Bessel's Equation: is it possible to solve the Bessel Equation without adding the Gamma function to the solution?

[mario99]

$$x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + (x^2 - \lambda^2)y = 0$$
$$y(x) = c_1J_{\lambda}(x) + c_2Y_{\lambda}(x)$$

where $J_{\lambda}(x)$ is the Bessel function of the first kind and $Y_{\lambda}(x)$ is the Bessel function of the second kind, both of order $\lambda$.
​

In the last 30 days, I have been working hard to understand the solution of Bessel Equation to be able to solve some of the partial differential equations. I tried to solve it from scratch as I did successfully before with Airy Equation (Now I can even represent the solution of Airy Equation in terms of the Bessel functions), but I failed. While solving the Bessel Equation, everything was fine until they added to the solution the annoying Gamma function. I don't understand why and how it fitted perfectly in the solution. In someway, I felt like we can replace it with something else. My question is, is it possible to solve the Bessel Equation without adding the Gamma function to the solution?

 

[topsquark]

$$x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + (x^2 - \lambda^2)y = 0$$
$$y(x) = c_1J_{\lambda}(x) + c_2Y_{\lambda}(x)$$

where $J_{\lambda}(x)$ is the Bessel function of the first kind and $Y_{\lambda}(x)$ is the Bessel function of the second kind, both of order $\lambda$.
​

In the last 30 days, I have been working hard to understand the solution of Bessel Equation to be able to solve some of the partial differential equations. I tried to solve it from scratch as I did successfully before with Airy Equation (Now I can even represent the solution of Airy Equation in terms of the Bessel functions), but I failed. While solving the Bessel Equation, everything was fine until they added to the solution the annoying Gamma function. I don't understand why and how it fitted perfectly in the solution. In someway, I felt like we can replace it with something else. My question is, is it possible to solve the Bessel Equation without adding the Gamma function to the solution?

Where did you find a relationship between Bessel functions and Gamma functions? I'm not recalling anything there. Now Beta functions...

Anyway, what problem were you working on? I don't see why the Bessel functions themselves wouldn't be enough. There's a vast amount of literature out there on them.

-Dan

 

[mario99]

Thank you topsquark for helping me.

The solution is very long and somehow complicated. I am not going to write every step, rather will be focusing on some important parts of it where the magic happens.

I am sure that you remember how to solve the Airy Equation using the power series $\sum_{n=0}^{\infty}a_nx^n$.

Solving the Bessel equation is similar but the solution will look like $\sum_{n=0}^{\infty}a_nx^{n+m}$ where $m$ is the indicial exponent, something related to Frobenius series if you recall that. It happens that we look for a solution including the indicial exponent in the power series due to regular singular points issues.

With this new form of the power series, we solve normally, taking the first and second derivatives, and substiting the results in the original Bessel differential equation.

After that, we get the recurrence relation

$a_n = -\frac{1}{n(2\lambda+n)}a_{n-2}$

From that, we see that odd coefficients equal zero ($a_1 = a_3 = a_5 = ........ = 0$)

And we focus on even coefficients

$a_2 = -\frac{1}{1 \cdot (\lambda + 1)} \cdot \frac{1}{2^2}a_0$

With a general formula this becomes,

$a_{2n} = -\frac{1}{n \cdot (\lambda + n)} \cdot \frac{1}{2^2}a_{2n - 2}$

Don't ask me how, but this magically becomes,

$a_{2n} = \frac{(-1)^n}{n!(\lambda + 1)(\lambda + 2).....(\lambda + n)}\cdot \frac{1}{2^{2n}}a_0$

With all this the solution becomes,

$y(x) = a_0 x^{\lambda} + \sum_{n=1}^{\infty}a_{2n} x^{\lambda + 2n}$

After that, wihtout any further explanation, they say, the coefficient $a_0$ can have any nonzero value, and they chose,

$a_0 = \frac{1}{2^{\lambda}\Gamma(1+\lambda)}$

And from here, the annoying Gamma function appeared. Wouldn't it be better to choose $1$ instead of Gamma function since $1$ is simple and nonzero value?

After that, the other coefficient magically becomes,

$a_{2n} = \frac{(-1)^n}{2^{\lambda+2n} \ n! \ \Gamma(n + \lambda + 1)}$

$y(x) = \frac{x^{\lambda}}{2^{\lambda}\Gamma(1+\lambda)} + \sum_{n=1}^{\infty}\frac{(-1)^nx^{\lambda + 2n}}{2^{\lambda+2n} \ n! \ \Gamma(n + \lambda + 1)}$

And this is in fact one of the independent solutions of the Bessel equation, namely, the the Bessel function of the first kind.

$J_{\lambda}(x) = \frac{x^{\lambda}}{2^{\lambda}\Gamma(1+\lambda)} + \sum_{n=1}^{\infty}\frac{(-1)^nx^{\lambda + 2n}}{2^{\lambda+2n} \ n! \ \Gamma(n + \lambda + 1)}$

 

[topsquark]

Thank you topsquark for helping me.

The solution is very long and somehow complicated. I am not going to write every step, rather will be focusing on some important parts of it where the magic happens.

I am sure that you remember how to solve the Airy Equation using the power series $\sum_{n=0}^{\infty}a_nx^n$.

Solving the Bessel equation is similar but the solution will look like $\sum_{n=0}^{\infty}a_nx^{n+m}$ where $m$ is the indicial exponent, something related to Frobenius series if you recall that. It happens that we look for a solution including the indicial exponent in the power series due to regular singular points issues.

With this new form of the power series, we solve normally, taking the first and second derivatives, and substiting the results in the original Bessel differential equation.

After that, we get the recurrence relation

$a_n = -\frac{1}{n(2\lambda+n)}a_{n-2}$

From that, we see that odd coefficients equal zero ($a_1 = a_3 = a_5 = ........ = 0$)

And we focus on even coefficients

$a_2 = -\frac{1}{1 \cdot (\lambda + 1)} \cdot \frac{1}{2^2}a_0$

With a general formula this becomes,

$a_{2n} = -\frac{1}{n \cdot (\lambda + n)} \cdot \frac{1}{2^2}a_{2n - 2}$

Don't ask me how, but this magically becomes,

$a_{2n} = \frac{(-1)^n}{n!(\lambda + 1)(\lambda + 2).....(\lambda + n)}\cdot \frac{1}{2^{2n}}a_0$

With all this the solution becomes,

$y(x) = a_0 x^{\lambda} + \sum_{n=1}^{\infty}a_{2n} x^{\lambda + 2n}$

After that, wihtout any further explanation, they say, the coefficient $a_0$ can have any nonzero value, and they chose,

$a_0 = \frac{1}{2^{\lambda}\Gamma(1+\lambda)}$

And from here, the annoying Gamma function appeared. Wouldn't it be better to choose $1$ instead of Gamma function since $1$ is simple and nonzero value?

After that, the other coefficient magically becomes,

$a_{2n} = \frac{(-1)^n}{2^{\lambda+2n} \ n! \ \Gamma(n + \lambda + 1)}$

$y(x) = \frac{x^{\lambda}}{2^{\lambda}\Gamma(1+\lambda)} + \sum_{n=1}^{\infty}\frac{(-1)^nx^{\lambda + 2n}}{2^{\lambda+2n} \ n! \ \Gamma(n + \lambda + 1)}$

And this is in fact one of the independent solutions of the Bessel equation, namely, the the Bessel function of the first kind.

$J_{\lambda}(x) = \frac{x^{\lambda}}{2^{\lambda}\Gamma(1+\lambda)} + \sum_{n=1}^{\infty}\frac{(-1)^nx^{\lambda + 2n}}{2^{\lambda+2n} \ n! \ \Gamma(n + \lambda + 1)}$

Okay, now I remember the gammas. You are talking about the coefficients. I had thought you were talking about as a function of x.

First, the recursion. Just work it through, n by n:
$a_{2n} = \dfrac{-1}{4n(n+\lambda)} \cdot a_{2n - 2}$

From this we know that
$a_{2n-2} = \dfrac{-1}{4(n-1)(n-1+\lambda)} \cdot a_{2n - 4}$

so putting that into the $a_{2n}$ equation
$a_{2n} = \dfrac{-1}{4n(n+\lambda)} \left ( \dfrac{-1}{4(n-1)(n-1+\lambda} \right ) \cdot a_{2n - 4}$

and continuing:
$a_{2n} = \dfrac{-1}{4n(n+\lambda)} \left ( \dfrac{-1}{4(n-1)(n-1+\lambda)} \right ) \left ( \dfrac{-1}{4(n-2)(n-2+\lambda)} \right )\cdot a_{2n - 6}$

Doing this n-1 times gives us
$a_{2n} = \dfrac{-1}{4n(n+\lambda)} \left ( \dfrac{-1}{4(n-1)(n-1+\lambda)} \right ) \left ( \dfrac{-1}{4(n-2)(n-2+\lambda)} \right ) \dots \left ( \dfrac{-1}{4(1)(\lambda)} \right )\cdot a_0$

or
$a_{2n} = \dfrac{(-1)^n}{4^n n(n-1) \dots 1 \cdot (n + \lambda)(n + \lambda - 1) \dots \lambda}$

No magic to it. You should practice this. The solution of recursion relations isn't terribly well explained to Calculus students, but most of the time they aren't too bad to solve. You can usually spot the patterns in the derivations of the most common expansions pretty easily. (I'd advise you to learn how to solve at least 2nd order linear homogeneous recursions with constant coefficients. The theory is pretty similar to the corresponding ordinary differential equations. Inhomogeneous recursions can get nasty quickly, but there's a way to deal with those, too, though they are tougher than inhomogeneous differential equations.)

Now, $n(n-1) \dots 1 = n!$, but here's where the problem comes in: nobody told us that $\lambda$ is a positive integer. The best we can do is say $(n + \lambda)(n + \lambda - 1) \dots \lambda = \Gamma(n + \lambda + 1)/\Gamma(\lambda + 1)$. There's no way around it unless we are lucky enough to get a whole number for $\lambda$ and, at least for a Physics problem, you can bet it won't be.

You get used to them after a while.

By the way, I'm not too keen on the rest of the derivation. We certain have the freedom to choose $a_0$ to be anything we want, and getting rid of one of the gamma functions is nice, but I think it's better to say we are going to absorb any constants into the arbitrary constants of the Bessel functions in the general solution form for y(x). It's simpler. (And that way we get rid of that $2^{\lambda}$. I'm not sure why that's in there, either.)

-Dan

 

[mario99]

I like the way in how you manipulated the coefficient $a_{2n}$ to change its form to the final state. Fair enough, it is not magic. Of course, I would get back to your steps and suggestions whenever I needed help in dealing with similar forms of coefficients.

For the sake of perfection of this thread, I will focus on the main issue. You are telling me that it is possible to get rid of the Gamma function and write the solution without it. Or at least getting rid of one Gamma function if many occurred.

Wherever you go to see the derivation of the solution of the Bessel Equation, they always derived it with Gamma function. They can get rid of it, but they don't. Let me guess, they don't do that because of reasons that you don't know, I don't know, and nobody else knows! Correct me if I was wrong.

 

[topsquark]

I like the way in how you manipulated the coefficient $a_{2n}$ to change its form to the final state. Fair enough, it is not magic. Of course, I would get back to your steps and suggestions whenever I needed help in dealing with similar forms of coefficients.

For the sake of perfection of this thread, I will focus on the main issue. You are telling me that it is possible to get rid of the Gamma function and write the solution without it. Or at least getting rid of one Gamma function if many occurred.

Wherever you go to see the derivation of the solution of the Bessel Equation, they always derived it with Gamma function. They can get rid of it, but they don't. Let me guess, they don't do that because of reasons that you don't know, I don't know, and nobody else knows! Correct me if I was wrong.

You can "define away" the $\Gamma(\lambda + 1)$ because it is a constant. But $\Gamma(n + \lambda + 1)$ depends on $n$... we need it because it varies with which term in the expansion we are taking. We need to have this one in there.

-Dan

 

[mario99]

You can "define away" the $\Gamma(\lambda + 1)$ because it is a constant. But $\Gamma(n + \lambda + 1)$ depends on $n$... we need it because it varies with which term in the expansion we are taking. We need to have this one in there.

-Dan

Finally, I got the idea.

The conclusion is that the solution of the Bessel equation must include the Gamma function that depends on the index of the summation, in our case, $n$, any other Gammas are optional.

Thank you topsquark for the help.