# Bessel's ODE: Why does taking ν≥0 matter? It is squared anyway.

1. Dec 1, 2013

### gikiian

In Bessel's ODE  $x^{2}y''+xy''+(x^{2}-\nu^{2})y=0$, why must $\nu$ not be less than zero?

I have looked it up, but I do not find a satisfying answer anywhere.

Last edited: Dec 1, 2013
2. Dec 1, 2013

### AlephZero

In general $\nu$ can be any complex number.

But in the most common applications, it is a real integer or half-integer.

3. Dec 1, 2013

### gikiian

My instructor told me that $\nu$ has to be a non-negative real number (i.e. $\nu$ ≥ 0). Do you have any idea why he said that?

4. Dec 1, 2013

### HallsofIvy

Here, you are assuming that the coefficients, including $\nu[itex], are real. Because [itex]\nu$ is squared, it does not matter whether it is positive or negative, we must get the same solutions. It is simplest to assume it is positive, knowing that if it is, in fact, negative we will get the same solutions anyway.

5. Dec 1, 2013

### AlephZero

If the equation is modeling some physical system, it could be because of the physics, or the boundary conditions for the solutions that you want to find.

But there is no mathematical reason why $\nu$ has to be real, rather than complex.

6. Dec 3, 2013

### yungman

I thought $\nu$ can be +ve or -ve. You have Bessel Function in negative order.