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Orthogonality condition for disimilar Bessel functions

  1. Aug 22, 2013 #1
    As per orthogonality condition this equation is valid:

    [itex]\int_0^b xJ_0(\lambda_nx)J_0(\lambda_mx)dx = 0[/itex] for [itex] m\not=n[/itex]

    I want to know the outcome of the following:

    [itex]\int_0^b xJ_0(\lambda_nx)Y_0(\lambda_mx)dx = 0[/itex]

    for two cases:
    [itex] m\not=n[/itex]
    [itex] m=n[/itex]
     
  2. jcsd
  3. Aug 23, 2013 #2

    fzero

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    I didn't have access to Gradshteyn and Ryzhik earlier. Integral 5.54 claims that, for ##X,Z## any of the Bessel functions (##J,Y,\ldots##),

    $$\begin{split} \int x X_p(\alpha x) Z_p(\beta x) & = \frac{x}{\alpha^2-\beta^2} \left( \alpha X_{p+1}(\alpha x) Z_p(\beta x) -\beta X_{p}(\alpha x) Z_{p+1}(\beta x) \right) \\
    & = \frac{x}{\alpha^2-\beta^2} \left( \beta X_{p}(\alpha x) Z_{p-1}(\beta x) -\alpha X_{p-1}(\alpha x) Z_{p}(\beta x) \right) .\end{split} $$

    You should be able to work out your integral by setting the appropriate integration limits and using other identities as necessary.
     
  4. Aug 23, 2013 #3
    Thanks alot!

    I can now solve the problem. But there is still one more thing left. Is it safe to assume that above integral will be equal to zero in case


    [itex] \int x X_p(\alpha x) Z_q(\beta x) = ?

    \\

    when p\not = q [/itex]
     
  5. Aug 23, 2013 #4

    fzero

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    I am unable to find precisely that integral. The integral I gave above can be derived from the expressions for the derivative of the Bessel functions (which involves specific factors of ##x##), using integration by parts. The intermediate result that applies to ##p\neq q## is

    attachment.php?attachmentid=61093&stc=1&d=1377277967.png

    So the part involving the integral you want involves ##x^{-1}## instead of ##x##.

    However, the universality of the formula seems to agree with my suggestion that the orthogonality relations of ##J## with ##Y## are precisely analogous to ##J## with ##J##. I urge you to try to get your hands on a copy of Gradshteyn and Rhyzik, at least from a library, since you might note a useful result that I have overlooked.
     

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  6. Aug 24, 2013 #5
    I managed to get Gradshteyn and Ryzhik today. The book is really useful.


    Now that we have got the formula.....

    [itex]\begin{split} \int x X_p(\alpha x) Z_p(\beta x) & = \frac{x}{\alpha^2-\beta^2} \left( \alpha X_{p+1}(\alpha x) Z_p(\beta x) -\beta X_{p}(\alpha x) Z_{p+1}(\beta x) \right) \\
    & = \frac{x}{\alpha^2-\beta^2} \left( \beta X_{p}(\alpha x) Z_{p-1}(\beta x) -\alpha X_{p-1}(\alpha x) Z_{p}(\beta x) \right) .\end{split}[/itex]

    ...... I would like to evaluate the constant [itex]c_n[/itex] for the following equation:

    [itex] T = \sum^{∞}_{n=1} c_n e^{-\alpha \lambda^{2}_{n}t}(\frac{-Y_0(\lambda_nb)J_0(\lambda_nr)}{J_0(\lambda_nb)} +Y_0(\lambda_nr))[/itex]

    Applying initial condition leads to:
    [itex]T(r,0) = X [/itex]

    [itex] X = \sum^{∞}_{n=1} c_n (\frac{-Y_0(\lambda_nb)J_0(\lambda_nr)}{J_0(\lambda_nb)} +Y_0(\lambda_nr))[/itex]

    Multiply both sides by [itex]rJ_0(\lambda_mr)[/itex] and integrate from r=a to r=b.

    [itex] ∫^{b}_{a}rJ_0(\lambda_mr)Xdr = \sum^{∞}_{n=1} c_n (∫^{b}_{a}\frac{-Y_0(\lambda_nb)rJ_0(\lambda_mr)J_0(\lambda_nr)}{J_0(\lambda_nb)}dr +∫^{b}_{a}rJ_0(\lambda_mr)Y_0(\lambda_nr)dr)[/itex]

    The first term on the right side of equation will be taken care of by using the orthogonality condition:

    [itex]\int_a^b xJ_0(\lambda_nx)J_0(\lambda_mx)dx = 0[/itex]
    for [itex]n\not=m[/itex]
    So for first term on right side all integrals of the series would vanish except for the case n =m.

    The second term on right side will be evaluated using the formula that you mentioned in your post.

    Since for our case, the below equation is not equal to zero:

    [itex]\begin{split} \int x X_p(\alpha x) Z_p(\beta x) & = \frac{x}{\alpha^2-\beta^2} \left( \alpha X_{p+1}(\alpha x) Z_p(\beta x) -\beta X_{p}(\alpha x) Z_{p+1}(\beta x) \right) \\
    & = \frac{x}{\alpha^2-\beta^2} \left( \beta X_{p}(\alpha x) Z_{p-1}(\beta x) -\alpha X_{p-1}(\alpha x) Z_{p}(\beta x) \right) .\end{split}\not=0[/itex]


    .... the summation sign would stay intact and [itex]c_n[/itex] cannot be evaluated.


    Would you please help me get rid of the summation sign. I want to evaluate [itex]c_n[/itex].
     
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