Let $-\Delta u=\lambda u$ in $\Omega\subset\mathbb{R}^n$, $\lambda>0$. Suppose $u=0$ in a ball $B\subset\Omega$. I want to prove that $u\equiv0$ in $\Omega$.

If $\lambda\leq 0$, integrating by part solves this problem. But if $\lambda>0$, things become more subtle.

Besides, I think $\Omega$ must be connected. (am I right?)

Any hints will be appreciated a lot!

  • 1
    $\begingroup$ Of course $\Omega$ must be connected. If $\Omega=\Omega_1\cup B$,where $B$ is a ball and the union is disconnected, then if $u=0$ in $B$ the equation $-\Delta u = \lambda u$ is automatically satisfied on $B$ for whatever value of $\lambda$. $\endgroup$ Commented Sep 22, 2022 at 11:17

1 Answer 1


This follows from regularity theory. Indeed, elliptic regularity theory tells us that solutions to $-\Delta u = \lambda u $ must be analytic in $\Omega$ (since all the coefficients in the elliptic operator $Lv=-\Delta v -\lambda v$ are analytic). Hence, by unique continuation, if $u=0$ in a ball $B\subset \Omega$ then $u =0$ in the connected component of $\Omega$ containing $B$ . If $\Omega$ is connected then there is only one connected component, so $u=0$ in $\Omega$.

Inspired by @Guiseppe Negro's comment, here is a more 'elementary' solution. I put elementary in air quotes because it is certainly not as easy or as elegant as the above solution, but it is elementary in the sense that it only need the mean-value formula for solutions to $-\Delta u = \lambda u$.

Here is the idea: If $u\in C^2(\Omega)$ and $B_\rho(x_0) \subset \Omega$ then it can be shown that $$ \tag{$\ast$}\frac 1 {\vert B_\rho \vert}\int_{B_\rho(x_0)} u(x) \, d x = \alpha_n u(x_0) \lambda^{-n/4} \rho^{-n/2} J_{n/2}(\rho) $$ where $\alpha_n = n \Gamma(n/2) \cdot 2^{n/2-1}$ and $J_\beta$ is the Bessel function of order $\beta$. This is the analogue of the classical mean-value formula for harmonic functions. You can obtain also obtain it in a similar way: you let $\phi(\rho)$ be the average $u$ over $B_\rho(x_0)$ (as above) and $\psi(\rho)$ be the average of $u$ over $\partial B_\rho(x_0)$. Following the proof of the classical mean-value formula you can find an ODE in terms of $\phi$ and $\psi$ which, paired with a second ODE obtained from polar coordinates, you can solve to get the above formula.

For the sake of contradiction, suppose that $\Omega \setminus \{u =0\} \neq \varnothing$. Then either $\{u>0\}$ or $\{u<0\}$ is non-empty - assume without loss of generality that it is $\{u>0\} $. Moreover, we may assume without loss of generality that $\{u>0\}$ must touch {u=0} (i.e. that $\partial \{ u>0\} \cap \partial \{u=0\} \neq \varnothing$).

Now let $x_0\in \{u=0\}$ and $\rho>0$ be such that $B_\rho(x_0) \subset \{u\geqslant 0\}$ with $B_\rho(x_0) \cap \{u>0\} \neq \varnothing$. I believe this is possible since we know that $B\subset \{u=0\}$, so $\{u=0\}$ is not $(n-1)$-dimensional (or lower dimension). Now we use ($\ast$) to obtain that$$0<\frac 1 {\vert B_\rho \vert}\int_{B_\rho(x_0)} u(x) \, d x = \alpha_n u(x_0) \rho^{-n/2} J_{n/2}(\rho) =0 $$ which is a contradiction. Hence, our initial assumption that $\Omega \setminus \{u =0\} \neq \varnothing$ was false.

  • $\begingroup$ +1. I wonder if you can find a solution that is more elementary than this one, though. Something more direct. $\endgroup$ Commented Sep 22, 2022 at 11:18
  • $\begingroup$ Thanks for you help! Besides, can you provide with an ref on the unique continuation principle? It seems that the proof of this principle is important for my question. Thanks! $\endgroup$
    – Kimura Leo
    Commented Sep 22, 2022 at 12:31
  • $\begingroup$ @GiuseppeNegro I agree. I've updated my answer with another proof which you may find interesting $\endgroup$
    – JackT
    Commented Sep 22, 2022 at 12:41
  • $\begingroup$ @Frankie See the section "Analytic varieties and unique continuation" in encyclopediaofmath.org/wiki/Real_analytic_function . In this case, the analytic function $u$ and the analytic function 0 coincide in $\Omega$, so they must be the same. I'm sure it could be found in many real analysis textbooks - it's a very standard result $\endgroup$
    – JackT
    Commented Sep 22, 2022 at 12:43
  • $\begingroup$ Right, exactly, I was hoping for something exactly like that. It is perhaps less elegant but it gives you a better feel for what is going on, and it is more robust, you can hope to apply it to other equations. $\endgroup$ Commented Sep 22, 2022 at 12:44

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