Solve the following distributional differential equation: $(xT_f)' \equiv H$ 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.
 A: There is a familiar function whose distributional derivative is $H$: namely, $g(x)=(x+|x|)/2$. (By the way, the value $H(0)=1/2$ is irrelevant, since changing the value at one point does not change the distribution at all.) So the problem becomes $(xT_f-g)'=0$, which implies $xT_f-g = c$. From here, $T_f$ should be $((|x|+x)/2+c)/x$, which is more precisely written as  $$T_f = \frac12 + \frac12 \operatorname{sign}x + c\operatorname{PV}\frac1x$$ 
These manipulations were not really rigorous (no test functions were involved), but the solution can be justified by testing it against a test function $\phi$: 
$$
- \langle xT_f ,\phi'\rangle = 
-\int \left(\frac12 + \frac12\operatorname{sign}x\right) x\phi(x)'\,dx - c\lim_{r\to 0}\int_{|x|\ge r}\frac1x x \phi(x)'\,dx \\ = -\int \frac{x+|x|}{2}\phi'(x)\,dx - c \int \phi(x)'\,dx = \int H(x)\phi(x)\,dx$$
as required by the definition of distributional derivative: $-\langle T,\phi'\rangle = \langle T', \phi\rangle$.
A: $(xT_f)'=H(x)$, where  $H(x)$ is the Heaviside distribution.
Integrating both members, it remains:
$xT_f=xH(x)+C_1$
This is a distributional algebraic equation.
To solve it, we find the general solution of the homogeneous and a particular solution of the complete, as in the ODEs.
Solution of the homogeneous equation $xT_f=0$:
$T_f homog=C_2\delta(x)$
Particular solution of the complete $xT_f=xH(x)+C_1$:
$T_f part=PV(\frac{1}{x})xH(x)+C_1PV(\frac{1}{x})=H(x)+C_1PV(\frac{1}{x})$
Finally, we get $T_f$:
$T_f=T_f part+T_f homog=H(x)+C_1PV(\frac{1}{x})+C_2\delta(x)$
Checking the solution:
$(xT_f)'= [x(H(x)+C_1PV(\frac{1}{x})+C_2\delta(x))]'$
$(xT_f)'= [xH(x)+C_1xPV(\frac{1}{x})+C_2x\delta(x)]'$
$(xT_f)'= [xH(x)+C_11+0]'$
$(xT_f)'= H(x)+x\delta(x)=H(x)$
