- #1
Gregg
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Homework Statement
I have boundary conditions on my heat equation ## \dot{T}(x,t) = T''(x,t) ##
## T(0,t) = T(L,t), ##
## \frac{\partial T(0,t)}{\partial x} = \frac{\partial T(L,t)}{\partial x} ##
Then
at ## T= 0##
## T(x,0) = 1 ## for ## 0<x<L/4 ##
## T(x,0) = 0 ## for ## L/4<x<L ##
Homework Equations
None.
The Attempt at a Solution
I have done the usual separation of variables, then getting two ODEs equal to a constant. I've called it ## \alpha^2 ## I have eliminated the possibility that ##\alpha^2 = 0. ## Will this always happen?
Now do I need to consider ## \alpha^2 < 0 ## and ## \alpha^2 >0 ## rather than just ##\alpha^2\ne 0 ##?
I find the fact that the boundary conditions run over an interval to be off putting. Am I to use the differential boundary conditions instead for now? I have deduced that:
##T(x,t) = c_1(e^{\alpha x} - e^{-\alpha x})c_2 e^{\alpha^2 t} ##
Of course if ##T_i(x,t) ## is a solution then ##T(x,t) = \displaystyle \sum_i T_i(x,t) ##
I have found that
## T(x,t) = \sum_{n=1}^{\infty} C_n \sin ({n \pi x \over l}) e^{n^2 \pi^2 t \over l^2} ##
So I need to find the Fourier coefficients of that function that is constant on the two different intervals. This is where I get stuck.
##C_n = \frac{1}{l} \int_0^{2l} f(x) \sin ({n \pi x \over l}) dx = \frac{1}{l} \int_0^{l/4} 0\cdot dx + \int_{l/4}^{l} \sin ({n \pi x \over l}) dx+ \frac{1}{l} \int_{l}^{5l/4} 0\cdot dx + \int_{5l/4}^{2l} \sin ({n \pi x \over l}) dx ##
I don't think this the correct form for the coefficient:
## C_n = \frac{1}{n \pi} \left[ -(-1)^n + 1 +0 + \cos({4 n \pi \over 5})\right] ##
So is it just integration over ## [0,l] ## ?
This would give
## T(x,t) = \sum_{n=1}^{\infty} 2 \sin ({(2n-1) \pi x \over l}) e^{n^2 \pi^2 t \over l^2} ##
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