Existence of a positive solution for problems with $(p,q)$-Laplacian and convection term in $\mathbb{R}^{N}$

Abstract

This paper provides a positive solution for the $(p,q)$-Laplace equation in $\mathbb{R}^{N}$ with a nonlinear term depending on the gradient. The solution is constructed as the limit of positive solutions in bounded domains. Strengthening the growth condition, it is shown that the solution is also bounded. The positivity of the solution is obtained through a new comparison principle. Finally, under a stronger growth condition, we show the existence of a vanishing at infinity solution.

1 Introduction

In this paper, we study the existence of a (positive) solution for the following quasi-linear elliptic equation:

$$ (\mathrm{P})\quad \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -\varDelta _{p} u+{\lambda_{1}}|u|^{p-2}u-\mu \varDelta _{q}u +\mu\lambda_{2}|u|^{q-2}u=f(x,u,\nabla u) &\mbox{in } \mathbb{R}^{N}, \\ u>0&\mbox{in } \mathbb{R}^{N}. \end{array}\displaystyle \right . $$

In the left-hand side of the equation in (P), we have the p-Laplacian $\varDelta _{p}$ and the q-Laplacian $\varDelta _{q}$ with $1< q< p<+\infty$ and the constants $\mu\geq0$, $\lambda_{1}>0$, and $\lambda_{2}> 0$. The problem covers the corresponding statement with p-Laplacian in the principal part, for which it is sufficient to take $\mu=0$. Here $-\varDelta _{p}$ is regarded as the operator $-\varDelta _{p}:W^{1,p}(\mathbb{R}^{N})\to W^{-1,p'}(\mathbb{R}^{N})$, where $\frac{1}{p}+\frac{1}{p'}=1$, defined by

$$\begin{aligned} \langle-\varDelta _{p}u,v\rangle= \int_{\mathbb{R}^{N}}|\nabla u|^{p-2}\nabla u\nabla v \,dx \quad \mbox{for all }u,v\in W^{1,p}\bigl(\mathbb{R}^{N}\bigr). \end{aligned}$$

The right-hand side of the equation in (P) is in the form of convection term, meaning a nonlinearity that depends on the point x in $\mathbb{R}^{N}$, on the solution u, and on its gradient ∇u.

The existence of positive solutions for problems with p-Laplacian and convection term on a bounded domain has been studied in [1–3]. In the case where the principal part of the equation is driven by the $(p, q)$-Laplacian operator with $1 < q < p$ and by a nonhomogeneous operator, the existence of a positive solution of elliptic problems with convex term on a bounded domain has been investigated in [4] and [5], respectively. Results of this type when the principal part of the equation is expressed through a general Leray-Lions operator can be found in [6]. Essential features of the present work are the dependence on the gradient ∇u, which prevents the use of variational methods, and the unboundedness of the domain, which produces lack of compactness.

We assume that $f: \mathbb{R}^{N}\times\mathbb{R}\times\mathbb {R}^{N}\to \mathbb{R}$ is a continuous function satisfying the growth condition:

• (F0) $f(x,0,\xi)\equiv0$ for all $x\in\mathbb{R}^{N}, \xi\in\mathbb{R}^{N}$;

• (F1) there exist constants $r_{1},r_{2}\in(0,p-1)$ and continuous nonnegative functions $a_{0}\in L^{p^{\prime}}(\mathbb{R}^{N})$ and $a_{i}\in L^{\tilde{r}_{i}}(\mathbb{R}^{N})$ ($i=1,2$), where $1/p+1/p^{\prime}=1$ and $\tilde{r}_{i}=p/(p-r_{i}-1)=(p/(r_{i}+1))^{\prime}$ ($i=1,2$), such that

  $$ \bigl|f(x,t,\xi)\bigr|\leq a_{0}(x)+a_{1}(x)|t|^{r_{1}}+a_{2}(x)| \xi|^{r_{2}} $$

  (1)

  for all $(x,t,\xi)\in\mathbb{R}^{N}\times\mathbb{R}\times\mathbb{R}^{N}$.

In this setting, by a solution of problem (P) we mean any function $u\in W^{1,p}(\mathbb{R}^{N})\cap W^{1,q}(\mathbb{R}^{N})$ when $\mu>0$ or $u\in W^{1,p}(\mathbb{R}^{N})$ when $\mu=0$ such that $u(x)>0$ for a.e. $x\in\mathbb{R}^{N}$ and

$$\begin{aligned} & \int_{\mathbb{R}^{N}} \bigl(|\nabla u|^{p-2}+\mu|\nabla u|^{q-2}\bigr)\nabla u \nabla\varphi \,dx+ \int_{\mathbb{R}^{N}} \bigl(\lambda_{1} |u|^{p-2}+\mu \lambda_{2} |u|^{q-2}\bigr)u\varphi \,dx \\ &\quad= \int_{\mathbb{R}^{N}}f(x,u,\nabla u)\varphi \,dx \quad\mbox{for all } \varphi\in C_{0}^{\infty}\bigl(\mathbb{R}^{N}\bigr). \end{aligned}$$

In order to show the positivity of a solution, we will need an additional growth condition when t is small:

• (F2) there exist constants $\delta_{0}>0$ and $r_{0}\in(0,p-1)$ if $\mu=0$ or $r_{0}\in(0,q-1)$ if $\mu>0$ and a continuous positive function $b_{0}$ such that

  $$ b_{0}(x)t^{r_{0}}\leq f(x,t,\xi) \quad\mbox{for all } 0< t\le\delta_{0}, x\in\mathbb{R}^{N}, \xi\in \mathbb{R}^{N}. $$

  (2)

We mention that the fact that condition (F2) is supposed only for $t>0$ small is a significant improvement with respect to all the previous works. A direct consequence is that f is allowed to change sign.

For example, the following nonlinearity satisfies our assumptions $(\mathrm{F}0)\sim(\mathrm{F}2)$:

$$f(x,t,\xi)=a_{1}(x)|t|^{r_{0}} +a_{2}(x)| \xi|^{r_{2}} \sin t , $$

with $a_{1}\in L^{p^{\prime}}(\mathbb{R}^{N})\cap L^{\tilde{r}_{1}}(\mathbb{R}^{N})$ and $a_{2}\in L^{\tilde{r}_{2}}(\mathbb{R}^{N})$, where $\tilde{r}_{i}=p/(p-r_{i}-1)$ ($i=1,2$).

Our main result provides the existence of a (positive) solution for problem (P).

Theorem 1

Under assumptions (F0)-(F2), problem (P) admits a (positive) solution $u\in W^{1,p}(\mathbb{R}^{N})\cap W^{1,q}(\mathbb{R}^{N})\cap C^{1}_{\mathrm{loc}}(\mathbb{R}^{N})$ if $\mu>0$ and $u\in W^{1,p}(\mathbb{R}^{N})\cap C^{1}_{\mathrm{loc}}(\mathbb{R}^{N})$ if $\mu=0$.

The proof is based on a priori estimates obtained through hypotheses (F0)-(F2) for approximate solutions on bounded domains and the use of comparison arguments. In this respect, we establish several comparison principles that ultimately determine the positivity of solutions.

Under a stronger version of the growth condition (F1), we show that any positive solution disappear at infinity.

Theorem 2

Assume (F0)-(F2). If one of the following conditions holds, then any (positive) solution u of problem (P) satisfies that $u(x)\to0$ as $|x|\to\infty$:

• (i) $N\le p$;

• (ii) $N>p$ and $p^{2}< p^{*}$ (if and only if $p< N< p^{2}/(p-1)$);

• (iii) $N\ge p^{2}/(p-1)$, and for the functions $a_{i}$ ($i=0,1,2$) in (F1), there exist $R_{*}>0$ and $\gamma_{i}$ such that

  $$\begin{aligned}& \gamma_{0}, \gamma_{1} >\frac{p^{*}}{p^{*}-p},\qquad \gamma_{2}>\frac{pp^{*}}{(p-r_{2})(p^{*}-p)}, \end{aligned}$$

  (3)
  $$\begin{aligned}& \sup_{x_{0}\in\mathbb{R}^{N}} \int_{B(x_{0},2R_{*})} \bigl|a_{i}(x)\bigr|^{\gamma_{i}}\,dx< \infty \end{aligned}$$

  (4)

  for $i=0,1,2$, where $p^{*}:=pN/(N-p)$ if $N> p$.

Since we are looking for positive solutions of problem (P), without any loss of generality, we will suppose in the sequel that $f(x,t,s)\equiv0$ for all $t\leq0$ and $(x,s)\in \mathbb{R}^{N}\times\mathbb{R}^{N}$.

The rest of the paper is organized as follows. In Section 2, we present comparison principles related to problem (P). Section 3 deals with approximate solutions on bounded domains. Section 4 is devoted to the proof of Theorem 1. In Section 5, we give a proof of Theorem 2 after we show the boundedness of a solution.

2 Comparison principles

In this section, we assume that D is a bounded domain in $\mathbb{R}^{N}$. We consider the operator denoted $-\varDelta _{p}$ from $W^{1,p}(D)$ to $W^{1,p}(D)^{*}$ defined by

$$\begin{aligned} \langle-\varDelta _{p}u,v\rangle= \int_{D}|\nabla u|^{p-2}\nabla u\nabla v \,dx \quad \mbox{for all }u,v\in W^{1,p}(D). \end{aligned}$$

First we recall the following result.

Lemma 1

([7], Lemma 2.1)

Let $w_{1}, w_{2}\in L^{\infty}(D)$ satisfy $w_{i}\geq0$ a.e. on D, $w_{i}^{1/q}\in W^{1,p}(D)$ for $i=1,2$, and $w_{1}=w_{2}$ on ∂D. If $w_{1}/w_{2}, w_{2}/w_{1}\in L^{\infty}(D)$, then

$$\begin{aligned} 0\le \biggl\langle -\varDelta _{p} w_{1}^{1/q}- \mu \varDelta _{q} w_{1}^{1/q}, \frac{w_{1}-w_{2}}{w_{1}^{(q-1)/q}} \biggr\rangle - \biggl\langle -\varDelta _{p} w_{2}^{1/q}-\mu \varDelta _{q} w_{2}^{1/q}, \frac{w_{1}-w_{2}}{w_{2}^{(q-1)/q}} \biggr\rangle . \end{aligned}$$

(5)

Lemma 1 leads to a comparison principle for a subsolution and a supersolution of the problem

$$ \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -\varDelta _{p} u+\lambda_{1} |u|^{p-2}u-\mu \varDelta _{q}u+\mu\lambda_{2} |u|^{q-2}u=g(u)& \mbox{in } D,\\ u=0& \mbox{on } \partial D, \end{array}\displaystyle \right . $$

(6)

where $g:\mathbb{R}\rightarrow\mathbb{R}$ is a continuous function.

We say that $u_{1}\in W^{1,p}({D})$ is a subsolution of problem (6) if $u_{1}\leq0$ on ∂D and

$$\begin{aligned}[b] &\int_{D}(|\nabla u_{1}|^{p-2}\nabla u_{1}\nabla\varphi+ \lambda_{1} |u_{1}|^{p-2}u_{1} \varphi+\mu|\nabla u_{1}|^{q-2}\nabla u_{1}\nabla \varphi+ \mu\lambda_{2} |u_{1}|^{q-2}u_{1} \varphi \,dx \\ &\quad \leq \int_{D} g(u_{1})\varphi \,dx \end{aligned} $$

for all $\varphi\in W^{1,p}_{0}({D})$ with $\varphi\geq0$ in D, provided that the integral $\int_{D} g(u_{1})\varphi \,dx$ exists. We say that $u_{2}\in W^{1,p}({D})$ is a supersolution of (6) if the reversed inequalities are satisfied with $u_{2}$ in place of $u_{1}$ for all $\varphi\in W^{1,p}_{0}({D})$ with $\varphi\geq0$ in D.

Theorem 3

Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $t^{1-q}g(t)$ is nonincreasing for $t>0$ if $\mu>0$ and $t^{1-p}g(t)$ is nonincreasing for $t>0$ if $\mu=0$. Assume that $u_{1}$ and $u_{2}$ are a positive subsolution and a positive supersolution of problem (6), respectively. If $u_{2}(x)>u_{1}(x)=0$ for all $x\in\partial D$ and $u_{i}\in C^{1}(\overline{D})$ for $i=1,2$, then $u_{2}\geq u_{1}$ in D.

Proof

We prove the result only for $\mu>0$ because the case $\mu=0$ is easier. Suppose by contradiction that the set ${D}_{0}=\{x\in{D}: u_{1}(x)>u_{2}(x)\}$ is nonempty. Let U be a connected component of ${D}_{0}$. Noting that $\inf_{\overline{D}}u_{2}>0$, we see that $\overline{U}\subset D$, $u_{1}=u_{2}$ on ∂U, and $u_{i}/u_{j}\in L^{\infty}(U)$ for $i,j=1,2$. So, $(u_{1}^{q}-u_{2}^{q})/u_{1}^{q-1}, (u_{1}^{q}-u_{2}^{q})/u_{2}^{q-1}\in W_{0}^{1,p}(U)$, and extending by 0 on $D\setminus U$, we can take them as test functions in the above definitions of subsolution and supersolution for problem (6). It follows that

$$\begin{aligned} & \biggl\langle -\varDelta _{p} u_{1}-\mu \varDelta _{q} u_{1}, \frac{u_{1}^{q}-u_{2}^{q}}{u_{1}^{q-1}} \biggr\rangle - \biggl\langle - \varDelta _{p} u_{2}-\mu \varDelta _{q} u_{2}, \frac{u_{1}^{q}-u_{2}^{q}}{u_{2}^{q-1}} \biggr\rangle \\ &\quad\le \int_{U} \bigl(-\lambda_{1} u_{1}^{p-1}- \mu\lambda_{2} u_{1}^{q-1}+g(u_{1})\bigr) \frac{u_{1}^{q}-u_{2}^{q}}{u_{1}^{q-1}}\,dx \\ &\qquad{}- \int_{U} \bigl(-\lambda_{1} u_{2}^{p-1}- \mu\lambda_{2} u_{2}^{q-1}+g(u_{2})\bigr) \frac{u_{1}^{q}-u_{2}^{q}}{u_{2}^{q-1}}\,dx \\ &\quad=-\lambda_{1} \int_{U} \bigl(u_{1}^{p-q}-u_{2}^{p-q} \bigr) \bigl(u_{1}^{q}-u_{2}^{q}\bigr)\,dx + \int_{U} \biggl(\frac{g(u_{1})}{u_{1}^{q-1}}-\frac{g(u_{2})}{u_{2}^{q-1}} \biggr) \bigl(u_{1}^{q}-u_{2}^{q}\bigr) \\ &\quad< 0. \end{aligned}$$

The last inequality is obtained through our assumption that $g(t)/t^{q-1}$ is nonincreasing for $t>0$ and $\lambda_{1}>0$.

On the other hand, note that we can apply Lemma 1 with $w_{i}=u_{i}^{q}$ ($i=1,2$) and U in place of D. The conclusion provided by (5) in Lemma 1 contradicts the above inequality in the case where U is nonempty. Therefore, ${D}_{0}=\emptyset$, which completes the proof. □

The next theorem points out that the condition of supersolution in Theorem 3 can be relaxed to a weaker notion of supersolution directly related to the given subsolution.

Theorem 4

Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be continuous function such that $t^{1-q}g(t)$ is nonincreasing for $t>0$ if $\mu>0$ and $t^{1-p}g(t)$ is nonincreasing for $t>0$ if $\mu=0$. Assume that $u_{1}\in C^{1}(\overline{D})$ is a positive subsolution of problem (6) and that $h:\overline{D}\times\mathbb{R}\times \mathbb{R}^{N}\rightarrow\mathbb{R}$ is continuous function such that $h(x,t,\xi)\ge g(t)$ for all $x\in\mathbb{R}^{N}$, $\xi\in\mathbb{R}^{N}$, and $t\in(0,\|u_{1}\|_{L^{\infty}(D)} ]$. If $u_{2}\in C^{1}(\overline{D})$ is a (positive) solution of

$$ \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -\varDelta _{p} u+\lambda_{1} |u|^{p-2}u-\mu \varDelta _{q}u+\mu\lambda_{2} |u|^{q-2}u=h(x,u,\nabla u)& \textit{in } D,\\ u>0& \textit{in } D \end{array}\displaystyle \right . $$

(7)

such that $u_{2}(x)>u_{1}(x)=0$ for all $x\in\partial D$, then $u_{2}\geq u_{1}$ in D.

Proof

The conclusion can be achieved following the same argument as in Theorem 3. Indeed, arguing by contradiction, let us assume that the set $D_{0}:=\{x\in D : u_{1}(x)>u_{2}(x)\}$ is not empty. Let U be a connected component of $D_{0}$. Then $u_{1}=u_{2}$ on ∂U and $g(u_{2})\le h(x,u_{2},\nabla u_{2})$ in U. Proceeding as in the proof of Theorem 3, we have

$$\begin{aligned} \biggl\langle -\varDelta _{p} u_{1}-\mu \varDelta _{q} u_{1}, \frac{u_{1}^{q}-u_{2}^{q}}{u_{1}^{q-1}} \biggr\rangle - \biggl\langle - \varDelta _{p} u_{2}-\mu \varDelta _{q} u_{2}, \frac{u_{1}^{q}-u_{2}^{q}}{u_{2}^{q-1}} \biggr\rangle < 0. \end{aligned}$$

This leads to a contradiction by applying Lemma 1. □

In the case where $u_{1}$ and $u_{2}$ satisfy the homogeneous Dirichlet boundary condition we can state the following:

Theorem 5

Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $t^{1-q}g(t)$ is nonincreasing for $t>0$ if $\mu>0$ and $t^{1-p}g(t)$ is nonincreasing for $t>0$ if $\mu=0$. Assume that $u_{1}\in C_{0}^{1}(\overline{D})$ is a positive subsolution of problem (6) and that $h:\overline{D}\times\mathbb{R}\times \mathbb{R}^{N}\rightarrow\mathbb{R}$ is continuous and such that $h(x,t,\xi)\ge g(t)$ for all $x\in\mathbb{R}^{N}$, $\xi\in\mathbb{R}^{N}$, and $t\in(0, \|u_{1}\|_{L^{\infty}(D)} ]$. If $u_{2}\in C^{1}_{0}(\overline{D})$ is a (positive) solution of problem (7) such that $u_{1}/u_{2}\in L^{\infty}(D)$ and $u_{2}/u_{1}\in L^{\infty}(D)$, then $u_{2}\geq u_{1}$ in D.

Proof

Due to the assumptions $u_{1}/u_{2}\in L^{\infty}(D)$ and $u_{2}/u_{1}\in L^{\infty}(D)$, it turns out that $(u_{1}^{q}-u_{2}^{q})/u_{1}^{q-1}, (u_{1}^{q}-u_{2}^{q})/u_{2}^{q-1}\in W_{0}^{1,p}(U)$ with U introduced in the proof of Theorem 3. Then we can conclude as in the proof of Theorem 4. □

3 Solution on a bounded domain

In this section, we assume that D is a bounded domain in $\mathbb{R}^{N}$ with $C^{2}$ boundary ∂D. For $r\geq1$, we denote by $\|u\|_{L^{r}(D)}$ the usual norm on the space $L^{r}(D)$. We endow $W_{0}^{1,p}(D)$ with the norm $\|u\|_{D}^{p}=\|\nabla u\|_{L^{p}(D)}^{p}+\lambda_{1}\|u\|_{L^{p}(D)}^{p}$, which is equivalent to the usual one.

We focus on the existence of a (positive) solution for the problem

$$ (\mathrm{PD})\quad \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -\varDelta _{p} u+\lambda_{1} |u|^{p-2}u-\mu \varDelta _{q}u+\mu\lambda_{2} |u|^{q-2}u =f(x,u,\nabla u) &\mbox{in } D,\\ u>0&\mbox{in } D,\\ u(x)= 0 &\mbox{on }\partial D. \end{array}\displaystyle \right . $$

Here we impose the following hypotheses: $f: \overline{D}\times \mathbb{R}\times\mathbb{R}^{N}\to\mathbb{R}$ is a continuous function satisfying

$(\widetilde{\mathrm{F}0})$ :

$f(x,0,\xi)\equiv0$ for all $x\in D, \xi\in\mathbb{R}^{N}$;

$(\widetilde{\mathrm{F}1})$ :

there exist constants $r_{1},r_{2}\in(0,p-1)$ and continuous nonnegative functions $a_{i}$ ($i=0, 1,2$) on D̅ such that

$$ \bigl|f(x,t,\xi)\bigr|\leq a_{0}(x)+a_{1}(x)|t|^{r_{1}}+a_{2}(x)| \xi|^{r_{2}} $$

(8)

for all $(x,t,\xi)\in D\times\mathbb{R}\times\mathbb{R}^{N}$;

$(\widetilde{\mathrm{F}2})$ :

there exist constants $\delta_{0}>0$ and $r_{0}\in(0,p-1)$ if $\mu=0$ or $r_{0}\in(0,q-1)$ if $\mu>0$ and a continuous function $b_{0}$ such that $\inf_{x\in D} b_{0}(x)>0$ and

$$ b_{0}(x)t^{r_{0}}\leq f(x,t,\xi) \quad\mbox{for all } 0< t\le\delta_{0}, x\in D, \xi\in\mathbb{R}^{N}. $$

(9)

We say that $u\in W_{0}^{1,p}(D)$ is a solution of (PD) if $u(x)>0$ for a.e. $x\in D$ and

$$\begin{aligned} & \int_{D} \bigl(|\nabla u|^{p-2}+\mu|\nabla u|^{q-2}\bigr)\nabla u\nabla\varphi \,dx + \int_{D} \bigl(\lambda_{1} |u|^{p-2}+\mu \lambda_{2}|u|^{q-2}\bigr)u\varphi \,dx \\ &\quad= \int_{D} f(x,u,\nabla u)\varphi \,dx\quad \mbox{for all } \varphi \in W_{0}^{1,p}(D). \end{aligned}$$

The existence of a solution for problem (PD) is stated as follows.

Theorem 6

Under assumptions $(\widetilde{\mathrm{F}0})$-$(\widetilde{\mathrm{F}2})$, problem (PD) admits a (positive) solution $u\in C^{1}_{0}(\overline{D})$ such that $\partial u/\partial\nu<0$ on ∂D, where ν stands for the outer normal to ∂D.

In the proof of Theorem 6, we utilize the following approximate equation:

$$ (\mathrm{PD}_{\varepsilon})\quad \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -\varDelta _{p} u+\lambda_{1} |u|^{p-2}u-\mu \varDelta _{q}u+\mu\lambda_{2}|u|^{q-2}u =f(x,u,\nabla u)+\varepsilon\psi&\mbox{in } D,\\ u>0&\mbox{in } D,\\ u(x)= 0 &\mbox{on }\partial D, \end{array}\displaystyle \right . $$

with $\varepsilon>0$ and a nonnegative function $0\not\equiv \psi\in C(\overline{D})$.

Lemma 2

Under $(\widetilde{\mathrm{F}0})$-$(\widetilde{\mathrm{F}2})$, for any $\varepsilon>0$ and a nonnegative function $0\not\equiv\psi\in C(D)$, problem ($\mathrm{PD}_{\varepsilon}$) admits a (positive) solution $u_{\varepsilon}\in C^{1}_{0}(\overline{D})$ such that $\partial u_{\varepsilon}/\partial\nu<0$ on ∂D.

Proof

We argue as in [5], Proposition 8, and [7]. Fix $\varepsilon>0$ and consider a Schauder basis $\{e_{1},\ldots,e_{m},\ldots\}$ of $W^{1,p}_{0}(D)$ (refer to [4, 8] for its existence). For each $m\in\mathbb{N}$, we define the m-dimensional subspace $V_{m}:=\operatorname{span}\{e_{1},\ldots, e_{m}\}$ of $W^{1,p}_{0}(D)$. The map $T_{m}\colon\mathbb{R}^{m} \to V_{m}$ defined by $T_{m}(\xi_{1},\ldots,\xi_{m})=\sum_{i=1}^{m}\xi_{i}e_{i}$ is a linear isomorphism. Let $T_{m}^{*}:V_{m}^{*}\to(\mathbb{R}^{m})^{*}$ be the dual map of $T_{m}$. Identifying $\mathbb{R}^{m}$ and $(\mathbb{R}^{m})^{*}$, we may regard $T_{m}^{*}$ as a map from $V_{m}^{*}$ to $\mathbb{R}^{m}$. Define the maps $A_{m}$ and $B_{m}$ from $V_{m}$ to $V_{m}^{*}$ as follows:

$$\begin{aligned} \bigl\langle A_{m}(u), v \bigr\rangle := \int_{D} \bigl(|\nabla u|^{p-2}+\mu|\nabla u|^{q-2} \bigr) \nabla u\nabla v \,dx \end{aligned}$$

and

$$\begin{aligned} \bigl\langle B_{m}(u), v \bigr\rangle :=- \int_{D} \bigl(\lambda_{1}|u|^{p-2}+\mu \lambda_{2} |u|^{q-2} \bigr)uv \,dx + \int_{D} \bigl(f(x,u,\nabla u)+\varepsilon\psi \bigr) v \,dx \end{aligned}$$

for all u, $v\in V_{m}$.

By $(\widetilde{\mathrm{F}1})$ and Hölder’s inequality, we have

$$\begin{aligned} &\bigl\langle A_{m}(u)-B_{m}(u), u \bigr\rangle \\ &\quad\ge \|u\|_{D}^{p}-d\bigl(\|u\|_{L^{1}(D)}+\|u \|_{L^{r_{1}+1}(D)}^{r_{1}+1} +\|u\|_{D}^{r_{2}+1}\bigr)- \varepsilon\|\psi\|_{L^{\infty}(D)}\|u\| _{L^{1}(D)} \end{aligned}$$

(10)

for all $u\in V_{m}$, where d is a positive constant independent of m and u. Because of $r_{1}+1< p$ and $r_{2}+1< p$, we easily see that $A_{m}-B_{m}$ is coercive on $V_{m}$, whence $T_{m}^{*}\circ(A_{m}-B_{m})\circ T_{m}$ is coercive on $\mathbb{R}^{m}$. By a well-known consequence of Brouwer’s fixed point theorem it follows that there exists $y_{m}\in\mathbb{R}^{m}$ such that $(T_{m}^{*}\circ(A_{m}-B_{m})\circ T_{m} )(y_{m})=0$, and hence $A_{m}(u_{m})-B_{m}(u_{m})=0$ with $u_{m}=T_{m}(y_{m})\in V_{m}$.

Writing (10) with $u=u_{m}\in W^{1,p}_{0}(D)$ shows the boundedness of the sequence $\|u_{m}\|_{D}$. Thus, along a subsequence, $u_{m}$ converges to some $u_{0}$ weakly in $W^{1,p}_{0}(D)$ and strongly in $L^{p}(D)$.

We claim that

$$ u_{m}\to u_{0} \quad\mbox{in } W_{0}^{1,p}(D) \mbox{ as } m\to\infty. $$

(11)

Let $P_{m}$ denote the projection onto $V_{m}$, that is, $P_{m} u=\sum_{i=1}^{m}\xi_{i}e_{i}$ for $u=\sum_{i=1}^{\infty}\xi_{i}e_{i}$. Since $u_{m}$, $P_{m}u_{0}\in V_{m}$ and $A_{m}(u_{m})-B_{m}(u_{m})=0$ in $V_{m}^{*}$, we obtain

$$\begin{aligned} &\bigl\langle A_{m}(u_{m}), u_{m}-P_{m}u_{0} \bigr\rangle \\ &\quad=\bigl\langle B_{m}(u_{m}), u_{m}-P_{m}u_{0} \bigr\rangle \\ &\quad=\bigl\langle B_{m}(u_{m}), u_{m}-u_{0} \bigr\rangle +\bigl\langle B_{m}(u_{m}), u_{0}-P_{m}u_{0} \bigr\rangle \\ &\quad= \int_{D} \bigl(-\lambda_{1}|u_{m}|^{p-2}u_{m}- \mu\lambda_{2} |u_{m}|^{q-2}u_{m}+f(x,u_{m}, \nabla u_{m}) \bigr) (u_{m}-P_{m}u_{0}) \,dx \\ &\qquad{}+ \int_{D}\varepsilon\psi(u_{m}-P_{m}u_{0}) \,dx \\ &\quad\to0\quad \mbox{as } m\to\infty, \end{aligned}$$

where we use $(\widetilde{\mathrm{F}1})$, the boundedness of $\|u_{m}\|_{D}$, $u_{m}\to u_{0}$ in $L^{p}(\Omega)$, and $P_{m}u_{0}\to u_{0}$ in $W^{1,p}_{0}(D)$. This leads to

$$ \lim_{m\to\infty} \int_{D}|\nabla u_{m}|^{p-2}\nabla u_{m}\nabla (u_{m}-u_{0})\,dx=0. $$

In view of the $(S_{+})$-property of $-\varDelta _{p}$ (see, e.g., [9], Proposition 3.5, or refer to (22) in the proof of Theorem 1), we obtain (11).

Now let us prove that $u_{0}$ is a solution of ($\mathrm{PD}_{\varepsilon}$). Fix $l\in\mathbb{N}$ and $\varphi\in V_{l}$. For each $m\ge l$, letting $m\to\infty$ in $\langle A_{m}(u_{m}),\varphi\rangle=\langle B_{m}(u_{m}),\varphi\rangle$ and making use of (11), we have

$$\begin{aligned} & \int_{D} \bigl(|\nabla u_{0}|^{p-2}+\mu| \nabla u_{0}|^{q-2}\bigr)\nabla u_{0} \nabla \varphi \,dx+ \int_{D} \bigl(\lambda_{1} |u_{0}|^{p-2}+ \mu\lambda_{2} |u_{0}|^{q-2}\bigr)u_{0} \varphi \,dx \\ &\quad= \int_{D} \bigl(f(x,u_{0},\nabla u_{0})+\varepsilon\psi \bigr) \varphi \,dx. \end{aligned}$$

(12)

Since l is arbitrary, equality (12) holds for every $\varphi\in\bigcup_{l\ge1} V_{l}$. In fact, the density of $\bigcup_{l\ge 1} V_{l}$ in $W^{1,p}_{0}(D)$ guarantees that (12) holds for every $\varphi\in W^{1,p}_{0}(D)$. This means that $u_{0}$ is a solution of ($\mathrm{PD}_{\varepsilon}$). Acting with $-u_{0}^{-}$ (where $u_{0}^{-}:=\max\{0,-u_{0}\}$) and taking into account that $\psi\ge0$ and $f(x,t,\xi)=0$ for $t\le0$, we see that

$$\begin{aligned} &\bigl\| u_{0}^{-}\bigr\| _{D}^{p}+\mu\bigl\| \nabla u_{0}^{-}\bigr\| _{L^{q}(D)}^{q} +\mu\lambda_{2} \bigl\| u_{0}^{-}\bigr\| _{L^{q}(D)}^{q} \\ &\quad= \int_{u_{0}< 0} \bigl(f(x,u_{0},\nabla u_{0})+\varepsilon\psi \bigr) u_{0} \,dx =\varepsilon \int_{u_{0}< 0}\psi u_{0} \,dx \le 0, \end{aligned}$$

whence $u_{0}\ge0$ a.e. in D. Moreover, $u_{0}\not\equiv0$ because we assumed that $\psi\not\equiv0$ and $\varepsilon>0$. Next, we observe that hypothesis $(\widetilde{\mathrm{F}1})$ allows us to refer to [10], Theorem 7.1 (see also [11] and [5]), from which we infer that $u\in L^{\infty}(D)$. Furthermore, the regularity result up to the boundary in [12], Theorem 1, and [13], p.320, ensures that $u\in C_{0}^{1,\beta}(\overline{D})$ with some $\beta\in(0,1)$. Applying the strong maximum principle in [14], Theorem 5.4.1, and the boundary point lemma in [14], Theorem 5.5.1 (note that $f(x,t,\xi)\ge0$ for $0\le t\le \delta_{0}$) entails that $u>0$ in D and $\partial u/\partial\nu<0$ on ∂D. Altogether, we have established that the conclusion of lemma is fulfilled for $u_{\varepsilon}=u_{0}$. □

We will also need the following result.

Lemma 3

Let $1< q< p<+\infty$, $\lambda_{1}>0$, $\lambda_{2}\ge0$, and $\mu\geq0$. For any constants $b>0$ and $0< r< p-1$ with $0< r< q-1$ if $\mu>0$, the problem

$$ \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -\varDelta _{p} u+\lambda_{1}|u|^{p-2}u-\mu \varDelta _{q}u+\mu\lambda_{2} |u|^{q-2}u=bu^{r} &\textit{in }D,\\ u>0&\textit{in } D,\\ u=0 & \textit{on } \partial D \end{array}\displaystyle \right . $$

(13)

admits a solution $u_{b}\in C^{1}_{0}(\overline{D})$ satisfying $\lambda_{1}\|u_{b}\|_{L^{\infty}(D)}^{p-r-1}\le b$ and $\partial u_{b}/\partial\nu<0$ on ∂D.

Proof

We can proceed for the existence of a solution of (13) along the lines of the proof of [7], Lemma 3. For readers’ convenience, we outline the proof in the case where $\mu>0$. Given the constants $b>0$ and $0< r< q-1$, we define the functional $I:W^{1,p}_{0}(D)\to\mathbb{R}$ by

$$\begin{aligned} I(u) =\frac{1}{p} \int_{D}\bigl(|\nabla u|^{p}+\lambda_{1} |u|^{p}\bigr)\,dx +\frac{\mu}{q} \int_{D}\bigl(|\nabla u|^{q}+\lambda_{2} |u|^{q}\bigr)\,dx -\frac{b}{r+1} \int_{D}\bigl(u^{+}\bigr)^{r+1}\,dx \end{aligned}$$

for all $u\in W^{1,p}_{0}(D)$, where $u^{+}=\max\{0,u\}$. Notice that I is of class $C^{1}$. By using the Sobolev embedding theorem we have the estimate

$$I(u)\geq\frac{1}{p}\|u\|^{p}_{D}-c\|u \|^{r+1}_{D} \quad \mbox{for all }u\in W^{1,p}_{0}(D) $$

with a constant $c>0$ independent of u. Since $p>r+1$, I is bounded from below and coercive. Having that I is sequentially weakly lower semicontinuous too, there exists $u_{b}\in W^{1,p}_{0}(D)$ such that

$$I(u_{b})=\inf_{u\in W^{1,p}_{0}(D)}I(u) $$

(see, e.g., [15], Theorems 1.1, 1.2). In addition, because of $r+1< q< p$, taking any positive smooth function v and a sufficiently small $t>0$, we have

$$\begin{aligned} I(tv)&=t^{r+1} \biggl( \frac {t^{p-r-1}}{p}\|v\|_{D} + \frac{\mu t^{q-r-1}}{q} \bigl(\|\nabla v\|_{L^{q}(D)}^{q} + \lambda_{2}\|v\|_{L^{q}(D)}^{q} \bigr) - \frac{\|v\|_{L^{r+1}(D)}^{r+1}}{r+1} \biggr) < 0. \end{aligned}$$

This ensures that $\inf_{u\in W^{1,p}_{0}(D)}I(u) <0$, and hence $u_{b}$ is a nontrivial critical point of I. By the regularity theory we infer that $u_{b}\in C^{1}_{0}(\overline{D})$. Taking $-u_{b}^{-}$ as a test function in the equation $I'(u_{b})=0$, we see that $u_{b}\ge0$. Then the strong maximum principle enables us to derive that $u_{b}>0$ in D, so $u_{b}$ is a solution of problem (13), and $\partial u_{b}/\partial\nu<0$ on ∂D.

Taking $u_{b}^{\alpha+1}$ with $\alpha>0$ as a test function in (13), by using Hölder’s inequality and that $r+1< p$ we get

$$\lambda_{1}\|u_{b}\|_{L^{p+\alpha}(D)}^{p+\alpha} \le b \int_{D} u_{b}^{r+\alpha+1}\,dx \le b \|u_{b}\|_{L^{p+\alpha}(D)}^{r+\alpha+1}|D|^{(p-r-1)/(p+\alpha)}, $$

where $|D|$ denotes the Lebesgue measure of D, and hence

$$\lambda_{1}\|u_{b}\|_{L^{p+\alpha}(D)}^{p-r-1} \le b|D|^{(p-r-1)/(p+\alpha)}. $$

Letting $\alpha\to\infty$, we conclude that $\lambda_{1} \|u_{b}\|_{L^{\infty}(D)}^{p-r-1} \le b$. □

Proof of Theorem 6

Using the data $\delta_{0}$, $r_{0}$, and $b_{0}$ in $(\widetilde{\mathrm{F}2})$, we fix a positive constant b such that

$$ b\le\min \Bigl\{ \inf_{x\in D}b_{0}(x), \lambda_{1}\delta^{p-r_{0}-1} \Bigr\} . $$

(14)

Then, according to Lemma 3, there exists a (positive) solution $u_{b}$ of

$$ \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -\varDelta _{p} u+\lambda_{1}|u|^{p-2}u-\mu \varDelta _{q}u+\mu\lambda_{2} |u|^{q-2}u=bu^{r_{0}} &\mbox{in }D,\\ u>0&\mbox{in } D,\\ u=0 & \mbox{on } \partial D, \end{array}\displaystyle \right . $$

satisfying

$$ \|u_{b}\|_{L^{\infty}(D)}\le \biggl(\frac{b}{\lambda_{1}} \biggr)^{\frac {1}{p-r_{0}-1}}\le \delta_{0} $$

(15)

(note (14)). Let $u_{\varepsilon}$ ($\varepsilon>0$) be a positive solution of problem ($\mathrm{PD}_{\varepsilon}$) obtained by Lemma 2. Let us observe that $u_{b}/u_{\varepsilon}, u_{\varepsilon}/u_{b} \in L^{\infty}(D)$ because $u_{b}$ and $u_{\varepsilon}$ are positive functions belonging to $C^{1}_{0}(\overline{D})$ and satisfying $\partial u_{i}/\partial\nu<0$ on ∂D ($i=b$ and $i=\varepsilon$). On the basis of (15), we are able to apply Theorem 5 with $u_{1}=u_{b}$, $u_{2}=u_{\varepsilon}$, $g(t)=b t^{r_{0}}$, and $h(x,t,\xi)=f(x,t,\xi)+\varepsilon\psi$ because for any $0< t\le\|u_{b}\|_{L^{\infty}(D)}$, we have that

$$ h(x,t,\xi)=f(x,t,\xi)+\varepsilon\psi\ge f(x,t,\xi) \ge bt^{r_{0}}=g(t) $$

by (14) and $(\widetilde{\mathrm{F}2})$. In this way, we see that $u_{\varepsilon}\ge u_{b}$ in D for every $\varepsilon>0$.

Using the growth condition (8) of f, taking $u_{\varepsilon}$ as a test function in ($\mathrm{PD}_{\varepsilon}$), we obtain the inequality

$$\begin{aligned} \|u_{\varepsilon}\|_{D}^{p} \le& \int_{D} \bigl(a_{0} u_{\varepsilon}+a_{1} u_{\varepsilon}^{r_{1}+1} +a_{2} |\nabla u_{\varepsilon}|^{r_{2}}u_{\varepsilon}\bigr)\,dx +\varepsilon\|\psi\|_{L^{p^{\prime}}(D)}\|u_{\varepsilon}\|_{L^{p}(D)} \\ \le& \|a_{0}\|_{L^{p^{\prime}}(D)}\|u_{\varepsilon}\|_{L^{p}(D)} +\|a_{1}\|_{L^{\tilde{r}_{1}}(D)}\|u_{\varepsilon}\|_{L^{p}(D)}^{r_{1}+1} \\ &{}+\|a_{2}\|_{L^{\tilde{r}_{2}}(D)}\|\nabla u_{\varepsilon}\|_{L^{p}(D)}^{r_{2}} \|u_{\varepsilon}\|_{L^{p}(D)} + \varepsilon\|\psi\|_{L^{p^{\prime}}(D)}\|u_{\varepsilon}\|_{L^{p}(D)} \\ \le& \lambda_{1}^{-1/p}\|a_{0}\|_{L^{p^{\prime}}(D)} \|u_{\varepsilon}\|_{D} +\lambda_{1}^{-(r_{1}+1)/p} \|a_{1}\|_{L^{\tilde{r}_{1}}(D)}\|u_{\varepsilon}\|_{D}^{r_{1}+1} \\ &{}+\lambda_{1}^{-1/p} \|a_{2}\|_{L^{\tilde{r}_{2}}(D)} \|u_{\varepsilon}\|_{D}^{r_{2}+1} +\lambda_{1}^{-1/p} \varepsilon\|\psi\|_{L^{p^{\prime}}(D)}\| u_{\varepsilon}\|_{D} \end{aligned}$$

for every $\varepsilon>0$. This shows the boundedness of $\{u_{\varepsilon}\}_{\varepsilon\in(0,1]}$ in $W_{0}^{1,p}(D)$ because $p>r_{1}+1, r_{2}+1$ (note $0<\varepsilon\le1$). Thus, we can find a sequence $\varepsilon_{n}\to0^{+}$ such that $u_{n}:=u_{\varepsilon_{n}}$ is weakly convergent to some u in $W^{1,p}_{0}(D)$ and strongly in $L^{r}(D)$ (for all $r\in[1,p^{*})$). On the other hand, taking $u_{n}-u$ as a test function, we easily see that

$$\begin{aligned} U_{n} := & \int_{D} \bigl(|\nabla u_{n}|^{p-2}+\mu| \nabla u_{n}|^{q-2}\bigr) \nabla u_{n} \nabla(u_{n}-u)\,dx \\ \le& -\lambda_{1} \int_{D} u_{n}^{p-1}(u_{n}-u) \,dx -\mu\lambda_{2} \int_{D} u_{n}^{q-1}(u_{n}-u) \,dx \\ &{} + \int_{D} \bigl(a_{0}+a_{1} u_{n}^{r_{1}}+a_{2} |\nabla u_{n}|^{r_{2}} \bigr) (u_{n}-u)\,dx \\ &{}+\varepsilon_{n} \|\psi\|_{L^{p^{\prime}}(D)} \|u_{n}-u \|_{L^{p}(D)} \\ \le& \lambda_{1}\|u_{n}\|_{L^{p}(D)}^{p-1} \|u_{n}-u\|_{L^{p}(D)} +\mu\lambda_{2} \|u_{n}\|_{L^{q}(D)}^{q-1}\|u_{n}-u \|_{L^{q}(D)} \\ &{}+\|a_{0}\|_{L^{p^{\prime}}(D)}\|u_{n}-u\|_{L^{p}(D)} +\|a_{1}\|_{L^{\infty}(D)}\|u_{n}\|_{L^{p^{\prime}r_{1}}(D)}^{r_{1}} \|u_{n}-u\|_{L^{p}(D)} \\ &{} +\|a_{2}\|_{L^{\tilde{r}_{2}}(D)}\|\nabla u_{n} \|_{L^{p}(D)}^{r_{2}} \|u_{n}-u\|_{L^{p}(D)} + \varepsilon\|\psi\|_{L^{p^{\prime}}(D)}\|u_{n}-u\|_{L^{p}(D)} \\ \to&0\quad \mbox{as } n\to\infty. \end{aligned}$$

Hence, $\limsup_{n\to\infty} U_{n}\le0$. According to the $(S_{+})$ property of $-\varDelta _{p}$, this ensures that $u_{n}$ is strongly convergent to u in $W^{1,p}_{0}(D)$ (refer to (22)). Hence, u is a solution of (PD). Since we already know that $u_{n}\ge u_{b}$ in D for every n, in the limit, we obtain that $u\ge u_{b}$ in D. This completes the proof. □

4 Proof of Theorem 1

In this section, we denote $B_{n}:=B_{n}(0)$ the open ball with center at the origin and radius n. The spaces $W^{1,p}(B_{n})$ and $W^{1,q}(B_{n})$ are equipped with the norms

$$\|u\|_{p,n}^{p} := \int_{B_{n}} \bigl(|\nabla u|^{p}+\lambda _{1}|u|^{p} \bigr)\,dx $$

and

$$\|u\|_{q,n}^{q}:= \int_{B_{n}} \bigl(|\nabla u|^{q}+ \lambda_{2} |u|^{q} \bigr)\,dx, $$

respectively.

Proof of Theorem 1

By applying Theorem 6 with $D=B_{n}$ ($n\in\mathbb{N}$) we obtain a (positive) solution $v_{n}\in C_{0}^{1}(\overline{B_{n}})$ of the problem

$$ (\mathrm{P}_{n})\quad \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -\varDelta _{p} u+\lambda_{1} |u|^{p-2}u-\mu \varDelta _{q}u+\mu\lambda_{2} |u|^{q-2}u =f(x,u,\nabla u) &\mbox{in } B_{n},\\ u>0&\mbox{in } B_{n},\\ u(x)= 0 &\mbox{on }\partial B_{n}. \end{array}\displaystyle \right . $$

We claim that there exists a positive constant C such that

$$ \|v_{n}\|_{p,n} \le C \quad\mbox{for all } n\in\mathbb{N} $$

(16)

and

$$ \|v_{n}\|_{q,n}\le C \quad\mbox{for all } n\in\mathbb{N}, \mbox{provided that } \mu>0. $$

(17)

Indeed, acting with $v_{n}$ in ($\mathrm{P}_{n}$) as a test function, through assumption (F1) and Hölder’s and Young’s inequalities we obtain

$$\begin{aligned} \|v_{n}\|_{p,n}^{p}+\mu\|v_{n} \|_{q,n}^{q} =& \int_{B_{n}} f(x,v_{n},\nabla v_{n})v_{n} \,dx \\ \le& \|a_{0}\|_{L^{p^{\prime}}(B_{n})}\|v_{n}\|_{L^{p}(B_{n})} +\|a_{1}\|_{L^{\tilde{r}_{1}}(B_{n})}\|v_{n}\|_{L^{p}(B_{n})}^{r_{1}+1} \\ &{} +\|a_{2}\|_{L^{\tilde{r}_{2}}(B_{n})}\|v_{n} \|_{L^{p}(B_{n})} \|\nabla v_{n}\|_{L^{p}(B_{n})}^{r_{2}} \\ \leq& \frac{\lambda_{1}}{2}\|v_{n}\|_{L^{p}(B_{n})}^{p}+ \frac{1}{2} \|\nabla v_{n}\|_{L^{p}(B_{n})}^{p} \\ &{} +C \bigl(\|a_{0}\|_{L^{p^{\prime}}(B_{n})}^{p^{\prime}} +\|a_{1}\|_{L^{\tilde{r}_{1}}(B_{n})}^{\tilde{r}_{1}} +\|a_{2} \|_{L^{\tilde{r}_{2}}(B_{n})}^{\tilde{r}_{2}} \bigr), \end{aligned}$$

where C is a positive constant independent of n. It turns out that

$$\begin{aligned} \frac{1}{2}\|v_{n}\|_{p,n}^{p}+\mu \|v_{n}\|_{q,n}^{q} \le&C \bigl( \|a_{0}\|_{L^{p^{\prime}}(B_{n})}^{p^{\prime}} +\|a_{1} \|_{L^{\tilde{r}_{1}}(B_{n})}^{\tilde{r}_{1}} +\|a_{2}\|_{L^{\tilde{r}_{2}}(B_{n})}^{\tilde{r}_{2}} \bigr) \\ \le& C \bigl(\|a_{0}\|_{L^{p^{\prime}}(\mathbb{R}^{N})}^{p^{\prime}} + \|a_{1}\|_{L^{\tilde{r}_{1}}(\mathbb{R}^{N})}^{\tilde{r}_{1}} +\|a_{2} \|_{L^{\tilde{r}_{2}}(\mathbb{R}^{N})}^{\tilde{r}_{2}} \bigr), \end{aligned}$$

whence (16) and (17) follow.

Fix $m\in\mathbb{N}$. If $n\geq m+1$, then by (16) we have

$$ \|v_{n}\|_{p,m+1}\leq\|v_{n} \|_{p,n}\le C. $$

(18)

Therefore, there exists $v\in W^{1,p}(B_{m+1})$ such that

$$\begin{aligned}& v_{n} \rightharpoonup v \quad \mbox{in } W ^{1,p}(B_{m+1}), W^{1,q}(B_{m+1}), \end{aligned}$$

(19)

$$\begin{aligned}& v_{n}\to v \quad\mbox{in } L^{p}(B_{m+1}), L^{q}(B_{m+1}), \end{aligned}$$

(20)

$$\begin{aligned}& v_{n}(x)\rightarrow v(x) \quad \mbox{for a.e. } x\in B_{m+1} \end{aligned}$$

(21)

as $n\to\infty$.

Let us show that $v_{n}$ converges to v strongly in $W ^{1,p}(B_{m})$ and $W^{1,q}(B_{m})$. To this end, fix $l\in\mathbb{N}$ and choose a smooth function $\psi_{l}$ satisfying $0\le\psi_{l} \le1$, $\psi_{l}(r)=1$ if $r\le m$, and $\psi_{l}(r)=0$ if $r\ge m+1/l$. Setting $\eta_{l}(x):=\psi_{l}(|x|)$, we note that $(v_{n}-v)\eta_{l}\in W_{0}^{1,p}(B_{m+1})\subset W_{0}^{1,p}(B_{n})$ for any $n\ge m+1$. Denote

$$V_{n}= \int_{B_{m}} \bigl(|\nabla v_{n}|^{p-2}+\mu| \nabla v_{n}|^{q-2} \bigr)\nabla v_{n} (\nabla v_{n}-\nabla v)\,dx. $$

Using $(v_{n}-v)\eta_{l}$ as a test function in $(\mathrm{P}_{n})$ and invoking the growth condition (F1), we obtain

$$\begin{aligned} V_{n} =& \int_{|x|< m+1/l} \bigl(f(x,v_{n},\nabla v_{n})- \lambda_{1}v_{n}^{p-1} -\mu\lambda_{2} v_{n}^{q-1} \bigr) (v_{n}-v)\eta_{l} \,dx \\ &{}- \int_{m\le|x|< m+1/l} \bigl(|\nabla v_{n}|^{p-2}+\mu| \nabla v_{n}|^{q-2}\bigr)\nabla v_{n} \nabla(v_{n}- v))\eta_{l}\,dx \\ &{}- \int_{m\le|x|< m+1/l} \bigl(|\nabla v_{n}|^{p-2}+\mu| \nabla v_{n}|^{q-2}\bigr)\nabla v_{n} \nabla \eta_{l} (v_{n}-v)\,dx \\ \le& \int_{B_{m+1}} \bigl(a_{0}(x)+a_{1}(x)v_{n}^{r_{1}}+a_{2}(x)| \nabla v_{n}|^{r_{2}} +\lambda_{1} v_{n}^{p-1}+\mu\lambda_{2} v_{n}^{q-1} \bigr) |v_{n}-v|\,dx \\ &{} + \int_{m\le|x|< m+1/l} \bigl(|\nabla v_{n}|^{p-1}+\mu| \nabla v_{n}|^{q-1} \bigr)|\nabla v|\,dx \\ &{}+d_{l} \int_{m\le|x|< m+1/l} \bigl(|\nabla v_{n}|^{p-1}+\mu| \nabla v_{n}|^{q-1} \bigr)|v_{n}-v|\,dx \\ \equiv& I_{n}^{1} +I_{n}^{2} +I_{n}^{3}, \end{aligned}$$

where $d_{l}:=\sup_{|x|< m+1/l}|\nabla\eta_{l}(x)|$.

By Hölder’s inequality, (16) and (17), we have

$$\begin{aligned} I_{n}^{1} \le& \|v_{n}-v\|_{L^{p}(B_{m+1})} \bigl\{ \|a_{0}\|_{L^{p^{\prime}}(B_{m+1})} +\|a_{1}\|_{L^{\tilde{r}_{1}}(B_{m+1})} \|v_{n}\|_{L^{p}(B_{m+1})}^{r_{1}} \\ &{} +\|a_{2}\|_{L^{\tilde{r}_{2}}(B_{m+1})}\|\nabla v_{n} \|_{L^{p}(B_{m+1})}^{r_{2}} +\lambda_{1}\|v_{n} \|_{L^{p}(B_{m+1})}^{p-1} +\mu\lambda_{2} \|v_{n} \|_{L^{p^{\prime}(q-1)}(B_{m+1})}^{q-1} \bigr\} \\ \le& C_{1}\|v_{n}-v\|_{L^{p}(B_{m+1})}, \end{aligned}$$

where $C_{1}$ is a positive constant independent of $v_{n}$, n, m, and l. Again by Hölder’s inequality the following estimates follow:

$$\begin{aligned}& \begin{aligned}[b] I_{n}^{2} \le{}& \|\nabla v_{n}\|_{L^{p}(B_{m+1})}^{p-1} \biggl( \int_{m\le |x|< m+1/l} |\nabla v|^{p} \,dx \biggr)^{1/p} \\ &{} +\mu\|\nabla v_{n}\|_{L^{q}(B_{m+1})}^{q-1} \biggl( \int_{m\le |x|< m+1/l} |\nabla v|^{q} \,dx \biggr)^{1/q}, \end{aligned} \\& I_{n}^{3} \le d_{l}\|v_{n}-v \|_{L^{p}(B_{m+1})}\|\nabla v_{n}\|_{L^{p}(B_{m+1})}^{p-1} +d_{l}\mu\|v_{n}-v\|_{L^{q}(B_{m+1})}\|\nabla v_{n}\|_{L^{q}(B_{m+1})}^{q-1}. \end{aligned}$$

Thereby, from (16), (17), and (20) we derive

$$\begin{aligned} &\limsup_{n\to\infty} V_{n} \\ &\quad\le C^{p-1} \biggl( \int_{m\le|x|< m+1/l} |\nabla v|^{p} \,dx \biggr)^{1/p} +\mu C^{q-1} \biggl( \int_{m\le|x|< m+1/l} |\nabla v|^{q} \,dx \biggr)^{1/q} \end{aligned}$$

for all $l\in\mathbb{N}$. Thus, letting $l\to\infty$, we obtain that $\limsup_{n\to\infty} V_{n}\le0$. As known from (19), $v_{n}$ weakly converges to v in $W^{1,p}(B_{m})$ and $W^{1,q}(B_{m})$, so we may write

$$\begin{aligned} V_{n}+o(1) =& \int_{B_{m}} \bigl(|\nabla v_{n}|^{p-2}\nabla v_{n}-|\nabla v|^{p-2}\nabla v \bigr) (\nabla v_{n}-\nabla v)\,dx \\ &{} +\mu \int_{B_{m}} \bigl(|\nabla v_{n}|^{q-2}\nabla v_{n}-|\nabla v|^{q-2}\nabla v \bigr) (\nabla v_{n}-\nabla v)\,dx \\ \ge& \bigl(\|\nabla v_{n}\|_{L^{p}(B_{m})}^{p-1}-\|\nabla v\| _{L^{p}(B_{m})}^{p-1} \bigr) \bigl(\|\nabla v_{n} \|_{L^{p}(B_{m})}-\|\nabla v\|_{L^{p}(B_{m})} \bigr) \\ &{} +\mu \bigl(\|\nabla v_{n}\|_{L^{q}(B_{m})}^{p-1}-\| \nabla v\|_{L^{q}(B_{m})}^{q-1} \bigr) \bigl(\|\nabla v_{n} \|_{L^{q}(B_{m})}-\|\nabla v\|_{L^{q}(B_{m})} \bigr) \\ \ge& 0. \end{aligned}$$

(22)

What we have shown entails $\lim_{n\to\infty} V_{n}=0$, $\lim_{n\to \infty}\|\nabla v_{n}\|_{L^{p}(B_{m})}=\|\nabla v\|_{L^{p}(B_{m})}$ and $\lim_{n\to\infty}\|\nabla v_{n}\|_{L^{q}(B_{m})}=\|\nabla v\|_{L^{q}(B_{m})}$ if $\mu>0$. This implies that $v_{n}$ converges to v strongly in $W^{1,p}(B_{m})$ and $W^{1,q}(B_{m})$ because the spaces $W^{1,p}(B_{m})$ and $W^{1,q}(B_{m})$ are uniformly convex.

Recalling that $v_{n}>0$ in $B_{m}$, we infer that v is a nonnegative solution of the problem

$$-\varDelta _{p} v+\lambda_{1} |v|^{p-2}v-\mu \varDelta _{q}v+\mu \lambda_{2} |v|^{q-2}v=f(x,v,\nabla v) \quad\mbox{in } B_{m}, v\ge 0 \mbox{ on } \partial B_{m}. $$

Now, by a diagonal argument and (21) there exist a relabeled subsequence of $\{v_{n}\}$ and a function $v\in W^{1,p}(\mathbb{R}^{N})$ such that

$$\begin{aligned}& v_{n} \to v \quad \mbox{in } W^{1,p}_{\mathrm{loc}}\bigl( \mathbb{R}^{N}\bigr), \\& v_{n}(x)\rightarrow v(x) \quad\mbox{for a.e } x\in \mathbb{R}^{N}. \end{aligned}$$

These convergence properties ensure that v is a solution of problem (P).

The next step in the proof is to show that v does not vanish in Ω. To do this, we fix $m\in\mathbb{N}$ and a positive constant $b_{m}$ satisfying $b_{m}\le\inf_{x\in B_{m}} b_{0}(x)$, where the function $b_{0}$ appears in assumption (F2). Moreover, choosing $b_{m}$ even smaller, Lemma 3 provides a solution $u_{m}$ of the problem

$$ \left \{ \textstyle\begin{array}{@{}l@{\quad}ll} -\varDelta _{p} u+\lambda_{1} |u|^{p-2}u-\mu \varDelta _{q}u+\mu\lambda_{2} |u|^{q-2}u =b_{m} u^{r_{0}} &\mbox{in } B_{m},\\ u>0&\mbox{in } B_{m},\\ u(x)= 0 &\mbox{on }\partial B_{m} \end{array}\displaystyle \right . $$

such that $\|u_{m}\|_{L^{\infty}(B_{m})}\le\delta_{0}$, where $\delta_{0}$ is given in assumption (F2). It follows from hypothesis (F2) that if $t\le\|u_{m}\|_{L^{\infty}(B_{m})}$, then $f(x,t,\xi)\ge b_{0}(x)t^{r_{0}}$ for all $x\in\mathbb{R}^{N}$, $\xi\in\mathbb{R}^{N}$. We are thus in a position to apply Theorem 4 to the functions $u_{m}$ and $v_{n}$ with $n>m$ in place of $u_{1}=u_{m}$ and $u_{2}=v_{n}$, respectively, which renders $v_{n}\ge u_{m}$ in $B_{m}$ for every $n>m$. This enables us to deduce that $v\geq u_{m}$ in $B_{m}$, so $v(x)>0$ for almost every $x\in\mathbb{R}^{N}$ because m was arbitrary.

Furthermore, since $\lambda_{1}>0$ and $v_{n}$ weakly converges to v in $W^{1,p}(\mathbb{R}^{N})$ (we can extend $v_{n}(x)=0$ if $|x|\ge n$ (so $\|v_{n}\|_{p,n}=\|\nabla v_{n}\|_{L^{p}(\mathbb{R}^{N})} +\lambda_{1} \|v_{n}\|_{L^{p}(\mathbb{R}^{N})}$)), by means of (16) and (17), we can check that $v\in W^{1,p}(\mathbb{R}^{N})\cap W^{1,q}(\mathbb{R}^{N})$ if $\mu>0$ and $v\in W^{1,p}(\mathbb{R}^{N})$ if $\mu=0$. According to the iteration process, it is proved that v is bounded on any bounded sets (see Section 5.1). Hence, the regularity theory as in [7] leads to $v\in C^{1}_{\mathrm{loc}}(\mathbb{R}^{N})$. The proof of Theorem 1 is complete. □

5 Proof of Theorem 2

Throughout this section, we fix any (positive) solution v of (P) (belonging to $W^{1,p}(\mathbb{R}^{N})$). Define $v_{M}:=\max\{v,M\}$ for $M>0$. Here, we choose $\overline{p}^{*}$ satisfying $p^{2}<\overline{p}^{*}$ if $N\le p$ and set $\overline{p}^{*}=p^{*}=Np/(N-p)$ if $N>p$. For $R^{\prime}>R>0$, we take a smooth function $\eta_{R,R^{\prime}}$ such that $0\le\eta_{R,R^{\prime}}\le1$, $\|\eta_{R,R^{\prime}}^{\prime}\|_{\infty}\le2/(R^{\prime}-R)$, $\eta_{R,R^{\prime}}(t)=1$ if $t\le R$, and $\eta_{R,R^{\prime}}=0$ if $t\ge R^{\prime}$.

5.1 Boundedness of solutions

Lemma 4

Let $x_{0}\in\mathbb{R}^{N}$, $M>0$, $R^{\prime}>R>0$, $\tilde{p}>1$, $\gamma_{i}>1$, and $1/\gamma_{i}+1/\gamma_{i}^{\prime}=1$ $(i=0, 1)$. Denote $\eta(x):=\eta_{R,R^{\prime}}(|x-x_{0}|)$. Assume that $\gamma_{i}^{\prime}\le\tilde{p}$ $(i=0,1)$ and $v\in L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))$ with $\alpha\ge0$. Then:

$$\begin{aligned}& \int_{B(x_{0},R^{\prime})} a_{0}vv_{M}^{\alpha} \eta^{p}\,dx \le\|a_{0}\|_{L^{\gamma_{0}}(B(x_{0},R^{\prime}))} \|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{1+\alpha} B_{R^{\prime}}, \end{aligned}$$

(23)

$$\begin{aligned}& \int_{B(x_{0},R^{\prime})} a_{1} v^{r_{1}+1}v_{M}^{\alpha} \eta^{p}\,dx \le\|a_{1}\|_{L^{\gamma_{1}}(B(x_{0},R^{\prime}))} \|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{r_{1}+1+\alpha} B_{R^{\prime}}, \end{aligned}$$

(24)

where $B_{R^{\prime}} :=(1+|B(0,R^{\prime})|)$, and $|B(0,R^{\prime})|$ denotes the Lebesgue measure of the ball $B(0,R^{\prime})$. Moreover, if $\gamma_{2}>p/(p-r_{2})$ and $\gamma_{3}:=\frac{(p-r_{2})\gamma_{2}}{(p-r_{2})\gamma_{2}-p}\le\tilde{p}$, then

$$\begin{aligned} \int_{B(x_{0},R^{\prime})} a_{2} |\nabla v|^{r_{2}}vv_{M}^{\alpha} \eta^{p}\,dx \le&\frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha} \eta^{p}\,dx \\ &{}+ 4^{\frac{r_{2}}{p-r_{2}}}\|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac {p}{p-r_{2}}} \|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\frac {p}{p-r_{2}}+\alpha} B_{R^{\prime}}. \end{aligned}$$

(25)

Proof

According to Hölder’s inequality, we easily show our assertions (23) and (24). So, we prove (25) only. By Young’s inequality and recalling that $r_{2}< p-1$ and $\eta^{p}\ge\eta ^{p^{2}/r_{2}}$, we have

$$\begin{aligned} \int_{B(x_{0},R^{\prime})} a_{2} |\nabla v|^{r_{2}}vv_{M}^{\alpha} \eta^{p}\,dx \le& \frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha} \eta^{p}\,dx \\ &{}+4^{\frac{r_{2}}{p-r_{2}}} \int_{B(x_{0},R^{\prime})} a_{2}^{\frac{p}{p-r_{2}}}v^{\frac{p}{p-r_{2}}} v_{M}^{\alpha}\,dx. \end{aligned}$$

Moreover, because of $\gamma_{2}>p/(p-r_{2})$, $p>p/(p-r_{2})$, and $\tilde{p}\ge\gamma_{3}$, applying Hölder’s inequality, we obtain

$$\begin{aligned} \int_{B(x_{0},R^{\prime})} a_{2}^{\frac{p}{p-r_{2}}}v^{\frac{p}{p-r_{2}}} v_{M}^{\alpha}\,dx \le& \|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac{p}{p-r_{2}}} \|v\|_{L^{\gamma_{3}(\frac{p}{p-r_{2}}+\alpha)}(B(x_{0},R^{\prime}))} ^{\frac{p}{p-r_{2}}+\alpha} \\ \le& \|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac{p}{p-r_{2}}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))} ^{\frac{p}{p-r_{2}}+\alpha}\bigl(1+\bigl|B\bigl(0,R^{\prime}\bigr)\bigr|\bigr). \end{aligned}$$

Hence, (25) follows. □

Lemma 5

Let $x_{0}\in\mathbb{R}^{N}$, $R^{\prime}>R>0$, $\tilde{p}>1$, $\gamma_{i}>1$, and $1/\gamma_{i}+1/\gamma_{i}^{\prime}=1$ $(i=0,1)$. Assume that $\gamma_{i}^{\prime}\le\tilde{p}$ $(i=0,1)$, $\gamma_{2}>p/(p-r_{2})$, and $\gamma_{3}:=\frac{(p-r_{2})\gamma_{2}}{(p-r_{2})\gamma_{2}-p}\le\tilde{p}$. If $v\in L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))$ with $\alpha\ge 0$, then

$$\begin{aligned} &\|v\|_{L^{\frac{\overline{p}^{*}}{p}(p+\alpha )}(B(x_{0},R))}^{p+\alpha} \le2^{p}(p+\alpha)^{p}C_{*}^{p}B_{R^{\prime}}(C_{R^{\prime}}+D_{R,R^{\prime}}) \max\bigl\{ 1,\|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}\bigr\} ^{p+\alpha} \end{aligned}$$

(26)

with

$$\begin{aligned}& B_{R^{\prime}} :=\bigl(1+\bigl|B\bigl(0,R^{\prime}\bigr)\bigr|\bigr), \\& C_{R^{\prime}} := \bigl\{ \|a_{0}\|_{L^{\gamma_{0}}(B(x_{0},R^{\prime}))}+ \|a_{1}\|_{L^{\gamma _{1}}(B(x_{0},R^{\prime}))} +4^{\frac{r_{2}}{p-r_{2}}}\|a_{2} \|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac {p}{p-r_{2}}} \bigr\} , \\& D_{R,R^{\prime}} := \biggl\{ \frac{2^{3p-2}p^{p}+2^{p-1}}{(R^{\prime}-R)^{p}} +\frac{\mu2^{3q-2}p^{q}}{(R^{\prime}-R)^{q}} \biggr\} , \end{aligned}$$

where $C_{*}$ is the positive constant from embedding from $W^{1,p}(\mathbb{R}^{N})$ to $L^{\overline{p}^{*}}(\mathbb{R}^{N})$.

Proof

Taking $vv_{M}^{\alpha}\eta^{p}\in W^{1,p}_{0}(B(x_{0},R^{\prime}))$ (for $M>0$) as a test function, where $\eta(x)=\eta_{R,R^{\prime}}(|x-x_{0}|)$, by Lemma 4 and (F1) we obtain

$$\begin{aligned} &\|a_{0}\|_{L^{\gamma_{0}}(B(x_{0},R^{\prime}))} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{1+\alpha }B_{R^{\prime}} \\ &\qquad{}+\|a_{1}\|_{L^{\gamma_{1}}(B(x_{0},R^{\prime}))} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{r_{1}+1+\alpha }B_{R^{\prime}} \\ &\qquad{} +4^{\frac{r_{2}}{p-r_{2}}}\|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}\|v\| _{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\frac{p}{p-r_{2}}+\alpha} B_{R^{\prime}} +\frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx \\ &\quad\ge \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx +\lambda_{1} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\eta^{p} \,dx -\frac{2p}{R^{\prime}-R} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p-1} v_{M}^{\alpha}v\eta^{p-1}\,dx \\ &\qquad{}+\mu \biggl\{ \int_{B(x_{0},R^{\prime})} |\nabla v|^{q}v_{M}^{\alpha}\eta^{p}\,dx -\frac{2p}{R^{\prime}-R} \int_{B(x_{0},R^{\prime})} |\nabla v|^{q-1} v_{M}^{\alpha}v\eta^{p-1}\,dx \biggr\} , \end{aligned}$$

(27)

where we use $|\nabla\eta|\le2/(R^{\prime}-R)$. According to Young’s and Hölder’s inequalities, for $j=p,q$, we see that

$$\begin{aligned} &\frac{2p}{R^{\prime}-R} \int_{B(x_{0},R^{\prime})} |\nabla v|^{j-1} v_{M}^{\alpha}v\eta^{p-1}\,dx \\ &\quad\le \frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{j} v_{M}^{\alpha}\eta^{p}\,dx +\frac{2^{j}p^{j}4^{j-1}}{(R^{\prime}-R)^{j}} \int_{B(x_{0},R^{\prime})}v^{j+\alpha}\eta^{p-j}\,dx \\ &\quad\le \frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{j} v_{M}^{\alpha}\eta^{p}\,dx +\frac{2^{3j-2}p^{j}}{(R^{\prime}-R)^{j}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{j+\alpha} B_{R^{\prime}}. \end{aligned}$$

(28)

Consequently, because of $\mu\ge0$ and $p+\alpha>r_{1}+1+\alpha,p/(p-r_{2})+\alpha$, it follows from (27) and (28) that

$$\begin{aligned} &B_{R^{\prime}} \biggl(C_{R^{\prime}} +\frac{2^{3p-2}p^{p}}{(R^{\prime}-R)^{p}} + \frac{\mu2^{3q-2}p^{q}}{(R^{\prime}-R)^{q}} \biggr) \max\bigl\{ 1,\|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}\bigr\} ^{p+\alpha} \\ &\quad\ge \frac{1}{2} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx +\lambda_{1} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\eta^{p} \,dx. \end{aligned}$$

(29)

Moreover, by using

$$\begin{aligned} &\bigl\| \nabla\bigl(v_{M}^{1+\alpha/p}\eta\bigr)\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p}\\ &\quad\le2^{p-1} \bigl\{ \bigl\| \eta\nabla\bigl(v_{M}^{1+\alpha/p} \bigr)\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p} +\bigl\| v_{M}^{1+\alpha/p}\nabla \eta\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p} \bigr\} \\ &\quad\le 2^{p-1} \biggl(1+\frac{\alpha}{p} \biggr)^{p} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx +\frac{2^{2p-1}}{(R^{\prime}-R)^{p}} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\,dx \end{aligned}$$

and Hölder’s inequality, due to the embedding from $W^{1,p}(\mathbb{R}^{N})$ to $L^{\overline{p}^{*}}(\mathbb{R}^{N})$, we have

$$\begin{aligned} &\frac{1}{2} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha}\eta^{p}\,dx +\lambda_{1} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\eta^{p} \,dx \\ &\quad\ge2^{-p}p^{p}(p+\alpha)^{-p} \bigl\{ \bigl\| \nabla\bigl(v_{M}^{1+\alpha/p}\eta\bigr)\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p} +\lambda_{1}\bigl\| v_{M}^{1+\alpha/p}\eta\bigr\| _{L^{p}(\mathbb{R}^{N})}^{p} \bigr\} \\ &\qquad{} -\frac{2^{p-1}p^{p}}{(p+\alpha)^{p}(R^{\prime}-R)^{p}} \int_{B(x_{0},R^{\prime})}v_{M}^{p+\alpha}\,dx \\ &\quad\ge2^{-p}p^{p}(p+\alpha)^{-p} \bigl\| v_{M}^{1+\alpha/p}\eta\bigr\| _{W^{1,p}(\mathbb{R}^{N})}^{p} \\ &\qquad{} -\frac{2^{p-1}}{(R^{\prime}-R)^{p}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{p+\alpha }\bigl(1+\bigl|B \bigl(0,R^{\prime}\bigr)\bigr|\bigr) \\ &\quad\ge 2^{-p}p^{p}(p+\alpha)^{-p}C_{*}^{-p} \bigl\| v_{M}^{1+\alpha/p}\eta\bigr\| _{L^{\overline{p}^{*}}(\mathbb{R}^{N})}^{p} - \frac{2^{p-1}}{(R^{\prime}-R)^{p}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{p+\alpha }B_{R^{\prime}} \\ &\quad\ge2^{-p}p^{p}(p+\alpha)^{-p}C_{*}^{-p} \|v_{M}\|_{L^{\overline{p}^{*}(p+\alpha)/p}(B(x_{0},R))}^{p+\alpha} \\ &\qquad{} -\frac{2^{p-1}}{(R^{\prime}-R)^{p}} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{p+\alpha} B_{R^{\prime}}. \end{aligned}$$

(30)

Therefore, (29) and (30) lead to

$$\begin{aligned} &2^{-p}p^{p}(p+\alpha)^{-p}C_{*}^{-p} \|v_{M}\|_{L^{\overline{p}^{*}(p+\alpha)/p}(B(x_{0},R))}^{p+\alpha} \\ &\quad\le B_{R^{\prime}}(C_{R^{\prime}}+D_{R,R^{\prime}}) \max\bigl\{ 1,\|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}\bigr\} ^{p+\alpha}. \end{aligned}$$

(31)

Applying Fatou’s lemma and letting $M\to\infty$ in (31), our conclusion follows. □

Proposition 1

Under the assumptions in Theorem 2, we have that $v\in L^{\infty}(\mathbb{R}^{N})$.

Proof

First, in the case of $N>p$, we note that

$$\begin{aligned}& \gamma_{j}^{\prime}< \frac{p^{*}}{p} \quad\Longleftrightarrow \quad \gamma_{j}>\frac{p^{*}}{p^{*}-p}\quad (j=0,1), \\& \gamma_{2}>\frac{p}{p-r_{2}}\quad \mbox{and} \quad \gamma_{3}:=\frac{(p-r_{2})\gamma_{2}}{(p-r_{2})\gamma_{2}-p}< \frac{p^{*}}{p} \quad \Longleftrightarrow\quad \gamma_{2}>\frac{pp^{*}}{(p-r_{2})(p^{*}-p)}. \end{aligned}$$

In the cases of (i) and (ii) (case $p<\overline{p}^{*}/p$), we take $\gamma_{0}=p^{\prime}$ and $\gamma_{j}=\tilde{r}_{j}$ ($j=1,2$). Then, we have $\gamma_{0}^{\prime}=p$, $\gamma_{1}^{\prime}=\tilde{r}_{1}^{\prime}=p/(r_{1}+1)\le p$, $\gamma_{2}=\tilde{r}_{2}=p/(p-r_{2}-1)>p/(p-r_{2})$, and $\gamma_{3}=(p-r_{2})\tilde{r_{2}}/((p-r_{2})\tilde{r}_{2}-p)=p-r_{2}\le p$. Choose p̃ such that

$$\begin{aligned}& \bigl(\max\bigl\{ \gamma_{0}^{\prime},\gamma_{1}^{\prime}, \gamma_{3}\bigr\} = \bigr) \tilde{p}=p \biggl(< \frac{\overline{p}^{*}}{p}\biggr) \mbox{ in the cases of (i) and (ii)}, \\& \max\bigl\{ \gamma_{0}^{\prime},\gamma_{1}^{\prime}, \gamma_{3}\bigr\} \le \tilde{p}< \frac{p^{*}}{p} \mbox{ in the case of (iii)}. \end{aligned}$$

Let $R_{*}$ be the positive constant satisfying (4) in the case of (iii) and any positive constant in the cases of (i) and (ii). Put

$$ A_{i}:=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \|a_{i}\|_{L^{\gamma_{i}}(\mathbb{R}^{N})} & \mbox{if (i) and (ii)}, \\ \sup_{x\in\mathbb{R}^{N}} \|a_{i}\|_{L^{\gamma_{i}}(B(x,2R_{*}))} &\mbox{if (iii)} \end{array}\displaystyle \right . $$

for $i=0,1,2$. Define the sequences $\{\alpha_{n}\}$, $\{R_{n}^{\prime}\}$, and $\{R_{n}\}$ by

$$\begin{aligned}& \alpha_{0}:=\frac{\overline{p}^{*}}{\tilde{p}}-p>0,\qquad \tilde{p}(p+ \alpha_{n+1})=\frac{\overline{p}^{*}}{p} (p+\alpha_{n}), \\& R_{n}^{\prime}:=\bigl(1+2^{-n}\bigr)R_{*}, \qquad R_{n}:=R_{n+1}^{\prime}. \end{aligned}$$

Recall that $v\in W^{1,p}(\mathbb{R}^{N})$, and using the embedding of $W^{1,p}(\mathbb{R}^{N})$ to $L^{\overline{p}^{*}}(\mathbb{R}^{N})$, we see that $v\in L^{\overline{p}^{*}}(\mathbb{R}^{N})=L^{\tilde{p}(p+\alpha _{0})}(\mathbb{R}^{N})$.

Fix any $x_{0}\in\mathbb{R}^{N}$. Then Lemma 5 guarantees that if $v\in L^{\tilde{p}(p+\alpha_{n})}(B(x_{0},R_{n}^{\prime}))$, then $v\in L^{\frac{\overline{p}^{*}}{p}(p+\alpha_{n})}(B(x_{0},R_{n})) =L^{\tilde{p}(p+\alpha_{n+1})}(B(x_{0},R_{n+1}^{\prime}))$. Noting that

$$\begin{aligned}& B_{R^{\prime}_{n}} \le\bigl(1+\bigl|B(0,2R_{*})\bigr|\bigr)=:B_{0}, \\& C_{R_{n}^{\prime}} \le A_{0}+A_{1}+4^{\frac{r_{2}}{p-r_{2}}}A_{2}^{\frac {p}{p-r_{2}}}+1=:C_{0}, \\& D_{R_{n},R_{n}^{\prime}} \le\frac{(1+p^{p})2^{p(n+3)}}{R_{*}^{p}} +\frac{\mu q^{q}2^{q(n+3)}}{R_{*}^{q}}=:D_{n} \le C^{\prime}2^{p(n+3)} \end{aligned}$$

for any $n\ge0$ with sufficiently large $C^{\prime}$ independent of n and setting

$$b_{n}:=\max\bigl\{ 1,\|v\|_{L^{\tilde{p}(p+\alpha_{n})(B(x_{0},R_{n}^{\prime}))}}\bigr\} , $$

by Lemma 5 we obtain

$$ b_{n+1}\le C^{\frac{1}{p+\alpha_{n}}}(p+\alpha_{n})^{\frac{p}{p+\alpha_{n}}} (C_{0}+D_{n})^{\frac{1}{p+\alpha_{n}}}b_{n} $$

(32)

for every $n\ge0$ with $C:=2^{p}(C_{*}+1)^{p}B_{0}$. Put $P:=\tilde{p}p/\overline{p}^{*}<1$. Then, because of $p+\alpha_{n+1}=(p+\alpha_{n})/P$, $\alpha_{n+1}>\alpha_{n}/P>\alpha_{0}(1/P)^{n+1}\to\infty$ as $n\to \infty$. Moreover, we see that

$$\begin{aligned} &S_{1} :=\sum_{n=0}^{\infty}\frac{1}{p+\alpha_{n}}= \frac{1}{p+\alpha_{0}}\sum_{n=0}^{\infty}P^{n} =\frac{1}{(p+\alpha_{0})(1-P)}< \infty, \\ &S_{2} :=\ln \prod_{n=0}^{\infty}(p+ \alpha_{n})^{\frac{p}{p+\alpha_{n}}} =\frac{p}{p+\alpha_{0}}\sum _{n=0}^{\infty}P^{n} \bigl(\ln (p+ \alpha_{0})+n\ln P^{-1} \bigr)< \infty, \end{aligned}$$

and

$$\begin{aligned} S_{3}&:=\ln \prod_{n=0}^{\infty}(C_{0}+D_{n})^{\frac{1}{p+\alpha_{n}}} =\sum_{n=0}^{\infty}\frac{P^{n}}{p+\alpha_{0}} \ln (C_{0}+D_{n}) \\ &\le\sum_{n=0}^{\infty}\frac{P^{n}}{p+\alpha_{0}} p(n+3)\ln \bigl(C_{0}+C^{\prime}\bigr)2< \infty. \end{aligned}$$

As a result, by iteration in (32) and the equality $\tilde{p}(p+\alpha_{0})=\overline{p}^{*}$ we obtain

$$\|v\|_{L^{\frac{\overline{p}^{*}}{p}(p+\alpha_{n})}(B(x_{0},R_{*}))} \le b_{n} \le C^{S_{1}}e^{S_{2}}e^{S_{3}} \max\bigl\{ 1,\|v\|_{L^{\overline{p}^{*}}(B(x_{0},2R_{*}))}\bigr\} $$

for every $n\ge1$. Letting $n\to\infty$, this ensures that

$$ \|v\|_{L^{\infty}(B(x_{0},R_{*}))} \le C^{S_{1}}e^{S_{2}}e^{S_{3}} \max\bigl\{ 1,\|v\|_{L^{\overline{p}^{*}}(B(x_{0},2R_{*}))}\bigr\} . $$

(33)

Recalling that $v\in W^{1,p}(\mathbb{R}^{N})$ and using the embedding of $W^{1,p}(\mathbb{R}^{N})$ to $L^{\overline{p}^{*}}(\mathbb{R}^{N})$, (33) yields that

$$\begin{aligned} \|v\|_{L^{\infty}(B(x_{0},R_{*}))} &\le C^{S_{1}}e^{S_{2}}e^{S_{3}} \max\bigl\{ 1,\|v\|_{L^{\overline{p}^{*}}(\mathbb{R}^{N})}\bigr\} \\ &\le C^{S_{1}}e^{S_{2}}e^{S_{3}} \max\bigl\{ 1,C_{*}\|v \|_{W^{1,p}(\mathbb{R}^{N})}\bigr\} , \end{aligned}$$

whence v is bounded in $\mathbb{R}^{N}$ because $x_{0}\in\mathbb{R}^{N}$ is arbitrary and the constant $C^{S_{1}}e^{pS_{2}}e^{S_{3}}$ is independent of $x_{0}$. □

5.2 Proof of Theorem 2

Proof of Theorem 2

Since v is bounded in $\mathbb{R}^{N}$ by Proposition 1, we put $M_{0}:=\|v\|_{L^{\infty}(\mathbb{R}^{N})}$. Then, as in Lemma 4, we see that

$$\begin{aligned} & \int_{B(x_{0},R^{\prime})} a_{0}vv_{M}^{\alpha} \eta^{p}\,dx \le\|a_{0}\|_{L^{\gamma_{0}}(B(x_{0},R^{\prime}))} M_{0} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha} B_{R^{\prime}}, \end{aligned}$$

(34)

$$\begin{aligned} & \int_{B(x_{0},R^{\prime})} a_{1} v^{r_{1}+1}v_{M}^{\alpha} \eta^{p}\,dx \le\|a_{1}\|_{L^{\gamma_{1}}(B(x_{0},R^{\prime}))} M_{0}^{1+r_{1}}\|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha} B_{R^{\prime}}, \end{aligned}$$

(35)

$$\begin{aligned} &\begin{aligned}[b] & \int_{B(x_{0},R^{\prime})} a_{2} |\nabla v|^{r_{2}}vv_{M}^{\alpha} \eta^{p}\,dx \\ &\quad \le\frac{1}{4} \int_{B(x_{0},R^{\prime})} |\nabla v|^{p}v_{M}^{\alpha} \eta^{p}\,dx + 4^{\frac{r_{2}}{p-r_{2}}}\|a_{2}\|_{L^{\gamma_{2}}(B(x_{0},R^{\prime}))}^{\frac {p}{p-r_{2}}} M_{0}^{\frac{p}{p-r_{2}}}\|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha} B_{R^{\prime}}, \end{aligned} \end{aligned}$$

(36)

and

$$ \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{j+\alpha}\le M_{0}^{j}\|v \|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha}\quad (j=p,q) . $$

(37)

Fix any $x_{0}\in\mathbb{R}^{N}$. It follows from the argument as in the proof of Lemma 5 with (34), (35), (36), and (37) that

$$\begin{aligned} &\|v\|_{L^{\frac{\overline{p}^{*}}{p}(p+\alpha )}(B(x_{0},R))}^{p+\alpha} \le2^{p}(p+\alpha)^{p}C_{*}^{p}B_{R^{\prime}}(C_{R^{\prime}}+D_{R,R^{\prime}}) (M_{0}+1)^{p} \|v\|_{L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))}^{\alpha}, \end{aligned}$$

(38)

provided that $v\in L^{\tilde{p}(p+\alpha)}(B(x_{0},R^{\prime}))$. Choose $\gamma_{i}$ $(i=0,1,2)$ and p̃ and define the sequences $\{\alpha_{n}\}$, $\{R_{n}^{\prime}\}$, and $\{R_{n}\}$ as in the proof of Proposition 1. Set

$$V_{n}:=\|v\|_{L^{\tilde{p}(p+\alpha_{n})}(B(x_{0},R^{\prime}_{n}))}^{\alpha_{n}}. $$

Then, by the same argument as in the proof of Proposition 1 with (38) we obtain

$$ V_{n}^{\frac{p+\alpha_{n-1}}{\alpha_{n}}} \le C (p+\alpha_{n-1})^{p}(C_{0}+D_{n-1}) V_{n-1} $$

(39)

with $C:=2^{p}C_{*}^{p}B_{0}(M_{0}+1)^{p}$. Recall that

$$\alpha_{n}+p=P^{-1}(p+\alpha_{n-1}) \quad \mbox{and}\quad \frac{p}{p+\alpha_{0}}=P. $$

Define

$$Q_{n}:=\prod_{k=2}^{n+1} \biggl(1+\frac{P^{k}}{1-P^{k}} \biggr) = \prod_{k=2}^{n+1} \bigl(1-P^{k} \bigr)^{-1} \quad \mbox{and}\quad W_{n}:=(C_{0}+D_{n}). $$

Then, inequality (39) leads to

$$\begin{aligned} \ln V_{n} \le{}&\frac{\alpha_{n}}{p+\alpha_{n-1}} \bigl(\ln V_{n-1}+\ln C(p+\alpha_{n-1})^{p}+\ln W_{n-1} \bigr) \\ ={}&P^{-1} \bigl(1-P^{n+1} \bigr) \bigl(\ln V_{n-1}+ p\ln C P^{-n+1}(p+\alpha_{0})+\ln W_{n-1} \bigr) \\ \le{}& P^{-1} \bigl(1-P^{n+1} \bigr)\ln V_{n-1} +pP^{-1}\ln (C+1) P^{-n+1}(p+\alpha_{0})+P^{-1} \ln W_{n-1} \\ \le{}& P^{-n} \Biggl( \prod _{k=1}^{n} \bigl(1-P^{k+1} \bigr) \Biggr) \ln V_{0} +p\sum_{k=1}^{n} P^{-k}\ln (C+1) P^{-n+k}(p+\alpha_{0}) \\ &{} +\sum_{k=1}^{n} P^{-k}\ln W_{n-k} \\ ={}& P^{-n}Q_{n}^{-1} \ln V_{0} +p \sum_{k=1}^{n} P^{-k}\ln (C+1) P^{-n+k}(p+\alpha_{0}) +\sum_{k=1}^{n} P^{-k}\ln W_{n-k} \end{aligned}$$

for every n because of $\ln (C+1) P^{-n+1}(p+\alpha_{0})> 0$ and $\ln W_{n}>0$ for all n. Therefore, we have

$$\begin{aligned} &\ln \|v\|_{L^{\tilde{p}(p+\alpha_{n})}(B(x_{0},R^{\prime}_{n}))} \\ &\quad=\frac{\ln V_{n}}{\alpha_{n}}=\frac{P^{n}\ln V_{n}}{p+\alpha_{0}-pP^{n}} \\ &\quad\le\frac{Q_{n}^{-1}\ln V_{0}}{p+\alpha_{0}-pP^{n}} +\frac{\sum_{l=0}^{n-1} P^{l} \ln (C+1) P^{-l}(p+\alpha _{0})}{p+\alpha_{0}-pP^{n}} +\frac{\sum_{l=0}^{n-1} P^{l}\ln W_{l}}{p+\alpha_{0}-pP^{n}}. \end{aligned}$$

(40)

Here, taking a sufficiently large positive constant $C^{\prime}$ independent of n, we see that

$$\sum_{l=0}^{n-1} P^{l} \ln (C+1) P^{-l}(p+\alpha_{0}) \le C^{\prime}\sum _{l=0}^{\infty}P^{l} (l+1)=:S_{1}< \infty $$

and

$$\sum_{l=0}^{n-1} P^{l}\ln W_{l}\le C^{\prime}\sum_{l=0}^{n-1} P^{l} (l+3)\le C^{\prime}\sum_{l=0}^{\infty}P^{l} (l+3) =:S_{2}< \infty. $$

Next, we shall show that $\{Q_{n}\}$ is a convergent sequence. It is easy see that $\{Q_{n}\}$ is increasing. Moreover, setting $d_{k}:=\ln (1+\frac{P^{k}}{1-P^{k}} )$, we see that

$$\lim_{k\to\infty}\frac{d_{k+1}}{d_{k}} = \lim_{k\to\infty} \frac{\ln (1-P^{k+1})}{\ln (1-P^{k})} =\lim_{k\to\infty}\frac{1-P^{k}}{1-P^{k+1}} P=P< 1 $$

by L’Hospital’s rule. This implies that

$$\ln Q_{n}=\sum_{k=2}^{n+1} \ln \biggl(1+\frac {P^{k}}{1-P^{k}} \biggr) \le\sum_{k=1}^{\infty}\ln \biggl(1+\frac{P^{k}}{1-P^{k}} \biggr) < \infty. $$

Therefore, $\{Q_{n}\}$ is bounded from above, whence $\{Q_{n}\}$ converges, and

$$1< \frac{1}{1-P^{2}}=Q_{1}\le Q_{\infty}:=\lim _{n\to\infty}Q_{n}< \infty. $$

Consequently, letting $n\to\infty$ in (40), we have

$$\|v\|_{L^{\infty}(B(x_{0},R_{*}))} \le(pS_{1}S_{2})^{\frac{1}{p+\alpha_{0}}} \|v \|_{L^{\overline{p}^{*}}(B(x_{0},2R_{*}))}^{\frac{\alpha_{0}}{(p+\alpha _{0})Q_{\infty}}}. $$

This yields our conclusion since $\|v\|_{L^{\overline{p}^{*}}(B(x_{0},2R_{*}))}\to0$ as $|x_{0}|\to \infty$, $\alpha_{0}>0$, and the constant $pS_{1}S_{2}$ is independent of $x_{0}$. □