# 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.

• Solve it as you would an ordinary ode and integrate both sides from 0 to x. – cauchyproblem May 22 '16 at 17:21
• I edited the post to use correct angle brackets: \langle and \rangle, which are different from inequality signs. – user147263 May 22 '16 at 18:52

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: $-\langle T,\phi'\rangle = \langle T', \phi\rangle$.

• Great, thanks a lot for your answer. Using $g'=H$ is quite nice, I didn't think of that. – Fritz May 22 '16 at 17:41
• One question: Wouldn't the distributional derivative to $g(x)=|x|$ be the jump function $$j(x)=\begin{cases} 1, x\geq 0 \\ -1, x<0 \end{cases}$$ I should pick $g(x)=x^{+}=\max\{0,x\}$, shouldn't I? This is still locally integrable, so this choice is okay, I assume. – Fritz May 22 '16 at 18:39
• That is correct, I updated the answer. – user147263 May 22 '16 at 18:48
• The function $g$ can also be written as $g(x) = x \, H(x)$ which makes the division by $x$ trivial. So, $$T_f = H(x) + c \operatorname{PV}\frac1x.$$ – md2perpe Aug 17 '18 at 23:10

$$(xT_f)'=H(x)$$, 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:

$$(xT_f)'= [x(H(x)+C_1PV(\frac{1}{x})+C_2\delta(x))]'$$

$$(xT_f)'= [xH(x)+C_1xPV(\frac{1}{x})+C_2x\delta(x)]'$$

$$(xT_f)'= [xH(x)+C_11+0]'$$

$$(xT_f)'= H(x)+x\delta(x)=H(x)$$