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I am interested in the existence of a vector valued solution $y = y(x, t) \in\mathbb{R}^n$ to a system of $2n$ equations: there are twice more equations than unknowns. More precisely:

Let $A$ and $B$ be matrix valued functions $A, B \in C^1([0, L]\times [0, T]; \mathbb{R}^{n \times n})$. Does the system \begin{equation} \begin{aligned} \partial_x y(x, t) = A(x, t) y(x, t) \\ \partial_t y(x, t) = B(x, t) y(x, t) \end{aligned} \qquad \text{in } [0, L]\times [0, T] \end{equation} with initial and boundary conditions (with $g \in C^1([0, T]; \mathbb{R}^n)$ and $y_0 \in C^1([0, L]; \mathbb{R}^n)$) \begin{equation} y(x, 0) = y_0(x) \quad \text{for }x \in [0, L]\\ y(0, t ) = g(t) \quad \text{ for }t \in [0, T] \end{equation} have a solution $y \in C^1([0, L]\times [0, T]; \mathbb{R}^n)$?

(Here, $\mathbb{R}^{n \times n}$ denotes the square matrices of size $n \times n$ with real coefficients.)

Writing the equations in integral form, $y^1$ is solution to $\partial_x y^1 = A y^1$ if and only if
\begin{align*} y^1(x,t) =& y^1(0, 0) + \int_0^t \partial_t y^1(0, s) ds \\ &+ \int_0^x \partial_x y^1(\xi, 0) d\xi + \int_0^x \int_0^t ((\partial_t A) y^1 + A \partial_t y^1) d\xi ds \end{align*} while $y^2$ is solution to $\partial_t y^2 = B y^2$ if and only if \begin{align*} y^2(x,t) =& y^2(0, 0) + \int_0^x \partial_x y^2(\xi, 0) d\xi \\ &+ \int_0^t \partial_t y^2(0, s) ds + \int_0^x \int_0^t ((\partial_x B) y^2 + B \partial_x y^2) d\xi ds. \end{align*}

Assume that the following compatibility conditions hold: \begin{align*} \partial_t A + A B = \partial_x B + B A,\qquad \text{in } [0, L]\times [0, T] \end{align*} and $g(0) = y_0(0)$.

In this case, if $y$ is solution to both systems, then the expressions for $y^1$ and $y^2$ coincide and are equal to \begin{align*} y(x,t) =& f(0) + \int_0^x \frac{\mathrm{d}}{\mathrm{d}x}y_0(x) d\xi + \int_0^t \frac{\mathrm{d}}{\mathrm{d}t}g(s) ds \\ &+ \int_0^x \int_0^t ((\partial_x B) y + B A y) d\xi ds \end{align*}

But how can we show that there exists $y$ solution to both $\partial_x y = A y$ and $\partial_t y = By$? Which method(s) would you suggest? Do we need to add assumptions to obtain existence?

Also asked on Mathematics Stack Exchange: https://math.stackexchange.com/questions/3384407/uniqueness-but-no-existence

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    $\begingroup$ Your equation is just a local coordinate expression of the condition that $y(x,t)$ is parallel with respect to a linear connection. What Deane Yang described is essentially parallel transport wrt this connection. Parallel transport is independent of the path precisely when the connection is flat (zero curvature, which is the same as your compatibility condition). Check any standard reference for details (e.g., see link). $\endgroup$ – Igor Khavkine Apr 14 at 23:09
  • $\begingroup$ @IgorKhavkine, that's a nice way to reformulate the question. $\endgroup$ – Deane Yang Apr 15 at 15:30
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COMMENT: The answer below is just the proof of the Frobenius theorem (https://en.wikipedia.org/wiki/Frobenius_theorem_(differential_topology) applied to this specific case. The arguments below are also illustrative of how the proof of the Cartan-Kähler theorem works for an overdetermined system of first order linear PDEs.

NEW ANSWER:

First, observe that if $y$ is a solution, then $$ 0 = [\partial_t - B, \partial_x - A]y = (BA-AB + \partial_xB - \partial_tA)y. $$

Now assume that, for any $y_0$, there exists a solution $y$ such that $y(0,0) = y_0$. By the existence and uniqueness theorem of a linear system of ODEs, it follows that, given any $(x,t) \in [0,L]\times[0,T]$, the map from $y_0$ to $y(x,t)$ is a linear isomorphism. That implies that, at each $(x,t)$, $$ (BA-AB + \partial_xB - \partial_tA)v = 0,\ \forall\ v \in \mathbb{R}^n,.$$

Conversely, assume this equation holds. Then, given any $y_0$, we can solve for $y$ as follows:

1) There exists a unique solution to $\partial_xy = Ay$ and $y(0,0) = y_0$ on $[0,L]\times\{0\}$.

2) There exists a unique solution to $\partial_ty = By$ on $[0,L]\times[0,T]$, where $y(\cdot,0)$ is given by 1).

Since $\partial_xy-Ay = 0$ on $[0,L]\times\{0\}$ and, on $[0,L]\times[0,T]$, $$ (\partial_t - B)(\partial_xy-Ay) = (\partial_x-A)(\partial_t-B)y + (BA-AB + \partial_xB - \partial_tA)y = 0. $$ it follows that $\partial_xy-Ay = 0$ on the full domain. Therefore, $y$ is solution to the full system.

OLD ANSWER:

The value of $y$ at a single point uniquely determines the solution. If you specify, for example, the value of $y(0,0)$, the first equation uniquely determines the value of $y(x,0)$ for $x \in [0,L]$. Then the second equation uniquely determines the value of $y$ on the entire domain. If, in fact, you solve for a function $y$ this way, then the only thing that needs to be checked is that $y$ really does solve the first equation on the whole domain. This can be done by observing that the compatibility condition implies, on the full domain, $$ \partial_t(\partial_xy - Ay) = \partial_x(\partial_ty - By) = 0 $$

This can also be shown using the Frobenius theorem, which encodes the argument above in its proof.

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  • $\begingroup$ The compatibility condition is equivalent to $[\partial_t -B,\partial_x-A] = 0$. $\endgroup$ – Deane Yang Nov 29 '19 at 18:28
  • $\begingroup$ Thank you very much for your reply. From what I understood, you choose as a candidate solution $y(x, t)=y(0, 0)+\int_0^xA(\xi, 0)y(\xi, 0)d\xi+\int_0^tB(x,s)y(x,s)ds$, and want to use the compatibility condition to show that this $y$ in fact satisfies the first equation for any $(x, t)$. However, for the last argument, I don't understand why the compatibility condition implies that $\partial_t(\partial_xy-Ay)=\partial_x(\partial_ty-By)=0$ without assuming that $y$ satisfies both equations. $\endgroup$ – user344045 Dec 2 '19 at 15:52
  • $\begingroup$ Yes, my explanation was too sketchy. I'll add details, when I get a chance. $\endgroup$ – Deane Yang Dec 2 '19 at 15:56
  • $\begingroup$ Thank you. Just one question, what is the argument behind the fact that if $\partial_x y - A y = 0$ on $[0, L] \times \{0\}$ and if $(\partial_t - B)(\partial_x y - A y) = 0$ in $[0, L]\times[0, L]$, then one has $\partial_x y - A y = 0$ in $[0, L]\times[0, L]$? $\endgroup$ – user344045 Apr 15 at 8:48
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    $\begingroup$ If you set $u = \partial_xy - Ay$, then $u=0$ when $t = 0$ and $\partial_tu + Bu = 0$ on the full domain. By the uniqueness theorem for a first order system of ODEs, $u$ must vanish on the full domain. $\endgroup$ – Deane Yang Apr 15 at 14:39

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