Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 71 / 1 / d

Fix $T>t$. In the representation from (c), the finite-horizon extension of a boundary transform allows $G_0(k,t)$ to be replaced by $G_0(k,T)$. To see this directly, their difference contributesIt has an holomorphic integrand inside $D_{+}$; there $\mathrm{Re}\omega <0$, and the factors $e^{\omega (s-t)}$ and $e^{ikx}$ allow the contour to be closed. Its integral is zero for $x>0$. This replacement puts the observation-time dependence entirely in $e^{ikx-\omega t}$.

Every such exponential satisfies the linear dispersive Stokes equation, becauseConsequently the representation satisfies the partial differential equation. The stipulated smoothness and decay justify the differentiated oscillatory integrals, or equivalently their regularized limits. The fixed horizon is useful here: no artificial source term arises from differentiating the endpoint of a time transform.

At $t=0$, use the original $G_0(k,0)=0$ version. The $F$ integral vanishes after closing inside $D_{+}$, since $F$ is holomorphic there and $x>0$. Fourier inversion of the initial half-line transform therefore givesFor the boundary trace use the fixed-horizon version, with $0<t<T$. First consider the initial-data terms at $x=0$. Differentiating ${\nu }_j^2+k{\nu }_j+k^2-1=0$ givesThus changing variables separately in the two pieces of the $F$ integral giveswhere ${\Gamma }_{-}$ goes from lower-left infinity to $-1/\sqrt{3}$, across to $1/\sqrt{3}$, and then to lower-right infinity along $\mathrm{Re}\omega =0$. For completeness, on the upper-right curved edge one root maps onto the real ray $(-\infty ,-2/\sqrt{3}]$ and the other onto the lower-right curved edge. On the upper-left edge the corresponding images are $[2/\sqrt{3},\infty )$ and the lower-left edge. The middle segment maps to the remaining real intervals $[1/\sqrt{3},2/\sqrt{3}]$ and $[-2/\sqrt{3},-1/\sqrt{3}]$. These real images cancel the outer pieces of the original real-axis integral, leaving exactly ${\Gamma }_{-}$ with the stated orientation.

This contour bounds the lower-central region $\mathrm{Im}k<0$, $\mathrm{Re}\omega >0$. The spatial transform is holomorphic there and $e^{-\omega t}$ decays on its closing arcs for $t>0$. Its integral is therefore zero by the Cauchy integral theorem. The initial-data part of the boundary trace vanishes.

In the remaining boundary integral, use $d\omega =i(1-3k^2)dk$. Along the oriented $\partial D_{+}$, $\omega$ traverses the imaginary axis once from $-i\infty$ to $i\infty$. On its real segment, the imaginary part runs from $-2/(3\sqrt{3})$ to $2/(3\sqrt{3})$; the two curved edges supply the complementary intervals. It follows thatThe last equality is ordinary temporal Fourier inversion, after $\omega =i\xi$. Because $t$ is strictly inside $(0,T)$ it gives the full value, not a truncated-transform endpoint half-value. ThusCompatibility $q_0(0)=g_0(0)$ makes the two traces agree at the corner. Spatial decay follows from the upper-contour exponential and nonstationary oscillatory estimates on the real integral as $x\to \infty$, under the assumed data decay.

The decaying smooth solution is also unique. For a zero-data difference $w$, integration by parts in the linear dispersive Stokes equation givesusing $w(0,t)=0$ and decay at infinity. The initial norm is zero, so $w=0$. This completes verification of the initial value, boundary value and governing equation.