Constructing Green’s Function from Eigenfunctions
Continuing from the last tutorial, we can also develop the Green’s function notion by superposing the eigenfunctions of the corresponding Hermitian operators. Let the general ODE be
\begin{align}
\mathcal{L}[y](x) = f(x) \tag{1}
\end{align}
where \( \mathcal{L} \) is a Hermitian operator (including the Sturm-Liouville case) as before. From the results in this tutorial, we know that the eigenvalues and eigenfunctions are related by
\begin{align}
\mathcal{L}[\varphi_j] = \lambda_j \rho(x) \varphi_j \tag{2}
\end{align}
and are real and orthogonal, and with the premise that each of the \( \varphi_j \) satisfies the (homogeneous) boundary conditions. Furthermore, by the Spectral Theorem in this tutorial, we know that we can express \( y(x) \) in (1), if it also obeys the same boundary condition, as the superposition of the eigenfunctions:
\begin{align}
y &= \sum_{j=1}^{\infty} c_j\varphi_j = c_1\varphi_1 + c_2\varphi_2 + \cdots \tag{3} \\
\end{align}
Substituting (3) into (1), using (2), and then taking the inner product with \( \varphi_k \), we have
\begin{align}
\mathcal{L}[\sum_{j=1}^{\infty} c_j\varphi_j](x) = \sum_{j=1}^{\infty} c_j\mathcal{L}[\varphi_j] &= f \\
\sum_{j=1}^{\infty} c_j\lambda_j \rho \varphi_j &= f \\
\langle \sum_{j=1}^{\infty} c_j\lambda_j \rho \varphi_j, \varphi_k \rangle &= \langle f, \varphi_k \rangle \tag{4}
\end{align}
where we have passed the now continuous operator \(\mathcal{L}\) into the infinite sum, and now we are going to do the same for the inner product operation:
\begin{align}
\sum_{j=1}^{\infty} c_j\lambda_j \langle \rho \varphi_j, \varphi_k \rangle &= \langle f, \varphi_k \rangle \\
c_k\lambda_k \langle \rho \varphi_k, \varphi_k \rangle &= \langle f, \varphi_k \rangle \\
c_k &= \frac{1}{\lambda_k} \frac{\langle f, \varphi_k \rangle}{\langle \rho \varphi_k, \varphi_k \rangle} \tag{5}
\end{align}
The second line is due to the weighted orthogonality of the eigenfunctions (Theorem 2 in this tutorial). Hence plugging this into (3) and expressing it in terms of integrals, we arrive at the solution
\begin{align}
y &= \sum_{j=1}^{\infty} \frac{1}{\lambda_j} \frac{\langle f, \varphi_j \rangle}{\langle \rho \varphi_j, \varphi_j \rangle} \varphi_j(x) \\
&= \sum_{j=1}^{\infty} \frac{1}{\lambda_j} \frac{\int_I f(z)\varphi_j^*(z) dz}{\int_I \rho(z)\varphi_j(z)\varphi_j^*(z) dz} \varphi_j(x) \tag{6}
\end{align}
We can make it even simpler if we assume the eigenfunctions \( \varphi_j \) have been normalized so that the weighted norm \( \langle \rho \varphi_k, \varphi_k \rangle \) in the denominator is \(1\):
\begin{align}
y &= \sum_{j=1}^{\infty} (\frac{1}{\lambda_j} \int_I f(z)\varphi_j^*(z) dz) \varphi_j(x) \\
&= \int_I [\sum_{j=1}^{\infty} \frac{1}{\lambda_j} \varphi_j(x)\varphi_j^*(z)] f(z)dz \tag{7}
\end{align}
Here we have moved the summation under the integral sign, allowed by the \( L^2 \) convergence. By comparing (7) to (5) in this post, we immediately identify that
\begin{align}
G(x,z) = \sum_{j=1}^{\infty} \frac{1}{\lambda_j} \varphi_j(x)\varphi_j^*(z) \tag{8}
\end{align}
is the Green’s function for (1). Clearly, this also shows that the Green’s function of a Hermitian operator is conjugate-symmetric (in \(x,z\)).
A variant of (1) will be
\begin{align}
\mathcal{L}[y](x) -\mu\rho(x)y(x) = f(x) \tag{9}
\end{align}
so that the equation is “shifted” by a fixed parameter \( \mu \) times the dependent variable \( y \) itself (and the same weighting \( \rho \) as in (2)). The above procedure can be readily generalized, where (4) and (5) become
\begin{align}
\sum_{j=1}^{\infty} (c_j\mathcal{L}[\varphi_j] -c_j\mu\rho\varphi_j) &= \sum_{j=1}^{\infty} \langle f, \rho \varphi_j \rangle \varphi_j \\
\sum_{j=1}^{\infty} c_j(\lambda_j-\mu)\rho \varphi_j &= \sum_{j=1}^{\infty} \langle f, \rho\varphi_j \rangle \varphi_j \\
\langle \sum_{j=1}^{\infty} c_j(\lambda_j-\mu) \rho \varphi_j, \varphi_k \rangle &= \langle \sum_{j=1}^{\infty} \langle f, \rho \varphi_j \rangle \varphi_j, \varphi_k \rangle \\
\sum_{j=1}^{\infty} c_j(\lambda_j-\mu) \langle \rho \varphi_j, \varphi_k \rangle &= \langle f, \varphi_k \rangle \\
c_k(\lambda_k-\mu) \langle \rho \varphi_k, \varphi_k \rangle &= \langle f, \varphi_k \rangle \\
c_k &= \frac{1}{\lambda_k-\mu} \frac{\langle f, \varphi_k \rangle}{\langle \rho \varphi_k, \varphi_k \rangle} \tag{10}
\end{align}
i.e. the eigenvalues \( \lambda_k \) is replaced by \( \lambda_k -\mu\). Hence (7) and (8) will also correspondingly be
\begin{align}
y &= \int_I [\sum_{j=1}^{\infty} \frac{1}{\lambda_j-\mu} \varphi_j(x)\varphi_j^*(z)] f(z)dz \tag{11} \\
G(x,z) &= \sum_{j=1}^{\infty} \frac{1}{\lambda_j-\mu} \varphi_j(x)\varphi_j^*(z) \tag{12}
\end{align}
Example
Solve the following ODE
\begin{align}
y^{\prime\prime} = x \tag{13}
\end{align}
with the boundary conditions \( y(0) = y(\pi) = 0 \), via constructing the Green’s function from eigenfunctions.
The appropriate Hermitian operator is \( \mathcal{L} = d^2/dx^2 \), and the eigenvalue-eigenfunction relation (2) can be seen to be
\begin{align}
\mathcal{L}[y] = y^{\prime\prime} = \lambda y \tag{14}
\end{align}
This has the familiar eigenfunctions of \( \varphi_m = \sin(mx) \) that satisfy the boundary conditions and the respective eigenvalues are \( \lambda = -m^2, m = 1,2,\ldots\). By (9), the Green’s function is then
\begin{align}
G(x,z) &= \sum_{j=1}^{\infty} \frac{1}{\lambda_j} \varphi_j(x)\varphi_j^*(z) \\
&= \sum_{m=1}^{\infty} -\frac{1}{m^2} \sin(mx)\sin(mz) \tag{15}
\end{align}
The solution is therefore
\begin{align}
y &= \int_0^{\pi} G(x,z)zdz \\
&= \int_0^{\pi} \sum_{m=1}^{\infty} -\frac{1}{m^2} \sin(mx)\sin(mz) z dz \\
&= \sum_{m=1}^{\infty} -\frac{1}{m^2} (\int_0^{\pi} \sin(mz) z dz) \sin(mx) \\
&= \sum_{m=1}^{\infty} -\frac{1}{m^2} ([-\frac{1}{m}\cos(mz)z]_0^{\pi} -\int_0^{\pi} -\frac{1}{m}\cos(mz)dz) \sin(mx) \\
&= \sum_{m=1}^{\infty} -\frac{1}{m^2} (-\frac{\pi}{m}\cos(m\pi) + [\frac{1}{m^2}\sin(mz)]_0^{\pi}) \sin(mx) \\
&= \sum_{m=1}^{\infty} (\frac{\pi}{m^3} \cos(m\pi) -(0))\sin(mx) \\
&= \sum_{m=1}^{\infty} \frac{(-1)^m\pi}{m^3} \sin(mx) \tag{16}
\end{align}
This can be verified to be equivalent to the Fourier series of \( y = \frac{1}{6} (x^3 -\pi^2 x) \) over \( [0,\pi] \).
Exercise
Revisit the exercise in the last tutorial from what we have just learnt. Use this perspective to explain why the Green’s function method failed in that scenario.
Answer
In the context of (9) and (12), \( \mathcal{L} = d^2/dx^2 \) where the eigenfunctions are \( \varphi_m = \cos(mx) \) with the eigenvalues of \( \lambda_m = -m^2, m = 0, 1, 2, \ldots \), and \( \mu = -1 \). When \( m = 1 \), the denominator of (12) is \( 1/(\lambda_m -\mu) = 1/[(-1) -(-1)] = 1/0 \) which leads to infinity and hence the Green’s function blows up.
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