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

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

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

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