Global $L^{2}$ solution of Laplace equation in $\mathbb{R}^{3}$.

Question

Does there exists a global and non-trivial solution $u\in L^{2}(\mathbb{R}^{3})$ to the Laplace equation $\Delta u=0$?

Using the spherical symmetry of the problem, one can consider the usual solution obtained by seperation of variables which yields something like this $$u(r,\theta,\varphi)=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}c_{ml}r^{l}Y_{ml}(\theta,\varphi)$$ where $Y_{ml}$ are the spherical harmonics and $c_{ml}$ coefficients. This sum is well defined as long as $r$ is smaller then a critical radius $R$ which depends on the choice of coefficients $c_{ml}$. However, even if we manage to find coefficients such that $R\to \infty$, its seems to me that the solutions will never be in $L^{2}$ due to the $r^{l}$ terms. Is there any other Ansatz one can think of?

Answer 1

Let $x \in \mathbb R^n$ be arbitrary. Since $u$ is harmonic, by the mean value formula,$$ u(x) = \frac 1 { \omega_n r^n} \int_{B_r(x)} u(y) \, dy$$ where $\omega_n:= \vert B_1 \vert$. Hence, by Hölder’s inequality, \begin{align*} \vert u(x) \vert &\leq \frac 1 { \omega_n r^n} \int_{B_r(x)} \vert u(y) \vert \, dy\\ &\leq \frac {\vert B_r \vert^{1/2}}{\omega_n r^n} \| u \|_{L^2(B_r(x))}\\ &\leq \frac 1{\omega_n ^{1/2}r^{n/2}} \| u \|_{L^2(\mathbb R^n)}. \end{align*} Sending $r \to + \infty$ implies that $u$ must be zero.

Answer 2

Also, in addition to @JackT's apt answer: we can view this as a generalization of the Liouville theorem for entire functions: if we ask for a tempered distribution (a pretty general thing) $u$ such that $\Delta u=0$, then, taking Fourier transform, $r^2\cdot \widehat{u}=0$. That is, $\widehat{u}$ is a tempered distribution supported at $0$. Thus, it is a finite linear combination of derivatives of Dirac deltas (and more can be said). Thus, the original $u$ must have been a polynomial... and none of these is square-integrable.