# Fourier transform to solve inhomogeneous PDE

• Oct 8th 2011, 05:02 AM
math2011
Fourier transform to solve inhomogeneous PDE
Express the solution to the inhomogeneous equation
$\frac{\partial u}{\partial t} = \kappa \frac{\partial^2u}{\partial x^2} + S(x,t)$
satisfying the initial condition
$u(x,0) = 0$, $-\infty < x < \infty$,
as an integral involving the source term $S(x,t)$.

The given solution is $\frac{1}{\sqrt{2\pi}} \int^t_0 \left( \int^\infty_{-\infty} \frac{e^{-(x-x')^2/(4\kappa(t - t'))}}{\sqrt{2\kappa(t-t')}} S(x',t') dx' \right) dt'$.

Attempt:

Take Fourier transform with respect to $x$.
$\frac{\partial U}{\partial t} = - \kappa k^2 U + \mathcal{F} \{S(x,t)\}$

Solve the ODE by integrating factor.
$U = \frac{1}{e^{\kappa k^2 t}} \int^t_0 e^{\kappa k^2 t'} \mathcal{F} \{ S(x,t) \} dt'$
$U = \int^t_0 e^{-\kappa k^2 (t - t')} \mathcal{F} \{ S(x,t) \} dt'$

Find inverse transform of $U$.
$u = \frac{1}{\sqrt{2\pi}} \int^\infty_{-\infty} \int^t_0 e^{-\kappa k^2 (t - t')} \mathcal{F} \{ S(x,t) \} dt' e^{-ikx} dk$
$u = \frac{1}{\sqrt{2\pi}} \int^t_0 \int^\infty_{-\infty} e^{-\kappa k^2 (t - t')} \mathcal{F} \{ S(x,t) \} e^{-ikx} dk dt'$

Have I got above correct? Why should I use definite integral $\int^t_0$ instead of indefinite integral $\int$ for solving $U$? What should I do next? (I know I am using $k$ instead of $x'$ in the given solution but I don't see how this is even close to the solution).

Should I expand $\mathcal{F} \{ S(x,t) \}$ like below?
$u = \frac{1}{\sqrt{2\pi}} \int^t_0 \int^\infty_{-\infty} e^{-\kappa k^2 (t - t')} \int^\infty_{-\infty} S(x,t) e^{ikx} dx e^{-ikx} dk dt'$
• Oct 14th 2011, 02:08 PM
Jose27
Re: Fourier transform to solve inhomogeneous PDE
I think it would be easier if you used Duhamel's principle. Assuming you know how to solve the homogenous case.
• Oct 16th 2011, 05:01 AM
math2011
Re: Fourier transform to solve inhomogeneous PDE
Quote:

Originally Posted by Jose27
I think it would be easier if you used Duhamel's principle. Assuming you know how to solve the homogenous case.

Thank you. But I have not studied Duhamel's principle before and it is not in my lecture notes. There must be another way to solve the problem.