10.8: Fourier-Bessel Series
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So how can we determine in general the coefficients in the Fourier-Bessel series
f(ρ)=∞∑j=1CjJν(αjρ)?
The corresponding self-adjoint version of Bessel’s equation is easily found to be (with Rj(ρ)=Jν(αjρ))
(ρR′j)′+(α2jρ−ν2ρ)Rj=0.
Where we assume that f and R satisfy the boundary condition
b1f(c)+b2f′(c)=0b1Rj(c)+b2R′j(c)=0
From Sturm-Liouville theory we do know that ∫c0ρJν(αiρ)Jν(αjρ)=0if i≠j,
Let us use the self-adjoint form of the equation, and multiply with 2ρR′, and integrate over ρ from 0 to c,
∫c0[(ρR′j)′+(α2jρ−ν2ρ)Rj]2ρR′jdρ=0.
This can be brought to the form (integrate the first term by parts, bring the other two terms to the right-hand side)
∫c0ddρ(ρR′j)2dρ=2ν2∫c0RjR′jdρ−2α2j∫c0ρ2RjR′jdρ(ρR′j)2|c0=ν2R2j|c0−2α2j∫c0ρ2RjR′jdρ.
The last integral can now be done by parts: 2∫c0ρ2RjR′jdρ=−2∫c0ρR2jdρ+ρR2j|c0.
So we finally conclude that
2α2j∫c0ρR2jdρ=[(α2jρ2−ν2)R2j+(ρR′j)2|c0.
In order to make life not too complicated we shall only look at boundary conditions where f(c)=R(c)=0. The other cases (mixed or purely f′(c)=0) go very similar! Using the fact that Rj(r)=Jν(αjρ), we find R′j=αjJ′ν(αjρ).
2α2j∫c0ρ2R2jdρ=[(ραjJ′ν(αjρ))2|c0=c2α2j(J′ν(αjc))2=c2α2j(ναjcJν(αjc)−Jν+1(αjc))2=c2α2j(Jν+1(αjc))2,
We thus finally have the result ∫c0ρ2R2jdρ=c22J2ν+1(αjc).
Consider the function
f(x)={x30<x<100x>10
Expand this function in a Fourier-Bessel series using J3.
Solution
From our definitions we find that
f(x)=∞∑j=1AjJ3(αjx),
with
Aj=2100J4(10αj)2∫100x3J3(αjx)dx=2100J4(10αj)21α5j∫10αj0s4J3(s)ds=2100J4(10αj)21α5j(10αj)4J4(10αj)ds=200αjJ4(10αj).
Using αj=…, we find that the first five values of Aj are 1050.95,−821.503,703.991,−627.577,572.301. The first five partial sums are plotted in Fig. 10.8.1.
