[proofplan]
We prove both directions. For the forward direction (compatibility implies zero curvature), we compute $v_{xt}$ and $v_{tx}$ from the two equations of the overdetermined system, equate them using the equality of mixed partial derivatives, and extract the zero curvature condition as the coefficient of $v$. For the converse (zero curvature implies compatibility), we invoke the Frobenius theorem: the zero curvature condition is precisely the integrability condition for the system of first-order PDEs.
[/proofplan]
[step:Compute $v_{xt}$ by differentiating $v_x = Uv$ with respect to $t$]
Starting from $v_x = Uv$, we differentiate both sides with respect to $t$. On the left-hand side, this produces $v_{xt}$. On the right-hand side, we apply the product rule: the matrix $U = U(x, t; \lambda)$ depends on $t$, and the vector $v$ depends on $t$. This gives
\begin{align*}
v_{xt} = \frac{\partial U}{\partial t}\, v + U\, v_t.
\end{align*}
Now substitute $v_t = Vv$ from the second equation of the system:
\begin{align*}
v_{xt} = \frac{\partial U}{\partial t}\, v + U V\, v.
\end{align*}
[/step]
[step:Compute $v_{tx}$ by differentiating $v_t = Vv$ with respect to $x$]
Starting from $v_t = Vv$, we differentiate both sides with respect to $x$. The product rule gives
\begin{align*}
v_{tx} = \frac{\partial V}{\partial x}\, v + V\, v_x.
\end{align*}
Now substitute $v_x = Uv$ from the first equation:
\begin{align*}
v_{tx} = \frac{\partial V}{\partial x}\, v + V U\, v.
\end{align*}
[/step]
[step:Equate $v_{xt} = v_{tx}$ and extract the zero curvature equation]
For the system to admit a solution $v$, the mixed partial derivatives must be equal: $v_{xt} = v_{tx}$. Equating the expressions from the previous two steps:
\begin{align*}
\frac{\partial U}{\partial t}\, v + U V\, v = \frac{\partial V}{\partial x}\, v + V U\, v.
\end{align*}
Rearranging all terms to one side and factoring out $v$:
\begin{align*}
\left(\frac{\partial U}{\partial t} - \frac{\partial V}{\partial x} + UV - VU\right)v = 0.
\end{align*}
Since $UV - VU = [U, V]$ is the matrix commutator, this reads
\begin{align*}
\left(\frac{\partial U}{\partial t} - \frac{\partial V}{\partial x} + [U, V]\right)v = 0.
\end{align*}
Therefore compatibility of the system requires
\begin{align*}
\frac{\partial U}{\partial t} - \frac{\partial V}{\partial x} + [U, V] = 0.
\end{align*}
[/step]
[step:Prove the converse: the zero curvature condition guarantees existence of a solution]
Conversely, suppose the zero curvature equation
\begin{align*}
\frac{\partial U}{\partial t} - \frac{\partial V}{\partial x} + [U, V] = 0
\end{align*}
holds. We must show that the overdetermined system admits a solution $v(x, t)$ for any initial condition $v(x_0, t_0) = v_0$.
Rewrite the system in the language of differential forms. Define the matrix-valued $1$-form
\begin{align*}
\omega = U\, dx + V\, dt.
\end{align*}
The system $dv = \omega\, v$ is a total differential equation. The obstruction to integrability is the curvature $2$-form
\begin{align*}
\Omega = d\omega - \omega \wedge \omega = \left(\frac{\partial U}{\partial t} - \frac{\partial V}{\partial x} + [U, V]\right) dx \wedge dt,
\end{align*}
where we have used $d\omega = (\partial_t U - \partial_x V)\, dx \wedge dt$ and $\omega \wedge \omega = -[U, V]\, dx \wedge dt$.
The Frobenius theorem states that a total differential system $dv = \omega\, v$ is locally integrable (i.e., admits a solution through every point with every initial value) if and only if the curvature $\Omega$ vanishes identically. Since the zero curvature equation is precisely $\Omega = 0$, the Frobenius theorem guarantees the existence of a solution.
[/step]
