For any \(\epsilon>0\), let \(f_{\epsilon}(x, t)=f(x, t)+\epsilon p|t|^{p-2}t\) and
$$ \Phi_{\epsilon}(u) = \Phi(u)-\epsilon\|u\|_{p}^{p},\quad \forall u\in H_{0}^{1}(\Omega). $$
(3.1)
Similarly, we define
$$\begin{aligned}& \mathcal{M}_{\epsilon}:= \bigl\{ u\in H_{0}^{1}( \Omega): u^{\pm}\ne0, \bigl\langle \Phi_{\epsilon}'(u), u^{+}\bigr\rangle =\bigl\langle \Phi_{\epsilon}'(u), u^{-}\bigr\rangle =0 \bigr\} , \end{aligned}$$
(3.2)
$$\begin{aligned}& \mathcal{N}_{\epsilon}:= \bigl\{ u\in H_{0}^{1}( \Omega): u\ne0, \bigl\langle \Phi_{\epsilon}'(u), u\bigr\rangle =0 \bigr\} , \end{aligned}$$
(3.3)
and
$$ m_{\epsilon}:=\inf_{u\in\mathcal{M}_{\epsilon}} \Phi_{\epsilon }(u), \qquad c_{\epsilon}:=\inf_{u\in\mathcal{N}_{\epsilon}} \Phi_{\epsilon}(u). $$
(3.4)
Lemma 3.1
Assume that (F1)–(F4) hold. Then there exists a constant
\(\alpha>0\)
which does not depend on
\(\epsilon\in(0, 1]\)
such that
$$ \Phi_{\epsilon}(u)\ge\alpha, \quad\forall u\in\mathcal{N}_{\epsilon }, \epsilon\in(0, 1]. $$
(3.5)
Proof
By (F1) and (F2), there exists a constant \(C_{2}>0\) such that
$$ F(x, t)\le\frac{1}{4\gamma_{2}^{2}}t^{2}+C_{2}|t|^{p},\quad \forall(x, t)\in \Omega\times \mathbb {R}. $$
(3.6)
From (3.1), (3.6), and Corollary 2.3, one has
$$\begin{aligned} \Phi_{\epsilon}(u) & = \max_{t\ge0}\Phi_{\epsilon}(tu) =\max_{t\ge0} \biggl[\frac{t^{2}}{2}\|u\|^{2}- \int_{\Omega}F(x, tu)\, \mathrm{d}x-\epsilon t^{p}\|u \|_{p}^{p} \biggr] \\ & \ge \max_{t\ge0} \biggl[\frac{t^{2}}{4}\|u \|^{2}-(C_{2}+1)\gamma _{p}^{p}t^{p} \|u\|^{p} )\biggr] \\ & = \frac{p-2}{4p[2(C_{2}+1)\gamma_{p}^{p}p]^{2/(p-2)}}:=\alpha>0, \quad\forall u\in\mathcal{N}_{\epsilon}, \epsilon\in(0, 1].\end{aligned} $$
□
Proof of Theorem 1.1
Under the conditions of Theorem 1.1, for \(\epsilon>0\), \(f_{\epsilon}\) satisfies (F1)–(F3) and (Ne). In view of Lemmas 2.6 and 2.7, there exists \(u_{\epsilon}\in\mathcal{M}_{\epsilon}\) such that \(\Phi_{\epsilon}(u_{\epsilon})=m_{\epsilon}\) and \(\Phi_{\epsilon }'(u_{\epsilon})=0\).
By (F1)–(F3), one can easily prove that \(\mathcal{M}_{0}\ne\emptyset \). Let \(u_{0}\in\mathcal{M}_{0}\). Then \(\Phi(u_{0}):=c^{*}> 0\) and \(\langle\Phi'(u_{0}), u_{0}^{\pm}\rangle=0\). By Lemma 2.4, there exist \(s_{\epsilon}>0\) and \(t_{\epsilon}>0\) such that \(s_{\epsilon}u_{0}^{+}+t_{\epsilon }u_{0}^{-} \in\mathcal{M}_{\epsilon}\). Hence, from Corollary 2.2 and Lemma 3.1, we have
$$ \begin{aligned}[b] c^{*} & = \Phi(u_{0}) \\ & \ge \Phi\bigl(s_{\epsilon}u_{0}^{+}+t_{\epsilon}u_{0}^{-} \bigr)\ge\Phi _{\epsilon}\bigl(s_{\epsilon}u_{0}^{+}+t_{\epsilon}u_{0}^{-} \bigr) \\ & \ge m_{\epsilon}\ge\hat{\kappa},\quad \forall\epsilon\in(0, 1).\end{aligned} $$
(3.7)
Hence, we can choose a sequence \(\{\epsilon_{n}\}\) such that \(\epsilon _{n}\searrow0\) as \(n\to\infty\), and
$$ u_{\epsilon_{n}}\in\mathcal{M}_{\epsilon_{n}},\qquad \Phi_{\epsilon _{n}}(u_{\epsilon_{n}})=m_{\epsilon_{n}}\rightarrow\bar{m}, \qquad\Phi _{\epsilon_{n}}'(u_{\epsilon_{n}})=0. $$
(3.8)
First, we prove that \(\{u_{\epsilon_{n}}\}\) is bounded in \(H_{0}^{1}(\Omega )\). Arguing by contradiction, suppose that \(\|u_{\epsilon_{n}}\| \to\infty\). Let \(v_{n}=u_{\epsilon_{n}}/\| u_{\epsilon_{n}}\|\), then \(\|v_{n}\|=1\). By Sobolev embedding theorem, passing to a subsequence, we may assume that \(v_{n}\rightarrow v\) in \(L^{s}(\Omega)\), \(2\le s<2^{*}\), \(v_{n}\rightarrow v\) a.e. on Ω.
If \(v=0\), then \(v_{n}\rightarrow0\) in \(L^{s}(\Omega)\) for \(2\le s<2^{*}\). Fix \(R>[2(1+\bar{m})]^{1/2}\). By (F1) and (F2), there exists \(C_{3}>0\) such that
$$ \limsup_{n\to\infty} \int_{\Omega}F(x, Rv_{n})\,\mathrm{d}x\le R^{2}\lim_{n\to\infty}\|v_{n} \|_{2}^{2} +C_{3}R^{p}\lim _{n\to\infty}\|v_{n}\|_{p}^{p}=0. $$
(3.9)
Let \(t_{n}=R/\|u_{\epsilon_{n}}\|\). Hence, using (3.1), (3.8), (3.9), and Corollary 2.3, one has
$$\begin{aligned} m_{\epsilon_{n}} & = \Phi_{\epsilon_{n}}(u_{\epsilon_{n}})\ge \Phi_{\epsilon _{n}}(t_{n}u_{\epsilon_{n}}) \\ & = \frac{t_{n}^{2}}{2}\|u_{\epsilon_{n}}\|^{2}- \int_{\Omega}\bigl[F(x, t_{n}u_{\epsilon_{n}})+ \epsilon_{n}|t_{n}u_{\epsilon_{n}}|^{p}\bigr]\, \mathrm{d}x \\ & = \frac{R^{2}}{2}- \int_{\Omega}\bigl[F(x, Rv_{n})+\epsilon_{n}R^{p}|v_{n}|^{p} \bigr]\, \mathrm{d}x \\ & = \frac{R^{2}}{2}+o(1) > \bar{m}+1+o(1),\end{aligned} $$
which is a contradiction. Thus \(v\ne0\).
For \(x\in\{z\in \mathbb {R}^{N} : v(z)\ne0\}\), we have \(\lim_{n\to\infty }|u_{\epsilon_{n}}(x)|=\infty\). Hence, it follows from (F3), (F4), (3.8), and Fatou’s lemma that
$$\begin{aligned} 0 & = \lim_{n\to\infty}\frac{m_{\epsilon_{n}}}{\|u_{\epsilon_{n}}\|^{2}} = \lim _{n\to\infty}\frac{\Phi_{\epsilon_{n}}(u_{\epsilon_{n}})}{\| u_{\epsilon_{n}}\|^{2}} \\ & = \lim_{n\to\infty} \biggl[\frac{1}{2}\|v_{n} \|^{2} - \int_{\Omega}\frac{F(x, u_{\epsilon_{n}})+\epsilon_{n}|u_{\epsilon _{n}}|^{p}}{u_{\epsilon_{n}}^{2}}v_{n}^{2}\, \mathrm{d}x \biggr] \\ & \le \frac{1}{2}-\liminf_{n\to\infty} \int_{\Omega}\frac{F(x, u_{\epsilon_{n}})}{u_{\epsilon_{n}}^{2}}v_{n}^{2}\, \mathrm{d}x \le\frac{1}{2}- \int_{\Omega}\liminf_{n\to\infty}\frac{F(x, u_{\epsilon_{n}})}{u_{\epsilon_{n}}^{2}}v_{n}^{2} \,\mathrm{d}x \\ & = -\infty.\end{aligned} $$
This contradiction shows that \(\{u_{\epsilon_{n}}\}\) is bounded in \(H_{0}^{1}(\Omega)\). Hence, there exists a subsequence of \(\{\epsilon_{n}\}\) still denoted by \(\{\epsilon_{n}\}\) and \(u_{0}\in H_{0}^{1}(\Omega)\) such that \(u_{\epsilon_{n}}\rightharpoonup u_{0}\) in \(H_{0}^{1}(\Omega)\).
Second, we prove that \(\Phi'(u_{0})=0\) and \(\Phi(u_{0})=m_{0}\). By Sobolev embedding theorem, \(u_{\epsilon_{n}}\rightarrow u_{0}\) in \(L^{s}(\Omega)\), \(2\le s<2^{*}\), \(u_{\epsilon_{n}}\rightarrow u_{0}\) a.e. on Ω. Then, from (2.2), (3.1), and (3.8), one has
$$\begin{aligned} \bigl\langle \Phi'(u_{0}), \varphi\bigr\rangle & = (u_{0},\varphi)- \int_{\Omega}f(x, u_{0})\varphi\,\mathrm{d}x \\ & = \lim_{n\to\infty} \biggl[(u_{\epsilon_{n}}, \varphi) - \int_{\Omega}\bigl[f(x, u_{\epsilon_{n}})+\epsilon_{n}p|u_{\epsilon _{n}}|^{p-2}u_{\epsilon_{n}} \bigr]\varphi\,\mathrm{d}x \biggr] \\ & = \lim_{n\to\infty}\bigl\langle \Phi_{\epsilon_{n}}'(u_{\epsilon _{n}}), \varphi\bigr\rangle =0, \quad\forall\varphi\in C_{0}^{\infty}(\Omega).\end{aligned} $$
This shows \(\Phi'(u_{0})=0\). Since \(u_{\epsilon_{n}}\rightarrow u_{0}\) in \(L^{s}(\Omega)\), \(2\le s<2^{*}\), by (3.1) and (3.8), we have
$$ \begin{aligned}[b] & \|u_{\epsilon_{n}}-u_{0}\|^{2} \\ &\quad = \bigl\langle \Phi_{\epsilon_{n}}'(u_{\epsilon_{n}})- \Phi'(u_{0}), u_{\epsilon_{n}}-u_{0}\bigr\rangle +\epsilon_{n}p \int_{\Omega} \bigl(|u_{\epsilon_{n}}|^{p-2}u_{\epsilon _{n}}-|u_{0}|^{p-2}u_{0} \bigr) (u_{\epsilon_{n}}-u_{0})\,\mathrm{d}x \\ & \qquad{} + \int_{\Omega}\bigl[f(x, u_{\epsilon_{n}})-f(x, u_{0}) \bigr](u_{\epsilon _{n}}-u_{0})\,\mathrm{d}x \rightarrow0, \quad\mbox{as } n \rightarrow\infty,\end{aligned} $$
(3.10)
which implies \(u_{\epsilon_{n}}\rightarrow u_{0}\) in \(H_{0}^{1}(\Omega)\), and so \(u_{\epsilon_{n}}^{\pm}\rightarrow u_{0}^{\pm}\) in \(H_{0}^{1}(\Omega)\). Consequently, it follows from (3.1) and (3.8) that \(\Phi(u_{0})=\bar{m}\). Again from (3.1) and (3.5), one has
$$ \begin{aligned}[b] \int_{\Omega} \biggl[\frac{1}{2}f\bigl(x, u_{0}^{\pm}\bigr)-F\bigl(x, u_{0}^{\pm } \bigr) \biggr]\,\mathrm{d}x & = \lim_{n\to\infty} \int_{\Omega} \biggl[\frac{1}{2}f\bigl(x, u_{\epsilon_{n}}^{\pm}\bigr)-F\bigl(x, u_{\epsilon_{n}}^{\pm} \bigr) +\frac{(p-2)\epsilon_{n}}{2}\big|u_{\epsilon_{n}}^{\pm}\big|^{p} \biggr] \, \mathrm{d}x \\ & = \lim_{n\to\infty} \biggl[\Phi_{\epsilon_{n}} \bigl(u_{\epsilon _{n}}^{\pm}\bigr) -\frac{1}{2}\bigl\langle \Phi_{\epsilon_{n}}'\bigl(u_{\epsilon_{n}}^{\pm}\bigr), u_{\epsilon_{n}}^{\pm}\bigr\rangle \biggr] \\ & = \lim_{n\to\infty}\Phi_{\epsilon_{n}}\bigl(u_{\epsilon_{n}}^{\pm } \bigr)\ge\alpha>0.\end{aligned} $$
(3.11)
This, together with (2.4) (\(t=0\)), shows \(u_{0}^{\pm}\ne0\). Thus \(u_{0}\in\mathcal{M}\) and \(\bar{m}=\Phi(u_{0})\ge m_{0}\). Next, we prove \(\Phi(u_{0})=m_{0}\). Let ε be any positive number. Then there exists \(v_{\varepsilon}\in\mathcal{M}\) such that \(\Phi(v_{\varepsilon})< m_{0}+\varepsilon\). Then (F3) implies that there exists \(K_{\varepsilon}>0\) such that, for \(s\ge K_{\varepsilon}\) or \(t\ge K_{\varepsilon}\),
$$ \begin{aligned}[b] \Phi_{\epsilon_{n}}\bigl(sv_{\varepsilon}^{+}+tv_{\varepsilon}^{-} \bigr) = {}& \frac{s^{2}}{2}\big\| v_{\varepsilon}^{+}\big\| ^{2} - \int_{\Omega}F\bigl(x, sv_{\varepsilon}^{+}\bigr)\, \mathrm{d}x-\epsilon _{n}s^{p}\big\| v_{\varepsilon}^{+} \big\| _{p}^{p} \\ & +\frac{t^{2}}{2}\big\| v_{\varepsilon}^{-}\big\| ^{2} - \int_{\Omega}F\bigl(x, tv_{\varepsilon}^{-}\bigr)\, \mathrm{d}x-\epsilon _{n}t^{p}\big\| v_{\varepsilon}^{-} \big\| _{p}^{p} \\ \le{}& \frac{s^{2}}{2}\big\| v_{\varepsilon}^{+}\big\| ^{2} - \int_{\Omega}F\bigl(x, sv_{\varepsilon}^{+}\bigr)\, \mathrm{d}x +\frac{t^{2}}{2}\big\| v_{\varepsilon}^{-}\big\| ^{2}- \int_{\Omega}F\bigl(x, tv_{\varepsilon}^{-}\bigr)\, \mathrm{d}x < 0.\end{aligned} $$
(3.12)
In view of Lemma 2.4, there exists a pair \((s_{n}, t_{n})\) of positive numbers such that \(s_{n}v_{\varepsilon}^{+} +t_{n}v_{\varepsilon}^{-}\in\mathcal{M}_{\epsilon_{n}}\), which, together with (3.12) and \(c_{\epsilon_{n}}>0\), implies \(0< s_{n}, t_{n}< K_{\varepsilon}\). Hence, from (2.3), (3.1), and \(\langle\Phi'(v_{\varepsilon}), v_{\varepsilon}^{\pm}\rangle=0\), we have
$$\begin{aligned} m_{0}+\varepsilon & > \Phi(v_{\varepsilon}) = \Phi_{\epsilon_{n}}(v_{\varepsilon})+\epsilon_{n}\|v_{\varepsilon}\| _{p}^{p} \\ & \ge \Phi_{\epsilon_{n}}\bigl(s_{n}v_{\varepsilon}^{+}+t_{n}v_{\varepsilon}^{-} \bigr) +\frac{1-s_{n}^{2}}{2}\bigl\langle \Phi_{\epsilon_{n}}'(v_{\varepsilon}), v_{\varepsilon}^{+}\bigr\rangle +\frac{1-t_{n}^{2}}{2}\bigl\langle \Phi_{\epsilon_{n}}'(v_{\varepsilon}), v_{\varepsilon}^{-} \bigr\rangle \\ & \ge m_{\epsilon_{n}}-\frac{1+K_{\varepsilon}^{2}}{2}\big|\bigl\langle \Phi _{\epsilon_{n}}'(v_{\varepsilon}), v_{\varepsilon}^{+}\bigr\rangle \big| -\frac{1+K_{\varepsilon}^{2}}{2}\big|\bigl\langle \Phi_{\epsilon _{n}}'(v_{\varepsilon}), v_{\varepsilon}^{-} \bigr\rangle \big| \\ & = m_{\epsilon_{n}}-\frac{(1+K_{\varepsilon}^{2})p\epsilon_{n}}{2}\big\| v_{\varepsilon}^{+} \big\| _{p}^{p} -\frac{(1+K_{\varepsilon}^{2})p\epsilon_{n}}{2}\big\| v_{\varepsilon}^{-} \big\| _{p}^{p},\end{aligned} $$
which yields
$$ \bar{m}=\lim_{n\to\infty}m_{\epsilon_{n}}\le m_{0}+\varepsilon. $$
(3.13)
Since \(\varepsilon>0\) is arbitrary, one has \(\bar{m}\le m_{0}\). Thus, \(\bar{m}= m_{0}\), i.e., \(\Phi(u_{0})=m_{0}\).
Finally, we show that \(u_{0}\) has exactly two nodal domains. Let \(u_{0}=u_{1}+u_{2}+u_{3}\), where
$$\begin{aligned}& \begin{gathered}u_{1}\ge0,\qquad u_{2}\le0,\qquad \Omega_{1}\cap \Omega_{2}=\emptyset,\\ u_{1}|_{\Omega \setminus(\Omega_{1}\cup\Omega_{2})} =u_{2}|_{\Omega\setminus(\Omega_{1}\cup\Omega_{2})}=u_{3}|_{\Omega_{1}\cup \Omega_{2}}=0,\end{gathered} \end{aligned}$$
(3.14)
$$\begin{aligned}& \Omega_{1}:=\bigl\{ x\in\Omega: u_{1}(x)>0 \bigr\} ,\qquad \Omega_{2}:= \bigl\{ x\in\Omega: u_{2}(x)< 0 \bigr\} , \end{aligned}$$
(3.15)
and \(\Omega_{1}\), \(\Omega_{2}\) are connected open subsets of Ω.
Setting \(v=u_{1}+u_{2}\), we see that \(v^{+}=u_{1}\) and \(v^{-}=u_{2}\), i.e., \(v^{\pm}\ne0\). Note that \(\Phi'(u_{0})=0\), by a simple computation, one has
$$ \bigl\langle \Phi'(v), v^{+} \bigr\rangle =\bigl\langle \Phi'(v), v^{-} \bigr\rangle =0. $$
(3.16)
From (2.1), (2.2), (2.3), (3.14), and (3.16), we have
$$\begin{aligned} m_{0} & = \Phi(u_{0})-\frac{1}{2}\bigl\langle \Phi'(u_{0}), u_{0} \bigr\rangle \\ & = \Phi(v)+\Phi(u_{3}) -\frac{1}{2} \bigl[\bigl\langle \Phi'(v), v \bigr\rangle +\bigl\langle \Phi'(u_{3}), u_{3} \bigr\rangle \bigr] \\ & \ge \sup_{s, t\ge0}\Phi\bigl(sv^{+}+tv^{-} \bigr)+\Phi(u_{3})-\frac {1}{2}\bigl\langle \Phi'(u_{3}), u_{3} \bigr\rangle \\ & \ge m_{0}+ \int_{\Omega} \biggl[\frac{1}{2}f(x, u_{3})u_{3}-F(x, u_{3}) \biggr]\,\mathrm{d}x,\end{aligned} $$
which, together with (1.6), shows \(u_{3}=0\). Therefore, \(u_{0}\) has exactly two nodal domains. □
Proof of Theorem 1.2
In view of Theorem 1.1, there exists \(u_{0}\in\mathcal{M}\) such that \(m_{0}=\Phi(u_{0})\). Since \(u^{\pm}\in\mathcal{N}\), then one has
$$m_{0}=\Phi(u_{0})=\Phi\bigl(u_{0}^{+} \bigr)+\Phi\bigl(u_{0}^{-}\bigr)\ge2c_{0}. $$
□