# One Dimensional Diffusion Equation

1. Mar 19, 2007

### NeoDevin

1. The problem statement, all variables and given/known data
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

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

3. 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?

2. Mar 20, 2007

### mjsd

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

3. Mar 20, 2007

### HallsofIvy

Staff Emeritus
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