The Laplace Equation and Harmonic Functions
Given a scalar field j , the Laplace equation in Cartesian coordinates is
This equation arises in many important physical applications, such as potential fields in gravitation and electro-statics, velocity potential fields in fluid dynamics, etc. Typically we are given a set of boundary conditions and we need to solve for the (unique) scalar field j that is a solution of the Laplace equation and that satisfies those boundary conditions. A solution of Laplace's equation is called a "harmonic function" (for reasons explained below). Since the Laplace equation is linear, the sum of two or more individual solutions is also a solution.
The usual approach to solving the Laplace equation is to seek a "separable" solution given by the product of independent function of x, y, and z, as follows
Every solution can be expressed as a sum of (possibly infinitely many) solutions of this form. Inserting this into the Laplace equation and evaluating the derivatives gives
Dividing through by the product A(x)B(y)C(z), this can be written in the form
Since x, y, and z can be varied independently, this equation can be identically satisfied only if each of the three terms is a constant, and these three constants sum to zero. It will be convenient to denote these constants by squared quantities, a2, b2, g2 respectively. Thus we have three separate ordinary differential equations
with the condition a2 + b2 + g2 = 0. Taking just the first of these, we can re-write it in the form
which the familiar equation of harmonic motion. This is the reason solutions of Laplace's equation are called "harmonic functions". The characteristic polynomial is
s2 - a 2 = 0
whose roots are a , so the general solution of the differential equation is
for constants a1 and a2. We arrive at similar solutions for B(y) and C(z), so the general solution of the Laplace equation of the form A(x)B(y)C(z) is
where the ai and bi are arbitrary constants, and a , b , g are any complex constants such that a2 + b2 + g2 = 0. By summing terms of this general form using the techniques of Fourier analysis we can determine the values of these constants necessary to match any given boundary conditions.
The above analysis is useful when the boundary conditions are easily expressible in terms of Cartesian coordinates, but we often encounter problems that are easier to express in terms of other, possibly curvilinear, coordinate systems. The general form of the Laplacian operator for an arbitrary (differentiable) coordinate system xk, k = 1,2,3, with the metric tensor gij is
where g is the determinant of the covariant metric tensor. If the basis vectors of the coordinate system are mutually orthogonal at each point, the fundamental line element has the diagonal form
where 
. In this case we have 
and gij = 0 for all i j. Also, the non-zero components of the contravariant metric are gii = 1/gii , so the Laplacian expressed in terms of orthogonal coordinates has the form
For example, ordinary spherical coordinates r, q, f where
are orthogonal, i.e., the basis vectors in the r, q and f directions are mutually orthogonal at each point, so we can express the Laplacian in terms of spherical coordinates by means of the preceding expression. The line element for spherical coordinates is found by simply evaluating the derivatives ds/dr, etc., along the coordinate axes. This gives
and we have hr = 1, hq = r, hf = rsin(q). Inserting these into the general expression for the Laplacian in orthogonal coordinates gives
We are often interested in situations with axial symmetry, such as when determining the potential flow around a sphere. In such cases we can take the axis of symmetry over the angle f, so we have j/f = 0, and Laplace's equation reduces to
As in the case of Cartesian rectangular coordinates, we can focus on separable solutions, i.e., solutions of the form j(r,q) = A(r)B(q). Evaluating the Laplacian of this, equating this to zero, and dividing through by A(r)B(q) gives
where dots signify derivatives of the function with respect to the argument. Since r and q are independently variable, this equation can be identically satisfied only if the sum of the first two terms is constant, and the sum of the last two terms is constant. Denoting these constant by m,n, we have the two separate ordinary linear differential equations
with the condition m + n = 0. We could solve these explicitly, but notice that the first is satisfied by setting A(r) = rn, provided we set m = n(n+1). Also, if we set B(q) = cos(q) the second equation is satisfied with n = 0. Hence the function j (r,q) = rn cos(q) is a solution of Laplace's equation provided n(n+1) = 0, which implies either n = 0 or n = -1.
More generally, we can consider solutions that are linear combinations of functions of the form
where Pn(x) is a polynomial. Inserting this into the axially symmetric form of Laplace's equation gives
where dotted functions signify derivatives of Pn(x) with respect to x. Dividing through by rn, replacing cos(q) with "x", and noting that sin(q)2 = 1-cos(q)2, this becomes
There is a unique (up to a constant factor) polynomial Pn for each value of n(n+1) = k, which implies that each such polynomial is compatible with two distinct values of n, i.e., the two roots n and n' of n2 + n = k. Of course, n need not be an integer, but if it is, we can easily determine the polynomial Pn(x) that satisfies the above equation, simply by inserting a polynomial with undetermined coefficients into the equation and solving for the coefficients. The results for the first few values of n are as shown below:
Up to a scale factor, these are the Legendre polynomials (discussed further in the note on Inverse Square Forces and Orthogonal Polyomials). We can express an axially symmetrical solution of the Laplace equation as a linear combination of these individual solutions as follows
where the ai and bi are arbitrary constants. The factors of the form Pn(cos(q )) in this expansion are called zonal harmonics. As an example, the velocity potential of the flow of an ideal, irrotational, frictionless fluid past a stationary sphere of radius R is given by the second zonal harmonic with a1 = U and b1 = UR3/2, where U is the free-stream velocity of the fluid moving in the positive x direction (assuming the x axis is our axis of symmetry). Thus we have
The first term is obviously just Ux in Cartesian coordinates, which represents a uniform flow field of constant velocity U in the positive x direction. The second term is a dipole potential, proportional to x/r3. This is the form of the "far field" potential associated with two nearby oppositely charged particles in electrostatics (aligned along the axis of symmetry). It's interesting that the superposition of these two simple fields gives the potential flow field around a sphere, as shown by the fact that the radial velocity vanishes at r = R. (See the note on Potential Flow and d'Alembert's Paradox for more.)
It's useful to notice that if a continuous function j(x,y,z) is a solution of Laplace's equation, then the mean value of j on any sphere centered on the point p is equal to the value of j at the point p. One proof of this is presented in Differential Operators and the Divergence Theorem. Another way of seeing this is to expand j(x,y,z) into a power series up to second order about the point p, after shifting the coordinates to place p at the origin, as follows
Recall that if q denotes the geographical latitude (zero at the equator) and f the longitude, then the mean value of a function f(q,f) over the surface of a sphere of radius R is given by the double integral
Making the substitutions x = Rcos(f)cos(q), y = Rsin(f)cos(q), z = Rsin(q) into the expression for j and evaluating the mean value over this sphere gives
Since we have placed the point p at the origin, the value of j at p is just c0, whereas this shows that the average value of j on the sphere (up to second order) has an additional term equal to the average of the three coefficients of the squared coordinates times the square of the radius. But if we insert our expansion for j(x,y,z) into Laplace's equation we find that c7 + c8 + c9 = 0, and therefore jmean = j(xp,yp,zp).
Several useful facts follow directly from this simple proposition. For example, it's clear that a continuous differentiable harmonic function j (i.e., a solution of Laplace's equation) cannot contain a local maximum or minimum, because by definition such points have values of j properly greater then (resp. less than) the values of j for all the surrounding points, which would make it impossible for the average on the surface of a sphere surrounding the point to equal the value of j at that point.
From this it follows that if the value of a harmonic function j has the constant value c over an entire surface that completely encloses a region of space, then the value of j must be c throughout that region. This is because if the value was anything other than c at any point in the interior of the region, then the region must contain a local maximum or minimum, which is already ruled out.
Having established this, we can also see that the solution of Laplace's equation satisfying a complete set of boundary conditions is unique. To prove this, suppose two distinct harmonic functions j1 and j2 have the same values on a closed surface, but have different values in the interior of the enclosed region. We know that harmonic functions are additive, so j1 - j2 is also a harmonic function, and it's value is zero over the closed surface. But this implies the value of this difference is zero throughout the interior region enclosed by the surface (by the preceding proposition), so j1 and j2 are identical throughout the region.