How to find the coefficients of this expansion?

In summary, the conversation discusses solving a differential equation and finding the orthogonality of functions to determine the value of B. The source mentioned is Jackson Classical Electrodynamics 3rd Edition, which guarantees the orthogonality of functions. An alternative expansion using a series of functions is also mentioned.
  • #1
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I have solved a differential equation whose solutions is $$u = B + \sum_{n=1} C_{n }e^{-\lambda_{n}² q² t} J_{0}(\lambda_{n}r)$$

Where ##(\lambda_{n}r)## is such that ##J_{0}'(\lambda_{n}a) = 0##. So i should now try to satisfy the condition that, at t=0, u = ##f(r)##.

The problem is that i don't know what is the orthogonality here. If ##\lambda_{n}## were such that ##J_{0}(\lambda_{n}a)=0##, i would use the normal orthogonaltiy generally used, namely ##\int_{0}^{a} r J_{0}(\lambda_{n}r)J_{0}(\lambda_{q}r)##

But this dosen't work here, since ##\lambda_{q}a## is not a zero of J0, but it is a zero in fact of its derivative.

Not just it, what orthogornality i would use to find the B?
 
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  • #2
Believe it or not, orthogonality, exactly the way you wrote it, is already guaranteed. My source is Jackson Classical Electrodynamics 3rd Edition, bottom of page 115:

But it will be noted that an alternative expansion is possible in a series of functions ##\sqrt{\rho} J_\nu (y_{\nu n}\rho / a)## where ##y_{\nu n}## is the ##n##th root of the equation ##[dJ_\nu(x)]/dx = 0##. The reason is that, in proving the orthogonality of functions, all that is demanded is that thte quantity ##[\rho J_\nu (k\rho) (d/d\rho) J_\nu (k' \rho) - \rho J_\nu (k' \rho) (d/d\rho) J_\nu (k\rho)]## vanish at the endpoints ##\rho = 0## and ##\rho = a##.
 

FAQ: How to find the coefficients of this expansion?

What is the purpose of finding the coefficients of an expansion?

The coefficients of an expansion are used to represent the relative importance of each term in the expansion. They also help in simplifying complex expressions and solving equations.

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