# Question on zeros of a Bessel function.

1. Dec 28, 2009

### 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

2. Dec 28, 2009

### Astronuc

Staff Emeritus
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 Jo, for solid cylindrical geometry or Yo 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

Jn(z) and Yn(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 Cn are constants (or coefficients), and xn are zeros. Multiplying both sides by $$J_0(\frac{x_mz}{R})$$ (xm 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 Jn for instance.

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

Last edited: Dec 29, 2009
3. Dec 29, 2009

### 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

4. Dec 29, 2009

### Astronuc

Staff Emeritus
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., $$0\,\leq\,x\,\leq\,1$$

5. Dec 30, 2009

### 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.

6. Dec 30, 2009

### Astronuc

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

then the zeros of J1/2(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 J1/2(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 Jν

7. Dec 30, 2009

### yungman

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

8. Dec 30, 2009

### Astronuc

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

9. Dec 30, 2009

### yungman

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

Sincerely

Alan

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