I'm trying to solve the equation
$(1-|x|^2)T = 0$, where $T$ is a tempered distribution. I know how to do this (it is a common exercise) in dimension $1$. How can I solve it in higher dimensions?
Thank you very much.
2
-
$\begingroup$ There's a very different question with the exact same title at m.se, math.stackexchange.com/questions/2113226/… $\endgroup$– Gerry MyersonFeb 5, 2017 at 11:17
-
$\begingroup$ Do you mean to find at least a solution or all solutions? What if you assume $T$ radially symmetric? $\endgroup$– Pietro MajerFeb 5, 2017 at 13:07
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
The solution is the direct product of $\delta(|x|-1)$ and any distribution on the sphere.
answered Apr 1, 2011 at 19:54
-
$\begingroup$ Where could I learn why this is the solution? $\endgroup$– John HApr 2, 2011 at 0:01
-
$\begingroup$ First prove that T has compact support on the sphere. Any test function localized near the sphere can be split into a part that vanishes on the sphere and a part that is radially constant near the sphere. Now consider how T acts on each part. $\endgroup$ Apr 5, 2011 at 19:50
$\begingroup$
$\endgroup$
This is not a different answer. I only give here the details of Michael Renardy's solution.
Theorem. A tempered distribution $T$ on $\mathbb{R}^n$ solves $(1-|x|^2)T=0$ if and only if for some distribution $V$ on the unit sphere $S^{n-1}=\{x \in \mathbb{R}^n : |x|=1 \}$, we have $T(\phi)=V(\phi_{|S^{n-1}})$, for every $\phi \in \mathcal{S}(\mathbb{R}^n)$.
Proof. If for some distribution $V$ on $S^{n-1}$, we have $T(\phi)=V(\phi_{|S^{n-1}})$, for every $\phi \in \mathcal{D}(\mathbb{R}^n)$, then clearly $T$ is tempered, that is it can be extended to a continuous linear functional on $\mathcal{S}(\mathbb{R}^n)$ by putting $T(\phi)=V(\phi_{|S^{n-1}})$, for every $\phi \in \mathcal{S}(\mathbb{R}^n)$. Moreover, we clearly have $T((1-|x|^2) \phi(x))=V(0)=0$ for every $\phi \in \mathcal{S}(\mathbb{R}^n)$.
Now, let $T$ any distribution (tempered or not) such that $(1-|x|^2)T=0$. Then the support of $T$ is contained in $S^{n-1}$. Let $\epsilon \in \left(0,\frac{1}{2} \right)$ and $\xi \in \mathcal{D}(\mathbb{R}^n)$ a function such that $\xi=1$ on $\{x \in \mathbb{R}^n: 1 - \epsilon < |x| < 1 + \epsilon \}$, and such that the support of $\xi$ is constrained in $\{x \in \mathbb{R}^n: 1 - 2 \epsilon < |x| < 1 + 2 \epsilon \}$. Then we have $T(\xi \phi)=T(\phi)$ for every $\phi \in \mathcal{D}(\mathbb{R}^n)$. Now let $\phi \in \mathcal{D}(\mathbb{R}^n)$, and define \begin{equation} \alpha(x)= \xi(x) \phi \left( \frac{x}{|x|} \right), \end{equation} \begin{equation} \beta(x)= \xi(x) \left[\phi(x) - \phi \left( \frac{x}{|x|} \right)\right]. \end{equation} Then $\alpha$ and $\beta$ clearly belong to $\mathcal{D}(\mathbb{R}^n)$ and we have $\xi \phi= \alpha + \beta$. Set \begin{equation} \chi(x) = \begin{cases} \frac{\beta(x)}{1-|x|^2} & \textit{if } |x| \neq 1,\\ - \frac{1}{2} (\nabla \phi)(x) \cdot \frac{x}{|x|} & \textit{if } |x|=1,\end{cases} \end{equation} It is easy to see by induction that $\chi$ and all its partial derivatives are differentiable, so that $\chi \in \mathcal{D}(\mathbb{R}^n)$. From the equation \begin{equation} \xi(x) \phi(x)= \alpha(x) + (1-|x|^2) \chi(x), \end{equation} we conclude that $T(\phi)=T(\alpha(x))$. So, if we define for any smooth function $\gamma$ on $S^{n-1}$ \begin{equation} V(\gamma)=T\left( \xi(x) \gamma\left( \frac{x}{|x|} \right) \right), \end{equation} clearly $V$ is a distribution on $S^{n-1}$, and we have $T(\phi)=V(\phi_{|S^{n-1}})$ for any $\phi \in \mathcal{D}(\mathbb{R}^n)$.
QED
