# Solution to Laplace Equation across Boundaries

1. Aug 21, 2008

### Apteronotus

Hi everyone,

I read somewhere that solutions to Laplace's Equation must agree at a shared boundary.
So for example if $$\Phi_{1}$$ and $$\Phi_{2}$$ are two solutions to the Laplace equation in two different regions which share a boundary, then on the boundary $$\Phi_{1}$$ = $$\Phi_{2}$$
Is this true?
Can you help me see why?

2. Aug 21, 2008

### HallsofIvy

Staff Emeritus
In order that a function satisfy Laplace's equation, it must be differentiable. Otherewise the derivatives don't make sense! In particular, that means it must be continuous.

3. Aug 21, 2008

### Apteronotus

Though I dont disagree with your statement HallsofIvy, it wouldn't necessarily apply in this situation. One function [tex]\Phi_{1}[tex] is continuous in region 1 and the other [tex]\Phi_{2}[tex] is continuous in region 2, where region 1 and 2 only share a boundary.

Would it be true that these two different functions have the same value on this boundary, given that they both satisfy the Laplace equation?

4. Aug 22, 2008

### nicksauce

If this true? Let f : S1->R = 37. Let g: S2->R = 92. Both f and g satisfy Laplace's equation on S1 and S2 respectively. Suppose S1 = {(x,y) | x^2 + y^2 <= 2}, and S2={(x,y) | 1< x^2 + y^2 <= 2}.
S1 and S2 share a boundary, namely x^2 + y^2 = 2, but f and g are different functions on this boundary.

5. Aug 22, 2008

### Ben Niehoff

No, there is NO requirement that solutions of Laplace's equations in different regions (with shared boundary) coincide on the boundary. Rather, Laplace's equation itself makes no requirement. Within a region, a solution must be differentiable, but at the boundaries, it is usually not, and can even be discontinuous.

The question as to whether two adjoining solutions must be "stitched together" in a continuous way depends entirely on the physical problem at hand. In electromagnetism, for example, the electric potential must be continuous in most cases; therefore, we must choose Laplace solutions that match up on the boundaries. Even so, the potential will usually NOT be differentiable at the boundaries (consider the potential of a conducting shell, which is constant inside, but falls as 1/r outside).

There are some cases, however, in which the potential is also discontinuous: e.g., a dipole layer. A dipole layer is a surface which is considered to have a certain dipole moment per unit area (perpendicular to the surface). Across a dipole layer, the potential makes a discontinuous jump in proportion to the local dipole moment.

6. Aug 26, 2008

### Apteronotus

Thank you kindly for your replies.