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

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

2009-11-10T04:41:37Z

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 $f_2$.

Answer by Theo Johnson-Freyd for In an n-dimensional linear 2nd-order ODE, why is the transpose-inverse to a system of solutions also a solution?

2009-11-10T04:58:47Z

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.

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

2009-11-10T08:42:22Z

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.

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

2009-11-19T20:18:19Z

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.