The Method of Eigenfunction Expansion

While most introductory PDE courses spend a great deal of time investigating solutions using separation of variables, a powerful method that is good to know for studies in theoretical physics is the method of eigenfunction expansion. This method is useful because in many cases, equations cannot be solved using separation of variables. To understand the method, in what follows you can think in terms of expanding the function which solves the equation in a Fourier series. We then solve mode by mode-turning a single partial differential equation into a set of elementary ordinary differential equations that can be solved by inspection. Rather than trying to explain the techniques in words or with difficult to understand abstract theorems, let’s demonstrate how to use the method with two concrete examples.

In our first attempt to tackle the method of eigenfunction expansion, we will consider a standard example from quantum mechanics, a particle trapped in an infinite square well of width a. Inside the well, the potential is zero and therefore the Schrodinger equation is written as

For this problem we will take . The particle cannot be found outside the boundaries, therefore the boundary conditions for the problem are

As can be seen in any standard quantum mechanics text like Griffiths, this problem is easily solved using separation of variables. While that elementary option is available in this case, we will use the method of eigenfunction expansion so that we can demonstrate the method using a simple equation.

Step One: Identify a differential operator

The first step is to identify a differential operator L that can be used to solve an eigenvalue equation. In problems that involve space and time, its a good bet to choose spatial derivative operators as L. In this case the choice is immediately obvious. We let

Step Two: Find the eigenvalues and eigenfunctions of the operator

We denote the eigenfunctions of L by and the eigenvalues by . The eigenvalue equation is then written as

For convenience, we set (the reader can later verify that this would follow automatically). The differential equation to be solved is then written as

This part of the solution will be familiar to those who have had quantum mechanics. Basically, we are solving the time-independent Schrodinger equation. Students will no doubt recognize the above equation as the harmonic oscillator equation with solutions

The boundary conditions in the problem tell us that. This applies to the eigenfunctions as well. Therefore, we find that

Now since , this means that . Next we apply the second boundary condition, . This gives

This can only be true if , where . In other words, we can write the solution as

In quantum theory, the constants can be found using normalization. However that isn’t necessary to understand the solution method being explored here, so we will ignore it from here on out. The reader familiar with quantum theory will also note that the eigenvalues are related to the energy.

Step Three: Expand the function in a series and solve for the time-dependent coefficients of the expansion

Next, we write the function in an expansion of the form

And then apply the partial differential equation. In our case, recall that we had

Inserting the series expansion for we obtain, on the left side

And on the right

Now we found that , and so we can write this as

Now let’s equate both terms from the left hand side and the right hand side. The Schrodinger equation becomes

Step Four: Solve mode-by-mode

We now have two series expansions using the same eigenfunctions that are equal, therefore each individual term in the series on the left must equal the corresponding term on the right. This means that the coefficient functions satisfy the following ordinary differential equation

Cross multiplying constant terms gives us

where we have denoted the particle energy by . This can be readily integrated giving

Normally, the constant of integration would be determined using initial conditions stated in the problem. In this case once again normalization determines the constants in this problem. We will ignore what those values are and simply write the solution, which is