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Heat equation with a Fourier Series on an infinitely long rod

  • Thread starter nyt
  • Start date
  • #1
nyt
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0

Homework Statement



The heat equation for an infinitely long rod is shown as:

[tex]
\alpha^2 \frac{\partial^2}{\partial x^2}u(x,t) = \frac{\partial}{\partial t}u(x,t)
[/tex]

[tex]
u(0,t) = u(L,t) = 0,\ \forall \ t > 0
[/tex]

[tex]
u(x,0) = sin(\pi x) \ \forall \ 1 < x < 2
[/tex]
[tex]
u(x,0) = 0\ \forall \ otherwise
[/tex]

[tex]
\alpha^2 = 0.1
[/tex]


2. The attempt at a solution

I know that the solution of this problem is a Fourier sine series:

u(x,t)= sum (n=0 to infinity) B_n * sin ((n pi x)/L)

However, I am having problem trying to determine the coefficient Bn:

[tex]
B_n = \frac{2}{L}\int_{0}^{L} sin(\pi x) sin(\frac{n \pi x}{L}) dx
[/tex]

Since the function u(x,0)= sin(pi x) for 10<x<11 I'm not sure if I should aproach this as:
[tex]
B_n = \frac{2}{L}\int_{1}^{2} sin(\pi x) sin(\frac{n \pi x}{L}) dx
[/tex]
where L is a large number

OR

[tex]
B_n = \frac{2}{2}\int_{1}^{2} sin(\pi x) sin(\frac{n \pi x}{2}) dx
[/tex]


Do I go ahead with the second one and ignore that the rod is infinitely long since u(x,0) = 0 for all other values or is this a mistake?

I apologise that some of the equations of the post look like this but I couldn't get my tex brackets to work.
 
Last edited:

Answers and Replies

  • #2
nyt
2
0
I believe I've figured this out.

How can I edit/remove posts?

Thanks.
 
  • #3
1
0
nyt,

im working on the same problem but im not having much success, i was wonder how u solved the problem
 
  • #4
119
0
nyt,

im working on the same problem but im not having much success, i was wonder how u solved the problem
What is your struggle exactly?
 

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