Required math: calculus
Required physics: electrostatics
Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Sec 3.3.
Laplace’s equation governs the electric potential in regions where there is no charge. Its form is
We’ve seen that, for a particular set of boundary conditions, solutions to Laplace’s equation are unique. That fact can be used to invent the method of images, in which a complex problem can be solved by inventing a simpler problem that has the same boundary conditions.
However, the method of images works only in a few special (and fairly contrived) situations. In the more general case, we need a way of solving Laplace’s equation directly.
A method which we have already met in quantum mechanics when solving Schrödinger’s equation is that of separation of variables. In general, the potential is a function of all three spatial coordinates: . We try to find a solution by assuming that is a product of three functions, each of which is a function of only one spatial coordinate:
Substituting this into Laplace’s equation, we get
We can divide through by to get
The key point in this equation is that each term in the sum is a function of only one of the three independent variables , and . The fact that these variables are independent is important, for it means that the only way this equation can be satisfied is if each term in the sum is a constant. Suppose this wasn’t true; for example, suppose the first term in the sum was some function that actually does vary with . Then we could hold and constant and vary , causing this first term to vary. In this case we cannot satisfy the overall equation, since if we found some value of for which the sum of the three terms was zero, changing would change the first term but not the other two, so the overall sum would no longer be zero.
Thus we can say that
where the three constants satisfy
Equations of this form have one of two types of solution (well, three, if we consider the constant to be zero, but that’s not usually very interesting), depending on whether the constant is positive or negative. For example, if , we can write it as and the solution has the form
for some constants and .
If , we can write it as and the solution has the form
for some constants and . The constants in each case must be determined from the boundary conditions. Similar solutions exist for and .
Now you might be wondering whether the assumption that the potential is the product of three separate functions is valid. After all, it does seem to be a rather severe restriction on the solution. It’s easiest to see whether this assumption is valid by considering a particular example.
The key consideration in any Laplace problem is the specification of the boundary conditions. As a first example, suppose we have the following setup. We have two semi-infinite conducting plates that lie parallel to the plane, with their edges lying on the axis (that is, at ). One plate is at and the other is at . Both plates are grounded, so their potential is constant at .
The strip between the plates at is filled with another substance (not a conductor, so the potential can vary across it) that is insulated from the two plates, and its potential is some function . Solve Laplace’s equation to find the potential between the plates.
It’s important to note what the boundary conditions are here. The two plates are held at so provide boundary conditions at and :
The strip at provides another boundary condition
Finally, we can impose the condition that the potential drops to zero as we get infinitely far from the strip at so we have
The first thing to notice is that none of these boundary conditions depends on , so we can take so that above. This means that the problem effectively reduces to a two-dimensional problem with the condition
Now we must make a choice as to which of the constants is positive and which is negative. Suppose we chose . Then we would get
Looking at the boundary conditions above, we see as we need . But since is the sum of two oscillating functions, this can’t happen unless or , which isn’t a valid solution since that would mean that everywhere, and that violates the condition at .
So we can try the other choice: . This gives
Now as the negative exponential term drops to zero, so we need only require that and we get
From this choice, we know that and
From the condition when we get
Finally, from when we get
where is a positive integer. It must be non-zero, since again gives us everywhere. It must not be negative, since that would give us a negative which would give the wrong behaviour for .
So our solution so far is
At this stage, you might think we’ve solved ourself into a corner, since we haven’t used the final boundary condition which is that . From our solution so far, we have
so unless we choose to be one of those sine functions, we’re stuffed. Does this mean that the separation of variables method doesn’t work here?
Not quite. The crucial point is that Laplace’s equation is linear (the derivatives occur to the first power only), so any number of separate solutions can be added together to give another solution. That is, if and are solutions, then so is . The separation of variables method has actually given us an infinite number of solutions (one for each value of ) so we can create yet more solutions by adding together any combination of these individual solutions. In particular, we can say
for some choice of coefficients . (Here, we’ve simply combined the two constants and for each value of to give the constant .)
How can we find these coefficients? In general, this can be fairly tricky, but for certain boundary conditions, it turns out to be fairly straightforward. For the boundary condition we have here, , things aren’t too bad. We have
Some readers might recognize this as a Fourier series, and there is a clever technique that can be used to find the in such a case. We multiply through by and integrate from 0 to :
The integrals in the sum on the right are fairly straightforward, and we get
As usual for physicists, the problem of proving that a Fourier series exists and converges for any given function is left to the mathematicians, but for pretty well any function of physical relevance, this technique works. Although the example here has a clean solution, many other problems do not. If the boundaries are of some exotic shape, then it becomes impossible to specify things in such a way that we have a clean Fourier series to work with. As usual in such cases, we need to resort to numerical solution of Laplace’s equation, and for that we need a computer.