Finding 2 solutions of this Bessel's function using a power series

[happyparticle]

Homework Statement

find 2 solutions of this Bessel's function using a power series.

Relevant Equations

$x^2 d^2y/dx^2 + x dy/dx + (x^2 -9/4)y = 0$

I have to find 2 solutions of this Bessel's function using a power series.

$x^2 d^2y/dx^2 + x dy/dx+ (x^2 -9/4)y = 0$

I'm using Frobenius method.

What I did so far

I put the function in the standard form and we have a singularity at x=0. Then using $y(x) = (x-x_0)^p \sum(a_n)(x-x_0)^n$

I got $\sum[(n+p)(n+p-1)a_n x^{n+p}] + \sum[(n+p)a_n x^{n+p}] + \sum[a_n x^{n+p+2}] - 9/4 \sum[a_n x^{n+p}] = 0$

Thus, the smallest power for n=0 is $x^p$, so I found $p = \pm 3/2$

To have $x^{m+p}$

I got $(m+p)(m+p-1)a_m + (m+p)a_m + a_{m-2} - 9/4 a_m = 0=> a_{m-2} = [(m+p(m+p-1) + (m+p) -9/4)] a_m$

For p = 3/2 I have $a_{m-2} = -m(m+3)a_m$

So far I'm stuck here, since I can't find a series with this above and then by plugging the solution in the standard form I should get some cos and sin from the hint. I guess I made an mistake somewhere but I don't know where.

Any help will be really appreciate.

 

[FactChecker]

Is that a typo in the equation? Do you mean the second term to be $x (dy/dx)$?

 

[happyparticle]

Is that a typo in the equation? Do you mean the second term to be $x dy/dx$?

yes, my bad.

 

[pasmith]

Homework Statement:: find 2 solutions of this Bessel's function using a power series.
Relevant Equations:: $x^2 d^2y/dx^2 + x dy/dx + (x^2 -9/4)y = 0$

I have to find 2 solutions of this Bessel's function using a power series.

$x^2 d^2y/dx^2 + x dy/dx+ (x^2 -9/4)y = 0$

I'm using Frobenius method.

What I did so far

I put the function in the standard form and we have a singularity at x=0. Then using $y(x) = (x-x_0)^p \sum(a_n)(x-x_0)^n$

I got $\sum[(n+p)(n+p-1)a_n x^{n+p}] + \sum[(n+p)a_n x^{n+p}] + \sum[a_n x^{n+p+2}] - 9/4 \sum[a_n x^{n+p}] = 0$

Thus, the smallest power for n=0 is $x^p$, so I found $p = \pm 3/2$

What do you get if you set n = 1? Does that restrict a_1, or are you free to choose it as you wish?

For n \geq 2 I agree that you have <br /> a_{n}\left(n(n+2p)\right) + a_{n-2} = 0 where p = \pm\frac32.

The form of the recurrence allows you to treat the even terms n = 2k and the odd terms n = 2k+1 separately. This leads you to linear recurrences of the form <br /> b_{k} = f(k)b_{k-1} where b_{k} = a_{2k} or a_{2k+1} as appropriate. In solving these it may assist you to note that <br /> \prod_{r=0}^{k-1}(r + p) = \frac{\Gamma(k+p)}{\Gamma(p)} where \Gamma is the Gamma function.

So far I'm stuck here, since I can't find a series with this above and then by plugging the solution in the standard form I should get some cos and sin from the hint. I guess I made an mistake somewhere but I don't know where.

Any help will be really appreciate.

Please post the hint.

 

[happyparticle]

Please post the hint.

Using the standard form and my solutions I should get
$\sqrt{\frac{2}{\pi x}} (\frac{sin x}{x} - cos x)$ and $-\sqrt{\frac{2}{\pi x}} (\frac{cos x}{x} + sin x)$

I'm not sure to understand how you get $b_{k} = f(k)b_{k-1}$

I though by adding some terms to $a_{m-2} = -m(m+3)a_m$ I could find the serie.

 

[pasmith]

Using the standard form and my solutions I should get
$\sqrt{\frac{2}{\pi x}} (\frac{sin x}{x} - cos x)$ and $-\sqrt{\frac{2}{\pi x}} (\frac{cos x}{x} + sin x)$

The general solution of <br /> x^2 \frac{d^2y}{dx^2} + x\frac{dy}{dx} + \left(x^2 - \frac94\right)y = 0<br /> is y(x) = AJ_{3/2}(x) + BJ_{-3/2}(x). Now it happens that \begin{split}<br /> j_1(x) &amp;= \frac{\sin x}{x^2} - \frac{\cos x}{x} = \sqrt{\frac{\pi}{2x}} J_{3/2}(x) \\<br /> y_1(x) &amp;= -\frac{\cos x}{x^2} - \frac{\sin x}{x} = \sqrt{\frac{\pi}{2x}} J_{-3/2}(x)<br /> \end{split} where j_1 and y_1 are spherical bessel functions, which are the solutions of <br /> x^2 \frac{d^2y}{dx^2} + x\frac{dy}{dx} + (x^2 - 2)y = 0. So the solutions given in your hint are <br /> \begin{split}<br /> \sqrt{\frac{\pi x}{2}}j_1(x) &amp;= \frac{\pi}{2} J_{3/2}(x),\mbox{ and} \\<br /> \sqrt{\frac{\pi x}{2}}y_1(x) &amp;= \frac{\pi}{2} J_{-3/2}(x).\end{split}

You have yet to find the series for J_{\pm3/2}. Once you have found them, you can attempt to recognise them as the series of the functions given in the hint. To that end, it is best to use <br /> \prod_{k=0}^{n-1} \left(k + \frac32 \right) = \frac{1}{2^n} \prod_{k=0}^{n-1} (2k + 3) rather than the expresssion involving Gamma functions.

Go back and look again at the result of applying Frobeniu's method. Setting n = 0 told you the possible values of p, because a_0 is constrained to be non-zero. What happens if you set n = 1? Does that tell you anything about the possible values of a_1?

Have you not encountered first order linear recurrences before? It surprises me - although perhaps it no longer should - that they are not regularly taught alongside series solution of differential equations, since you can't really do one without the other.

 

[happyparticle]

I'm following the method used in the textbook mathematics for physicist by lea.
Here's the example I'm following. Maybe you'll understand what I'm trying to do.
If I'm correct I found the recursion relation and then I plugged p = 3/2.

I found $a_m = -\frac{1}{(m)(m+3)} \cdot -\frac{a_{m-4}}{(m-2)(m+2)}$
$a_m = \frac{-1^{m/2 + 1}}{(m+2)!(m+5)!}a_0$
Should I find the odd solution and add to the even solution to have the full solution for p = 3/2 ?

 

[happyparticle]

From $a_m = \frac{-1^{m/2}}{m!(m+3)!} a_0$
I found for m even
$a_{2k} = \frac{-1^{k}}{2k!(2k+3)!} a_0$

Now I'm wondering if I can find the odd series using only the recurrence relation and can I add up both series?
Furthermore, I don't see any cos or sin series yet.

 

[happyparticle]

I think I'm not that far since I guess a solution should be in this form $x^{1/2} \sum\limits_{n = 0}^\infty {a_n x^{2n} }$ but I don't see how to have (2n+1)! to the dominator.