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Coefficients using orthogonality relation

  1. May 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Have a solution for the temperature u(x,t) of a heated rod, now using the orthogonality relation below show that the coefficients [itex]a_n , n = 0,1,2,...[/itex] can be expressed as:

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

    2. Relevant equations

    [tex]\int_{0}^{L} cos\frac{n\pi x}{L} cos\frac{m\pi x}{L} dx = \left\{\begin{matrix}
    L & n=m=0 \\
    L\delta_{nm}/2 & otherwise
    \end{matrix}\right.[/tex]

    [tex]u(x,t)=\frac{a_{0}}{2} + \sum_{n=1}^{inf} a_n cos(\frac{n \pi x}{L})exp[-\alpha (\frac{n\pi}{L})^2 t][/tex]

    [tex]u(x,0) = f(x)[/tex]

    [tex]a_n = \frac{<cos(\frac{n\pi x}{L}),f(x))>}{{\left \| cos(\frac{n\pi x}{L} \right \|}^2}[/tex]



    3. The attempt at a solution



    So I've tried doing this using the relation involving inner products above, I've also tried multiplying both sides of the equation [itex]f(x) = \frac{a_{0}}{2} + \sum_{n=1}^{inf} a_n cos(\frac{n \pi x}{L})[/itex] by [itex]cos(\frac{m \pi x}{L})[/itex] and integrating from 0 to L.

    So going the inner product way, I think my problem is with the value of the norm squared because a similar issue turns up the other way (even though it is essentially the same thing). I get my expression for the norm and note that there are two terms with n so n = m therefore I consider the case where n = 0 therefore n = m = 0 and I get L as my norm squared. Then I consider the case where n > 0 then n = m != 0 , I then get L/2 as my norm squared. This is what doesn't make sense to me. I can get the right answer if I just consider n = m != 0 but I get a second one which isn't the answer if I consider n = m = 0.

    Thank You.
     
  2. jcsd
  3. May 18, 2013 #2
    I think this issue has now been solved, thank you.
     
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