Heat equation with a Fourier Series on an infinitely long rod

nyt
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Homework Statement



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

<br /> \alpha^2 \frac{\partial^2}{\partial x^2}u(x,t) = \frac{\partial}{\partial t}u(x,t) <br />

<br /> u(0,t) = u(L,t) = 0,\ \forall \ t &gt; 0<br />

<br /> u(x,0) = sin(\pi x) \ \forall \ 1 &lt; x &lt; 2<br />
<br /> u(x,0) = 0\ \forall \ otherwise<br />

<br /> \alpha^2 = 0.1<br />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:

<br /> B_n = \frac{2}{L}\int_{0}^{L} sin(\pi x) sin(\frac{n \pi x}{L}) dx <br />

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

OR

<br /> B_n = \frac{2}{2}\int_{1}^{2} sin(\pi x) sin(\frac{n \pi x}{2}) dx<br />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:
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I believe I've figured this out.

How can I edit/remove posts?

Thanks.
 
nyt,

im working on the same problem but I am not having much success, i was wonder how u solved the problem
 
justiz1 said:
nyt,

im working on the same problem but I am not having much success, i was wonder how u solved the problem

What is your struggle exactly?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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