1. ## Eigenfunctions and PDE's

Solve,

$\displaystyle \frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}=F(x,t)$

$\displaystyle u(0,t)=0, \frac{\partial u}{\partial x}(\pi,t )=0, u(x,0)=0$

where $\displaystyle F(x,t)$ is a given function.

(a) Determine the x-dependent eigenfunctions appropriate to the solution of the equation with given boundary conditions.

(b) Solve the method of eigenfunction expansions to obtain a series solution for the the problem in terms of the eigenfunctions determined in (a)

(c) Simplify for the particular case $\displaystyle F(x,t)=1$

Any help appreciated. Thanks

2. Originally Posted by featherbox
Solve,

$\displaystyle \frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}=F(x,t)$

$\displaystyle u(0,t)=0, \frac{\partial u}{\partial x}(\pi,t )=0, u(x,0)=0$

where $\displaystyle F(x,t)$ is a given function.

(a) Determine the x-dependent eigenfunctions appropriate to the solution of the equation with given boundary conditions.

(b) Solve the method of eigenfunction expansions to obtain a series solution for the the problem in terms of the eigenfunctions determined in (a)

(c) Simplify for the particular case $\displaystyle F(x,t)=1$

Any help appreciated. Thanks
When you have a forced equation problem you can use Duhamel's principle
Duhamel's principle - Wikipedia, the free encyclopedia

So we need to solve the associated problem

$\displaystyle \phi(x,t;s)$ such that

$\displaystyle \phi_t-\phi_{xx}=0$

$\displaystyle \phi(0,s)=0=\phi_{x}(\pi,s) \quad \phi(x,s)=F(x,s)$

Separating variables as usual gives

$\displaystyle \phi=XT \iff \frac{X''}{X}=\frac{\dot{T}}{T}=-\lambda^2$

$\displaystyle T(t;s)=e^{-\lambda^2(t-s)} \quad X(x)=c_1\cos(\lambda x)+x_2\sin(\lambda x)$

imposing boundary conditions gives $\displaystyle \lambda =\frac{1}{2}(2n+1)$

$\displaystyle X(x)=\sin\left( \lambda x\right)$

$\displaystyle \phi_n(x,t;s)=A_ne^{-\lambda^2(t-s)}\sin(\lambda)x$

Now using the initial condition gives

$\displaystyle F(x,s)=A_n\sin(\lambda x) \implies A_n = \frac{2}{\pi}\int_{0}^{\pi}F(x,s)\sin(\lambda x)dx$

So phi has the form

$\displaystyle \phi_n(x,t;s)=A_ne^{-\lambda^2(t-s)}\sin(\lambda x)$

So now by Duhamel's Principle the final solution is

$\displaystyle u(x,t)=\sum_{n=1}^{\infty}\int_{0}^{t}\phi_n ds$

$\displaystyle u(x,t)=\sum_{n=1}^{\infty}\int_{0}^{t}\left( A_ne^{-\lambda^2(t-s)}\sin(\lambda x) \right)ds$