# Partial Differential Equations

• Sep 2nd 2009, 06:08 AM
sssouljah
Partial Differential Equations
• Sep 2nd 2009, 11:27 AM
Coomast
Quote:

Originally Posted by sssouljah

This is a nice exercise on the Fourier series solution of PDE. Do you know this theory because it is absolutely necessary if you want to find the solution. If so, can you explain where exactly you are stuck?

Coomast
• Sep 2nd 2009, 04:58 PM
luobo
Quote:

Originally Posted by sssouljah

$
u(x,t)=X(x)\;T(t)
$

$
u_t=XT_t,\ \ \ \ u_{xx}=TX_{xx}, \ \ \ \ u_t-Du_{xx}-\alpha u=0
$

$
\frac{T_t}{T}-D\frac{X_{xx}}{X}-\alpha = 0
$

$
D\frac{X_{xx}}{X}=\frac{T_t}{T}-\alpha=-C\text{ (constant)}
$

$
u(x,t)=e^{(\alpha-C)t}\left[A\cos\left(\sqrt{\frac{C}{D}}\;x\right)+B\sin\left (\sqrt{\frac{C}{D}}\;x\right)\right]
$

$
u_x(0,t)=0 \Rightarrow B=0
$

$
u_x(L,t)=0 \Rightarrow \sin\left(\sqrt{\frac{C}{D}}\;L\right) = 0 \Rightarrow C_n=\frac{n^2\pi^2D}{L^2}, \ n=1, 2, 3, ...
$

$
u(x,0)=\cos\left(\frac{\pi x}{L}\right)+\cos\left(\frac{2\pi x}{L}\right) \Rightarrow n=1, 2, \ A_1=A_2 =1
$

- $
\boxed{u(x,t)=e^{\left(\alpha-\tfrac{\pi^2D}{L^2}\right)t}\cos\left(\frac{\pi x}{L}\;\right)+e^{\left(\alpha-\tfrac{4\pi^2}{L^2}\right)t}\cos\left(\frac{2\pi x}{L}\;\right)}
$