Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 61 / 3

Put $\theta =kx-4k^{3}t$. Conjugation by $e^{i\theta {\sigma }_3}$ is the meaning of $e^{i\theta {\hat{\sigma }}_3}$. The normalized eigenfunctions have the Volterra integral equation representationChoose ${\gamma }_1$ from $(0,T)$ down the time boundary and then right, ${\gamma }_2$ from $(0,0)$ up the time boundary and then right, and ${\gamma }_3$ horizontally from infinity. Compatibility of the Lax pair makes the resulting construction path-independent.

Let $a=\mathrm{Im}k$ and $b=\mathrm{Im}{k}^3$. The nontrivial exponential in column one has modulus $e^{2a(x-x^{\prime })-8b(t-t^{\prime })}$; column two has its reciprocal. For ${\gamma }_1$, $x-x^{\prime }\ge 0$ and $t-t^{\prime }\le 0$, giving signs $a,b\le 0$ for column one and $a,b\ge 0$ for column two. For ${\gamma }_2$, both differences are nonnegative, giving $(a\le 0,b\ge 0)$ and $(a\ge 0,b\le 0)$. For ${\gamma }_3$, $x-x^{\prime }\le 0$ and the time difference is zero, giving upper and lower half-planes, respectively.

Define $D_1:(a>0,b>0)$, $D_2:(a>0,b<0)$, $D_3:(a<0,b>0)$ and $D_4:(a<0,b<0)$. The bounded analytic column domains are thereforeThese are the Volterra analyticity sectors for reverse-dispersion mKdV, shown in the right panel of the figure. $D_1$ is the pair of upper outer sectors, $D_2$ the upper middle sector, $D_3$ the lower middle sector and $D_4$ the lower outer pair. Their boundaries are the six rays $\arg k=m\pi /3$. The Volterra integral equation converges and differentiates analytically where these kernels are bounded, assuming the usual sufficiently integrable spatial decay and smoothness needed by these normalizations. If decay is interpreted as merely pointwise convergence to zero without an integrability condition, existence and the stated large-$k$ normalization need additional hypotheses. A finite-path eigenfunction may be entire for fixed coordinates, but boundedness for large $k$ holds in the displayed domains; entire continuation does not enlarge those bounded sectors.

Remove the commutator gauge by setting ${\Psi }_j={\mu }_j{e}^{i\theta {\sigma }_3}$. These matrices all satisfy the ordinary first-order systemsBoth coefficient matrices are traceless, so the normalization implies $\det {\Psi }_j=\det {\mu }_j=1$. For any two solutions, differentiating ${\Psi }_2^{-1}{\Psi }_j$ shows that it is independent of both $x$ and $t$. Calling these constant matrices $s$ and $S$ givesAt $(0,0)$, ${\mu }_2=I$, so $s={\mu }_3(0,0,k)$. More generally,At $(0,T)$ the other normalization is ${\mu }_1=I$, soThus $s$ is determined by the initial spatial Volterra integral equation, while $S$ is determined by the boundary time equation evaluated at $T$. The latter uses all boundary values occurring in $\tilde{Q}$, including the initially unknown second-derivative trace. It is the global relation, rather than a claim of independence of those traces, that supplies their constraint.

For real $q$, write $\bar{f}(k)=\overline{f(\bar{k})}$. The reduction symmetry and unit determinants give the scattering data parametrizationsTo see the symmetry, conjugate the coefficient matrices by $C=\left(\begin{array}{cc}0& 1\\ \lambda & 0\end{array}\right)$ after taking complex conjugates at $\bar{k}$; $C{\sigma }_3{C}^{-1}=-{\sigma }_3$ and the real-$q$ reduction restore the original Lax pair. DefineIf ${\mu }_j^{(l)}$ denotes column $l$, bounded columns produce the following normalized matrices:Each divisor is exactly the determinant of the unscaled column pair, so $\det M_j=1$. The eigenfunctions and divisors tend to $I$ and one, respectively, giving $M\to I$ at infinity. The contour for the sectorial half-line mKdV Riemann-Hilbert reconstruction isIt is six rays, not just the real axis or a four-quadrant contour. For the displayed meromorphic construction assume the divisors have no zeros on the contour; contour zeros require an indented or otherwise regularized formulation.

Here are explicit jump data without ambiguity about which side is used. DefineThe connection formulas give $M_j={\mu }_2{e}^{i\theta {\hat{\sigma }}_3}{C}_j$. Orient each ray outwards and let plus mean its left side. If that ray separates domains $D_i$ on the minus side and $D_j$ on the plus side, its jump isThis specifies every jump using only the spectral connection matrices; each jump has determinant one. Boundedness and consistency at the junction are inherited from the original eigenfunctions when the spectral data satisfy their global constraint.

If $a,\bar{a},d_{+}$ or $d_{-}$ has zeros in its corresponding domain, their simple zeros generate poles and require residue conditions. For example, at a simple zero $k_0$ of $a$ in $D_3$,At a simple zero of $d_{+}$ in $D_1$, the analogous coefficient for the first column is $\lambda \bar{b}{e}^{-2i\theta }/(A{d}_{+}^{\prime })$ times the second column; in $D_4$ it is $b{e}^{2i\theta }/(\bar{A}{d}_{-}^{\prime })$ for the second column. The $\bar{a}$ residue in $D_2$ is the conjugate first-column relation. These formulas assume the other displayed denominators are nonzero there; multiple or coincident zeros need their corresponding higher-order conditions. In a sufficiently small-data zero-free regime, the normalized Riemann-Hilbert problem has only the jumps. Finally the order-one term in ${\mu }_x-ik[{\sigma }_3,\mu ]=Q\mu$ givesThe column normalizations do not change this off-diagonal leading coefficient.

Eliminate ${\mu }_2$ from the two connection formulas:At $(0,T)$, where ${\mu }_1=I$ and $\theta =-4k^{3}T$, this is the half-line mKdV spectral global relationTaking its upper-right entry gives a particularly useful scalar form,The second column of ${\mu }_3$ shows that $c$ is analytic in the lower half-plane and has the usual $O(k^{-1})$ normalization. The opposite off-diagonal equation follows from the real-$q$ reduction. The exponential sign here follows from conjugating a $12$ entry, which multiplies it by $e^{8i{k}^{3}T}$; reversing it would give the wrong linear global relation and boundary count. This relation constrains the initial and boundary scattering data; they cannot be chosen independently.

Take the linear limit and write $g_j(t)={\partial }_x^{j}q(0,t)$. The spatial and temporal Volterra integral equations give, to first order,while $a=A=1$ to first order and $c(k,T)=-\hat{q}(2k,T)$. The minus sign in $b$ comes from the spatial path running from infinity to zero. For $B$, the inverse in $S=e^{4i{k}^{3}T{\hat{\sigma }}_3}{\mu }_2(0,T{)}^{-1}$ supplies the minus sign; its conjugation cancels the final-time exponential in the temporal kernel. Substitution in the global relation therefore yieldsThis is the backward-sign Airy half-line global relation, because the linear limit is $q_t-q_{xxx}=0$, independently of $\lambda$.

Its constraint is especially transparent after a time Laplace transform. Let $\mathrm{Re}p>0$, choose the principal root $r=p^{1/3}$ with positive real part, and set $k=-ir/2$ in the lower transform half-plane. Then $8i{k}^3=-p$. Taking the infinite-horizon Laplace limit for a solution of sufficient exponential order givesThis proves that the second-derivative trace is fixed by the initial data and two boundary traces; it is not a third freely prescribable datum. Finite-horizon versions retain the terminal transform and express the same relation on $0<t<T$.

To see that two are necessary, rather than merely sufficient to eliminate a transform, the homogeneous transformed spatial equation is ${\partial }_x^{3}u-pu=0$. Its roots are $r$, $e^{2\pi i/3}r$, and $e^{4\pi i/3}r$. Since $-\pi /6<\arg r<\pi /6$, exactly two roots have negative real part. Decay at infinity retains two independent spatial modes. Their value/first-derivative boundary matrix has determinant equal to the difference of those two roots, which is nonzero. Thus prescribed $g_0,g_1$ fix both modes; one condition leaves a free mode, while three arbitrary traces overdetermine them.

Two independent admissible boundary conditions are needed at $x=0$, for example $q(0,t)$ and $q_x(0,t)$. This is the two boundary traces for reverse-dispersion Airy flow count and hence the perturbative count for the nonlinear modified Korteweg-De Vries equation. The argument determines the small-norm boundary count; it does not claim a global nonlinear existence theorem for arbitrary large boundary data or arbitrary incompatible choices of traces.