Existence of positive solutions for nonlocal problems with indefinite nonlinearity

Abstract

In this paper, we consider the following new nonlocal problem:

$$\left \{ \textstyle\begin{array}{l@{\quad}l} - (a-b\int_{\varOmega} \vert \nabla u \vert ^{2}\,dx )\Delta u=\lambda f(x) \vert u \vert ^{p-2}u, & x\in\varOmega,\\u=0, & x\in\partial\varOmega, \end{array}\displaystyle \right . $$

where Ω is a smooth bounded domain in $\mathbb{R}^{3}$, $a,b>0$ are constants, $3< p<6$, and the parameter $\lambda>0$. Under some assumptions on the sign-changing function f, we obtain the existence of positive solutions via variational methods.

1 Introduction and main resluts

In this paper, we are concerned with the existence of positive solutions for the following new nonlocal problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} - (a-b\int_{\varOmega} \vert \nabla u \vert ^{2}\,dx )\Delta u=\lambda f(x) \vert u \vert ^{p-2}u, & x\in\varOmega,\\u=0, & x\in\partial\varOmega, \end{array}\displaystyle \right . $$

(1.1)

where Ω is a smooth bounded domain in $\mathbb {R}^{3}$, $a,b>0$ are constants, $3< p<6$, and the parameter $\lambda >0$. $f(x)$ is sign changing in Ω, which is the reason why we call it indefinite nonlinearity in the title.

In recent years, the Kirchhoff type problems in a bounded domain

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} - (a+b\int_{\varOmega} \vert \nabla u \vert ^{2}\,dx )\Delta u=f(x,u), & x\in\varOmega,\\u=0, & x\in\partial\varOmega, \end{array}\displaystyle \right . $$

(1.2)

have been investigated by a lot of scholars under different assumptions on $f(x,u)$, see e.g. [1–9]. There are also many studies for problem (1.2) on the whole space $\mathbb {R}^{N}$ ($N\ge3$). For this case, we refer the interested readers to [10–17]. In particular, Chen [10] obtained the existence and multiplicity of positive solutions to the Kirchhoff problem with indefinite nonlinearity by using mountain pass theorem and minimization argument.

In problem (1.2), if we replace $a+b\int_{\varOmega} |\nabla u|^{2}\,dx$ by $a-b\int_{\varOmega} |\nabla u|^{2}\,dx$, it turns to be the following new nonlocal problem:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} - (a-b\int_{\varOmega} \vert \nabla u \vert ^{2}\,dx )\Delta u=f(x,u), & x\in\varOmega,\\u=0, & x\in\partial\varOmega. \end{array}\displaystyle \right . $$

(1.3)

So far, there are only a few works about such a nonlocal problem. Yin and Liu [18] first studied this kind of nonlocal one and obtained two nontrivial solutions for problem (1.3) when $f(x,u)=|u|^{p-2}u$ with $2< p<2^{*}$ ($2^{*}={\frac{2N}{N-2}}$ when $N\ge 3$ and $2^{*}=+\infty$ when $N=1,2$). Lei et al. [19] proved that problem (1.3) has at least two positive solutions in the case of $f(x,u)=f_{\lambda }(x)|u|^{q-2}u$, where $1< q<2$ and $f_{\lambda }(x)\in L^{\infty}(\varOmega)$ changes sign. Generalization of the result of [18] and [19] was done by Duan et al. [20]. In [20], the authors proved the existence and multiplicity results for problem (1.3) in the case of $f(x,u)=f(x)|u|^{p-2}u$, where $1\le p<2^{*}$ and $f(x)\in L^{\frac{2^{*}}{2^{*}-p}}(\varOmega)\setminus\{0\}$ is nonnegative. Subsequently, Lei et al. [21] also obtained the multiplicity of nontrivial solutions for problem (1.3) with singularity. On the whole space $\mathbb {R}^{4}$, Wang et al. [22] researched problem (1.3) when $f(x,u)=|u|^{2}u+\mu f(x)$. Under the assumption $f(x)\in L^{4/3}(\mathbb {R}^{4})$, they showed the existence of multiple positive solutions for the considered problem.

From the works described above, we do not see any existence result of positive solution to problem (1.1) in the case that $3< p<6$, the nonlocal term $a-b\int_{\varOmega} |\nabla u|^{2}\,dx$, and indefinite nonlinearity. Thus, it is natural to ask what the case would be. Our purpose of this paper is to show some existence of positive solutions for problem (1.1).

In order to state our main results, we first make the following hypotheses on f:

($f_{1}$):

$f(x)\in L^{\infty}(\varOmega)$;

($f_{2}$):

$|\varSigma_{0}|=0$ and the set $\varSigma_{+}$ has an interior point, where $\varSigma_{0}=\{x\in\varOmega: f(x)=0\}$, $\varSigma_{+}=\{x\in\varOmega: f(x)>0\}$, and $\varSigma_{-}=\{ x\in\varOmega: f(x)<0\}$;

($f'_{1}$):

$f(x)\in L^{p_{*}}(\varOmega)$, where $p_{*}= {\frac{6}{6-p}}$;

($f'_{2}$):

there is $d_{0}>0$ such that the set $\varSigma_{d_{0}}=\{ x\in\varOmega: f(x)\ge d_{0}\}$ has an interior point.

Our main results can be stated as follows.

Theorem 1.1

Assume that hypotheses ($f_{1}$), ($f_{2}$) hold and$4\le p<6$. Then, for any$\lambda >0$, problem (1.1) has at least a positive solution.

Theorem 1.2

Assume that hypotheses ($f'_{1}$), ($f'_{2}$) hold and$3< p<4$. Then there exists$\lambda _{\ast}>0$such that, for any$\lambda \in(0,\lambda _{\ast})$, problem (1.1) has at least a positive solution.

Problem (1.1) is variational in nature. Note that, for any $u\in H_{0}^{1}(\varOmega)$, the functional

$$I(u)= \frac{a}{2} \Vert u \Vert ^{2} - \frac{b}{4} \Vert u \Vert ^{4}- \frac{\lambda }{p} \int _{\varOmega}f(x) \vert u \vert ^{p}\,dx $$

is well defined, where $\|u\|^{2}=\int_{\varOmega}|\nabla u|^{2}\,dx$. Furthermore, it belongs to $C^{1}(H^{1}_{0}(\varOmega),\mathbb {R})$ and its critical points are precisely the weak solutions of problem (1.1). Here, we say $u \in H^{1}_{0}(\varOmega)$ is a weak solution to problem (1.1) if, for any $v \in H^{1}_{0}(\varOmega)$, it holds

$$\biggl(a-b \int_{\varOmega} \vert \nabla u \vert ^{2}\,dx \biggr) \int_{\varOmega} \nabla u \nabla {v}\,dx - \lambda \int_{\varOmega}f(x) \vert u \vert ^{p-2}u v \,dx=0. $$

In the proof of Theorem 1.1, the main difficulty is to prove the boundedness of the Palais–Smale ($(\mathit{PS})$ for short) sequence, which cannot be proved directly in the case of $4\le p<6$ and indefinite nonlinearity. We will prove the boundedness of $(\mathit{PS})$ sequence indirectly, which is inspired by [10]. In the proof of Theorem 1.2, since ($f'_{2}$) without assuming $|\varSigma_{0}|=0$, we cannot get the compactness property for the corresponding energy functional as in the proof of Theorem 1.1. We overcome this difficulty by pulling the energy level down below some critical level. However, we now face the situation with superlinear and indefinite nonlinearity, it is difficult to achieve the critical energy level. To this end, we choose a suitable test function presented in [23] (see also [12]). At this point, we need to require the restrictive condition $p>3$. Without this condition, we do not know if the result is still right.

Throughout this paper, we make use of the following notations. $H^{1}_{0}(\varOmega)$ and $L^{q}(\varOmega)$ are standard Sobolev spaces with the usual norm $\|u\|^{2}=\int_{\varOmega}|\nabla u|^{2}\,dx$, $|u|^{q}_{q}=\int_{\varOmega}|u|^{q}\,dx$. $f_{+}=\max\{f,0\}$ and $f_{-}=\min\{f,0\}$. $B_{r}(x)$ denotes an open ball centered at x with radius $r>0$. → and ⇀ denote strong and weak convergence, respectively. C and $C_{i}$ denote various positive constants whose values may vary from line to line. $o(1)$ denotes a quantity such that $o(1)\to0$ as $n\to+\infty$. All limitations hold as $n\to\infty$ unless otherwise stated. Let S be the best Sobolev constant for the embedding of $H^{1}_{0}(\varOmega)$ into $L^{6}(\varOmega )$, that is,

$$ S = \inf_{u\in H^{1}_{0}(\varOmega)\setminus\{0\}}{\frac{\int_{\varOmega } \vert \nabla u \vert ^{2}\,dx}{ (\int_{\varOmega} \vert u \vert ^{6}\,dx )^{1/3}}}. $$

(1.4)

The paper contains two more sections, one is devoted to the proof of Theorem 1.1, the other is dedicated to the proof of Theorem 1.2.

2 Proof of Theorem 1.1

Lemma 2.1

Under the assumptions of Theorem1.1, for any$\lambda >0$, there exists a sequence$\{u_{n}\}\subset H^{1}_{0}(\varOmega)$such that$I(u_{n})\to c_{\ast}<{\frac{a^{2}}{4b}}$and$I'(u_{n})\to0$.

Proof

By ($f_{1}$), (1.4), and Hölder’s inequality, one has

$$\begin{aligned} I(u)&= \frac{a}{2} \Vert u \Vert ^{2} - \frac{b}{4} \Vert u \Vert ^{4} - \frac{\lambda }{p} \int _{\varOmega}f(x) \vert u \vert ^{p}\,dx \\ &\ge{\frac{a}{2}} \Vert u \Vert ^{2} - \frac{b}{4} \Vert u \Vert ^{4}- \frac{\lambda }{p} \vert f \vert _{\infty} \biggl( \int_{\varOmega}1\,dx \biggr)^{\frac{6-p}{6}} \biggl( \int _{\varOmega} \vert u \vert ^{p\cdot{\frac{6}{p}}}\,dx \biggr)^{\frac{p}{6}} \\ &\ge{\frac{a}{2}} \Vert u \Vert ^{2} - \frac{b}{4} \Vert u \Vert ^{4}- \frac{\lambda }{p} \vert f \vert _{\infty} \vert \varOmega \vert ^{\frac{6-p}{6}}S^{-p/2} \Vert u \Vert ^{p}. \end{aligned}$$

Thanks to $4\le p<6$, the above inequality implies that, for any $\lambda >0$, there are $\alpha, \delta>0$ such that $I(u)\ge\alpha$ for all $\|u\|=\delta$.

Moreover, by ($f_{2}$), we know that there is $u\in H^{1}_{0}(\varOmega)$ satisfying $\int_{\varOmega} f(x)|u|^{p}\,dx>0$. Then we have that

$$\lim_{t\to+\infty}I(tu)=\lim_{t\to+\infty} \biggl[{ \frac{a}{ 2}}t^{2} \Vert u \Vert ^{2}-{ \frac{b}{4}}t^{4} \Vert u \Vert ^{4}-{ \frac{\lambda }{p}}t^{p} \int f(x) \vert u \vert ^{p}\,dx \biggr]=-\infty, $$

which means that there is $t_{0}$ large enough such that $\|e\|>\rho$ and $I(e)<0$, where $e=t_{0}u$.

Thus, we can use the mountain pass theorem without $(\mathit{PS})$ condition [24] to obtain a sequence $\{u_{n}\}\subset H^{1}_{0}(\varOmega)$ satisfying $I(u_{n})\to c_{*}$ and $I'(u_{n})\to0$ for

$${c}_{\ast}:=\inf_{\gamma\in\varGamma}\max_{t\in[0,1]}I \bigl(\gamma(t) \bigr)\ge\alpha>0, $$

where

$$\varGamma:=\bigl\{ \gamma\in C \bigl([0,1],H^{1}_{0}( \varOmega) \bigr): \gamma(0)=0 \text{ and } \gamma(1)=e \bigr\} . $$

To complete the proof of the lemma, it suffices to prove that $c_{*}<{\frac{a^{2}}{4b}}$. By easy calculations, one has

$$\begin{aligned} \max_{t\ge0}I(te)&=\max_{t\ge0} \biggl\{ \frac{a}{2}t^{2} \Vert e \Vert ^{2} - \frac{b}{4}t^{4} \Vert e \Vert ^{4}- \frac{\lambda }{p}t^{p} \int_{\varOmega }f(x) \vert e \vert ^{p}\,dx \biggr\} \\ &< \max_{t\ge0} \biggl\{ \frac{a}{2}t^{2} \Vert e \Vert ^{2} - \frac{b}{4}t^{4} \Vert e \Vert ^{4} \biggr\} \\ &\le{\frac{a^{2}}{4b}}. \end{aligned}$$

This together with the definition of $c_{*}$ implies $c_{*}<{\frac{a^{2}}{ 4b}}$. □

Lemma 2.2

Under the assumptions of Theorem1.1, the functionalIsatisfies the$(\mathit{PS})_{c}$condition with$c<{\frac{a^{2}}{4b}}$.

Proof

Let $\{u_{n}\}\subset H^{1}_{0}(\varOmega)$ be a $(PS)_{c}$ sequence for I with $c<{\frac{a^{2}}{4b}}$, that is,

$$ I(u_{n})\to c \quad\text{and} \quad I'(u_{n}) \to0. $$

(2.1)

By (2.1), we have that

$$ c+o(1)=I(u_{n})={\frac{a}{2}} \int_{\varOmega} \vert \nabla u_{n} \vert ^{2}\, dx-{\frac{b}{4}} \biggl( \int_{\varOmega} \vert \nabla u_{n} \vert ^{2} \,dx \biggr)^{2}-{\frac{\lambda }{p}} \int_{\varOmega}f(x) \vert u_{n} \vert ^{p} \,dx, $$

(2.2)

and for any $\psi\in H^{1}_{0}(\varOmega)$,

$$\begin{aligned} o(1) \Vert \psi \Vert &=\bigl\langle I'(u_{n}), \psi\bigr\rangle \\ &=a \int_{\varOmega} \nabla u_{n} \nabla \psi\,dx -b \biggl( \int_{\varOmega } \vert \nabla u_{n} \vert ^{2} \,dx \biggr) \int_{\varOmega} \nabla u_{n} \nabla \psi\,dx \\ &\quad-\lambda \int _{\varOmega}f(x) \vert u_{n} \vert ^{p-2}u_{n} \psi\,dx. \end{aligned}$$

(2.3)

First, we show that $\{u_{n}\}$ is bounded in $H_{0}^{1}(\varOmega)$. Inspired by [10], arguing by contradiction, we define $t_{n}=\| u_{n}\|$ and assume that $t_{n}\to+\infty$. Set $v_{n}=u_{n}/t_{n}$. Then we have

$$ \Vert v_{n} \Vert ={\frac{ \Vert u_{n} \Vert }{t_{n}}}=1\quad \text{for each } n\in\mathbb{N}. $$

(2.4)

Taking $\psi=u_{n}$ in (2.3), then dividing (2.2) and (2.3) by $\|u_{n}\|^{2}$, we obtain

$$ o(1)={\frac{c+o(1)}{ \Vert u_{n} \Vert ^{2}}}={\frac{a}{2}}-{ \frac{b}{4}} \int _{\varOmega} \vert \nabla u_{n} \vert ^{2} \,dx \int_{\varOmega} \vert \nabla v_{n} \vert ^{2} \,dx-{\frac{\lambda }{p}} \int_{\varOmega}f(x) \vert u_{n} \vert ^{p-2}v_{n}^{2}\,dx $$

(2.5)

and

$$ o(1)={\frac{o(1) \Vert u_{n} \Vert }{ \Vert u_{n} \Vert ^{2}}}=a-b \int_{\varOmega} \vert \nabla u_{n} \vert ^{2} \,dx \int_{\varOmega} \vert \nabla v_{n} \vert ^{2} \,dx-\lambda \int_{\varOmega }f(x) \vert u_{n} \vert ^{p-2}v_{n}^{2}\,dx. $$

(2.6)

We distinguish two cases. In case $p=4$, multiplying (2.5) by 4 and combining with (2.6), we obtain

$$\lim_{n\to\infty}a=0, $$

which is impossible, since $a>0$ is constant. In case $4< p<6$, we also obtain

$$\lim_{n\to\infty} \int_{\varOmega}f(x) \vert u_{n} \vert ^{p-2}v_{n}^{2}\, dx={\frac{ap}{\lambda (4-p)}}. $$

Combining this with (2.6) and $\|u_{n}\|\to\infty$, we must have

$$\lim_{n\to\infty} \int_{\varOmega} \vert \nabla v_{n} \vert ^{2} \,dx=0, $$

which is a contradiction to (2.4).

In conclusion, either of the two cases above will lead to a contradiction. Thus, we obtain that $\{u_{n}\}$ is bounded in $H_{0}^{1}(\varOmega)$. Then, up to a subsequence (still denoted by $\{ u_{n}\}$), we may assume that

$$ \textstyle\begin{cases} u_{n}\rightharpoonup u, & \text{in } H^{1}_{0}(\varOmega),\\ u_{n}\to u, & \text{in } L^{r}(\varOmega),1\le r< 6,\\ u_{n}\to u, & \text{a.e. in } \varOmega. \end{cases} $$

(2.7)

By using Hölder’s inequality, we obtain from (2.7) that

$$\biggl\vert \int_{\varOmega}f(x) \vert u_{n} \vert ^{p-2}u_{n}(u_{n}-u)\,dx \biggr\vert \le \vert f \vert _{\infty} \vert u_{n} \vert _{p}^{p-1} \vert u_{n}-u \vert _{p} \to0. $$

It then follows from (2.1) that

$$ o(1)=\bigl\langle I'(u_{n}),(u_{n}-u) \bigr\rangle = \bigl(a-b \Vert u_{n} \Vert ^{2} \bigr) \int _{\varOmega} \nabla u_{n}\nabla (u_{n}-u)+o(1), $$

(2.8)

that is,

$$a-b \Vert u_{n} \Vert ^{2}\to0 \quad\text{or}\quad \int_{\varOmega} \nabla u_{n}\nabla (u_{n}-u)\to0. $$

Next, we show that the former alternative above does not occur. If, to the contrary, namely,

$$ \Vert u_{n} \Vert ^{2}\to{ \frac{a}{b}}. $$

(2.9)

Let us define a functional by

$$\phi({{w}})= \frac{1}{p} \int_{\varOmega}f(x) \vert {{w}} \vert ^{p}\,dx,\quad {{w}}\in H^{1}_{0}(\varOmega). $$

Consequently,

$$\bigl\langle \phi'({{w}}),v\bigr\rangle = \int_{\varOmega }f(x) \vert {{w}} \vert ^{p-2}{{w}}v\,dx, \quad\forall v\in H^{1}_{0}(\varOmega). $$

By using Hölder’s inequality again, we obtain

$$\begin{aligned} & \biggl\vert \int_{\varOmega}f(x) \bigl( \vert u_{n} \vert ^{p-2}u_{n}- \vert u \vert ^{p-2}u \bigr)v\,dx \biggr\vert \\ &\quad\le \vert f \vert _{\infty} \int_{\varOmega} \bigl\vert \vert u_{n} \vert ^{p-2}u_{n}- \vert u \vert ^{p-2}u \bigr\vert \vert v \vert \,dx \\ &\quad\le \vert f \vert _{\infty} \biggl( \int_{\varOmega} \bigl\vert \vert u_{n} \vert ^{p-2}u_{n}- \vert u \vert ^{p-2}u \bigr\vert ^{\frac{p}{p-1}}\,dx \biggr)^{\frac{p-1}{p}} \biggl( \int_{\varOmega} \vert v \vert ^{p}\,dx \biggr)^{\frac{1}{p}} \\ &\quad\le \vert f \vert _{\infty} \bigl\vert |u_{n} \vert ^{p-2}u_{n}- \vert u \vert ^{p-2}u\bigr|_{\frac{p}{p-1}} \vert \varOmega \vert ^{\frac{6-p}{6}}S^{-p/2} \Vert v \Vert ^{p}. \end{aligned}$$

Hence, by using (2.7), we obtain

$$ \bigl\Vert \phi'(u_{n})- \phi'(u) \bigr\Vert \le \vert f \vert _{\infty} \bigl\vert \vert u_{n} \vert ^{p-2}u_{n}- \vert u \vert ^{p-2}u \bigr\vert _{\frac{p}{p-1}} \vert \varOmega \vert ^{\frac{6-p}{6}}S^{-p/2}\to0. $$

(2.10)

Moreover, we also have

$$o(1)=\bigl\langle I'(u_{n}),v\bigr\rangle =\bigl(a-b \Vert u_{n} \Vert ^{2}\bigr) \int_{\varOmega} \nabla u_{n}\nabla v \,dx - \lambda \bigl\langle \phi'(u_{n}),v\bigr\rangle , \quad\forall v\in H^{1}_{0}(\varOmega). $$

This with (2.9) implies that $\phi'(u_{n}) \to0$. Thus, we deduce from (2.10) that

$$\bigl\langle \phi'(u),v\bigr\rangle = \int_{\varOmega}f(x) \vert u \vert ^{p-2}uv \, dx=0,\quad \forall v\in H^{1}_{0}(\varOmega). $$

Then we can apply the variational method fundamental lemma [25] to obtain

$$f(x) \bigl\vert u(x) \bigr\vert ^{p-2}u(x)=0,\quad\text{a.e. } x\in \varOmega. $$

Combining this with assumption ($f_{2}$) that $|\varSigma_{0}|=0$, we obtain $u=0$.

By ($f_{1}$) and (2.7), we have

$$\lim_{n\to\infty} \int_{\varOmega}f(x) \vert u_{n} \vert ^{p} \,dx= \int _{\varOmega}f(x) \vert u \vert ^{p}\,dx, $$

and hence,

$$\phi(u_{n})= \frac{1}{p} \int_{\varOmega}f(x) \vert u_{n} \vert ^{p} \,dx\to\frac {1}{p} \int_{\varOmega}f(x) \vert u \vert ^{p}\,dx=0. $$

This together with (2.9) yields

$$I(u_{n})= \frac{a}{2} \Vert u_{n} \Vert ^{2} - \frac{b}{4} \Vert u_{n} \Vert ^{4}- \frac{\lambda }{p} \int_{\varOmega}f(x) \vert u_{n} \vert ^{p} \,dx\to{\frac{a^{2}}{4b}}, $$

which is a contradiction to $I(u_{n})\to c<{\frac{a^{2}}{4b}}$. Therefore, we obtain $\int_{\varOmega} \nabla u_{n}\nabla (u_{n}-u)\,dx\to0$, which gives $\|u_{n}\|\to\|u\|$. This and the weak convergence of $\{ u_{n}\}$ in $H^{1}_{0}(\varOmega)$ imply that $u_{n}\to u$ in $H^{1}_{0}(\varOmega )$. The proof of Lemma 2.2 is complete. □

Now, we are in a position to prove Theorem 1.1.

Proof of Theorem 1.1

In view of Lemma 2.1, there is a sequence $\{u_{n}\}\subset {{H^{1}_{0}(\varOmega)}}$ such that $I(u_{n})\rightarrow c_{*}<{\frac{a^{2}}{ 4b}}$ and $I'(u_{n})\rightarrow0$ for any $\lambda >0$. It follows from Lemma 2.2 that, along a subsequence, $u_{n}\rightarrow u$ in $H^{1}_{0}(\varOmega)$, and u is a weak solution of problem (1.1). Furthermore, if we replace the functional I with the following one:

$${\tilde{I}}(u)= \frac{a}{2} \Vert u \Vert ^{2} - \frac{b}{4} \Vert u \Vert ^{4}- \frac{\lambda }{p} \int_{\varOmega}f(x) (u_{+})^{p}\,dx. $$

It is clear that all the above calculations can be repeated word by word. Thus, there is a nontrivial critical point $u\in H^{1}_{0}(\varOmega )$ of Ĩ. Then we have $\langle{\tilde{I}}'(u),u_{-} \rangle=0$, and hence

$$\bigl(a-b \Vert u \Vert ^{2} \bigr) \Vert u_{-} \Vert ^{2}=0. $$

Due to $u_{n}\to u$ and the fact $\|u_{n}\|^{2}\to{\frac{a}{b}}$ is false, we deduce $\|u_{-}\|^{2}=0$. In turn, we deduce $u\ge0$. By the standard elliptic regularity argument and the strong maximum principle, we get $u>0$, namely, u is a positive solution of problem (1.1). This finishes the proof. □

3 Proof of Theorem 1.2

In order to prove Theorem 1.2, we need the following four lemmas.

Lemma 3.1

Suppose that ($f'_{1}$) holds, then the functional defined by

$$\chi: u\in H^{1}_{0}(\varOmega) \to \int_{\varOmega}f(x) \vert u \vert ^{p}\,dx $$

is weakly continuous.

Proof

It is easy to verify that the functional χ is well defined. Assume that $u_{n}\rightharpoonup u$ in $H^{1}_{0}(\varOmega)$, then we have $u_{n}\to u$ in $L^{r}(\varOmega)$, $2\le r <6$. Hence, $u_{n} \to u$ a.e. in Ω. Since $\{u_{n}\}$ is bounded in $H^{1}_{0}(\varOmega)$, we have $\{u_{n}\}$ is bounded in $L^{6}(\varOmega)$, and so $\{|u_{n}|^{p}\}$ is bounded in $L^{{6}/{p}}(\varOmega)$. Furthermore, we obtain that

$$\vert u_{n} \vert ^{p}\rightharpoonup \vert u \vert ^{p} \quad\text{in } L^{{6}/{p}}(\varOmega), $$

and consequently, we can apply ($f'_{1}$) to deduce that

$$\lim_{n\to+\infty} \int_{\varOmega}f(x) \vert u_{n} \vert ^{p} \,dx = \int_{\varOmega}f(x) \vert u \vert ^{p}\,dx. $$

Thus, the proof is complete. □

Lemma 3.2

Under the assumptions of Theorem1.2, the functionalIsatisfies the mountain-pass geometry:

• (i) there exist$\rho,\beta>0$such that$I(u)\ge\beta>0$for all$\|u\|=\rho$;

• (ii) there exists$e\in H^{1}_{0}(\varOmega)$with$\|e\|>\rho$such that$I(e)<0$.

Proof

(i) By ($f'_{2}$), Hölder’s inequality, and (1.4), we have that

$$\begin{aligned} I(u)&={\frac{a}{2}} \Vert u \Vert ^{2}-{\frac{b}{4}} \Vert u \Vert ^{4}-{\frac{\lambda }{ p}} \int_{\varOmega} f(x) \vert u \vert ^{p}\,dx \\ &\ge{\frac{a}{2}} \Vert u \Vert ^{2}-{ \frac{b}{4}} \Vert u \Vert ^{4}-{\frac{\lambda }{ p}} \int_{\varOmega} f_{+}(x) \vert u \vert ^{p} \,dx \\ &\ge{\frac{a}{2}} \Vert u \Vert ^{2}-{ \frac{b}{4}} \Vert u \Vert ^{4}-{\frac{\lambda }{ p}} \vert f_{+} \vert _{p_{*}} \vert u \vert _{6}^{p} \\ &\ge{\frac{a}{2}} \Vert u \Vert ^{2}-{ \frac{b}{4}} \Vert u \Vert ^{4}-{\frac{\lambda }{ p}} \vert f_{+} \vert _{p_{*}}S^{-p/2} \Vert u \Vert ^{p}. \end{aligned}$$

As $3< p<4$, there exist $\rho,\beta>0$ such that $I(u)\ge\beta>0$ for all $\|u\|=\rho$.

(ii) The proof is similar to Lemma 2.1. □

Lemma 3.3

Suppose that the assumptions of Theorem1.2hold. Let$\{{u}_{n}\} \subset H^{1}_{0}(\varOmega)$be a$(PS)_{c}$sequence for the functionalI, then

• (i) there exists$m>0$such that$\|u_{n}\|\le m$for each$n\in\mathbb {N}$;

• (ii) $\{{u}_{n}\}$has a convergent subsequence if$c<{\frac{a^{2}}{ 4b}}-D_{0}\lambda$, where$D_{0}= ({\frac{1}{p}}-{\frac{1}{4}} )|f_{+}|_{p_{*}}S^{-p/2}m^{p}$.

Proof

(i) Let $\{{u}_{n}\}\subset H^{1}_{0}(\varOmega)$ be a $(\mathit{PS})_{c}$ sequence for I, i.e.,

$$ \lim_{n\to+\infty}I(u_{n}) =c \quad\text{and} \quad\lim_{n\to +\infty}I'(u_{n}) =0. $$

(3.1)

By (3.1), we have that, for $3< p<4$,

$$\begin{aligned} c+1+o(1) \Vert {u}_{n} \Vert &\ge I({u}_{n})-{ \frac{1}{p}}\bigl\langle I'({u}_{n}),{u}_{n} \bigr\rangle \\ &=a \biggl( {\frac{1}{2}} - {\frac{1}{p}} \biggr) \Vert {u}_{n} \Vert ^{2}-b \biggl( {\frac{1}{4}} - { \frac{1}{p}} \biggr) \Vert {u}_{n} \Vert ^{4} \\ &\ge a \biggl( {\frac{1}{2}} - {\frac{1}{p}} \biggr) \Vert {u}_{n} \Vert ^{2}, \end{aligned}$$

which implies that $\{{u}_{n}\}$ is bounded in $H^{1}_{0}(\varOmega)$, that is, there exists $m>0$ such that $\|u_{n}\|\le m$ for each $n\in\mathbb {N}$.

(ii) After passing to a subsequence, we may assume that

$$ \textstyle\begin{cases} u_{n}\rightharpoonup u_{*}, & \text{in } H^{1}_{0}(\varOmega),\\ u_{n}\to u_{*}, & \text{in } L^{r}(\varOmega),1\le r< 6,\\ u_{n}\to u_{*}, & \text{a.e. in } \varOmega. \end{cases} $$

(3.2)

First, we show that

$$ \lim_{n\rightarrow\infty} \int_{\varOmega} f(x) \vert u_{n} \vert ^{p-2}u_{n}v = \int_{\varOmega} f(x) \vert u_{*} \vert ^{p-2}u_{*}v, \quad \forall v \in H^{1}_{0}(\varOmega). $$

(3.3)

To prove (3.3), let us consider $\varphi\in C_{0}^{\infty}(\varOmega)$. Recalling that $p_{*}={\frac{6}{6-p}}$, we can choose $r_{0}<6$ and near 6 satisfying

$${\frac{6}{6-p}}>{\frac{r_{0}}{(r_{0}+1)-p}}. $$

As a consequence, there is $r_{*}>1$ such that

$${\frac{1}{p_{*}}}+{\frac{1}{r_{0}/(p-1)}}+{\frac{1}{r_{*}}}=1. $$

The strong convergence $u_{n}\to u_{*}$ in $L^{r_{0}}(\varOmega)$ gives $\phi_{r_{0}}\in L^{r_{0}}(\varOmega)$ satisfying $|u_{n}(x)|\le\phi _{r_{0}}(x)$ a.e. in Ω. Therefore, using Young’s inequality, we obtain

$$\begin{aligned} \bigl\vert f(x) \bigl\vert u_{n}(x) \bigr\vert ^{p-2}u_{n}(x)\varphi(x) \bigr\vert &\le C \bigl( \bigl\vert f(x) \bigr\vert ^{p_{*}}+ \bigl\vert u_{n}(x) \bigr\vert ^{r_{0}}+ \bigl\vert \varphi(x) \bigr\vert ^{r_{*}} \bigr) \\ &\le C \bigl( \bigl\vert f(x) \bigr\vert ^{p_{*}}+ \bigl\vert \phi_{r_{0}}(x) \bigr\vert ^{r_{0}}+ \bigl\vert \varphi (x) \bigr\vert ^{r_{*}} \bigr), \end{aligned}$$

a.e. in Ω. Since $\varphi(x)$ is smooth, the right-hand side above belongs to $L^{1}(\varOmega)$. Then we can infer from the Lebesgue theorem that

$$\lim_{n\rightarrow\infty} \int_{\varOmega} f(x) \vert u_{n} \vert ^{p-2}u_{n}\varphi= \int_{\varOmega} f(x) \vert u_{*} \vert ^{p-2}u_{*}\varphi. $$

By density, we further obtain (3.3) holds.

Set $v_{n}=u_{n}-u_{*}$ and we claim that $\|v_{n}\|\to0$. If not, we may suppose $\|v_{n}\| \to l$ with $l>0$. From (3.1), it follows that $\langle I'(u_{n}),u_{*} \rangle =o(1)$. Combining this with (3.2) and (3.3), we obtain

$$ 0=a \Vert u_{*} \Vert ^{2}-b\bigl(l^{2}+ \Vert u_{*} \Vert ^{2}\bigr) \Vert u_{*} \Vert ^{2} - \lambda \int_{\varOmega} f(x) \vert u_{*} \vert ^{p}\,dx. $$

(3.4)

Moreover, by $\langle I'(u_{n}),u_{n} \rangle=o(1)$, we can use (3.2) and Lemma 3.1 to obtain

$$ \begin{aligned}[b] 0&=a\bigl( \Vert v_{n} \Vert ^{2}+ \Vert u_{*} \Vert ^{2}\bigr)-b\bigl( \Vert v_{n} \Vert ^{4}+2 \Vert v_{n} \Vert ^{2} \Vert u_{*} \Vert ^{2}+ \Vert u_{*} \Vert ^{4}\bigr)\\&\quad- \lambda \int_{\varOmega} f(x) \vert u_{*} \vert ^{p}\,dx +o(1).\end{aligned} $$

(3.5)

Combining (3.4) and (3.5), we have

$$ o(1)=a \Vert v_{n} \Vert ^{2}-b \Vert v_{n} \Vert ^{4}-b \Vert v_{n} \Vert ^{2} \Vert u_{*} \Vert ^{2}. $$

(3.6)

Passing the limit as $n\to\infty$, we get that

$$l^{2} \bigl(a-bl^{2}-b \Vert u_{*} \Vert ^{2} \bigr)=0, $$

that is,

$$ l^{2}={\frac{a}{b}}- \Vert u_{*} \Vert ^{2}. $$

(3.7)

By (3.4) and Hölder’s inequality, we obtain for $3< p<4$

$$\begin{aligned} I(u_{*})&=\frac{a}{2} \Vert u_{*} \Vert ^{2}- \frac{b}{4} \Vert u_{*} \Vert ^{4} -\frac{\lambda}{p} \int_{\varOmega} f(x) \vert u_{*} \vert ^{p}\,dx \\ &=\frac{a}{4} \Vert u_{*} \Vert ^{2}+ \frac{b}{4} l^{2} \Vert u_{*} \Vert ^{2}-\lambda \biggl({\frac{1}{p}}-{\frac{1}{4}} \biggr) \int_{\varOmega} f(x) \vert u_{*} \vert ^{p}\,dx \\ &\ge\frac{a}{4} \Vert u_{*} \Vert ^{2}+ \frac{b}{4} l^{2} \Vert u_{*} \Vert ^{2}-\lambda \biggl({\frac{1}{p}}-{\frac{1}{4}} \biggr) \int_{\varOmega} f_{+}(x) \vert u_{*} \vert ^{p}\,dx \\ &\ge\frac{a}{4} \Vert u_{*} \Vert ^{2}+ \frac{b}{4} l^{2} \Vert u_{*} \Vert ^{2}-\lambda \biggl({\frac{1}{p}}-{\frac{1}{4}} \biggr) \vert f_{+} \vert _{p_{*}}S^{-p/2} \Vert u_{*} \Vert ^{p} \\ &\ge\frac{a}{4} \Vert u_{*} \Vert ^{2}+ \frac{b}{4} l^{2} \Vert u_{*} \Vert ^{2}-\lambda \biggl({\frac{1}{p}}-{\frac{1}{4}} \biggr) \vert f_{+} \vert _{p_{*}}S^{-p/2}m^{p}. \end{aligned}$$

(3.8)

Furthermore, we can use (3.6)–(3.8) and Lemma 3.1 to obtain

$$\begin{aligned} c+o(1)&=I(u_{n}) \\ &= {\frac{a}{2} \Vert u_{n} \Vert ^{2}- \frac{b}{4} \Vert u_{n} \Vert ^{4} - \frac{\lambda}{p} \int f(x) \vert u_{n} \vert ^{p}\,dx} \\ &= {\frac{a}{2} \Vert u_{*} \Vert ^{2}- \frac{b}{4} \Vert u_{*} \Vert ^{4} -\frac{\lambda}{p} \int f(x) \vert u_{*} \vert ^{p}\,dx} \\ &\quad +\frac{a}{2} \Vert v_{n} \Vert ^{2}- \frac{b}{4} \Vert v_{n} \Vert ^{4} - \frac{b}{2} \Vert v_{n} \Vert ^{2} \Vert u_{*} \Vert ^{2}+o(1) \\ &= {I(u_{*})+ \frac{a}{2} \Vert v_{n} \Vert ^{2}- \frac{b}{4} \Vert v_{n} \Vert ^{4} -\frac{b}{2} \Vert v_{n} \Vert ^{2} \Vert u_{*} \Vert ^{2}+o(1)} \\ &= {I(u_{*})+ \frac{a}{2} \Vert v_{n} \Vert ^{2}- \frac{1}{4} \bigl(a \Vert v_{n} \Vert ^{2}-b \Vert v_{n} \Vert ^{2} \Vert u_{*} \Vert ^{2} \bigr) -\frac{b}{2} \Vert v_{n} \Vert ^{2} \Vert u_{*} \Vert ^{2}+o(1)} \\ &= {I(u_{*})+ \frac{a}{4} \Vert v_{n} \Vert ^{2}- \frac{b}{4} \Vert v_{n} \Vert ^{2} \Vert u_{*} \Vert ^{2}+o(1)} \\ &= {I(u_{*})+ \frac{a}{4} l^{2}- \frac{b}{4} l^{2} \Vert u_{*} \Vert ^{2}+o(1)} \\ &= {I(u_{*})+ \frac{a}{4} \biggl({\frac{a}{b}}- \Vert u_{*} \Vert ^{2} \biggr)- \frac{b}{4} l^{2} \Vert u_{*} \Vert ^{2}+o(1)} \\ &= {I(u_{*})+ \frac{a^{2}}{4b}- \frac{a}{4} \Vert u_{*} \Vert ^{2}- \frac{b}{4} l^{2} \Vert u_{*} \Vert ^{2}+o(1)} \\ &\ge{ \frac{a^{2}}{4b}-\lambda \biggl({\frac{1}{p}}-{ \frac{1}{ 4}} \biggr) \vert f_{+} \vert _{p_{*}}S^{-p/2}m^{p}}, \end{aligned}$$

which is a contradiction with our assumption $c<{\frac{a^{2}}{ 4b}}-D_{0}{\lambda }$. Thus, the claim follows, that is, $u_{n} \to u_{*}$ in $H^{1}_{0}(\varOmega)$. We complete the proof of Lemma 3.3. □

By condition ($f'_{2}$), we can choose $x_{0} \in\operatorname {int}(\varSigma _{d_{0}})$ and $\eta> 0$ small enough such that $B_{2\eta}(x_{0}) \subset \varSigma_{d_{0}}$. Define a cutoff function $\varphi(x)$ satisfying $\varphi(x) \equiv1$ in $B_{\eta}(x_{0})$, $\varphi(x) \equiv0$ outside $B_{2\eta}(x_{0})$ and $0 \leq\varphi\leq1$. Inspired by [23] (see also [12]), we consider the following test function:

$$u_{\varepsilon}(x)= \varphi(x) {\frac{1}{ (\varepsilon^{2} + \vert x-x_{0} \vert ^{2} )^{1/2}}} . $$

Without loss of generality, we may assume $x_{0} = 0$.

Lemma 3.4

Under the assumptions of Theorem1.2, there exists$\lambda _{\ast}>0$such that, for any$\lambda \in (0,\lambda _{\ast})$,

$$\sup_{t>0}I(tu_{\varepsilon})< {\frac{a^{2}}{4b}}-D_{0}{ \lambda }, $$

where$D_{0}$is defined as in Lemma3.3.

Proof

Let $\lambda _{\ast}={\frac{a^{2}}{4bD_{0}}}$ and note that, for any $\lambda \in(0,\lambda _{\ast})$, there holds

$${\frac{a^{2}}{4b}}-D_{0}\lambda >0. $$

It is clear that $\lim_{t\to0^{+}}I(tu_{\varepsilon})=0$, and so there is $t_{1}>0$ such that, for any $\lambda \in(0,\lambda _{*})$,

$$ \sup_{0< t\le t_{1}}I(tu_{\varepsilon})< { \frac{a^{2}}{4b}}-D_{0}{\lambda }. $$

(3.9)

In the following, we discuss the case $t>t_{1}$. Noting that $u_{\varepsilon}=0$ in $\varOmega\setminus\varSigma_{d_{0}}$ and $f_{-}(x)=0$ in $\varSigma_{d_{0}}$, we have

$$\int_{\varOmega} f_{-}(x) \vert u_{\varepsilon} \vert ^{p}\,dx=0. $$

Set

$$R_{0}=\min \biggl\{ \eta, 2^{\frac{2-p}{2(p-3)}}t_{1}^{\frac{p}{p-3}} \biggl({\frac{\pi d_{0}}{3pD_{0}}} \biggr)^{\frac{1}{p-3}} \biggr\} $$

and let $\varepsilon< R_{0}$, then by ($f'_{2}$) we obtain

$$\begin{aligned} \int_{\varOmega} f(x) \vert u_{\varepsilon} \vert ^{p} \,dx &= \int_{\varSigma _{d_{0}}} f_{+}(x) \vert u_{\varepsilon} \vert ^{p}\,dx \\ &\ge \int_{B_{R_{0}}(0)} f_{+}(x){\frac{1}{ (\varepsilon^{2} + \vert x \vert ^{2} )^{p/2}}}\,dx \\ &\ge{\frac{1}{(2R_{0}^{2})^{p/2}}} \int_{B_{R_{0}}(0)} f_{+}(x)\,dx \\ &\ge{\frac{1}{2^{p/2}}} {\frac{4\pi d_{0}}{3R_{0}^{p-3}}}. \end{aligned}$$

(3.10)

By (3.10), we have that, for any $\lambda \in(0,\lambda _{*})$ and $t>t_{1}$,

$$\begin{aligned} I(tu_{\varepsilon})&= {\frac{a}{2}}t^{2} \Vert u_{\varepsilon} \Vert ^{2}-{\frac{b}{4}}t^{4} \Vert u_{\varepsilon} \Vert ^{4} -\lambda {\frac{t^{p}}{p}} \int _{\varOmega} f(x)u_{\varepsilon}^{p}\,dx \\ &\le\sup_{t>0} \biggl\{ \frac{{a} {2}}{t}^{2} \Vert u_{\varepsilon} \Vert ^{2}-{\frac{b}{4}}t^{4} \Vert u_{\varepsilon} \Vert ^{4} \biggr\} -\lambda {\frac{t_{1}^{p}}{p}} \int_{\varOmega} f(x)u_{\varepsilon}^{p}\,dx \\ &\le{\frac{a^{2}}{4b}}-{\frac{t_{1}^{p}}{p}} {\frac{1}{2^{p/2}}} { \frac{4\pi d_{0}}{3R_{0}^{p-3}}}\lambda \\ &\le{\frac{a^{2}}{4b}}-2D_{0}\lambda \\ &< {\frac{a^{2}}{4b}}-D_{0}\lambda . \end{aligned}$$

(3.11)

Combining (3.9) and (3.11), we have that, for any $\lambda \in(0,{\lambda }_{*})$,

$$\sup_{t>0}I(tu_{\varepsilon})< {\frac{a^{2}}{4b}}-D_{0}{ \lambda }. $$

This completes the proof of Lemma 3.4. □

Now, we are ready to prove Theorem 1.2.

Proof of Theorem 1.2

Define the minimax level of the mountain pass theorem as in Lemma 2.1. By the definition of $c_{*}$ and Lemma 3.4, we know there is $\lambda _{\ast}>0$ such that $c_{*}<{\frac{a^{2}}{4b}}-D_{0}\lambda$ for any $\lambda \in(0,\lambda _{\ast})$. According to Lemma 3.3, we further obtain the compactness property on the level $c_{*}$. Then we can argue as in the proof of Theorem 1.1 to conclude that there exists $u_{*}\in H^{1}_{0}(\varOmega)$ such that $u_{*}$ is a positive solution of problem (1.1). This completes the proof. □