Uniqueness of weak solution L[u]=0 Suppose L is a partial differential operator of arbitrary order with constant coefficients.
If u is in $L^p(\mathbb{R}^n)$ and Lu=0 in distributions, is it necessarily the case that u=0?  Does the answer depend on p?  
Also, if u is a compactly supported distribution in $\mathbb{R}^n$ with Lu=0 (in the usual sense, i.e. strongly), is it necessarily the case that u=0?
(Suggested reference material appreciated) 
 A: If $Lu=0$, then the Fourier transform of $u$ must have its support on the manifold where the symbol of $L$ is zero. Hence the Fourier transform of $u$ cannot be a function. This rules out $u\in L^p$ for $p\le 2$; it also rules out a compactly supported distribution. The Bessel function $J_0(\sqrt{x^2+y^2})$ satisfies $\Delta u+u=0$, and it is in $L^p$ for every $p>4$. You can generalize this example to $n$ dimensions, and you find that $u\in L^p$ for every $p>2n/(n-1)$.
A: Strichartz estimates show that the solution $u$ of the Cauchy problem for several equations of physical interest do belong to an $L^p_t(L^q_x)$ if the initial data is appropriate. When $p=q$, this just means that $u\in L^p$.
For instance, consider the wave equation
$$\partial_t^2u-\Delta_xu=0,\qquad t\in\mathbb R,x\in\mathbb R^d,$$
in which $n=d+1$. Say that $d\ge3$. Let the initial data be
$$u(0,x)=a(x),\qquad \partial_tu(0,x)=b(0,x),$$
where $a\in H^1(\mathbb R^d)$ and $b\in L^2(\mathbb R^d)$. Then $u\in L^p(\mathbb R^{1+d})$ with
$$p=2\frac{d+1}{d-2}.$$
There are variants of this result, but this is too a rich topic to be developped here.
Edit. This phenomenon is called a dispersion effect. It is related to the fact that the curvature of the characteristic cone of $L$ (here $\xi_0^2=\xi_1^2+\cdots+\xi_d^2$) is non-zero.
