Existence of a positive solution for quasilinear elliptic equations with nonlinearity including the gradient

Abstract

We provide the existence of a positive solution for the quasilinear elliptic equation

$$-div\left(a(x,|\mathrm{\nabla }u|)\mathrm{\nabla }u\right)=f(x,u,\mathrm{\nabla }u)$$

in Ω under the Dirichlet boundary condition. As a special case ($a(x,t)=t^{p-2}$), our equation coincides with the usual p-Laplace equation. The solution is established as the limit of a sequence of positive solutions of approximate equations. The positivity of our solution follows from the behavior of $f(x,t\xi )$ as t is small. In this paper, we do not impose the sign condition to the nonlinear term f.

MSC:35J92, 35P30.

1 Introduction

In this paper, we consider the existence of a positive solution for the following quasilinear elliptic equation:

where $\mathrm{\Omega }\subset {\mathbb{R}}^N$ is a bounded domain with $C^2$ boundary ∂ Ω. Here, $A:\overline{\mathrm{\Omega }}\times {\mathbb{R}}^N\to {\mathbb{R}}^N$ is a map which is strictly monotone in the second variable and satisfies certain regularity conditions (see the following assumption (A)). Equation (P) contains the corresponding p-Laplacian problem as a special case. However, in general, we do not suppose that this operator is $(p-1)$-homogeneous in the second variable.

Throughout this paper, we assume that the map A and the nonlinear term f satisfy the following assumptions (A) and (f), respectively.

• (A) $A(x,y)=a(x,|y|)y$, where $a(x,t)>0$ for all $(x,t)\in \overline{\mathrm{\Omega }}\times (0,+\mathrm{\infty })$, and there exist positive constants $C_0$, $C_1$, $C_2$, $C_3$, $0<t_0\le 1$ and $1<p<\mathrm{\infty }$ such that

• (i) $A\in C^0(\overline{\mathrm{\Omega }}\times {\mathbb{R}}^N,{\mathbb{R}}^N)\cap C^1(\overline{\mathrm{\Omega }}\times ({\mathbb{R}}^N\setminus \{0\}),{\mathbb{R}}^N)$;

• (ii) $|D_{y}A(x,y)|\le C_1{|y|}^{p-2}$ for every $x\in \overline{\mathrm{\Omega }}$, and $y\in {\mathbb{R}}^N\setminus \{0\}$;

• (iii) $D_{y}A(x,y)\xi \cdot \xi \ge C_0{|y|}^{p-2}{|\xi |}^2$ for every $x\in \overline{\mathrm{\Omega }}$, $y\in {\mathbb{R}}^N\setminus \{0\}$ and $\xi \in {\mathbb{R}}^N$;

• (iv) $|D_{x}A(x,y)|\le C_2(1+{|y|}^{p-1})$ for every $x\in \overline{\mathrm{\Omega }}$, $y\in {\mathbb{R}}^N\setminus \{0\}$;

• (v) $|D_{x}A(x,y)|\le C_3{|y|}^{p-1}(-log|y|)$ for every $x\in \overline{\mathrm{\Omega }}$, $y\in {\mathbb{R}}^N$ with $0<|y|<t_0$.

• (f) f is a continuous function on $\mathrm{\Omega }\times [0,\mathrm{\infty })\times {\mathbb{R}}^N$ satisfying $f(x,0,\xi )=0$ for every $(x,\xi )\in \mathrm{\Omega }\times {\mathbb{R}}^N$ and the following growth condition: there exist $1<q<p$, $b_1>0$ and a continuous function $f_0$ on $\mathrm{\Omega }\times [0,\mathrm{\infty })$ such that

  $$-b_1(1+t^{q-1})\le f_0(x,t)\le f(x,t,\xi )\le b_1(1+t^{q-1}+{|\xi |}^{q-1})$$

  (1)

for every $(x,t,\xi )\in \mathrm{\Omega }\times [0,\mathrm{\infty })\times {\mathbb{R}}^N$.

In this paper, we say that $u\in W_0^{1,p}(\mathrm{\Omega })$ is a (weak) solution of (P) if

$${\int }_{\mathrm{\Omega }}A(x,\mathrm{\nabla }u)\mathrm{\nabla }\phi dx={\int }_{\mathrm{\Omega }}f(x,u,\mathrm{\nabla }u)\phi dx$$

for all $\phi \in W_0^{1,p}(\mathrm{\Omega })$.

A similar hypothesis to (A) is considered in the study of quasilinear elliptic problems (see [[1], Example 2.2.], [2–5] and also refer to [6, 7] for the generalized p-Laplace operators). From now on, we assume that $C_0\le p-1\le C_1$, which is without any loss of generality as can be seen from assumptions (A)(ii), (iii).

In particular, for $A(x,y)={|y|}^{p-2}y$, that is, $divA(x,\mathrm{\nabla }u)$ stands for the usual p-Laplacian ${\mathrm{\Delta }}_{p}u$, we can take $C_0=C_1=p-1$ in (A). Conversely, in the case where $C_0=C_1=p-1$ holds in (A), by the inequalities in Remark 3(ii) and (iii), we see that $a(x,t)={|t|}^{p-2}$ whence $A(x,y)={|y|}^{p-2}y$. Hence, our equation contains the p-Laplace equation as a special case.

In the case where f does not depend on the gradient of u, there are many existence results because our equation has the variational structure (cf. [1, 4, 8]). Although there are a few results for our equation (P) with f including ∇u, we can refer to [7, 9] and [10] for the existence of a positive solution in the case of the $(p,q)$-Laplacian or m-Laplacian ($1<m<N$). In particular, in [9] and [7], the nonlinear term f is imposed to be nonnegative. The results in [7] and [10] are applied to the m-Laplace equation with an $(m-1)$-superlinear term f w.r.t. u. Here, we mention the result in [9] for the p-Laplacian. Faria, Miyagaki and Motreanu considered the case where f is $(p-1)$-sublinear w.r.t. u and ∇u, and they supposed that $f(x,u,\mathrm{\nabla }u)\ge c{u}^r$ for some $c>0$ and $0<r<p-1$. The purpose of this paper is to remove the sign condition and to admit the condition like $f(x,u,\mathrm{\nabla }u)\ge \lambda u^{p-1}+o(u^{p-1})$ for large $\lambda >0$ as $u\to 0+$. Concerning the condition for f as $|u|\to 0$, Zou in [10] imposed that there exists an $L>0$ satisfying $f(x,u,\mathrm{\nabla }u)=L{u}^{m-1}+o({|u|}^{m-1}+{|\mathrm{\nabla }u|}^{m-1})$ as $|u|,|\mathrm{\nabla }u|\to 0$ for the m-Laplace problem. Hence, we cannot apply the result of [10] and [9] to the case of $f(x,u,\mathrm{\nabla }u)=\lambda m(x)u^{p-1}+(1-u^{p-1}){|\mathrm{\nabla }u|}^{r-1}+o(u^{p-1})$ as $u\to 0+$ for $1<r<p$ and $m\in L^{\mathrm{\infty }}(\mathrm{\Omega })$ (admitting sign changes), but we can do our result if $\lambda >0$ is large.

In [9], the positivity of a solution is proved by the comparison principle. However, since we are not able to do it for our operator in general, after we provide a non-negative and non-trivial solution as a limit of positive approximate solutions (in Section 2), we obtain the positivity of it due to the strong maximum principle for our operator.

1.1 Statements

To state our first result, we define a positive constant $A_p$ by

$$A_p:=\frac{C_1}{p-1}{\left(\frac{C_1}{C_0}\right)}^{p-1}\ge 1,$$

(2)

which is equal to 1 in the case of $A(x,y)={|y|}^{p-2}y$ (i.e., the case of the p-Laplacian) because we can choose $C_0=C_1=p-1$. Then, we introduce the hypothesis (f1) to the function $f_0(x,t)$ in (f) as t is small.

(f1) There exist $m\in L^{\mathrm{\infty }}(\mathrm{\Omega })$ and $b_0>{\mu }_1(m)A_p$ such that the Lebesgue measure of $\{x\in \mathrm{\Omega };m(x)>0\}$ is positive and

$$\underset{t\to 0+}{lim\hspace{0.17em}inf}\frac{f_0(x,t)}{t^{p-1}}\ge b_{0}m(x)\text{uniformly in }x\in \mathrm{\Omega },$$

(3)

where $f_0$ is the continuous function in (f) and ${\mu }_1(m)$ is the first positive eigenvalue of the p-Laplacian with the weight function m obtained by

$${\mu }_1(m):=inf\{{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }u|}^{p}dx;u\in W_0^{1,p}(\mathrm{\Omega })\text{ and }{\int }_{\mathrm{\Omega }}m{|u|}^{p}dx=1\}.$$

(4)

Theorem 1 Assume (f1). Then equation (P) has a positive solution $u\in intP$, where

$$\begin{array}{c}P:=\{u\in C_0^1(\overline{\mathrm{\Omega }});u(x)\ge 0\mathit{\text{in}}\mathrm{\Omega }\},\hfill \\ intP:=\{u\in C_0^1(\overline{\mathrm{\Omega }});u(x)>0\mathit{\text{in}}\mathrm{\Omega }\mathit{\text{and}}\partial u/\partial \nu <0\mathit{\text{on}}\partial \mathrm{\Omega }\},\hfill \end{array}$$

and ν denotes the outward unit normal vector on ∂ Ω.

Next, we consider the case where A is asymptotically $(p-1)$-homogeneous near zero in the following sense:

(AH0) There exist a positive function $a_0\in C(\overline{\mathrm{\Omega }},(0,+\mathrm{\infty }))$ and ${\tilde{a}}_0(x,t)\in C(\overline{\mathrm{\Omega }}\times [0,+\mathrm{\infty }),\mathbb{R})$ such that

$$A(x,y)=a_0(x){|y|}^{p-2}y+{\tilde{a}}_0(x,|y|)y\text{for every }x\in \mathrm{\Omega },y\in {\mathbb{R}}^N\text{and}$$

(5)

$$\underset{t\to 0+}{lim}\frac{{\tilde{a}}_0(x,t)}{t^{p-2}}=0\text{uniformly in }x\in \overline{\mathrm{\Omega }}.$$

(6)

Under (AH0), we can replace the hypothesis (f1) with the following (f2):

(f2) There exist $m\in L^{\mathrm{\infty }}(\mathrm{\Omega })$ and $b_0>{\lambda }_1(m)$ such that (3) and the Lebesgue measure of $\{x\in \mathrm{\Omega };m(x)>0\}$ is positive, where ${\lambda }_1(m)$ is the first positive eigenvalue of $-div(a_0(x){|\mathrm{\nabla }u|}^{p-2}\mathrm{\nabla }u)$ with a weight function m obtained by

$${\lambda }_1(m):=inf\{{\int }_{\mathrm{\Omega }}{a}_0(x){|\mathrm{\nabla }u|}^{p}dx;u\in W_0^{1,p}(\mathrm{\Omega })\text{ and }{\int }_{\mathrm{\Omega }}m{|u|}^{p}dx=1\}.$$

(7)

Theorem 2 Assume (AH0) and (f2). Then equation (P) has a positive solution $u\in intP$.

Throughout this paper, we may assume that $f(x,t,\xi )=0$ for every $t\le 0$, $x\in \mathrm{\Omega }$ and $\xi \in {\mathbb{R}}^N$ because we consider the existence of a positive solution only. In what follows, the norm on $W_0^{1,p}(\mathrm{\Omega })$ is given by $\parallel u\parallel :={\parallel \mathrm{\nabla }u\parallel }_p$, where ${\parallel u\parallel }_q$ denotes the usual norm of $L^q(\mathrm{\Omega })$ for $u\in L^q(\mathrm{\Omega })$ ($1\le q\le \mathrm{\infty }$). Moreover, we denote $u_{\pm }:=max\{\pm u,0\}$.

1.2 Properties of the map A

Remark 3 The following assertions hold under condition (A):

• (i) for all $x\in \overline{\mathrm{\Omega }}$, $A(x,y)$ is maximal monotone and strictly monotone in y;

• (ii) $|A(x,y)|\le \frac{C_1}{p-1}{|y|}^{p-1}$ for every $(x,y)\in \overline{\mathrm{\Omega }}\times {\mathbb{R}}^N$;

• (iii) $A(x,y)y\ge \frac{C_0}{p-1}{|y|}^p$ for every $(x,y)\in \overline{\mathrm{\Omega }}\times {\mathbb{R}}^N$,

where $C_0$ and $C_1$ are the positive constants in (A).

Proposition 4 ([[3], Proposition 1])

Let $A:W_0^{1,p}(\mathrm{\Omega })\to W_0^{1,p}{(\mathrm{\Omega })}^{\ast }$ be a map defined by

$$〈A(u),v〉={\int }_{\mathrm{\Omega }}A(x,\mathrm{\nabla }u)\mathrm{\nabla }vdx$$

for $u,v\in W_0^{1,p}(\mathrm{\Omega })$. Then A is maximal monotone, strictly monotone and has ${(S)}_{+}$ property, that is, any sequence $\{u_n\}$ weakly convergent to u with ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈A(u_n),u_n-u〉\le 0$ strongly converges to u.

2 Constructing approximate solutions

Choose a function $\psi \in P\setminus \{0\}$. In this section, for such ψ and $\epsilon >0$, we consider the following elliptic equation:

In [7], the case $\psi \equiv 1$ in the above equation is considered.

Lemma 5 Suppose (f1) or (f2). Then there exists ${\lambda }_0>0$ such that $f(x,t,\xi )t+{\lambda }_0{t}^p\ge 0$ for every $x\in \mathrm{\Omega }$, $t\ge 0$ and $\xi \in {\mathbb{R}}^N$.

Proof From the growth condition of $f_0$ and (3), it follows that

$$f_0(x,t)t\ge -b_0{\parallel m\parallel }_{\mathrm{\infty }}{t}^p-b_1^{\mathrm{\prime }}{t}^p\text{for every }(x,t)\in \mathrm{\Omega }\times [0,\mathrm{\infty })$$

holds, where $b_1^{\mathrm{\prime }}$ is a positive constant independent of $(x,t)$. Therefore, for ${\lambda }_0\ge b_0{\parallel m\parallel }_{\mathrm{\infty }}+b_1^{\mathrm{\prime }}$, we easily see that $f(x,t,\xi )t+{\lambda }_0{t}^p\ge f_0(x,t)t+{\lambda }_0{t}^p\ge 0$ for every $x\in \mathrm{\Omega }$, $t\ge 0$ and $\xi \in {\mathbb{R}}^N$ holds. □

Proposition 6 If $u_{\epsilon }\in W_0^{1,p}(\mathrm{\Omega })$ is a non-negative solution of ($P;\epsilon$) for $\epsilon \ge 0$, then $u_{\epsilon }\in L^{\mathrm{\infty }}(\mathrm{\Omega })$. Moreover, for any ${\epsilon }_0>0$, there exists a positive constant $D>0$ such that ${\parallel u_{\epsilon }\parallel }_{\mathrm{\infty }}\le Dmax\{1,\parallel u_{\epsilon }\parallel \}$ holds for every $\epsilon \in [0,{\epsilon }_0]$.

Proof Set ${\overline{p}}^{\ast }=Np/(N-p)$ if $N>p$, and in the case of $N\le p$, ${\overline{p}}^{\ast }>p$ is an arbitrarily fixed constant. Let $u_{\epsilon }$ be a non-negative solution of ($P;\epsilon$) with $0\le \epsilon \le {\epsilon }_0$ (some ${\epsilon }_0>0$). For $r>0$, choose a smooth increasing function $\eta (t)$ such that $\eta (t)=t^{r+1}$ if $0\le t\le 1$, $\eta (t)=d_{0}t$ if $t\ge d_1$ and ${\eta }^{\mathrm{\prime }}(t)\ge d_2>0$ if $1\le t\le d_1$ for some $0<d_2<1<d_0,d_1$. Define ${\xi }_M(u):=M^{r+1}\eta (u/M)$ for $M>1$.

If $u_{\epsilon }\in L^{r+p}(\mathrm{\Omega })$, then by taking ${\xi }_M(u_{\epsilon })$ as a test function (note that ${\eta }^{\mathrm{\prime }}$ is bounded), we have

$$\begin{array}{r}\frac{C_0}{p-1}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }{u}_{\epsilon }|}^p{\xi }_M^{\mathrm{\prime }}(u_{\epsilon })dx\\ \le {\int }_{\mathrm{\Omega }}A(x,\mathrm{\nabla }{u}_{\epsilon })\mathrm{\nabla }{u}_{\epsilon }{\xi }_M^{\mathrm{\prime }}(u_{\epsilon })dx\\ ={\int }_{\mathrm{\Omega }}(f(x,u_{\epsilon },\mathrm{\nabla }{u}_{\epsilon })+\epsilon \psi ){\xi }_M(u_{\epsilon })dx\\ \le b_1{\int }_{\mathrm{\Omega }}(1+u_{\epsilon }^{q-1}+{\epsilon }_0{\parallel \psi \parallel }_{\mathrm{\infty }})M^{r+1}\eta (u_{\epsilon }/M)dx+b_1{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }{u}_{\epsilon }|}^{q-1}{\xi }_M(u_{\epsilon })dx\\ \le d_0{d}_1(2b_1+{\epsilon }_0{\parallel \psi \parallel }_{\mathrm{\infty }})({\parallel u_{\epsilon }\parallel }_{r+q}^{r+q}+{\parallel u_{\epsilon }\parallel }_{r+1}^{r+1})+b_1{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }{u}_{\epsilon }|}^{q-1}{\xi }_M(u_{\epsilon })dx\end{array}$$

(8)

due to Remark 3(iii) and $M^{r+1}\eta (t/M)\le d_0{d}_1{t}^{r+1}$. Putting $\beta :=p/(p-q+1)<p$, we see that $({\xi }_M(u_{\epsilon }))/{({\xi }_M^{\mathrm{\prime }}(u_{\epsilon }))}^{(q-1)/p}=u_{\epsilon }^{r+1}/{((r+1)u_{\epsilon }^r)}^{(q-1)/p}\le u_{\epsilon }^{1+r/\beta }$ provided $0<u_{\epsilon }<M$ (note $r>0$). Similarly, if $M\le u_{\epsilon }\le d_{1}M$, then $({\xi }_M(u_{\epsilon }))/{({\xi }_M^{\mathrm{\prime }}(u_{\epsilon }))}^{(q-1)/p}\le d_0{d}_1{M}^{r+1}/{(d_2{M}^r)}^{(q-1)/p}=d_0{d}_1{d}_2^{(1-q)/p}{M}^{1+r/\beta }\le d_0{d}_1{d}_2^{(1-q)/p}{u}_{\epsilon }^{1+r/\beta }$, and if $u_{\epsilon }>d_{1}M$, then $({\xi }_M(u_{\epsilon }))/{({\xi }_M^{\mathrm{\prime }}(u_{\epsilon }))}^{(q-1)/p}=d_0^{1/\beta }{M}^{r/\beta }{u}_{\epsilon }\le d_0^{1/\beta }{u}_{\epsilon }^{1+r/\beta }$ (note $d_1>1$). Thus, according to Young’s inequality, for every $\delta >0$, there exists $C_{\delta }>0$ such that

$$\begin{array}{rl}{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }{u}_{\epsilon }|}^{q-1}{\xi }_M(u_{\epsilon })dx& \le \delta {\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }{u}_{\epsilon }|}^p{\xi }_M^{\mathrm{\prime }}(u_{\epsilon })dx+C_{\delta }{\int }_{u_{\epsilon }>0}\frac{{({\xi }_M(u_{\epsilon }))}^{\beta }}{{({\xi }_M^{\mathrm{\prime }}(u_{\epsilon }))}^{(q-1)\beta /p}}dx\\ \le \delta {\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }{u}_{\epsilon }|}^p{\xi }_M^{\mathrm{\prime }}(u_{\epsilon })dx+C_{\delta }{d}_3{\int }_{\mathrm{\Omega }}{u}_{\epsilon }^{r+\beta }dx,\end{array}$$

(9)

where $\beta :=p/(p-q+1)<p$ and $d_3=max\{d_0{d}_1{d}_2^{(1-q)/p},d_0^{1/\beta }\}$ (>1). As a result, because of $r+p>r+q,r+\beta$, according to Hölder’s inequality and the monotonicity of $t^r$ with respect to r on $[1,\mathrm{\infty })$, taking a $0<\delta <C_0/b_1(p-1)$ and setting $u_{\epsilon }^M(x):=min\{u_{\epsilon }(x),M\}$, we obtain

$$\begin{array}{rl}{b}_4{\left(r^{\mathrm{\prime }}\right)}^{p}max\{1,{\parallel u_{\epsilon }\parallel }_{r+p}^{r+p}\}& \ge {\left(r^{\mathrm{\prime }}\right)}^p{\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }{u}_{\epsilon }|}^p{\xi }_M^{\mathrm{\prime }}(u_{\epsilon })dx\ge {\left(r^{\mathrm{\prime }}\right)}^p{\int }_{\mathrm{\Omega }}{\left|\mathrm{\nabla }{u}_{\epsilon }^M\right|}^p{\left(u_{\epsilon }^M\right)}^{r}dx\\ ={\parallel {\left(u_{\epsilon }^M\right)}^{r^{\mathrm{\prime }}}\parallel }^p\ge C_{\ast }{\parallel {\left(u_{\epsilon }^M\right)}^{r^{\mathrm{\prime }}}\parallel }_{{\overline{p}}^{\ast }}^p=C_{\ast }{\parallel u_{\epsilon }^M\parallel }_{{\overline{p}}^{\ast }{r}^{\mathrm{\prime }}}^{r+p}\end{array}$$

(10)

provided $u_{\epsilon }\in L^{r+p}(\mathrm{\Omega })$ by (8) and (9), where $r^{\mathrm{\prime }}=1+r/p$, $C_{\ast }$ comes from the continuous embedding of $W_0^{1,p}(\mathrm{\Omega })$ into $L^{{\overline{p}}^{\ast }}(\mathrm{\Omega })$ and $d_4$ is a positive constant independent of $u_{\epsilon }$, ε and r. Consequently, Moser’s iteration process implies our conclusion. In fact, we define a sequence ${\{r_m\}}_m$ by $r_0:={\overline{p}}^{\ast }-p$ and $r_{m+1}:={\overline{p}}^{\ast }(p+r_m)/p-p$. Then, we see that $u_{\epsilon }\in L^{{\overline{p}}^{\ast }(p+r_m)/p}(\mathrm{\Omega })=L^{p+r_{m+1}}(\mathrm{\Omega })$ holds if $u_{\epsilon }\in L^{p+r_m}(\mathrm{\Omega })$ by applying Fatou’s lemma to (10) and letting $M\to \mathrm{\infty }$. Here, we also see $r_{m+1}={\overline{p}}^{\ast }{r}_m/p+{\overline{p}}^{\ast }-p\ge {({\overline{p}}^{\ast }/p)}^{m+1}{r}_0\to \mathrm{\infty }$ as $m\to \mathrm{\infty }$. Therefore, by the same argument as in Theorem C in [4], we can obtain $u_{\epsilon }\in L^{\mathrm{\infty }}(\mathrm{\Omega })$ and ${\parallel u_{\epsilon }\parallel }_{\mathrm{\infty }}\le Dmax\{1,\parallel u_{\epsilon }\parallel \}$ for some positive constant D independent of $u_{\epsilon }$ and ε. □

Lemma 7 Suppose (f1) or (f2). If $u_{\epsilon }\in W_0^{1,p}(\mathrm{\Omega })$ is a solution of ($P;\epsilon$) for $\epsilon >0$, then $u_{\epsilon }\in intP$.

Proof Taking $-{(u_{\epsilon })}_{-}$ as a test function in ($P;\epsilon$), we have

$$\frac{C_0}{p-1}{\parallel \mathrm{\nabla }{(u_{\epsilon })}_{-}\parallel }_p^p\le {\int }_{\mathrm{\Omega }}A(x,\mathrm{\nabla }{u}_{\epsilon })(-\mathrm{\nabla }{(u_{\epsilon })}_{-})dx=-\epsilon {\int }_{\mathrm{\Omega }}\psi {(u_{\epsilon })}_{-}dx\le 0$$

because of $f(x,t,\xi )=0$ if $t\le 0$ and by Remark 3(iii). Hence, $u_{\epsilon }\ge 0$ follows. Because Proposition 6 guarantees that $u_{\epsilon }\in L^{\mathrm{\infty }}(\mathrm{\Omega })$, we have $u_{\epsilon }\in C_0^{1,\alpha }(\overline{\mathrm{\Omega }})$ (for some $0<\alpha <1$) by the regularity result in [11]. Note that $u_{\epsilon }\not\equiv 0$ because of $\epsilon >0$ and $\psi \not\equiv 0$. In addition, Lemma 5 implies the existence of ${\lambda }_0>0$ such that $-divA(x,\mathrm{\nabla }{u}_{\epsilon })+{\lambda }_0{u}_{\epsilon }^{p-1}\ge 0$ in the distribution sense. Therefore, according to Theorem A and Theorem B in [4], $u_{\epsilon }>0$ in Ω and $\partial u_{\epsilon }/\partial \nu <0$ on ∂ Ω, namely, $u_{\epsilon }\in intP$. □

The following result can be shown by the same argument as in [[9], Theorem 3.1].

Proposition 8 Suppose (f1) or (f2). Then, for every $\epsilon >0$, ($P;\epsilon$) has a positive solution $u_{\epsilon }\in intP$.

Proof Fix any $\epsilon >0$ and let $\{e_1,\dots ,e_m,\dots \}$ be a Schauder basis of $W_0^{1,p}(\mathrm{\Omega })$ (refer to [12] for the existence). For each $m\in \mathbb{N}$, we define the m-dimensional subspace $V_m$ of $W_0^{1,p}(\mathrm{\Omega })$ by $V_m:=\text{lin.sp.}\{e_1,\dots ,e_m\}$. Moreover, set a linear isomorphism $T_m:{\mathbb{R}}^m\to V_m$ by $T_m({\xi }_1,\dots ,{\xi }_m):={\sum }_{i=1}^m{\xi }_i{e}_i\in V_m$, and let $T_m^{\ast }:V_m^{\ast }\to {({\mathbb{R}}^m)}^{\ast }$ be a dual map of $T_m$. By identifying ${\mathbb{R}}^m$ and ${({\mathbb{R}}^m)}^{\ast }$, we may consider that $T_m^{\ast }$ maps from $V_m^{\ast }$ to ${\mathbb{R}}^m$. Define maps $A_m$ and $B_m$ from $V_m$ to $V_m^{\ast }$ as follows:

$$〈A_m(u),v〉:={\int }_{\mathrm{\Omega }}A(x,\mathrm{\nabla }u)\mathrm{\nabla }vdx\text{and}〈B_m(u),v〉:={\int }_{\mathrm{\Omega }}f(x,u,\mathrm{\nabla }u)vdx+\epsilon {\int }_{\mathrm{\Omega }}\psi vdx$$

for u, $v\in V_m$. We claim that for every $m\in \mathbb{N}$, there exists $u_m\in V_m$ such that $A_m(u_m)-B_m(u_m)=0$ in $V_m^{\ast }$. Indeed, by the growth condition of f, Remark 3(iii) and Hölder’s inequality, we easily have

$$\begin{array}{r}〈A_m(u)-B_m(u),u〉\\ \ge \frac{C_0}{p-1}{\parallel u\parallel }^p-b_1({\parallel u\parallel }_1+{\parallel u\parallel }_q^q+{\parallel \mathrm{\nabla }u\parallel }_p^{q-1}{\parallel u\parallel }_{\beta })-\epsilon {\parallel \psi \parallel }_{\mathrm{\infty }}{\parallel u\parallel }_1\end{array}$$

(11)

for every $u\in V_m$, where $\beta =p/(p-q+1)<p$. This implies that $A_m-B_m$ is coercive on $V_m$ by $q<p$. Set a homotopy $H_m(t,y):=ty+(1-t)T_m^{\ast }(A_m(T_m(y))-B_m(T_m(y)))$ for $t\in [0,1]$ and $y\in {\mathbb{R}}^m$. By recalling that $A_m-B_m$ is coercive on $V_m$, we see that there exists an $R>0$ such that $(H_m(t,y),y)>0$ for every $t\in [0,1]$ and $|y|\ge R$ because $\parallel \cdot \parallel$ and the norm of ${\mathbb{R}}^m$ are equivalent on $V_m$. Therefore, we have

$$\begin{array}{rl}1& =deg(I_m,B_R(0),0)=deg(H_m(1,\cdot ),B_R(0),0)\\ =deg(H_m(0,\cdot ),B_R(0),0)=deg(T_m^{\ast }\circ (A_m-B_m)\circ T_m,B_R(0),0),\end{array}$$

where $I_m$ is the identity map on ${\mathbb{R}}^m$, $B_R(0):=\{y\in {\mathbb{R}}^m;|y|<R\}$ and $deg(g,B,0)$ denotes the degree on ${\mathbb{R}}^m$ for a continuous map $g:B\to {\mathbb{R}}^m$ (cf. [13]). Hence, this yields the existence of $y_m\in {\mathbb{R}}^m$ such that $(T_m^{\ast }\circ (A_m-B_m)\circ T_m)(y_m)=0$, and so the desired $u_m$ is obtained by setting $u_m=T_m(y_m)\in V_m$ since $T_m^{\ast }$ is injective.

Because (11) with $u=u_m\in W_0^{1,p}(\mathrm{\Omega })$ leads to the boundedness of $\parallel u_m\parallel$ by $q<p$, we may assume, by choosing a subsequence, that $u_m$ converges to some $u_0$ weakly in $W_0^{1,p}(\mathrm{\Omega })$ and strongly in $L^p(\mathrm{\Omega })$. Let $P_m$ be a natural projection onto $V_m$, that is, $P_{m}u={\sum }_{i=1}^m{\xi }_i{e}_i$ for $u={\sum }_{i=1}^{\mathrm{\infty }}{\xi }_i{e}_i$. Since $u_m,P_m{u}_0\in V_m$ and $A_m(u_m)-B_m(u_m)=0$ in $V_m^{\ast }$, by noting that $A_m=A$ on $V_m$ for a map A defined in Proposition 4, we obtain

$$\begin{array}{r}〈A(u_m),u_m-u_0〉+〈A(u_m),u_0-P_m{u}_0〉\\ =〈A_m(u_m),u_m-P_m{u}_0〉=〈B_m(u_m),u_m-P_m{u}_0〉\\ ={\int }_{\mathrm{\Omega }}(f(x,u_m,\mathrm{\nabla }{u}_m)+\epsilon \psi )(u_m-u_0)dx\\ +{\int }_{\mathrm{\Omega }}(f(x,u_m,\mathrm{\nabla }{u}_m)+\epsilon \psi )(u_0-P_m{u}_0)dx\to 0\end{array}$$

as $m\to \mathrm{\infty }$, where we use the boundedness of $\parallel u_m\parallel$, the growth condition of f and $u_m\to u_0$ in $L^p(\mathrm{\Omega })$. In addition, since ${\parallel A(u_m)\parallel }_{W_0^{1,p}{(\mathrm{\Omega })}^{\ast }}$ is bounded, by the boundedness of $\parallel u_m\parallel$, we see that $〈A(u_m),u_0-P_m{u}_0〉\to 0$ as $m\to \mathrm{\infty }$, whence $〈A(u_m),u_m-u_0〉\to 0$ as $m\to \mathrm{\infty }$ holds. As a result, it follows from the ${(S)}_{+}$ property of A that $u_m\to u_0$ in $W_0^{1,p}(\mathrm{\Omega })$ as $m\to \mathrm{\infty }$.

Finally, we shall prove that $u_0$ is a solution of ($P;\epsilon$). Fix any $l\in \mathbb{N}$ and $\phi \in V_l$. For each $m\ge l$, by letting $m\to \mathrm{\infty }$ in $〈A_m(u_m),\phi 〉=〈B_m(u_m),\phi 〉$, we have

$${\int }_{\mathrm{\Omega }}A(x,\mathrm{\nabla }{u}_0)\mathrm{\nabla }\phi dx={\int }_{\mathrm{\Omega }}f(x,u_0,\mathrm{\nabla }{u}_0)\phi dx+\epsilon {\int }_{\mathrm{\Omega }}\psi \phi dx.$$

(12)

Since l is arbitrary, (12) holds for every $\phi \in {\bigcup }_{l\ge 1}{V}_l$. Moreover, the density of ${\bigcup }_{l\ge 1}{V}_l$ in $W_0^{1,p}(\mathrm{\Omega })$ guarantees that (12) holds for every $\phi \in W_0^{1,p}(\mathrm{\Omega })$. This means that $u_0$ is a solution of ($P;\epsilon$). Consequently, our conclusion $u_0\in intP$ follows from Lemma 7. □

3 Proof of theorems

Lemma 9 Let $\phi ,u\in intP$. Then

$${\int }_{\mathrm{\Omega }}A(x,\mathrm{\nabla }u)\mathrm{\nabla }\left(\frac{{\phi }^p}{u^{p-1}}\right)dx\le A_p{\parallel \mathrm{\nabla }\phi \parallel }_p^p$$

holds, where $A_p$ is the positive constant defined by (2).

Proof Because of $\phi ,u\in intP$, there exist ${\delta }_1>{\delta }_2>0$ such that ${\delta }_{1}u\ge \phi \ge {\delta }_{2}u$ in $\overline{\mathrm{\Omega }}$. Thus, ${\delta }_1\ge \phi /u\ge {\delta }_2$ and $1/{\delta }_2\ge u/\phi \ge 1/{\delta }_1$ in Ω. Hence, $u/\phi ,\phi /u\in L^{\mathrm{\infty }}(\mathrm{\Omega })$ hold. Therefore, we have

$$\begin{array}{rl}A(x,\mathrm{\nabla }u)\mathrm{\nabla }\left(\frac{{\phi }^p}{u^{p-1}}\right)=& p{\left(\frac{\phi }{u}\right)}^{p-1}A(x,\mathrm{\nabla }u)\mathrm{\nabla }\phi -(p-1){\left(\frac{\phi }{u}\right)}^{p}A(x,\mathrm{\nabla }u)\mathrm{\nabla }u\\ \le & \frac{p{C}_1}{p-1}{\left(\frac{\phi }{u}\right)}^{p-1}{|\mathrm{\nabla }u|}^{p-1}|\mathrm{\nabla }\phi |-C_0{\left(\frac{\phi }{u}\right)}^p{|\mathrm{\nabla }u|}^p\\ =& {\{{\left(\frac{p{C}_0}{p-1}\right)}^{1/p}\frac{\phi }{u}|\mathrm{\nabla }u|\}}^{p-1}{\left(\frac{p}{p-1}\right)}^{1/p}{C}_1{C}_0^{(1-p)/p}|\mathrm{\nabla }\phi |\\ -C_0{\left(\frac{\phi }{u}\right)}^p{|\mathrm{\nabla }u|}^p\le A_p{|\mathrm{\nabla }\phi |}^p\end{array}$$

(13)

in Ω by (ii) and (iii) in Remark 3 and Young’s inequality. □

Lemma 10 Assume that $a_0\in C(\overline{\mathrm{\Omega }},[0,\mathrm{\infty }))$ and let $\phi ,u\in intP$. Then

$${\int }_{\mathrm{\Omega }}{a}_0(x){|\mathrm{\nabla }\phi |}^{p-2}\mathrm{\nabla }\phi \mathrm{\nabla }\left(\frac{{\phi }^p-u^p}{{\phi }^{p-1}}\right)dx-{\int }_{\mathrm{\Omega }}{a}_0(x){|\mathrm{\nabla }u|}^{p-2}\mathrm{\nabla }u\mathrm{\nabla }\left(\frac{{\phi }^p-u^p}{u^{p-1}}\right)dx\ge 0$$

holds.

Proof First, we note that $u/\phi ,\phi /u\in L^{\mathrm{\infty }}(\mathrm{\Omega })$ hold by the same reason as in Lemma 9. Applying Young’s inequality to the second term of the right-hand side in (14) (refer to (13) with $C_0=C_1=p-1$), we obtain

$$\begin{array}{r}{a}_0(x){|\mathrm{\nabla }\phi |}^{p-2}\mathrm{\nabla }\phi \mathrm{\nabla }\left(\frac{{\phi }^p-u^p}{{\phi }^{p-1}}\right)\\ \ge a_0(x)({|\mathrm{\nabla }\phi |}^p-p{\left(\frac{u}{\phi }\right)}^{p-1}{|\mathrm{\nabla }\phi |}^{p-1}|\mathrm{\nabla }u|+(p-1){\left(\frac{u}{\phi }\right)}^p{|\mathrm{\nabla }\phi |}^p)\end{array}$$

(14)

$$\ge a_0(x)({|\mathrm{\nabla }\phi |}^p-{|\mathrm{\nabla }u|}^p)$$

(15)

in Ω. Similarly, we also have

$$a_0(x){|\mathrm{\nabla }u|}^{p-2}\mathrm{\nabla }u\mathrm{\nabla }\left(\frac{{\phi }^p-u^p}{u^{p-1}}\right)\le a_0(x)({|\mathrm{\nabla }\phi |}^p-{|\mathrm{\nabla }u|}^p)\text{in }\mathrm{\Omega }.$$

(16)

The conclusion follows from (15) and (16). □

Under (f1) or (f2), we denote a solution $u_{\epsilon }\in intP$ of ($P;\epsilon$) for each $\epsilon >0$ obtained by Proposition 8.

Lemma 11 Assume (f1) or (f2). Let $I:=(0,1]$. Then ${\{u_{\epsilon }\}}_{\epsilon \in I}$ is bounded in $W_0^{1,p}(\mathrm{\Omega })$.

Proof Taking $u_{\epsilon }$ as a test function in ($P;\epsilon$), we have

$$\begin{array}{rl}\frac{C_0}{p-1}{\parallel \mathrm{\nabla }{u}_{\epsilon }\parallel }_p^p& \le {\int }_{\mathrm{\Omega }}A(x,\mathrm{\nabla }{u}_{\epsilon })\mathrm{\nabla }{u}_{\epsilon }dx={\int }_{\mathrm{\Omega }}f(x,u_{\epsilon },\mathrm{\nabla }{u}_{\epsilon })u_{\epsilon }dx+\epsilon {\int }_{\mathrm{\Omega }}\psi u_{\epsilon }dx\\ \le b_1({\parallel u_{\epsilon }\parallel }_1+{\parallel u_{\epsilon }\parallel }_q^q+{\parallel \mathrm{\nabla }{u}_{\epsilon }\parallel }_p^{q-1}{\parallel u_{\epsilon }\parallel }_{\beta })+{\parallel \psi \parallel }_{\mathrm{\infty }}{\parallel u_{\epsilon }\parallel }_1\\ \le b_1^{\mathrm{\prime }}(\parallel u_{\epsilon }\parallel +{\parallel u_{\epsilon }\parallel }^q)\end{array}$$

by Remark 3(iii), the growth condition of f, Hölder’s inequality and the continuity of the embedding of $W_0^{1,p}(\mathrm{\Omega })$ into $L^p(\mathrm{\Omega })$, where $\beta =p/(p-q+1)$ (<p) and $b_1^{\mathrm{\prime }}$ is a positive constant independent of $u_{\epsilon }$. Because of $q<p$, this yields the boundedness of $\parallel u_{\epsilon }\parallel$ ($={\parallel \mathrm{\nabla }{u}_{\epsilon }\parallel }_p$). □

Lemma 12 Assume (f1) or (f2). Then $|\mathrm{\nabla }{u}_{\epsilon }|/u_{\epsilon }\in L^p(\mathrm{\Omega })$ and ${\parallel |\mathrm{\nabla }{u}_{\epsilon }|/u_{\epsilon }\parallel }_p^p\le {\lambda }_0|\mathrm{\Omega }|/C_0$ hold for every $\epsilon >0$, where $|\mathrm{\Omega }|$ denotes the Lebesgue measure of Ω, and where $C_0$ and ${\lambda }_0$ are positive constants as in (A) and Lemma 5, respectively.

Proof Fix any $\epsilon >0$ and choose any $\rho >0$. By taking ${(u_{\epsilon }+\rho )}^{1-p}$ as a test function, we obtain

$$\begin{array}{rl}(1-p){\int }_{\mathrm{\Omega }}\frac{A(x,\mathrm{\nabla }{u}_{\epsilon })\mathrm{\nabla }{u}_{\epsilon }}{{(u_{\epsilon }+\rho )}^p}dx& ={\int }_{\mathrm{\Omega }}\frac{f(x,u_{\epsilon },\mathrm{\nabla }{u}_{\epsilon })+\epsilon \psi }{{(u_{\epsilon }+\rho )}^{p-1}}dx\ge -{\lambda }_0{\int }_{\mathrm{\Omega }}\frac{u_{\epsilon }^{p-1}}{{(u_{\epsilon }+\rho )}^{p-1}}dx\\ \ge -{\lambda }_0|\mathrm{\Omega }|,\end{array}$$

(17)

by Lemma 5 and $\epsilon \psi \ge 0$. On the other hand, by Remark 3(iii) and $1-p<0$, we have

$$(1-p){\int }_{\mathrm{\Omega }}\frac{A(x,\mathrm{\nabla }{u}_{\epsilon })\mathrm{\nabla }{u}_{\epsilon }}{{(u_{\epsilon }+\rho )}^p}dx\le -C_0{\int }_{\mathrm{\Omega }}\frac{{|\mathrm{\nabla }{u}_{\epsilon }|}^p}{{(u_{\epsilon }+\rho )}^p}dx.$$

(18)

Therefore, (17) and (18) imply the inequality ${\int }_{\mathrm{\Omega }}{|\mathrm{\nabla }{u}_{\epsilon }|}^p/{(u_{\epsilon }+\rho )}^{p}dx\le {\lambda }_0|\mathrm{\Omega }|/C_0$ for every $\rho >0$. As a result, by letting $\rho \to 0+$, our conclusion is shown. □

Lemma 13 Assume (f2) and (AH0). Let $\phi \in intP$. If $u_{\epsilon }\to 0$ in $C_0^1(\overline{\mathrm{\Omega }})$ as $\epsilon \to 0+$, then

$$\underset{\epsilon \to 0+}{lim}\left|{\int }_{\mathrm{\Omega }}{\tilde{a}}_0(x,|\mathrm{\nabla }{u}_{\epsilon }|)\mathrm{\nabla }{u}_{\epsilon }\mathrm{\nabla }\left(\frac{{\phi }^p-u_{\epsilon }^p}{u_{\epsilon }^{p-1}}\right)dx\right|=0$$

holds, where ${\tilde{a}}_0$ is a continuous function as in (AH0).

Proof Note that $u_{\epsilon }/\phi ,\phi /u_{\epsilon }\in L^{\mathrm{\infty }}(\mathrm{\Omega })$ hold (as in the proof of Lemma 9). Because we easily see that $|{\int }_{\mathrm{\Omega }}{\tilde{a}}_0(x,|\mathrm{\nabla }u|){|\mathrm{\nabla }u|}^{2}dx|\le C{\parallel \mathrm{\nabla }u\parallel }_p^p$ for every $u\in W_0^{1,p}(\mathrm{\Omega })$ with some $C>0$ independent of u (see (6)), it is sufficient to show $|{\int }_{\mathrm{\Omega }}{\tilde{a}}_0(x,|\mathrm{\nabla }{u}_{\epsilon }|)\mathrm{\nabla }{u}_{\epsilon }\mathrm{\nabla }({\phi }^p/u_{\epsilon }^{p-1})dx|\to 0$ as $\epsilon \to 0+$. Here, we fix any $\delta >0$. By the property of ${\tilde{a}}_0$ (see (6)) and because we are assuming that $u_{\epsilon }\to 0$ in $C_0^1(\overline{\mathrm{\Omega }})$ as $\epsilon \to 0+$, we have $|{\tilde{a}}_0(x,|\mathrm{\nabla }{u}_{\epsilon }|)|\le \delta {|\mathrm{\nabla }{u}_{\epsilon }|}^{p-2}$ for every $x\in \mathrm{\Omega }$ provided sufficiently small $\epsilon >0$. Therefore, for such sufficiently small $\epsilon >0$, we obtain

$$\begin{array}{r}\left|{\int }_{\mathrm{\Omega }}{\tilde{a}}_0(x,|\mathrm{\nabla }{u}_{\epsilon }|)\mathrm{\nabla }{u}_{\epsilon }\mathrm{\nabla }\left(\frac{{\phi }^p}{u_{\epsilon }^{p-1}}\right)dx\right|\\ \le p{\int }_{\mathrm{\Omega }}\frac{|{\tilde{a}}_0(x,|\mathrm{\nabla }{u}_{\epsilon }|)||\mathrm{\nabla }{u}_{\epsilon }||\mathrm{\nabla }\phi |{\phi }^{p-1}}{u_{\epsilon }^{p-1}}dx+(p-1){\int }_{\mathrm{\Omega }}\frac{|{\tilde{a}}_0(x,|\mathrm{\nabla }{u}_{\epsilon }|)|{|\mathrm{\nabla }{u}_{\epsilon }|}^2{\phi }^p}{u_{\epsilon }^p}dx\\ \le \delta {\parallel \phi \parallel }_{C_0^1(\overline{\mathrm{\Omega }})}^p\{p{\int }_{\mathrm{\Omega }}{\left(\frac{|\mathrm{\nabla }{u}_{\epsilon }|}{u_{\epsilon }}\right)}^{p-1}dx+(p-1){\int }_{\mathrm{\Omega }}{\left(\frac{|\mathrm{\nabla }{u}_{\epsilon }|}{u_{\epsilon }}\right)}^{p}dx\}\\ \le \delta {\parallel \phi \parallel }_{C_0^1(\overline{\mathrm{\Omega }})}^p|\mathrm{\Omega }|(p{({\lambda }_0/C_0)}^{1-1/p}+(p-1)({\lambda }_0/C_0))\end{array}$$

because of $|\mathrm{\nabla }{u}_{\epsilon }|/u_{\epsilon }\in L^p(\mathrm{\Omega })$ by Lemma 12. Since $\delta >0$ is arbitrary, our conclusion is shown. □

3.1 Proof of main results

Proof of Theorems

Let $\epsilon \in (0,1]$. Due to Proposition 6 and Lemma 11, we have ${\parallel u_{\epsilon }\parallel }_{\mathrm{\infty }}\le M$ for some $M>0$ independent of $\epsilon \in (0,1]$. Hence, there exist $M^{\mathrm{\prime }}>0$ and $0<\alpha <1$ such that $u_{\epsilon }\in C_0^{1,\alpha }(\overline{\mathrm{\Omega }})$ and ${\parallel u_{\epsilon }\parallel }_{C_0^{1,\alpha }(\overline{\mathrm{\Omega }})}\le M^{\mathrm{\prime }}$ for every $\epsilon \in (0,1]$ by the regularity result in [11]. Because the embedding of $C_0^{1,\alpha }(\overline{\mathrm{\Omega }})$ into $C_0^1(\overline{\mathrm{\Omega }})$ is compact and by $u_{\epsilon }\in intP$, there exists a sequence $\{{\epsilon }_n\}$ and $u_0\in P$ such that ${\epsilon }_n\to 0+$ and $u_n:=u_{{\epsilon }_n}\to u_0$ in $C_0^1(\overline{\mathrm{\Omega }})$ as $n\to \mathrm{\infty }$. If $u_0\ne 0$ occurs, then $u_0\in intP$ by the same reason as in Lemma 7, and hence our conclusion is proved. Now, we shall prove $u_0\ne 0$ by contradiction for each theorem. So, we suppose that $u_0=0$, whence $u_n\to 0$ in $C_0^1(\overline{\mathrm{\Omega }})$ as $n\to \mathrm{\infty }$.

Proof of Theorem 1 Let $\phi \in intP$ be an eigenfunction corresponding to the first positive eigenvalue ${\mu }_1(m)$ (cf. [14, 15], it is well known that we can obtain φ as the minimizer of (4)), namely, φ is a positive solution of $-{\mathrm{\Delta }}_{p}u={\mu }_1(m)m(x){|u|}^{p-2}u$ in Ω and $u=0$ on ∂ Ω. Since p-Laplacian is $(p-1)$-homogeneous, we may assume that φ satisfies ${\int }_{\mathrm{\Omega }}m(x){\phi }^{p}dx=1$, and hence ${\parallel \mathrm{\nabla }\phi \parallel }_p^p={\mu }_1(m){\int }_{\mathrm{\Omega }}m(x){\phi }^{p}dx={\mu }_1(m)$ holds by taking φ as a test function. Choose $\rho >0$ satisfying $b_0-A_p{\mu }_1(m)>\rho {\parallel \phi \parallel }_p^p$ (note that $b_0-A_p{\mu }_1(m)>0$ as in (f1)). Due to (f1), there exists a $\delta >0$ such that $f_0(x,t)\ge (b_{0}m(x)-\rho )t^{p-1}$ for every $0\le t\le \delta$ and $x\in \mathrm{\Omega }$. Since we are assuming $u_n\to 0$ in $C_0^1(\overline{\mathrm{\Omega }})$ as $n\to \mathrm{\infty }$, ${\parallel u_n\parallel }_{\mathrm{\infty }}\le \delta$ occurs for sufficiently large n. Then, for such sufficiently large n, according to Lemma 9, (1) and $\psi \ge 0$, we obtain

$$\begin{array}{rl}{A}_p{\mu }_1(m)& =A_p{\parallel \mathrm{\nabla }\phi \parallel }_p^p\ge {\int }_{\mathrm{\Omega }}A(x,\mathrm{\nabla }{u}_n)\mathrm{\nabla }\left(\frac{{\phi }^p}{u_n^{p-1}}\right)dx={\int }_{\mathrm{\Omega }}\frac{f(x,u_n,\mathrm{\nabla }{u}_n)+\epsilon \psi }{u_n^{p-1}}{\phi }^{p}dx\\ \ge {\int }_{\mathrm{\Omega }}\frac{f_0(x,u_n)}{u_n^{p-1}}{\phi }^{p}dx\ge b_0{\int }_{\mathrm{\Omega }}m(x){\phi }^{p}dx-\rho {\parallel \phi \parallel }_p^p=b_0-\rho {\parallel \phi \parallel }_p^p>A_p{\mu }_1(m).\end{array}$$

This is a contradiction. □

Proof of Theorem 2 Since $\mathrm{\infty }>{sup}_{x\in \mathrm{\Omega }}{a}_0(x)\ge {inf}_{x\in \mathrm{\Omega }}{a}_0(x)>0$ holds, by the standard argument as in the p-Laplacian, we see that ${\lambda }_1(m)>0$ and it is the first positive eigenvalue of $-div(a_0(x){|\mathrm{\nabla }u|}^{p-2}\mathrm{\nabla }u)=\lambda m(x){|u|}^{p-2}u$ in Ω and $u=0$ on ∂ Ω. Therefore, by the well-known argument, there exists a positive eigenfunction ${\phi }_1\in intP$ corresponding to ${\lambda }_1(m)$ (we can obtain ${\phi }_1$ as the minimizer of (7)). Hence, by taking ${\phi }_1$ as a test function, we have $0<{\int }_{\mathrm{\Omega }}{a}_0(x){|\mathrm{\nabla }{\phi }_1|}^{p}dx={\lambda }_1(m){\int }_{\mathrm{\Omega }}m(x){\phi }_1^{p}dx$. Thus, ${\int }_{\mathrm{\Omega }}m(x){\phi }_1^{p}dx>0$ follows. Because $u_n\in intP$ is a solution of $(P;{\epsilon }_n)$ and ${\phi }_1\in intP$ is an eigenfunction corresponding to ${\lambda }_1(m)$, according to Lemma 11 and Lemma 13 (note $A(x,y)=a_0{|y|}^{p-2}y+{\tilde{a}}_0(x,|y|)y$ as in (AH0)), we obtain

$$\begin{array}{rl}0\le & {\int }_{\mathrm{\Omega }}{a}_0(x){|\mathrm{\nabla }{\phi }_1|}^{p-2}\mathrm{\nabla }{\phi }_1\mathrm{\nabla }\left(\frac{{\phi }_1^p-u_n^p}{{\phi }_1^{p-1}}\right)dx-{\int }_{\mathrm{\Omega }}{a}_0(x){|\mathrm{\nabla }{u}_n|}^{p-2}\mathrm{\nabla }{u}_n\mathrm{\nabla }\left(\frac{{\phi }_1^p-u_n^p}{u_n^{p-1}}\right)dx\\ \le & {\lambda }_1(m){\int }_{\mathrm{\Omega }}m({\phi }_1^p-u_n^p)dx-{\int }_{\mathrm{\Omega }}\frac{f_0(x,u_n)}{u_n^{p-1}}{\phi }_1^{p}dx\\ +{\int }_{\mathrm{\Omega }}{\tilde{a}}_0(x,|\mathrm{\nabla }{u}_n|)\mathrm{\nabla }{u}_n\mathrm{\nabla }\left(\frac{{\phi }_1^p-u_n^p}{u_n^{p-1}}\right)dx+{\int }_{\mathrm{\Omega }}f(x,u_n,\mathrm{\nabla }{u}_n)u_{n}dx+{\epsilon }_n{\int }_{\mathrm{\Omega }}\psi u_{n}dx\\ =& -{\int }_{\mathrm{\Omega }}(\frac{f_0(x,u_n)}{u_n^{p-1}}-b_{0}m(x)){\phi }_1^{p}dx-(b_0-{\lambda }_1(m))\int m(x){\phi }_1^{p}dx+o(1)\end{array}$$

(19)

as $n\to \mathrm{\infty }$ since we are assuming $u_n\to 0$ in $C_0^1(\overline{\mathrm{\Omega }})$, where we use the facts that $\psi \ge 0$ and ${\phi }_1>0$ in Ω. Furthermore, by Fatou’s lemma and (3), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\int }_{\mathrm{\Omega }}(\frac{f_0(x,u_n)}{u_n^{p-1}}-b_{0}m(x)){\phi }_1^{p}dx\ge 0.$$

As a result, by taking a limit superior with respect to n in (19), we have $0\le -(b_0-{\lambda }_1(m))\int m(x){\phi }_1^{p}dx<0$. This is a contradiction. □