# Solving one dimensonal PDE

I have solve my PDE. But I just want to check my answer with you because I'm not sure whether im right or wrong. So could you just help me out ??

QUESTION :

Given PDE : $U_{xx}=(1/k)(U_{t})$

BCs : $U_{x}(0,t)=0$, $U_{x}(N,t)=0$

IC: $U(x,0)$=x

Assuming we only use separation constant value, $-p^2$

My Solution:

Solving my PDE using separation of variables I got

$$\frac{X''(x)}{X(x)}=(\frac{1}{k})(\frac{T'(t)}{T(t)})=-p^2$$

Solving this I got :

$X(x)=Acospx+Bsinpx$

$T(t)=Ce^{-kp^2t}$

Step 2 : Apply BCs

$U_{x}(0,t)=0$,

$U_{x}(N,t)=0$,

I got $X'(0)=0$ and $X'(N)=0$

Then I got B=0 and p=($\frac{n\pi}{N}$)

Subsituting into equation:

$X(x)$=$Acos(\frac{nx\pi}{N})$

$T(t)$=$Ce^{-k\frac{(n\pi)^2}{N^2}t}$

So U(x,t)=$$\sum_{n=1}^{\infty}B_{n}cos(\frac{nx\pi}{N})e^{-k\frac{(n\pi)^2}{N^2}t} And applying IC and Fourier Series I got B_{n}=\frac{2N}{\pi^2n^2}[(-1)^n-1] So is my answer for PDE and B_{n} correct??? - ## 1 Answer @Maxwell: I think you did a great job here. Only one thing comes to mind: what happened to the n=0 term? It doesn't vanish:$$\begin{align} \lim_{n \rightarrow 0} \frac{1-(-1)^n}{\pi^2 n^2} &= \lim_{n \rightarrow 0} \frac{1-\cos{\pi n}}{\pi^2 n^2} \\ &= \frac{1}{2} \end{align} $$- Thanks Sir for your help and the idea on n=0. Yes, I actually forgot about that thing. I usually start n=0 to n= infinity. Again thanks Sir. Now I have confidence on PDE. – maxwell Feb 1 '13 at 11:13 You're welcome, and keep up the good work! – Ron Gordon Feb 1 '13 at 11:18 So it is like this right what you said U(x,t)=$$\sum_{n=0}^{\infty}\frac{2N}{\pi^2n^2}[$(-1)^n-1$]$cos(\frac{nx\pi}{N‌​})e^{-k\frac{(n\pi)^2}{N^2}t}$ –  maxwell Feb 1 '13 at 11:28
Right, but now you can simplify. I mean, who wants that ghastly $(-1)^n-1$ term in their series anyway? You should rewrite your sum to ignore those zero terms. Think of it as reducing a fraction. –  Ron Gordon Feb 1 '13 at 11:37