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I've been mulling over a problem from Friedlander's Introduction to Distribution Theory for a few days now: in Chapter 3 (on distributions with compact support), it asks to solve the differential equation $x\partial u-\lambda u=0$ for arbitrary $\lambda\in\mathbb{C}$.

I've seen it stated that a distribution $u\in\mathbb{R}^n$ is homogeneous of degree $\lambda$ iff it satisfies Euler's equation, i.e. $\sum x_i\partial_i u=\lambda u$, but these just refer to Gel'fand and Shilov's book which I don't have access to. Moreover, this approach doesn't seem the author's intended one, given the chapter this problem is in.

My method was basically to try and use "integrating factors", but of course we have only defined multiplication by smooth functions, which $x_+^\lambda$ is generally not. I've basically seen it without proof via online searching that $u=Ax_+^\lambda+Bx_-^\lambda$ is the general answer, but I can't seem to crack it, or find anything in this problem with compact support. Any help would be much appreciated.

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I'll outline several ideas that come into mind and give some remarks concerning the validity of the reasoning.

First, I assume that we work in $\Bbb R$.

Let's for instant suppose that $\lambda< -1$. We take a test function $\phi$, multiply it by $x^{-\lambda-1}$ (therefore, it's still a test function), and then apply to it both parts of our equation. After some easy modifications, we arrive to $$\langle(x^{-\lambda}u)',\phi\rangle=0,$$ which yields $$x^{-\lambda}u=const.$$ Now we have a particular solution (notably, so called $\mathrm{pf.}x^{\lambda}_+$ and $\mathrm{pf.}x^{\lambda}_+$, partie finie d'Hadamard in France, don't know the name in English). There're also general solutions, i.e. solving $$x^{-\lambda}u=0.$$ It's easy to see that the support of $u$ in this case (cf. your reference book, the corresponding lemma should be there; or just prove, it's an exercise, really) is exactly $\{0\}$. By another classic theorem, the family of solutions is described by $$\sum_{j=0}^{[\lambda]}c_j\delta_{0}^{(j)},$$ where $c_j$ are constants and $[\lambda]$ is the integer part of $\lambda$.

What doesn't work this smoothly in other cases: if $\lambda$ is complex, then $x^\lambda$ is a multivalued function. Then again, we can't directly apply the same method if $\lambda\ge-1$.

You can explicitly find the solutions for $\lambda=0$ and $\lambda=-1$.

On the other hand, if a differential equation has a solution in a classical sense, then this solution is a solution in the sense of distributions.

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