I am working on the following exercise in Partial Differential Equations: An Introduction, 2nd Edition by Strauss:
Show that the boundary conditions $X(b)=\alpha X(a)+\beta X'(a)$ and $X'(b)=\gamma X(a)+\delta X'(a)$ are symmetric on the interval $a\leq x\leq b$ symmetric if and only if $\alpha\delta-\beta\gamma=1$.
I feel like the forward direction is not necessarily true. Assume the boundary conditions are symmetric. Let $f(x), g(x)$ be two eigenfunctions that satisfy the boundary conditions. Then $f'(x)g(x)-g'(x)f(x)\Big|_{x=a}^{x=b}=0$.
$f'(b)g(b)-g'(b)f(b)-f'(a)g(a)+g'(a)f(a)=[\gamma f(a)+\delta f'(a)][\alpha g(a)+\beta g'(a)]-[\gamma g(a)+\delta g'(a)][\alpha f(a)+\beta f'(a)]-f'(a)g(a)+g'(a)f(a)=[-\beta\gamma+\alpha\delta-1]f'(a)g(a)+[\beta\gamma-\alpha\delta+1]f(a)g'(a)=0$
But this does not imply that $\alpha\delta-\beta\gamma=1$? It is possible that $f'(a)=g'(a)=0$. How can we conclude that $\alpha\delta-\beta\gamma=1$?
It has to be true for all $f$ and $g$ satisfying the boundary conditions to be symmetric. So you just have to find an example where $f(a)$, $g(a)$, $f'(a)$, and $g'(a)$ are non-zero.