= Solution
Use a common invariant operator domain on which the stated differentiations and <commutator> are valid. For a differentiable normalized <eigenfunction>, the Hellmann-Feynman calculation and self-adjointness give
$$
\lambda'=\langle f,L_tf\rangle=\langle f,LAf-ALf\rangle
=\lambda\langle f,Af\rangle-\lambda\langle f,Af\rangle=0.
$$
The <Lax equation preserves the spectrum>. Differentiating $Lf=\lambda f$ now gives $(L-\lambda)(f_t+Af)=0$, so $f_t+Af$ is in the same eigenspace, or is zero. For a nondegenerate <eigenvalue> it equals $c(t,\lambda)f$. Multiplying $f$ by $\exp[-\int c(t,\lambda)\,dt]$ produces $\widehat f$ with $\widehat f_t+A\widehat f=0$ while preserving its <eigenvalue>.
Write $b=u_x+a_0$ and $d=u-\lambda+a_1$. The spatial <eigenvalue> equation gives $\widehat f_{xx}=(u-\lambda)\widehat f$, so $A\widehat f=b\widehat f+d\widehat f_x$. For $F=(\widehat f,\widehat f_x)^T$ the two first-order systems are therefore
$$
\boxed{U=\begin{pmatrix}0&1\\u-\lambda&0\end{pmatrix},\qquad
V=\begin{pmatrix}-b&-d\\-b_x-d(u-\lambda)&-b-d_x\end{pmatrix}.}
$$
The second row of $V$ follows by differentiating the first time equation in $x$ and using the spatial equation once more. Conversely these <matrix> equations imply both scalar equations, so the formulations are equivalent.
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