Define the functional \(I: H_{0}^{1}(\Omega)\to R\) as follows:

$$ I(u)=\frac{b}{4}\|u\|^{4}+\frac{a}{2}\|u \|^{2}- \int_{\Omega}F(x,u)\,dx. $$

(10)

From (8) (or (9)), by a standard argument, the functional \(I\in C^{1}( H_{0}^{1}(\Omega), R)\), and a weak solution of problem (1) is a critical point of the functional *I* in \(H_{0}^{1}(\Omega)\).

Recall that a sequence \(\{u_{n}\}\subset H_{0}^{1}(\Omega)\) is called a \((PS)_{c}\) sequence for any \(c\in R\) of the functional *I* on \(H_{0}^{1}(\Omega)\) if \(I(u_{n})\to c\) and \(I'(u_{n})\to0\) as \(n\to\infty\). The functional *I* is called to satisfy the \((PS)_{c}\) condition if any \((PS)_{c}\) sequence has a convergent subsequence. We will prove our theorems by using the following three-critical-point theorem related to local linking due to Brezis and Nirenberg (see Theorem 4 in [1]).

### Theorem A

*Let*
*X*
*be a Banach space with a direct sum decomposition*
\(X=X_{1}\oplus X_{2}\)
*with*
\(\dim X_{1}<\infty\). *Let*
*I*
*be a*
\(C^{1}\)
*function on*
*X*
*with*
\(I(0)=0\)
*satisfying the*
\((PS)\)
*condition*, *and assume that*, *for some*
\(R>0\),

$$\textstyle\begin{cases} I(u)\leq0 \quad \textit{for }u\in X_{1}, \|u\|\leq R, \\ I(u)\geq0 \quad \textit{for }u\in X_{2}, \|u\|\leq R. \end{cases} $$

*Assume also that*
*I*
*is bounded below and*
\(\inf_{X} I<0\). *Then*
*I*
*has at least two nonzero critical points*.

### Proof of Theorem 1

(a) The functional *I* satisfies the local linking at zero with respect to \((V_{k}, V_{k}^{\bot})\), where \(V_{k}=\bigoplus_{i=1}^{k}\ker(-\Delta-\lambda_{i}(m_{0}))\) and \(V_{k}^{\bot}=\bigoplus _{i=k+1}^{+\infty}\ker(-\Delta-\lambda_{i}(m_{0}))\) such that \(H_{0}^{1}(\Omega )=V_{k}\oplus V_{k}^{\bot}\).

In fact, from (5), for any \(\varepsilon>0\), there is a positive constant \(L_{0}\) such that

$$\bigl\vert 2F(x,t)-am_{0}(x)t^{2} \bigr\vert \leq a \varepsilon t^{2} \quad \mbox{for any }x\in\Omega \mbox{ and } \vert t \vert \leq L_{0}. $$

Combining the continuity of *F*, (8), and the above inequality, there is \(M_{0}=M_{0}(\varepsilon)>0\) such that

$$\begin{aligned}& F(x,t)\geq\frac{a}{2}m_{0}(x)t^{2}- \frac{a\varepsilon}{2}t^{2}-M_{0}|t|^{p}\quad \mbox{for any } (x,t)\in\Omega\times R,\quad \mbox{and} \end{aligned}$$

(11)

$$\begin{aligned}& F(x,t)\leq\frac{a}{2}m_{0}(x)t^{2}+ \frac{a\varepsilon}{2}t^{2}+M_{0}|t|^{p} \quad \mbox{for any } (x,t)\in\Omega\times R. \end{aligned}$$

(12)

For any \(u\in V_{k}\), from (2), (10), and (11), it follows that

$$\begin{aligned} I(u) \leq&\frac{b}{4}\|u\|^{4}+\frac{a}{2}\|u \|^{2}-\frac{a}{2} \int_{\Omega}m_{0}(x)|u|^{2}\,dx+ \frac{a\varepsilon}{2} \int_{\Omega}|u|^{2}\,dx+M_{0} \int_{\Omega}|u|^{p}\,dx \\ \leq&\frac{a}{2} \biggl(1-\frac{1}{\lambda_{k}(m_{0})}+\varepsilon\tau _{2}^{2} \biggr)\|u\|^{2}+\frac{b}{4}\|u \|^{4}+M_{0} \tau_{p}^{p}\|u \|^{p}. \end{aligned}$$

(13)

On the other hand, for any \(u\in V_{k}^{\bot}\), from (2), (10), and (12), we obtain

$$\begin{aligned} I(u) \geq&\frac{b}{4}\|u\|^{4}+\frac{a}{2}\|u \|^{2}-\frac{a}{2} \int_{\Omega}m_{0}(x)|u|^{2}\,dx- \frac{a\varepsilon}{2} \int_{\Omega}|u|^{p}\,dx-M_{0} \int_{\Omega}|u|^{p}\,dx \\ \geq&\frac{a}{2} \biggl(1-\frac{1}{\lambda_{k+1}(m_{0})}-\varepsilon\tau ^{2}_{2} \biggr)\|u\|^{2}+\frac{b}{4}\|u \|^{4}-M_{0} \tau_{p}^{p}\|u \|^{p}. \end{aligned}$$

(14)

Noting that \(\lambda_{k}(m_{0})<1<\lambda_{k+1}(m_{0})\) and \(4< p<2^{*}\), (13) and (14), let \(\varepsilon=\min\{(1-\lambda _{k}(m_{0}))/\lambda_{k}(m_{0}), (\lambda_{k+1}(m_{0})-1)/\lambda_{k+1}(m_{0})\}/2\tau^{2}_{2}\), there is a constant \(r_{0}>0\) such that

$$\begin{aligned}& I(u)< 0\quad \mbox{for any } u\in V_{k}\mbox{ with }0< \|u\|\leq r_{0}, \\& I(u)> 0\quad \mbox{for any } u\in V_{k}^{\bot}\mbox{ with } 0< \|u\|\leq r_{0}. \end{aligned}$$

(b) The functional *I* satisfies the \((PS)\) condition. To the end, it suffices to say the functional *I* is coercive on \(H^{1}_{0}(\Omega)\), i.e., \(I(u)\to+\infty\) as \(\|u\|\to\infty\).

If \(\mu_{1}(m_{\infty})>1\), by (6), for any \(\varepsilon>0\), there is \(L_{1}>0\) such that

$$\bigl\vert 4F(x,t)-b m_{\infty}(x)t^{4} \bigr\vert \leq b \varepsilon t^{4} \quad \mbox{for any }x\in \Omega\mbox{ and } \vert t \vert \geq L_{1}. $$

Hence, from the continuity of *F*, there exists \(M_{1}=M_{1}(\varepsilon )>0\) such that

$$ F(x,t)\leq\frac{b}{4}m_{\infty}(x)t^{4}+ \frac{b\varepsilon}{4}t^{4}+M_{1}\quad \mbox{for any } (x,t)\in \Omega\times R. $$

(15)

From (2), (10), and (15), we obtain

$$\begin{aligned} I(u) \geq&\frac{b}{4}\|u\|^{4}+\frac{a}{2}\|u \|^{2}-\frac{b}{4} \int_{\Omega }m_{\infty}(x)|u|^{4}\,dx - \frac{b\varepsilon}{4} \int_{\Omega}|u|^{4}\,dx-M_{1}|\Omega| \\ \geq&\frac{b}{4} \biggl(1-\frac{1}{\mu_{1}(m_{\infty})} -\varepsilon \tau_{4}^{4} \biggr)\|u\|^{4}-M_{1}| \Omega|, \end{aligned}$$

where \(|\Omega|\) denotes the Lebesgue measure of Ω. Hence, for \(\varepsilon>0\) small enough, it follows that the functional *I* is coercive on \(H^{1}_{0}(\Omega)\).

If \(\mu_{1}(m_{\infty})=1\) and (7) hold, let

$$H(x,t)=F(x,t)-\frac{b}{4}m_{\infty}(x)t^{4}. $$

By a simple computation, it follows that

$$H'(x,t)t-4H(x,t)=f(x,t)t-4F(x,t). $$

From (7), for any \(M_{2}>0\), there is \(L_{2}>0\) such that

$$H'(x,t)t-4H(x,t)\geq M_{2}\quad \mbox{for any }x\in \Omega\mbox{ and }|t|\geq L_{2}. $$

Hence, we have

$$\frac{d}{ds} \biggl(\frac{H(x,s)}{s^{4}} \biggr)= \frac{H'(x,s)s-4H(x,s)}{s^{5}}\geq \frac{M_{2}}{s^{5}}\quad \mbox{for any }x\in \Omega\mbox{ and }|s|\geq L_{2}. $$

Integrating the above expression over the interval \([t,T]\subset [L_{2},\infty)\), we obtain

$$\frac{H(x,t)}{t^{4}}\leq\frac{H(x,T)}{T^{4}}+\frac{M_{2}}{4} \biggl( \frac {1}{T^{4}}-\frac{1}{t^{4}} \biggr). $$

Noting that \(\lim_{|T|\to\infty}H(x,T)/T^{4}=0\), let \(T\to+\infty\), we obtain \(H(x,t)\leq-M_{2}/4\) for \(t\geq L_{2}\) and \(x\in\Omega\). Similarly, \(H(x,t)\leq-M_{2}/4\) for \(t\leq-L_{2}\) and \(x\in\Omega\). Hence, from the arbitrariness of \(M_{2}(>0)\), we have

$$\lim_{|t|\to\infty}H(x,t)=-\infty\quad \mbox{uniformly in }x\in\Omega. $$

Moreover, from the continuity of *F*, there is a positive constant \(M_{3}\) such that

$$ H(x,t)< M_{3}\quad \mbox{for any } (x,t)\in\Omega\times R. $$

(16)

If the functional *I* is not coercive on \(H^{1}_{0}(\Omega)\), there are a sequence \(\{u_{n}\}\subset H^{1}_{0}(\Omega)\) and a positive constant \(M_{4}\) such that \(\|u_{n}\|\to\infty\) as \(n\to\infty\) and \(I(u_{n})\leq M_{4}\). By the definition of \(\mu_{1}(m_{\infty})\) and \(\mu_{1}(m_{\infty})=1\), we have that \(\int_{\Omega}m_{\infty}(x)|u_{n}|^{4}\,dx\leq\|u_{n}\|^{4}\). Hence, from (16), it follows that

$$\begin{aligned} M_{4}\geq I(u_{n}) =&\frac{b}{4}\|u_{n} \|^{4}+\frac{a}{2}\|u_{n}\|^{2}- \frac {b}{4} \int_{\Omega}m_{\infty}(x)|u_{n}|^{4} \,dx- \int_{\Omega}H(x,u_{n})\,dx \\ \geq&\frac{b}{4}\|u_{n}\|^{4}+\frac{a}{2} \|u_{n}\|^{2}-\frac{b}{4} \int_{\Omega}m_{\infty}(x)|u_{n}|^{4} \,dx-M_{3}|\Omega| \\ \geq&\frac{a}{2}\|u_{n}\|^{2}-M_{3}| \Omega| \\ \to& +\infty \quad \mbox{as } n\to\infty, \end{aligned}$$

which is a contradiction, and the conclusion is proved.

(c) From (b), we have that the functional *I* is bounded from below. From the fact that \(I(u)<0\) for any \(u\in V_{k}\) with \(0<\|u\|\leq r_{0}\), we have \(\inf_{u\in H^{1}_{0}(\Omega)}I(u)<0\). Moreover, \(I(0)=0\). Therefore, Theorem 1 is proved by Theorem A. □

### Proof of Theorem 2

First of all, from (a) of the proof of Theorem 1, we have that the functional *I* satisfies the local linking at zero with respect to \((V_{k}, V_{k}^{\bot})\). And then, we know from (9) that \(f(x,t)\) is 3-sublinear at infinity, which implies that the functional *I* is coercive on \(H^{1}_{0}(\Omega)\) by a standard argument. We obtain that the functional *I* is bounded from below and satisfies the \((PS)\) condition for \(N=1,2,3\). In the following, we only prove that the functional *I* also satisfies the \((PS)\) condition for \(p=2^{*}\) (\(N\geq4\)), where \(f(x,t)\) is not only 3-sublinear at infinity, but also is asymptotically critical growth at infinity.

In fact, let \(\{u_{n}\}\) be a \((PS)\) sequence of *I*, that is,

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

(17)

Noting that the functional *I* is coercive on \(H^{1}_{0}(\Omega)\), we obtain that \(\{u_{n}\}\) is bounded in \(H^{1}_{0}(\Omega)\). Going if necessary to a subsequence, we can assume \(u_{n}\rightharpoonup u\) in \(H^{1}_{0}(\Omega )\), and by the Rellich theorem, \(u_{n}\to u\) in \(L^{r}(\Omega)\) (\(1\leq r<2^{*}\)). From (17) and the boundedness of \(\{u_{n}\}\), we have

$$ \bigl\langle I'(u_{n}),u_{n}-u \bigr\rangle =\bigl(a+b\|u_{n}\|^{2}\bigr) \int_{\Omega}\nabla u_{n}(\nabla u_{n}-\nabla u)\,dx+ \int_{\Omega}f(x,u_{n}) (u_{n}-u)\,dx\to0 $$

(18)

as \(n\to\infty\). From (9), for any \(\varepsilon>0\), there is \(M_{5}>0\) such that

$$\bigl\vert f(x,t) \bigr\vert \leq\varepsilon|t|^{p-1}+M_{5} \quad \mbox{for any }(x,t)\in\Omega \times R. $$

Hence, from Hölder’s inequality, (2), the boundedness of \(\{ u_{n}\}\), and the arbitrariness of *ε*, we have

$$\begin{aligned} \biggl\vert \int_{\Omega}f(x,u_{n}) (u_{n}-u)\,dx \biggr\vert \leq& \int_{\Omega}\bigl(\varepsilon|u_{n}|^{p-1}+M_{5} \bigr)|u_{n}-u|\,dx \\ \leq&\varepsilon \int_{\Omega}\bigl(|u_{n}|^{p}+|u_{n}|^{p-1}|u| \bigr)\,dx+M_{5} \Vert u_{n}-u \Vert _{L^{1}} \\ \leq&\varepsilon \Vert u_{n} \Vert _{L^{p}}^{p-1} \bigl( \Vert u_{n} \Vert _{L^{p}}+ \Vert u \Vert _{L^{p}}\bigr)+M_{5} \Vert u_{n}-u \Vert _{L^{1}} \\ \to& 0 \quad \mbox{as }n\to\infty. \end{aligned}$$

Combining with (18), we have

$$\int_{\Omega}\nabla u_{n}(\nabla u_{n}-\nabla u)\,dx\to0 \quad \mbox{as }n\to\infty. $$

Since \(u_{n}\rightharpoonup u\) weakly in \(H^{1}_{0}(\Omega)\), we have

$$\int_{\Omega}\nabla u(\nabla u_{n}-\nabla u)\,dx\to0 \quad \mbox{as } n\to\infty. $$

Then \(u_{n}\rightarrow u\) strongly in \(H^{1}_{0}(\Omega)\) as \(n\rightarrow \infty\).

At last, similar to (c) of the proof of Theorem 1, Theorem 2 is proved. □