# More PDE fun

1. Dec 4, 2005

### stunner5000pt

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

$u(0,t) = u(\pi,t) = 0$
$u(x,0) = x (\pi - x)$

OK i know the boring part of getting u(x,t) = X(x) T(t)
the infinite series part is hard part
the coefficient $$c_{n} = \frac{2}{\pi} \int_{0}^{\pi} \sin(n(\pi -x)) \sin(nx) = \frac{-2}{\pi} \left(\frac{-1 + (-1)^n}{n}\right) \sin(nx)$$
the -1^n is from the Cos n pi term taht would coem from the integration

thus n must be odd
$$c_{n} = \frac{4}{\pi} \frac{1}{2n-1} \sin(nx)$$
is this good so far?

2. Dec 4, 2005

### harsh

I am confused with your IC's. Is it u(x,0) = x(pi - x) or sin(n(pi - x))?

3. Dec 4, 2005

### stunner5000pt

x(pi -x ) is the initial condition
and i think i have mistaken this for X(pi -x)
wich is not the same thing

4. Dec 4, 2005

### harsh

Ok, so if x(pi -x) is your IC, then you will have to integrals to do. xpi* sin(nx)
and x^2 sin (nx). Its a bit more involved than the last one.

- harsh