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Let $u$ be smooth, non-trivial function on $B(0,1)$ s.t. for some $\lambda \in \mathbb C$:

$$ \begin{align} \Delta u &= \lambda u &\text{ in } B(0,1)\\ u&=0 &\text{ in }\partial B(0,1) \end{align} $$

The task is to show that $\lambda \in \mathbb R $ and $\lambda \leq 0 $. I thought that the shortest way to prove this is the following:

  1. Extend $u$ with $0$ to the whole space $\mathbb R^n$.
  2. Take the Fourier transform: $\mathcal 0 = \mathcal F(0) = \mathcal F[(-\Delta + \lambda)u](k) = (|k|^2 + \lambda)\mathcal F(u)$. Since the function $u$ is non-trivial, there exists $k \in \mathbb R ^n$ s.t. $\lambda = -|k|^2$ (evidently Fourier transform of $u$ must be trivial up to $\lambda = -|k|^2$). Therefore $\lambda \in \mathbb R$ and $\lambda \leq 0$.

However, operating with Fourier transform is sometimes tricky and requires rigor, so I kind of have a feeling I missed something which would invalidate this proof. So the question is, is this approach alright, or should I go some other way?

Thanks in advance!

3 Answers 3

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This approach cannot work because $k$ is a variable. When you take the Fourier transform you get that $$ (|k|^2 + \lambda) \mathcal{F}(u)(k) =0. $$ Since your extension of $u$ to $\mathbb{R}^n$ results in a square-integrable function, i.e. $u \in L^2(\mathbb{R}^n)$ we know that $\mathcal{F}(u) \in L^2$ as well. This means that we conclude that $|k|^2 + \lambda =0$ for almost every $k$, and this is nonsense.

If you drop the $L^2$ condition then you find that $\mathcal{F}(u)$ must be supported on the set $$\{k \in \mathbb{R}^n : |k|^2 =-\lambda \},$$ but this can never result in a corresponding $u$ that vanishes outside a ball.

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    Why is this nonsense that $|k|^2 + \lambda = 0$ for almost every $k$, for example $\mathcal F (1) = \delta(k)$ which is supported at $0$, so it vanishes for almost every $k$, so we have non-trivial function whose Fourier transform is supported by a zero set. – 2017-01-19
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    The function $u(x)=1$ is not square integrable, nor does it vanish outside a ball as in OP's post. In my answer I have used the inclusion $\mathcal{F}(u) \in L^2$ to eliminate objects such as the delta funciton. There are certainly tempered distributions whose support is contained in a sphere of any radius (including the degenerate radius leading to a point), but these do not correspond to square integrable functions. – 2017-01-19
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The trick is to integrate by parts

$$\int_B |\nabla u|^2 \, dx = -\int_B \bar{u}\Delta u\, dx = -\lambda\int_B |u|^2\, dx.$$

Your approach does not work because the extension by zero does not solve your PDE on all of $\mathbb{R}^n$ (the boundary of the ball is problematic).

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If $u$ satisfies your equations, then the divergence theorem gives \begin{align} \lambda \int_{B}u^2dx & = \int_{B}(\nabla^2u)udx \\ & =\int_{B}\nabla\cdot((\nabla u)u)-|\nabla u|^2dx \\ & =\int_{\partial B}u\frac{\partial u}{\partial n}dS-\int_{B}|\nabla u|^2dx \\ & = -\int_{B}|\nabla u|^2dx. \end{align} Therefore $\lambda\int_{B}u^2dx \le 0$. Because $u$ is assumed to be non-trivial, then $\lambda \le 0$..