most recent 30 from http://mathoverflow.net 2013-05-19T10:35:43Z http://mathoverflow.net/feeds/question/103394 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103394/fundamental-solutions-with-compact-support-distributions Peadar Coyle 2012-07-28T15:52:45Z 2012-07-29T00:20:42Z

<p>Assume that we have a differential operator such as $-\frac{\partial}{\partial x^2} + id$ on $\mathbb{R}^1$ We also then argue that if a fundamental solution has compact support, then it is supported on the origin. My follow up question is how can one then show assuming that the fundamental solution is compactly supported - that the differential operator must be of order 0?</p>

http://mathoverflow.net/questions/103394/fundamental-solutions-with-compact-support-distributions/103408#103408 timur 2012-07-28T21:32:28Z 2012-07-28T21:32:28Z

<p>If the fundamental solution is supported at the origin, it must be a finite combination of derivatives of the Dirac distribution. This means that your original operator was of nonpositive order.</p>

http://mathoverflow.net/questions/103394/fundamental-solutions-with-compact-support-distributions/103417#103417 Shanlin Huang 2012-07-29T00:20:42Z 2012-07-29T00:20:42Z

<p>For a constant coefficient partial differential operator P(D), the fundamental solution of P can never belong to $\epsilon'(\mathbb{R}^{n})$,i.e.have compact support.</p> <p>In fact,assume we have $P(D)u=f$,where u is a distribution,then u have compact support $\Leftrightarrow$ $\frac{f}{P(\xi)}$ is analytic(The result can be found in Hormander's ALPDO, volume 1,ch7.)</p> <p>Now,if we have $$P(D)u=\delta$$,obviously $\frac{1}{P(\xi)}$ is never an analytic function for a polynomia P.So the fundamental solution of P can not be compact supported.</p>