Delay Functional Equation: $f(t) - f(t-\tau) = u(t)$

Question

Up front, I do not have any background in functional equations. Maybe this is an easy problem, maybe it is an impossible problem; I do not know.

I am trying to solve the following equation for $f(t)$: \begin{equation} f(t) - f(t-\tau) = u(t), \end{equation} where $\tau$ is some constant and $u$ is an arbitrary smooth function of $t$. I read this post which suggests that there is a homogeneous and a particular component to a solution. I am only interested in the particular part. Are there any relevant approaches to find particular solutions of functional equations?

Answer 1

Hint.

Using the Laplace transform we have

$$ F(s)-e^{-s\tau}F(s)-e^{-s\tau}\int_{-\tau}^0e^{-s\eta}f(\eta)d\eta = U[s) $$

Assuming that $f(\eta) = 0$ for $-\tau\le \eta \le 0$ we have

$$ F(s) = U(s)(1-e^{-s\tau})^{-1} $$

Answer 2

I think methods for finding solutions depend on the quality of $u$. For example set $$ f(t)=\sum_{k=0}^\infty u(t-k\tau) $$ in case that this series is convergent for each $t \in \mathbb{R}$. Then $$ f(t)-f(t-\tau) $$ $$ =(u(t)+u(t-\tau)+ u(t-2\tau) + \dots) - (u(t-\tau)+u(t-2\tau)+ u(t-3\tau) + \dots) = u(t). $$ This works for example if $u$ has compact support or for $u(t)=\exp(-t^2)$.