Uniqueness of solution of nonlinear PDE Let $\Omega \subseteq \mathbb{R}^n$ be bounded open set and $\partial \Omega \in C^1$. Consider the problem
$$  \begin{cases} -\Delta u =  \lambda u^2 (1-u ) \text{ in } \Omega, \\ u=0 \text{ on } \partial \Omega, \end{cases} $$ where $\lambda >0$ is a constant.
I want to show that $u \equiv 0 $ is the unique solution as $\lambda \to 0$. The hint of the exercise is to use Poincaré's Inequality.
My attempt:
Multiplying by $u$ in $-\Delta u =  \lambda u^2 (1-u )$, we have $-u\,\Delta u =  \lambda u^3 (1-u )$. Integrating over $\Omega$, using the Green's Identity and $u=0$ on $\partial \Omega$, we have: $$ - \int_{\Omega} \lambda u^3 (1-u ) \, dx= \int_\Omega u \, \Delta u \, dx = \int_\Omega |\nabla u |^2 \, dx.$$
Now by Poincaré's Inequality $$\int_\Omega |\nabla u |^2 \, dx \ge C \int_\Omega  u^2 \, dx,$$
for some contant $C$, so that
$$ C \int_\Omega u^2 \, dx \le - \lambda \int_{\Omega} u^3 (1-u ) \, dx.$$
It is not hard to show that $ 0 \le u(x) \le 1$ for all $ x \in \bar{\Omega}$.
Since $ 0 \le u(x) \le 1$ for all $ x \in \Omega$, we have $u^3(1-u) \ge 0$. So $$ 0 \le C \int_\Omega u^2 \, dx \le -\lambda \int_\Omega u^3 (1-u ) \, dx \le 0.$$ Therefore, the unique solution is $u \equiv 0$ when $\lambda \to 0$ since $u \equiv 1$ is not a solution for the problem.
I'm not sure it's right especially in the last step. That's right?
Thanks!
 A: Ok, I will sums up what we said in comment and discussion here
Let u be a solution for some $\lambda$.
So we multiply by $u$ in $-\Delta u =  \lambda u^2 (1-u )$, and get  by integrating over $\Omega$, using the Green's Identity and =0 on $\partial\Omega$, we have:
$$\int_{\Omega} \lambda u^3 (1-u ) \, dx= -\int_\Omega u \, \Delta u \, dx = \int_\Omega |\nabla u |^2 \, dx.$$
Now by Poincaré's Inequality, there is some constant C>0 such that :
$$\int_\Omega |\nabla u |^2 \, dx \ge C \int_\Omega  u^2 \, dx,$$
So here we have that :
$$C \int_\Omega u^2 \, dx \le  \lambda \int_{\Omega} u^3 (1-u ) \, dx \le \lambda (\int_{\Omega}u^3\, dx-\int_{\Omega}u^4\, dx). $$
Finally, $u(x) \le 1$ follows analysing the sign of $\Delta u (x_0)$ for $x_0$ such that $u(x_0) > 1$ assuming (WLOG) that $x_0$ is a maximum of $u$.
So by integrating the relation $u^3 \le u^2$, and using that $u^4\geq0$ we finally get that
$$C \int_\Omega u^2 \, dx \le \lambda (\int_{\Omega}u^3\, dx-\int_{\Omega}u^4\, dx).\le \lambda \int_{\Omega}u^2\, dx $$
So for $\lambda<C$, such an u must have an $L^2$ norm nul, and so must be the null fonction $0$.
