Existence and multiplicity of solutions for Kirchhof-type problems with Sobolev–Hardy critical exponent

Abstract

In this paper, we discuss a class of Kirchhof-type elliptic boundary value problem with Sobolev–Hardy critical exponent and apply the variational method to obtain one positive solution and two nontrivial solutions to the problem under certain conditions.

1 Introduction and main results

In this paper, we investigate the following Kirchhof-type problem with Soblev–Hardy critical exponent:

$$ \textstyle\begin{cases} \{a+b [ {\int_{\Omega} ( \vert \nabla u \vert ^{2}-\mu\frac {u^{2}}{ \vert x \vert ^{2}} ) \,\mathrm{d}x} ]^{\frac{2-s}{2}} \} (-\Delta u-\mu\frac {u}{ \vert x \vert ^{2}} ) \\ \quad= \frac{ \vert u \vert ^{2^{*}(s)-2}u}{ \vert x \vert ^{s}}+f(x,u) &\text{in } \Omega,\\ u=0, &\text{on } \partial\Omega, \end{cases} $$

(1.1)

where Ω is a smooth bounded domain in $\mathbb{R}^{3}$, $0\in \Omega$, $a, b>0$, $0\le s<2$, $\mu\in[0,1/4)$, $2^{*}(s)=6-2s$ is the Sobolev–Hardy critical exponent, and $f(x,t): \overline{\Omega }\times\mathbb{R}$ is a continuous real function.

Equation (1.1) is called a Schrödinger equation of Kirchhoff type due to the presence of the term $b [\int_{\Omega} (|\nabla u|^{2}-\mu u^{2}|x|^{-2} )\,\mathrm{d}x ]^{(2-s)/2}$. When $\mu =0$ and $s=1$, it appears in the following classical Kirchhoff type equation:

$$ \textstyle\begin{cases} -(a+b {\int_{\Omega} \vert \nabla u \vert ^{2} \,\mathrm{d}x})\Delta u=k(x,u) &\text{in } \Omega,\\ u=0&\text{on } \partial\Omega, \end{cases} $$

(1.2)

related to the stationary analogue of the equation

$$ u_{tt}- \biggl(a+b { \int_{\Omega} \vert \nabla u \vert ^{2} \, \mathrm{d}x} \biggr)\Delta u=f(x,t), $$

which was first proposed by Kirchhoff [1] as an extension of the classical D’Alembert wave equation for free vibrations of elastic strings. Equation (1.2) has aroused widespread concern after the work of Lions [2], which proposes a function analysis framework. After that, many interesting results have been obtained such as [3–9]. For instance, Xu and Chen [10] studied Kirchhoff-type equations with a general nonlinearity in the critical growth. Under certain conditions, the existence of ground state solutions were proved by using variational methods. In particular, they do not use the classical Ambrosetti–Rabinowitz condition. Fiscella et al. [11] dealt with the existence of nontrivial solutions for critical Hardy–Schrd̈inger–Kirchhoff systems driven by the fractional p-Laplacian operator. The existence was derived as an application of the mountain pass theorem and the Ekeland variational principle. The authors extend the existence results recently obtained for fractional systems to entire solutions with critical nonlinear terms and generalized the systems driven by the p-Laplacian operator to the fractional Hardy–Schrödinger–Kirchhoff case. Xiang and Vicentiu [12] investigated the existence of solutions for critical Schrödinger–Kirchhof-type systems driven by nonlocal integro-differential operators. By applying the mountain pass theorem and Ekeland’s variational principle, the existence and asymptotic behavior of solutions for the problem under some suitable assumptions were obtained. A distinguished feature of their paper is that the systems are degenerate, that is, the Kirchhoff function could vanish at zero. This is the first time of exploiting the existence of solutions for fractional Schrödinger–Kirchhoff systems involving critical nonlinearities in $R^{N}$.

In the case $k(x,u)=f(x,u)+u^{5}$, Xie et al. [6] studied the nondegenerate and degenerate cases and proved the existence and multiplicity of solutions by using the Brezis–Lieb lemma and mountain pass theorem. Naimen [8] further discussed this problem in the case of $k(x,u)=\mu g(x,u)+u^{5}$ under different conditions of $g(x,u)$ and $\mu\in\mathbb{R}$. In the meantime, the results were expanded in [6] by establishing the existence and nonexistence of positive solutions by using the second concentration compactness lemma and mountain pass theorem.

Problem (1.1) in the case of $a=1$ and $b=0$ can be reduced to the classic semilinear elliptic problem with critical exponents, for which the existence and multiplicity of solutions was proved by Ding and Tang [9].

Inspired by the results of the above paper, the purpose of this paper is to consider the existence and multiplicity of solutions to problem (1.1). The main difficulty in this paper is that it contains the Sobolev–Hardy critical exponent term, which leads to the energy functional not satisfying the Palais–Smale condition.

In order to state our main results, let $F(x,u)=\int_{0}^{u} f(x,t)\,\mathrm{d}t$. We introduce the following assumptions:

$(f_{1})$:

$f\in C(\Omega\times\mathbb{R^{+}},\mathbb{R})$, and $\lim_{t\to0^{+}}\frac{f(x,t)}{t}=0$, $\lim_{t\to+\infty}\frac {f(x,t)}{t^{5-2s}}=0$ uniformly for a.e. $x\in\Omega$.

$(f_{2})$:

There exists a constant $\rho>\max\{\frac{6}{1+\sqrt {1-4\mu}},6-\sqrt{1-4\mu}\}$ such that $0<\rho F(x,t)\le f(x,t)t$ for all $x\in\overline{\Omega}$, $t\in\mathbb{R^{+}}\setminus\{0\}$.

$(f_{3})$:

$f\in C(\Omega\times\mathbb{R},\mathbb{R})$, and $\lim_{t\to0}\frac{f(x,t)}{t}=0$, $\lim_{t\to+\infty}\frac {f(x,t)}{t^{5-2s}}=0$ uniformly for a.e. $x\in\Omega$.

$(f_{4})$:

There exists a constant $\rho>\max\{\frac{6}{1+\sqrt {1-4\mu}},6-\sqrt{1-4\mu}\}$ such that $0<\rho F(x,t)\le f(x,t)t$ for all $x\in\overline{\Omega}$, $t\in\mathbb{R}\setminus\{0\}$.

Now our main results are as follows.

Theorem 1.1

Let $f(x,t)$ satisfy $(f_{1})$ and $(f_{2})$. Then problem (1.1) has at least one positive solution.

Theorem 1.2

Let $f(x,t)$ satisfy $(f_{3})$ and $(f_{4})$. Then problem (1.1) has at least two distinct nontrivial solutions.

Remark 1.1

We added the Hardy and Sobolev–Hardy terms in equation (1.1) on the basis of [6]. We overcome the compactness problem with concentration compactness principle. Lei [7] studied another special case of problem (1.1) with $f(x,u)=\lambda f(x)|u|^{q-2}u|x|^{-\beta}$ for a suitable function $f(x)$ and $1< q<2$. By using the Nehari manifold and fibering maps, they obtained two positive solutions. We observe that the term $\lambda f(x)|u|^{q-2}u|x|^{-\beta}$ has to be a homogeneous function; however, it does not satisfy the assumptions we give in this paper.

The rest of this paper is organized as follows. In Sect. 2, we give some preliminary results. In Sect. 3, we establish the proofs of our main results.

2 Preliminaries

In this part, we give some information to support this paper. Otherwise stated, $C, C_{0}, C_{1},\dots$ represent positive constants, and “→” and “⇀” represent the strong convergence and weak convergence in the corresponding space, respectively. Let $H^{1}_{0}(\Omega)$ be the usual Hilbert space endowed with the usual inner product and norm

$$ (u,v)_{H^{1}_{0}(\Omega)}= \int_{\Omega}\nabla u\nabla v\,\mathrm{d}x \quad \text{and} \quad \Vert u \Vert _{H^{1}_{0}(\Omega)}= \biggl( \int_{\Omega} \vert \nabla u \vert ^{2}\, \mathrm{d}x \biggr)^{\frac{1}{2}}. $$

By the well-known Hardy inequality [13]

$$ \int_{\Omega}\frac{u^{2}}{ \vert x \vert ^{2}} \,\mathrm{d}x\le4 \int_{\Omega} \vert \nabla u \vert ^{2}\, \mathrm{d}x, $$

we deduce that

$$ (u,v)= \int_{\Omega}\biggl(\nabla u\nabla v-\mu\frac{uv}{ \vert x \vert ^{2}} \biggr)\, \mathrm{d}x \quad \text{and}\quad \Vert u \Vert = \biggl[ \int_{\Omega}\biggl( \vert \nabla u \vert ^{2}- \mu\frac{u^{2}}{ \vert x \vert ^{2}} \biggr)\, \mathrm{d}x \biggr]^{\frac{1}{2}}, $$

respectively, which are equivalent to the usual inner product and norm on $H^{1}_{0}(\Omega)$ for any $\mu\in[0,1/4)$.

We also define the best Sobolev–Hardy constant

$$ S\triangleq\inf_{u\in H^{1}_{0}(\Omega)\setminus\{0\}}\frac{\int_{\Omega}( \vert \nabla u \vert ^{2}-\mu\frac{u^{2}}{ \vert x \vert ^{2}} )\,\mathrm{d}x}{ (\int _{\Omega}\frac{ \vert u \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm{d}x )^{\frac{2}{2^{*}(s)}}}. $$

(2.1)

From Lemma 2.2 in [14] we find that S is independent of Ω, and when $\Omega=\mathbb {R}^{3}$, it is obtained by the functions

$$ y_{\epsilon}(x)= \biggl[\frac{2\epsilon(3-s)(\bar{\mu}-\mu)}{\sqrt{\bar {\mu}}} \biggr]^{\frac{{\sqrt{\bar{\mu}}}}{2-s}} \Big/ \bigl[ \vert x \vert ^{\sqrt{\bar{\mu}}-\sqrt{\bar{\mu}-\mu}} \bigl(\epsilon+ \vert x \vert ^{\frac{(2-s)\sqrt{\bar{\mu}-\mu}}{\sqrt{\bar{\mu }}}} \bigr)^{\frac{3-s}{2-s}} \bigr] $$

for all $\epsilon>0$ and $\bar{\mu}=1/4$. In addition, the function $y_{\epsilon}(x)$ is the solution to the equation

$$ -\Delta u-\mu\frac{u}{ \vert x \vert ^{2}}=\frac{ \vert u \vert ^{2^{*}(s)-2}u}{ \vert x \vert ^{s}} \quad \text{in } \mathbb {R}^{3}\setminus\{0\}. $$

Since $0\in\Omega$, let $R_{0}$ be a positive constant such that $B_{2R_{0}}(0)\subset\Omega$. We take a cut-off function $\eta(x)\in C^{\infty}_{0}(\Omega)$ such that $\eta(x)=1$ for $|x|\le R_{0}$, $\eta (x)=0$ for $|x|>2R_{0}$, and $0\le\eta(x)\le1$ otherwise. Let $C_{\epsilon}= [\frac{2\epsilon(3-s)\sqrt{\bar{\mu}-\mu}}{\sqrt{\bar {\mu}}} ]^{\frac{\sqrt{\bar{\mu}}}{2-s}}$, $\gamma_{1}=\sqrt{\bar{\mu }}-\sqrt{\bar{\mu}-\mu}$, $\gamma_{2}=\sqrt{\bar{\mu}}+\sqrt{\bar{\mu}-\mu }$, and $U_{\epsilon}(x)=\frac{y_{\epsilon}(x)}{C_{\epsilon}}$. Suppose

$$ \begin{aligned} &U_{\epsilon}(x)=\frac{1}{ [ \vert x \vert ^{\frac{2-s}{2}\frac{\gamma_{1}}{\sqrt {\bar{\mu}}}} + \vert x \vert ^{\frac{2-s}{2}\frac{\gamma_{2}}{\sqrt{\bar{\mu}}}} ]^{\frac {2\sqrt{\bar{\mu}}}{2-s}}}, \\ &u_{\epsilon}(x)=\eta(x)U_{\epsilon}(x)=\frac{\eta(x)}{ [\epsilon \vert x \vert ^{\frac{2-s}{2}\frac{\gamma_{1}}{\sqrt{\bar{\mu}}}} + \vert x \vert ^{\frac{2-s}{2}\frac{\gamma_{2}}{\sqrt{\bar{\mu}}}} ]^{\frac {2\sqrt{\bar{\mu}}}{2-s}}}, \\ &v_{\epsilon}(x)=\frac{u_{\epsilon}(x)}{ (\int_{\Omega}\frac { \vert u_{\epsilon}(x) \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm{d}x )^{\frac{1}{2^{*}(s)}}}, \end{aligned} $$

so that $\|v_{\epsilon}(x)\|^{2^{*}(s)}_{L^{2^{*}(s)}(\Omega,|x|^{-s})}=\int _{\Omega}|v_{\epsilon}(x)|^{2^{*}(s)}|x|^{-s}\,\mathrm{d}x=1$. Then we have the following results [14]:

$$\begin{aligned}& \bigl\Vert v_{\epsilon}(x) \bigr\Vert ^{2}=S+O \bigl(\epsilon^{\frac{1}{2-s}} \bigr), \end{aligned}$$

(2.2)

$$\begin{aligned}& \int_{\Omega} \vert v_{\epsilon} \vert ^{q} \,\mathrm{d}x=\textstyle\begin{cases} O (\epsilon^{\frac{q\sqrt{\bar{\mu}}}{2-s}} ),& 1\le q< \frac {3}{\sqrt{\bar{\mu}}+\sqrt{\bar{\mu}-\mu}}, \\ O (\epsilon^{\frac{q\sqrt{\bar{\mu}}}{2-s}} \vert \ln\epsilon \vert ),&q= \frac{3}{\sqrt{\bar{\mu}}+\sqrt{\bar{\mu}-\mu}}, \\ O (\epsilon^{\frac{\sqrt{\bar{\mu}}(3-q\sqrt{\bar{\mu }})}{(2-s)\sqrt{\bar{\mu}-\mu}}} ),& \frac{3}{\sqrt{\bar{\mu }}+\sqrt{\bar{\mu}-\mu}}< q< 6. \end{cases}\displaystyle \end{aligned}$$

(2.3)

Now we define the functional I on $H^{1}_{0}(\Omega)$ by

$$ I(u)=\frac{a}{2} \Vert u \Vert ^{2}+ \frac{b}{4-s} \Vert u \Vert ^{4-s}-\frac{1}{2^{*}(s)} \int _{\Omega}\frac{u^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm{d}x- \int_{\Omega}F(x,u)\,\mathrm{d}x. $$

(2.4)

Obviously, the functional I belongs to the class $C^{1} (H^{1}_{0}(\Omega),\mathbb {R} )$. Furthermore,

$$\begin{aligned} \bigl\langle I'(u),v \bigr\rangle ={}&a \int_{\Omega}\biggl(\nabla u\nabla v-\mu\frac {uv}{ \vert x \vert ^{2}} \biggr)\,\mathrm{d}x+b \Vert u \Vert ^{2-s} \int_{\Omega}\biggl(\nabla u\nabla v-\mu\frac{uv}{ \vert x \vert ^{2}} \biggr)\,\mathrm{d}x \\ &{} - \int_{\Omega}\frac{u^{2^{*}(s)-1}v}{ \vert x \vert ^{s}}\,\mathrm{d}x- \int_{\Omega}f(x,u)v\,\mathrm{d}x,\quad \forall v\in H_{0}^{1}(\Omega). \end{aligned}$$

3 Proofs of our main results

In this section, we consider the existence and multiplicity of solutions to problem (1.1). We first verify that the functional $I(u)$ satisfies the local $(PS)$ condition.

Lemma 3.1

Let $f(x,t)$ satisfy $(f_{1})$ and $(f_{2})$. Suppose $c\in(0,\Lambda_{0})$, where

$$ \begin{aligned} \Lambda_{0}={}&\frac{a(2-s)}{2(3-s)}S \biggl[\frac{bS^{\frac{4-s}{2}}+\sqrt {b^{2}S^{4-s}+4aS}}{2} \biggr]^{\frac{2}{2-s}} \\ &{} +\frac{b(2-s)}{2(3-s)(4-s)}S \biggl[\frac{bS^{\frac{4-s}{2}}+\sqrt {b^{2}S^{4-s}+4aS}}{2} \biggr]^{\frac{4-s}{2-s}}. \end{aligned} $$

Then $I(u)$ satisfies the local $(PS)_{c}$ condition.

Proof

Suppose that $\{u_{n}\}$ is a $(PS)_{c}$ sequence. Then, for $c\in (0,\Lambda_{0})$,

$$ I(u_{n})\to c, \qquad I'(u_{n}) \to0 \quad \text{as } n\to\infty. $$

(3.1)

First, we prove that $\{u_{n}\}$ is a bounded sequence. From (3.1) we have

$$ \begin{aligned} 1+c+o(1) \Vert u_{n} \Vert \ge {}&I(u_{n})-\frac{1}{\theta} \bigl\langle I'(u_{n}),u_{n} \bigr\rangle \\ ={}& \biggl(\frac{1}{2}-\frac{1}{\theta} \biggr)a \Vert u_{n} \Vert ^{2}+ \biggl(\frac{1}{4-s}- \frac{1}{\theta} \biggr)b \Vert u_{n} \Vert ^{4-s} \\ &{} + \biggl(\frac{1}{\theta}-\frac{1}{2^{*}(s)} \biggr) \int_{\Omega}\frac { \vert u \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm{d}x+ \int_{\Omega}\biggl(\frac{1}{\theta }f(x,u_{n})u_{n}-F(x,u_{n}) \biggr)\,\mathrm{d}x \\ \ge{}& \biggl(\frac{1}{2}-\frac{1}{\theta} \biggr)a \Vert u_{n} \Vert ^{2}, \end{aligned} $$

where $\theta=\min\{\rho,2^{*}(s)\}$. Thus we conclude that $\{u_{n}\}$ is a bounded sequence in $H^{1}_{0}(\Omega)$. By the continuity of embedding we have $|u_{n}|^{2^{*}(s)}_{2^{*}(s)}\le C_{1}<\infty$ (denoting the usual $L^{p}(\Omega)$ norm by $|\cdot|_{p}$). Up to subsequences if necessary, there exists $u\in H^{1}_{0}(\Omega)$ such that

$$ \begin{aligned} &u_{n}\rightharpoonup u\quad \text{weakly in } H^{1}_{0}(\Omega), \\ &u_{n}\to u \quad \text{in } L^{q}(\Omega) \text{ for } q\in \bigl[1,2^{*}(s) \bigr], \\ &u_{n}\to u\quad \mbox{a.e. in } \Omega. \end{aligned} $$

Then we prove that $u_{n}\to u$ in $H^{1}_{0}(\Omega)$.

By $(f_{1})$, for any $\epsilon>0$, there exists $a(\epsilon)$ such that

$$ f(x,t)\le\frac{1}{2C_{1}}\epsilon t^{5-2s}+a(\epsilon). $$

Let $\delta_{1}=\frac{\epsilon}{2a(\epsilon)}$. When $E\subset\Omega$ and mes $E<\delta_{1}$, we have

$$ \begin{aligned} \biggl\vert \int_{E}f(x,u_{n})u_{n}\, \mathrm{d}x \biggr\vert &\le \int_{E} \bigl\vert f(x,u_{n})u_{n} \bigr\vert \,\mathrm{d}x \\ &\le \int_{E} a(\epsilon)\,\mathrm{d}x+\frac{1}{2C_{1}} \epsilon \int _{E} \vert u_{n} \vert ^{2^{*}(s)}\,\mathrm{d}x \\ &\le a(\epsilon)\operatorname{mes} E+\frac{1}{2C_{1}}\epsilon C_{1} \\ &\le\epsilon. \end{aligned} $$

(3.2)

Hence $\{\int_{\Omega}f(x,u_{n})u_{n}\,\mathrm{d}x, n\in N\}$ is equiabsolutely continuous. It is easy to get the following from the Vitali convergence theorem:

$$ \int_{\Omega}f(x,u_{n})u_{n}\, \mathrm{d}x\to \int_{\Omega}f(x,u)u\,\mathrm {d}x \quad \text{as } n\to \infty. $$

(3.3)

Similarly, we can prove that

$$ \int_{\Omega}F(x,u_{n})\,\mathrm{d}x\to \int_{\Omega}F(x,u)\,\mathrm{d}x \quad \text{as } n\to \infty. $$

Further, by the concentration compactness principle [15] there exist a countable set Γ, a set of different points $\{x_{j}\} \subset\Omega\setminus\{0\}$, nonnegative real numbers $\mu_{x_{j}},\nu _{x_{j}}$ for $j\in\Gamma$, and nonnegative real numbers $\mu_{0}$, $\gamma _{0}$, $\nu_{0}$ such that

$$ \begin{aligned} & \vert \nabla u_{n} \vert ^{2}\rightharpoonup\mathrm {d}\widetilde{\mu}\ge \vert \nabla u \vert ^{2}+\sum_{j\in\Gamma} \mu_{x_{j}}\delta_{x_{j}}+\mu_{0} \delta_{0}, \\ & u^{2}_{n} \vert x \vert ^{-2} \rightharpoonup\mathrm {d}\gamma=u^{2} \vert x \vert ^{-2}+\gamma _{0}\delta_{0}, \\ & \vert u_{n} \vert ^{2^{*}(s)} \vert x \vert ^{-s}\rightharpoonup\mathrm {d}\nu = \vert u \vert ^{2^{*}(s)} \vert x \vert ^{-s}+\sum _{j\in\Gamma}\nu_{x_{j}}\delta_{x_{j}}+\nu _{0}\delta_{0}, \end{aligned} $$

where $\delta_{x}$ is the Dirac mass at $x\in\Omega$. For any $\epsilon >0$, we let $x_{j}\notin B_{\epsilon}(0)$ for all $j\in\Gamma$ and choose ϕ to be s smooth cut-off function such that $0\le\phi\le 1$, $\phi\equiv0$ for $x\in B^{c}_{\epsilon}(0)$, $\phi\equiv1$ for $x\in B_{\epsilon/2}(0)$, and $|\nabla\phi|\le4/\epsilon$. Then

$$\begin{aligned} &\lim_{\epsilon\to0}\lim_{n\to\infty} \int_{\Omega} \vert \nabla u_{n} \vert ^{2}\phi\, \mathrm{d}x=\lim_{\epsilon\to0} \int_{\Omega}\phi\mathrm {d}\widetilde{\mu }\ge \mu_{0}, \\ &\lim_{\epsilon\to0}\lim_{n\to\infty} \int_{\Omega}u^{2}_{n} \vert x \vert ^{-2}\phi\, \mathrm{d}x=\lim_{\epsilon\to0} \int_{\Omega}\phi\mathrm {d}\gamma=\gamma _{0}, \\ &\lim_{\epsilon\to0}\lim_{n\to\infty} \int_{\Omega} \vert u_{n} \vert ^{2^{*}(s)} \vert x \vert ^{-s}\phi\,\mathrm{d}x=\lim _{\epsilon\to0} \int_{\Omega}\phi\mathrm {d}\nu=\nu_{0}, \\ &\lim_{\epsilon\to0}\lim_{n\to\infty} \int_{\Omega}(\nabla u_{n}\nabla\phi )u_{n}\,\mathrm{d}x=0, \end{aligned}$$

(3.4)

$$\begin{aligned} &\lim_{\epsilon\to0}\lim_{n\to\infty} \int_{\Omega}f(x,u_{n})u_{n}\phi\, \mathrm{d}x=0. \end{aligned}$$

(3.5)

The proofs of (3.4) and (3.5) are similar to that of Theorem 2.3 in [8] and are omitted here. Since $\{u_{n}\}$ is bounded, by (3.1) we have

$$\begin{aligned} 0={}&\lim_{\epsilon\to0}\lim_{n\to\infty} \bigl\langle I'(u_{n}),u_{n}\phi \bigr\rangle \\ ={}&\lim_{\epsilon\to0}\lim_{n\to\infty} \biggl\{ \bigl(a+b \Vert u_{n} \Vert ^{2-s} \bigr) \biggl[ \int_{\Omega}\biggl( \vert \nabla u_{n} \vert ^{2}\phi+(\nabla u_{n}\nabla\phi)u_{n}-\mu \frac {u^{2}_{n}}{ \vert x \vert ^{2}}\phi \biggr)\,\mathrm{d}x \biggr] \\ &{} - \int_{\Omega}\frac{ \vert u_{n} \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\phi\,\mathrm{d}x - \int _{\Omega}f(x,u_{n})u_{n}\phi\, \mathrm{d}x \biggr\} \\ \ge{}& a(\mu_{0}-\mu\gamma_{0})+b( \mu_{0}-\mu\gamma_{0})^{(4-s)/2}- \nu_{0}. \end{aligned}$$

Combining this with (2.1), we have that $S^{2^{*}(s)/2}\nu_{0}\le(\mu _{0}-\mu\gamma_{0})^{2^{*}(s)/2}$, and we deduce that

$$ S^{-3+s}(\mu_{0}-\mu\gamma_{0})^{2-s}-b( \mu_{0}-\mu\gamma_{0})^{(2-s)/2}-a\ge 0, $$

which implies

$$ (\mu_{0}-\mu\gamma_{0})\ge S \biggl[ \frac{bS^{\frac{4-s}{2}}+\sqrt {b^{2}S^{4-s}+4aS}}{2} \biggr]^{\frac{2}{2-s}}. $$

Therefore we get

$$\begin{aligned} c+o(1)={}&I(u_{n})-\frac{1}{4-s} \bigl\langle I'(u_{n}),u_{n} \bigr\rangle \\ ={}&\frac{a(2-s)}{2(4-s)} \Vert u_{n} \Vert ^{2}+ \frac{2-s}{2(3-s)(4-s)} \int_{\Omega}\frac { \vert u_{n} \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm{d}x \\ &{} +\frac{1}{4-s} \int_{\Omega}f(x,u_{n})u_{n}\, \mathrm{d}x- \int_{\Omega}F(x,u_{n})\,\mathrm{d}x \\ \ge{}&\frac{a(2-s)}{2(4-s)}(\mu_{0}-\mu\gamma_{0})+ \frac{2-s}{2(3-s)(4-s)}\nu _{0} \\ \ge{}&\frac{a(2-s)}{2(4-s)}(\mu_{0}-\mu\gamma_{0})+ \frac {2-s}{2(3-s)(4-s)} \bigl[a(\mu_{0}-\mu\gamma_{0})+b( \mu_{0}-\mu\gamma _{0})^{(4-s)/2} \bigr] \\ \ge{}&\Lambda, \end{aligned}$$

a contradiction. Thus we obtain

$$ \int_{\Omega}\frac{ \vert u_{n} \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm{d}x\to \int_{\Omega}\frac { \vert u \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm{d}x. $$

Combining this with (3.2), we find

$$\begin{aligned} o(1)&= \bigl\langle I'(u_{n})-I'(u),u_{n}-u \bigr\rangle \\ &= \bigl(a+b \Vert u_{n} \Vert ^{2-s} \bigr) (u_{n},u_{n}-u)- \bigl(a+b \Vert u \Vert ^{2-s} \bigr) (u,u_{n}-u)+o(1) \\ &= \bigl(a+b \Vert u_{n} \Vert ^{2-s} \bigr) (u_{n}-u,u_{n}-u)+b \bigl( \Vert u_{n} \Vert ^{2-s}- \Vert u \Vert ^{2-s} \bigr) (u,u_{n}-u)+o(1) \\ &\ge a \Vert u_{n}-u \Vert ^{2}+o(1), \end{aligned}$$

which shows that $u_{n}\to u$ in $H^{1}_{0}(\Omega)$. The proof is completed. □

Lemma 3.2

If $f(x,t)$ satisfies $(f_{1})$ and $(f_{2})$, then there exists $u_{0}\in H^{1}_{0}(\Omega)$ such that

$$ \sup_{t\ge0}I(tu_{0})< \Lambda. $$

Proof

We consider the functions

$$ \begin{aligned} &g(t)=\frac{at^{2}}{2} \Vert v_{\epsilon} \Vert ^{2}+\frac{bt^{4-s}}{4-s} \Vert v_{\epsilon} \Vert ^{4-s}-\frac{t^{2^{*}(s)}}{2^{*}(s)}- \int_{\Omega}F(x,tv_{\epsilon})\,\mathrm {d}x, \\ &g_{0}(t)=\frac{at^{2}}{2} \Vert v_{\epsilon} \Vert ^{2}+\frac{bt^{4-s}}{4-s} \Vert v_{\epsilon} \Vert ^{4-s}-\frac{t^{2^{*}(s)}}{2^{*}(s)}. \end{aligned} $$

From

$$ 0=g'(t_{\epsilon})=at_{\epsilon} \Vert v_{\epsilon} \Vert ^{2}+bt_{\epsilon}^{3-s} \Vert v_{\epsilon} \Vert ^{4-s}-t_{\epsilon}^{2^{*}(s)-1}- \int_{\Omega}f(x,t_{\epsilon}v_{\epsilon})v_{\epsilon}\,\mathrm{d}x $$

we derive

$$ a \Vert v_{\epsilon} \Vert ^{2}+bt_{\epsilon}^{2-s} \Vert v_{\epsilon} \Vert ^{4-s}=t_{\epsilon}^{4-2s}+\frac{1}{t_{\epsilon}} \int_{\Omega}f(x,t_{\epsilon}v_{\epsilon})v_{\epsilon}\,\mathrm{d}x\ge t_{\epsilon}^{4-2s}. $$

(3.6)

Since $4-2s=2(2-s)$, we have

$$ t_{\epsilon}\le \biggl[\frac{b \Vert v_{\epsilon} \Vert ^{4-s}+\sqrt{b^{2} \Vert v_{\epsilon} \Vert ^{2(4-s)}+4a \Vert v_{\epsilon} \Vert ^{2}}}{2} \biggr]^{\frac{1}{2-s}} \triangleq t_{0}. $$

By $(f_{1})$, obviously,

$$ f(x,t)\le\epsilon t^{5-2s}+d(\epsilon)t,\quad d(\epsilon)>0. $$

Therefore we obtain

$$ a \Vert v_{\epsilon} \Vert ^{2}+bt_{\epsilon}^{2-s} \Vert v_{\epsilon} \Vert ^{4-s}\le t_{\epsilon}^{4-2s}+\epsilon \int_{\Omega} \vert t_{\epsilon} \vert ^{4-2s} \vert v_{\epsilon} \vert ^{2^{*}(s)}\, \mathrm{d}x+d(\epsilon) \int_{\Omega} \vert v_{\epsilon} \vert ^{2} \,\mathrm{d}x $$

(3.7)

and

$$ t_{\epsilon}^{4-2s}+\epsilon \int_{\Omega} \vert t_{\epsilon} \vert ^{4-2s} \vert v_{\epsilon} \vert ^{2^{*}(s)}\, \mathrm{d}x=t_{\epsilon}^{4-2s} \biggl(1+\epsilon \int_{\Omega} \vert v_{\epsilon} \vert ^{2^{*}(s)}\,\mathrm{d}x \biggr)\le\frac{3}{2}t_{\epsilon}^{4-2s}. $$

(3.8)

Thanks to (2.3), when ϵ is small enough, we conclude from $d(\epsilon)\int_{\Omega}|v_{\epsilon}|^{2}\,\mathrm{d}x\to0$ as $\epsilon\to0$ that

$$ d(\epsilon) \int_{\Omega} \vert v_{\epsilon} \vert ^{2} \,\mathrm{d}x\le a \Vert v_{\epsilon} \Vert ^{2}. $$

(3.9)

From (3.7)–(3.9) we get

$$ a \Vert v_{\epsilon} \Vert ^{2}+bt_{\epsilon}^{2-s} \Vert v_{\epsilon} \Vert ^{4-s}\le\frac {3}{2}t_{\epsilon}^{4-2s}+a \Vert v_{\epsilon} \Vert ^{2}. $$

Combining this with (2.2), we obtain

$$ bS^{\frac{4-s}{2}}\le b \Vert v_{\epsilon} \Vert ^{4-s}\le \frac{3}{2}t_{\epsilon}^{2-s}, $$

which implies

$$ t_{\epsilon}\ge \biggl(\frac{2bS^{\frac{4-s}{2}}}{3} \biggr)^{\frac{1}{2-s}}. $$

Consequently, the function $g_{0}(t)$ attains its maximum at $t_{0}$ and continuously increases in the interval $[0,t_{0}]$. From this, together with (2.2) and the inequality $F(x,t)\ge C_{2}|t|^{\rho}$, which is directly obtained from $(f_{2})$, we derive that

$$\begin{aligned} g(t)\le{}& g_{0}(t_{0})- \int_{\Omega}F(x,t_{\epsilon}v_{\epsilon})\, \mathrm{d}x \\ ={}&\frac{a(2-s)}{2(3-s)}t^{2}_{0} \Vert v_{\epsilon} \Vert ^{2}+\frac {b(2-s)}{2(3-s)(4-s)}t^{4-s}_{0} \Vert v_{\epsilon} \Vert ^{4-s}- \int_{\Omega}F(x,t_{0} v_{\epsilon})\, \mathrm{d}x \\ \le{}&\frac{a(2-s)}{2(3-s)} \Vert v_{\epsilon} \Vert ^{2} \biggl[\frac{b \Vert v_{\epsilon} \Vert ^{4-s}+\sqrt{b^{2} \Vert v_{\epsilon} \Vert ^{2(4-s)}+4a \Vert v_{\epsilon} \Vert ^{2}}}{2} \biggr]^{\frac{2}{2-s}} \\ &{} +\frac{b(2-s)}{2(3-s)(4-s)} \Vert v_{\epsilon} \Vert ^{4-s} \biggl[\frac{b \Vert v_{\epsilon} \Vert ^{4-s} +\sqrt{b^{2} \Vert v_{\epsilon} \Vert ^{2(4-s)}+4a \Vert v_{\epsilon} \Vert ^{2}}}{2} \biggr]^{\frac{4-s}{2-s}} \\ &{} -C_{2} \int_{\Omega}t^{\rho}_{\epsilon} \vert v_{\epsilon} \vert ^{\rho}\,\mathrm{d}x \\ \le{}&\Lambda+O \biggl(\frac{1}{2(2-s)} \biggr)-C_{2} \biggl( \frac{2bS^{\frac {4-s}{2}}}{3} \biggr)^{\frac{\rho}{2-s}} \int_{\Omega} \vert v_{\epsilon} \vert ^{\rho}\, \mathrm{d}x. \end{aligned}$$

In addition, from (2.3) it follows that

$$ \int_{\Omega} \vert v_{\epsilon} \vert ^{\rho}\,\mathrm{d}x=O \bigl(\epsilon^{\frac {\sqrt{\bar{\mu}}(3-\rho\sqrt{\bar{\mu}})}{(2-s)\sqrt{\bar{\mu}-\mu }}} \bigr). $$

Thanks to $(f_{2})$, we have

$$ \frac{1}{2(2-s)}>\frac{\sqrt{\bar{\mu}}(3-\rho\sqrt{\bar{\mu }})}{(2-s)\sqrt{\bar{\mu}-\mu}}. $$

Choosing ϵ small enough, we conclude

$$ \sup_{t\ge0}I(tv_{\epsilon})=g(t_{\epsilon})< \Lambda. $$

This completes the proof of Lemma 3.2. □

Next, we prove that the functional $I(u)$ satisfies the mountain pass geometry.

Lemma 3.3

Suppose that $(f_{1})$ and $(f_{2})$ hold. Then we have:

• (i) there exist $r,\beta>0$ such that $\inf_{\|u\|=r}I(u)\ge\beta>0$,

• (ii) there exists a nonnegative function $e\in H^{1}_{0}(\Omega)$ such that $\|e\|>r$ and $I(e)<0$.

Proof

(i) By $(f_{1})$, for any $\epsilon>0$, there exists $C_{3}$ such that

$$ \bigl\vert f(x,t) \bigr\vert \le\epsilon t+C_{3}t^{5-2s} $$

for all $t\in\mathbb {R}^{+}$ and $x\in\overline{\Omega}$. By the definition of $F(x,u)$ we get

$$ \bigl\vert F(x,t) \bigr\vert \le\frac{1}{2}\epsilon t^{2}+C_{4}t^{2^{*}(s)} $$

for all $t\in\mathbb {R}^{+}$ and $x\in\overline{\Omega}$, where $C_{5}=\frac {1}{2^{*}(s)}C_{3}$. Then by (2.1),we get

$$\begin{aligned} I(u)&=\frac{a}{2} \Vert u \Vert ^{2}+ \frac{b}{4-s} \Vert u \Vert ^{4-s}-\frac{1}{2^{*}(s)} \int _{\Omega}\frac{ \vert u \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm{d}x- \int_{\Omega}F(x,u)\, \mathrm{d}x \\ &\ge\frac{a}{2} \Vert u \Vert ^{2}- \frac{1}{2^{*}(s)}S^{-\frac{2^{*}(s)}{2}} \Vert u \Vert ^{2^{*}(s)}- \frac{1}{2}\epsilon \vert u \vert ^{2}_{2}-C_{5} \vert u \vert ^{2^{*}(s)}_{2^{*}(s)} \\ &\ge\frac{a}{2} \Vert u \Vert ^{2}- \frac{1}{2^{*}(s)}S^{-\frac{2^{*}(s)}{2}} \Vert u \Vert ^{2^{*}(s)}- \frac{C}{2}\epsilon \Vert u \Vert ^{2}_{2}-CC_{5} \Vert u \Vert ^{2^{*}(s)}_{2^{*}(s)} \end{aligned}$$

for ϵ small enough. Hence there exists $\beta>0$ such that $I(u)\ge\beta$ for all $\|u\|=r$, where $r>0$ is small enough.

By Lemma 3.2, there exists $u_{0}\in H^{1}_{0}(\Omega)$, $u_{0}\not\equiv0$ such that

$$ \sup_{t\ge0}I(tu_{0})< \Lambda. $$

It follows from the nonnegativity of $F(x,t)$ that

$$ \begin{aligned} I(tu_{0})&=\frac{at^{2}}{2} \Vert u_{0} \Vert ^{2}+\frac{bt^{4-s}}{4-s} \Vert u_{0} \Vert ^{4-s}-\frac {t^{2^{*}(s)}}{2^{*}(s)} \int_{\Omega}\frac{ \vert u_{0} \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm {d}x- \int_{\Omega}F(x,tu_{0})\,\mathrm{d}x \\ &\le\frac{at^{2}}{2} \Vert u_{0} \Vert ^{2}+ \frac{bt^{4-s}}{4-s} \Vert u_{0} \Vert ^{4-s}- \frac {t^{2^{*}(s)}}{2^{*}(s)} \int_{\Omega}\frac{ \vert u_{0} \vert ^{2^{*}(s)}}{ \vert x \vert ^{s}}\,\mathrm{d}x, \end{aligned} $$

which shows that $\lim_{t\to+\infty}I(tu_{0})\to-\infty$. Therefore we can choose $t_{0}$ such that $\|t_{0}u_{0}\|>r$ and $I(t_{0}u_{0})\le0$. The proof of Lemma 3.3 is completed. □

Proof of Theorem 1.1

By the mountain pass theorem in [16] there is a sequence $\{u_{n}\}\subset H^{1}_{0}(\Omega)$ satisfying

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

where

$$ c=\inf_{\gamma\in\tau}\max_{t\in[0,1]}I \bigl(\gamma(t) \bigr), $$

$$ \tau= \bigl\{ \gamma\in C \bigl([0,1],H^{1}_{0}(\Omega) \bigr)|\gamma(0)=0,\gamma(1)=t_{0}u_{0} \bigr\} . $$

Therefore

$$ 0< \beta\le c=\inf_{\gamma\in\tau}\max_{t\in[0,1]}I \bigl( \gamma(t) \bigr)\le\max_{t\in[0,1]}I(tt_{0}u_{0}) \le\sup_{t\ge0}I(tu_{0})< \Lambda. $$

Applying this inequality and Lemma 3.1, we can obtain a critical point $u_{1}$ of the functional I. From the continuity of $I'$ we deduce that $u_{1}$ is a weak solution of problem (1.1). Then $\langle I'(u_{1}),u_{1}^{-}\rangle=0$, where $u_{1}^{-}=\min\{u,0\}$. Thus $u_{1}\ge0$ and $u_{1}\not\equiv0$. By the strong maximum principle there is $u_{1}>0$ that is a positive solution of problem (1.1). Thus Theorem 1.1 holds. □

Proof of Theorem 1.2

Theorem 1.2 can be proved similarly. □