Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
I myself am confused by the context in which an argument proceeds by saying that if a fundamental solution for a differential operator were compactly supported then it would have support $\{0\}$, leading to deducing that the operator is order $0$, so apparently is a multiplication operator? Or is this meant to be about pseudo-differential operators? Would you clarify the context? –  paul garrett Jul 28 '12 at 19:01
    
I'm not sure if this includes pseudo-differential operators. The question wasn't well stated to me, I got the impression it only involved differential operators (with constant coefficients). Apparently this can be shown by an appeal to the Fourier transform of distributions... –  Peadar Coyle Nov 7 '12 at 0:52

2 Answers 2

up vote 1 down vote accepted

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.

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.)

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.