Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 61 / 3

Let $D_1,D_2,D_3,D_4$ denote the open quadrants, counterclockwise from the first. Write $\theta =kx+2k^{2}t$. The Lax pair has the Volterra integral equation kernel $e^{-i[k(x-x^{\prime })+2k^2(t-t^{\prime })]{\hat{\sigma }}_3}$, obtained by integrating its closed matrix differential form from a normalization point.

In column one, the off-diagonal factor is $e^{2i[k(x-x^{\prime })+2k^2(t-t^{\prime })]}$, whose modulus is $e^{-2\mathrm{Im}[k(x-x^{\prime })+2k^2(t-t^{\prime })]}$. Column two has the reciprocal factor. For ${\mu }_2$, choose the path $(0,0)\to (0,t)\to (x,t)$, so both differences are nonnegative. Column one requires $\mathrm{Im}k\ge 0$ and $\mathrm{Im}{k}^2\ge 0$, giving $D_1$; column two requires the opposite inequalities, giving $D_4$.

For ${\mu }_1$, the time path starts at $T$ and runs backward, while the spatial path runs rightward. Its column-one conditions are $\mathrm{Im}k\ge 0$, $\mathrm{Im}{k}^2\le 0$, giving $D_2$; column two is bounded in $D_3$.

Finally, ${\mu }_3$ integrates spatially from infinity to $x$, where $x-x^{\prime }\le 0$. Column one is bounded in the lower half-plane and column two in the upper half-plane. Normalization at $(\infty ,T)$ is equivalent to normalization at $(\infty ,t)$ because the decaying potential vanishes on the vertical path at infinity. ThusThe Volterra integral equation Neumann series converges locally uniformly under the usual integrable smooth scattering assumptions, proving holomorphic dependence in the open domains and continuous boundary values. These are the Volterra analyticity sectors for a half-line NLS Lax pair; at $x=0$, absence of a spatial segment allows the larger time-only domains used for $A,B$.

Undo the commutator gauge by putting ${\psi }_j={\mu }_j{e}^{-i\theta {\sigma }_3}$. Then ${\psi }_{j,x}=(-ik{\sigma }_3+Q){\psi }_j$ and ${\psi }_{j,t}=(-2i{k}^2{\sigma }_3+V){\psi }_j$, with $V=2kQ-i{Q}_x{\sigma }_3-i|q{|}^2{\sigma }_3$. The trace of each coefficient matrix is zero, so all normalized fundamental matrices have determinant one. Their ratios ${\psi }_2^{-1}{\psi }_j$ are independent of both $x$ and $t$, because differentiation cancels the common left-hand coefficient.

Call these constant matrices $s$ for $j=3$ and $S$ for $j=1$. Returning to the commutator gauge givesHere $e^{\alpha {\hat{\sigma }}_3}C=e^{\alpha {\sigma }_3}C{e}^{-\alpha {\sigma }_3}$. Evaluation at $(0,0)$ gives $s={\mu }_3(0,0,k)$, computed solely by the spatial Volterra integral equation for the initial potential. At $(0,T)$ the normalization of ${\mu }_1$ givesThus $S$ is computed by integrating the time Lax pair for ${\mu }_2(0,t,k)$ from zero to $T$. It uses both boundary traces $q(0,t)$ and $q_x(0,t)$ before their compatibility is enforced.

Define $F^{♯}(k)=\overline{F(\bar{k})}$. The defocusing reduction of the Lax pair is $\mu (k)={\sigma }_1\overline{\mu (\bar{k})}{\sigma }_1$, so the half-line NLS spectral functions have the formPut $d=a{A}^{♯}-b{B}^{♯}$. Select bounded analytic columns and divide by their determinants to defineFor example, $[{\mu }_3{]}_2=b{e}^{-2i\theta }[{\mu }_2{]}_1+a[{\mu }_2{]}_2$, so the first pair has determinant $a$ before division. The corresponding determinant for the second pair is $d$. The other two are their reflected counterparts. Hence $\det M=1$ and $M\to I$ at spectral infinity.

Here are explicit connection matrices furnishing every jump. Write $E=e^{-2i\theta }$; wherever the boundary values coexist, $M_j={\mu }_2{C}_j$, withOrient each ray away from zero; the plus side is its left side. Then $M_{+}=M_{-}J$, with $J=C_{-}^{-1}{C}_{+}$, entirely determined by the spectral functions. For instance, on the positive real axis,The relevant Riemann-Hilbert problem contour is thereforeIt is a cross because time analyticity involves the sign of $\mathrm{Im}{k}^2$, not just $\mathrm{Im}k$. Zeros of the denominators require residue conditions or small contour detours; they may not simply be ignored. The half-line NLS Riemann-Hilbert reconstruction recovers $q=2i{\lim }_{k\to \infty }k{M}_{12}$, by matching the constant term $i[{\sigma }_3,M_1]=Q$ in the spatial Lax pair.

Eliminate ${\mu }_2$ between the two connection relations and evaluate at $(0,T)$, where ${\mu }_1=I$. This yields the matrix global relationIts upper off-diagonal entry is especially useful. Since $S^{-1}=\left(\begin{array}{cc}A& -B\\ -B^{♯}& A^{♯}\end{array}\right)$, it isThe function $c_T$ is analytic in the upper half-plane and is $O(1/k)$ there by the spatial Volterra integral equation. This is the half-line NLS global relation. It expresses compatibility between initial and boundary scattering and the final state. At finite $T$ the right-hand side is generally nonzero; setting it to zero would silently impose an extra assumption.

To first order in the potential, $a=A=1+O(q^2)$, while the spatial and temporal Volterra integral equations giveThus the linear limit of the global relation is exactlyfor the linear equation $i{q}_t+q_{xx}=0$. This ties the two boundary traces together, rather than allowing them to be independent data.

The count is made explicit by a temporal Laplace transform. For $\mathrm{Re}p>0$, the transformed equation is ${\tilde{q}}_{xx}+ip\tilde{q}=i{q}_0$. Put $\alpha =\sqrt{-ip}$ with positive real part. Of the two homogeneous solutions $e^{\pm \alpha x}$, only $e^{-\alpha x}$ decays at infinity. There is therefore exactly one free complex coefficient to be fixed at the endpoint. In particular, prescribed Dirichlet trace ${\tilde{g}}_0$ gives the definite Neumann traceThis follows by differentiating the decaying Dirichlet Green kernel $-(e^{-\alpha |x-y|}-e^{-\alpha (x+y)})/(2\alpha )$ at $x=0$. Thus one complex boundary condition is needed, such as Dirichlet, Neumann or a suitable Robin combination. Prescribing both traces arbitrarily would overdetermine the problem. In a Sobolev space with the required trace regularity, the cubic term is a locally Lipschitz perturbation, so the small-data perturbative problem retains the same boundary-condition count; corner compatibility still must be imposed for a classical solution.

We derive the Robin spectral elimination while retaining the finite-time global relation. This distinguishes the actual ratio $B/A$ from the equivalent, time-independent boundary coefficient needed by the Riemann-Hilbert problem.

At $x=0$ the time Lax pair coefficient has off-diagonal entries $(2k+ic)q$ and $(2k-ic)\bar{q}$. LetDirect multiplication shows $V(0,t,k)=K(k)V(0,t,-k)K(k{)}^{-1}$. Because the commutator term depends on $k^2$ and $K$ commutes with ${\sigma }_3$, uniqueness of the time Volterra integral equation gives $S(k)=K(k)S(-k)K(k{)}^{-1}$. ThereforeThe exceptional parameter points of $K$ are treated by analytic continuation. These identities express linearizable Robin scattering for defocusing NLS.

For finite horizon $T$, the exact upper-half-plane ratio isConsequently the often used answer $B/A=b/a$ is not a literal finite-$T$ identity. Its effective time-independent replacement, in the two bounded time sectors, isThe second branch follows from the Robin symmetry; its argument $-k$ lies in the upper half-plane where the initial scattering functions are known. We now justify this replacement for the inverse problem without assuming that boundary values decay as $t\to \infty$.

Only $\Gamma =B^{♯}/(ad)$ is needed for the connection across the upper imaginary ray. The conjugate global relation at $-k$ and the Robin symmetry giveFor upper-half-plane $k$, the notation $b^{♯}(-k)=\overline{b(-\bar{k})}$ uses only values of $b$ in its legitimate upper-half-plane domain. PutThe effective boundary coefficient isThis Robin spectral determinant for defocusing NLS is explicitly computed from the initial data and $c$.

For completeness, the finite-horizon defect can be removed algebraically, not by an unproved limit. If $r=B^{♯}/A^{♯}$, then $\Gamma =r/[a(a-br)]$. Subtracting the same expression with $r_0=-h{b}^{♯}(-k)/a^{♯}(-k)$ givesIts jump multiplier isThe exponential decays in $D_2$ for $x>0$ and $t<T$, since $\mathrm{Im}k>0$ and $\mathrm{Im}{k}^2<0$. The coefficient is holomorphic away from the stated determinant zeros and is appropriately decaying at infinity. From the explicit matrices in part (c), $C_2=C_{1}L(\Gamma e^{2i\theta })$, where $L(z)=\left(\begin{array}{cc}1& 0\\ z& 1\end{array}\right)$. Hence replace the sector-two solution byThe reflected upper-triangular change is made in $D_3$. These analytic changes remove the finite-time defect, leave $M_1,M_4$ unchanged, and leave the reconstruction from the second column in $D_2$ unchanged. At apparent poles, the old $d$-dependent singularities cancel by the connection formula; genuine poles of the new determinant remain as explicit residue data. Equivalently, small circles can first isolate all zeros and the same algebra can be performed on the regularized contours. This is finite-horizon Robin spectral elimination.

Every modified jump is now explicitly known: keep $C_1,C_4$ from part (c) and replaceAll jumps are again $C_{-}^{-1}{C}_{+}$, now involving only $a,b,c$. If $\mathcal{D}$ has a simple zero $k_j$ giving a genuine pole, its coefficient iswith the reflected second-column condition at ${\bar{k}}_j$. Common numerator/denominator zeros are checked for removability, and zeros on the cross require contour indentation or local circle jumps. Thus no unknown boundary norming data remain. One must not assume away such zeros for every real $c$; attractive Robin parameters can support localized modes.

It follows that the equivalent normalized Riemann-Hilbert problem is specified entirely by $q_0$ and $c$, including its pole data. When two solutions exist, their quotient has no jump, its prescribed pole singularities cancel, and it tends to $I$ at infinity; Liouville theorem makes the quotient $I$, proving uniqueness. Smooth corner evolution additionally requires $q_0^{\prime }(0)=c{q}_0(0)$ and higher compatibility conditions at higher regularity. Without that compatibility, the same initial data are interpreted in the weak evolution sense, rather than as a smooth solution through the corner.