Let $D$ be the unit disk in $\mathbb R^2$ centered at the origin. Given any $\lambda \in \mathbb R$, let $u_\lambda$ be the unique solution to the semilinear elliptic equation $$ -\Delta u + u^3=0 \quad \text{on $D$},$$ subject to the constant Dirichlet data $u|_{\partial D} =\lambda$. Prove that $u(0)$ is uniformly bounded in $|\lambda|$ and that more generally given any $h\in (0,1)$, the solution $u(x)$, with $|x|<h$ is uniformly bounded in $|\lambda|$.

  • $\begingroup$ I think this is true (maybe you already know it) since the radial ODE (with $u'(0)=0$, I guess) explodes for $u(0)$ large before reaching $r=1$. Do you have the same feeling? By the way, where does it come from? $\endgroup$ Dec 15, 2022 at 19:18
  • $\begingroup$ That’s exactly my intuition for this problem. As for where it comes from I remember reading this in a paper but can not locate it anymore. $\endgroup$
    – Ali
    Dec 15, 2022 at 19:54
  • 2
    $\begingroup$ I just checked the details...maybe I should write an answer, it seems to work in any dimension $\endgroup$ Dec 15, 2022 at 19:58
  • $\begingroup$ does maximum not show that the maximum is on the boundary? $\endgroup$
    – Math604
    Dec 16, 2022 at 0:01

1 Answer 1


Let me give a positive answer perhaps omitting some details.

Fact 1. Let $u'' \geq ku^\alpha$ in $[c,\ell[$ with $k>0, \alpha>1$ and $u,u' \geq 0$. Let $A=u(c)$, then $ \ell \to c$ as $A \to \infty$.

This follows by multiplying by $u'$ and integrating. One obtains $$u' \geq k \sqrt{u^{\alpha+1}-A^{\alpha+1}} \geq k \sqrt{(u-A)u^\alpha}$$ and integrating again $$ \ell-c \leq k^{-1}\int_A^\infty \frac{ds}{\sqrt{(s-A)s^\alpha}} =k^{-1} \int_0^\infty \frac{dt}{(t+A)^{\frac \alpha 2}\sqrt t} \to 0 $$ as $A \to \infty$.

Assume now that $u''+\frac{N-1}{r}u'=u^3$ with $u'(0)=0$ and $u(1)=\lambda >0$. The maximum principle yields $u \geq 0$ and $r^{1-N} (r^{N-1} u')'=u^3$ with $u'(0)=0$ also $u' \geq 0$. Multiplying the equation by $u'$ and integrating $$ \frac 12 u'^2(r)+(N-1)\int_0^r \frac{u'^2}{t} dt= \frac 14 (u^4(r)-u(0)^4) $$ and then $u'^2 \leq \frac 12 u^4$. If $A=u(0)$ is sufficiently large, then $u(r) \geq A$ gives $u'' \geq \frac 12 u^3$ for $r \geq \frac 12$ and one concludes by Fact 1 with initial point $\frac 12$ that the solution explodes before reaching $r=1$.


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