Solving the Heat Equation: Investigating an Error

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catcherintherye
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Homework Statement



I am solving the heat equation [tex]1/a^2\theta_t = \theta_x[/tex]

boundary conditions are [tex]\theta(0,t) = \theta(L,t) = 0 t > 0[/tex]

initial conditions are [tex]\theta(x,0) = T_0sin(x\pi/L)[/tex]


now I have derived the steady solution to be 0 and I have derived that the general solution will be of the form

[tex]\theta(x,t) = \sum_{n=1}^\infty\B_n sin\frac{n\pix}{L}exp\frac{-a^2n^\pi^2}{L}t[/tex]

I am next required to determine the sequence [tex]{B_n} \\<br /> <br /> \mbox{now usually I would proceed in the following manner, using the initial conditions}\\<br /> \theta(x,0) = \sum_{n=1}^\infty B_n sinn\pix/L = T_0sin(\frac{\pix}{L})[/tex]

I would then multiply both sides by [tex]\int_{0}^{L}sin\frac{mx\pi}{L}[/tex]

but instead i decided first to separted term [tex]B_1[/tex] and cancel term sin(mx\pi/L)

so I proceeded using the cancellation by the orthogonality conditions and derived the following

[tex]B_mL/2 = (T_0 -B_1)\int_{0}^{L}sin\frac{mx\pi}{L}\\<br /> <br /> = \frac{(B_1 -T_0)}{m\pix}\left[cosmx\pi/L\right]_{0}^{L}\\<br /> = \frac{B_1-T_0}{m\pi}\left[(-1)^m -1][/tex]

you see the problem is I end up with [tex]B_1[/tex] undetermined
[tex]B_1 = \frac{-4T_0}{m\pi -4}[/tex]

where did I go wrong/ what should I have done differently??:frown:





Homework Equations





The Attempt at a Solution

 
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You didn't do anything wrong except that you guarenteed you would not be able to determine B1 when you "decided first to separted term [itex]B_1[/itex] and cancel term sin(mx\pi/L)"!

There are many different ways to do this but I would probably expand [itex]T_0 sin(\pi x/L)[/itex] in a Fourier sine series in [itex]\pi x/L[/itex] and equate coefficients.
 
mmm...I am not familiar with this method, you see the reason I took out term B1 was because I knew it would allow me to proceed by multplying by

[tex]sin\frac{m\pix}{L}[/tex]and then integrating from 0 to L. Without taking out term B1 this would eventually lead me to do the following integral
[tex]L/2B_m = T_0\int_{0}^{L} sin(\frac{m\pix}{L})sin(\frac{pix}{L})dx[/tex]

is this also a valid method of solution and if so How do I do the integral? Do I have to use some sort of trig identity
 
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