= Solution
If $Lf=\lambda f$ and $f_t+Af=0$, <differentiation> and the <Isospectral Lax equation>[Lax equation] give $(L-\lambda)f_t=-A(L-\lambda)f$, so $\lambda_t=0$ after projection onto the normalized <eigenfunction> for a <simple eigenvalue>[nondegenerate eigenvalue].
For $\Psi=(f,f_x)^T$,
$$
M_L=\begin{pmatrix}0&1\\u-\lambda&0\end{pmatrix},
$$
$$
M_A=\begin{pmatrix}-u_x-b&-(u-\lambda+a)\\-u_{xx}-b_x-(u-\lambda+a)(u-\lambda)&-2u_x-b-a_x\end{pmatrix}.
$$
The compatibility $\Psi_{xt}=\Psi_{tx}$ is precisely $\partial_tM_L-\partial_xM_A-[M_A,M_L]=0$. If all coefficients are $x$-independent, this becomes $\dot M_L=[M_A,M_L]$, and cyclicity of trace gives $d\operatorname{Tr}(M_L^k)/dt=0$.
Back to article page
