= Solution
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 transform>
$$
F_j(k,t)=\int_0^t e^{ik^3s}f_j(s)\,ds.
$$
Temporarily let $f_j=\partial_x^jq(0,t)$ also denote the missing trace. Three <integrations by parts> in $x$ give
$$
\int_0^\infty e^{-ikx}q_{xxx}\,dx=-f_2-ikf_1+k^2f_0-ik^3\widehat q.
$$
Consequently the <backward-sign Airy half-line global relation> is
$$
\partial_t\widehat q+ik^3\widehat q=k^2f_0-ikf_1-f_2,\qquad
e^{ik^3t}\widehat q(k,t)=Q(k)+k^2F_0(k,t)-ikF_1(k,t)-F_2(k,t).
$$
The spatial transform is <holomorphic> below the real axis, while each finite-time $F_j$ is an <entire function>. Moreover, $F_j(\alpha k,t)=F_j(\alpha^2k,t)=F_j(k,t)$ because $\alpha^3=1$.
Suppose the missing <derivative> has order $j$, and put $r=2-j$. Its contribution to the right-hand side is $a_jk^rF_j$, where $a_0=1$, $a_1=-i$, $a_2=-1$. Apply the inversion from the preceding part at time $t$ and select
$$
c_1=-\alpha^{-2r},\qquad c_2=-\alpha^{-r}.
$$
The rotated coefficients of that missing trace now equal minus its real-line coefficient. Its entire contribution is therefore
$$
\frac{a_j}{2\pi}\left(\int_{\mathbb R}-\int_{\partial E}-\int_{\partial D}\right)
e^{ikx-ik^3t}k^rF_j(k,t)\,dk.
$$
The 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 $\operatorname{Im}k^3\leq0$, and
$$
e^{-ik^3t}F_j(k,t)=\int_0^t e^{-ik^3(t-s)}f_j(s)\,ds
$$
is 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 are
$$
\begin{array}{c|c|c|c}
\text{prescribed traces}&H(k,t)&c_1&c_2\\ \hline
f_0,f_1&k^2F_0-ikF_1&-1&-1\\
f_0,f_2&k^2F_0-F_2&-\alpha&-\alpha^2\\
f_1,f_2&-ikF_1-F_2&-\alpha^2&-\alpha
\end{array}
$$
For each row the solution is
$$
\boxed{\begin{aligned}
q(x,t)=\frac1{2\pi}\int_{\mathbb R}e^{ikx-ik^3t}S(k,t)\,dk
&+\frac{c_1}{2\pi}\int_{\partial E}e^{ikx-ik^3t}S(\alpha^2k,t)\,dk\\
&+\frac{c_2}{2\pi}\int_{\partial D}e^{ikx-ik^3t}S(\alpha k,t)\,dk.
\end{aligned}}
$$
All 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 $e^{-ik^3(t-s)}$ 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: \b[each of the three prescribed pairs determines the solution uniquely in the smooth decaying class].
Back to article page
