Question on zeros of a Bessel function.

[yungman]

A typical BVP of Bessel function is approximation of f(x) by a Bessel series expansion with y(0)=0 and y(a)=0, 0<x<a.

For example if we use J_{\frac{1}{2}} to approximate f(x) on 0<x<1. Part of the answer contain

J_{\frac{1}{2}}=\sqrt{\frac{2}{\pi x}}sin(\alpha_{j}x), j=1,2,3...

This will give \alpha_{j}=\pi,2\pi,3\pi...for j=1,2,3...


But in the books, they always have a table of \alpha_{j} for each order of J. For example for J_{1},\alpha_{1}=3.83171,\alpha_{2}=7.01559,\alpha_{3}=10.1735 etc. This mean \alpha_{j} is a constant for each order of the Bessel series. THis mean the zeros are a constant on the x-axis.

You see the above two example is contradicting each other. The series expansion show \alpha_{j} depends on the boundaries, the second show \alpha_{j} are constants!

Please tell me what did I miss.

Thanks
Alan

 

[Astronuc]

Bessel's equation is:
\frac{d^2\phi}{dx^2}\,+\,\frac{1}{x}\frac{d\phi}{dx}\,+\big(\alpha^2\,-\,\frac{n^2}{x^2}\big)\,\phi\,=\,0 ,

where \alpha and n are constants. It's encountered when doing problems in polar (2D) or cylindrical (3D) where the partial differential equations are separable, and the dimension x is actual the radial dimension (r). They represent standing wave patterns in circular memberanes, e.g., drums and neutron diffusion in homogenous systems. There can be multiple zeros and negative values of J for membranes where J represents the oscillation of the membrane. In the case of neutron diffusion solutions, there cannot be negative fluxes or quantities of neutrons, therefore the solution is limited to J_o, for solid cylindrical geometry or Y_o for annular geometry.
http://physics.usask.ca/~hirose/ep225/animation/drum/anim-drum.htm

If n is not an integer or zero, the two independent solutions to the equation are given by

\phi\,=\, J_n(\alpha{x}) and \phi\,=\, J_{-n}(\alpha{x})

n = 1/2 is a special case.

If n is an integer or zero, the two functions are NOT independent and the two solutions are written

\phi\,=\, J_n(\alpha{x}) and \phi\,=\, Y_n(\alpha{x}), where

J_n(z) and Y_n(z) are ordinary Bessel functions of the first and second kind.

The coefficients for the solutions would depend on the boundary conditions.

The boundary conditions are used to solving for coefficients of series where are well-behaved function is expanded in a series of Bessel functions.

Let f(z)\,=\,\sum_{n=1}^\infty\,C_nJ_0(\frac{x_nz}{R}), \,0\,\leq\,z\,\leq\,R

where C_n are constants (or coefficients), and x_n are zeros. Multiplying both sides by J_0(\frac{x_mz}{R}) (x_m are zeros) and integrating from 0 to R, yields

C_n\,=\,\frac{2}{R^2J^2_1(x_n)}\int^R_0\,f(z)J_0(\frac{x_nz}{R})z dz


Then the question is which intervals are appropriate for expanding f(z). In many, probably most cases, it will be the first interval between 0 and the first zero, in the case of J_n for instance.

http://mathworld.wolfram.com/BesselFunctionZeros.html
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html
http://mathworld.wolfram.com/BesselFunctionoftheSecondKind.html

 

[yungman]

Thanks for the reply. My main question is whether \alpha_{j} is a constant or not in the case I presented above:

J_{\frac{1}{2}}(\lambda_{j}x)=\sqrt{\frac{2}{\pi x}}sin(\lambda x)=0 for 0<x<a=1


\Rightarrow sin(\frac{\alpha_{j}x}{a})=0 \Rightarrow \alpha_{j}=j\pi at x=a.


After I post this thread, I was still thinking about this and I think I start to understand this:

I want to confirm that for J_{\frac{1}{2}}(\lambda_{j}x), \lambda_{j} is not a constant, because \lambda_{j}=\frac{\alpha_{j}}{a}

but \alpha_{j} is always a constant and \alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}...=\pi,2\pi,3\pi,4\pi... for j=1,2,3...no mater what the value a is.

Please let me know I am right or not.

Many thanks

Alan

 

[Astronuc]

I believe it is customary to apply a differential equation or partial differential equation to a particular interval, as well as the boundary conditions. In a physical system, the equation and boundary conditions reflect the physics of the problem, i.e. the differential equation reflects the behavior (or something about the behavior) of the physical system.

In the case one cited, with n = 1/2 and λ would infer the differential equation

\frac{d^2\phi}{dx^2}\,+\,\frac{1}{x}\frac{d\phi}{d x}\,+\big(\lambda^2\,-\,\frac{1}{4x^2}\big)\,\phi\,=\,0 ,

and one would apply that to 0\,\leq\,x\,\leq\,1 or if one was modeling over an interval 0 to a, one would choose a transformation x = z/a, where z = 0 for x = 0, and x = 1 for z = a.

Certainly to take advantage of orthogonality, one must select the appropriate interval, e.g., 0\,\leq\,x\,\leq\,1

 

[yungman]

Thanks for the reply, I understand all that, I just want to verify:

For J_{\frac{1}{2}}(\alpha_{j} x)

\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}...=\pi,2\pi,3\pi,4\pi... for j=1,2,3...no mater what a is as long as the boundary condition y(a)=0.

 

[Astronuc]

Since J_{\frac{1}{2}}(\lambda_{j}x)=\sqrt{\frac{2}{\pi x}}sin(\lambda x)

then the zeros of J_1/2(x) are simply the zeros of sin x. If we use sin x to expand a function, we take 0\,\leq\,x\,\leq\,n\pi, but if use sin (nπx) the we use 0\,\leq\,x\,\leq\,1

To use the property of orthgonality, one must choose the appropriate interval! Review the expansion of functions in terms of sines and cosines.

In the case of J_1/2(x), one would use 0\,\leq\,x\,\leq\,\alpha_j or if one wants to use the argument \alpha_j x, then use the interval 0\,\leq\,x\,\leq\,1.


Wolfram's Mathworld uses the example for orthogonality

\int_0^a J_\nu\large(a_{\nu m}\frac{\rho}{a}\large) J_\nu\large(a_{\nu n}\frac{\rho}{a}\large) \rho d\rho\,=\,\frac{1}{2}a^2\,[J_{\nu+1}(a_{\mu m})]^2 \delta_{mn}

where a_{\nu m} is the mth zero of J_ν

 

[yungman]

Since J_{\frac{1}{2}}(\lambda_{j}x)=\sqrt{\frac{2}{\pi x}}sin(\lambda x)

then the zeros of J_1/2(x) are simply the zeros of sin x. [/itex]

So \lambda_{j}=\frac{\alpha_{j}}{a}} \Rightarrow \alpha_{j}=\pi,2\pi,3\pi...for j=0,1,2,3...Means it is a constant.

That's all I want to verify.

Thanks

 

[Astronuc]

So \lambda_{j}=\frac{\alpha_{j}}{a}} \Rightarrow \alpha_{j}=\pi,2\pi,3\pi...for j=0,1,2,3...Means it is a constant.

That's all I want to verify.

Thanks

Yes, one would use α_jx, with x = x[0,1], or α_jx/a, with x = x[0,a].

 

[yungman]

Yes, one would use α_jx, with x = x[0,1], or α_jx/a, with x = x[0,a].

Thanks a million, that is music in my ears, now I can move on.

Sincerely

Alan