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>



Existence for an overdetermined system of PDEs
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.


