Unique Solutions to Laplace Eq. on Half-Open Cylinder

[Logarythmic]

I know that if the potential on the boundary of a sphere is zero, then the potential inside the sphere is zero. I also know that if the potential on the boundary is constant, then there exist a solution to the laplace equation inside the sphere and the solution is unique. But what if the boundary is not closed?
For exemple, if the potential is constant on the surface of a half-open cylinder (with one disc removed). Does a solution exist but not a unique one?

 

[OlderDan]

I know that if the potential on the boundary of a sphere is zero, then the potential inside the sphere is zero. I also know that if the potential on the boundary is constant, then there exist a solution to the laplace equation inside the sphere and the solution is unique. But what if the boundary is not closed?
For exemple, if the potential is constant on the surface of a half-open cylinder (with one disc removed). Does a solution exist but not a unique one?

If you had an open conducting cylinder (constant potential) with excess charges, the charges would distribute themselves as needed to achieve the constant potential. The charge configuration that does this is unique. The potential everywhere else is unique.

 

[Logarythmic]

So a solution exists and it is unique?

 

[OlderDan]

So a solution exists and it is unique?

I would have to say yes. That does not mean you can find an analytic solution. Maybe the best you can do is reduce the problem to an integral that would have to be done numerically.

 

[Logarythmic]

But for a closed boundary, say a box, there exists one and only one solution for the potential. If we remove one side of the box there exists infinitely many solutions. Shouldn't this be the case for a half-open cylinder as well?

 

[TMFKAN64]

I know that if the potential on the boundary of a sphere is zero, then the potential inside the sphere is zero. I also know that if the potential on the boundary is constant, then there exist a solution to the laplace equation inside the sphere and the solution is unique. But what if the boundary is not closed?
For exemple, if the potential is constant on the surface of a half-open cylinder (with one disc removed). Does a solution exist but not a unique one?

I think you are confusing physical boundaries with the boundaries of a boundary value problem.

Consider your cylinder... if it has a top, and you ground the whole thing, there is a unique solution to what is going on inside. If you remove the top, but I tell you the potential on that surface is still zero, there is *still* a unique solution inside. However, if you don't have any information about what is going on on that upper surface, you really can't say what is going on inside the cylinder. (I might have brought a large charge very close to the top... if you don't know what potential it is producing on the top, you don't know anything about what is going on inside the can either.)

 

[OlderDan]

But for a closed boundary, say a box, there exists one and only one solution for the potential. If we remove one side of the box there exists infinitely many solutions. Shouldn't this be the case for a half-open cylinder as well?

I assumed we were talking about a static configuration. There is something I only vaguely remember the time-varying case, where there is a transformation between the electric and magnetic potentials. If that is what you are talking about, then you might want to look a this.

http://www.mathpages.com/home/kmath563/kmath563.htm

If you are talking about static charge configurations, I think you are talking a unique poential, except of course for the arbitrary reference potential of zero at infinity. I could of course be wrong. It has happened before.

 

[Logarythmic]

This is my problem:

"Consider the Laplace equation with the boundary condition $$V = V_0$$ (constant) on the surface of a half-open cylinder (with one disc removed). Does a solution exist, and if so, is it unique?"

My knowledge tells me that if the potential on the surface of a closed box is zero, then the potential inside the box is zero because a non-zero potential would have a maximum or a minimum and therefore violate Laplace's equation. If the potential on the surface is constant and non-zero, one can show that the potential inside the box would have a unique solution.
However, if the potential is given only on part of the boundary, there is no unique solution. If we specify the potential on a region which is not a boundary, there is an infinity of possible solutions to the PDE under consideration.
So the question here reduces to if the boundary of the half-open cylinder is a closed or an open boundary.

 

[OlderDan]

This is my problem:

"Consider the Laplace equation with the boundary condition $$V = V_0$$ (constant) on the surface of a half-open cylinder (with one disc removed). Does a solution exist, and if so, is it unique?"

My knowledge tells me that if the potential on the surface of a closed box is zero, then the potential inside the box is zero because a non-zero potential would have a maximum or a minimum and therefore violate Laplace's equation. If the potential on the surface is constant and non-zero, one can show that the potential inside the box would have a unique solution.
However, if the potential is given only on part of the boundary, there is no unique solution. If we specify the potential on a region which is not a boundary, there is an infinity of possible solutions to the PDE under consideration.
So the question here reduces to if the boundary of the half-open cylinder is a closed or an open boundary.

Perhaps I need to revise my thinking on this. I was thinking in terms of a physical conductor like a can made out of metal, but I guess that would be a closed surface (inner and outer walls with one end connected by a disk and the "open end" inner and outer walls connected by a thin ring or washer. That is probably mathematically quite different from a cylindrically shaped sheet of charge. I don't have an answer to your question readily available. Maybe a better mathematician than I is lurking around here and can help you out.

 

[TMFKAN64]

This is my problem:

"Consider the Laplace equation with the boundary condition $$V = V_0$$ (constant) on the surface of a half-open cylinder (with one disc removed). Does a solution exist, and if so, is it unique?"

... However, if the potential is given only on part of the boundary, there is no unique solution. If we specify the potential on a region which is not a boundary, there is an infinity of possible solutions to the PDE under consideration.
So the question here reduces to if the boundary of the half-open cylinder is a closed or an open boundary.

Right. Basically, one unique solution exists within almost any continuous, differentiable potential on a closed surface that you care to define. If it isn't a closed surface though, you can't really say anything because there are too many possible solutions!

 

[OlderDan]

Right. Basically, one unique solution exists within almost any continuous, differentiable potential on a closed surface that you care to define. If it isn't a closed surface though, you can't really say anything because there are too many possible solutions!

Some of the cobwebs are starting to clear on this. I understand that if the top disk is gone from the cylinder, then there is no boundary condition for the region of space that disk would occupy. But I am having trouble getting my head around the physical implication of this.

Lets make the geometry even simpler and consider a planar disk of charge (one surface). If I know how the charge is distributed on the disk, I can find the potential from that charge distribution everywhere in space.

Now suppose instead of a disk of charge, I have a disk shaped region of space and I want that region to have a constant potential. I can distribute the charges in space in any way needed to achieve that constant potential. Are we saying here that there are infinitely many ways to distribute charge in space to achieve that potential, but that if the disk is replaced by a cylinder (all surfaces) then the charge distribution required to achieve constant potential over the entire surface is unique?

 

[TMFKAN64]

Now suppose instead of a disk of charge, I have a disk shaped region of space and I want that region to have a constant potential. I can distribute the charges in space in any way needed to achieve that constant potential. Are we saying here that there are infinitely many ways to distribute charge in space to achieve that potential, but that if the disk is replaced by a cylinder (all surfaces) then the charge distribution required to achieve constant potential over the entire surface is unique?

Part 1: Yes, there are an infinite number of ways to put the disk at constant potential. For the sake of argument, let's take the constant to be zero . Clearly if you put a charge +Q at a distance H above the center of the disk and a charge -Q at a distance H below the center of the disk, the disk is at zero potential for any Q and H. (If you want any other constant, let me bring in my infinite plane of charge as well, and I'd be happy to accommodate you.)

Part 2: That's not really what Laplace's equation is about. Laplace's equation gives us the potential within a region without charge, and the solutions are unique in various situations. (Two I can remember off the top of my head, solutions are unique if the potential is specified on the boundary, and solutions are unique if the boundary is composed of conductors with a fixed total charge on each of them.) I'm not sure if there is more than one way to generate a constant potential on the cylinder or not... but what I *will* tell you is that whatever distributions of charges outside manages this, the potential *inside* the cylinder is the same for all of them.

 

[OlderDan]

Part 1: Yes, there are an infinite number of ways to put the disk at constant potential. For the sake of argument, let's take the constant to be zero . Clearly if you put a charge +Q at a distance H above the center of the disk and a charge -Q at a distance H below the center of the disk, the disk is at zero potential for any Q and H. (If you want any other constant, let me bring in my infinite plane of charge as well, and I'd be happy to accommodate you.)

Part 2: That's not really what Laplace's equation is about. Laplace's equation gives us the potential within a region without charge, and the solutions are unique in various situations. (Two I can remember off the top of my head, solutions are unique if the potential is specified on the boundary, and solutions are unique if the boundary is composed of conductors with a fixed total charge on each of them.) I'm not sure if there is more than one way to generate a constant potential on the cylinder or not... but what I *will* tell you is that whatever distributions of charges outside manages this, the potential *inside* the cylinder is the same for all of them.

Your example of the two point charges is a good one, and cleary there could be an infinite set of charge pairs along the axial line with any linear charge distribution you can conceive of and still have zero potntial on the disk. I have a harder time conceiving of a variety of charge distributions leading to the same potential on the open or closed cylinder, but then I don't feel compelled to find even one, so I'll let this one rest for a while.

Thanks four your input.