How to solve the following integral equation?

Question

$g(s) = \frac{\lambda A}{1-e^{-\lambda T}}s$ for $s\in [0, T]$.

$g(s) = \int_{0}^{T} \lambda e^{-\lambda t} g(s-t) dt + A$. for $s \ge T$.

$T, A, \lambda$ are constant. I want to get the closed-form of $g$ for $s > T$.

Answer 1

I think we have to solve it iteratively. First on [T,2T] and then on larger intervals with the same procedure.

For $\epsilon\in [0,T] $ we have $$ g(T+\epsilon) = \int_{\epsilon}^T \lambda e^{-\lambda t} g(T+\epsilon-t) dt + \int_0^\epsilon \lambda e^{-\lambda t} g(T+\epsilon-t) dt + A $$ Since you have a explicit and nice representation for $g$ on $[0,T]$ that you can insert into the first integral, the first integral can be explicitly computed, say it has the value $f(\epsilon)$. Its derivative can be also explicitly computed. Writing $h(\epsilon) = g(T+\epsilon)$ and differentiating yields $$ h'(\epsilon) = f'(\epsilon)+ \lambda e^{\lambda\epsilon }h(0) + \int_0^\epsilon \lambda e^{-\lambda t} h'(\epsilon-t) d t. $$ Integration by parts yields $$ h'(\epsilon) =f'(\epsilon)+ \lambda h(\epsilon) -\lambda \int_0^\epsilon \lambda e^{-\lambda t} h(\epsilon-t) dt. $$

Inserting the first equation yields $$ h'(\epsilon) = f'(\epsilon) + \lambda f(\epsilon) + \lambda A, h(0) = g(T), $$ so that the fundamental theorem of calculus implies $$ g(T+\epsilon ) = h(\epsilon)= g(T) + f(\epsilon) -f(0) + \lambda \int_0^\epsilon f(s) d s + \lambda A \epsilon. $$