A distribution differential equation

Question

Problem. Find a particular solution $v \in \mathcal{D}'(\mathbb{R})$ such that $$xv=\delta$$ knowing just the basic operations on distributions.

Answer 1

Suppose that we want to solve the distributional equation $xv=u,$ where $u$ is a given distribution and $v$ is unknown, the one we want to find. Then $$ \langle u, \phi \rangle = \langle xv, \phi \rangle = \langle v, x\phi \rangle. $$ Thus, if $\psi \in C^\infty_c$ has the form $\psi=x\phi$ for some $\phi\in C^\infty_c$ then we can just set $\langle v, \psi \rangle = \langle u, \phi \rangle.$ But what if $\psi$ does not have that form?

Take $\rho\in C^\infty_c$ such that $\rho(0)=1$ and set $\tilde\psi = \psi - \psi(0) \rho.$ Then $\tilde\psi(0) = 0$ and $\tilde\psi=x\phi$ for some $\phi\in C^\infty_c.$ We conclude that $$ \langle v, \psi \rangle = \langle v, \tilde\psi + \psi(0)\rho \rangle = \langle v, \tilde\psi \rangle + \psi(0) \langle v, \rho \rangle = \langle u, \phi \rangle + C \langle \delta, \psi \rangle, $$ where $C=\langle v, \rho \rangle$ is a constant w.r.t. $\psi.$

Note now that a Maclaurin expansion of $\tilde\psi$ gives $\psi(x) = \psi(0)+\psi'(0) x + O(x^2)$ so $\tilde\psi(x)=\psi'(0)x + O(x^2)=x\left(\psi'(0)+O(x)\right)$ giving $\phi=\psi'(0)+O(x).$ For $u=\delta$ we therefore get $$ \langle u, \phi \rangle = \phi'(0) = \langle \delta, \phi' \rangle = -\langle \delta', \phi \rangle, $$ so $v=-\delta' + C\delta.$