Partial integro-differential equation using Laplace transform

Question

Is it possible to solve the linear PDE analytically \begin{equation} \frac{\partial u}{\partial z} + a \frac{\partial u}{\partial t} + \int_{0}^{t} e^{-\beta (t-t')} u(z,t') dt'=f(z,t), \end{equation} subject to the conditions \begin{equation} u(z,0) = 0; \quad u(L,t) = u_0(t), \end{equation} by applying the Laplace transform in time to reduce this to a regular inhomogeneous 1st order ODE in space?

Answer 1

Applying the Laplace transform we get

$$ U_z(z,s) +a(s U(z,s)-u(z,0))+\frac{1}{s+\beta}U(z,s) = F(z,s) $$

or

$$ U_z(z,s)+\left(a s+\frac{1}{s+\beta}\right)U(z,s) = F(z,s) $$

with the condition $U(L,s) = U_0(s)$

or

$$ U(z,s) = e^{-\frac{z (s (\beta +s)+1)}{\beta +s}} \left(\int_1^z e^{\frac{\zeta (s (\beta +s)+1)}{\beta +s}} F(\zeta ,s) \, d\zeta -\int_1^L e^{\frac{\zeta (s (\beta +s)+1)}{\beta +s}} F(\zeta ,s) \, d\zeta +U_0(s) e^{L \left(\frac{1}{\beta +s}+s\right)}\right) $$

After that you can find the anti-transform using residue theory.