Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 86 / 1

Extend $q$ by zero to the negative half-line. Its ordinary Fourier transform is then exactly the Half-range Fourier transform $\hat{q}(k)$. The Fourier inversion theorem gives the real-line term for $x>0$; no endpoint convention at $x=0$ is needed.

It remains to show that each extra contour contributes zero, independently of its coefficient. For $\mathrm{Im}k<0$, the integral defining $\hat{q}$ is holomorphic. With $\alpha =e^{2\pi i/3}$, multiplication by ${\alpha }^2$ rotates the sector $E$ into $-2\pi /3\le \arg ({\alpha }^{2}k)\le -\pi /3$, and multiplication by $\alpha$ rotates $D$ into $-2\pi /3\le \arg (\alpha k)\le -\pi /3$. Thus the two rotated transforms are holomorphic on their respective upper sectors.

Orient the contours as in the PDF: $\partial E$ runs from infinity on the $\pi /3$ ray into zero and then out along the positive real axis; $\partial D$ runs from negative real infinity into zero and then out along the $2\pi /3$ ray. Both have their sector on their left. Smooth decay and integration by parts give $\hat{q}(k)=O(1/k)$ in closed lower-half-plane sectors. The factor $e^{ikx}$ decays in the upper half-plane for $x>0$. Close each sector by a large arc; the arc integral vanishes by Jordan lemma, including its short portion near the real axis. The Cauchy integral theorem therefore gives the rotated null contours for half-range Fourier inversion:Adding arbitrary constant multiples of these two zero integrals to ordinary Fourier inversion proves the asserted inversion formula. The constants $c_1,c_2$ are arbitrary in this part; selecting them later is what removes an unknown boundary trace.

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.