# One Dimensional Diffusion Equation

## Homework Statement

Solve:

$$\frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2}, 0<x<\pi, t>0$$

with initial condition

$$u(x,0)=f(x)=\left\{\begin{array}{cc} 1,& 0\leq x< \pi/2 \\ 0, &\pi/2 \leq x < \pi \end{array}\right$$

and with non-homogeneous boundary conditions

$$u(0,t) = 1, u(\pi,t)=0$$

## The Attempt at a Solution

I've tried this one a couple different ways, I tried separation of variables, and fourier series. I can get a solution (or infinitely many) to the equation, but I can't seem to make them fit the boundary/initial conditions both.

My solution for the fourier series method is:

$$u(x,0)=\frac{A_0}{2} + \sum^{\infty}_{n=1} e^{-kn^2t}(A_n \cos(nx) + B_n \sin(nx)$$

Any ideas how to make this (or any other solution) match both the boundary and initial conditions? Or is the problem incosistent?

mjsd
Homework Helper
it would be easier to see this if you express the function f(x), the initial condition, as a Fourier series.

HallsofIvy
Homework Helper
There is a standard way of handling problems with non-homogeneous boundary conditions: "homogenize" them!

The simple linear function $1- \frac{1}{\pi}x$ satisfies the boundary conditions. Now let $v(x,t)= u(x,t)- 1+ \frac{1}{\pi}x$
It is easy to see that v(x,t) also satisfies
$$\frac{\partial v}{\partial t} = k\frac{\partial^2 v}{\partial x^2}$$
but the boundary conditions for v are v(0, t)= 0, $v(\pi,t)= 0$.
Of course, you have to change the initial condition to
$$v(x,0)=\left\{\begin{array}{cc} \frac{1}{\pi}x,& 0\leq x< \pi/2 \\ \frac{1}{\pi}x-1, &\pi/2 \leq x < \pi \end{array}\right$$

Now you can write v in a purely sine series:
$$v(x,t)=\sum^{\infty}_{n=1} e^{-kn^2t}B_n \sin(nx)$$

Of course, as mjsd said, you will eventually have to write the initial condition as a Fourier series in order to find the Bn