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Let $\mathcal{S}$ denote the space of Schwartz functions on $\mathbb{R}^n$, and $\mathcal{S}'$ the space of tempered distributions. Let $L$ denote a linear, constant-coefficient, partial differential operator. I would like to know if there is a "simple" proof of solvability for $L$; $\textit{i.e.}$ of the result that for any $f \in \mathcal{S}'$ there is some $u \in \mathcal{S}'$ such that $Lu = f$. I am interested in the result for general $L$, but would be satisfied with a proof that worked for the classical Laplace, wave and heat operators.

Here is what I know so far:

1) If $L$ has a fundamental solution, $G \in \mathcal{S}'$ such that $LG = \delta$ (the Dirac delta), then $Lu = f$ is solvable for any $f$ such that the convolution $f \ast G$ exists (in $\mathcal{S}')$. This is not obviously equivalent to solvability for arbitrary $f$, though a "simple" proof of such equivalence, even for the classical operators, would answer my question.

2) Hormander and Lojasiewicz answered the general problem of solvability affirmatively in 1958. (Any "simple" proof therefore has to be simpler than Hormander's.)

3) Bernstein, in the 1970's, used analytic continuation to show the existence of a fundamental solution $G \in \mathcal{S}'$ for any $L$.

4) There have subsequently been several "simple" proofs of the existence of fundamental solutions, though these are often located in the space $\mathcal{D}'$ of general distributions.

A possible re-statement of my original question is therefore: can Bernstein's method of analytic continuation, or any of the other "simple' proofs of the existence of a fundamental solution, be extended to answer the general solvability problem in $\mathcal{S}'$?

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For your constant coefficient operator $L(D)$, you want a fundamental solution $G$ such that $\hat G$ is a multiplier of $\mathcal S'$. This is not even true for the Laplace equation: the fundamental solution is, in more than 3D, $a_d\vert x\vert^{2-d}$ and its Fourier transform is $$ b_d\vert \xi\vert^{-2}=\hat G(\xi). $$ The map $\mathcal S\ni\phi\mapsto \phi \hat G\in L^1$ is well defined but $\hat G$ is not a multiplier of $\mathcal S$ or $\mathcal S'$.

On the other hand, if you consider $L(D)=-\Delta +m$ with $m>0$, you find $$ \hat G=c_d(m+\vert \xi\vert^2)^{-1} $$ which is indeed a multiplier of $\mathcal S$ and for this equation you have your solvability property.

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The Laplacian does not have a unique fundamental solution. –  Liviu Nicolaescu Jan 11 '13 at 14:53
    
Yes of course, but they are all of the form $a_d\vert x\vert^{2-d}$+ harmonic function on $\mathbb R^d$. –  Bazin Jan 11 '13 at 20:21
    
It is true that if I want to solve $Lu = f$ for arbitrary $f$ by convolution with a fundamental solution, that fundamental solution must be a multiplier. It is nevertheless true that $Lu = f$ is always solvable; I can solve $\Delta u = f$ for any $f$, for example, even though the fundamental solution of the Laplacian is not a multiplier. (H\"{o}rmander's proof simplifies considerably in this case.) The interest in the question is in fact entirely for those situations when a solution can not be found by convolution. –  Kevin McLeod Jan 17 '13 at 4:58
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