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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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 $p>2$.p>4$. |
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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. Clearly The Bessel function $u\in L^\infty$ is possibleJ_0(\sqrt{x^2+y^2})$ satisfies $\Delta u+u=0$, as trivial examples with and it is in $u=1$ show. I suspect L^p$ for every $p<\infty$ is not possible, but I cannot immediately think of a proof.p>2$. |
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