 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 giveThe 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 $\hat{f}$ with ${\hat{f}}_t+A\hat{f}=0$ while preserving its eigenvalue.

Write $b=u_x+a_0$ and $d=u-\lambda +a_1$. The spatial eigenvalue equation gives ${\hat{f}}_{xx}=(u-\lambda )\hat{f}$, so $A\hat{f}=b\hat{f}+d{\hat{f}}_x$. For $F=(\hat{f},{\hat{f}}_x{)}^T$ the two first-order systems are thereforeThe 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.