Classical Electromagnetism

Siméon Denis Poisson, 1781 to 1840

Pierre-Simon Laplace, 1749 to 1827

Three Dimensional Boundary Value Problems in Rectangular Coordinates

Laplace's equation for the potential in the absence of sources is:

Ñ²Φ = 0

Ѳ in rectangular coordinates is:

	∂2		∂2		∂2
Ñ2 =		+		+
	∂x2		∂y2		∂z2

We'll presume that our solution is separable, that is Φ( x, y, z ) = X(x)Y(y)Z(z). We can check the separability assumption later. Laplace's equation is a second order differential equation, so if we find two unique solutions we've done it. Generally speaking, if the boundary conditions are separable, there's a good chance the solution is separable. If the boundary conditions are not separable, most likely we're hosed. Now,

ѲΦ(x,y,z) = ѲX(x)Y(y)Z(z).

X is independent of y and z, so it can be pulled through the y and z partial derivatives. Similarly Y is independent of x and z, and Z is independent of x and y, so they can be pulled through the corresponding partial derivatives.

∂2

∂2

∂2

∂2X

∂2Y

∂2Z

Ñ2XYZ =

XYZ +

XYZ +

XYZ =

YZ

+ XZ

+ XY

= 0

∂x2

∂y2

∂z2

∂x2

∂y2

∂z2

Now divide the entire equation by XYZ:

	1	∂2X		1	∂2Y		1	∂2Z
Ñ2XYZ =			+			+			= 0
	X	∂x2		Y	∂y2		Z	∂z2

We now have three seperate expressions, one that depends only on X, one that depends only on Y, and one that depends only on Z. These three expressions must sum to 0 for any x, y, z, so they must be equal to constants that sum to 0. Since these are second-order equations, we'll anticipate and set the X equation equal to -α², the Y equation equal to -β², and the Z expression = γ² = α² + β².

	1	∂2X		1	∂2Y		1	∂2Z
			= -α2			= -β2			= α2 + β2
	X	∂x2		Y	∂y2		Z	∂z2

We can solve these immediately.

X = A_αsin( αx ) + B_α cos( αx );
Y = A_βsin( βy ) + B_β cos( βy );
Z = A_γsinh( γz ) + B_γ cosh( γz );
Φ(x,y,z) = X(x)Y(y)Z(z).

Notice that it was arbitrary which dimension got sines and cosines and which got sinhs and coshs - the only requirement is that at least one dimension gets sines and cosines, and at least one dimension gets sinhs and coshs. It could have been one dimension getting sines and cosines and two dimensions getting sinhs and coshs. The choice must be made based on the boundary conditions.

A Worked Problem

Suppose we have a box with known potentials. The interior of the box is 0 < x < a;   0 < y < b;  0 < z < c. The box has a voltage V(x,y) on the top face (z = c) and zero on all other faces.

Figure 4.1: A boundary value problem

We have six boundary conditions.

1) the face for x = 0 has potential 0. This implies we need the sine solution for X.

2) the face for y = 0 has potential 0. This implies we need the sine solution for Y.

3) the face for z = 0 has potential 0. This implies we need the sinh solution for Z.

4) the face for x = a has potential 0. This implies that α = nπ/a.

5) the face for y = b has potential 0. This implies that β = mπ/b.

Now our solution is of the form:

Φ(x,y,z) = A_nmsin( nπx/a ) sin( mπy/b ) sinh( γ_nmz )   where γ_nm = π √ ( n²/a² + m²/b² )

The 6th boundary condition is used to find the A_nm via a double fourier series.