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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)

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Just a comment, the two examples given below by Michael and Denis are illustrations of "Fourier restriction theorems". See, e.g., Chapter 8, section 3 of Stein's Harmonic Analysis (the big orange PUP book). –  Willie Wong Mar 16 '11 at 16:05
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2 Answers

up vote 3 down vote accepted

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 $p>2n/(n-1)$.

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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.

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