How does derivative transfer to test function?

Question

I set out to find a fundamental solution $E$ for the Poisson equation, i.e. a distribution $\mathscr D'(\Bbb R^d)$ such that $\Delta E = \delta$; I'm almost done. The only thing I have left to do is to calculate $$\langle \Delta u_{|\ \cdot\ |^{2-d}}, \sigma \rangle, $$ where $u_\varphi$ is the usual distribution generated by a locally integrable function $\varphi$, and $\sigma \in \mathscr D(\Bbb R^d)$ is such that $\sigma\equiv 1$ on the unit ball, and it is radially symmetric, i.e. $\sigma = \sigma_r \otimes 1$ for some smooth $\sigma_r$ and $1$ is the constant function s.t. $\Bbb S^{d-1} \ni \Omega \mapsto 1$. Because it is invariant under the action of $\operatorname{SO}(3)$, the laplacian can be nicely split in hyperspherical coordinates as $$\Delta = \partial_r^2 + \frac{d-1}{r} \partial_r + \frac{1}{r^2}\Delta_{\Bbb S^{d-1}}, $$ where $\Delta_{\Bbb S^{d-1}}$ contains only angular derivatives. Calling $\Delta_r^{(d)}$ the radial part, then, we find $$\require{cancel}\begin{split} \langle \Delta u_{|\ \cdot\ |^{2-d}},\sigma\rangle &= \left\langle (\Delta_r^{(d)}u_{(\ \cdot\ )^{2-d}}) \otimes u_1 + \cancel{u_{(\ \cdot\ )^{2-d}}\otimes (\Delta_{\Bbb S^{d-1}} u_1)}, \sigma_r\otimes 1 \right\rangle \\ &= \left\langle\Delta_r^{(d)}u_{(\ \cdot\ )^{2-d}}, \sigma_r \right\rangle \left\langle u_1,1\right\rangle; \end{split} $$ at this point, I should discharge the distributional derivative on the test function using the usual adjoint-like formula $$\left\langle \partial^\alpha u, f\right\rangle = (-1)^{|\alpha|}\left\langle u,\partial^\alpha f\right\rangle \tag{$\ast$}$$ to obtain $$\begin{split} \langle \Delta u_{|\ \cdot\ |^{2-d}},\sigma\rangle &= \left\langle u_{(\ \cdot\ )^{2-d}}, \partial_r^2\sigma_r \color{red}- \frac{d-1}{r}\partial_r \sigma_r \right\rangle \left\langle u_1,1\right\rangle\\ &= \int_0^\infty \frac{1}{r^{d-2}} \left(\partial_r^2\sigma_r(r) \color{red}- \frac{d-1}{r}\partial_r \sigma_r(r) \right) r^{d-1}\ dr \cdot \int_{\Bbb S^{d-1}} j(\Omega)\ d^{d-1}\Omega. \end{split} $$ The angular integral is just the surface of the $d$-dimensional ball ($j$ just contains the angular terms of the Jacobian determinant), while the radial integral should evaluate to $(2-d)$; however, I am finding that, by integrating by parts, $$ \begin{split} \int_0^\infty r \left(\partial_r^2\sigma_r(r) \color{red}- \frac{d-1}{r}\partial_r \sigma_r(r) \right) dr &= \cancel{(r\partial_r \sigma_r)|_0^\infty} - \int_0^\infty \partial_r r\ \partial_r\sigma_r\ dr \color{red}- (d-1)\int_0^\infty \partial_r \sigma_r\ dr \\ &= (-d)\int_0^\infty \partial_r \sigma \ dr = (-d)[\sigma_r(\infty)-\sigma_r(0)] = d. \end{split} $$ This issue is generated by the red minus sign $\color{red}-$: if it were a $+$, I would be finding the correct result. However, the $\color{red}-$ seems perfectly justified in terms of $(\ast)$. Why should it be a $+$? What am I missing?

Answer 1

I think the issue lays in the fact that you have exploited the expression of the Laplacian in hyperspherical coordinates when it was applied to the distribution $u$. But regrettably that formula is valid when we deal with ordinary functions (at least $f\in C^2(\mathbb{R}^d)$). Instead, the initial Laplacian has to be considered in distributional sense, i.e. in the way it is suggested in formula $(\ast)$: $$\langle (w\!-\!\Delta) u, \sigma\rangle=\langle u,\Delta\sigma\rangle.$$ Where we made use of the definition of the distributional Laplacian $w\!-\!\Delta:=\sum_{i=1}^{d}\partial_i^2$. Once you have thrown the Laplacian against the test-function you can use the hyperspherical Laplacian formula in order to decompose the integral in radial and angular part. As we can see here, differential operators in distributional sense do not transform like the ordinary derivative ones, but they follow rules that can be proved as the way provided above.