Verifying the integral form of the Bessel equation by substitution

[tjackson3]

Homework Statement

The following is an integral form of the Bessel equation of order n:

[tex]J_n(x) = \frac{1}{\pi}\int_0^{\pi}\ \cos(x\sin(t)-nt)\ dt[/tex]

Show by substitution that this satisfies the Bessel equation of order n.

Homework Equations

Bessel equation of order n: [itex]x^2y'' + xy' + (x^2-n^2)y = 0[/itex]

The Attempt at a Solution

I tried simply plugging this into the Bessel equation, but it didn't really help. You end up with:

[tex]\int_0^{\pi} \left[-x^2\sin^2 t\cos(x\sin t - nt) - x\sin t\sin(x\sin t - nt) + (x^2-n^2)\cos(x\sin t - nt)\right]\ dt = 0[/tex]

This is all assuming you can interchange the order of integration and differentiation, but it seems like you should be able to, since the integration is wrt t, while the differentiation is wrt x.

Thanks!

 

[mighty2000]

Thank you for the interesting question. To show that the integral form satisfies the Bessel equation, we need to use the fundamental theorem of calculus and differentiate both sides of the integral equation with respect to x. This gives us:

\frac{d}{dx} J_n(x) = \frac{d}{dx} \left(\frac{1}{\pi}\int_0^{\pi}\ \cos(x\sin(t)-nt)\ dt\right)

Using the fundamental theorem of calculus, we can interchange the order of differentiation and integration, giving us:

\frac{d}{dx} J_n(x) = \frac{1}{\pi}\int_0^{\pi}\ \frac{d}{dx} \cos(x\sin(t)-nt)\ dt

Now, using the chain rule, we can evaluate the derivative of the cosine function and simplify to get:

\frac{d}{dx} J_n(x) = \frac{1}{\pi}\int_0^{\pi}\ -\sin(x\sin(t)-nt)\cdot \frac{d}{dx} (x\sin(t)-nt)\ dt

= \frac{1}{\pi}\int_0^{\pi}\ -\sin(x\sin(t)-nt)\cdot \sin(t)\ dt

= \frac{-1}{\pi}\int_0^{\pi}\ \sin^2(t)\cos(x\sin(t)-nt)\ dt

= -J_n(x)

Therefore, we have shown that the integral form of the Bessel equation satisfies the Bessel equation of order n. I hope this helps clarify things for you. Let me know if you have any further questions. Keep up the good work in your studies!


, Scientist.