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Given an equation $P(D)u=0$, where $P$ is a polynomial (not equal to a constant). Here are some basic information about the distributional solution $u$:

  1. If $P$ has at least one real root, then there exist a (non zero) solution $u \in S'$(temperate distribution);
  2. It never has a (non zero) solution in $\epsilon'$(distribution with compact support).

These two properties can be checked easily by Fourier transform.

My question is why it always has a distributional solution(non zero) and a $C^{\infty}$ solution(non zero)? For the distributional solution,it's only needed to check that $P(D)C_{0}^{\infty }(\mathbb{R}^{n})$ is a strictly smaller than $C_{0}^{\infty}(\mathbb{R}^{n})$,than Hahn-Banach theorem can be used. For the $C^{\infty}$ solution, I don't know how to deal with it.

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If $P$ has a real root $r=(r_1,\dots,r_n)$, then $e^{r\cdot x}$ is a $C^{\infty}$ solution. –  Davide Giraudo Aug 12 '12 at 13:46
    
right,but for general P,it's still true .(I think the point here is fourier transform can't be used,maybe it needs some topological statement ) –  sun Aug 12 '12 at 14:23
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1 Answer

up vote 2 down vote accepted

Let $P(D)u=0$ with $u\in \mathcal{D}'$. Pick $\phi\in\mathcal{D}$. Then $\phi*u\in C^\infty$ and $$ P(D)(\phi*u) = \phi*P(D)u = \phi*0=0. $$

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thanks very much! –  sun Aug 13 '12 at 0:10
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@ShanLinHuang: Don't forget to accept and upvote the answer. –  timur Aug 13 '12 at 0:16
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