0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.