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

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Discussion Overview

The discussion revolves around the conditions for the parameter ν in Bessel's ordinary differential equation (ODE), specifically why ν is often taken to be non-negative. Participants explore the implications of this condition in both mathematical and physical contexts.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions why ν must be non-negative in Bessel's ODE, seeking a satisfying explanation.
  • Another participant states that ν can generally be any complex number, but in common applications, it is typically a real integer or half-integer.
  • A participant mentions their instructor's assertion that ν should be a non-negative real number, prompting inquiries into the reasoning behind this requirement.
  • It is suggested that since ν is squared in the equation, the sign of ν does not affect the solutions, and it may be simpler to assume it is positive.
  • Another participant notes that if the equation models a physical system, the non-negativity of ν might relate to physical constraints or boundary conditions, rather than a strict mathematical necessity.
  • A later reply points out that Bessel functions can indeed exist for negative orders, indicating that ν can be both positive and negative.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of ν being non-negative, with some supporting the instructor's position while others argue that negative values are also valid. The discussion remains unresolved regarding the implications of ν's sign.

Contextual Notes

Participants acknowledge that the assumptions about the coefficients being real may influence the discussion, and the implications of boundary conditions in physical applications are also noted as relevant but not fully explored.

Who May Find This Useful

This discussion may be of interest to students and researchers studying differential equations, particularly in the context of Bessel functions and their applications in physics and engineering.

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

I have looked it up, but I do not find a satisfying answer anywhere.
 
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In general ##\nu## can be any complex number.

But in the most common applications, it is a real integer or half-integer.
 
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?
 
Here, you are assuming that the coefficients, including [itex]\nu[itex], are real. <b>Because</b> [itex]\nu[/itex] 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.[/itex][/itex]
 
gikiian said:
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.
 
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I thought [itex]\nu[/itex] can be +ve or -ve. You have Bessel Function in negative order.
 

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