Existence for an overdetermined system of PDEs 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
 A: 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.
