6.3: The Eigenfunction Expansion Method

In section 4.3.2 we saw generally how one can use the eigenfunctions of a differential operator to solve a nonhomogeneous boundary value problem. In this chapter we have seen that Sturm-Liouville eigenvalue problems have the requisite set of orthogonal eigenfunctions. In this section we will apply the eigenfunction expansion method to solve a particular nonhomogenous boundary value problem.

Recall that one starts with a nonhomogeneous differential equation

$$\mathcal{L} y=f \nonumber$$

where $y(x)$ is to satisfy given homogeneous boundary conditions. The method makes use of the eigenfunctions satisfying the eigenvalue problem

$$\mathcal{L} \phi_{n}=-\lambda_{n} \sigma \phi_{n} \nonumber$$

subject to the given boundary conditions. Then, one assumes that $y(x)$ can be written as an expansion in the eigenfunctions,

$$y(x)=\sum_{n=1}^{\infty} c_{n} \phi_{n}(x) \nonumber$$

and inserts the expansion into the nonhomogeneous equation. This gives

$$f(x)=\mathcal{L}\left(\sum_{n=1}^{\infty} c_{n} \phi_{n}(x)\right)=-\sum_{n=1}^{\infty} c_{n} \lambda_{n} \sigma(x) \phi_{n}(x). \nonumber$$

The expansion coefficients are then found by making use of the orthogonality of the eigenfunctions. Namely, we multiply the last equation by $\phi_{m}(x)$ and integrate. We obtain

$$\int_{a}^{b} f(x) \phi_{m}(x) d x=-\sum_{n=1}^{\infty} c_{n} \lambda_{n} \int_{a}^{b} \phi_{n}(x) \phi_{m}(x) \sigma(x) d x \nonumber$$

Orthogonality yields

$$\int_{a}^{b} f(x) \phi_{m}(x) d x=-c_{m} \lambda_{m} \int_{a}^{b} \phi_{m}^{2}(x) \sigma(x) d x \nonumber$$

Solving for $c_{m}$, we have

$$c_{m}=-\dfrac{\int_{a}^{b} f(x) \phi_{m}(x) d x}{\lambda_{m} \int_{a}^{b} \phi_{m}^{2}(x) \sigma(x) d x} \nonumber$$

Example $\PageIndex{1}$

As an example, we consider the solution of the boundary value problem

$$\left(x y^{\prime}\right)^{\prime}+\dfrac{y}{x} =\dfrac{1}{x}, \quad x \in[1, e], \label{6.23}$$

$$y(1) =0=y(e) . \label{6.24}$$

Solution

This equation is already in self-adjoint form. So, we know that the associated Sturm-Liouville eigenvalue problem has an orthogonal set of eigenfunctions. We first determine this set. Namely, we need to solve

$$\left(x \phi^{\prime}\right)^{\prime}+\dfrac{\phi}{x}=-\lambda \sigma \phi, \quad \phi(1)=0=\phi(e) . \label{6.25}$$

Rearranging the terms and multiplying by $x$, we have that

$$x^{2} \phi^{\prime \prime}+x \phi^{\prime}+(1+\lambda \sigma x) \phi=0 . \nonumber$$

This is almost an equation of Cauchy-Euler type. Picking the weight function $\sigma(x)=\dfrac{1}{x}$, we have

$$x^{2} \phi^{\prime \prime}+x \phi^{\prime}+(1+\lambda) \phi=0 . \nonumber$$

This is easily solved. The characteristic equation is

$$r^{2}+(1+\lambda)=0 . \nonumber$$

One obtains nontrivial solutions of the eigenvalue problem satisfying the boundary conditions when $\lambda>-1$. The solutions are

$$\phi_{n}(x)=A \sin (n \pi \ln x), \quad n=1,2, \ldots \nonumber$$

where $\lambda_{n}=n^{2} \pi^{2}-1$
It is often useful to normalize the eigenfunctions. This means that one chooses $A$ so that the norm of each eigenfunction is one. Thus, we have

$$\begin{aligned} 1 &=\int_{1}^{e} \phi_{n}(x)^{2} \sigma(x) d x \\[4pt] &=A^{2} \int_{1}^{e} \sin (n \pi \ln x) \dfrac{1}{x} d x \\[4pt] &=A^{2} \int_{0}^{1} \sin (n \pi y) d y=\dfrac{1}{2} A^{2} \end{aligned} \label{6.26}$$

Thus, $A=\sqrt{2}$
We now turn towards solving the nonhomogeneous problem, $\mathcal{L} y=\dfrac{1}{x}$. We first expand the unknown solution in terms of the eigenfunctions,

$$y(x)=\sum_{n=1}^{\infty} c_{n} \sqrt{2} \sin (n \pi \ln x) \nonumber$$

Inserting this solution into the differential equation, we have

$$\dfrac{1}{x}=\mathcal{L} y=-\sum_{n=1}^{\infty} c_{n} \lambda_{n} \sqrt{2} \sin (n \pi \ln x) \dfrac{1}{x} \nonumber$$

Next, we make use of orthogonality. Multiplying both sides by $\phi_{m}(x)= \sqrt{2} \sin (m \pi \ln x)$ and integrating, gives

$$\lambda_{m} c_{m}=\int_{1}^{e} \sqrt{2} \sin (m \pi \ln x) \dfrac{1}{x} d x=\dfrac{\sqrt{2}}{m \pi}\left[(-1)^{m}-1\right] \nonumber$$

Solving for $c_{m}$, we have

$$c_{m}=\dfrac{\sqrt{2}}{m \pi} \dfrac{\left[(-1)^{m}-1\right]}{m^{2} \pi^{2}-1} . \nonumber$$

Finally, we insert our coefficients into the expansion for $y(x)$. The solution is then

$$y(x)=\sum_{n=1}^{\infty} \dfrac{2}{n \pi} \dfrac{\left[(-1)^{n}-1\right]}{n^{2} \pi^{2}-1} \sin (n \pi \ln (x)) . \nonumber$$