## 4 Weak Solutions for the First Problem
Then,  
\[\int\limits_{\overline{\Omega}_{\delta}}\
abla\underline{u}\
abla\phi dx\leq \int\limits_{\overline{\Omega}_{\delta}}c\left(\underline{u}\right)\phi dx. \tag{17}\]  
Next, on \(\Omega\setminus\overline{\Omega}_{\delta}\), we have \(\phi_{1}\geq d\) for some \(d>0\). By \(\left(H_{3}\right)\) and by the definition of \(\underline{u}\), it follows that:  
\[\int\limits_{\Omega\setminus\overline{\Omega}_{\delta}}c\left( \underline{u}\right)\phi dx \geq\int\limits_{\Omega\setminus\overline{\Omega}_{\delta}}\sigma \phi dx\] \[\lim\limits_{u\rightarrow+\infty}\frac{c\left(u\right)}{u}=0.\]0 \[\lim\limits_{u\rightarrow+\infty}\frac{c\left(u\right)}{u}=0.\]1  
From (17) and (18), we deduce that  
\[\lim\limits_{u\rightarrow+\infty}\frac{c\left(u\right)}{u}=0.\]2  
for any \(\phi\in H_{0}^{1}\left(\Omega\right)\). That is, \(\underline{u}\) is a weak LS of problem (1). Next, we shall construct a WUS of (1). Let \(e\) be the solution of the following problem  
\[\left\{ \begin{array}{ll}\Delta e=1&\text{in }\Omega,\\ e=0,&\text{on }\partial\Omega.\end{array} \right. \tag{19}\]  
Let \(\overline{u}=Ce\), where \(C\) is a positive real number which will be chosen later. We will verify that \(\overline{u}\) is a weak US (1). Let \(\phi\in H_{0}^{1}\left(\Omega\right)\) with \(\phi>0\) in \(\Omega\). Then, from (19), we get  
\[\int\limits_{\Omega}\
abla\overline{u}\
abla\phi dx =C\int\limits_{\Omega}\
abla e\
abla\phi dx\] \[=C\int\limits_{\Omega}\phi dx.\]  
By \(\left(H_{3}\right)\), we can choose \(C\) large enough so that  
\[C\geq c\left(C\left\|e\right\|_{\infty}\right).\]Therefore,  
\[\int\limits_{\Omega}\
abla\overline{u}\
abla\phi dx \geq c\left(C\left\|e\right\|_{\infty}\right)\int\limits_{\Omega} \phi dx\] \[\geq\int\limits_{\Omega}c\left(C\left\|e\right\|_{\infty}\right) \phi dx\] \[\geq\int\limits_{\Omega}c\left(\overline{u}\right)\phi dx.\]  
that is, \(\overline{u}\) a weak US of (1) for a large enough \(C\). In order to obtain a weak solution of (1), we define the sequence  
\[\left\{u_{n}\right\}\subset E=H_{0}^{1}\left(\Omega\right)\cap C\left(\Omega \right),\]  
as \(u_{0}=\overline{u}\subset E=H_{0}^{1}\left(\Omega\right)\cap C\left(\Omega \right),\) and \(u_{n}\) is the unique solution of the problem  
\[\left\{ \begin{array}{l}-\Delta u_{n}=c\left(u_{n-1}\right),\text{ in } \Omega,\\ u=0,\text{ in }\partial\Omega.\end{array} \right. \tag{20}\]  
If \(u_{n-1}\in E\) is given, the right-hand-side of (20) is independent of \(u_{n}\) since \(c\left(u_{0}\right)\in C\left(\Omega\right)\subset L^{2}\left(\Omega\right)\) in \(x\) and from Lemma 6. Then, (20) with \(n=1\) has unique solution \(u_{1}\in H_{0}^{1}\left(\Omega\right)\). We deduce from Lemma 5 that \(u_{1}\in C\left(\Omega\right)\). Consequently, we conclude that \(u_{1}\in E\). In the same way, we construct the following elements \(u_{n}\in E\) of our sequence. From (20), and by the fact that \(u_{0}\) is a weak US of (1), we have  
\[\left\{ \begin{array}{l}-\Delta u_{0}\geq c\left(u_{0}\right)=-\Delta u_{1}, \text{ in }\Omega,\\ u=0,\text{ in }\partial\Omega,\end{array} \right.\]  
from which and Lemma 7, we have \(u_{0}\geq u_{1}\). Since \(u_{0}\geq\underline{u}\) and by the monotonicity of \(c\), we have  
\[-\Delta u_{1}=c\left(u_{0}\right)\geq c\left(\underline{u}\right)\geq-\Delta \underline{u}.\]  
from which and Lemma 7, we deduce that \(u_{1}\geq\underline{u}\). From (20), and by the monotonicity of \(c\), \(u_{1}\geq\underline{u}\) and \(u_{0}\geq u_{1}\), for \(u_{2}\), we write  
\[-\Delta u_{1}=c\left(u_{0}\right)\geq c\left(u_{1}\right)=-\Delta u_{2},\]  
and  
\[-\Delta u_{2}=c\left(u_{1}\right)\geq c\left(\underline{u}\right)\geq-\Delta \underline{u}.\]  
Then, thanks to the Lemma 7, we get \(u_{1}\geq u_{2}\) and \(u_{2}\geq\underline{u}\).