show/hide this revision's text 3 Corrected error caught by Darsh

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) - C(t)^{\rm T} = B'(t)$, where $C^{\rm T}$ is symmetric. 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 $\mathcal D[(g_a)^{\rm T}] = 0$, where $(g_a)^{\rm T}(t)$ is the transpose of $g_a(t)$..

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 $f_2$.
show/hide this revision's text 2 fixed an error

In an n-dimensional linear 2nd-order ODE, why is the inverse 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 $C(t)$ is symmetric. 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 $\mathcal D[g_aD[(g_a)^{\rm T}] = 0$., where $(g_a)^{\rm T}(t)$ is the transpose of $g_a(t)$.

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 $f_2$.
show/hide this revision's text 1

In an n-dimensional linear 2nd-order ODE, why is the 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 $C(t)$ is symmetric. 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 $\mathcal D[g_a] = 0$.

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 $f_2$.