We consider the following elliptic equation with Dirichlet boundary value condition

$ \left\{\begin{array}{ll} -\Delta u=f(x,u),& x\in\Omega,\\ u=0,& x\in\partial\Omega, \end{array}\right. $

(1.1)

where $\Omega$ is a bounded domain in $\mathbb{R}^{n}(n\ge 3)$ with smooth boundary $\partial\Omega$ and $f\in C(\overline{\Omega}\times \mathbb{R}, \mathbb{R})$.

Many authors were interested in studying the existence of nontrivial solution of (1.1) via variational methods, for example, see [1-14].

Ambrosetti, Rabinowitz [1] established the existence of nontrivial solution of (1.1) by applying mountain pass lemma under the following conditions (also see [3]).

(f$_1)$ There exist positive constants $a$, $b$ and $s\in\big(0,\frac{n+2}{n-2}\big)$ such that for $(x,u)\in\Omega\times\mathbb{R}$,

$\left|f(x,u)\right|\le a+b\left|u\right|^{s};$

(f$_{2})$ $\liminf\limits_{u\to +\infty}\frac{f(x,u)}{u}>\lambda_1$ uniformly in $x\in\overline{\Omega}$;

(f$_{3})$ There exist positive constants $\mu>2$ and $r>0$ such that for $x\in\Omega$ and $\left|u\right|\ge r$,

$0<\mu F(x,u)\le uf(x,u),$

where $F(x,u)=\int_0^uf(x,t)dt$ and $\lambda_1$ is the first eigenvalue of the Laplacian $(-\Delta)$ on $\Omega$ with zero Dirichlet boundary condition;

(f$_{4})$ $\limsup\limits_{u\to 0}\frac{f(x,u)}{u}<\lambda_1$ uniformly in $x\in\overline{\Omega}$.

As is well known, (f$_3)$ is so-called Ambrosetti-Rabinowitz condition (see [1]), (AR) for short, which guarantees that Palais-Smale sequence of the Euler-Lagrange functional is bounded. Actually, integrating (AR) it follows that $\liminf\limits_{u\to +\infty}\frac{f(x,u)}{u}=+\infty$ uniformly in $x\in\overline{\Omega}$, and hence condition (f$_{2})$ can be eliminated in [3, Theorem 4.8.13]. Recently, Mavinga and Nkashama in [15] provided a new method to ensure that (PS) condition is satisfied. Motivated by above references, we in this paper prove the existence of solution for (1.1) by using mountain pass lemma, and replace (f$_1)$-(f$_3)$ with

(f$_{1}')$ $\lambda_{1}<\liminf\limits_{|u|\to \infty} \frac{f(x,u)}{u}\le \limsup\limits_{|u|\to \infty} \frac{f(x,u)}{u}<\lambda_{2}$ uniformly in $x\in\overline{\Omega}$, where $\lambda_2$ is the second eigenvalue of the Laplacian $(-\Delta)$ on $\Omega$ with zero Dirichlet boundary condition.

We also establish the existence of solution for (1.1) by the least action principle under the conditions:

(f$_{2}')$ $\limsup\limits_{\left|u\right|\to \infty} \frac{f(x,u)}{u}<\lambda_1$ uniformly in $x\in\overline{\Omega}$;

(f$_{3}')$ $\liminf\limits_{u\to 0^+}\frac{f(x,u)}{u}>\lambda_1$ uniformly in $x\in\overline{\Omega}$.

Let the norm of $u$ in Sobolev space $W_{0}^{1,2}(\Omega)$ be ${\|u\|}_{1,2}=(\int_\Omega |Du|^{2}dx)^{\frac{1}{2}}$ and $\|u\|_2=(\int_\Omega|u|^2dx)^{\frac{1}{2}}$ stand for the usual $L^2$-norm. In addition, the $W_{0}^{1,2}(\Omega)$-inner product is defined as $[u,v]=\int_{\Omega}Du\cdot Dvdx$ and we denote the $n$-dimensional Lebesgue measure of $\Omega$ by $|\Omega|$. From [10] we know that under (f$_1)$ the Euler-Lagrange functional

$I(u)=\frac{1}{2}\int_\Omega\ \left|Du\right|^{2}dx-\int_{\Omega}F(x,u)dx,\ \forall u\in W_{0}^{1,2}(\Omega)$

belongs to $C^1(W_{0}^{1,2}(\Omega),\mathbb{R})$ and

$(I^{\prime}(u),\varphi)=\int_\Omega\ Du\cdot D\varphi dx-\int_{\Omega}f(x,u)\varphi dx,\ \forall u,\varphi\in W_{0}^{1,2}(\Omega).$

Thus the critical points of $I$ are the weak solutions to (1.1).

From [5] we have the following facts. The Laplacian $(-\Delta)$ on $\Omega$ with zero Dirichlet boundary condition has a sequence of eigenvalues $0<\lambda_{1}<\lambda_{2}\le\lambda_{3}\le\cdots \le\lambda_{j}\le\cdots \to \infty$ as $j\to \infty$. The first eigenvalue

$\lambda_{1}=\min\limits_{u\in W_0^{1,2}(\Omega),\|u\|_2\neq0}\frac{[u,u]}{\|u\|_2^2} =\min\limits_{u\in W_0^{1,2}(\Omega),\|u\|_2\neq0}\frac{\|u\|_{1,2}^2}{\|u\|_2^2}$

(1.2)

is simple and there exists an eigenfunction $\varphi_1\in W_0^{1,2}(\Omega)\cap C^2(\overline{\Omega})$ (see [2, Theorem 1.16]) corresponding to $\lambda_1$ such that $\varphi_1(x)>0$ in $\Omega$ and

$\lambda_1\|\varphi_1\|_2^2=\|\varphi_1\|_{1,2}^2.$

(1.3)

The eigenspace corresponding to $\lambda_1$ can be expressed as $V={\rm span}\{\varphi_1\}=\{t\varphi_1:t\in\mathbb{R}\}$ and

$\lambda_2=\min\limits_{u\in W_0^{1,2}(\Omega)\cap V^{\perp},\|u\|_2\neq0}\frac{\|u\|_{1,2}^2}{\|u\|_2^2}.$

(1.4)

Now we introduce some auxiliary results which will be need in the sequel.

Let $E$ be a real Banach space. For $I\in C^{1}(E,\mathbb{R})$, we say $I$ satisfies the Palais-Smale condition ((PS) for short) if any sequence $\{u_{m}\}\subset E$ for which $\{I(u_{m})\}$ is bounded and $I^{\prime}(u_{m})\to 0$ as $m\to\infty$ possesses a convergent subsequence.

Lemma 1  [10] (Mountain pass lemma) Let $I\in C^{1}(E,\mathbb{R})$ satisfy (PS). Suppose $I(0)=0$ and

(I$_{1})$ there are constants $\rho,\alpha>0$ such that $I\left|_{\partial B_\rho}\right.\ge \alpha$, and

(I$_{2})$ there is an $e\in E\backslash\overline{B}_\rho$ such that $I(e)\le0$.

Then $I$ possesses a critical value $c\ge\alpha$. Moreover $c$ can be characterized as

$c=\inf\limits_{g\in\Gamma}\max\limits_{u\in g([0,1])}I(u),$

where $\Gamma=\{g\in C([0,1],E)\left|\right.g(0)=0,g(1)=e\}$.

Lemma 2  [11] (The least action principle) Suppose $E$ is a reflexive Banach space and $I:E\to\mathbb{R}$ is coercive and (sequentially) weakly lower semi-continuous, that is, the following conditions are fulfilled:

(ⅰ) $I(u)\to\infty$ as $\|u\|\to\infty$;

(ⅱ) For any $u\in E$, any sequence $\{u_m\}$ in $E$ such that $u_m\to u$ weakly in $E$ there holds

$I(u)\le\liminf\limits_{m\to\infty}I(u_m).$

Then $I$ is bounded from below and attains its infimum which is a critical value if $I\in C^{1}(E,\mathbb{R})$.

Theorem 1  If (f$_{1}')$ and (f$_4)$ are satisfied, then (1.1) has at least one nontrivial weak solution $u\in W_{0}^{1,2}(\Omega)$.

Proof  We know easily from (f$_{1}')$ that (f$_1)$ is fulfilled. By (f$_{1}')$ we take $\varepsilon>0$ such that $\lambda_{1}+\varepsilon<\liminf\limits_{|u|\to \infty} \frac{f(x,u)}{u}\le \limsup\limits_{|u|\to \infty}\frac{f(x,u)}{u}<\lambda_{2}-\varepsilon$. Then there exists $r>0$ such that for $|u|\ge r$,

$\lambda_{1}+\varepsilon\le \frac{f(x,u)}{u}\le \lambda_{2}-\varepsilon,\ \forall x\in \overline{\Omega}.$

(2.1)

First we prove that the functional $I$ satisfies (PS). By (2.1) we denote $\tau:\overline{\Omega}\times\mathbb{R}\to\mathbb{R}$ by

$ \tau(x,u)=\left\{\begin{array}{ll} \frac{f(x,u)}{u},&\left|u\right|\ge r,\\ \frac{f(x,r)+f(x,-r)}{2r^{2}}u+\frac{f(x,r)-f(x,-r)}{2r},&\left|u\right|<r. \end{array}\right. $

(2.2)

Since $f$ is continuous, $\tau$ is continuous in $\overline{\Omega}\times\mathbb{R}$. It is easy to see from (2.1) that

$ \lambda_1+\varepsilon\le\tau(x,u)\le\lambda_2-\varepsilon,\ \forall (x,u)\in\overline{\Omega}\times\mathbb{R}. $

(2.3)

Define $l:\overline{\Omega}\times\mathbb{R}\to\mathbb{R}$ by $l(x,u)=f(x,u)-\tau(x,u)u$. Then it follows from the continuity of $f$ and $\tau$ that there exists a constant $k>0$ such that

$ |l(x,u)|\le k$

(2.4)

for all $(x,u)\in\overline{\Omega}\times\mathbb{R}$.

Now suppose $\{u_{m}\}\subset W_{0}^{1,2}(\Omega)$ for which $\{I(u_{m})\}$ is bounded and $\lim\limits_{m\to \infty}I^{\prime}(u_{m})=0$. Let $u_{m}=v_{m}+w_{m}$, where $v_{m}\in V,w_{m}\in X=V^{\perp}$.

Since $\lim\limits_{m\to \infty}I^{\prime}(u_{m})=0$, there exists $N>0$ such that$(I^{\prime}(u_{m}),w_{m}-v_{m})\le\varepsilon{\|w_{m}-v_{m}\|}_{1,2}$ for all $m\ge N$. According to the orthogonality of $w_{m}$ and $v_{m}$ in $W_{0}^{1,2}(\Omega)$, we have

$\begin{eqnarray*} &&(I^{\prime}(u_{m}),w_{m}-v_{m})=\int_{\Omega}Du_{m}\cdot D(w_{m}-v_{m})dx-\int_{\Omega}f(x,u_{m})(w_{m}-v_{m})dx\\ &=&\int_{\Omega}D(v_{m}+w_{m})\cdot D(w_{m}-v_{m})dx-\int_{\Omega}f(x,u_{m})(w_{m}-v_{m})dx\\ &=&\int_{\Omega}\left|Dw_{m}\right|^{2}dx-\int_{\Omega}\left|Dv_{m}\right|^{2}dx-\int_{\Omega}l(x,u_{m})w_{m}dx+\int_{\Omega}l(x,u_{m})v_{m}dx\\ &&-\int_{\Omega}\tau(x,u_{m})w^{2}_{m}dx+\int_{\Omega}\tau(x,u_{m})v^{2}_{m}dx. \end{eqnarray*}$

Thus

$\begin{eqnarray*} &&{\|w_{m}\|}_{1,2}^{2}-{\|v_{m}\|}_{1,2}^{2}-\int_{\Omega}\tau(x,u_{m})w^{2}_{m}dx+\int_{\Omega}\tau(x,u_{m})v^{2}_{m}dx\\ &\le&\varepsilon{\|w_{m}-v_{m}\|}_{1,2}+\int_{\Omega}l(x,u_{m})w_{m}dx-\int_{\Omega}l(x,u_{m})v_{m}dx. \end{eqnarray*}$

It follows from (2.3), (1.4) and (1.3) that

$\begin{eqnarray*} &&{\|w_{m}\|}_{1,2}^{2}-{\|v_{m}\|}_{1,2}^{2}-\int_{\Omega}\tau(x,u_{m})w^{2}_{m}dx+\int_{\Omega}\tau(x,u_{m})v^{2}_{m}dx\\ &\ge&{\|w_{m}\|}_{1,2}^{2}-{\|v_{m}\|}_{1,2}^{2}-(\lambda_2-\varepsilon)\int_{\Omega}w^{2}_{m}dx+(\lambda_1+\varepsilon)\int_{\Omega}v^{2}_{m}dx\\ &\ge&{\|w_{m}\|}_{1,2}^{2}-{\|v_{m}\|}_{1,2}^{2}-\frac{\lambda_2-\varepsilon}{\lambda_{2}}{\|w_{m}\|}_{1,2}^{2} +\frac{\lambda_1+\varepsilon}{\lambda_{1}}{\|v_{m}\|}_{1,2}^{2}=\frac{\varepsilon}{\lambda_{2}}\|w_{m}\|_{1,2}^{2} +\frac{\varepsilon}{\lambda_{1}}\|v_{m}\|_{1,2}^{2}, \end{eqnarray*}$

and from (2.4) and (1.2) that

$\begin{eqnarray*} &&\varepsilon{\|w_{m}-v_{m}\|}_{1,2}+\int_{\Omega}l(x,u_{m})w_{m}dx-\int_{\Omega}l(x,u_{m})v_{m}dx\\ &\le&\varepsilon({\|w_{m}\|}_{1,2}+{\|v_{m}\|}_{1,2})+k\int_{\Omega}\left|w_{m}\right|dx+k\int_{\Omega}\left|v_{m}\right|dx\\ &\le&\varepsilon({\|w_{m}\|}_{1,2}+{\|v_{m}\|}_{1,2})+k|\Omega|^{\frac{1}{2}}\|w_{m}\|_2+k|\Omega|^{\frac{1}{2}}\|v_{m}\|_2\\ &\le& (\varepsilon+k|\Omega|^{\frac{1}{2}}\lambda_1^{-\frac{1}{2}})({\|w_{m}\|}_{1,2}+{\|v_{m}\|}_{1,2}). \end{eqnarray*}$

Therefore,

$\frac{\varepsilon}{\lambda_{2}}\|w_{m}\|_{1,2}^{2}+\frac{\varepsilon}{\lambda_{1}}\|v_{m}\|_{1,2}^{2} \le(\varepsilon+k|\Omega|^{\frac{1}{2}}\lambda_1^{-\frac{1}{2}})(\|w_{m}\|_{1,2}+\|v_{m}\|_{1,2}).$

By the orthogonality of $w_{m}$ and $v_{m}$, we have

$\frac{\varepsilon}{\lambda_{2}}\|u_{m}\|_{1,2}^{2}=\frac{\varepsilon}{\lambda_{2}}(\|w_{m}\|_{1,2}^{2}+\|v_{m}\|_{1,2}^{2}) \le\frac{\varepsilon}{\lambda_{2}}\|w_{m}\|_{1,2}^{2}+\frac{\varepsilon}{\lambda_{1}}\|v_{m}\|_{1,2}^{2},$

and $\|w_{m}\|_{1,2}+\|v_{m}\|_{1,2}\le2\|u_{m}\|_{1,2}$. Hence

${\|u_{m}\|}_{1,2}^{2}\le2(\varepsilon+k|\Omega|^{\frac{1}{2}}\lambda_1^{-\frac{1}{2}})\frac{\lambda_{2}}{\varepsilon}\|u_{m}\|_{1,2},$

which implies that $\{u_{m}\}$ is bounded in $W_{0}^{1,2}(\Omega)$. By [10, Proposition B.35], $I$ satisfies (PS).

By means of mountain pass lemma, the rest of proof is similar to the proofs in [3, Theorem 4.8.13] and [2, Theorem 8.11].

Theorem 2  If (f$_{1})$, (f$_{2}')$ and (f$_{3}')$ are satisfied, then (1.1) has at least one nontrivial weak solution $u\in W_{0}^{1,2}(\Omega)$.

Proof  It follows from (f$_{1})$ and [10, Proposition B.10] that $I$ is weakly lower semi-continuous. We will prove that $I$ is coercive.

By (f$_{2}')$ and (f$_3')$, we can take $0<\varepsilon<\lambda_1$ and there exists $0<r<R$ such that for $|u|>R$,

$\frac{f(x,u)}{u}\le\lambda_1-\varepsilon,\forall x\in\overline{\Omega},$

(2.5)

and for $0<u<r$,

$f(x,u)\ge(\lambda_1+\varepsilon)u,\forall x\in\overline{\Omega}.$

(2.6)

From (2.5) we have that for $|u|>R$, $F(x,u)\le\frac{1}{2}(\lambda_1-\varepsilon)u^{2},\forall x\in\overline{\Omega}$. For $|u|\le R$, it is easy to see from (f$_1)$ that $F(x,u)\le aR+\frac{b}{s+1}R^{s+1}\triangleq C,\forall x\in\overline{\Omega}$. Then for every $(x,u)\in\overline{\Omega}\times \mathbb{R}$,

$F(x,u)\le\frac{1}{2}(\lambda_1-\varepsilon)u^{2}+C.$

(2.7)

Thus for $u\in W_{0}^{1,2}(\Omega)$, we have from (2.7) and (1.2) that

$ I(u)\ge\frac{1}{2}{\|u\|}_{1,2}^{2}-\frac{1}{2}(\lambda_1-\varepsilon){\|u\|}_{2}^{2}-C|\Omega| \ge\frac{\varepsilon}{2\lambda_1}\|u\|_{1,2}^{2}-C|\Omega|$

and $I$ is coercive. It follows from Lemma 2 that $I$ has a critical point in $u\in W_{0}^{1,2}(\Omega)$ such that $I(u)=\inf\limits_{v\in W_{0}^{1,2}(\Omega)}I(v)$.

Now we show that it is nontrivial. In fact, let $\varphi_1\in W_0^{1,2}(\Omega)\cap C^2(\overline{\Omega})$ be the eigenfunction corresponding to $\lambda_1$ with $0<\varphi_1(x)< r$ in $\Omega$. Hence by (2.6) and (1.3) we have

$\begin{eqnarray*} && I(\varphi_1)=\displaystyle\frac{1}{2}\int_{\Omega}|D\varphi_1|^{2}dx-\int_{\Omega}\Big(\int_{0}^{\varphi_1}f(x,t)dt\Big)dx\\ &\le & \displaystyle\frac{1}{2}\|\varphi_1\|_{1,2}^{2}-\frac{1}{2}(\lambda_{1}+\varepsilon)\|\varphi_1\|_{2}^{2}=-\frac{\varepsilon}{2\lambda_{1}}\|\varphi_1\|_{1,2}^{2}<0. \end{eqnarray*}$

Therefore,

$I(u)=\inf\limits{v\in W_{0}^{1,2}(\Omega)}I(v)\le I(\varphi_1)<0.$

The proof is completed.