# Uniqueness theorem for Laplace's equation

Hi all.

Suppose that U1 is the solution of the Laplace's equation for a given set of boundary conditions and U2 is the the solution for the same set plus one extra boundary condition. Thus U2 satisfies the Laplace's equation and the boundary conditions of the first problem, so it's a solution of the first problem.

I know that the above argument must be wrong according to the uniqueness theorem, but what's wrong with it?

Thanks in advance.

## Answers and Replies

It seems to me that there are several possibilities here. Either the original boundary conditions are underspecified - in which case U1 will actually be a whole class of solutions and U2 will be one specific element of that class - or, U1 happens to obey the added condition anyway. Otherwise, U2 will be overspecified and will, generically, not exist.

Thank you Parlyne.

My problem is that I think boundary conditions could be anything. This thought is coming from the boundary value problems I saw in electromagnetics. We had a potential function satisfying Laplace's equation and the boundary conditions were a bunch of conducting surfaces held at a fixed potential. There was no restriction on number of boundary conditions, since every set of boundary conditions was corresponding to a real world problem.

...There was no restriction on number of boundary conditions, since every set of boundary conditions was corresponding to a real world problem.

Actually, there is always a restriction on the number of boundary conditions, even in real-world problems, or you would over-specify a problem. There is actually a specific number of boundary conditions you must have to have a unique solution. A second-order differential equation, such as the Laplace equation, in three dimensions must always have 6 boundary conditions (3x2) to give a unique solution. In some coordinate systems, all of the boundary conditions may not be obvious, so you may be tempted into thinking you don't need all six, but you still do.

For instance, consider a hollow sphere with some fixed electrostatic potential V(θ,φ) on its surface and you want to know the potential everywhere inside the sphere. It would seem there is only one boundary condition, on the sphere, but there is actually six:

(1) V(θ,φ) at the sphere's surface
(2) finite potential at the sphere's center
(3) finite potential at the north pole
(4) finite potential at the south pole
(5) and (6) single-valued potential in the azimuthal angle

You may not realize you are doing it, but you have to use all six boundary conditions to uniquely determine the potential.

Thank you chrisbaird.

So the boundaries are actually the boundaries of the region that we are solving for, right?

Suppose that there is a square shaped conducting surface inside that sphere with a fixed potential and the potential of the sphere is the same V(θ,φ). (Not possible?) Now what are the boundaries?

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This is a little different than what you first implied. If you're talking about adding a physically distinct boundary, you need to remember that Laplace's equation will no longer be true at the location of the new boundary. Therefore, your original solution will no longer be correct. (Strictly, you could probably frame the new solution as a superposition of the old one and a new one that corrects for the new part of the boundary.)

Thank you chrisbaird.

So the boundaries are actually the boundaries of the region that we are solving for, right?

Suppose that there is a square shaped conducting surface inside that sphere with a fixed potential and the potential of the sphere is the same V(θ,φ). (Not possible?) Now what are the boundaries?

Such a problem has two different geometries, so the Laplace equation is not seperable. The Laplace equation and the uniqueness theorem still hold true, but they cannot be solved analytically. To solve such a problem numerically, you could split up the space between the square plate and spherical shell into a bunch of cubic regions all stacked together to fill the irregularly shaped volume. Then each cubic region would have its own solutions and you would link all the regions with boundary conditions on each cubic face.