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The free Schrodinger equation describes linear dispersive evolution without a potential term. In this normalization a spectral mode is $e^{ikx-i{k}^{2}t}$. On a half-line, boundary data enter through the global relation for the half-line free Schrodinger equation.

Use the half-line Fourier transform and finite-time boundary transformsThe half-line Fourier transform is analytic for $\mathrm{Im}k<0$ under spatial decay. All spectral integrals below have their usual oscillatory, or vanishing Gaussian damping, interpretation until absolute convergence is established.

The free Schrodinger equation has the local divergence identityIntegrating in $x,t$ gives the global relation for the half-line free Schrodinger equationLet $D_{+}=\{\mathrm{Re}k>0,\mathrm{Im}k>0\}$. Orient its boundary from $i\infty$ down to $0$, and then from $0$ to $+\infty$, so that $D_{+}$ lies on the left. Fourier inversion followed by contour integration in the second quadrant yieldsThis is a complex spectral representation involving the initial trace and both boundary traces. The contour deformation works because each boundary-time integrand contains $e^{-i{k}^2(t-s)}$ with $s\le t$, which decays in the second quadrant, as well as $e^{ikx}$ for $x>0$. This fixes both the quadrant and the orientation; changing either without changing the signs would produce an incorrect representation.

The unknown Neumann boundary condition can be removed by the Fokas method. Evaluate the global relation for the half-line free Schrodinger equation at $-k$:Consequently $k{G}_0-i{G}_1=2k{G}_0-{\hat{q}}_0(-k)+e^{i{k}^{2}t}\hat{q}(-k,t)$. In the representation from part (a), the last term has integral ${\int }_{\partial D_{+}}{e}^{ikx}\hat{q}(-k,t)dk=0$, by the Cauchy integral theorem and Jordan lemma. It is analytic in the upper half-plane, and the exponential decays on the closing first-quadrant arc. Hence an expression involving only the given data isOne may replace $t$ by $T$ in the upper limit defining $G_0$: the contribution of boundary times $s>t$ closes to zero in $D_{+}$, since $e^{-i{k}^2(t-s)}$ then decays there. Using $t$ makes causality transparent.

For an explicit proof of uniform convergence, it is useful to apply boundary lifting before inversion. Set $h(x)=e^{-x}$, $c=g_0(0)$, $w_0(x)=q_0(x)-ch(x)$, andThe compatibility condition gives $w_0(0)=0$. The Fourier sine transform of $w=q-g_{0}h$ satisfies $S_t+i{k}^{2}S=Hf$, with initial value $S_0$. The integrating factor therefore gives the equivalent representationThis last integral is absolutely and uniformly convergent for $x\ge 0$, $0\le t\le T$ under concrete sufficient hypotheses $w_0,w_0^{\prime },w_0^{\prime \prime }\in L^1$, decay of the boundary terms, and $g_0\in C^2[0,T]$. Indeed, two integrations by parts give $|S_0(k)|\le \Vert w_0^{\prime \prime }{\Vert }_1/k^2$ for $k\ge 1$. Another integration by parts, this time in $s$, givesThus the initial term is uniformly $O(k^{-2})$ and the forcing term uniformly $O(k^{-3})$. For $0<k<1$, use $|S_0(k)|\le \Vert w_0{\Vert }_1$ and the bounded time integral. An integrable majorant proves the claimed uniform convergence and permits evaluation at both boundaries.

At $t=0$, Fourier sine inversion gives $q(x,0)=c{e}^{-x}+w_0(x)=q_0(x)$. At $x=0$, the integral vanishes, giving $q(0,t)=g_0(t)$, including the compatible corner. To verify the equation, note that $h^{\prime \prime }=h$ and the transformed equation impliesAdding the lifted part gives $i{q}_t+q_{xx}=0$. Under the stated smoothness, this holds classically in the interior; differentiated spectral integrals can first be Gaussian-regularized, or read in the sine-transform $L^2$ sense and then identified with the smooth solution. Uniform convergence of $q$ itself does not require claiming uniform convergence of every differentiated integral at the corner.

Differentiate the spatial equation with respect to $t$ and the temporal equation with respect to $x$. Substitution ofinto ${\psi }_{xt}-ik{\psi }_t$ cancels every term involving $\psi$, $u$, and $u_x$, leavingThe left side is $u_t$ by differentiating ${\psi }_x-ik\psi =u$, so consistency is exactly $u_t=i{u}_{xx}$. This is a scalar Lax pair formulation.

Taking the half-line Fourier transform and integrating twice by parts givesMultiplication by $e^{i{k}^{2}t}$ and integration in time yields

Replace $k$ by $-k$ and subtract. The unknown Neumann boundary value cancels:For $x>0$, Fourier inversion of the zero extension givesInverting the preceding identity therefore givesThusThis elimination is the global relation for the half-line free Schrodinger equation.