Find distribution $f(t)$ given integral equation $\int_0 ^\infty e^{-u} f(t-u) du = H(t)$

Question

Find a distribution function$f(t)$ that solves the integral solution $$\int_0 ^\infty e^{-u} f(t-u) du = H(t)$$ for $-\infty < t<\infty$ and $t\neq 0$. $H(t)$ is Heaviside step function.

How to solve this? I have no idea. I think here distribution means general solutions.

Answer 1

Distributions means topological dual of the $\mathcal{C}^{\infty}_c$ functions. This allows for very general solutions (as Dirac masses for instance). (In this case however, the expression is ill-defined for some distributions...)

In this case, there is a solution following these steps :

1. change the variable in the integral, so that $f$ does not depend on $t$ anymore in the integrand
2. differentiate the equality with respect to $t$
3. find $f$

I find : $f(t) = \delta(t) + H(t)$ where $\delta$ is the Dirac mass (at $0$).

You can check indeed that if $t < 0$ :

$\int_0^{\infty} e^{-u} (\delta(t-u) + H(t-u)) du = 0 = H(t)$

and if $t > 0$ :

$\int_0^{\infty} e^{-u} (\delta(t-u) + H(t-u)) du = e^{-t} + (1 - e^{-t}) = 1 = H(t)$

Answer 2

Differentiate both sides of the given equation $$H(t)=\int_0 ^\infty e^{-u} f(t-u) du$$ with respect to $t$ \begin{align} H’(t)=&\int_0 ^\infty e^{-u} \frac{d}{dt}f(t-u) du =- \int_0 ^\infty e^{-u} \frac{d}{du}f(t-u) du \\ \overset{ibp}=&\>-e^{-u} f(t-u) \bigg|_0^\infty -\int_0 ^\infty e^{-u} f(t-u) du =f(t)-H(t) \end{align} Thus $$f(t) = H’(t)+H(t)=\delta(t)+H(t)$$