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Direct differentiation givesConsequently this transformation takes solutions of the modified Korteweg-De Vries equation to solutions of the Korteweg-De Vries equation in these sign conventions. Both summands in $u$ have the same scaling weight, so the modified-equation scaling induces $(x,t,u)\mapsto (e^{s}x,e^{3s}t,e^{-2s}u)$.

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.

For the modified Korteweg-De Vries equation with linear part $q_t-q_{xxx}$, the Volterra integral equation kernel is $e^{i[k(x-x^{\prime })-4k^3(t-t^{\prime })]{\hat{\sigma }}_3}$. In column one its off-diagonal modulus is $e^{2\mathrm{Im}k(x-x^{\prime })-8\mathrm{Im}{k}^3(t-t^{\prime })}$; column two has the reciprocal modulus. Define $D_1$ by signs $(+,+)$, $D_2$ by $(+,-)$, $D_3$ by $(-,+)$ and $D_4$ by $(-,-)$. The normalization points $(0,T)$, $(0,0)$ and $(\infty ,t)$ give column domains $(D_4,D_1)$, $(D_3,D_2)$ and $({\mathbb{C}}_{+},{\mathbb{C}}_{-})$, respectively. Finite-path eigenfunctions can be entire at fixed coordinates while failing to be bounded outside these spectral sectors.