How Can a 3D PDE Be Simplified to a 1D Equation in Theta?

[Ido]

I got the following PDE:
Laplasian[F]+a*d(F)/d(teta)=E*F

I worked with cylindrical coordinates (r,teta,z)
(teta is the angle between the x-axis and the r vector (in xy plane))
a,E are constants

I got the constrains: z=0 r=a , so the whole problem is on a simple ring

How can I make the 3D PDE just 1D in teta?
Can I use the Laplasian in cylindrical coordinates, and throw away the d/dr and d/dz?

I hope someone can help me.

 

[Palindrom]

Without separation of variables (which I know you don't want to do), I don't see a way.

 

[thepatientmental]

The given PDE can be simplified to a 1D PDE in the teta direction by making use of the constraints given. Since the whole problem is on a simple ring, we can assume that the function F is only dependent on the teta coordinate. Therefore, the Laplacian operator in cylindrical coordinates can be simplified to just the second derivative with respect to teta, i.e. Laplacian[F] = (1/r)*d/dr(r*dF/dteta).

Furthermore, since the constraints state that z=0 and r=a, any terms containing the z or r coordinates can be eliminated. This leaves us with the simplified 1D PDE:

(1/a)*d/dteta(a*dF/dteta) + a*E*F = 0

This can be solved using standard methods for solving 1D PDEs. It is important to note that the solution obtained will only be valid for the specific constraints given, i.e. for the problem on a simple ring. If the constraints were to change, the PDE would need to be re-evaluated accordingly.

In summary, to make the 3D PDE just 1D in teta, we can use the Laplacian in cylindrical coordinates and eliminate any terms containing the z or r coordinates based on the given constraints. This will result in a simplified 1D PDE that can be solved using appropriate methods.