I'm at a sticky spot in my research. Namely, I have a particular fact, and it ought to have a short proof, but the only way I know how to show it is long and drawn out, and I don't like it and worry I might have an error. I'm hoping one of y'all will either see a short proof or respond with "all your questions are answered in [link]". And I'm hoping this isn't too close to "homework question".


I have a linear second-order differential operator $\mathcal D$ on $C^\infty( [0,1], \mathbb R^n)$, where $\mathbb R^n$ has its usual metric, and of the following form:
$$ \mathcal D = \frac{d^2}{dt^2} + B(t) \frac{d}{dt} + C(t) $$
where $B,C$ are $n\times n$ matrix-valued functions on $[0,1]$, $B(t)$ is antisymmetric for each $t$, and $C(t) - C(t)^{\rm T} = B'(t)$, where $C^{\rm T}$ is the transpose of $C$. I happen to know a lot of solutions to $\mathcal D[f] = 0$. In particular, I have two matrix-valued functions $f\\_1(t)$ and $f\\_2(t)$, which satisfy $\mathcal D[f\\_a] = 0$, and also $f\\_1(0) = \delta = f\\_2(1)$ and $f\\_2(0) = 0 = f\\_1(1)$, where $\delta$ is the unit $n\times n$ matrix.


(Incidentally, this implies that the columns of the $f\\_a$ are a basis for the space of solutions of $\mathcal D[f]=0$, so that there are no nonzero solutions with $f(0) = 0 = f(1)$. Indeed, any solution with $\mathcal D[f] = 0$, $f(0) = 0$ is determined by the derivative $f'(0)$, so that there is a linear map $\mathbb R^n \to \mathbb R^n$ sending $v$ to the value $f(1)$ where $f'(0) = v$. But $f\\_2(1) = \delta$, and so $f\\_2'(0)$ is full-rank, and so if $f$ solves the differential equation with $f(0) = 0$, then $f(t) = f\\_2(t)\left(f\\_2'(0)\right)^{-1}f'(0)$.)


Anyhoo, so my question is this. Let $g\\_1(t),g\\_2(t)$ be matrix-valued functions such that:
$$ f\\_1g\\_1 + f\\_2g\\_2 = 0 \text{ and } f\\_1' g\\_1 + f\\_2' g\\_2 = \delta$$
**Prove that $\mathcal D[(g\\_a)^{\rm T}] = 0$.**


For example, when $n=1$, $B(t) = 0$ because there are no antisymmetric $1\times 1$ matrices, and then by Abel's formula the determinant of the matrix $\left(\begin{smallmatrix} f\\_1 & f\\_2 \\\ f\\_1' & f\\_2' \end{smallmatrix}\right)$ is constant. Therefore, $g\\_2$, which is the lower-right corner of the inverse of this matrix, is a constant times $f\\_1$, and $g\\_1$, which is the upper right-hand-corner of the inverse, is a constant times $f\\_2$.



In an n-dimensional linear 2nd-order ODE, why is the transpose-inverse to a system of solutions also a solution?
I promised an answer, so I'll sketch it here, but I hope someone can give a better one.


The operator $\mathcal D$ is *self-adjoint* in the following sense. Let $\langle f,g\rangle = \int\_0^1 f(t) \cdot g(t) dt$ be the usual inner-product on $C^\infty([0,1],\mathbb R^n)$. Then if $f(0) = g(0) = 0 = f(1) = g(1)$, we have $\langle \mathcal D[f],g\rangle = \langle f, \mathcal D[g]\rangle$. This follows from integration by parts and the (anti)symmetry of $B,C$.


Define a Green's function to be a matrix-valued function $G(t,s)$ on the square $[0,1]\times [0,1]$ such that $\mathcal D\\_t[\mathcal G] = \delta(t-s)$ (times the identity matrix), and also satisfying $G(0,s) = 0 = G(1,s)$. In particular, $\mathcal G$ is smooth away from the diagonal, and has a corner like $|s-t|$ at the diagonal. $\mathcal G$ is unique if it exists. Because $\mathcal D$ is self-adjoint, switching $G$ is *symmetric* in the sense that $G(t,s)$ is the transpose of $G(s,t)$.


By the usual variation-of-parameters mumbo-jumbo, I can explicitly write down a formula for $G$. Namely, $G(t,s) = f\\_1(t)g\\_1(s) \Theta(t-s) - f\\_2(t)g\\_2(s) \Theta(s-t)$, or something similar. Then use symmetry: $G(s,t) = f\\_1(s)g\\_1(t) \Theta(s-t) - f\\_2(s)g\\_2(t) \Theta(t-s) = (G(t,s))^{\rm T}$, and so $f\\_1(s)g\\_1(t) = (f\\_2(t)g\\_2(s))^{\rm T}$. But $\mathcal D[f\\_1] = 0$, so $(g\\_1)^{\rm T}$ must be a solution as well.


The problems with this argument are:


* Actually going through the variation-of-parameters is tedious and unenlightening.
* I'm not completely sure I believe that $\mathcal G$ is symmetric in the way that I said it is. I mean, it should be, and I believed it until I started doubting myself.


