Let $\Omega\subset R^N$ be open bounded. Define \begin{cases} \Delta u = f(u) &x\in\Omega\\ u=1 & x\in\partial \Omega \end{cases} Q1: Suppose $f(u)=u^m$ where $m$ is odd. Prove that if there is a solution $u\in C^2(\Omega)\cap C^0(\bar{\Omega})$ then $0\leq u\leq 1$ and $u$ is unique.
For show that $0\leq u\leq 1$, I proved it by using maximal principle. We could apply maximal principle because $m$ is odd. So the argument in Evans Chapter 6.4 will work in the same way.
But I have no idea of proving $u$ is unique. Since this PDE is non-linear and hence we can not apply the method that assume $u_1$ $u_2$ and take $u_1-u_2$.
Q2: Suppose $f(u)=u-\frac{1}{u}$ and prove that $u\equiv 1$ is the only solution... I have no idea how to get start...
Any help is really welcome!
Put $u,v$ solutions of the equation and $w=u-v$, then you can check that $w$ satisfies the equation $\Delta w= a(x) w$, where $$ a(x)=u^{m-1}+ u^{m-2}v +\ldots+ uv^{m-2}+v^{m-1} $$ and therefore $a\geq 0$. The maximum principle now implies $w=0$.
First notice that $u(x)\neq 0$ for every $x\in \Omega$, and therefore is positive everywhere. We consider $\Omega_1=\{ x \in \Omega : u(x)<1\}$, then it's easy to check that $\Delta u<0$ in $\Omega_1$ and $u=1$ on $\partial \Omega_1$, therefore the minimum principle for superharmonic functions implies that $u \geq 1$ in $\Omega_1$ a contradiction. Therefore $\Omega_1=\emptyset$. Similarly the set where $u>1$ is empty and therefore $u=1$