Partial Differential Equation using separation of variables

[Hendrick]

Homework Statement

Solve the heat flow problem using the method of separation of variables:

Homework 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&lt;x&lt;L/4
{1, L/4&lt;x&lt;3L/4
{0, 3L/4&lt;x&lt;L
(Piecewise IC)

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

 

[HallsofIvy]

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 C_n 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.)