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While considering the zero curvature equation $U_t - V_x + [U, V] = 0$, I developed a similar problem, albeit one that discards time dependence entirely. For a given $U(x)$ and $C(x)$, find $V(x)$ such that $[U, V] - V_x = C(x)$ holds. After considering this question, I believe that the system would either have a single solution or an infinite class of solutions depending on the choice of $C$ and $U$. In the question, I found the method involving the Kronecker product to be more suited for matrix functions as I ultimately hope to prove that the compatibility condition is isospectral. (I do not want to assume that the eigenvalues are constant across space, as I believe that it would produce a circular argument.)

I had also wondered if the system could be transformed by finding $\gamma(x)$ and $\eta(x)$ such that $[U + \gamma, V + \eta] = [U,V] - V_x$. Ultimately, I realized such an attempt would reduce to solving $[\eta, \gamma] + [V, \gamma] + [\eta, U] = V_x$, which is no simpler than the original equation. Most of the difficulty arises from the lack of commutativity of the matrix product.

Question: For a given $U(x)$ and $C(x)$, can one solve for $V(x)$ such that $[U, V] - V_x = C(x)$ holds?

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Yes, you can always do this, as follows:

First, consider the equation $M_x = -M U$ with the initial condition $M(0) = I_n$. This linear equation with initial condition has a unique solution and $M(x)$ will be invertible for all $x$. Now, set $V = M^{-1}AM$ for some matrix $A$ and consider the equation $$ V_x - UV + VU + C = 0. $$ This expands to $$ M^{-1}A_xM + C = 0, $$ or $$ A_x = -MCM^{-1} $$ which clearly has a unique solution for any given value of $A(0)$. Thus, there is always an $n^2$-dimensional family of solutions for any given $U(x)$ and $C(x)$.

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  • $\begingroup$ Thank you for your excellent answer. How would one go about finding a solution to $M_x = -MU$? I originally hoped to write $X(x) = Ce^{-\int M(x) dx}$ but was later told that the lack of commutativity for matrix multiplication causes the chain rule to break down. Would I have to use a Magnus series or consider the system as a time-ordered exponential? Thank you again for all of your help. $\endgroup$
    – Talmsmen
    Commented May 18, 2021 at 3:58
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    $\begingroup$ @JPwin: Unless $U$ takes values in a solvable Lie subalgebra of ${\frak{gl}}(n,\mathbb{R})$ there is no algorithmic way to solve the equation $M_x = - MU$, unfortunately. The so-called 'time-ordered exponential' is just the formalization of the limit of Euler's method for solving the equation numerically. For a specific $U$, one can sometimes (though not very often, it has to be admitted) find an explicit solution using differential Galois theory. $\endgroup$
    – Robert Bryant
    Commented May 18, 2021 at 9:26

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