Radial solutions to laplace equation

In summary, the solution to the laplace equation or poisson equation can be found when you only consider (in polar coordinates) \frac{\partial u^2} {\partial^2 r} + \frac{1}{r} \frac{\partial u} {\partial r} = f(r,\theta) instead of \frac{\partial u^2} {\partial^2 r} + \frac{1}{r} \frac{\partial u} {\partial r} + \frac{1}{r^2}\frac{\partial u^2} {\partial^2 \theta} = f(r,\theta)
  • #1
wumple
60
0
Hi,

When can I assume that the solution the laplace equation (or poisson equation) is radial? That is, when can I look only at (in polar coordinates)

[tex]\frac{\partial u^2} {\partial^2 r} + \frac{1}{r} \frac{\partial u} {\partial r} = f(r,\theta)[/tex]

instead of

[tex]\frac{\partial u^2} {\partial^2 r} + \frac{1}{r} \frac{\partial u} {\partial r} + \frac{1}{r^2}\frac{\partial u^2} {\partial^2 \theta} = f(r,\theta)[/tex]

With appropriate boundary conditions of course. My guess is any time that the boundary conditions are constant in theta and f depends only on r? But I'm not really sure.

Thanks for any help!

-wumple
 
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  • #2
wumple said:
Hi,

When can I assume that the solution the laplace equation (or poisson equation) is radial? That is, when can I look only at (in polar coordinates)

[tex]\frac{\partial u^2} {\partial^2 r} + \frac{1}{r} \frac{\partial u} {\partial r} = f(r,\theta)[/tex]

instead of

[tex]\frac{\partial u^2} {\partial^2 r} + \frac{1}{r} \frac{\partial u} {\partial r} + \frac{\partial u^2} {\partial^2 \theta} = f(r,\theta)[/tex]

With appropriate boundary conditions of course. My guess is any time that the boundary conditions are constant in theta and f depends only on r? But I'm not really sure.

Thanks for any help!

-wumple

Look at [tex]\frac{\partial^2u} {\partial\theta^2}[/tex] When would this term vanish from the equation...
 
  • #3
kjohnson said:
Look at [tex]\frac{\partial^2u} {\partial\theta^2}[/tex] When would this term vanish from the equation...

when [tex] u = u(r) [/tex] only?
 
  • #4
Yes if u(r) only then it would surely drop out. But that is not the only way... For a moment ignore the partial and just think of it like a normal derivative, more specifically notice it is a second derivative. When would the second derivative of some function equal zero?
 
  • #5
when it's linear in theta? How do you know when the solution is linear in theta?
 
  • #6
Yes, when its linear in theta. How do you know when its linear in theta...thats a good question. I will have to do a little thinking on that one, this stuff is kinda new to me as well ;). Hopefully that gives you at least some idea and if I come up with something I will keep you updated.
 
  • #7
I think it has to do with the laplacian being rotationally invariant - hence no theta dependence. See sec 8.3/8.4:

http://www.math.ucsb.edu/~grigoryan/124B/lecs/lec8.pdf

could anyone else explain this a bit better? Is that true - if a PDE is rotationally invariant, its solutions will have no theta dependence (or some of them do anyway)?
 
  • #8
After doing some thinking I don't know if 'u' can be linear in theta (at least physically). Because if it was linear in theta then at theta=0 'u' would have some value, while at theta=2pi 'u' would have some different value despite the fact that theta=0 and theta=2pi correspond to the same physical point. Thus if it was linear in 'u' there would be a discontinuity. That being said the only thing that would satisfy it is u=u(r)
 
  • #9
wumple, the intuition is really all that you need- if we assume that f has no dependence on theta, we get a differential equation only involving r, so we can find solutions involving only r. Compare that with this proposed differential equation:
[tex] \theta \frac{\partial f}{\partial r} + f = 0[/tex]

If you assume that f doesn't depend on theta, then you still have a theta in your differential equation and things get confusing.

In terms of when your solution is going to be rotationally invariant - if your boundary condition is rotationally invariant and the shape of your domain is rotationally invariant then your solution will be rotationally invariant as well - otherwise, rotating the whole problem yields the same exact problem but a different solution (this is assuming that things are specified to the point where there is a unique solution)
 

What is the Laplace equation?

The Laplace equation is a partial differential equation that describes the behavior of a scalar field in a given region. It is often used in physics and engineering to model various physical phenomena such as heat flow, electrostatics, and fluid dynamics.

What are radial solutions to the Laplace equation?

Radial solutions to the Laplace equation are solutions that depend only on the distance from the center of a given region. In other words, these solutions have a spherical symmetry and do not vary in different directions.

What are some applications of radial solutions to the Laplace equation?

Radial solutions to the Laplace equation have various applications in physics and engineering. For example, they are used to model the behavior of electric charges in a spherical region, the distribution of temperature in a spherical object, and the flow of fluids in a spherical container.

How do you solve for radial solutions to the Laplace equation?

To solve for radial solutions to the Laplace equation, one can use separation of variables and assume a solution of the form u(r,θ,φ) = R(r)Θ(θ)Φ(φ), where r is the distance from the center, θ is the polar angle, and φ is the azimuthal angle. This leads to a set of ordinary differential equations that can be solved to find the radial solution.

Why are radial solutions important in physics?

Radial solutions to the Laplace equation are important in physics because they simplify the mathematical models used to describe physical phenomena. In many cases, the behavior of a system can be accurately captured by a radial solution, making it easier to analyze and understand the system's behavior.

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