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For the backward-sign Airy half-line global relation, suppose exactly one trace of derivative order $j\in \{0,1,2\}$ is unknown. Its spectral multiplier is a constant times $k^r$, $r=2-j$. Use the rotated null contours for half-range Fourier inversion with the displayed coefficients. The unknown contribution becomes an integral over the boundary of the middle upper sector $\pi /3<\arg k<2\pi /3$. There $\mathrm{Im}{k}^3<0$, sois bounded, while $e^{ikx}$ decays exponentially. The Cauchy integral theorem therefore removes that entire unknown contribution. The resulting solution depends on the initial transform and the two prescribed boundary transforms only.

// Target: analysis.bigb

For the three corner-normalized Lax pair eigenfunctions, let $s={\mu }_3(0,0,k)$ and $S=e^{4i{k}^{3}T{\hat{\sigma }}_3}{\mu }_2(0,T,k{)}^{-1}$. Their constant connection matrices give the displayed global relation. If their second columns are $(b,a{)}^T$ and $(B,A{)}^T$, its upper-right entry is $Ab-Ba=e^{8i{k}^{3}T}{\mu }_{3,12}(0,T,k)$. The right side is analytic in the lower half-plane. In the linear limit this is the backward-sign Airy half-line global relation at Fourier parameter $2k$, coupling all three boundary traces rather than allowing three independent boundary conditions.

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.

Define the initial Half-range Fourier transform $Q(k)={\int }_0^{\infty }{e}^{-ikx}{q}_0(x)dx$, and, for each available boundary trace, its finite-time spectral boundary transformTemporarily let $f_j={\partial }_x^{j}q(0,t)$ also denote the missing trace. Three integrations by parts in $x$ giveConsequently the backward-sign Airy half-line global relation isThe spatial transform is holomorphic below the real axis, while each finite-time $F_j$ is an entire function. Moreover, $F_j(\alpha k,t)=F_j({\alpha }^{2}k,t)=F_j(k,t)$ because ${\alpha }^3=1$.

Suppose the missing derivative has order $j$, and put $r=2-j$. Its contribution to the right-hand side is $a_j{k}^r{F}_j$, where $a_0=1$, $a_1=-i$, $a_2=-1$. Apply the inversion from the preceding part at time $t$ and selectThe rotated coefficients of that missing trace now equal minus its real-line coefficient. Its entire contribution is thereforeThe real-axis pieces cancel. What remains is the boundary of the middle upper sector, outward on $\arg k=\pi /3$ and inward on $\arg k=2\pi /3$. In this sector $\mathrm{Im}{k}^3\le 0$, andis bounded by ${\int }_0^t|f_j(s)|ds$. The factor $e^{ikx}$ decays exponentially throughout the closing arc, overcoming $k^r$. There are no poles. Thus the Cauchy integral theorem proves the cubic-dispersion elimination of one missing boundary trace and removes every occurrence of the unknown $f_j$.

For a compact explicit answer, use only the two prescribed traces to form $S(k,t)=Q(k)+H(k,t)$. The three choices areFor each row the solution isAll quantities in this formula are transforms of the given data. On the contour rays $k^3$ is real, so the time exponential is oscillatory; the integrals use the usual limiting interpretation of Fourier inversion. The decay estimate used for elimination belongs to the middle sector, not to $D$ or $E$, where $e^{-i{k}^3(t-s)}$ can grow.

At $t=0$, $F_j=0$ and the preceding inversion recovers $q_0$. The spectral exponential satisfies the Airy equation with the required minus sign. Differentiating the finite-time amplitudes produces only derivatives of the Dirac delta supported at $x=0$ and zero rotated closed-contour contributions, so the interior equation also holds. Finally, two solutions with the same data have $Q=H=0$ for their difference, so this representation gives zero: each of the three prescribed pairs determines the solution uniquely in the smooth decaying class.