8.4: Series Representations of Green's Functions

There are times that it might not be so simple to find the Green's function in the simple closed form that we have seen so far. However, there is a method for determining the Green's functions of Sturm-Liouville boundary value problems in the form of an eigenfunction expansion. We will finish our discussion of Green's functions for ordinary differential equations by showing how one obtains such series representations. (Note that we are really just repeating the steps towards developing eigenfunction expansion which we had seen in Chapter 6.)

We will make use of the complete set of eigenfunctions of the differential operator, \(\mathcal{L}\), satisfying the homogeneous boundary conditions:

\[\mathcal{L}\left[\phi_{n}\right]=-\lambda_{n} \sigma \phi_{n}, \quad n=1,2, \ldots \nonumber \]

We want to find the particular solution \(y\) satisfying \(\mathcal{L}[y]=f\) and homogeneous boundary conditions. We assume that

\[y(x)=\sum_{n=1}^{\infty} a_{n} \phi_{n}(x) . \nonumber \]

Inserting this into the differential equation, we obtain

\[\mathcal{L}[y]=\sum_{n=1}^{\infty} a_{n} \mathcal{L}\left[\phi_{n}\right]=-\sum_{n=1}^{\infty} \lambda_{n} a_{n} \sigma \phi_{n}=f . \nonumber \]

This has resulted in the generalized Fourier expansion

\[f(x)=\sum_{n=1}^{\infty} c_{n} \sigma \phi_{n}(x) \nonumber \]

with coefficients

\[c_{n}=-\lambda_{n} a_{n} . \nonumber \]

We have seen how to compute these coefficients earlier in the text. We multiply both sides by \(\phi_{k}(x)\) and integrate. Using the orthogonality of the eigenfunctions,

\[\int_{a}^{b} \phi_{n}(x) \phi_{k}(x) \sigma(x) d x=N_{k} \delta_{n k}, \nonumber \]

one obtains the expansion coefficients (if \(\lambda_{k} \neq 0\) )

\[a_{k}=-\dfrac{\left(f, \phi_{k}\right)}{N_{k} \lambda_{k}}, \nonumber \]

where \(\left(f, \phi_{k}\right) \equiv \int_{a}^{b} f(x) \phi_{k}(x) d x\).

As before, we can rearrange the solution to obtain the Green's function. Namely, we have

\[y(x)=\sum_{n=1}^{\infty} \dfrac{\left(f, \phi_{n}\right)}{-N_{n} \lambda_{n}} \phi_{n}(x)=\int_{a}^{b} \underbrace{\sum_{n=1}^{\infty} \dfrac{\phi_{n}(x) \phi_{n}(\xi)}{-N_{n} \lambda_{n}}}_{G(x, \xi)} f(\xi) d \xi \nonumber \]

Therefore, we have found the Green's function as an expansion in the eigenfunctions:

\[G(x, \xi)=\sum_{n=1}^{\infty} \dfrac{\phi_{n}(x) \phi_{n}(\xi)}{-\lambda_{n} N_{n}} . \label{8.72} \]

Example 8.10. Eigenfunction Expansion Example

We will conclude this discussion with an example. Consider the boundary value problem

\[y^{\prime \prime}+4 y=x^{2}, \quad x \in(0,1), \quad y(0)=y(1)=0 . \nonumber \]

The Green's function for this problem can be constructed fairly quickly for this problem once the eigenvalue problem is solved. We will solve this problem three different ways in order to summarize the methods we have used in the text.
The eigenvalue problem is

\[\phi^{\prime \prime}(x)+4 \phi(x)=-\lambda \phi(x), \nonumber \]

where \(\phi(0)=0\) and \(\phi(1)=0\). The general solution is obtained by rewriting the equation as

\[\phi^{\prime \prime}(x)+k^{2} \phi(x)=0 \nonumber \]

where

\[k^{2}=4+\lambda . \nonumber \]

Solutions satisfying the boundary condition at \(x=0\) are of the form

\[\phi(x)=A \sin k x . \nonumber \]

Forcing \(\phi(1)=0\) gives

\[0=A \sin k \Rightarrow k=n \pi, \quad k=1,2,3 \ldots \nonumber \]

So, the eigenvalues are

\[\lambda_{n}=n^{2} \pi^{2}-4, \quad n=1,2, \ldots \nonumber \]

and the eigenfunctions are

\[\phi_{n}=\sin n \pi x, \quad n=1,2, \ldots \nonumber \]

We need the normalization constant, \(N_{n}\). We have that

\[N_{n}=\left\|\phi_{n}\right\|^{2}=\int_{0}^{1} \sin ^{2} n \pi x=\dfrac{1}{2} . \nonumber \]

We can now construct the Green’s function for this problem using Equation (8.72).

\[G(x, \xi)=2 \sum_{n=1}^{\infty} \dfrac{\sin n \pi x \sin n \pi \xi}{\left(4-n^{2} \pi^{2}\right)} . \label{8.73} \]

We can use this Green’s function to determine the solution of the boundary value problem. Thus, we have

\[\begin{aligned}
y(x) &=\int_{0}^{1} G(x, \xi) f(\xi) d \xi \\
&=\int_{0}^{1}\left(2 \sum_{n=1}^{\infty} \dfrac{\sin n \pi x \sin n \pi \xi}{\left(4-n^{2} \pi^{2}\right)}\right) \xi^{2} d \xi \\
&=2 \sum_{n=1}^{\infty} \dfrac{\sin n \pi x}{\left(4-n^{2} \pi^{2}\right)} \int_{0}^{1} \xi^{2} \sin n \pi \xi d \xi \\
&=2 \sum_{n=1}^{\infty} \dfrac{\sin n \pi x}{\left(4-n^{2} \pi^{2}\right)}\left[\dfrac{\left(2-n^{2} \pi^{2}\right)(-1)^{n}-2}{n^{3} \pi^{3}}\right]
\end{aligned} \label{8.74} \]

We can compare this solution to the one one would obtain if we did not employ Green’s functions directly. The eigenfunction expansion method for solving boundary value problems, which we saw earlier proceeds as follows. We assume that our solution is in the form

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

Inserting this into the differential equation \(\mathcal{L}[y]=x^{2}\) gives

\[\begin{aligned}
x^{2} &=\mathcal{L}\left[\sum_{n=1}^{\infty} c_{n} \sin n \pi x\right] \\
&=\sum_{n=1}^{\infty} c_{n}\left[\dfrac{d^{2}}{d x^{2}} \sin n \pi x+4 \sin n \pi x\right] \\
&=\sum_{n=1}^{\infty} c_{n}\left[4-n^{2} \pi^{2}\right] \sin n \pi x
\end{aligned} \label{8.75} \]

We need the Fourier sine series expansion of \(x^{2}\) on \([0,1]\) in order to determine the \(c_{n}\)'s. Thus, we need

\[\begin{aligned}
b_{n} &=\dfrac{2}{1} \int_{0}^{1} x^{2} \sin n \pi x \\
&=2\left[\dfrac{\left(2-n^{2} \pi^{2}\right)(-1)^{n}-2}{n^{3} \pi^{3}}\right], \quad n=1,2, \ldots
\end{aligned} \label{8.76} \]

Thus,

\[x^{2}=2 \sum_{n=1}^{\infty}\left[\dfrac{\left(2-n^{2} \pi^{2}\right)(-1)^{n}-2}{n^{3} \pi^{3}}\right] \sin n \pi x . \nonumber \]

Inserting this in Equation (8.75), we find

\[2 \sum_{n=1}^{\infty}\left[\dfrac{\left(2-n^{2} \pi^{2}\right)(-1)^{n}-2}{n^{3} \pi^{3}}\right] \sin n \pi x=\sum_{n=1}^{\infty} c_{n}\left[4-n^{2} \pi^{2}\right] \sin n \pi x . \nonumber \]

Due to the linear independence of the eigenfunctions, we can solve for the unknown coefficients to obtain

\[c_{n}=2 \dfrac{\left(2-n^{2} \pi^{2}\right)(-1)^{n}-2}{\left(4-n^{2} \pi^{2}\right) n^{3} \pi^{3}} . \nonumber \]

Therefore, the solution using the eigenfunction expansion method is

\[\begin{aligned}
y(x) &=\sum_{n=1}^{\infty} c_{n} \phi_{n}(x) \\
&=2 \sum_{n=1}^{\infty} \dfrac{\sin n \pi x}{\left(4-n^{2} \pi^{2}\right)}\left[\dfrac{\left(2-n^{2} \pi^{2}\right)(-1)^{n}-2}{n^{3} \pi^{3}}\right] .
\end{aligned} \label{8.77} \]

We note that this is the same solution as we had obtained using the Green's function obtained in series form.

One remaining question is the following: Is there a closed form for the Green's function and the solution to this problem? The answer is yes! We note that the differential operator is a special case of the example done is section 8.2.2. Namely, we pick \(\omega=2\). The Green's function was already found in that section. For this special case, we have

\[G(x, \xi)=\left\{\begin{array}{l}
-\dfrac{\sin 2(1-\xi) \sin 2 x}{2 \sin 2}, 0 \leq x \leq \xi \\
-\dfrac{\sin 2(1-x) \sin 2 \xi}{2 \sin 2}, \xi \leq x \leq 1
\end{array}\right. \label{8.78} \]

What about the solution to the boundary value problem? This solution is given by

\[\begin{aligned}
y(x) &=\int_{0}^{1} G(x, \xi) f(\xi) d \xi \\
&=-\int_{0}^{x} \dfrac{\sin 2(1-x) \sin 2 \xi}{2 \sin 2} \xi^{2} d \xi+\int_{x}^{1} \dfrac{\sin 2(\xi-1) \sin 2 x}{2 \sin 2} \xi^{2} d \xi \\
&=-\dfrac{1}{4 \sin 2}\left[-x^{2} \sin 2-\sin 2 \cos ^{2} x+\sin 2+\cos 2 \sin x \cos x+\sin x \cos x\right] . \\
&=-\dfrac{1}{4 \sin 2}\left[-x^{2} \sin 2+\left(1-\cos ^{2} x\right) \sin 2+\sin x \cos x(1+\cos 2)\right] . \\
&\left.=-\dfrac{1}{4 \sin 2}\left[-x^{2} \sin 2+2 \sin ^{2} x \sin 1 \cos 1+2 \sin x \cos x \cos ^{2} 1\right)\right] . \\
&=-\dfrac{1}{8 \sin 1 \cos 1}\left[-x^{2} \sin 2+2 \sin x \cos 1(\sin x \sin 1+\cos x \cos 1)\right] . \\
&=\dfrac{x^{2}}{4}-\dfrac{\sin x \cos (1-x)}{4 \sin 1} .
\end{aligned} \label{8.79} \]

In Figure 8.4 we show a plot of this solution along with the first five terms of the series solution. The series solution converges quickly.

Figure 8.4. Plots of the exact solution to Example 8.10 with the first five terms of the series solution.

As one last check, we solve the boundary value problem directly, as we had done in Chapter 4. Again, the problem is

\[y^{\prime \prime}+4 y=x^{2}, \quad x \in(0,1), \quad y(0)=y(1)=0 . \nonumber \]

The problem has the general solution

\[y(x)=c_{1} \cos 2 x+c_{2} \sin 2 x+y_{p}(x) \nonumber \]

where \(y_{p}\) is a particular solution of the nonhomogeneous differential equation. Using the Method of Undetermined Coefficients, we assume a solution of the form

\[y_{p}(x)=A x^{2}+B x+C . \nonumber \]

Inserting this in the nonhomogeneous equation, we have

\[2 A+4\left(A x^{2}+B x+C\right)=x^{2}, \nonumber \]

Thus, \(B=0,4 A=1\) and \(2 A+4 C=0\). The solution of this system is

\[A=\dfrac{1}{4}, \quad B=0, \quad C=-\dfrac{1}{8} . \nonumber \]

So, the general solution of the nonhomogeneous differential equation is

\[y(x)=c_{1} \cos 2 x+c_{2} \sin 2 x+\dfrac{x^{2}}{4}-\dfrac{1}{8} . \nonumber \]

We now determine the arbitrary constants using the boundary conditions.

We have

\[\begin{aligned}
0 &=y(0) \\
&=c_{1}-\dfrac{1}{8} \\
0 &=y(1) \\
&=c_{1} \cos 2+c_{2} \sin 2+\dfrac{1}{8}
\end{aligned} \label{8.80} \]

Thus, \(c_{1}=\dfrac{1}{8}\) and

\[c_{2}=-\dfrac{\dfrac{1}{8}+\dfrac{1}{8} \cos 2}{\sin 2} . \nonumber \]

Inserting these constants in the solution we find the same solution as before.

\[\begin{aligned}
y(x) &=\dfrac{1}{8} \cos 2 x-\left[\dfrac{\dfrac{1}{8}+\dfrac{1}{8} \cos 2}{\sin 2}\right] \sin 2 x+\dfrac{x^{2}}{4}-\dfrac{1}{8} \\
&=\dfrac{\cos 2 x \sin 2-\sin 2 x \cos 2-\sin 2 x}{8 \sin 2}+\dfrac{x^{2}}{4}-\dfrac{1}{8} \\
&=\dfrac{\left(1-2 \sin ^{2} x\right) \sin 1 \cos 1-\sin x \cos x\left(2 \cos ^{2} 1-1\right)-\sin x \cos x-\sin 1 \cos 1}{8 \sin 1 \cos 1}+\dfrac{x^{2}}{4} \\
&=-\dfrac{\sin ^{2} x \sin 1+\sin x \cos x \cos 1}{4 \sin 1}+\dfrac{x^{2}}{4} \\
&=\dfrac{x^{2}}{4}-\dfrac{\sin x \cos (1-x)}{4 \sin 1} .
\end{aligned} \label{8.81} \]