# Partial Differential Equation using separation of variables

1. Jun 10, 2009

### Hendrick

1. The problem statement, all variables and given/known data
Solve the heat flow problem using the method of separation of variables:

2. Relevant equations
PDE:$$\frac{\partial u}{\partial t}=k\frac{\partial^{2} u}{\partial t^{2}}$$
for 0<x<L, 0<t<$$\infty$$

BC's:$$\frac{\partial u}{\partial x}(0,t)=0,\frac{\partial u}{\partial x}(L,t)=0$$
for 0<t<$$\infty$$

IC's: $$u(x,0)=$$
{0, $$0<x<L/4$$
{1, $$L/4<x<3L/4$$
{0, $$3L/4<x<L$$
(Piecewise IC)

3. The attempt at a solution
I have separated the variables, then applied the boundary conditions. I am stuck on applying the initial conditions.

I have come up with a general product solution of $$u_{n}=F_{n}cos(\frac{n \pi x}{L}) e^{-k(\frac{n \pi x}{L})^{2}t}$$

Trying to combine all product solutions and match the initial data:
$$u(x,0)=f(x)$$
$$\sum^{\infty}_{n=1}u_{n}(x,0)=f(x)$$
$$\sum^{\infty}_{n=1}F_{n}cos(\frac{n \pi x}{L})=f(x)$$

I don't know how to apply the piecewise initial condition, any help would be appreciated. Thank you

2. Jun 10, 2009

### HallsofIvy

Staff Emeritus
Once you have
$$\sum^{\infty}_{n=1}F_{n}cos(\frac{n \pi x}{L})=f(x)$$
you do it exactly the same way you would if f were not "piecewise". Find the Fourier cosine coefficients by doing the appropriate integrals for C0 and Cn for n> 0. The only difference "piecewise" makes is that instead of integrating a single formula from 0 to L, you integrate using the given formulas from 0 to L/4, L/4 to 3L/4, 3L/4 to L and adding those integrals. (Which I notice now is just integrating from L/4 to 3L/4 since outside that the function is 0. The difference between this and just doing the problem on the interval from L/4 to 3L/4 is that you use the whole interval, of length L in determining the "normalization" constant.)