<s>[INST] Solving the eigenvalue from a set of coupled second order differential equation numerically I met a problem in solving a set of coupled differential equation. Given a set of equation as shown in below:
\begin{align*}
A_1\psi_1(z)+A_2\frac{d^2\psi_1(z)}{dz^2}+A_3\frac{d\psi_2(z)}{dz}&=\lambda\psi_1(z)\\
A_4\psi_2(z)+A_5\frac{d^2\psi_2(z)}{dz^2}+A_3\frac{d\psi_1(z)}{dz}&=\lambda\psi_2(z)
\end{align*}
with the following 4 boundary conditions:
\begin{align*}
\psi_1(0)=\psi_1(d)&=0\\
\psi_2(0)=\psi_2(d)&=0
\end{align*}
where $A_{i}$ is a constant coefficient.
I've been stuck in this question for a long time. As the coupling term $A_{3}$ present in the problem, we cannot simply solve it analytically. Without the boundary condition on the first order derivative, I was unable to determine the value of $\lambda$. Is the boundary value sufficient to tackle the problem? Is there any numerical method to solve this type of eigenvalue problem?
 [/INST] Hint.
Using the Laplace transform we have
$$
\cases{
(A_1+s^2A_2-\lambda)\Psi_1(s)+A_3 s \Psi_2(s) = A_2(s\psi_1(0)+\dot\psi_1(0))+A_3\psi_2(0)\\
(A_4+s^2A_5-\lambda)\Psi_2(s)+A_3 s \Psi_1(s) = A_5(s\psi_2(0)+\dot\psi_2(0))+A_3\psi_1(0)\\
}
$$
and then
$$
\cases{
\Psi_1(s) = \frac{A_3 (A_3 \psi_1(0) s-A_$ \psi_2(0)+A_5 \dot\psi_2(0)s+\lambda  \psi_2(0))-A_2 (\dot\psi_1(0)+\psi_1(0) s)
   \left(A_4+A_5 s^2-\lambda \right)}{\left(A_1+A_2s^2-\lambda \right) \left(\lambda-A_4-A_5 s^2 \right)+A_3^2s^2}\\
\Psi_2(s) = \frac{\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots}{\left(A_1+A_2s^2-\lambda \right) \left(\lambda-A_4-A_5 s^2 \right)+A_3^2s^2}
}
$$
so the eigenfunctions are linked to the denominator roots or
$$
\left(A_1+A_2s^2-\lambda \right) \left(\lambda-A_4-A_5 s^2 \right)+A_3^2
   s^2 = 0
$$
NOTE
Assuming numerical values to $A_k = 1$ and after Laplace inversion we obtain the following set of conditions
$$
\left(
\begin{array}{cccc}
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
 \frac{2 \cosh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & -\frac{2 \sinh \left(\frac{L}{2}\right)
   \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \cosh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda
   -3}\right)-\frac{\sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \frac{\cosh
   \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}}-\sinh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2}
   L \sqrt{4 \lambda -3}\right) \\
 -\frac{2 \sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \frac{2 \cosh \left(\frac{L}{2}\right)
   \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \frac{\cosh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4
   \lambda -3}\right)}{\sqrt{4 \lambda -3}}-\sinh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right) & \cosh
   \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)-\frac{\sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4
   \lambda -3}\right)}{\sqrt{4 \lambda -3}} \\
\end{array}
\right)\left(\begin{array}{c}\dot\psi_1(0)\\ \dot\psi_2(0)\\ \psi_1(0)\\ \psi_2(0)\end{array}\right) = 0
$$
and
$$
\det(M) = \frac{2 \left(\cosh \left(L \sqrt{4 \lambda -3}\right)-1\right)}{4 \lambda -3}
$$
</s>