= Solution
Differentiate the spatial equation with respect to $t$ and the temporal equation with respect to $x$. Substitution of
$$
\psi_x=ik\psi+u,
\qquad
\psi_t=-ik^2\psi+iu_x-ku
$$
into $\psi_{xt}-ik\psi_t$ cancels every term involving $\psi$, $u$, and $u_x$, leaving
$$
\psi_{xt}-ik\psi_t=iu_{xx}.
$$
The left side is $u_t$ by differentiating $\psi_x-ik\psi=u$, so consistency is exactly $u_t=iu_{xx}$. This is a <scalar> <Lax pair> formulation.
Taking the half-line Fourier transform and integrating twice by parts gives
$$
\partial_t\widehat u(k,t)+ik^2\widehat u(k,t)
=kh(t)-iu_x(0,t).
$$
Multiplication by $e^{ik^2t}$ and integration in time yields
$$
\boxed{
\widehat u(k,t)e^{ik^2t}-\widehat u_0(k)
=\int_0^t e^{ik^2\tau}
[kh(\tau)-iu_x(0,\tau)]\,d\tau.}
$$
Replace $k$ by $-k$ and subtract. The unknown Neumann boundary value cancels:
$$
e^{ik^2t}\bigl[\widehat u(k,t)-\widehat u(-k,t)\bigr]
=\widehat u_0(k)-\widehat u_0(-k)
+2k\int_0^t e^{ik^2\tau}h(\tau)\,d\tau.
$$
For $x>0$, Fourier inversion of the zero extension gives
$$
\int_{\mathbb R}e^{ikx}\widehat u(-k,t)\,dk=0.
$$
Inverting the preceding identity therefore gives
$$
\begin{aligned}
u(x,t)
={}&\frac1{2\pi}\int_{-\infty}^{\infty}
e^{-ik^2t+ikx}[\widehat u_0(k)-\widehat u_0(-k)]\,dk\\
&+\frac1\pi\int_{-\infty}^{\infty}\int_0^t
e^{-ik^2(t-\tau)+ikx}\,k h(\tau)\,d\tau\,dk.
\end{aligned}
$$
Thus
$$
\boxed{G(k,\tau)=k h(\tau).}
$$
This elimination is the <global relation for the half-line free Schrodinger equation>.
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