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?

  • $\begingroup$ Welcome to Math StackExchange! You might want to enclose formulae in $-signs to properly display them. Visit the MathJax tutorial page for further info. $\endgroup$
    – harry
    Commented Aug 18, 2021 at 9:13
  • 2
    $\begingroup$ As suggested in the previous comment, it is better to avoid using pictures for mathematical content. To help you get started, I have edited your post - please, edit it further, if needed. (And if the mathematical formulas look ok, you can remove the pictures.) $\endgroup$ Commented Aug 18, 2021 at 9:32
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    $\begingroup$ @MartinSleziak Thank you for your suggestion and correction. I have already edit my post. $\endgroup$
    – JensenPang
    Commented Aug 19, 2021 at 2:21

1 Answer 1



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

$$ \left(A_1+A_2s^2-\lambda \right) \left(\lambda-A_4-A_5 s^2 \right)+A_3^2 s^2 = 0 $$


Assuming numerical values to $A_k = 1$ and after Laplace inversion we obtain the following set of conditions

$$ \left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \frac{2 \cosh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & -\frac{2 \sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \cosh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)-\frac{\sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \frac{\cosh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}}-\sinh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right) \\ -\frac{2 \sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \frac{2 \cosh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} & \frac{\cosh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}}-\sinh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right) & \cosh \left(\frac{L}{2}\right) \cosh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)-\frac{\sinh \left(\frac{L}{2}\right) \sinh \left(\frac{1}{2} L \sqrt{4 \lambda -3}\right)}{\sqrt{4 \lambda -3}} \\ \end{array} \right)\left(\begin{array}{c}\dot\psi_1(0)\\ \dot\psi_2(0)\\ \psi_1(0)\\ \psi_2(0)\end{array}\right) = 0 $$


$$ \det(M) = \frac{2 \left(\cosh \left(L \sqrt{4 \lambda -3}\right)-1\right)}{4 \lambda -3} $$

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    $\begingroup$ Thank you for your hints and kindly help. For you comment I have a some question. Without knowing the value of $\dot{\Psi}_{1/2}(0)$ how can we evaluate the expression in the numerator then do a inverse transform ? It seems that the given boundary condition is not enough to do it. $\endgroup$
    – JensenPang
    Commented Aug 19, 2021 at 4:19
  • $\begingroup$ Please. See attached note. $\endgroup$
    – Cesareo
    Commented Aug 19, 2021 at 8:39

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