Sturm-Liouville system

Question

Let $A$ be $n \times n$ real matrix and consider the following Sturm-Liouville system $$y''(x)=Ay(x) \\ y(0)=y(1)=0,$$ where 0 is the null vector in $\mathbb{R}^n$. How can we solve this system explicitly? Is there any references about that?. Thank you.

Answer 1

$y = e^{rt} u$ is a solution of $y'' = A y$ if $r^2$ is an eigenvalue of $A$ for eigenvector $u$. Assuming for simplicity the eigenvalues $\lambda_j$ of $A$ are distinct and nonzero, with corresponding eigenvectors $u_j$, the general solution is then $\sum_j \left(a_j e^{\sqrt{\lambda_j} t} + b_j e^{-\sqrt{\lambda_j} t}\right) u_j$. In order for this to satisfy the boundary conditions, all $a_j$ and $b_j$ must be $0$ unless $\sinh(\sqrt{\lambda_j}) = 0$, in which case $b_j = -a_j$. $\sinh(\sqrt{\lambda_j}) = 0$ with $\lambda_j \ne 0$ requires $\lambda_j = -n^2 \pi^2$ for some positive integer $n$.

Answer 2

$y(x)=\sin(x\sqrt{A})y_0$ solves the equation $y''(x)=Ay(x)$ with $y(0)=0$, regardless of the choice of vector $y_0$, assuming you can find a square root matrix $\sqrt{A}$. To further have $y(1)=0$ would require choosing $y_0$ to be in the null space of $\sin(\sqrt{A})$. You will have a full solution of your problem if $y_0$ is in the null space of $\sqrt{A}$ because $\sin(\sqrt{A})=\frac{\sin(\lambda)}{\lambda}|_{\lambda=\sqrt{A}}\sqrt{A}$.

There are problems with $\sqrt{A}$ for a general matrix $A$ with non-zero null space. For example, $$ A = \left[\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right] $$ has no square root. If $B^2=A$ were to hold, then $B^4=0$ would imply that the minimal polynomial for $B$ would be $\lambda^n$ for some $n > 2$, which is a contradiction.