As stated in the title, I want to solve the distributional differential equation $(\star)$ $$(xT_f)' \equiv H $$
$T_f \in (C_0^\infty)^*$ is a distribution induced by an arbitrary $f \in L_{\text{loc}}^1$ via $$\langle T_f,\phi\rangle :=\int_{\mathbb{R}} f(x) \phi(x) \text{d}x, ~ \phi \in C_0^\infty$$
'$\equiv$' means equality in distributional sense, i.e. for all $\phi \in C_0^\infty$ $$ \langle (xT_f)',\phi\rangle = \langle H,\phi\rangle$$ $$ :\Leftrightarrow \int_{\mathbb{R}} (xf(x))' \phi(x) \text{d}x = \int_{\mathbb{R}} H(x) \phi(x) \text{d}x$$
The derivative is the distributional derivative, i.e. $\langle T_f',\phi\rangle:=-\langle T_f,\phi'\rangle$ for every test function $\phi$.
$H$ is the Heaviside function, i.e. $$H(x)=\begin{cases} 1, & x>0 \\ \frac{1}{2}, & x=0 \\ 0, & x<0 \end{cases}$$ Therefore $\int_{\mathbb{R}} H(x) \phi(x) dx=\int_{0}^\infty \phi(x) dx$ and $H' \equiv \delta$.
What I did so far:
Playing around brings me to: (nothing much) $$\langle (xT_f)',\phi\rangle=-\langle xT_f,\phi'\rangle=-\langle T_f,x\phi'\rangle$$
The homog. case (setting $H \equiv 0$) yields $$(xT_f)'\equiv 0 \Rightarrow 0=\langle (xT_f)',\phi\rangle=-\langle xT_f,\phi'\rangle \Rightarrow xf(x)=0 \Rightarrow x=0 \vee f(x)=0$$ Lets look at $x=0$: $f(0)=H(0)=\frac{1}{2}$. That means $\langle T_f,\phi\rangle=0=\langle 0,\phi\rangle \Rightarrow T_f\equiv0$.
- I am quite stuck in the inhomogeneous case, solving this integral equation. Variation of parameters seems not applicable.
Does someone have an idea for me? That would be great. I haven't dealt with distributional differential equations so far, so some things might not be correct, sorry for that.
\langleand\rangle, which are different from inequality signs. $\endgroup$