# 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!

• There are a couple of issues. I think you want $-\Delta u = \lambda u^2(1-u)$ instead of $\Delta u = \lambda u^2(1-u)$. If $\Delta u = \lambda u^2(1-u)$ and $0\leqslant u \leqslant 1$ then the maximum principle along with your boundary condition imply that $u=0$ which is what you want to prove. Jul 15, 2021 at 5:32
• @JackT, of course is $-\Delta u = \lambda u^2(1-u)$. Jul 15, 2021 at 12:46
• I think there is a problem with your equality $\int_\Omega u \, \Delta u \, dx = \int_\Omega |\nabla u |^2 \, dx$, since normally (because we are integrating by part and saying u=0 on the boundary) it would be $- \int_\Omega u \, \Delta u \, dx = \int_\Omega |\nabla u |^2 \, dx$ Jul 15, 2021 at 12:59
• To conclude from $0 \le C \int_\Omega u_{\lambda}^2 \, dx \le \lambda \int_\Omega u_{\lambda}^3 (1-u_{\lambda} ) \, dx$, think you just need to use Lp inclusion (since $\Omega$ is bounded) to say that $0 \le C \int_\Omega u_\lambda^2 \, dx \le \lambda C'\int_\Omega u_\lambda^2 \, dx .$, and conclude that when $\lambda C'<C$ we must have $u_{\lambda}=0$ Jul 15, 2021 at 13:11
• @pFRAUX You are right! I didn't understand the last step. $u_\lambda$? Jul 15, 2021 at 13:35

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 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, 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 $$\lambda, such an u must have an $$L^2$$ norm nul, and so must be the null fonction $$0$$.