## Monday Math 50

Let us consider Laplace’s equation,  in two dimensions. In Cartesian coordinates, this is . We can see by immediate inspection that linear functions f(x,y)=ax+by+c satisfy this. We also see that terms of the form axy satisfy the equation.

A useful method for finding general solutions is the separation of variables for partial differential equations technique. Here, we substitute in a function of the form f(x,y)=X(x)Y(y). Then Laplace’s equation becomes:

Dividing both sides by f(x,y)=X(x)Y(y), we have:

Now, we note that  is a function of x only, while  is a function of y only. By varying x while holding y fixed, and vice versa, we see that the sum of these two can be zero for all x and y if and only if both functions are constants, which sum to zero. Thus, we have ,  for some constant λ.
Both of these are second order differential equations:  has solution , while  has solution .
Combining these, we have solutions of the form , and we can sum any of these to produce another solution.

For a more specific example, let us choose the region 0≤x,y≤1 with boundary conditions (Dirichlet) , , , .
Then we have, from the above, solutions of the form , , . To get a solution , we want
,
for a function . For a nonzero solution which obeys , we need to be periodic in x (complex exponential, rather than real), so we need λ<0. Thus, let λ=-ω2. We then have
,
and  tells us that , and thus  for positive integer n, 
Thus, we have λ=-ω2=-n2π2, and .  tells us , which means
,
and so
,
and our solutions for these three parts of the boundary are .
Consider the Fourier series for the square wave function with period 2π , which is

Compare to 
With  for odd n and  for even n, we have when y=1 the terms of the Fourier series for . Thus, our last boundary is met, and our solution found, by
.

Here is a graph of the partial series with the first 25 terms:

Now, let us consider polar coordinates, useful for solving Laplace’s equation on the disk. In polar coordinates, the laplacian is
.
Trying a function of the form  in Laplace’s equation, we get:

Multiplying that equation by r2 and then dividing by , we have:

As the terms in the first set of parentheses is a function of r only, and the second set holds a function of θ only, these two functions are constants, and so we have , .
The latter equation is easily solved:  has general solution . Meanwhile, the second order linear differential equation
 is a second order Cauchy-Euler equation, which we solve by substituting a trial solution of the form rm:

for λ≠0. With λ=0, we have the equation: , which, with , gives the first order equation , which is separable:

So we have  for λ≠0, and  for λ=0.
Note that if our region covers all angles θ, then we have the requirement that  must be periodic, with . This tells us first that λ≥0; with λ=ω2, we have . Second, the requirement that  means that ω=n, n an integer:
.

Examining our radial solution for λ=ω2=n2, we have for positive n the radial function . The latter term has a singularity at the origin, so if the region over which we wish to solve contains the origin, we must have A2=0 for n>0. For n=0, we have radial function , and to avoid a singularity at the origin, we again set A2=0. Thus, the solutions for Laplace’s equation for a domain in polar coordinates containing the origin (and all angles θ) are of the form , n=0,1,2,3,…
For example, n=0 is the constant function; n=1 gives
; these confirm our early result of linear functions in x and y being solutions. n=2 gives us  and , this latter which we also noted earlier.

Suppose we want to solve Laplace’s equation on the unit disk with (Dirichlet) boundary condition . Then we see that we want to combine solutions of the form . As our boundary condition is an odd function of θ, the cosine terms disappear, and we have , plugging in r=1, and noting that our boundary condition is the square wave function we defined earlier: , we put , giving solution


Here is a graph of the partial series with the first 25 terms: