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Application of orthogonality condition

  1. Aug 22, 2013 #1
    1. The problem statement, all variables and given/known data

    I have applied separation of variables to a transient radial heat equation problem.

    T is a function of r and t.

    I have reached the following step:

    2. Relevant equations

    [itex]

    T_2(t,r) = \sum_{m=1}^ \infty c_m e^{-\alpha_2\lambda_m^2t}\left(\dfrac{-Y_0(\lambda_mb)J_0(\lambda_m r)}{J_0(\lambda_mb)}+Y_0(\lambda_mr)\right)

    [/itex]

    3. The attempt at a solution

    I need to find [itex]c_m[/itex] which is usually found using orthogonality condition. This is done by multiplying both sides with [itex]rJ_0(λ_nr)[/itex] and integrating both sides with respect to r. But here in above equation I can't apply the orthogonality condition due to presence of [itex]Y_0(λ_mr)[/itex] on right side of equation.

    How do I apply orthogonality condition in this case?

    I just want to find out the coefficient [itex]c_m[/itex] and somehow get rid of the summation sign.

    Please help!
     
  2. jcsd
  3. Aug 22, 2013 #2

    fzero

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    Have you tried using the definition of ##Y_0## as a limit of a linear combination of ##J_{\pm\alpha}##?
     
  4. Aug 22, 2013 #3
    @fzero:

    This doesn't work because if I put [itex]\alpha = 0[/itex] then result would become infinity.

    02c38389b2a52132a588e3def4871872.png
     
  5. Aug 22, 2013 #4

    fzero

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    ##Y_\alpha(x)## diverges at the origin, but the functions are finite elsewhere. The limit in the definition is an indeterminate form and care must be taken when computing it. However, this expression can be taken as a definition of ##Y_\alpha(x)## and should be useful for your computation.

    You should perform the integration before taking the limit. The resulting limit should involve fairly simple functions.
     
  6. Aug 22, 2013 #5
    Before I integrate, I should have same index for J, i.e. [itex]\alpha[/itex] should be there in both J's. Only that way I can apply orthogonality condition.

    Please help.
     
  7. Aug 22, 2013 #6

    fzero

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    I see, I was hoping that there was an appropriate integral formula to use there. I haven't been able to find a directly useful formula from tables, but I think that the definition of ##Y_\nu(x)## should be a clue. The whole point of defining the Bessel function of the 2nd kind is that for integer order, ##J_{\pm \nu}## are not linearly independent. So ##Y_\nu## is defined to be linearly independent to the ##J_\nu## for integral ##\nu##. Since ##J_0## and ##Y_0## have no common zeros, I believe the integrals you end up with should vanish. I haven't been able to prove this though, sorry.
     
  8. Aug 22, 2013 #7
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