In order to overcome the difficulty caused by the singular term and to obtain the second solution of problem (1.3) for \(\lambda >0\) small enough, in this section, we firstly consider the following perturbation problem:
$$ \textstyle\begin{cases} - ( a+b \Vert u \Vert ^{2} ) \Delta u+\phi_{u}u=\lambda hf(u+\alpha )+g(u), & \mbox{in } \Omega , \\ u\geqslant 0, &\mbox{in } \Omega , \\ u=0, &\mbox{on } \partial \Omega , \end{cases} $$
(3.1)
where \(\alpha >0\). We define the functional corresponding to problem (3.1)
$$\begin{aligned} &J_{\alpha }(u)=\frac{a}{2} \Vert u \Vert ^{2}+ \frac{b}{4} \Vert u \Vert ^{4}+\frac{1}{4} \int_{\Omega }\phi_{u}u^{2}-\lambda \int_{\Omega }h\bigl(F\bigl(u^{+}+\alpha \bigr)-F( \alpha )\bigr)- \int_{\Omega }G(u),\\ &\quad u\in H_{0}^{1}(\Omega ). \end{aligned}$$
It is obvious that \(J_{\alpha }\) is a \(C^{1}\) functional defined on \(H_{0}^{1}(\Omega )\). The solution of problem (3.1) corresponds to the critical point of the functional \(J_{\alpha }\). That is, if \(u\in H_{0}^{1}(\Omega )\) is a solution of problem (3.1), it satisfies
$$\begin{aligned} &\bigl(a+b \Vert u \Vert ^{2}\bigr) (u,\phi )+ \int_{\Omega }\phi_{u}u\phi -\lambda \int_{\Omega }hf\bigl(u^{+}+\alpha \bigr)\phi - \int_{\Omega }g(u)\phi =0, \\ &\quad \phi \in H_{0}^{1}( \Omega ). \end{aligned}$$
(3.2)
For any \(s>0\), since f is nonincreasing, we have
$$ F(s+\alpha )-F(\alpha )= \int_{\alpha }^{s+\alpha }f(t)\,dt= \int_{0}^{s}f( \tau +\alpha )\,d\tau \leqslant \int_{0}^{s}f(\tau )\,d\tau =F(s), $$
(3.3)
by \(F(s)=0\) if \(s\leqslant 0\), (3.3) holds for all \(s\in {\mathbb{R}}\). Then, for any \(u\in H_{0}^{1}(\Omega )\), we have
$$ J(u)\leqslant J_{\alpha }(u)\leqslant I(u), $$
(3.4)
where \(I(u)=\frac{a}{2} \Vert u \Vert ^{2}+\frac{b}{4} \Vert u \Vert ^{4}+\frac{1}{4} \int_{\Omega }\phi_{u}u^{2}-\int_{\Omega }G(u)\), \(u\in H_{0}^{1}(\Omega)\).
In order to show that \(J_{\alpha }\) satisfies the mountain pass geometry and to estimate the mountain pass critical level, we firstly consider the functional I. In fact, under the assumptions of (g1) and (g2), we have the following lemma.
Lemma 3.1
Under the assumptions of (g1) and (g2), there exist
\(r_{0},\rho_{0}>0\)
such that the functional
I
satisfies
-
(i)
\(I|_{S_{r_{0}}}\geqslant \rho_{0}\);
-
(ii)
there exists
\(u_{1}\in H_{0}^{1}(\Omega )\)
with
\(\Vert u_{1} \Vert >r_{0}\)
such that
\(I(u_{1})<0\).
Proof
(i) By (1.6) with \(\varepsilon >0\) small enough, for any \(u\in H_{0}^{1}(\Omega )\), we have
$$ I(u)\geqslant \biggl(\frac{a}{2}-c\varepsilon\biggr) \Vert u \Vert ^{2}-c_{1} \Vert u \Vert ^{6}-c_{2} \Vert u \Vert ^{p+1}, $$
it is obvious that the conclusion (i) holds.
(ii) It follows from (g1) and (g2), for any given \(M>0\), there exists \(R>0\) such that \(g(t)\geqslant Mt^{3}\), \(t>R\) and
$$ \lim_{t\to 0^{+}}\frac{g(t)-Mt^{3}}{t}=0. $$
Then there exists \(C>0\) such that \(g(t)-Mt^{3}\geqslant -Ct\), \(t\in [0,R]\) and \(g(t)\geqslant Mt^{3}-Ct\), \(t\geqslant 0\). For G, we also have
$$ G(t)\geqslant \frac{M}{4}t^{4}-\frac{Ct^{2}}{2},\quad t\in { \mathbb{R}}. $$
Thus, for any \(u\in H_{0}^{1}(\Omega )\setminus \{0\}\), \(\int_{\Omega }G(tu) \geqslant \frac{M}{4}t^{4} \vert u \vert _{4}^{4}-\frac{C}{2}t^{2} \vert u \vert _{2}^{2}\). It follows that
$$ \lim_{t\to \infty } \int_{\Omega }\frac{G(tu)}{t^{4}}=\infty . $$
(3.5)
Thus
$$\begin{aligned} I(tu) =&\frac{at^{2}}{2} \Vert u \Vert ^{2}+\frac{bt^{4}}{4} \Vert u \Vert ^{4}+\frac{t ^{4}}{4} \int_{\Omega }\phi_{u}u^{2}- \int_{\Omega }G(tu) \\ =&t^{4} \biggl( \frac{a}{2t^{2}} \Vert u \Vert ^{2}+ \frac{b}{4} \Vert u \Vert ^{4}+ \frac{1}{4} \int_{\Omega }\phi_{u}u^{2}- \int_{\Omega }\frac{G(tu)}{t ^{4}} \biggr) , \end{aligned}$$
it follows from (3.5) that \(\lim_{t\to \infty }I(tu)=-\infty \), hence, there exists \(t>0\) large enough such that \(\Vert u_{1} \Vert = \Vert tu \Vert >r _{0}\) and \(I(u_{1})<0\). □
Now, we define
$$ \Gamma =\bigl\{ \gamma \in C\bigl([0,1],H_{0}^{1}(\Omega ) \bigr):\gamma (0)=0,\gamma (1)=u_{1}\bigr\} ,\qquad c=\inf _{\gamma \in \Gamma }\max_{t\in [0,1]}I\bigl(\gamma (t)\bigr). $$
It follows from Lemma 3.1 that \(c\geqslant \rho_{0}>0\). For any given \(\alpha >0\), \(J_{\alpha }\) also has the mountain pass geometry. In fact, we have the following lemma.
Lemma 3.2
Assume
\(\lambda \in (0,\lambda^{*})\), under the assumptions of (f), (g1), (g2) and (h), for
\(r,\rho >0\) (where
\(\lambda^{*}\), r, ρ
are given in Lemma
2.2), the functional
\(J_{\alpha }\)
satisfies the following:
-
(i)
\(J_{\alpha }|_{S_{r}}\geqslant \rho \);
-
(ii)
there exists
\(v\in H_{0}^{1}(\Omega )\)
such that
\(J_{\alpha }(v)<0\).
Proof
(i) By (3.4) and Lemma 2.2, the conclusion holds.
(ii) From (3.4) and (ii) of Lemma 3.1, we choose \(v=u_{1}\) in Lemma 3.1 and the conclusion holds. □
We also can define the mountain pass critical level
$$ c_{\alpha }=\inf_{\gamma \in \Gamma }\max_{t\in [0,1]}J_{\alpha } \bigl( \gamma (t)\bigr). $$
From (3.4) and (i) of Lemma 3.2, for \(\lambda \in (0, \lambda^{*})\),
$$ 0< \rho \leqslant c_{\alpha }\leqslant c. $$
(3.6)
In the following, we give the existence of the mountain pass type solution to system (3.1).
Lemma 3.3
Suppose that (f), (g1)-(g3) and (h) hold, \(\lambda \in (0, \lambda^{*})\). Then there exists
\(u_{\alpha }\in H_{0}^{1}(\Omega )\)
such that
$$ J'_{\alpha }(u_{\alpha })=0,\qquad J_{\alpha }(u_{\alpha })=c_{\alpha }. $$
Proof
By Lemma 3.2 and the mountain pass lemma, there exists a sequence \(\{u_{n}\}\subset H_{0}^{1}(\Omega )\) such that \(J_{\alpha }(u _{n})\to c_{\alpha }\), \(J'_{\alpha }(u_{n})\to 0\). By (3.3), (1.4) and (g3), for n large enough, we have
$$\begin{aligned} c_{\alpha }+1+ \Vert u_{n} \Vert \geqslant &J_{\alpha }(u_{n})-\frac{1}{4}\bigl(J'_{ \alpha }(u_{n}),u_{n} \bigr) \\ =&\frac{a}{4} \Vert u_{n} \Vert ^{2}-\lambda \int_{\Omega }h \biggl( F\bigl(u_{n}^{+}+ \alpha \bigr)-F(\alpha )-\frac{1}{4}f\bigl(u_{n}^{+}+ \alpha \bigr)u_{n} \biggr) \\ &{}- \int_{\Omega } \biggl( G(u_{n})-\frac{1}{4}g(u_{n})u_{n} \biggr) \\ \geqslant &\frac{a}{4} \Vert u_{n} \Vert ^{2}- \lambda \int_{\Omega }hF\bigl(u_{n} ^{+}\bigr)+ \frac{\lambda }{4}f(\alpha ) \int_{\Omega }hu_{n}^{-} \\ \geqslant &\frac{a}{4} \Vert u_{n} \Vert ^{2}- \lambda^{*}\biggl(c_{0} \vert h \vert _{2} \Vert u_{n} \Vert +c_{2} \vert h_{1} \vert +\frac{1}{4}f(\alpha )c_{0} \vert h \vert _{2} \Vert u_{n} \Vert \biggr). \end{aligned}$$
Then \(\{u_{n}\}\) is bounded. Up to a subsequence, there exists \(u_{\alpha }\in H_{0}^{1}(\Omega )\) such that \(u_{n}\rightharpoonup u _{\alpha }\) in \(H_{0}^{1}(\Omega )\) and
$$\begin{aligned} &u_{n}\to u_{\alpha } \quad \mbox{in } L^{s}(\Omega ), s\in [1,6), \\ &u_{n}(x)\to u_{\alpha }(x) \quad \mbox{a.e. }x\in \Omega , \\ &\mbox{there exists }k_{1}\in L^{2}(\Omega ) \mbox{ such that for all }n, \bigl\vert u_{n}(x) \bigr\vert , \bigl\vert u_{\alpha }(x) \bigr\vert \leqslant k_{1}(x) \mbox{ a.e. in } \Omega . \end{aligned}$$
It follows from \(J'_{\alpha }(u_{n})\to 0\) that
$$\begin{aligned} 0 \leftarrow &\bigl(J'_{\alpha }(u_{n}),u_{n}-u_{\alpha }\bigr) \\ =&\bigl(a+b \Vert u_{n} \Vert ^{2}\bigr)(u_{n},u_{n}-u_{\alpha })+ \int_{\Omega }\phi_{u _{n}}u_{n}(u_{n}-u_{\alpha }) \\ &{}-\lambda \int_{\Omega }hf\bigl(u_{n}^{+}+\alpha \bigr)(u_{n}-u_{\alpha })- \int_{\Omega }g(u_{n})(u_{n}-u_{\alpha }). \end{aligned}$$
(3.7)
Since \(hf(u_{n}^{+}+\alpha )(u_{n}-u_{\alpha })\to 0\) a.e. in Ω and
$$ \bigl\vert hf\bigl(u_{n}^{+}+\alpha \bigr) (u_{n}-u_{\alpha }) \bigr\vert \leqslant 2f(\alpha )hk_{1} \in L^{1}(\Omega ), $$
by the dominated convergence theorem, we have
$$ \int_{\Omega }hf\bigl(u_{n}^{+}+\alpha \bigr) (u_{n}-u_{\alpha })\to 0, \quad \mbox{as }n\to \infty . $$
(3.8)
By (1.5), we can deduce that
$$\begin{aligned} \int_{\Omega }g(u_{n}) (u_{n}-u_{\alpha }) \leqslant{}& \varepsilon \bigl( \vert u_{n} \vert _{2} \vert u_{n}-u_{\alpha } \vert + \vert u_{n} \vert _{6}^{5} \vert u_{n}-u \vert _{6}\bigr) \\ &{} +C_{\varepsilon } \vert u_{n} \vert _{p+1}^{p} \vert u_{n}-u \vert _{p+1} \to 0, \end{aligned}$$
(3.9)
and
$$ \int_{\Omega }g(u_{n})u_{n}\to \int_{\Omega }g(u_{\alpha })u_{\alpha },\qquad \int_{\Omega }g(u_{n})\phi \to \int_{\Omega }g(u_{\alpha })\phi ,\quad \forall \phi \in H_{0}^{1}(\Omega ). $$
(3.10)
From (3.7), using (3.8), (3.9), Lemma 2.1(iv) and the boundedness of \(\{u_{n}\}\), we get \(\Vert u_{n} \Vert \to \Vert u_{\alpha } \Vert \). This combined with \(u_{n}\rightharpoonup u_{\alpha }\) implies that \(u_{n}\to u_{\alpha }\) in \(H_{0}^{1}(\Omega )\). Consequently, we have \(J_{\alpha }(u_{\alpha })=c_{\alpha }> \rho , J'_{\alpha }(u_{\alpha })=0\), that is, \(u_{\alpha }\) is a nontrivial solution to problem (3.1). Then \(u_{\alpha }\) satisfies (3.2), taking the test function \(\phi =u_{\alpha }^{-}\) in (3.2), it follows that \(\Vert u_{\alpha }^{-} \Vert =0\). Thus, we have \(u_{\alpha }\geqslant 0, u_{\alpha }\neq 0\) and \(J_{\alpha }(u_{ \alpha })=c_{\alpha }\geqslant \rho_{1}\). Hence, by the strong maximum principle, \(u_{\alpha }\) is a positive solution of the perturbation problem (3.1). □
In order to consider the convergence of \(\{u_{\alpha }\}\) as \(\alpha \to 0\) and to obtain the second solution of problem (1.3), we need the following result, which can be found in [13].
Lemma 3.4
Brezis and Nirenberg [13]
Let Ω be a bounded domain in
\({\mathbb{R}}^{n}\)
with smooth boundary
∂Ω. Let
\(u\in L_{\mathrm{loc}}^{1}(\Omega )\)
and assume that, for some
\(k\geqslant 0\), u
satisfies, in the sense of distributions,
$$ \textstyle\begin{cases} -\Delta u+ku\geqslant 0, &\textit{in } \Omega , \\ u\geqslant 0, &\textit{in } \Omega . \end{cases} $$
Then either
\(u\equiv 0\), or there exists
\(C>0\)
such that
$$ u(x)\geqslant C\operatorname{dist}(x,\partial \Omega ),\quad x\in \Omega . $$
Remark 3.5
By Lemma 3.3, (3.2) and Lemma 2.1(v), we have
$$ -\Delta u_{\alpha }+ Ku_{\alpha }\geqslant -\Delta u_{\alpha }+ \frac{ \phi_{u_{\alpha }}u_{\alpha }}{a+b \Vert u_{\alpha } \Vert ^{2}}=\frac{\lambda hf(u_{\alpha }+\alpha )+g(u_{\alpha })}{a+b \Vert u_{\alpha } \Vert ^{2}}\geqslant 0,\quad x\in \Omega , $$
where \(K>0\). Then, by Lemma 3.4, there exists \(C>0\) such that \(u_{\alpha }(x)\geqslant C\operatorname{dist}(x,\partial \Omega ), x\in \Omega \).
Finally, let \(\alpha \to 0\), we shall prove that the limit of a family of solutions \(\{u_{\alpha }\}\) of the perturbation problem (3.1) is the second solution of problem (1.3) with \(\lambda \in (0,\lambda^{*})\), where \(\lambda^{*}\) is defined in Lemma 2.2.
Theorem 3.6
Suppose that (f), (g1)-(g3) and (h) hold, \(\lambda \in (0, \lambda^{*})\). Then problem (1.3) has a solution
\(v_{0}\)
satisfying
\(J(v_{0})>0\).
Proof
Let \(\alpha \to 0\) and \(u_{\alpha }\geqslant 0\) is the solution of problem (3.1), that is, \(J_{\alpha }(u_{\alpha })=c_{\alpha }\), \(J'_{ \alpha }(u_{\alpha })=0\). Then, by (g3), (3.3), (3.6) and (1.4), we have
$$\begin{aligned} c>c_{\alpha } =&J_{\alpha }(u_{\alpha })-\frac{1}{4} \bigl(J'_{\alpha }(u _{\alpha }),u_{\alpha }\bigr) \\ =&\frac{a}{4} \Vert u_{\alpha } \Vert ^{2}-\lambda \int_{\Omega }h \biggl( F(u _{\alpha }+\alpha )-F(\alpha )- \frac{1}{4}f(u_{\alpha }+\alpha )u_{ \alpha } \biggr)\\ &{} - \int_{\Omega } \biggl( G(u_{\alpha })-\frac{1}{4}g(u_{ \alpha })u_{\alpha } \biggr) \\ \geqslant &\frac{a}{4} \Vert u_{\alpha } \Vert ^{2}- \lambda \int_{\Omega }h \bigl( F(u_{\alpha }) \bigr) \\ \geqslant &\frac{a}{4} \Vert u_{\alpha } \Vert ^{2}- \lambda^{*}\bigl(c_{0} \vert h \vert _{2} \Vert u_{\alpha } \Vert +c_{2} \vert h \vert _{1} \bigr), \end{aligned}$$
then \(\{u_{\alpha }\}\) is bounded in \(H_{0}^{1}(\Omega )\). Up to a subsequence, there exists \(v_{0}\in H_{0}^{1}(\Omega )\) such that \(u_{\alpha }\rightharpoonup v_{0}\) in \(H_{0}^{1}(\Omega )\) and
$$\begin{aligned} &u_{\alpha }\to v_{0} \quad \mbox{in } L^{s}(\Omega ), s\in [1,6), \\ &u_{\alpha }(x)\to v_{0}(x) \quad \mbox{a.e. }x\in \Omega , \\ &\mbox{there exists }k_{2}\in L^{2}(\Omega )\mbox{ such that for all }n, \bigl\vert u_{\alpha }(x) \bigr\vert , \bigl\vert v_{0}(x) \bigr\vert \leqslant k_{2}(x) \mbox{ a.e. in } \Omega . \end{aligned}$$
Firstly, we show that \(v_{0}(x)>0\) a.e. in Ω. For that purpose, we denote \(w_{\alpha }=u_{\alpha }-v_{0}\) and \(l=\lim_{\alpha \to 0} \Vert w_{\alpha } \Vert \). Taking \(\phi \in H_{0}^{1}(\Omega )\) with \(\phi \geqslant 0\) in (3.2), we have
$$ \lambda \int_{\Omega }hf(u_{\alpha }+\alpha )\phi = \bigl( a+b \Vert u_{ \alpha } \Vert ^{2} \bigr) (u_{\alpha },\phi )+ \int_{\Omega }\phi_{u_{ \alpha }}u_{\alpha }\phi - \int_{\Omega }g(u_{\alpha })\phi . $$
By using Fatou’s lemma, Lemma 2.1(iv) and (3.10), we have
$$ \lambda \int_{\Omega }hf(v_{0})\phi \leqslant \bigl( a+bl^{2}+b \Vert v_{0} \Vert ^{2} \bigr) (v_{0},\phi )+ \int_{\Omega }\phi_{v_{0}}v_{0}\phi - \int_{\Omega }g(v_{0})\phi . $$
Similar to the proof of \(u_{0}(x)>0\) in \(x\in \Omega \) in Theorem 2.5, we can show that \(v_{0}(x)>0\) a.e. in Ω.
Next, we show that \(u_{\alpha }\to v_{0}\) in \(H_{0}^{1}(\Omega )\) and \(v_{0}\) is the solution of problem (1.3), that is, we need to show that \(l=0\) and \(v_{0}\) satisfies (1.8).
We take \(\phi \in C_{0}^{\infty }(\Omega )\) with \(\operatorname{supp}\phi =\Omega _{1}\Subset \Omega \) in (3.2). By Remark 3.5, for \(x\in \Omega_{1}\), we have
$$\begin{aligned} \bigl\vert hf(u_{\alpha }+\alpha )\phi \bigr\vert &\leqslant \bigl\vert hf(u_{\alpha })\phi \bigr\vert \\ &\leqslant \bigl\vert hf\bigl(\operatorname{dist}(x,\partial \Omega )\bigr)\phi \bigr\vert \\ &\leqslant \bigl\vert hf(k_{0})\phi \bigr\vert \in L^{1}(\Omega ), \end{aligned}$$
where \(k_{0}=\min_{x\in \Omega_{1}}\operatorname{dist}(x,\partial \Omega )>0\). Since \(hf(u_{\alpha }+\alpha )\phi \to hf(v_{0})\phi \) a.e. in Ω, then by the dominant convergence theorem, we have
$$ \int_{\Omega }hf(u_{\alpha }+\alpha )\phi \to \int_{\Omega }hf(v_{0}) \phi \quad \mbox{as }\alpha \to 0. $$
By \((J'_{\alpha }(u_{\alpha }),\phi )=0\), using \(\int_{\Omega } \phi_{u_{\alpha }}u_{\alpha }\phi \to \int_{\Omega }\phi_{v_{0}}v_{0} \phi \) and (3.10), we get that
$$\begin{aligned} &\bigl( a+bl^{2}+b \Vert v_{0} \Vert ^{2} \bigr) (v_{0},\phi )+ \int_{\Omega } \phi_{v_{0}}v_{0}\phi =\lambda \int_{\Omega }hf(v_{0})\phi + \int_{ \Omega }g(v_{0})\phi , \\ &\quad \forall \phi \in C_{0}^{\infty }(\Omega ). \end{aligned}$$
(3.11)
Since \(C_{0}^{\infty }(\Omega )\) is dense in \(H_{0}^{1}(\Omega )\), then for \(\phi \in H_{0}^{1}(\Omega )\), there exists a sequence \(\{\phi _{n}\}\subset C_{0}^{\infty }(\Omega )\) such that \(\phi_{n}\to \phi \) as \(n\to \infty \). For \(n,m\in {\mathbb{N}}\) large enough, replacing ϕ with \(\phi_{n}-\phi_{m}\) in (3.11), we obtain that
$$\begin{aligned} &\bigl( a+bl^{2}+b \Vert v_{0} \Vert ^{2} \bigr) (v_{0},\phi_{n}-\phi_{m})+ \int_{\Omega }\phi_{v_{0}}v_{0}( \phi_{n}-\phi_{m}) \\ &\quad =\lambda \int_{ \Omega }hf(v_{0}) (\phi_{n}- \phi_{m})+ \int_{\Omega }g(v_{0}) (\phi_{n}- \phi_{m}). \end{aligned}$$
(3.12)
Since \(\phi_{n}\to \phi \), from (3.12), we can deduce that \(\{hf(v_{0})\phi_{n}\}\) is a Cauchy sequence in \(L^{1}(\Omega )\), hence there exists \(v\in L^{1}(\Omega )\) satisfying \(hf(v_{0})\phi_{n} \to v\) in \(L^{1}(\Omega )\), which means that \(hf(v_{0})\phi_{n}\to v\) in measure. By Riesz’s theorem, \(\{hf(v_{0})\phi_{n}\}\) has a subsequence, still denoted by \(\{hf(v_{0})\phi_{n}\}\), such that \(hf(v_{0})\phi_{n}\to v\) a.e. in Ω. On the other hand, \(hf(v_{0})\phi_{n}\to hf(v_{0})\phi \) a.e. in Ω. So \(v=hf(v_{0})\phi \), that is, \(\int_{\Omega }hf(v_{0})\phi_{n}\to \int_{\Omega }hf(v_{0})\phi \) as \(n\to \infty \). Then, taking the test function \(\phi_{n}\) in (3.11) and passing to the limit as \(n\to \infty \), we obtain that (3.11) holds for any \(\phi \in H_{0}^{1}(\Omega )\). We take \(\phi =v_{0}\) in (3.11) and obtain that
$$ \bigl( a+bl^{2}+b \Vert v_{0} \Vert ^{2} \bigr) \Vert v_{0} \Vert ^{2}+ \int_{\Omega } \phi_{v_{0}}v_{0}^{2}= \lambda \int_{\Omega }hf(v_{0})v_{0}+ \int_{ \Omega }g(v_{0})v_{0}. $$
(3.13)
On the other hand, by \((J'_{\alpha }(u_{\alpha }),u_{\alpha })=0\), we have
$$ \bigl(a+b \Vert u_{\alpha } \Vert ^{2}\bigr) \Vert u_{\alpha } \Vert ^{2}+ \int_{\Omega } \phi_{u_{\alpha }}u_{\alpha }^{2}= \lambda \int_{\Omega }hf(u_{\alpha }+\alpha )u_{\alpha }+ \int_{\Omega }g(u_{\alpha })u_{\alpha }. $$
(3.14)
Since f is nonincreasing on \((0,\infty )\), we have
$$ sf(s)\leqslant \int_{0}^{s} f(t)\,dt=F(s),\quad s\geqslant 0. $$
Then, by (1.4), we have
$$ hf(u_{\alpha }+\alpha )u_{\alpha }\leqslant hf(u_{\alpha })u_{\alpha } \leqslant hF(u_{\alpha })\leqslant h \bigl( c_{1} \vert u_{\alpha } \vert +c_{2} \bigr) \leqslant h ( c_{1}k_{2}+c_{2} ) \in L^{1}(\Omega ). $$
Then, combining with \(hf(u_{\alpha }+\alpha )\to hf(v_{0})v_{0}\) a.e. in Ω and using the dominant convergence theorem, we have
$$ \int_{\Omega }hf(u_{\alpha }+\alpha )u_{\alpha }\to \int_{\Omega }hf(v _{0})v_{0}. $$
Thus, from (3.14), by (3.10) and Lemma 2.1(iv), we obtain that
$$ \bigl( a+bl^{2}+b \Vert v_{0} \Vert ^{2} \bigr) \bigl(l^{2}+ \Vert v_{0} \Vert ^{2}\bigr)+ \int_{ \Omega }\phi_{v_{0}}v_{0}^{2}= \lambda \int_{\Omega }hf(v_{0})v_{0}+ \int_{\Omega }g(v_{0})v_{0}. $$
(3.15)
Combining with (3.13) and (3.15), we get
$$ \bigl( a+bl^{2}+b \Vert v_{0} \Vert ^{2} \bigr) l^{2}=0, $$
from \(a>0, b\geqslant 0\), it implies that \(l=0\), that is, \(u_{\alpha }\to v_{0}\). Hence, from (3.11) with \(l=0\) and \(\phi \in H_{0} ^{1}(\Omega )\), we get that \(v_{0}\) is the solution of problem (1.3) for \(\lambda \in (0,\lambda^{*})\) and \(J(v_{0})= \lim_{\alpha \to 0}J_{\alpha }(u_{\alpha })\geqslant \rho >0\). □
Proof of Theorem 1.1
By Theorems 2.5 and 3.6, for \(\lambda \in (0,\lambda^{*})\), there exist two solutions \(u_{0},v_{0}\in H_{0}^{1}(\Omega )\) to problem (1.3) with \(J(v_{0})>0>J(u_{0})\), that is, system (1.1) possesses at least two solutions for each \(\lambda \in (0,\lambda^{*})\). □