on Bessel function's orthogonality

by samuelandjw
Tags: bessel, function, orthogonality
samuelandjw is offline
Dec13-12, 10:16 PM
P: 23
Use the orthogonality relation of Bessel function to argue whether the following two integrals are zero or not:
[itex]\displaystyle\int_0^1J_1(k_1x)J_1(k_2x)dx[/itex], where [tex]k_1,k_2[/tex] are two distinct zeros of Bessel function of order 1.

The textbook we are using is Boas's Mathematical Methods in the Physical Sciences, so the formulas that are available to us are a set of recursion relations of Bessel function, and the orthogonality relations between Bessel function of the same order: http://upload.wikimedia.org/math/c/d...024cce03fd.png.

For the first integral, the two Bessel functions are of different order, and there is no zeros in the arguments of the two functions, so I have no idea how to link the first integral to the orthogonality relation of Bessel functions.

For the second integral, my argument is that since [itex]\displaystyle\int_0^1J_1(k_1x)xJ_1(k_2x)dx[/itex] is zero, the 2nd integral (note that there is no x between the two Bessel function) cannot be zero. I think this argument is quite weak. Would anyone give me a better argument?

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lurflurf is offline
Dec13-12, 11:25 PM
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The first one integral(positive function)=positive number
samuelandjw is offline
Dec13-12, 11:41 PM
P: 23
Quote Quote by lurflurf View Post
The first one integral(positive function)=positive number
Thanks for your reply. We can surely say that [tex]J_1(x),J_2(x)[/tex] are positive functions between x=0 and x=1 because the first nontrivial zeros are larger than 1. One has to somehow use the information of the location of zeros to reach this conclusion. Suppose we don't have this information, is it still possible to argue that the integral is non-zero?

vijay0 is offline
Dec14-12, 12:27 AM
P: 2

on Bessel function's orthogonality

See if this will help: http://math.stackexchange.com/questi...ssel-functions
samuelandjw is offline
Dec15-12, 01:33 AM
P: 23
Quote Quote by vijay0 View Post
Thanks for your reply. That post does not seem very useful though.

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