Bessel's Differential equation

Click For Summary
SUMMARY

The discussion focuses on solving Bessel's differential equation arising from a heat transfer problem involving a solid ball at 30 degrees Celsius placed in a refrigerator at 0 degrees Celsius. The user attempts to separate variables in the heat equation in spherical coordinates, leading to a Bessel equation for the radial component R(r). The coefficients of the Bessel equation are identified, but the user struggles with the solution process and the relationship to the Helmholtz equation. The discussion emphasizes the need for series methods to solve Bessel's equation effectively.

PREREQUISITES
  • Understanding of partial differential equations, specifically the heat equation.
  • Familiarity with spherical coordinates and their application in physics.
  • Knowledge of Bessel functions and their properties.
  • Basic concepts of separation of variables in differential equations.
NEXT STEPS
  • Study the series methods for solving Bessel's differential equation.
  • Learn about the Helmholtz equation and its applications in heat transfer problems.
  • Explore the derivation and properties of Bessel functions of the first kind.
  • Review examples of solving the heat equation in spherical coordinates.
USEFUL FOR

Students and professionals in applied mathematics, physics, and engineering, particularly those dealing with heat transfer and differential equations.

Fisherman87
Messages
6
Reaction score
0

Homework Statement



A solid ball at 30 degrees C with radius a=1 is placed in a refrigerator that maintains a constant temperature of 0 degrees C. Take c (speed)=1 and determine the temperature u(r,theta,phi,t) inside the ball

Homework Equations



partial differential heat equation in spherical coordinates

The Attempt at a Solution



In this case, I started out with setting all the partials of the function u (assuming u is the solution) with respect to theta and phi as 0 since the temperature is uniform and doesn't depend on either. I then set u(r,t)=R(r)T(t), and then plugged back into the partial equation in order to separate variables, and I got T(t)=exp(mu*t). However, R turns into a bessel equation that I can't figure out how to solve, with the coefficient of R double prime equal to r^2, the coefficient of R prime being 2r, and the coefficient of R being mu. Mu is the separation constant. The book is generally unhelpful. Can anyone tell me how to solve that equation?
 
Physics news on Phys.org
Fisherman87 said:

Homework Statement



A solid ball at 30 degrees C with radius a=1 is placed in a refrigerator that maintains a constant temperature of 0 degrees C. Take c (speed)=1 and determine the temperature u(r,theta,phi,t) inside the ball

Homework Equations



partial differential heat equation in spherical coordinates

The Attempt at a Solution



In this case, I started out with setting all the partials of the function u (assuming u is the solution) with respect to theta and phi as 0 since the temperature is uniform and doesn't depend on either. I then set u(r,t)=R(r)T(t), and then plugged back into the partial equation in order to separate variables, and I got T(t)=exp(mu*t). However, R turns into a bessel equation that I can't figure out how to solve, with the coefficient of R double prime equal to r^2, the coefficient of R prime being 2r, and the coefficient of R being mu. Mu is the separation constant. The book is generally unhelpful. Can anyone tell me how to solve that equation?

The Bessel's equation is solved with series methods. You can read about how to do it here:

http://www.ucl.ac.uk/~ucahhwi/MATH7402/handout9.pdf
 
Last edited by a moderator:
I guess I should've been more descriptive. I have all that information already (even though this is for spherical coordinates and not polar). The equation I got for the last term multiplied by R is mu*r^2. This is why I'm having trouble, since it doesn't really agree completely with the normal bessel equation, and I don't understand how the helmholtz equation is related to this equation, and why I should use it.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K