# In an n-dimensional linear 2nd-order ODE, why is the transpose-inverse to a system of solutions also a solution?

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 $f\_2$.

• Quick question: Your title says "transpose inverse", but you write $D[(g_\a)^T]$, which doesn't have an inverse in it. Is one of these a typo? – David E Speyer Nov 10 '09 at 11:34
• Oh, I get it. Never mind. – David E Speyer Nov 10 '09 at 12:09
• Sorry, I couldn't think of a better term. Of course, I mean half the columns of the inverse of the bigger matrix above. – Theo Johnson-Freyd Nov 10 '09 at 19:00

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.
• Sorry if I'm just being a doofus here, but I'm not sure I believe that D is self-adjoint in that inner product. I'm computing the adjoint to be g --> g'' + Bg' + (C + B')g, which isn't the same as D unless B is constant... – Darsh Ranjan Nov 10 '09 at 9:34
• Oh, ooh, good question, that might be the problem. A said I didn't believe my proof. Now I should probably go and check that my actual differential equation is what I said it is. If C is not symmetric, but has an antisymmetric part that cancels out half of B', then it works. And I know that the operator that's turning up in my project is self-adjoint. – Theo Johnson-Freyd Nov 10 '09 at 18:46
• I've corrected the error in the statement of the problem. – Theo Johnson-Freyd Nov 10 '09 at 18:56

It seems like when you get to the variation-of-parameters step, everything gets fuzzy. Have you tried doing variation-of-parameters and finding the form of that solution? Your solution is probably not the fastest way, but it may still be correct.

• Equivalently, I'd like to show that if $S(q_1,q_2)$ is Hamilton's principle function, then $\frac{\partial^2 S}{\partial q_1\partial q_2}$ is a symmetric matrix. But when there are magnetic terms, then I don't know how to do this. – Theo Johnson-Freyd Nov 10 '09 at 18:58

How about trying to pose the problem in a coordinate-free way, and if possible write the equation as a Hamiltonian equation? This may give some insight.