mδ)≤λn(m) (refer to Proposition 2.10). Because we can apply our fist assertion to mδ (note ∫Ω‍mδdx>0), we obtain limn→∞μn(m)≥limn→∞λn(m)≥limn→∞λn(mδ)=+∞.Proposition 2.10.
Let1<r<∞ if N≤p and p*/(p*-p)≤r<∞ if N>p. Then, the following assertions hold:  (i)
ifm′≥m in Ω, then μk(m′)≤μk(m);(ii)
iflimn→∞mn=m in Lr(Ω), then limsupn→∞μk(mn)≤μk(m);(iii)
if∫Ω‍mdx≠0 and limn→∞mn=m in Lr(Ω), then limn→∞μk(mn)=μk(m). Moreover, the same conclusion holds for λk(m).Proof.
We only treatμk(m) because we can give the proof for λk(m) similarly.(i)
Letm′≥m in Ω. Fix an arbitrary ɛ>0. Then, by the definition of μk(m), there exists a g∈ℱk(m) such that maxz∈Sk-1J(g(z))<μk(m)+ɛ. Set g~(z):=g(z)/(∫Ω‍m′|g(z)|pdx)1/p for z∈Sk-1 (note ∫Ω‍m′|g(z)|pdx≥∫Ω‍m|g(z)|pdx=1), then g~∈ℱk(m′) holds. Therefore, by the definition of μk(m′), we have
(2.10)μk(m′)≤max⁡z∈Sk-1J(g~(z))=max⁡z∈Sk-1J(g(z))∫Ω‍m′|g(z)|pdx≤max⁡z∈Sk-1J(g(z))<μk(m)+ɛ.because of∫Ω‍m′|g(z)|pdx≥∫Ω‍m|g(z)|pdx=1 for every z∈Sk-1. Since ɛ>0 is arbitrary, we obtain μk(m′)≤μk(m).(ii)
Letlimn→∞mn=m in Lr(Ω) and fix an arbitrary ɛ>0. By the definition of μk(m), there exists a g∈ℱk(m) such that maxz∈Sk-1J(g(z))<μk(m)+ɛ/2. Since g(Sk-1) is compact and pr′:=pr/(r-1)≤p*, we set M:=maxu∈g(Sk-1)∥u∥pr′. Then, due to Hölder's inequality and mn→m in Lr(Ω), there exists an n0∈ℕ such that
(2.11)∫Ω‍mn|u|pdx=1+∫Ω‍(mn-m)|u|pdx≥1-‖mn-m‖rMp>0for everyu∈g(Sk-1) and n≥n0. Therefore, by a similar argument to (i), we obtain
(2.12)μk(mn)≤max⁡z∈Sk-1J(g(z))∫Ω‍mn|g(z)|pdx≤μk(m)+ɛ/21-‖mn-m‖rMp<μk(m)+ɛfor sufficiently largen. Hence, limsupn→∞μk(mn)≤μk(m)+ɛ follows. Since ɛ>0 is arbitrary, our conclusion is proved.(iii)
Letlimn→∞mn=m in Lr(Ω) and ∫Ω‍mdx≠0. We fix an arbitrary ɛ>0. Due to our assertion (ii), there exists an n1∈ℕ such that μk(mn)≤μk(m)+ɛ/2. For every n≥n1, by the definition of μk(mn), we can take gn∈ℱk(mn) satisfying maxz∈Sk-1J(gn(z))<μk(mn)+ɛ/2.
Here, we will prove(2.13)sup⁡n≥n1max⁡{‖u‖p;u∈gn(Sk-1)}<∞.
Ifu∈gn(Sk-1) satisfies ∫Ω‍m|u|pdx≤0, then we obtain
(2.14)b(m,1)‖u‖pp≤J(u)-∫Ωm|u|pdx‍=J(u)-∫Ωmn|u|pdx‍+∫Ω(mn-m)|u|pdx‍≤μk(mn)+ɛ2-1+‖mn-m‖r‖u‖pr′p≤μk(m)+ɛ+C‖mn-m‖r‖u‖pp+CJ(u)‖mn-m‖rinf⁡Ωa∞≤(1+C‖mn-m‖rinf⁡Ωa∞)(μk(m)+ɛ)+C‖mn-m‖r‖u‖pp
by Lemma 2.2 and Hölder's inequality (note ∥∇u∥pp≤J(u)/infΩa∞ and μk(mn)≤μk(m)+ɛ/2), where C>0 is a constant (independent of n and u) obtained by the continuity of W1,p(Ω) into Lpr′(Ω). Therefore, if we take an n2≥n1 satisfying C∥mn-m∥r≤b(m,1)/2 for every n≥n2, then we obtain
(2.15)‖u‖pp≤2b(m,1)(1+b(m,1)2inf⁡Ωa∞)(μk(m)+ɛ)
for every u∈gn(Sk-1) provided ∫Ω‍m|u|pdx≤0 and n≥n2. Similarly, in the case where m changes sign, for every u∈gn(Sk-1) satisfying ∫Ω‍m|u|pdx>0, we have
(2.16)b(-m,1)‖u‖pp≤J(u)-∫Ω(-m)|u|pdx‍≤(1+C‖mn-m‖rinf⁡Ωa∞)(μk(m)+ɛ)+1+C‖mn-m‖r‖u‖pp.
Hence, by taking a sufficiently large n3≥n2, we get the inequality
(2.17)‖u‖pp≤2b(-m,1)(1+b(-m,1)2inf⁡Ωa∞)(μk(m)+ɛ+1),
for every u∈gn(Sk-1) with ∫Ω‍m|u|pdx>0 and n≥n3. In the case of m≥0 in Ω, by using Lemma 2.3 instead of Lemma 2.2, we have a similar inequality
(2.18)‖u‖pp≤2d(m,1)(1+d(m,1)2inf⁡Ωa∞)(μk(m)+ɛ+1),
for every u∈gn(Sk-1) provided n≥n4 (some sufficiently large n4≥n3). Consequently, our claim follows from (2.15), (2.17), and (2.18).
Let us return to the proof of (iii). Because(2.19)sup⁡{‖u‖pr′;u∈gn(Sk-1),n≥n1}=:R<+∞
holds by (2.13), J(u)≤μk(m)+ɛ/2 and the continuity of W1,p(Ω) into Lpr′(Ω), we see the inequality
(2.20)∫Ω‍m|u|pdx=1-∫Ω‍(mn-m)|u|pdx>1-‖mn-m‖rRp>0,
for every u∈gn(Sk-1) and n≥n5 (some sufficiently large n5≥n4). By considering g~n(·):=gn(·)/(∫Ω‍m|gn(·)|pdx)1/p∈ℱk(m), we obtain
(2.21)μk(m)≤max⁡z∈Sk-1J(g~n(z))≤max⁡z∈Sk-1⁡J(gn(z))1-‖mn-m‖rRp≤μk(mn)+ɛ/21-‖mn-m‖rRp.
Because of ∥mn-m∥r→0, we get μk(mn)≥μk(m)-ɛ for sufficiently large n, and hence our conclusion holds.Finally, we recall the second eigenvalue of(EV;m) obtained by the mountain pass theorem.
(2.22)Σ(m):={η∈C([0,1],S(m));η(0)∈P,η(1)∈(-P)},c(m):=inf⁡η∈Σ(m)max⁡t∈[0,1]J~(η(t)),
where P:={u∈W1,p(Ω);u(x)≥0fora.e.x∈Ω}.Since∞>supx∈Ωa∞(x)≥infx∈Ωa∞(x)>0 holds, the following result can be shown by the same argument as in [19] (although they handle the asymmetry case, it is sufficient to consider the case of m≡n in this paper). See [19, Theorem 3.2]  for the proof.Theorem 2.11.
c
(
m
) is an eigenvalue of (EV;m) which satisfies λ*(m)<c(m). Moreover, there is no eigenvalues of (EV;m) between λ*(m) and c(m).Now, we have the following result.Proposition 2.12.
(2.23)
λ
2
(
m
)
=
μ
2
(
m
)
=
c
(
m
)
holds, where c(m) is a minimax value defined by (2.22).Proof.
First, we prove the inequalityc(m)≥μ2(m). Because c(m) is an eigenvalue (note that the following equation is homogeneous), we can choose a solution u∈W1,p(Ω) with ∫Ω‍m|u|pdx=1 for
(2.24)-div⁡(a∞(x)|∇u|p-2∇u)=c(m)m(x)|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω.
Note that u is a sign-changing function because any eigenfunction associated with any eigenvalue greater than the principal eigenvalue changes sign (refer to [18, Proposition  4.3]). Thus, we have
(2.25)0<∫Ω‍a∞|∇u±|pdx=c(m)∫Ω‍mu±pdx
by taking ±u± as test function (recall that u±:=max{±u,0}). Hence, we may assume that ∫Ω‍mu±pdx=1 by the normalization. Set X:={su+-tu-;|s|p+|t|p=1}⊂S(m). Then, because X is homeomorphic to S1, there exists g∈ℱ2(m) such that g(S1)=X. Since the value of J is equal to c(m) on X, we obtain
(2.26)μ2(m)≤max⁡z∈S1J~(g(z))=c(m)
by the definition of μ2(m) and X.
Next, we will prove the inequalityc(m)≤λ2(m) by dividing into two cases: ∫Ω‍mdx≠0 and ∫Ω‍mdx=0.
Case of∫Ω‍mdx≠0: by way of contradiction, we assume that λ2(m)<c(m). Then, λ*(m)=λ1(m)=λ2(m) follows from Theorem 2.11. Note that J~ satisfies the Palais-Smale condition in this case (see Lemma 2.8), and hence we can apply the first deformation lemma to J~. Therefore, by the standard argument (cf. [22], [17, Lemma 5.6]), we see that γ(K)≥2, where K:={u∈S(m);J~′(u)=0,J~(u)=λ*(m)}. This means that K is an infinite set, that is, the following equation has infinite many solutions:
(2.27)-div⁡(a∞(x)|∇u|p-2∇u)=λ*(m)m(x)|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω
due to the Lagrange multiplier's rule. This contradicts to the fact described as in Proposition 2.4 that λ*(m) is simple. As a result, we have shown that c(m)=λ2(m)=μ2(m) holds in the case of ∫Ω‍mdx≠0 (note λn(m)≤μn(m)).
Case of∫Ω‍mdx=0: According to Proposition 2.10 (i) for λ2(m), we have λ2(m)≥λ2(m+ɛ)=c(m+ɛ) for every ɛ>0 since we can apply the first result to m+ɛ. Because we prove limɛ→0+c(m+ɛ)=c(m) by the same argument as in [6, Lemma 2.9] (for the case a∞≡1), our conclusion is proved by taking ɛ↓0 in the inequality λ2(m)≥c(m+ɛ).


## 2.1. Preliminaries

The following lemmas can be easily shown by way of contradiction because∫Ω‍a∞|∇u|pdx is equivalent to ∥∇u∥pp (note that a∞ is positive). Here, we omit the proofs (refer to [7]).Lemma 2.1.
Assume∫Ω‍mdx<0. Then, there exists a constant c>0 such that ∫Ω‍a∞|∇u|pdx≥c∥u∥pp for every u∈W1,p(Ω) with ∫Ω‍m|u|pdx>0.Lemma 2.2.
Assume that∫Ω‍mdx≠0 and ξ>0. Then, there exists a constant b(m,ξ)>0 such that
(2.1)∫Ω‍a∞|∇u|pdx-ξ∫Ω‍m|u|pdx≥b(m,ξ)∫Ω‍|u|pdx
for every u∈B(m):={u∈W1,p(Ω);∫Ω‍m|u|pdx≤0}.Lemma 2.3.
Assume thatm≥0 in Ω. Then, for every ξ>0 there existed d(m,ξ)>0 such that
(2.2)∫Ω‍a∞|∇u|pdx-ξ∫Ω‍m|u|pdx≥d(m,ξ)∫Ω‍|u|pdx
for every u∈W1,p(Ω).First, we recall the following principle eigenvalueλ*(m):
(2.3)λ*(m)≔inf⁡⁡{∫Ω‍a∞|∇u|pdx;u∈W1,p(Ω),∫Ω‍m|u|pdx=1}.
Because of ∞>supx∈Ωa∞(x)≥infx∈Ωa∞(x)>0, we have the following result as the same argument as in the case of the p-Laplacian.Proposition 2.4 (see [7, Proposition  2.2]).
The following assertions hold:(i)
If∫Ω‍mdx≥0 holds, then λ*(m)=0;(ii)
If∫Ω‍mdx<0 holds, then λ*(m)>0 is a simple eigenvalue and it admits a positive eigenfunction. In addition, the open interval (0,λ*(m)) contains no eigenvalues of (EV;m).Lemma 2.5.
Assume∫Ω‍mdx<0. Then, one has λ*(m+ɛ)<λ*(m)<λ*(m-ɛ′) for every ɛ>0 and ɛ′>0 with |{m-ɛ′>0}|>0.Proof.
We choose a minimizeru for λ*(m) because Proposition 2.4 guarantees the existence of it. Then, for every ɛ>0, we have
(2.4)λ*(m+ɛ)≤∫Ω‍a∞|∇u|pdx∫Ω‍(m+ɛ)|u|pdx<∫Ω‍a∞|∇u|pdx∫Ω‍m|u|pdx=∫Ω‍a∞|∇u|pdx=λ*(m)
by the definition of λ*(m+ɛ). By applying the same argument to a minimizer for λ*(m-ɛ), we obtain λ*(m)<λ*(m-ɛ′) for ɛ′>0 with |{m-ɛ′>0}|>0.


## 2.2. Other Eigenvalues

Here, we introduce two unbounded sequences{λn(m)}n and {μn(m)}n as follows:
(2.5)J(u):=∫Ω‍a∞|∇u|pdxforu∈W1,p(Ω),J~:=J∣S(m),S(m):={u∈W1,p(Ω);∫Ω‍m|u|pdx=1},Sn(m):={X⊂S(m);compact,symmetricandγ(X)≥n},Fn(m):={g∈C(Sn-1,S(m));gisodd},λn(m):=inf⁡X∈Sn(m)max⁡u∈XJ~(u),μn(m):=inf⁡g∈Fn(m)max⁡z∈Sn-1J~(g(z)),
where γ(X) denotes the Krasnoselskii genus of X (see [17, Definition 5.1] for the definition) and Sn-1 denotes the usual unit sphere in ℝn. We see that λn(m) is defined by Ljusternik-Schnirelman theory and it is known that the definition of μn(m) is introduced by Drábek and Robinson ([16]) under the p-Laplace Dirichlet problem with m≡1.Remark 2.6.
The following assertions can be shown easily:(i)
λ1(m)=μ1(m)=λ*(m);(ii)
𝒮n(m)≠∅ and ℱn(m)≠∅ for every n∈ℕ;(iii)
g(Sn-1)⊂𝒮n(m) for every g∈ℱn(m);(iv)
μn(m)≥λn(m) for every n∈ℕ;(v)
λn+1(m)≥λn(m) and μn+1(m)≥μn(m) for every n∈ℕ,see [18] for the proof of  (ii).Define aC1 functional Φm on W1,p(Ω) by Φm(u):=∫Ω‍m|u|pdx for u∈W1,p(Ω). Because 1∈ℝ is a regular value of Φm, it is well known that the norm of the derivative at u∈S(m) of the restriction of J to S(m) is defined as follows:
(2.6)‖J~′(u)‖*:=min⁡⁡{‖J′(u)-tΦm′(u)‖W1,p(Ω)*;t∈R}=sup⁡⁡{〈J′(u),v〉;v∈Tu(S(m)),‖v‖=1},
where Tu(S(m)) denotes the tangent space of S(m) at u, that is, Tu(S(m))={v∈W1,p(Ω);∫Ω‍m|u|p-2uvdx=0}. Here, we recall the definition of the Palais-Smale condition for J~.Definition 2.7.
J
~ is said to satisfy the  bounded Palais-Smale condition if any bounded sequence un∈S(m) such that ∥J~′(un)∥*→0 has a convergent subsequence. Moreover, we say that J~ satisfies the Palais-Smale condition at level c∈ℝ if any sequence un∈S(m) such that J~(un)→c and ∥J~′(un)∥*→0 as n→∞ has a convergent subsequence. In addition, we say that J~ satisfies the Palais-Smale condition if J~ satisfies the Palais-Smale condition for every c∈ℝ.The following result can be proved by the same argument as in [19, Proposition 3.3]  (which treats the case of the p-Laplacian, i.e., a∞≡1) because of ∞>supx∈Ωa∞(x)≥infx∈Ωa∞(x)>0. Here, we omit the proof.Lemma 2.8.
The following assertions hold:(i)
J~ satisfies the bounded Palais-Smale condition;(ii)
J~ satisfies the Palais-Smale condition provided ∫Ω‍mdx≠0.Proposition 2.9.
λ
n
(
m
) and μn(m) are eigenvalues of (EV;m) such that
(2.7)lim⁡n→∞λn(m)=lim⁡n→∞μn(m)=+∞.Proof.
In the case of∫Ω‍mdx≠0, since J~ satisfies the Palais-Smale condition, we can apply the first deformation lemma on C1 manifold (refer to [20]). Thus, by the standard argument, we can prove that λn(m) and μn(m) are critical values of J~. This means that λn(m) and μn(m) are eigenvalues of (EV;m) by the Lagrange multiplier rule. In addition, we can easily show limn→∞λn(m)=+∞ by the standard argument via the first deformation lemma on C1 manifold (refer to [21, Proposition  3.14.7], [22] or [17] in the case of a Banach space). Hence, limn→∞μn(m)=+∞ holds because of μn(m)≥λn(m) for every n∈ℕ.
In the case of∫Ω‍mdx=0, by the same argument as in [18], our conclusion can be proved. For readers' convenience, we give a sketch of the proof. For ɛ>0, we define Jɛ(u):=J(u)+ɛ∥u∥pp and J~ɛ:=Jɛ∣S(m). Moreover, we set minimax values λnɛ(m) and μnɛ(m) of J~ɛ by
(2.8)λnɛ(m):=inf⁡X∈Sn(m)max⁡u∈XJ~ɛ(u),μnɛ(m):=inf⁡g∈Fn(m)max⁡z∈Sn-1J~ɛ(g(z)).
Because any Palais-Smale sequence of J~ɛ is bounded, it is easily shown that J~ɛ satisfies the Palais-Smale condition (refer to [19, Proposition  3.3]) Hence, it can be proved that λnɛ(m) and μnɛ(m) are critical values of J~ɛ. Furthermore, it follows from the argument as in [18, Lemma  3.5] that λnɛ(m)→λn(m) and μnɛ(m)→μn(m) as ɛ→0+. Therefore, by noting that Jɛ is p-homogeneous, we can obtain a solution uɛ with ∥uɛ∥=1 for -div(a∞|∇u|p-2∇u)=cɛm|u|p-2u in Ω, ∂u/∂ν=0 on ∂Ω, where cɛ=λnɛ(m) or μnɛ(m). Because of ∥uɛ∥=1, it follows from the standard argument that uɛ has a subsequence strongly convergent to a solution u for
(2.9)-div⁡(a∞|∇u|p-2∇u)=cm|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω,
where c=limɛ→0+cɛ. Thus, λn(m) and μn(m) are eigenvalues of (EV;m). To prove limn→∞λn(m)=+∞, by considering a function mδ(x):=max{m(x),δ}  for δ>0, we have λn(mδ)≤λn(m) (refer to Proposition 2.10). Because we can apply our fist assertion to mδ (note ∫Ω‍mδdx>0), we obtain limn→∞μn(m)≥limn→∞λn(m)≥limn→∞λn(mδ)=+∞.Proposition 2.10.
Let1<r<∞ if N≤p and p*/(p*-p)≤r<∞ if N>p. Then, the following assertions hold:  (i)
ifm′≥m in Ω, then μk(m′)≤μk(m);(ii)
iflimn→∞mn=m in Lr(Ω), then limsupn→∞μk(mn)≤μk(m);(iii)
if∫Ω‍mdx≠0 and limn→∞mn=m in Lr(Ω), then limn→∞μk(mn)=μk(m). Moreover, the same conclusion holds for λk(m).Proof.
We only treatμk(m) because we can give the proof for λk(m) similarly.(i)
Letm′≥m in Ω. Fix an arbitrary ɛ>0. Then, by the definition of μk(m), there exists a g∈ℱk(m) such that maxz∈Sk-1J(g(z))<μk(m)+ɛ. Set g~(z):=g(z)/(∫Ω‍m′|g(z)|pdx)1/p for z∈Sk-1 (note ∫Ω‍m′|g(z)|pdx≥∫Ω‍m|g(z)|pdx=1), then g~∈ℱk(m′) holds. Therefore, by the definition of μk(m′), we have
(2.10)μk(m′)≤max⁡z∈Sk-1J(g~(z))=max⁡z∈Sk-1J(g(z))∫Ω‍m′|g(z)|pdx≤max⁡z∈Sk-1J(g(z))<μk(m)+ɛ.because of∫Ω‍m′|g(z)|pdx≥∫Ω‍m|g(z)|pdx=1 for every z∈Sk-1. Since ɛ>0 is arbitrary, we obtain μk(m′)≤μk(m).(ii)
Letlimn→∞mn=m in Lr(Ω) and fix an arbitrary ɛ>0. By the definition of μk(m), there exists a g∈ℱk(m) such that maxz∈Sk-1J(g(z))<μk(m)+ɛ/2. Since g(Sk-1) is compact and pr′:=pr/(r-1)≤p*, we set M:=maxu∈g(Sk-1)∥u∥pr′. Then, due to Hölder's inequality and mn→m in Lr(Ω), there exists an n0∈ℕ such that
(2.11)∫Ω‍mn|u|pdx=1+∫Ω‍(mn-m)|u|pdx≥1-‖mn-m‖rMp>0for everyu∈g(Sk-1) and n≥n0. Therefore, by a similar argument to (i), we obtain
(2.12)μk(mn)≤max⁡z∈Sk-1J(g(z))∫Ω‍mn|g(z)|pdx≤μk(m)+ɛ/21-‖mn-m‖rMp<μk(m)+ɛfor sufficiently largen. Hence, limsupn→∞μk(mn)≤μk(m)+ɛ follows. Since ɛ>0 is arbitrary, our conclusion is proved.(iii)
Letlimn→∞mn=m in Lr(Ω) and ∫Ω‍mdx≠0. We fix an arbitrary ɛ>0. Due to our assertion (ii), there exists an n1∈ℕ such that μk(mn)≤μk(m)+ɛ/2. For every n≥n1, by the definition of μk(mn), we can take gn∈ℱk(mn) satisfying maxz∈Sk-1J(gn(z))<μk(mn)+ɛ/2.
Here, we will prove(2.13)sup⁡n≥n1max⁡{‖u‖p;u∈gn(Sk-1)}<∞.
Ifu∈gn(Sk-1) satisfies ∫Ω‍m|u|pdx≤0, then we obtain
(2.14)b(m,1)‖u‖pp≤J(u)-∫Ωm|u|pdx‍=J(u)-∫Ωmn|u|pdx‍+∫Ω(mn-m)|u|pdx‍≤μk(mn)+ɛ2-1+‖mn-m‖r‖u‖pr′p≤μk(m)+ɛ+C‖mn-m‖r‖u‖pp+CJ(u)‖mn-m‖rinf⁡Ωa∞≤(1+C‖mn-m‖rinf⁡Ωa∞)(μk(m)+ɛ)+C‖mn-m‖r‖u‖pp
by Lemma 2.2 and Hölder's inequality (note ∥∇u∥pp≤J(u)/infΩa∞ and μk(mn)≤μk(m)+ɛ/2), where C>0 is a constant (independent of n and u) obtained by the continuity of W1,p(Ω) into Lpr′(Ω). Therefore, if we take an n2≥n1 satisfying C∥mn-m∥r≤b(m,1)/2 for every n≥n2, then we obtain
(2.15)‖u‖pp≤2b(m,1)(1+b(m,1)2inf⁡Ωa∞)(μk(m)+ɛ)
for every u∈gn(Sk-1) provided ∫Ω‍m|u|pdx≤0 and n≥n2. Similarly, in the case where m changes sign, for every u∈gn(Sk-1) satisfying ∫Ω‍m|u|pdx>0, we have
(2.16)b(-m,1)‖u‖pp≤J(u)-∫Ω(-m)|u|pdx‍≤(1+C‖mn-m‖rinf⁡Ωa∞)(μk(m)+ɛ)+1+C‖mn-m‖r‖u‖pp.
Hence, by taking a sufficiently large n3≥n2, we get the inequality
(2.17)‖u‖pp≤2b(-m,1)(1+b(-m,1)2inf⁡Ωa∞)(μk(m)+ɛ+1),
for every u∈gn(Sk-1) with ∫Ω‍m|u|pdx>0 and n≥n3. In the case of m≥0 in Ω, by using Lemma 2.3 instead of Lemma 2.2, we have a similar inequality
(2.18)‖u‖pp≤2d(m,1)(1+d(m,1)2inf⁡Ωa∞)(μk(m)+ɛ+1),
for every u∈gn(Sk-1) provided n≥n4 (some sufficiently large n4≥n3). Consequently, our claim follows from (2.15), (2.17), and (2.18).
Let us return to the proof of (iii). Because(2.19)sup⁡{‖u‖pr′;u∈gn(Sk-1),n≥n1}=:R<+∞
holds by (2.13), J(u)≤μk(m)+ɛ/2 and the continuity of W1,p(Ω) into Lpr′(Ω), we see the inequality
(2.20)∫Ω‍m|u|pdx=1-∫Ω‍(mn-m)|u|pdx>1-‖mn-m‖rRp>0,
for every u∈gn(Sk-1) and n≥n5 (some sufficiently large n5≥n4). By considering g~n(·):=gn(·)/(∫Ω‍m|gn(·)|pdx)1/p∈ℱk(m), we obtain
(2.21)μk(m)≤max⁡z∈Sk-1J(g~n(z))≤max⁡z∈Sk-1⁡J(gn(z))1-‖mn-m‖rRp≤μk(mn)+ɛ/21-‖mn-m‖rRp.
Because of ∥mn-m∥r→0, we get μk(mn)≥μk(m)-ɛ for sufficiently large n, and hence our conclusion holds.Finally, we recall the second eigenvalue of(EV;m) obtained by the mountain pass theorem.
(2.22)Σ(m):={η∈C([0,1],S(m));η(0)∈P,η(1)∈(-P)},c(m):=inf⁡η∈Σ(m)max⁡t∈[0,1]J~(η(t)),
where P:={u∈W1,p(Ω);u(x)≥0fora.e.x∈Ω}.Since∞>supx∈Ωa∞(x)≥infx∈Ωa∞(x)>0 holds, the following result can be shown by the same argument as in [19] (although they handle the asymmetry case, it is sufficient to consider the case of m≡n in this paper). See [19, Theorem 3.2]  for the proof.Theorem 2.11.
c
(
m
) is an eigenvalue of (EV;m) which satisfies λ*(m)<c(m). Moreover, there is no eigenvalues of (EV;m) between λ*(m) and c(m).Now, we have the following result.Proposition 2.12.
(2.23)
λ
2
(
m
)
=
μ
2
(
m
)
=
c
(
m
)
holds, where c(m) is a minimax value defined by (2.22).Proof.
First, we prove the inequalityc(m)≥μ2(m). Because c(m) is an eigenvalue (note that the following equation is homogeneous), we can choose a solution u∈W1,p(Ω) with ∫Ω‍m|u|pdx=1 for
(2.24)-div⁡(a∞(x)|∇u|p-2∇u)=c(m)m(x)|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω.
Note that u is a sign-changing function because any eigenfunction associated with any eigenvalue greater than the principal eigenvalue changes sign (refer to [18, Proposition  4.3]). Thus, we have
(2.25)0<∫Ω‍a∞|∇u±|pdx=c(m)∫Ω‍mu±pdx
by taking ±u± as test function (recall that u±:=max{±u,0}). Hence, we may assume that ∫Ω‍mu±pdx=1 by the normalization. Set X:={su+-tu-;|s|p+|t|p=1}⊂S(m). Then, because X is homeomorphic to S1, there exists g∈ℱ2(m) such that g(S1)=X. Since the value of J is equal to c(m) on X, we obtain
(2.26)μ2(m)≤max⁡z∈S1J~(g(z))=c(m)
by the definition of μ2(m) and X.
Next, we will prove the inequalityc(m)≤λ2(m) by dividing into two cases: ∫Ω‍mdx≠0 and ∫Ω‍mdx=0.
Case of∫Ω‍mdx≠0: by way of contradiction, we assume that λ2(m)<c(m). Then, λ*(m)=λ1(m)=λ2(m) follows from Theorem 2.11. Note that J~ satisfies the Palais-Smale condition in this case (see Lemma 2.8), and hence we can apply the first deformation lemma to J~. Therefore, by the standard argument (cf. [22], [17, Lemma 5.6]), we see that γ(K)≥2, where K:={u∈S(m);J~′(u)=0,J~(u)=λ*(m)}. This means that K is an infinite set, that is, the following equation has infinite many solutions:
(2.27)-div⁡(a∞(x)|∇u|p-2∇u)=λ*(m)m(x)|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω
due to the Lagrange multiplier's rule. This contradicts to the fact described as in Proposition 2.4 that λ*(m) is simple. As a result, we have shown that c(m)=λ2(m)=μ2(m) holds in the case of ∫Ω‍mdx≠0 (note λn(m)≤μn(m)).
Case of∫Ω‍mdx=0: According to Proposition 2.10 (i) for λ2(m), we have λ2(m)≥λ2(m+ɛ)=c(m+ɛ) for every ɛ>0 since we can apply the first result to m+ɛ. Because we prove limɛ→0+c(m+ɛ)=c(m) by the same argument as in [6, Lemma 2.9] (for the case a∞≡1), our conclusion is proved by taking ɛ↓0 in the inequality λ2(m)≥c(m+ɛ).


## 3. Proof of Theorem1.1

We define a functionalIλ,m on W1,p(Ω) as follows:
(3.1)Iλ,m(u)=∫ΩG(x,∇u)dx‍-λp∫Ωm|u|pdx‍-∫ΩF(x,u)dx‍-∫Ωhudx‍=1p∫Ωa∞|∇u|pdx‍+∫ΩG~(x,∇u)dx‍-λp∫Ωm|u|pdx‍-∫ΩF(x,u)dx‍-∫Ωhudx
for u∈W1,p(Ω) ((1.15) or (1.9) for the definition of G, G~, and F). It is easily seen that Iλ,m is well defined and class of C1 on W1,p(Ω) by (1.1), (1.16) and the continuity of W1,p(Ω)↪Lp(Ω).Remark 3.1.
Letu∈W1,p(Ω) be a critical point of Iλ,m, namely, u satisfies the equality
(3.2)∫Ω‍A(x,∇u)∇φdx=λ∫Ω‍m|u|p-2uφdx+∫Ω‍f(x,u)φdx+∫Ω‍hφdx
for every φ∈W1,p(Ω). Then, u∈L∞(Ω) by the Moser iteration process (refer to Theorem C in [4]). Therefore, u∈C1,α(Ω-) (0<α<1) follows from the regularity result in [23]. Furthermore, due to [24, Theorem 3], u satisfies (P;λ,m,h) in the distribution sense and the boundary condition
(3.3)0=∂u∂νA=A(⋅,∇u)ν=a(⋅,|∇u|)∂u∂νinW-1/q,q(∂Ω)
for every 1<q<∞ (see [24] for the definition of W-1/q,q(∂Ω)). Since u∈C1,α(Ω-) and a(x,t)>0 for every t≠0, u satisfies the Neumann boundary condition, that is, (∂u/∂ν)(x)=0 for every x∈∂Ω.


### 3.1. The Palais-Smale Condition in the Nonresonant Case

First, we recall the definition of the Palais-Smale condition.Definition 3.2.
AC1 functional Ψ on a Banach space X is said to satisfy the Palais-Smale condition at c∈ℝ if a Palais-Smale sequence {un}⊂X at level c, namely,
(3.4)Ψ(un)⟶c,‖Ψ′(un)‖X*⟶0asn⟶∞
has a convergent subsequence. We say that Ψ satisfies the Palais-Smale condition if Ψ satisfies the Palais-Smale condition at any c∈ℝ. Moreover, we say that Ψ satisfies the bounded Palais-Smale condition if any bounded sequence {un} such that {Ψ(un)} is bounded and ∥Ψ′(un)∥X*→0 as n→∞ has a convergent subsequence.Concerning the Palais-Smale condition, we state the following result developed from [6, Proposition  7].Proposition 3.3.
Ifλ is not an eigenvalue of (EV;m), then Iλ,m satisfies the Palais-Smale condition.Proof.
Let{un} be a Palais-Smale sequence of Iλ,m, namely,
(3.5)Iλ,m(un)⟶c,‖Iλ,m′(un)‖W1,p(Ω)*⟶0asn⟶∞
for some c∈ℝ. It is sufficient to prove only the boundedness of ∥un∥ because the operator A:W1,p(Ω)→W1,p(Ω)* described in Proposition 1.5 has the (S)+ property.
To prove the boundedness of∥un∥, it suffices to show that ∥un∥p is bounded because of the inequality |f(x,u)|≤C(|u|p-1+1) (obtained by (1.1)) and the following inequality:
(3.6)〈Iλ,m′(un),un〉+λ∫Ωm|un|pdx‍+∫Ωf(x,un)undx‍+∫Ωhundx‍,=∫ΩA(x,∇un)∇undx‍≥C0p-1‖∇un‖pp,
where we use Remark 1.4 (iii) in the last inequality. By way of contradiction, we may assume that ∥un∥p→∞ as n→∞ by choosing a subsequence if necessary. Set vn:=un/∥un∥p. Then, since the inequality (3.6) guarantees that {vn} is bounded in W1,p(Ω), we may suppose, by choosing a subsequence, that vn⇀v0 in W1,p(Ω) and vn→v0 in Lp(Ω) for some v0.
Here, we will prove that(3.7)lim⁡n→∞‖f(⋅,un)‖p′‖un‖pp-1=0,
where p′=p/(p-1). Fix an arbitrary ɛ>0. It follows from (1.1) that there exists a Cɛ>0 such that
(3.8)|f(x,u)|≤ɛ|u|p-1+Cɛforeveryu∈R,a.e.x∈Ω.
Then, we obtain
(3.9)∫Ω|f(x,un)|p′dx≤2p′∫Ω(ɛp′|un|p+Cɛp′)dx≤2p′ɛp′‖un‖pp+2p′Cɛp′|Ω|.
Since we are assuming that ∥un∥p→∞ as n→∞, there exists n0∈ℕ such that for every n≥n0(3.10)‖f(⋅,un)‖p′‖un‖pp-1≤4ɛ
holds. This shows that limn→∞∥f(·,un)∥p′/∥un∥pp-1=0 because ɛ>0 is arbitrary.
Here, we recall the following result proved in [6]:
(3.11)lim⁡n→∞∫Ω‍a~(x,|∇un|)∇un‖un‖pp-1∇(vn-v0)dx=lim⁡n→∞∫Ω‍a~(x,|∇un|)∇un‖un‖pp-1∇φdx=0,
for every φ∈W1,p(Ω). Thus, by considering
(3.12)o(1)=〈Iλ,m′(un),vn-v0〉‖un‖pp-1=∫Ω‍a∞|∇vn|p-2∇vn∇(vn-v0)dx+o(1),
we see that vn strongly converges to v0 in W1,p(Ω) (note that p-Laplacian has the (S)+ property). Therefore, by taking a limit in o(1)=〈Iλ,m′(un),φ〉/∥un∥pp-1 for any φ∈W1,p(Ω) and by noting (3.7) and (3.11), we know that v0 is a nontrivial solution (note ∥v0∥p=1) of
(3.13)-div⁡(a∞|∇u|p-2∇u)=λm|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω.
This means that λ is an eigenvalue of (EV;m). This is a contradiction. Hence, ∥un∥p is bounded.


### 3.2. Key Lemmas

To show the linking lemma, we define(3.14)Y(μ,m):={u∈W1,p(Ω);∫Ω‍a∞|∇u|pdx≥μ∫Ω‍m|u|pdx}forμ∈ℝ.Lemma 3.4.
Letg0∈C(Sk-1,W1,p(Ω)∖{0}) be odd and 0<μ≤μk+1(m). Then, g(S+k)∩Y(μ,m)≠∅ for every g∈C(S+k,W1,p(Ω)) with g∣Sk-1=g0, where Y(μ,m) is the set introduced in (3.14) and S+k is the upper hemisphere in ℝk+1 with boundary Sk-1.Proof.
Fix anyg∈C(S+k,W1,p(Ω)) such that g∣Sk-1=g0. If u∈g(S+k) satisfies ∫Ω‍m|u|pdx≤0, then u∈Y(μ,m) holds. So, we may assume that ∫Ω‍m|u|pdx>0 for every u∈g(S+k). Define g~∈ℱk+1(m) as follows:
(3.15)g~(z):={g(z)(∫Ω‍m|g(z)|pdx)1/pifz∈S+k,-g(-z)(∫Ω‍m|g(-z)|pdx)1/pifz∈S-k.
By the definition of μk+1(m), there exists z0∈Sk such that J~(g~(z0))≥μk+1(m). Since g~ is odd and J is even, we may suppose z0∈S+k. So, this yields the inequality J(g(z0))≥μk+1(m)∫Ω‍m|g(z0)|pdx≥μ∫Ω‍m|g(z0)|pdx, whence g(z0)∈Y(μ,m) holds.Lemma 3.5.
Letμk(m)<λ. Then, there exists g0∈ℱk(m) such that
(3.16)max⁡z∈Sk-1J(g0(z))<λ,max⁡z∈Sk-1Iλ,m(Tg0(z))⟶-∞as|T|⟶∞,
where μk(m) is defined by (2.5).Proof.
Chooseɛ0>0 such that μk(m)+ɛ0<λ. By the definition of μk(m), there exists g0∈ℱk(m) such that
(3.17)max⁡z∈Sk-1J(g0(z))<μk(m)+ɛ0.
Due to the compactness of g0(Sk-1), we put M:=maxz∈Sk-1∥g0(z)∥p. By the property of the function a~ as in (AH) and Young's inequality, for every ɛ>0 there exist constants Cɛ>0 and Cɛ′>0 such that
(3.18)|G~(x,y)|≤ɛ2|y|p+Cɛ|y|≤ɛ|y|p+Cɛ′≤ɛinf⁡Ωa∞a∞(x)|y|p+Cɛ′
for every x∈Ω and y∈ℝN. Moreover, the hypothesis (1.1) ensures that for every ɛ′>0 there exist constants Dɛ′>0 satisfying
(3.19)|F(x,u)|≤ɛ′2|u|p+Dɛ′|u|≤ɛ′|u|p+Dɛ′′
for every u∈ℝ and a.e. x∈Ω. Hence, we have
(3.20)Iλ,m(Tu)≤Tpp‍(1+pɛa_)∫Ωa∞|∇u|pdx-Tp(λ-pɛ′Mp)p+T‖h‖∞‖u‖1+C≤Tpp{(1+pɛa_)(μk(m)+ɛ0)-λ+pMpɛ′}+TM‖h‖∞|Ω|(p-1)/p+C
for every T>0, u∈g0(Sk-1), ɛ>0 and ɛ′>0 since g0(Sk-1)⊂S(m), (3.17), (3.18) and (3.19), where C=(Cɛ′+Dɛ′′)|Ω| and a_=infx∈Ωa∞(x)>0. By taking ɛ>0 and ɛ′>0 satisfying (1+pɛ/a_)(μk(m)+ɛ0)-λ+pMpɛ′<0, we show that maxz∈Sk-1Iλ,m(Tg0(z))→-∞ as T→+∞. Thus, our conclusion follows because g0(Sk-1) is symmetric.


### 3.3. The Case∫Ωmdx≠0

Lemma 3.6.
Let∫Ω‍mdx<0 and 0<λ<λ*(m). Then, Iλ,m is bounded from below, coercive and weakly lower semicontinuous (w.l.s.c.) on W1,p(Ω).Proof.
Φ
(
u
)
:
=
∫
Ω
‍
G
(
x
,
∇
u
)
d
x is w.l.s.c. on W1,p(Ω) because Φ is convex and continuous on W1,p(Ω) (cf. [25, Theorem 1.2]). Thus, Iλ,m is also w.l.s.c. on W1,p(Ω) since the inclusion from W1,p(Ω) to Lp(Ω) is compact.
Chooseɛ>0 such that pɛ<a_(1-λ/λ*(m)), where a_:=infΩa∞. By an easy estimation, (3.18) and (3.19) as in Lemma 3.5, we have
(3.21)Iλ,m(u)≥a_-ɛppa_∫Ωa∞|∇u|pdx‍-λp∫Ωm|u|pdx‍-ɛ′‖u‖pp-‖h‖∞‖u‖p|Ω|(p-1)/p-(Cɛ′+Dɛ′′)|Ω|
for every u∈W1,p(Ω) and ɛ′>0.
Letu∈W1,p(Ω) satisfy ∫Ω‍m|u|pdx≤0. Then, the following inequality follows from Lemma 2.2:
(3.22)D0∫Ω‍a∞|∇u|pdx-λ∫Ω‍m|u|pdx≥D02∫Ω‍a∞|∇u|pdx+b(m,ξ)‖u‖pp,
where b(m,ξ) is a positive constant independent of u with ξ=2λ/D0 and D0=(a_-ɛp)/a_.
For everyu∈W1,p(Ω) such that ∫Ω‍m|u|pdx>0, we obtain
(3.23)D0∫Ωa∞|∇u|pdx‍-λ∫Ωm|u|pdx‍≥(D0-λλ*(m))∫Ωa∞|∇u|pdx≥12(D0-λλ*(m))∫Ωa∞|∇u|pdx+c2(D0-λλ*(m))‖u‖pp
by the definition of λ*(m), Lemma 2.1 and D0-λ/λ*(m)>0, where c>0 is a constant obtained by Lemma 2.1.
Consequently, if we choose aɛ′>0 satisfying ɛ′<min{b(m,ξ)/p,c(D0-λ/λ*(m))/(2p)}, then we obtain positive constants d1 and d2 (independent of u) such that
(3.24)Iλ,m(u)≥d1‍∫Ωa∞|∇u|pdx+d2‖u‖pp-‖h‖∞‖u‖p|Ω|(p-1)/p-(Cɛ′+Dɛ′′)|Ω|≥min⁡⁡{a_d1,d2}‖u‖p-‖h‖∞‖u‖|Ω|(p-1)/p-(Cɛ′+Dɛ′′)|Ω|
for every u∈W1,p(Ω) by (3.21), (3.22), and (3.23). Because of p>1, our conclusion is shown.Lemma 3.7.
Letm≥0 in Ω and m≢0. If λ<0 holds, then Iλ,m is bounded from below, coercive and w.l.s.c. on W1,p(Ω).Proof.
First, as the same reason in Lemma3.6, it follows that Iλ,m is w.l.s.c. on W1,p(Ω). By a similar argument to Lemma 3.6, for every ɛ′>0 and 0<ɛ<a_/p where a_=infΩa∞, we obtain
(3.25)Iλ,m(u)≥a_-ɛppa_∫Ωa∞|∇u|pdx‍+|λ|p∫Ωm|u|pdx-ɛ′‖u‖pp‍-‖h‖∞‖u‖p|Ω|(p-1)/p-(Cɛ′+Dɛ′′)|Ω|
for every u∈W1,p(Ω) (note λ<0). Here, from Lemma 2.3,
(3.26)D0∫Ω‍a∞|∇u|pdx+|λ|∫Ω‍m|u|pdx≥D02∫Ω‍a∞|∇u|pdx+D02b(ξ,m)‖u‖pp
for every u∈W1,p(Ω) follows, where D0:=(a_-ɛp)/a_, ξ:=2|λ|/D0 and b(ξ,m) is a constant obtained in Lemma 2.3. Therefore, by choosing a ɛ′ such that 0<ɛ′<D0b(ξ,m)/2, we can prove our conclusion.Lemma 3.8.
Let∫Ω‍mdx≠0 and 0<λ<μ. Then, Iλ,m is bounded from below on Y(μ,m), where Y(μ,m) is the set introduced in (3.14).Proof.
Due to the same inequalities concerningG and F as in Lemma 3.5, for every ɛ>0 and ɛ′>0, there exists C=C(ɛ,ɛ′)>0 such that
(3.27)Iλ,m(u)≥a_-pɛpa_∫Ω‍a∞|∇u|pdx-λp∫Ω‍m|u|pdx-ɛ′‖u‖pp-‖h‖∞‖u‖1-C|Ω|
for every u∈W1,p(Ω), where a_:=infx∈Ωa∞(x). Choose positive constants ɛ and δ such that D0:=1-pɛ/a_>δ>λ/μ (note λ/μ<1).
First, we consider the case ofm≥0 in Ω. For every u∈Y(μ,m), we obtain
(3.28)D0∫Ωa∞|∇u|pdx‍-λ∫Ωm|u|pdx‍≥(D0-δ)∫Ωa∞|∇u|pdx‍+(δμ-λ)∫Ωm|u|pdx‍≥d(m,ξ1)(D0-δ)‖u‖pp
by Lemma 2.3 with ξ1=(δμ-λ)/(D0-δ) (note δμ-λ>0 and D0-δ>0).
Next, we handle with the case wherem changes sign. Let u∈W1,p(Ω) satisfy ∫Ω‍m|u|pdx≤0. Then, we have for such u(3.29)D0∫Ω‍a∞|∇u|pdx-λ∫Ω‍m|u|pdx≥b(m,ξ2)D0‖u‖pp
by Lemma 2.2, where D0=1-pɛ/a_ and ξ2:=λ/D0.
On the other hand, foru∈Y(μ,m) with ∫Ω‍m|u|pdx>0, the following inequality follows from Lemma 2.2:
(3.30)D0∫Ωa∞|∇u|pdx-λ∫Ωm|u|pdx‍≥(D0-δ)∫Ωa∞|∇u|pdx‍-(δμ-λ)∫Ω(-m)|u|pdx‍≥b(-m,ξ1)(D0-δ)‖u‖pp.
Consequently, by (3.27), (3.29), (3.28), and (3.30), there exists d>0 independent of u such that
(3.31)Iλ,m(u)≥(d-ɛ′)‖u‖pp-‖h‖∞‖u‖p|Ω|(p-1)/p-C|Ω|
for every u∈Y(μ,m). Hence, our conclusion is shown by taking ɛ′>0 satisfying ɛ′<d.Proof of Theorem1.1 in the Case ∫Ω‍mdx≠0.
First, if eitherm≥0 on Ω and λ<0 or 0<λ<λ*(m)=μ1(m) (i.e., ∫Ω‍mdx<0) holds, then Lemma 3.7 or Lemma 3.6 guarantees the existence of a global minimizer of Iλ,m, respectively (cf. [25, Theorem  1.1]). Hence, (P;λ,m,h) has a solution.
Sinceλ is an eigenvalue of (EV;m) if and only if -λ is one of (EV;-m), it suffices to consider the case of λ>λ*(m)≥0. Furthermore, by Proposition 2.9, Remark 2.6 (i), and our hypothesis that λ is not an eigenvalue of (EV;m), we may assume that there exists a k∈ℕ such that μk(m)<λ<μk+1(m). By Lemmas 3.5 and 3.8, we can choose T>0 and g0∈ℱk(m) satisfying
(3.32)max⁡z∈Sk-1Iλ,m(Tg0(z))<inf⁡{Iλ,m(u);u∈Y(μk+1(m),m)}=:α.
Put
(3.33)Σ≔{g∈C(S+k,W1,p(Ω));g∣Sk-1=Tg0},c≔inf⁡g∈Σmax⁡z∈S+kIλ,m(g(z)).
Then, it follows from Lemma 3.4 and (3.32) that c≥α>maxz∈Sk-1Iλ,m(Tg0(z)) holds. Since Iλ,m satisfies the Palais-Smale condition by Proposition 3.3, the minimax theorem guarantees (cf. [25, Theorem  4.6]) that c is a critical value of Iλ,m. Hence, (P;λ,m,h) has at least one solution.


### 3.4. The Case∫Ωmdx=0

First, we introduce an approximate functionalIλ,m,n+ as follows:
(3.34)Iλ,m,n+(u):=Iλ,m(u)+1pn‖u‖pp=Iλ,m-1/(λn)(u)foru∈W1,p(Ω).Lemma 3.9.
Let0<λ<μ. Then, there exists an n0∈ℕ such that for each n≥n0, Iλ,m,n+ is bounded from below on Y(μ,m-1/λn), where Y(μ,m-1/λn) is the set introduced in (3.14).Proof.
Choosen0∈ℕ such that 1/n0<λesssupx∈Ωm(x)/2. Then, for every n≥n0, Lemma 3.8 guarantees that Iλ,m,n+=Iλ,m-1/(λn) bounded from below on Y(μ,m-1/(λn)) because of ∫Ω‍(m-1/(λn))dx<0 and |{m-1/(λn)>0}|>0.Proof of Theorem1.1 in the Case ∫Ωmdx=0.
By noting thatλm=(-λ)(-m) and μ1(m)=λ*(m)=0, we may assume that μk(m)<λ<μk+1(m) for some k∈ℕ. Let n0 be a natural number obtained by Lemma 3.9. Due to Proposition 2.10 (i) and (ii), there exists an n1≥n0 such that
(3.35)μk(m)≤μk(m-1nλ)≤μk(m-1n1λ)<λ<μk+1(m)≤μk+1(m-1nλ)
for every n≥n1. Thus, for every n≥n1, we can take Tn>0 and gn∈ℱk(m-1/(nλ)) satisfying
(3.36)max⁡z∈Sk-1Iλ,m,n+(Tngn(z))<inf⁡{Iλ,m,n(u);u∈Y(μk+1(m-1(nλ)),m-1(nλ))}
by applying Lemmas 3.5 and 3.9 to Iλ,m,n+=Iλ,m-1/(nλ) (note (3.35)). Set
(3.37)Σn:={g∈C(S+k,W1,p(Ω));g∣Sk-1=Tngn},cn:=inf⁡g∈Σnmax⁡z∈S+kIλ,m,n+(g(z))
for each n≥n1. Then, for each n≥n1, we can obtain un satisfying
(3.38)|Iλ,m,n+(un)-cn|<1n,‖(Iλ,m,n+)′(un)‖W1,p(Ω)<1n
by applying Ekeland's variational principle to each Iλ,m,n+ (refer to [25, Theorem 4.3]). In addition, we can see that {un} is bounded in W1,p(Ω). Indeed, if there exists a subsequence {unl}l satisfying ∥unl∥p→∞ as l→∞, then we can show that λ is an eigenvalue of (EV;m) by the same argument as in Proposition 3.3. This contradicts to our assumption that λ is not an eigenvalue of (EV;m). Moreover, the boundedness of ∥∇un∥p follows from a similar inequality to (3.6) as in Proposition 3.3 under the boundedness of ∥un∥p.
Therefore, we may assume, by choosing a subsequence that{un} is a Palais-Smale sequence of Iλ,m since Iλ,m is bounded on a bounded set and according to the following inequality:
(3.39)‖Iλ,m′(un)‖(W1,p(Ω))*≤‖Iλ,m′(un)-(Iλ,m,n+)′(un)‖(W1,p(Ω))*+1n≤1n‖un‖pp-1+1n.
Therefore, because Iλ,m satisfies the Palais-Smale condition by Proposition 3.3, Iλ,m has a critical point, whence (P;λ,m,h) has at least one solution.


## 3.1. The Palais-Smale Condition in the Nonresonant Case

First, we recall the definition of the Palais-Smale condition.Definition 3.2.
AC1 functional Ψ on a Banach space X is said to satisfy the Palais-Smale condition at c∈ℝ if a Palais-Smale sequence {un}⊂X at level c, namely,
(3.4)Ψ(un)⟶c,‖Ψ′(un)‖X*⟶0asn⟶∞
has a convergent subsequence. We say that Ψ satisfies the Palais-Smale condition if Ψ satisfies the Palais-Smale condition at any c∈ℝ. Moreover, we say that Ψ satisfies the bounded Palais-Smale condition if any bounded sequence {un} such that {Ψ(un)} is bounded and ∥Ψ′(un)∥X*→0 as n→∞ has a convergent subsequence.Concerning the Palais-Smale condition, we state the following result developed from [6, Proposition  7].Proposition 3.3.
Ifλ is not an eigenvalue of (EV;m), then Iλ,m satisfies the Palais-Smale condition.Proof.
Let{un} be a Palais-Smale sequence of Iλ,m, namely,
(3.5)Iλ,m(un)⟶c,‖Iλ,m′(un)‖W1,p(Ω)*⟶0asn⟶∞
for some c∈ℝ. It is sufficient to prove only the boundedness of ∥un∥ because the operator A:W1,p(Ω)→W1,p(Ω)* described in Proposition 1.5 has the (S)+ property.
To prove the boundedness of∥un∥, it suffices to show that ∥un∥p is bounded because of the inequality |f(x,u)|≤C(|u|p-1+1) (obtained by (1.1)) and the following inequality:
(3.6)〈Iλ,m′(un),un〉+λ∫Ωm|un|pdx‍+∫Ωf(x,un)undx‍+∫Ωhundx‍,=∫ΩA(x,∇un)∇undx‍≥C0p-1‖∇un‖pp,
where we use Remark 1.4 (iii) in the last inequality. By way of contradiction, we may assume that ∥un∥p→∞ as n→∞ by choosing a subsequence if necessary. Set vn:=un/∥un∥p. Then, since the inequality (3.6) guarantees that {vn} is bounded in W1,p(Ω), we may suppose, by choosing a subsequence, that vn⇀v0 in W1,p(Ω) and vn→v0 in Lp(Ω) for some v0.
Here, we will prove that(3.7)lim⁡n→∞‖f(⋅,un)‖p′‖un‖pp-1=0,
where p′=p/(p-1). Fix an arbitrary ɛ>0. It follows from (1.1) that there exists a Cɛ>0 such that
(3.8)|f(x,u)|≤ɛ|u|p-1+Cɛforeveryu∈R,a.e.x∈Ω.
Then, we obtain
(3.9)∫Ω|f(x,un)|p′dx≤2p′∫Ω(ɛp′|un|p+Cɛp′)dx≤2p′ɛp′‖un‖pp+2p′Cɛp′|Ω|.
Since we are assuming that ∥un∥p→∞ as n→∞, there exists n0∈ℕ such that for every n≥n0(3.10)‖f(⋅,un)‖p′‖un‖pp-1≤4ɛ
holds. This shows that limn→∞∥f(·,un)∥p′/∥un∥pp-1=0 because ɛ>0 is arbitrary.
Here, we recall the following result proved in [6]:
(3.11)lim⁡n→∞∫Ω‍a~(x,|∇un|)∇un‖un‖pp-1∇(vn-v0)dx=lim⁡n→∞∫Ω‍a~(x,|∇un|)∇un‖un‖pp-1∇φdx=0,
for every φ∈W1,p(Ω). Thus, by considering
(3.12)o(1)=〈Iλ,m′(un),vn-v0〉‖un‖pp-1=∫Ω‍a∞|∇vn|p-2∇vn∇(vn-v0)dx+o(1),
we see that vn strongly converges to v0 in W1,p(Ω) (note that p-Laplacian has the (S)+ property). Therefore, by taking a limit in o(1)=〈Iλ,m′(un),φ〉/∥un∥pp-1 for any φ∈W1,p(Ω) and by noting (3.7) and (3.11), we know that v0 is a nontrivial solution (note ∥v0∥p=1) of
(3.13)-div⁡(a∞|∇u|p-2∇u)=λm|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω.
This means that λ is an eigenvalue of (EV;m). This is a contradiction. Hence, ∥un∥p is bounded.


## 3.2. Key Lemmas

To show the linking lemma, we define(3.14)Y(μ,m):={u∈W1,p(Ω);∫Ω‍a∞|∇u|pdx≥μ∫Ω‍m|u|pdx}forμ∈ℝ.Lemma 3.4.
Letg0∈C(Sk-1,W1,p(Ω)∖{0}) be odd and 0<μ≤μk+1(m). Then, g(S+k)∩Y(μ,m)≠∅ for every g∈C(S+k,W1,p(Ω)) with g∣Sk-1=g0, where Y(μ,m) is the set introduced in (3.14) and S+k is the upper hemisphere in ℝk+1 with boundary Sk-1.Proof.
Fix anyg∈C(S+k,W1,p(Ω)) such that g∣Sk-1=g0. If u∈g(S+k) satisfies ∫Ω‍m|u|pdx≤0, then u∈Y(μ,m) holds. So, we may assume that ∫Ω‍m|u|pdx>0 for every u∈g(S+k). Define g~∈ℱk+1(m) as follows:
(3.15)g~(z):={g(z)(∫Ω‍m|g(z)|pdx)1/pifz∈S+k,-g(-z)(∫Ω‍m|g(-z)|pdx)1/pifz∈S-k.
By the definition of μk+1(m), there exists z0∈Sk such that J~(g~(z0))≥μk+1(m). Since g~ is odd and J is even, we may suppose z0∈S+k. So, this yields the inequality J(g(z0))≥μk+1(m)∫Ω‍m|g(z0)|pdx≥μ∫Ω‍m|g(z0)|pdx, whence g(z0)∈Y(μ,m) holds.Lemma 3.5.
Letμk(m)<λ. Then, there exists g0∈ℱk(m) such that
(3.16)max⁡z∈Sk-1J(g0(z))<λ,max⁡z∈Sk-1Iλ,m(Tg0(z))⟶-∞as|T|⟶∞,
where μk(m) is defined by (2.5).Proof.
Chooseɛ0>0 such that μk(m)+ɛ0<λ. By the definition of μk(m), there exists g0∈ℱk(m) such that
(3.17)max⁡z∈Sk-1J(g0(z))<μk(m)+ɛ0.
Due to the compactness of g0(Sk-1), we put M:=maxz∈Sk-1∥g0(z)∥p. By the property of the function a~ as in (AH) and Young's inequality, for every ɛ>0 there exist constants Cɛ>0 and Cɛ′>0 such that
(3.18)|G~(x,y)|≤ɛ2|y|p+Cɛ|y|≤ɛ|y|p+Cɛ′≤ɛinf⁡Ωa∞a∞(x)|y|p+Cɛ′
for every x∈Ω and y∈ℝN. Moreover, the hypothesis (1.1) ensures that for every ɛ′>0 there exist constants Dɛ′>0 satisfying
(3.19)|F(x,u)|≤ɛ′2|u|p+Dɛ′|u|≤ɛ′|u|p+Dɛ′′
for every u∈ℝ and a.e. x∈Ω. Hence, we have
(3.20)Iλ,m(Tu)≤Tpp‍(1+pɛa_)∫Ωa∞|∇u|pdx-Tp(λ-pɛ′Mp)p+T‖h‖∞‖u‖1+C≤Tpp{(1+pɛa_)(μk(m)+ɛ0)-λ+pMpɛ′}+TM‖h‖∞|Ω|(p-1)/p+C
for every T>0, u∈g0(Sk-1), ɛ>0 and ɛ′>0 since g0(Sk-1)⊂S(m), (3.17), (3.18) and (3.19), where C=(Cɛ′+Dɛ′′)|Ω| and a_=infx∈Ωa∞(x)>0. By taking ɛ>0 and ɛ′>0 satisfying (1+pɛ/a_)(μk(m)+ɛ0)-λ+pMpɛ′<0, we show that maxz∈Sk-1Iλ,m(Tg0(z))→-∞ as T→+∞. Thus, our conclusion follows because g0(Sk-1) is symmetric.


## 3.3. The Case∫Ωmdx≠0

Lemma 3.6.
Let∫Ω‍mdx<0 and 0<λ<λ*(m). Then, Iλ,m is bounded from below, coercive and weakly lower semicontinuous (w.l.s.c.) on W1,p(Ω).Proof.
Φ
(
u
)
:
=
∫
Ω
‍
G
(
x
,
∇
u
)
d
x is w.l.s.c. on W1,p(Ω) because Φ is convex and continuous on W1,p(Ω) (cf. [25, Theorem 1.2]). Thus, Iλ,m is also w.l.s.c. on W1,p(Ω) since the inclusion from W1,p(Ω) to Lp(Ω) is compact.
Chooseɛ>0 such that pɛ<a_(1-λ/λ*(m)), where a_:=infΩa∞. By an easy estimation, (3.18) and (3.19) as in Lemma 3.5, we have
(3.21)Iλ,m(u)≥a_-ɛppa_∫Ωa∞|∇u|pdx‍-λp∫Ωm|u|pdx‍-ɛ′‖u‖pp-‖h‖∞‖u‖p|Ω|(p-1)/p-(Cɛ′+Dɛ′′)|Ω|
for every u∈W1,p(Ω) and ɛ′>0.
Letu∈W1,p(Ω) satisfy ∫Ω‍m|u|pdx≤0. Then, the following inequality follows from Lemma 2.2:
(3.22)D0∫Ω‍a∞|∇u|pdx-λ∫Ω‍m|u|pdx≥D02∫Ω‍a∞|∇u|pdx+b(m,ξ)‖u‖pp,
where b(m,ξ) is a positive constant independent of u with ξ=2λ/D0 and D0=(a_-ɛp)/a_.
For everyu∈W1,p(Ω) such that ∫Ω‍m|u|pdx>0, we obtain
(3.23)D0∫Ωa∞|∇u|pdx‍-λ∫Ωm|u|pdx‍≥(D0-λλ*(m))∫Ωa∞|∇u|pdx≥12(D0-λλ*(m))∫Ωa∞|∇u|pdx+c2(D0-λλ*(m))‖u‖pp
by the definition of λ*(m), Lemma 2.1 and D0-λ/λ*(m)>0, where c>0 is a constant obtained by Lemma 2.1.
Consequently, if we choose aɛ′>0 satisfying ɛ′<min{b(m,ξ)/p,c(D0-λ/λ*(m))/(2p)}, then we obtain positive constants d1 and d2 (independent of u) such that
(3.24)Iλ,m(u)≥d1‍∫Ωa∞|∇u|pdx+d2‖u‖pp-‖h‖∞‖u‖p|Ω|(p-1)/p-(Cɛ′+Dɛ′′)|Ω|≥min⁡⁡{a_d1,d2}‖u‖p-‖h‖∞‖u‖|Ω|(p-1)/p-(Cɛ′+Dɛ′′)|Ω|
for every u∈W1,p(Ω) by (3.21), (3.22), and (3.23). Because of p>1, our conclusion is shown.Lemma 3.7.
Letm≥0 in Ω and m≢0. If λ<0 holds, then Iλ,m is bounded from below, coercive and w.l.s.c. on W1,p(Ω).Proof.
First, as the same reason in Lemma3.6, it follows that Iλ,m is w.l.s.c. on W1,p(Ω). By a similar argument to Lemma 3.6, for every ɛ′>0 and 0<ɛ<a_/p where a_=infΩa∞, we obtain
(3.25)Iλ,m(u)≥a_-ɛppa_∫Ωa∞|∇u|pdx‍+|λ|p∫Ωm|u|pdx-ɛ′‖u‖pp‍-‖h‖∞‖u‖p|Ω|(p-1)/p-(Cɛ′+Dɛ′′)|Ω|
for every u∈W1,p(Ω) (note λ<0). Here, from Lemma 2.3,
(3.26)D0∫Ω‍a∞|∇u|pdx+|λ|∫Ω‍m|u|pdx≥D02∫Ω‍a∞|∇u|pdx+D02b(ξ,m)‖u‖pp
for every u∈W1,p(Ω) follows, where D0:=(a_-ɛp)/a_, ξ:=2|λ|/D0 and b(ξ,m) is a constant obtained in Lemma 2.3. Therefore, by choosing a ɛ′ such that 0<ɛ′<D0b(ξ,m)/2, we can prove our conclusion.Lemma 3.8.
Let∫Ω‍mdx≠0 and 0<λ<μ. Then, Iλ,m is bounded from below on Y(μ,m), where Y(μ,m) is the set introduced in (3.14).Proof.
Due to the same inequalities concerningG and F as in Lemma 3.5, for every ɛ>0 and ɛ′>0, there exists C=C(ɛ,ɛ′)>0 such that
(3.27)Iλ,m(u)≥a_-pɛpa_∫Ω‍a∞|∇u|pdx-λp∫Ω‍m|u|pdx-ɛ′‖u‖pp-‖h‖∞‖u‖1-C|Ω|
for every u∈W1,p(Ω), where a_:=infx∈Ωa∞(x). Choose positive constants ɛ and δ such that D0:=1-pɛ/a_>δ>λ/μ (note λ/μ<1).
First, we consider the case ofm≥0 in Ω. For every u∈Y(μ,m), we obtain
(3.28)D0∫Ωa∞|∇u|pdx‍-λ∫Ωm|u|pdx‍≥(D0-δ)∫Ωa∞|∇u|pdx‍+(δμ-λ)∫Ωm|u|pdx‍≥d(m,ξ1)(D0-δ)‖u‖pp
by Lemma 2.3 with ξ1=(δμ-λ)/(D0-δ) (note δμ-λ>0 and D0-δ>0).
Next, we handle with the case wherem changes sign. Let u∈W1,p(Ω) satisfy ∫Ω‍m|u|pdx≤0. Then, we have for such u(3.29)D0∫Ω‍a∞|∇u|pdx-λ∫Ω‍m|u|pdx≥b(m,ξ2)D0‖u‖pp
by Lemma 2.2, where D0=1-pɛ/a_ and ξ2:=λ/D0.
On the other hand, foru∈Y(μ,m) with ∫Ω‍m|u|pdx>0, the following inequality follows from Lemma 2.2:
(3.30)D0∫Ωa∞|∇u|pdx-λ∫Ωm|u|pdx‍≥(D0-δ)∫Ωa∞|∇u|pdx‍-(δμ-λ)∫Ω(-m)|u|pdx‍≥b(-m,ξ1)(D0-δ)‖u‖pp.
Consequently, by (3.27), (3.29), (3.28), and (3.30), there exists d>0 independent of u such that
(3.31)Iλ,m(u)≥(d-ɛ′)‖u‖pp-‖h‖∞‖u‖p|Ω|(p-1)/p-C|Ω|
for every u∈Y(μ,m). Hence, our conclusion is shown by taking ɛ′>0 satisfying ɛ′<d.Proof of Theorem1.1 in the Case ∫Ω‍mdx≠0.
First, if eitherm≥0 on Ω and λ<0 or 0<λ<λ*(m)=μ1(m) (i.e., ∫Ω‍mdx<0) holds, then Lemma 3.7 or Lemma 3.6 guarantees the existence of a global minimizer of Iλ,m, respectively (cf. [25, Theorem  1.1]). Hence, (P;λ,m,h) has a solution.
Sinceλ is an eigenvalue of (EV;m) if and only if -λ is one of (EV;-m), it suffices to consider the case of λ>λ*(m)≥0. Furthermore, by Proposition 2.9, Remark 2.6 (i), and our hypothesis that λ is not an eigenvalue of (EV;m), we may assume that there exists a k∈ℕ such that μk(m)<λ<μk+1(m). By Lemmas 3.5 and 3.8, we can choose T>0 and g0∈ℱk(m) satisfying
(3.32)max⁡z∈Sk-1Iλ,m(Tg0(z))<inf⁡{Iλ,m(u);u∈Y(μk+1(m),m)}=:α.
Put
(3.33)Σ≔{g∈C(S+k,W1,p(Ω));g∣Sk-1=Tg0},c≔inf⁡g∈Σmax⁡z∈S+kIλ,m(g(z)).
Then, it follows from Lemma 3.4 and (3.32) that c≥α>maxz∈Sk-1Iλ,m(Tg0(z)) holds. Since Iλ,m satisfies the Palais-Smale condition by Proposition 3.3, the minimax theorem guarantees (cf. [25, Theorem  4.6]) that c is a critical value of Iλ,m. Hence, (P;λ,m,h) has at least one solution.


## 3.4. The Case∫Ωmdx=0

First, we introduce an approximate functionalIλ,m,n+ as follows:
(3.34)Iλ,m,n+(u):=Iλ,m(u)+1pn‖u‖pp=Iλ,m-1/(λn)(u)foru∈W1,p(Ω).Lemma 3.9.
Let0<λ<μ. Then, there exists an n0∈ℕ such that for each n≥n0, Iλ,m,n+ is bounded from below on Y(μ,m-1/λn), where Y(μ,m-1/λn) is the set introduced in (3.14).Proof.
Choosen0∈ℕ such that 1/n0<λesssupx∈Ωm(x)/2. Then, for every n≥n0, Lemma 3.8 guarantees that Iλ,m,n+=Iλ,m-1/(λn) bounded from below on Y(μ,m-1/(λn)) because of ∫Ω‍(m-1/(λn))dx<0 and |{m-1/(λn)>0}|>0.Proof of Theorem1.1 in the Case ∫Ωmdx=0.
By noting thatλm=(-λ)(-m) and μ1(m)=λ*(m)=0, we may assume that μk(m)<λ<μk+1(m) for some k∈ℕ. Let n0 be a natural number obtained by Lemma 3.9. Due to Proposition 2.10 (i) and (ii), there exists an n1≥n0 such that
(3.35)μk(m)≤μk(m-1nλ)≤μk(m-1n1λ)<λ<μk+1(m)≤μk+1(m-1nλ)
for every n≥n1. Thus, for every n≥n1, we can take Tn>0 and gn∈ℱk(m-1/(nλ)) satisfying
(3.36)max⁡z∈Sk-1Iλ,m,n+(Tngn(z))<inf⁡{Iλ,m,n(u);u∈Y(μk+1(m-1(nλ)),m-1(nλ))}
by applying Lemmas 3.5 and 3.9 to Iλ,m,n+=Iλ,m-1/(nλ) (note (3.35)). Set
(3.37)Σn:={g∈C(S+k,W1,p(Ω));g∣Sk-1=Tngn},cn:=inf⁡g∈Σnmax⁡z∈S+kIλ,m,n+(g(z))
for each n≥n1. Then, for each n≥n1, we can obtain un satisfying
(3.38)|Iλ,m,n+(un)-cn|<1n,‖(Iλ,m,n+)′(un)‖W1,p(Ω)<1n
by applying Ekeland's variational principle to each Iλ,m,n+ (refer to [25, Theorem 4.3]). In addition, we can see that {un} is bounded in W1,p(Ω). Indeed, if there exists a subsequence {unl}l satisfying ∥unl∥p→∞ as l→∞, then we can show that λ is an eigenvalue of (EV;m) by the same argument as in Proposition 3.3. This contradicts to our assumption that λ is not an eigenvalue of (EV;m). Moreover, the boundedness of ∥∇un∥p follows from a similar inequality to (3.6) as in Proposition 3.3 under the boundedness of ∥un∥p.
Therefore, we may assume, by choosing a subsequence that{un} is a Palais-Smale sequence of Iλ,m since Iλ,m is bounded on a bounded set and according to the following inequality:
(3.39)‖Iλ,m′(un)‖(W1,p(Ω))*≤‖Iλ,m′(un)-(Iλ,m,n+)′(un)‖(W1,p(Ω))*+1n≤1n‖un‖pp-1+1n.
Therefore, because Iλ,m satisfies the Palais-Smale condition by Proposition 3.3, Iλ,m has a critical point, whence (P;λ,m,h) has at least one solution.


## 4. Proof of Theorem1.2

First, we will prove the following result concerning the Palais-Smale condition under the additional hypothesis (H±) or (HF±).Proposition 4.1.
Assume that one of the following conditions hold:(i)
λ=0 and (HF+) or (HF-);(ii)
λ≠0 and one of (H+), (H-), (HF+) and (HF-). Then, Iλ,m satisfies the Palais-Smale condition.Proof.
As the same reason in Proposition3.3, it suffices to prove the boundedness of a Palais-Smale sequence {un} such that Iλ,m(un)→c (for some c∈ℝ) and ∥Iλ,m′(un)∥W*→0 as n→∞. By way of contradiction, we may assume that ∥un∥p→∞ as n→∞ by choosing a subsequence. Set vn:=un/∥un∥p. Then, by the same argument as in Proposition 3.3, {vn} has a subsequence strongly convergent to v0 being a nontrivial solution of
(4.1)-div⁡(a∞(x)|∇u|p-2∇u)=λm(x)|u|p-2uinΩ,∂u∂ν=0on⁡∂Ω.
To simplify the notation, we denote the above subsequence strongly convergent to v0 by {vn}, again. Thus, |un(x)|→∞ as n→∞ for a.e. x∈Ω0:={x′∈Ω;v0(x′)≠0} (note ∥v0∥p=1).
Assume (HF+) or (HF-). Then, we can obtain
(4.2)(I):=∫Ω‍f(x,un)un-pF(x,un)‖un‖p1+qdx⟶±∞if(HF±),respectively.
Indeed, it follows from (HF+) that there exist R>0 and C>0 independent of n such that f(x,t)t-pF(x,t)≥0 if |t|≥R and a.e. x∈Ω, and |f(x,t)t-pF(x,t)|≤C for every |t|≤R and a.e. x∈Ω. Therefore, since |un(x)|→∞ a.e. x∈Ω0 and |Ω0|>0 (note ∥v0∥p=1), we have (4.2) if (HF+) holds, by applying Fatou's lemma to the following inequality:
(4.3)(I)≥∫Ω0‍f(x,un)un-pF(x,un)|un|1+q|vn|1+qdx-C|Ω∖Ω0|‖un‖p1+q.
In the case of (HF-), by considering -f instead of f as in the above argument, we can show our claim (4.2).
Furthermore, by Hölder's inequality, we have(4.4)(II):=∫ΩpG~(x,∇un)-a~(x,|∇un|)|∇un|2‖un‖p1+qdx‍≤H0∫Ω(|∇vn|1+q+1‖un‖p1+q)dx‍≤H0‖∇vn‖p1+q|Ω|(p-1-q)/p+o(1)≤H0‖∇v0‖p1+q|Ω|(p-1-q)/p+o(1)
in the case of (HF-) because vn→v0 in W1,p(Ω), where q∈[0,p-1] and H0>0 are constants as in (HF-). Similarly, we obtain
(4.5)(II)≥-H0‖∇v0‖p1+q|Ω|(p-1-q)/p+o(1)
in the case of (HF+).
Hence, we have a contradiction because of (4.2), (4.4), or (4.5) by taking a limit inferior or superior in the following equality:
(4.6)o(1)=pIλ,m(un)-〈Iλ,m′(un),un〉‖un‖p1+q=(II)+(I)+(1-p)∫Ω‍hvn‖un‖pqdx,
where we use the fact that ∥un∥/∥un∥p1+q=∥vn∥/∥un∥pq is bounded because of q≥0.
Assumeλ≠0 and (H+) or (H-): because v0 is a nontrivial solution of (4.1) with λ≠0, v0 is not a constant function, that is, ∥∇v0∥p>0. Therefore, we have |∇un(x)|→∞ as n→∞ for a.e. x∈Ω~0:={x′∈Ω;|∇v0(x′)|≠0}. Because of |Ω~0|>0, we can show
(4.7)∫Ω‍pG~(x,∇un)-a~(x,|∇un|)|∇un|2‖un‖p1+qdx⟶±∞if(H±),respectively,
by a similar argument to one for f in the above. In addition, we can easily obtain the following inequality:
(4.8)±∫Ω‍f(x,un)un-pF(x,un)‖un‖p1+qdx≥-H0‖vn‖1+q1+q+o(1)=-H0‖v0‖1+q1+q+o(1)
in the case of (H±), respectively. Hence, we have a contradiction by considering o(1)=(pIλ,m(un)-〈Iλ,m′(un),un〉)/∥un∥p1+q.By a similar way to the case∫Ω‍mdx=0, we introduce the following approximate functionals on W1,p(Ω):
(4.9)Iλ,m,n±(u):=Iλ,m(u)±1pn‖u‖ppforu∈W1,p(Ω).
Note Iλ,m,n±(u)=Iλ,m∓1/(λn)(u) on W1,p(Ω) provided λ≠0.Proposition 4.2.
If eitherλ≠0 and (H+) or (HF+)  (resp., either λ≠0 and (H-) or (HF-)) and {un} satisfies
(4.10)sup⁡n∈NIλ,m,n+(un)<+∞,lim⁡n→∞‖(Iλ,m,n+)′(un)‖W1,p(Ω)*=0,(4.11)(resp.inf⁡n∈NIλ,m,n-(un)>-∞,lim⁡n→∞‖(Iλ,m,n-)′(un)‖W1,p(Ω)*=0),
then {un} is bounded in W1,p(Ω).Proof.
First, we note that the boundedness of∥un∥p guarantees that ∥un∥ is bounded by limn→∞∥(Iλ,m,n±)′(un)∥W1,p(Ω)*=0 (refer to (3.6) as in the proof of Proposition 3.3). Moreover, because of the following equality:
(4.12)pIλ,m,n±(un)-〈(Iλ,m,n±)′(un),un〉‖un‖p1+q=(1-p)∫Ωhvn‖un‖pqdx,+∫ΩpG~(x,∇un)-a~(x,|∇un|)|∇un|2‖un‖p1+qdx‍+∫Ω‍f(x,un)un-pF(x,un)‖un‖p1+qdx,
we can prove the boundedness of ∥un∥p by the same argument as in Proposition 4.1.Proof of Theorem1.2.
Because ofλm=(-λ)(-m), we may assume λ≥0. In the case where ∫Ω‍mdx≠0 and μk(m)<λ<μk+1(m) for some k∈ℕ, the proof of Theorem 1.1 implies the existence of a critical point of Iλ,m because Iλ,m satisfies the Palais-Smale condition by Proposition 4.1. Concerning other cases, in the next section, we will prove the existence of a bounded sequence {un} satisfying (Iλ,m,n+)′(un)→0 or (Iλ,m,n-)′(un)→0 in W1,p(Ω)* as n→∞. Because Iλ,m is bounded on a bounded set, we may assume that Iλ,m(un) converges to some c∈ℝ by choosing a subsequence. In addition, by noting the inequality ∥Iλ,m′(un)∥W1,p(Ω)*≤∥(Iλ,m,n±)′(un)∥W1,p(Ω)*+∥un∥pp-1/n, we easily see that {un} is a bounded Palais-Smale sequence of Iλ,m. Therefore, Iλ,m has a critical point since Iλ,m satisfies the Palais-Smale condition by Proposition 4.1.


## 5. Construction of a Bounded Palais-Smale Sequence

In this section, due to the reason stated in the proof of Theorem1.2, we will construct a bounded sequence {un} satisfying (Iλ,m,n+)′(un)→0 or (Iλ,m,n-)′(un)→0 in W1,p(Ω)* as n→∞. It implies the existence of a bounded Palais-Smale sequence of Iλ,m.


### 5.1. The Caseλ=0

Assume (HF+)
In this c ase, we can show that for eachn∈ℕ, Iλ,m,n+ has a global minimizer un. Indeed, for 0<ɛ<1/(pn), there exists Cɛ>0 such that Iλ,m,n+(u)≥C0∥∇u∥pp/(p(p-1))+(1/(pn)-ɛ)∥u∥pp-∥h∥∞∥u∥1-Cɛ for every u∈W1,p(Ω) by (1.1), (1.16) and λ=0 (refer to the inequality as in the proof of Lemma 3.5). This means that Iλ,m,n+ is coercive and bounded from below on W1,p(Ω). Therefore, Iλ,m,n+ has a global minimizer un since Iλ,m,n+ is w.l.s.c. on W1,p(Ω) as the same reason in Lemma 3.6.
Furthermore, because of(Iλ,m,n+)′(un)=0 in W1,p(Ω)* and Iλ,m,n+(un)=minW1,p(Ω)Iλ,m,n+≤Iλ,m,n+(0)=0, it follows from Proposition 4.2 that {un} is bounded.Assume (HF-)
Choosen0∈ℕ such that 1/n0<c(1)=μ2(1), where c(1) is the second eigenvalue of (EV;1) (so the weight function m≡1 and see (2.22) for the definition). Then, by noting that I0,m,n0-=I1/n0,1, we have
(5.1)α:=inf⁡{I0,m,n0-(u);u∈Y(c(1),1)}>-∞
by Lemma 3.8, where Y(c(1),1) is a subset defined by (3.14) with the weight m≡1, that is,
(5.2)Y(c(1),1):={u∈W1,p(Ω);∫Ω‍a∞|∇u|pdx≥c(1)‖u‖pp}.
Moreover, inf{I0,m,n-(u);u∈Y(c(1),1)}≥α for every n≥n0 holds because I0,m,n-(u)≥I0,m,n0-(u) for every u∈W1,p(Ω). Since ∫Ω‍F(x,u)dx=o(1)∥u∥pp as ∥u∥p→∞ by (1.1), there exists Tn>0 such that I0,m,n-(±Tn)=-Tnp(|Ω|/(np)-o(1))<α-2.
Define(5.3)Σn≔{g∈C([0,1],W1,p(Ω));g(0)=Tn,g(1)=-Tn},cn≔inf⁡g∈Σnmax⁡t∈[0,1]I0,m,n-(g(t))
for n≥n0. By the definition of c(1), we easily see that g([0,1])∩Y(c(1),1)≠∅ for every g∈Σn (refer to [6] or Lemma 3.4). Hence,
(5.4)cn≥inf⁡{I0,m,n-(u);u∈Y(c(1),1)}≥α>I0,m,n(±Tn)
holds, whence cn is bounded from below. Moreover, by applying Ekeland's variational principle to each I0,m,n-, we can obtain a sequence {un} satisfying |I0,m,n-(un)-cn|<1/n and ∥(I0,m,n-)′(un)∥W1,p(Ω)*<1/n. Since cn is bounded from below, it follows from Proposition 4.2 that {un} is bounded. As a result, we can construct a bounded sequence {un} satisfying (I0,m,n-)′(un)→0 as n→∞ in W1,p(Ω)*.


### 5.2. The Caseλ=λ*(m)=μ1(m) with ∫Ω‍mdx<0

Assume (H+) or (HF+)
Since we see thatIλ,m,n+=Iλ,m-1/(nλ) and λ*(m-1/(nλ))>λ*(m)=λ>0 (according to Lemma 2.5), Iλ,m,n+ is coercive, bounded from below and w.l.s.c. on W1,p(Ω) by Lemma 3.6. Thus, we obtain a global minimizer un of Iλ,m,n+ for sufficiently large n such that |{m-1/(nλ)>0}|>0. Because of Iλ,m,n+(un)≤Iλ,m,n+(0)=0 for every n, Proposition 4.2 guarantees that {un} is bounded.Assume (H-) or (HF-)
First, we note thatIλ,m,n-=Iλ,m+1/(nλ) and 0<λ*(m+1/(nλ))<λ*(m)=λ by Lemma 2.5 for sufficiently large n such that ∫Ω‍(m+1/(nλ))dx<0. Moreover, it follows from Proposition 2.10 and μ1(m)<μ2(m) that there exists an n0∈ℕ satisfying ∫Ω‍m+1/(n0λ)dx<0 and
(5.5)λ*(m+1nλ)<λ=μ1(m)<μ2(m+1n0λ)≤μ2(m+1nλ)≤μ2(m)
for every n≥n0. By applying Theorem 1.1 to each case of a weight m+1/(nλ) (note that λ is not an eigenvalue of (EV;m+1/(nλ)) by (5.5), there exists un satisfying (Iλ,m,n-)′(un)=0 (note Iλ,m,n-=Iλ,m+1/(nλ)) and
(5.6)Iλ,m,n-(un)=cn≥inf⁡{Iλ,m,n-(u);u∈Y(μ2(mn0),mn0)},
where the last inequality follows from Lemma 3.4 with mn0:=m+1/(n0λ). On the other hand, because Iλ,m,n-(u)≥Iλ,m,n0-(u)=Iλ,mn0(u) for every u∈W1,p(Ω) and n≥n0, we have
(5.7)cn≥inf⁡{Iλ,mn0(u);u∈Y(μ2(mn0),mn0)}>-∞
for every n≥n0, where the last inequality follows from Lemma 3.8. Thus, cn is bounded from below. Hence, Proposition 4.2 guarantees the boundedness of {un}.


### 5.3. The Caseλ=μk+1(m) with ∫Ω‍mdx≠0

Assume (H+) or (HF+)
We may assumeμk(m)<μk+1(m)=λ by taking k anew if necessary (note that we have already proved the case of μk(m)<λ<μk+1(m) in Section 4). Here, we can choose an n0∈ℕ such that ∫Ω‍(m-1/(nλ))dx≠0, |{m-1/(nλ)>0}|>0 and
(5.8)μk(m-1nλ)≤μk(m-1n0λ)<λ-1n‖m‖∞<λ=μk+1(m)≤μk+1(m-1nλ)
for every n≥n0 by ∫Ω‍mdx≠0 and Proposition 2.10 (i), (iii). Note the following inequality:
(5.9)Iλ,m,n0+(u)≥Iλ,m,n+(u)≥Iλ-1/(n‖m‖∞),m(u)
for every u∈W1,p(Ω) and n≥n0, where the last inequality is obtained by ∥u∥pp≥∫Ω‍m|u|pdx/∥m∥∞. Let n≥n0. It follows from Lemma 3.8 and (5.8) that Iλ-1/(n∥m∥∞),m is bounded from below on Y(λ,m). Hence, (5.9) yields that Iλ,m,n+ is also bounded from below on Y(λ,m), namely,
(5.10)αn:=inf⁡{Iλ,m,n+(u);u∈Y(λ,m)}>-∞.
On the other hand, because of μk(m-1/(n0λ))<λ (see (5.8)), Lemma 3.5 guarantees the existence of g0∈ℱk(m-1/(n0λ)) satisfying
(5.11)max⁡z∈Sk-1Iλ,m,n0+(Tg0(z))=max⁡z∈Sk-1Iλ,m-1/(n0λ)(Tg0(z))⟶-∞as|T|⟶∞.
Thus, for each n≥n0, we can take Tn>0 such that
(5.12)max⁡z∈Sk-1Iλ,m,n+(Tng0(z))≤max⁡z∈Sk-1Iλ,m,n0+(Tng0(z))≤αn-1,
(note (5.9) for the first inequality). Set
(5.13)Σn≔{g∈C(S+k,W1,p(Ω));g∣Sk-1=Tng0},cn+≔inf⁡g∈Σnmax⁡z∈S+kIλ,m,n+(g(z))
for n≥n0. Since g(S+k)∩Y(λ,m)≠∅ for every g∈Σn by Lemma 3.4 and λ=μk+1(m), we have cn+≥αn>maxz∈Sk-1Iλ,m,n+(Tng0(z)). Therefore, Ekeland's variational principle (refer to [25, Theorem  4.3]) guarantees the existence of un satisfying |Iλ,m,n+(un)-cn|<1/n and ∥(Iλ,m,n+)′(un)∥W1,p(Ω)*<1/n.Finally, to show the boundedness of{un} due to Proposition 4.2, we will prove that cn+ is bounded from above. For each n≥n0, we define a continuous map gn from S+k to W1,p(Ω) by
(5.14)gn(z):={(1-zk+1)Tng0(z′1-zk+12)forz=(z′,zk+1)∈S+kwith0≤zk+1<1,0forz=(z′,zk+1)∈S+kwithzk+1=1.
Then, gn∈Σn holds. This leads to
(5.15)cn+≤sup⁡t≥0,z∈Sk-1Iλ,m,n+(tg0(z))≤sup⁡t≥0,z∈Sk-1Iλ,m,n0+(tg0(z))<+∞
because of (5.9), (5.11), and the compactness of g0(Sk-1).Assume (H-) or (HF-)
Because the case ofμ1(m)=λ*(m) is already shown (see Sections 5.1 and 5.2), We may assume (0<)μk(m)=λ<μk+1(m) for some k≥2 by taking k anew if necessary. Here, we can choose an n0∈ℕ such that ∫Ω‍(m+1/(nλ))dx≠0 and
(5.16)μk(m+1nλ)≤μk(m)=λ<μk+1(m+1n0λ)≤μk+1(m+1nλ)≤μk+1(m)
for every n≥n0 by ∫Ω‍mdx≠0 and Proposition 2.10 (i), (iii). Moreover, we note the following inequality:
(5.17)Iλ,m,n0-(u)≤Iλ,m,n-(u)=Iλ,m+1/(nλ)(u)≤Iλ+1/(n‖m‖∞),m(u)
for every u∈W1,p(Ω) and n≥n0. It follows from Lemma 3.8 and (5.16) (note (5.17) also) that Iλ,m,n0-=Iλ,m0 is bounded from below on Y(μk+1(m0),m0) with m0:=m+1/(n0λ). Hence, (5.17) implies
(5.18)inf⁡⁡{Iλ,m,n-(u);u∈Y(μk+1(m0),m0)}≥inf⁡⁡{Iλ,m,n0-(u);u∈Y(μk+1(m0),m0)}=:α0>-∞
for every n≥n0. Because of λ+1/(n∥m∥∞)>λ=μk(m), there exist gn∈ℱk(m) and Tn>0 such that
(5.19)max⁡z∈Sk-1Iλ,m,n-(Tngn(z))≤max⁡z∈Sk-1Iλ+1/(n‖m‖∞),m(Tngn(z))<α0-1
by Lemma 3.5. Define
(5.20)Σn≔{g∈C(S+k,W1,p(Ω));g∣Sk-1=Tngn},cn-≔inf⁡g∈Σnmax⁡z∈S+kIλ,m,n-(g(z))
for n≥n0. Then, cn-≥α0 occurs (see (5.18)) since g(S+k)∩Y(μk+1(m0),m0)≠∅ for every g∈Σn by Lemma 3.4. This means that cn- is bounded from below. Consequently, we can obtain a desired bounded sequence by the same argument as in Sections 5.1 and 5.2.


### 5.4. The Case (iii) as in Theorem1.2

First, note that we are assuming the hypothesis (H+) or (HF+) in this case (iii). In addition, as the reason in the proof of Theorem 1.2, it suffices to handle with λ>0.Letk∈ℕ satisfy μk(m)<λ≤μk+1(m). According to Proposition 2.10 (i) and (ii), we can take an n0∈ℕ such that |{m-1/(nλ)>0}|>0 and
(5.21)μk(m-12nλ)≤μk(m-1n0λ)<λ-12n‖m‖∞<λ≤μk+1(m)≤μk+1(m-12nλ)
for every n≥n0. The following inequality follows from the easy estimates:
(5.22)Iλ,m,n0+(u)≥Iλ,m,n+(u)=Iλ,m-1/(nλ)(u)≥Iλ-1/(2n‖m‖∞),m-1/(2nλ)(u)
for every u∈W1,p(Ω) and n≥n0. Let n≥n0 and set mn:=m-1/(2nλ). Because of (5.21), Lemma 3.8 implies that Iλ-1/(2n∥m∥∞),mn is bounded from below on Y(μk+1(mn),mn) with (note ∫Ω‍mndx≠0). Hence, (5.22) yields that
(5.23)αn:=inf⁡{Iλ,m,n+(u);u∈Y(μk+1(mn),mn)}>-∞
for each n≥n0. On the other hand, because of μk(m-1/(n0λ))<λ (see (5.21)), Lemma 3.5 guarantees the existence of g0∈ℱk(m-1/(n0λ)) satisfying
(5.24)max⁡z∈Sk-1Iλ,m,n0+(Tg0(z))=max⁡z∈Sk-1Iλ,m-1/(n0λ)(Tg0(z))⟶-∞asT⟶∞.
Therefore, for each n≥n0, we can choose Tn>0 such that
(5.25)max⁡z∈Sk-1Iλ,m,n+(Tng0(z))≤max⁡z∈Sk-1Iλ,m,n0+(Tng0(z))≤αn-1,
(note (5.22) for the first inequality). Set
(5.26)Σn≔{g∈C(S+k,W1,p(Ω));g∣Sk-1=Tng0},cn+≔inf⁡g∈Σnmax⁡z∈S+kIλ,m,n+(g(z))
for n≥n0. Since g(S+k)∩Y(μk+1(mn),mn)≠∅ for every g∈Σn by Lemma 3.4, we have cn+≥αn>maxz∈Sk-1Iλ,m,n+(Tng0(z)). Moreover, by the same argument as in Section 5.3 (note (5.24)), we have
(5.27)cn+≤sup⁡t≥0,z∈Sk-1Iλ,m,n+(tg0(z))≤sup⁡t≥0,z∈Sk-1Iλ,m,n0+(tg0(z))<+∞,
and hence our conclusion is shown.Remark 5.1.
If∫Ω‍mdx=0 holds, then we can not show the continuity of μk(m) with respect to m (refer to Proposition 2.10). Hence, we are not able to construct a bounded Palais-Smale sequence under (H-) or (HF-). However, if we have the additional information about the existence of a suitable m′∈L∞(Ω) such that m′≥m in Ω, ∫Ω‍m′dx≠0 and μk(m)≤λ<μk+1(m′) when μk(m)≤λ<μk+1(m) occurs, then we can still easily prove that equation (P;λ,m,h) has a solution in the case also where λ≠0, ∫Ω‍mdx=0 and (H-) or (HF-). In fact, let 0<μk(m)≤λ<μk+1(m′) for some k≥2. Note the following inequality:
(5.28)Iλ+1/(n‖m‖∞),m(u)≥Iλ,m,n-(u)≥Iλ,m′(u)-1np‖u‖pp=Iλ,m′-1/(nλ)(u)
for every u∈W1,p(Ω) and n. Fix n0∈ℕ such that ∫Ω‍m′-1/(n0λ)dx>0 and |{m′-1/(n0λ)>0}|>0. Set m0′:=m′-1/(n0λ). Because of λ<μk+1(m′)≤μk+1(m0′) (the last inequality follows from Proposition 2.10 (i)), Lemma 3.8 implies that Iλ,m0′ is bounded from below on Y(μk+1(m0′),m0′) (note ∫Ω‍m0′dx>0). By combining this fact and (5.28), we have
(5.29)inf⁡n≥n0⁡inf⁡⁡{Iλ,m,n-(u);u∈Y(μk+1(m0′),m0′)}≥inf⁡⁡{Iλ,m0′(u);u∈Y(μk+1(m0′),m0′)}>-∞.
Because of λ+1/(n∥m∥∞)>λ≥μk(m), for each n≥n0, we can take a gn∈ℱk(m) satisfying
(5.30)max⁡z∈Sk-1Iλ,m,n-(Tgn(z))≤max⁡z∈Sk-1Iλ+1/(n‖m‖∞),m(Tgn(z))⟶-∞
as T→∞ by Lemma 3.5.
Since any extensiong∈C(S+k,W1,p(Ω)) of Tgn (T>0) links Y(μk+1(m0′),m0′) by Lemma 3.4, we can construct a desired sequence by the same argument as in Section 5.3 under (H-) or (HF-).


## 5.1. The Caseλ=0

Assume (HF+)
In this c ase, we can show that for eachn∈ℕ, Iλ,m,n+ has a global minimizer un. Indeed, for 0<ɛ<1/(pn), there exists Cɛ>0 such that Iλ,m,n+(u)≥C0∥∇u∥pp/(p(p-1))+(1/(pn)-ɛ)∥u∥pp-∥h∥∞∥u∥1-Cɛ for every u∈W1,p(Ω) by (1.1), (1.16) and λ=0 (refer to the inequality as in the proof of Lemma 3.5). This means that Iλ,m,n+ is coercive and bounded from below on W1,p(Ω). Therefore, Iλ,m,n+ has a global minimizer un since Iλ,m,n+ is w.l.s.c. on W1,p(Ω) as the same reason in Lemma 3.6.
Furthermore, because of(Iλ,m,n+)′(un)=0 in W1,p(Ω)* and Iλ,m,n+(un)=minW1,p(Ω)Iλ,m,n+≤Iλ,m,n+(0)=0, it follows from Proposition 4.2 that {un} is bounded.Assume (HF-)
Choosen0∈ℕ such that 1/n0<c(1)=μ2(1), where c(1) is the second eigenvalue of (EV;1) (so the weight function m≡1 and see (2.22) for the definition). Then, by noting that I0,m,n0-=I1/n0,1, we have
(5.1)α:=inf⁡{I0,m,n0-(u);u∈Y(c(1),1)}>-∞
by Lemma 3.8, where Y(c(1),1) is a subset defined by (3.14) with the weight m≡1, that is,
(5.2)Y(c(1),1):={u∈W1,p(Ω);∫Ω‍a∞|∇u|pdx≥c(1)‖u‖pp}.
Moreover, inf{I0,m,n-(u);u∈Y(c(1),1)}≥α for every n≥n0 holds because I0,m,n-(u)≥I0,m,n0-(u) for every u∈W1,p(Ω). Since ∫Ω‍F(x,u)dx=o(1)∥u∥pp as ∥u∥p→∞ by (1.1), there exists Tn>0 such that I0,m,n-(±Tn)=-Tnp(|Ω|/(np)-o(1))<α-2.
Define(5.3)Σn≔{g∈C([0,1],W1,p(Ω));g(0)=Tn,g(1)=-Tn},cn≔inf⁡g∈Σnmax⁡t∈[0,1]I0,m,n-(g(t))
for n≥n0. By the definition of c(1), we easily see that g([0,1])∩Y(c(1),1)≠∅ for every g∈Σn (refer to [6] or Lemma 3.4). Hence,
(5.4)cn≥inf⁡{I0,m,n-(u);u∈Y(c(1),1)}≥α>I0,m,n(±Tn)
holds, whence cn is bounded from below. Moreover, by applying Ekeland's variational principle to each I0,m,n-, we can obtain a sequence {un} satisfying |I0,m,n-(un)-cn|<1/n and ∥(I0,m,n-)′(un)∥W1,p(Ω)*<1/n. Since cn is bounded from below, it follows from Proposition 4.2 that {un} is bounded. As a result, we can construct a bounded sequence {un} satisfying (I0,m,n-)′(un)→0 as n→∞ in W1,p(Ω)*.


## 5.2. The Caseλ=λ*(m)=μ1(m) with ∫Ω‍mdx<0

Assume (H+) or (HF+)
Since we see thatIλ,m,n+=Iλ,m-1/(nλ) and λ*(m-1/(nλ))>λ*(m)=λ>0 (according to Lemma 2.5), Iλ,m,n+ is coercive, bounded from below and w.l.s.c. on W1,p(Ω) by Lemma 3.6. Thus, we obtain a global minimizer un of Iλ,m,n+ for sufficiently large n such that |{m-1/(nλ)>0}|>0. Because of Iλ,m,n+(un)≤Iλ,m,n+(0)=0 for every n, Proposition 4.2 guarantees that {un} is bounded.Assume (H-) or (HF-)
First, we note thatIλ,m,n-=Iλ,m+1/(nλ) and 0<λ*(m+1/(nλ))<λ*(m)=λ by Lemma 2.5 for sufficiently large n such that ∫Ω‍(m+1/(nλ))dx<0. Moreover, it follows from Proposition 2.10 and μ1(m)<μ2(m) that there exists an n0∈ℕ satisfying ∫Ω‍m+1/(n0λ)dx<0 and
(5.5)λ*(m+1nλ)<λ=μ1(m)<μ2(m+1n0λ)≤μ2(m+1nλ)≤μ2(m)
for every n≥n0. By applying Theorem 1.1 to each case of a weight m+1/(nλ) (note that λ is not an eigenvalue of (EV;m+1/(nλ)) by (5.5), there exists un satisfying (Iλ,m,n-)′(un)=0 (note Iλ,m,n-=Iλ,m+1/(nλ)) and
(5.6)Iλ,m,n-(un)=cn≥inf⁡{Iλ,m,n-(u);u∈Y(μ2(mn0),mn0)},
where the last inequality follows from Lemma 3.4 with mn0:=m+1/(n0λ). On the other hand, because Iλ,m,n-(u)≥Iλ,m,n0-(u)=Iλ,mn0(u) for every u∈W1,p(Ω) and n≥n0, we have
(5.7)cn≥inf⁡{Iλ,mn0(u);u∈Y(μ2(mn0),mn0)}>-∞
for every n≥n0, where the last inequality follows from Lemma 3.8. Thus, cn is bounded from below. Hence, Proposition 4.2 guarantees the boundedness of {un}.


## 5.3. The Caseλ=μk+1(m) with ∫Ω‍mdx≠0

Assume (H+) or (HF+)
We may assumeμk(m)<μk+1(m)=λ by taking k anew if necessary (note that we have already proved the case of μk(m)<λ<μk+1(m) in Section 4). Here, we can choose an n0∈ℕ such that ∫Ω‍(m-1/(nλ))dx≠0, |{m-1/(nλ)>0}|>0 and
(5.8)μk(m-1nλ)≤μk(m-1n0λ)<λ-1n‖m‖∞<λ=μk+1(m)≤μk+1(m-1nλ)
for every n≥n0 by ∫Ω‍mdx≠0 and Proposition 2.10 (i), (iii). Note the following inequality:
(5.9)Iλ,m,n0+(u)≥Iλ,m,n+(u)≥Iλ-1/(n‖m‖∞),m(u)
for every u∈W1,p(Ω) and n≥n0, where the last inequality is obtained by ∥u∥pp≥∫Ω‍m|u|pdx/∥m∥∞. Let n≥n0. It follows from Lemma 3.8 and (5.8) that Iλ-1/(n∥m∥∞),m is bounded from below on Y(λ,m). Hence, (5.9) yields that Iλ,m,n+ is also bounded from below on Y(λ,m), namely,
(5.10)αn:=inf⁡{Iλ,m,n+(u);u∈Y(λ,m)}>-∞.
On the other hand, because of μk(m-1/(n0λ))<λ (see (5.8)), Lemma 3.5 guarantees the existence of g0∈ℱk(m-1/(n0λ)) satisfying
(5.11)max⁡z∈Sk-1Iλ,m,n0+(Tg0(z))=max⁡z∈Sk-1Iλ,m-1/(n0λ)(Tg0(z))⟶-∞as|T|⟶∞.
Thus, for each n≥n0, we can take Tn>0 such that
(5.12)max⁡z∈Sk-1Iλ,m,n+(Tng0(z))≤max⁡z∈Sk-1Iλ,m,n0+(Tng0(z))≤αn-1,
(note (5.9) for the first inequality). Set
(5.13)Σn≔{g∈C(S+k,W1,p(Ω));g∣Sk-1=Tng0},cn+≔inf⁡g∈Σnmax⁡z∈S+kIλ,m,n+(g(z))
for n≥n0. Since g(S+k)∩Y(λ,m)≠∅ for every g∈Σn by Lemma 3.4 and λ=μk+1(m), we have cn+≥αn>maxz∈Sk-1Iλ,m,n+(Tng0(z)). Therefore, Ekeland's variational principle (refer to [25, Theorem  4.3]) guarantees the existence of un satisfying |Iλ,m,n+(un)-cn|<1/n and ∥(Iλ,m,n+)′(un)∥W1,p(Ω)*<1/n.Finally, to show the boundedness of{un} due to Proposition 4.2, we will prove that cn+ is bounded from above. For each n≥n0, we define a continuous map gn from S+k to W1,p(Ω) by
(5.14)gn(z):={(1-zk+1)Tng0(z′1-zk+12)forz=(z′,zk+1)∈S+kwith0≤zk+1<1,0forz=(z′,zk+1)∈S+kwithzk+1=1.
Then, gn∈Σn holds. This leads to
(5.15)cn+≤sup⁡t≥0,z∈Sk-1Iλ,m,n+(tg0(z))≤sup⁡t≥0,z∈Sk-1Iλ,m,n0+(tg0(z))<+∞
because of (5.9), (5.11), and the compactness of g0(Sk-1).Assume (H-) or (HF-)
Because the case ofμ1(m)=λ*(m) is already shown (see Sections 5.1 and 5.2), We may assume (0<)μk(m)=λ<μk+1(m) for some k≥2 by taking k anew if necessary. Here, we can choose an n0∈ℕ such that ∫Ω‍(m+1/(nλ))dx≠0 and
(5.16)μk(m+1nλ)≤μk(m)=λ<μk+1(m+1n0λ)≤μk+1(m+1nλ)≤μk+1(m)
for every n≥n0 by ∫Ω‍mdx≠0 and Proposition 2.10 (i), (iii). Moreover, we note the following inequality:
(5.17)Iλ,m,n0-(u)≤Iλ,m,n-(u)=Iλ,m+1/(nλ)(u)≤Iλ+1/(n‖m‖∞),m(u)
for every u∈W1,p(Ω) and n≥n0. It follows from Lemma 3.8 and (5.16) (note (5.17) also) that Iλ,m,n0-=Iλ,m0 is bounded from below on Y(μk+1(m0),m0) with m0:=m+1/(n0λ). Hence, (5.17) implies
(5.18)inf⁡⁡{Iλ,m,n-(u);u∈Y(μk+1(m0),m0)}≥inf⁡⁡{Iλ,m,n0-(u);u∈Y(μk+1(m0),m0)}=:α0>-∞
for every n≥n0. Because of λ+1/(n∥m∥∞)>λ=μk(m), there exist gn∈ℱk(m) and Tn>0 such that
(5.19)max⁡z∈Sk-1Iλ,m,n-(Tngn(z))≤max⁡z∈Sk-1Iλ+1/(n‖m‖∞),m(Tngn(z))<α0-1
by Lemma 3.5. Define
(5.20)Σn≔{g∈C(S+k,W1,p(Ω));g∣Sk-1=Tngn},cn-≔inf⁡g∈Σnmax⁡z∈S+kIλ,m,n-(g(z))
for n≥n0. Then, cn-≥α0 occurs (see (5.18)) since g(S+k)∩Y(μk+1(m0),m0)≠∅ for every g∈Σn by Lemma 3.4. This means that cn- is bounded from below. Consequently, we can obtain a desired bounded sequence by the same argument as in Sections 5.1 and 5.2.


## 5.4. The Case (iii) as in Theorem1.2

First, note that we are assuming the hypothesis (H+) or (HF+) in this case (iii). In addition, as the reason in the proof of Theorem 1.2, it suffices to handle with λ>0.Letk∈ℕ satisfy μk(m)<λ≤μk+1(m). According to Proposition 2.10 (i) and (ii), we can take an n0∈ℕ such that |{m-1/(nλ)>0}|>0 and
(5.21)μk(m-12nλ)≤μk(m-1n0λ)<λ-12n‖m‖∞<λ≤μk+1(m)≤μk+1(m-12nλ)
for every n≥n0. The following inequality follows from the easy estimates:
(5.22)Iλ,m,n0+(u)≥Iλ,m,n+(u)=Iλ,m-1/(nλ)(u)≥Iλ-1/(2n‖m‖∞),m-1/(2nλ)(u)
for every u∈W1,p(Ω) and n≥n0. Let n≥n0 and set mn:=m-1/(2nλ). Because of (5.21), Lemma 3.8 implies that Iλ-1/(2n∥m∥∞),mn is bounded from below on Y(μk+1(mn),mn) with (note ∫Ω‍mndx≠0). Hence, (5.22) yields that
(5.23)αn:=inf⁡{Iλ,m,n+(u);u∈Y(μk+1(mn),mn)}>-∞
for each n≥n0. On the other hand, because of μk(m-1/(n0λ))<λ (see (5.21)), Lemma 3.5 guarantees the existence of g0∈ℱk(m-1/(n0λ)) satisfying
(5.24)max⁡z∈Sk-1Iλ,m,n0+(Tg0(z))=max⁡z∈Sk-1Iλ,m-1/(n0λ)(Tg0(z))⟶-∞asT⟶∞.
Therefore, for each n≥n0, we can choose Tn>0 such that
(5.25)max⁡z∈Sk-1Iλ,m,n+(Tng0(z))≤max⁡z∈Sk-1Iλ,m,n0+(Tng0(z))≤αn-1,
(note (5.22) for the first inequality). Set
(5.26)Σn≔{g∈C(S+k,W1,p(Ω));g∣Sk-1=Tng0},cn+≔inf⁡g∈Σnmax⁡z∈S+kIλ,m,n+(g(z))
for n≥n0. Since g(S+k)∩Y(μk+1(mn),mn)≠∅ for every g∈Σn by Lemma 3.4, we have cn+≥αn>maxz∈Sk-1Iλ,m,n+(Tng0(z)). Moreover, by the same argument as in Section 5.3 (note (5.24)), we have
(5.27)cn+≤sup⁡t≥0,z∈Sk-1Iλ,m,n+(tg0(z))≤sup⁡t≥0,z∈Sk-1Iλ,m,n0+(tg0(z))<+∞,
and hence our conclusion is shown.Remark 5.1.
If∫Ω‍mdx=0 holds, then we can not show the continuity of μk(m) with respect to m (refer to Proposition 2.10). Hence, we are not able to construct a bounded Palais-Smale sequence under (H-) or (HF-). However, if we have the additional information about the existence of a suitable m′∈L∞(Ω) such that m′≥m in Ω, ∫Ω‍m′dx≠0 and μk(m)≤λ<μk+1(m′) when μk(m)≤λ<μk+1(m) occurs, then we can still easily prove that equation (P;λ,m,h) has a solution in the case also where λ≠0, ∫Ω‍mdx=0 and (H-) or (HF-). In fact, let 0<μk(m)≤λ<μk+1(m′) for some k≥2. Note the following inequality:
(5.28)Iλ+1/(n‖m‖∞),m(u)≥Iλ,m,n-(u)≥Iλ,m′(u)-1np‖u‖pp=Iλ,m′-1/(nλ)(u)
for every u∈W1,p(Ω) and n. Fix n0∈ℕ such that ∫Ω‍m′-1/(n0λ)dx>0 and |{m′-1/(n0λ)>0}|>0. Set m0′:=m′-1/(n0λ). Because of λ<μk+1(m′)≤μk+1(m0′) (the last inequality follows from Proposition 2.10 (i)), Lemma 3.8 implies that Iλ,m0′ is bounded from below on Y(μk+1(m0′),m0′) (note ∫Ω‍m0′dx>0). By combining this fact and (5.28), we have
(5.29)inf⁡n≥n0⁡inf⁡⁡{Iλ,m,n-(u);u∈Y(μk+1(m0′),m0′)}≥inf⁡⁡{Iλ,m0′(u);u∈Y(μk+1(m0′),m0′)}>-∞.
Because of λ+1/(n∥m∥∞)>λ≥μk(m), for each n≥n0, we can take a gn∈ℱk(m) satisfying
(5.30)max⁡z∈Sk-1Iλ,m,n-(Tgn(z))≤max⁡z∈Sk-1Iλ+1/(n‖m‖∞),m(Tgn(z))⟶-∞
as T→∞ by Lemma 3.5.
Since any extensiong∈C(S+k,W1,p(Ω)) of Tgn (T>0) links Y(μk+1(m0′),m0′) by Lemma 3.4, we can construct a desired sequence by the same argument as in Section 5.3 under (H-) or (HF-).



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*Source: 568120-2012-07-09.xml*