Solution to Laplace’s Equation In Cartesian Coordinates Lecture 6

1 Introduction

We wish to solve the 2nd order, linear partial differential equation;

∇2V (x, y, z)=0

First consider a solution in Cartesian coordinates. The equation takes the form;

2 2 2 ∂ V + ∂ V + ∂ V = 0 ∂x2 ∂y2 ∂z2 Assume a solution using the method of separation of variables.

V (x, y, z) = X (x) Y(y) Z(z)

Thus the solution is written as a product of functions of each of the separate variables. Clearly this is not the only possible functional form, and indeed the assumption is highly restrictive. However, there is a uniqueness theorem which states that if we find a solution to the partial differential equation which also satisfies appropriate boundary conditions, then this solution must be a representation of the unique solution to the partial differential equa- tion.

2 Application of Separation of Variables

Substitute the above functional form into the differential equation. This gives;

d2X (x) d2Y(y) d2Z(z) Y(y) Z(z) + X (x) Z(z) + X (x) Y(y) = 0 dx2 dy2 dx2 Note the partial derivative is replaced by the total derivative as each function depends on only one variable. Divide the equation by V which gives;

2 2 2 1 d X (x) 1 d Y(y) 1 d Z(z) 2 + 2 + 2 = 0 X (x) dx Y(y) dy Z(z) dz Each term is a separate function of x, y, and z. Since the variables can change in arbitrary ways then;

1 d2X (x) d2Y(y) 1 = −[ 1 + X (x) dx2 Y(y) dy2 2 d Z(z) 2 1 ] = −α Z(z) dz2 where α is a constant. Continuing in the same way;

2 2 d Y(y) 2 d Z(z) 2 1 = α − ] = −β Y(y) dy2 dz2 where β2 is another constant. Define γ2 = α2 + β2. The remaining solution satisfies an ordinary differential equation ( ode) of the form;

2 d η(x) 2 ± c η(x)=0 dx2 Such solutions were presented in the last lecture, and they can be represented by the Fourier series involving harmonic eigenfunctions. The constants α and β are the separation con- stants which are determined by the boundary conditions. Write the general solution in the following form.

A eiax + B e−iax η = −  A eax + B e ax  By combining solutions, V is written as;

V ∝ e±iαx e±iβy e±γz

In the case when α = β = 0 the equation to be solved is;

2 d Z = 0 dz2 This has the solution, Z = A + B z, as obtained earlier for the potential between two parallel conducting planes.

3 Examples

Find the potential and field for an infinite set of parallel conducting plates, Figure 1. In this case by symmetry, the solution must be independent of the variables, (x, y). From the last section this means that the solution for the potential must have the form;

V (z) = A + B z

2 z a

V = Vo

y 0 V = 0

x

Figure 1: The geometry to find the potential between 2 parallel conducting plates held at different potential

Apply the boundary conditions to determine the constants. At z = 0 V = 0 and at z = a V = V0. The solution is;

V = (V0/a) z

The field is obtained from the gradient of the potential;

E~ = −∇~ V = V0/a

As another example, consider a hollow cube having sides of length, a. All sides are held at potential V = 0 except the side at z = a which is held at potential, V = V0(x, y). Look for the potential by solving Laplace’s equation within the cube using separation of vari- ables. The uniqueness theorem tells us that the solution must satisfy the partial differential equation and also satisfy the boundary conditions on the enclosed surface of the cube - ie Dirichlet conditions on a closed boundary, Figure 2. Use the general form for the solution obtained above.

V ∝ e±iαx e±iβy e±γz

The potential must vanish at the walls determined by x = 0, a and y = 0, a. To do this, re-write the complex exponential above in terms of harmonic functions. For the § function, choose sin(αx). This makes the value of X vanish ( and thus the potential) at x = 0. To vanish at x = a choose α = nπ/a. These are eigenvalues with eigenvectors, sin([nπ/a]x).

3 z

a

V = V0

V = 0 y a V = 0 V = 0

a x

Figure 2: The geometry to find the potential within a conducting cube with a potential, V = V0 placed on one side and the other sides grounded

The eigenvalues and eigenfunctions for the potential between 2 parallel conducting plates held at different potentials Y are similar. For the z direction, the complex exponentials are rewritten as hyperbolic sine and cosine functions.

Z = A sinh(γz) + B cosh(γz)

The final boundary conditions are that V = 0 for z = 0 and V = V0 for z = a. To satisfy these conditions, choose,

Z = A sinh(γz)

with γ = α2 + β2 . Then use the separated form of the initial function to write the po- tential; p

∞ V = Anm sin([nπ/a]x) sin([mπ/a]y) sinh(γz) =1 n,mP Note that the solution in terms of the two eigenvectors, sin([nπ.a]x) and sin([mπ.a]y), was obtained by choosing the two separation constants so that the eigenvectors satisfy the boundary conditions. There are 2 separation constants because there are (3 - 1) dimensions. However, the values of the expansion coefficients of the 2-D Fourier series must be found.

4 V(x)

a V(x) Vo

x a

Figure 3: An example of convergence at the boundary to produce the potential, V = V0

The remaining boundary condition at z = a is used along with the orthogonality of the eigen- functions in x and y. Multiply both sides of the above equation by sin([jπ/a]x) sin([lπ/a]y) and integrate over x and y, using orthogonality. For example integration over x gives;

a (a/2) if j = n dx sin([jπ/a]x) sin([nπ/a]x) = 0  0 if j =6 n  R Also the solution must have n, m as odd integers, otherwise the sine function vanishes. The value of the expansion coefficients is then;

a a 4 A = dx dyV0(x, y) sin[(nπ/a)x] sin[(mπ/a)y] nm a2 R0 R0 When V0(x, y) = V0 a constant then;

A = 16 V0 m, n odd nm n m π2 sinh(γa) In the later case, the final solution is;

∞ 16 V0 V = 2 sin([nπ/a]x) sin([mπ/a]y) sinh(γz) =1 n m π sinh(γa) n,mP Convergence to the potential V = V0 is shown in Figure 3 which is a typical Fourier con- vergence pattern.

5 z z z

a a a

V = 0 V = V0 V = V0

V = 0 y V = V0 y V = V0 y a a a V = 0 V = 0 V = 0 V = 0 V = 0 V = 0

a a a x x x

Figure 4: The geometry to find the potential within a conducting cube with a potential, V = V1 and V = V2 placed on two sides with the other sides grounded

y V = V1

a 8

V = V2 8 V = 0 x

Figure 5: The geometry of a slit infinite in the z direction and extending to + infinity for positive y, with different potentials on the conducting surfaces

Finally, how could one find a solution to the same problem, but with sides other than the one at z = a held at potentials other than 0? Such a solution is obtained by superposition. Look at Figure 4. The solution for the problem is obtained by superimposing solutions of the same form as in Figure 2 above. Note that Laplace’s equation is linear and the sum of two solutions is itself a solution (superposition). In his case the boundary conditions of the superimposed solution match those of the problem in question.

One should note that a solution for the potential outside a cube cannot be found using separation of variables, even though the Laplace equation is separable. Note also that the solution outside the cube could still be inside a closed surface (i.e. a closed surface at infinity). In this case, a boundary condition cannot be specified by a single separation function. In some cases, a mixed boundary condition, ie using Dirichlet conditions on part of the boundary and Neumann conditions on the remaining surface, yields a solution. In any event, the uniqueness conditions are still valid and can be used to guarantee the uniqueness of any representation of any proper solution. This is the basis of the finite element method, which is a standard computer code used in many, if not most, engineering solutions to Laplace’s equation. As a final example, find the solution for the potential within the infinite slit shown in Figure 5 where the three sides are held at different potentials. In this case, Laplace’s equation must

6 be independent of z. From the general solution above, choose eigenfunctions in y and z, but choose β = 0 to remove the z dependence. Exponential functions are needed so that the z dependence matches the boundary condition at x →∞. Thus choose;

Y(y) = A sin([nk/a]y)

Here k = π/2 and n must be odd. In this case Y is zero at y = 0 and A at y = a. Now choose X = B e−γx with γ = nk/a. Then X equals B at x = 0 and goes to zero as x →∞. The solution has the form;

∞ −γx V = Cn sin([nk/a]y) e + [V1/a]y nP odd In the above a term [V1/a]y was added to represent the potential of 2 infinite parallel plates held at potentials V1 and V0. This is the asymptotic condition as x →∞. Then match the boundary condition at x = 0.

∞ V2 = Cn sin([nk/a]y) + [V1/a]y nP odd Use orthogonality to obtain; a Cn = (a/2) dy (V2 − [V1/a]y) sin([nk/a]y) n odd R0

4 Application to the Schrodinger Equation of Quan- tum Mechanics

This example illustrates another aspect of the separation of variables technique. The Schrodinger 2 ~ 2 equation (−2m)∇ Ψ = EΨ is not Laplace’s equation. Actually it is a time suppressed ver- sion of the wave equation but the solution illustrates several points. First assume separation in the form;

Ψ = X YZ

Substitute into Schrodinger’s equation and divide by Ψ as previously. This gives;

2 2 2 1 d X 1 d Y 1 d Z 2 2 + 2 + 2 = −c X dx Y dy Z dz Here c2 = [2mE/~2]. then proceed to separate the equations obtaining the two separation constants α and β as previously.

7 2 d X + α2X = 0 dx2 2 d Y 2 2 + β X = 0 dy 2 d Z − γ2X = 0 dz2 Here γ2 = α2 + β2 − c2. The form of the solutions for the first two equations are har- monic eigenfunctions as previously. However, the last equation can be either exponential or harmonic depending on whether γ2 is positive or negative. Now if E is always positive, then for a finite solution it must be sufficient large so that a harmonic solution develops. Thus the lowest energy value, Emin cannot equal zero. Indeed if we put the particle in a box of radius a and require the solution to vanish at the box surfaces, the minimum value of E is;

2 2 π ~ 2 E = 2m (3/a )

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