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*=-*ω*^{2}. We then have

,

and tells us that , and thus for positive integer *n*,

Thus, we have *λ*=-*ω*^{2}=-*n*^{2}π^{2}, 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 *r*^{2} 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 *r ^{m}*:

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*=

*ω*

^{2}, we have . Second, the requirement that means that

*ω*=

*n*,

*n*an integer:

.

Examining our radial solution for

*λ*=

*ω*

^{2}=

*n*

^{2}, 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

*A*

_{2}=0 for

*n*>0. For

*n*=0, we have radial function , and to avoid a singularity at the origin, we again set

*A*

_{2}=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:

Tags: Boundary Conditions, Cartesian Coordinates, Dirichlet Problem, Laplace's Equation, Laplacian, Math, Monday Math, Polar Coordinates, Separation of Variables

December 22, 2008 at 2:15 am |

[…] Math 51 By twistedone151 Last week, we looked at solving Laplace’s equation, in two dimensions. In particular, I demonstrated […]