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$g(s) = \frac{\lambda A}{1-e^{-\lambda T}}s$ 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 $s > T$.

ftor
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    Solving for what ? What is known ? What is unknown ? – JJacquelin Feb 08 '20 at 16:26
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    This does not seem an integral equation. – Jon Feb 08 '20 at 17:04
  • @JJacquelin I edited the question. I can plot the $g$ function for $s>T$ but I don't know how to get its closed-form. – ftor Feb 08 '20 at 18:06
  • @Jon I edited the question. Is that clearer now? – ftor Feb 08 '20 at 18:07
  • @ftor Ok, but you gave $g(s)$ explicitly in the first row of your question and it does not seem an unknown. – Jon Feb 08 '20 at 18:12
  • @Jon Indeed this $g(.)$ function is known, but only on the domain $[0,T]$. But for the domain $[T, +\infty]$, we don't know what g(.) is. We are left with just an equation in the second row that defines $g(.)$. – ftor Feb 08 '20 at 18:17
  • @ftor the integral can be performed straightforwardly and the result can be prolonged to any value greater than $T$. – Jon Feb 08 '20 at 19:26
  • @Jon For example, I don't know how to express $g(T+\epsilon)$ where $\epsilon>0$ even it is small. – ftor Feb 08 '20 at 19:38

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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 $$ g(T+\epsilon ) = h(\epsilon)= g(T) + f(\epsilon) -f(0) + \lambda \int_0^\epsilon f(s) d s + \lambda A \epsilon. $$

Martinus Maximus
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