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

[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.

 

[AlephZero]

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

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

 

[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?

 

[HallsofIvy]

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

 

[AlephZero]

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?

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.

 

[yungman]

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