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So here is what i have for a solution to the heat eqn.

  1. Jan 20, 2014 #1

    joshmccraney

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    Gold Member

    1. The problem statement, all variables and given/known data
    i am solving the heat equation and so far i know what i have is correct. basically, i am down to this [tex]\sum_{n=0}^{\infty}A_n\cos(\frac{n\pi x}{L})=273+96(2L-4x)[/tex] where all i need is to solve for [itex]A_n[/itex]

    3. The attempt at a solution
    i was thinking about taking advantage of the orthogonality of the cosine function and multiplying both sides by [itex]\cos(\frac{m\pi x}{L})[/itex] and then integrate over the interval [itex][0,L][/itex]. my question is, if [itex]m\neq n[/itex] then i can move this cosine into the sum, integrate term wise, yet the left side equals zero ([itex]m\neq n[/itex]). Thus, [itex]m = n[/itex], and then if i multiply both sides by [itex]\cos(\frac{n\pi x}{L})[/itex] i cannot put this cosine term inside the sum, and thus i have lost the idea of how to solve for [itex]A_n[/itex]. any help/advice is awesome!

    for what it's worth, this is not a class i am in, i'm just doing the problem for fun. thanks for your help!!
     
  2. jcsd
  3. Jan 20, 2014 #2

    pasmith

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    Homework Helper

    This is the standard procedure.

    No. The left hand side is
    [tex]
    \int_0^L \cos(m\pi x/L) \sum_{n=0}^\infty A_n \cos(n \pi x /L)\,dx
    = \sum_{n= 0}^\infty \left(A_n \int_0^L \cos(m \pi x/L) \cos(n\pi x/L)\,dx\right) \\
    = A_m \int_0^L \cos^2(m \pi x/L)\,dx
    [/tex]
    Remember that [itex]n[/itex] varies in the summation, but [itex]m[/itex] is fixed. At some point [itex]n[/itex] must take the value [itex]m[/itex], and this is the only term in the sum which doesn't vanish on integration.
     
    Last edited: Jan 20, 2014
  4. Jan 20, 2014 #3

    joshmccraney

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    Gold Member

    thanks!
     
  5. Jan 20, 2014 #4

    joshmccraney

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    Gold Member

    i have another post in the pde/ode theory part on a similar topic. if youre not too busy, perhaps you could take a look?
     
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