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?


  • $\begingroup$ 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. $\endgroup$
    – JackT
    Jul 15, 2021 at 5:32
  • $\begingroup$ @JackT, of course is $-\Delta u = \lambda u^2(1-u)$. $\endgroup$
    – Mary E.
    Jul 15, 2021 at 12:46
  • 2
    $\begingroup$ 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 $- \int_\Omega u \, \Delta u \, dx = \int_\Omega |\nabla u |^2 \, dx$ $\endgroup$
    – p FRAUX
    Jul 15, 2021 at 12:59
  • $\begingroup$ 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 $u_{\lambda}=0$ $\endgroup$
    – p FRAUX
    Jul 15, 2021 at 13:11
  • $\begingroup$ @pFRAUX You are right! I didn't understand the last step. $u_\lambda$? $\endgroup$
    – Mary E.
    Jul 15, 2021 at 13:35

1 Answer 1


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


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