Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 61 (Boundary Value Problems for Integrable PDE's)

Set $\kappa =(a^2/3-b{)}^{1/2}>0$, $\sigma =-c+ab/3-2a^3/27$, and use the gauge reduction of a third-order half-line equationThe spatial differential operator acting on $q$ is obtained by replacing ${\partial }_X$ with $\kappa {\partial }_x-a/3$. Its second-derivative coefficient vanishes. The first-derivative coefficient is $\kappa (b-a^2/3)=-{\kappa }^3$, and the constant coefficient is $c-ab/3+2a^3/27=-\sigma$. The time derivative supplies $\sigma q+{\kappa }^3{q}_t$, so all constant terms cancel and the remaining equation is ${\kappa }^3(q_t+q_{xxx}-q_x)=0$.

The transformed initial condition and Dirichlet boundary data are thereforeCompatibility at the corner is preserved. Since $a<0$, this exponential gauge improves the spatial decay of the initial data, which supports convergence of its Half-range Fourier transform.

Put $w(k)=-i(k^3+k)$ and define the spectral fluxWriting $E=e^{-ikx+w(k)t}$, division of the proposed divergence form by $E$ gives $q_t+wq+X_x-ikX$. Direct differentiation yieldsThe last term cancels $wq$, so the divergence form is exactly the normalized partial differential equation. This calculation fixes both the sign of the dispersion relation and the sign of the boundary flux used below.

Use the negative-exponential Half-range Fourier transformFor the boundary traces $g_j(t)={\partial }_x^{j}q(0,t)$, define finite-time spectral boundary transforms $G_j(k,t)={\int }_0^t{e}^{w(k)s}{g}_j(s)ds$. Integrating the conservation law in $x$ and then time gives the global relationThe plus sign of the boundary contribution follows because the integrated flux is $-E(0,t)X(0,t,k)$.

Fourier inversion consequently gives a real-line initial integral plus a boundary integral. To put the latter in the complex plane, defineOrient $\partial D_{+}$ from the left upper infinity, through $i$, to the right upper infinity, with $D_{+}$ on the left. Each $G_j$ is entire, and in the region between the real axis and this contour, $\mathrm{Re}w\ge 0$. The causal factors $e^{-w(t-s)}$ are bounded there and $e^{ikx}$ decays for $x>0$. The Cauchy integral theorem, with the usual large-arc limiting argument, deforms the boundary integral to giveInitial smoothness, spatial decay and corner compatibility justify inversion; if necessary the integrals are first regularized before contour deformation. The unknown Neumann-type traces $g_1,g_2$ will be removed by the global relation.

The cubic dispersion symmetry with negative transport supplies two other roots of $w(\nu )=w(k)$:Both lie in the lower half-plane when $k\in D_{+}$. For example this is immediate at large positive imaginary $k$. A root could leave that half-plane only by becoming real, where $\mathrm{Re}w(\nu )=0$, contradicting $\mathrm{Re}w(k)<0$. Connectedness of $D_{+}$ completes the argument; a coalescence merely exchanges the two roots.

Evaluate the global relation at these roots. Because their dispersion values equal $w(k)$, the same boundary transforms occur, andInterpolate this linear polynomial in $\nu$ at ${\nu }_1,{\nu }_2$. Defineand similarly $\mathcal{T}\hat{q}(k,t)$. Substitution givesIndeed, interpolation of ${\nu }^2+1$ yields $({\nu }_1+{\nu }_2)k-{\nu }_1{\nu }_2+1=-2k^2$, which proves the coefficient $-(3k^2+1)$ after subtracting $k^2+1$.

The final-state term has zero contour integral. After cancelling $e^{wt}$ its integrand is $e^{ikx}\mathcal{T}\hat{q}(k,t)$, holomorphic in $D_{+}$ and decaying upon closing the contour upward for $x>0$. The possible square-root singularities are removable: the expression is unchanged by exchanging the roots, and when they coalesce at $\nu$, its limit is $F(\nu )+(k-\nu )F^{\prime }(\nu )$, with the same formula for the final-state transform. Thus the Cauchy integral theorem eliminates this unknown term.

We obtain the fully specified Dirichlet spectral formula for Airy flow with negative transportOnly the initial Fourier transform and the given boundary-value time transform remain.

First remove the observation time from the spectral amplitude. For any fixed horizon $H>t$, replace $G_0(k,t)$ by $G_0(k,H)$ in the formula of part (d). The difference in the boundary integral is proportional toFor every $s>t$, this integrand is holomorphic inside $D_{+}$, where $\mathrm{Re}w<0$, and decays on its large closing arcs. The Cauchy integral theorem makes the difference zero. This finite-horizon extension of a boundary transform therefore gives the time-independent spectraAll observation-time dependence is now in $e^{ikx-wt}$.

When the boundary history is sufficiently decaying, the natural long-time version isThe integral converges in $\mathrm{Re}w<0$ for bounded boundary data and on its boundary for integrable histories; otherwise one must state an Abel prescription or a suitable temporal growth assumption. Arbitrary smooth $g_0$ need not have such an infinite-time transform. The finite-horizon formula remains exact without that extra assumption, and the source alone does not prescribe a universal long-time limit.

A useful further decomposition, if $g_0(s)=g_{\infty }+r(s)$ with integrable $r$, isThis separates the persistent forcing from the transient spectrum. The pole at $k=i$, where $w=0$, accounts for the decaying stationary profile $g_{\infty }{e}^{-x}$; that profile also follows directly from $q_{xxx}-q_x=0$, spatial decay and its boundary value. For the remaining oscillatory integrals the phase is $i[kx+(k^3+k)t]$. At a fixed ray $x/t=\xi$, its stationary points satisfy $3k^2+1+\xi =0$, or $k=\pm i\sqrt{(1+\xi )/3}$. These time-independent amplitudes expose precisely the poles and saddle points used in the method of steepest descent; allowable deformations must also respect the analyticity strip of the initial transform.

The algebraic change of variables still works for $a>0$, but the transformed initial data are now multiplied by $e^{aX/3}$. Mere decay of $Q_0$ does not ensure existence of the Half-range Fourier transform used above. For example, $Q_0(X)=e^{-dX}$ with $0<d<a/3$ gives a growing transformed profile.

There is also a genuine stability distinction. On the whole line a real-frequency mode of the original equation evolves with factorIts modulus is $e^{(a{K}^2-c)T}$, so positive $a$ amplifies arbitrarily high frequencies rather than damping them. Thus the formal gauge equivalence does not justify the same unweighted Sobolev well-posedness conclusion. Weighted or more strongly decaying data and corresponding contour changes may allow particular solutions; the assumptions stated for the negative-$a$ construction cannot simply be reused.

Write $\chi =e^{-ikz-\lambda \bar{z}/(ik)}$. In terms of Wirtinger derivatives, the elliptic equation is $q_{z\bar{z}}=\lambda q$. If $W=Adz+Bd\bar{z}$, its coefficients are $A=\chi q_z$ and $B=-\lambda \chi q/(ik)$. Direct differentiation givesTherefore $dW=(B_z-A_{\bar{z}})dz\wedge d\bar{z}=0$, proving that $W$ is a closed differential one-form. The parameter restriction $k\ne 0$ makes its coefficients well defined.

Put $d=\ell /(2\sqrt{3})$ and $a=e^{2\pi i/3}$. Consistent with the actual PDF vertices, parametrize the three oriented sides byThey run counterclockwise, with unit tangent $i{\beta }_j$ and outward normal ${\beta }_j$. Consequently the normal derivative and tangential derivative satisfyLet $\kappa =k{\beta }_j$. On this side the exponential factor becomesPulling back the closed differential one-form therefore givesDefine ${\Phi }_j(\kappa )$ and ${\Psi }_j(\kappa )$ by integrating the Dirichlet terms and normal term in this expression. The Generalized Stokes theorem gives ${\int }_{\partial D}W=0$, soThe factor $2i$ follows by solving $\sum E\Phi +(i/2)\sum E\Psi =0$. The bars on the side-2 rotation are essential; replacing them with $a$ would use the wrong side orientation. The converted TeX also reverses the sign in the upper vertex's exponential, whereas the PDF puts that vertex above the real axis; the parametrization here follows the PDF.

Identical oriented Dirichlet boundary data imply a rotation-invariant solution whenever the Dirichlet problem is unique. For $\lambda \ge 0$, uniqueness follows directly: the difference $h$ of two solutions has zero boundary trace, and integration by parts gives ${\int }_D|\nabla h{|}^2+4\lambda {\int }_D|h{|}^2=0$. Thus $h=0$. The same conclusion holds for a nonresonant negative parameter by invertibility of the Dirichlet Laplacian shift. Then $q(az)=q(z)$, all three normal traces equal a common $q_N(s)$, and ${\Psi }_j=\Psi$.

SetDivide the global relation by $E(-iak)$. Using $1+a+\bar{a}=0$ yields $E(-ik)/E(-iak)=e(\bar{a}k)$ and $E(-i\bar{a}k)/E(-iak)=e(-k)$. Hencewhere the requested known function isFor an explicitly derivative-free form, put $H(k)={\int }_{-\ell /2}^{\ell /2}{e}^{(k+\lambda /k)s}f(s)ds$ and $f_{\ast }=f(-\ell /2)=f(\ell /2)$. An integration by parts givesThe uniqueness qualification cannot be removed for every real $\lambda$: at a Dirichlet Laplacian eigenvalue a homogeneous addition need not preserve rotation invariance. The scalar reduced relation then applies to the rotation-averaged solution, but not necessarily to the originally chosen solution. Part (d) treats this genuine exceptional case.

Use Schwarz conjugation of a spectral function, $F^{♯}(k)=\overline{F(\bar{k})}$. Reality of $f,q_N$ and $\lambda$ gives ${\Psi }^{♯}=\Psi$, ${\Phi }^{♯}=\Phi$ and $e^{♯}=e$. Conjugating part (c) yields the second relationwith $A^{♯}=e(ak)\Phi (k)+e(-k)\Phi (ak)+\Phi (\bar{a}k)$. The explicit $i$ changes sign, while $a$ and $\bar{a}$ are interchanged.

Choose sampling points satisfying $k_n+\lambda /k_n=2\pi in/\ell$. Put ${\epsilon }_n=(-1{)}^n$ and ${\chi }_n=\ell (\bar{a}{k}_n+\lambda /(\bar{a}{k}_n))/2$. Then $e(k_n)=e(-k_n)={\epsilon }_n$ and $e(k_n)e(\bar{a}{k}_n)e(a{k}_n)=1$. Multiply the first relation by ${\epsilon }_n$ and subtract the conjugate relation. The coefficient of $\Psi (a{k}_n)$ vanishes, leavingThus the Fourier coefficient $N_n=\Psi (k_n)={\int }_{-\ell /2}^{\ell /2}{e}^{2\pi ins/\ell }{q}_N(s)ds$ isThe endpoint terms in the derivative-free expression for $\Phi$ cancel, because the other two hyperbolic arguments sum to $-i\pi n$ and $\sinh \chi (k_n)=0$. Writing $h(k)=k-\lambda /k$, we obtain the entirely known expressionThis Dirichlet reconstruction in an equilateral triangle uses Fourier series inversion on the finite side, with convergence in the natural trace norm and pointwise away from endpoints under sufficient smoothness.

For $\lambda >0$, take $k_n=i[\pi n/\ell +\sqrt{(\pi n/\ell {)}^2+\lambda }]$. It is nonzero for every integer $n$. Moreover $\mathrm{Re}{\chi }_n=(\sqrt{3}\ell /4)[v+\lambda /v]>0$ when $k_n=iv$, so the denominator cannot vanish. For $\lambda =0$ use $k_n=2\pi in/\ell$ for $n\ne 0$; the remaining coefficient is $N_0=0$, since the divergence theorem gives $3\int q_N={\int }_D\Delta q=0$. This includes the constant harmonic solution.

For negative nonresonant parameters, use either root of the sampling equation and analytic continuation of the reconstruction. Any zero denominator at a parameter where the Dirichlet operator remains invertible is removable: the uniquely determined trace, and hence its Fourier coefficients, depend analytically on that parameter. Evaluate such values by limits rather than claiming a pole or dividing by zero.

At a true Dirichlet resonance in an equilateral triangle, the request for a uniquely determined normal trace is false. A concrete counterexample uses the triangle's affine barycentric coordinates ${\beta }_1+{\beta }_2+{\beta }_3=1$. The functionis positive inside and zero on the boundary. Since $|\nabla {\beta }_j|=2/(\sqrt{3}\ell )$, direct differentiation gives $\Delta h=-16{\pi }^{2}h/(3{\ell }^2)$. At $\lambda =-4{\pi }^2/(3{\ell }^2)$ both $q=0$ and $q=h$ have the same zero Dirichlet data, but different normal derivatives. On a side where ${\beta }_1=0$, the latter derivative is $-2\pi {\sin }^2(\pi {\beta }_2)/(\sqrt{3}\ell )$, not zero.

In general, if $-4\lambda$ is a Dirichlet eigenvalue, the Fredholm solvability condition is ${\int }_{\partial D}f{\partial }_n\varphi ds=0$ for every eigenfunction $\varphi$ at that eigenvalue. This follows from Green's second identity. When these conditions hold, the complete answer is $q=q_p+{\sum }_j{c}_j{\varphi }_j$ and $q_N^{(r)}=(q_p{)}_N^{(r)}+{\sum }_j{c}_j({\partial }_n{\varphi }_j){|}_r$, with arbitrary real coefficients $c_j$. A rotation-invariant particular solution may be used, but homogeneous additions need not be invariant. Such noninvariant eigenfunctions must exist: otherwise the complete Dirichlet eigenbasis would span only the proper rotation-invariant subspace of $L^2(D)$. Thus the source's unrestricted real parameter needs this resonance qualification in both (c) and (d).

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 $c$.