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

Use the linear dispersive Stokes equation with time variable $t$. Put $E_k=e^{-ikx+\omega (k)t}$. Expanding the required local relation givesChoose $X=q_{xx}+ik{q}_x+(1-k^2)q$. Direct differentiation then yieldsThe unwanted coefficient of $q$ cancels precisely when $\omega (k)=ik(1-k^2)$. HenceWith these choices, the local relation is exactly $E_k(q_t+q_x+q_{xxx})=0$ for every complex spectral parameter for a linear boundary value problem.

Define the negative-exponential Half-range Fourier transformswhich are holomorphic for $\mathrm{Im}k<0$. Write $g_j(s)={\partial }_x^{j}q(0,s)$ and define the finite-time spectral boundary transformsIntegrate the local relation over $0<x<\infty$, $0<s<t$. The spatial boundary term at infinity vanishes by decay, and the term at $x=0$ is subtracted. Consequently the half-line linear dispersive Stokes global relation isFourier inversion on the real axis gives an initial-data integral plus an integral of $B$. To put the boundary term on a complex contour, setFor $k=u+iv$, $\mathrm{Re}\omega =v(3u^2-v^2-1)$, so $D_{+}$ is the connected upper domain $v>0$, $v^2>3u^2-1$. Orient $\partial D_{+}$ with this domain on the left: from upper-left infinity down to $-1/\sqrt{3}$, along the real segment to $1/\sqrt{3}$, and then to upper-right infinity.

The finite-time spectral boundary transforms are entire functions. In the upper regions outside $D_{+}$, where $\mathrm{Re}\omega \ge 0$, their time-integrated factors satisfy $|e^{-\omega t}{e}^{\omega s}|\le 1$ for $0\le s\le t$. Together with $e^{ikx}$, this permits contour deformation of the two remaining real-axis pieces to the curved boundary of $D_{+}$. The resulting representation isThe contour orientation accounts for the plus sign in this convention. This formula still contains the two unknown derivative traces through $G_1,G_2$.

The dispersion relation has two additional arguments with the same value. Indeed,Thus use the cubic Stokes dispersion symmetryBoth roots are in the lower half-plane for $k\in D_{+}$. At $k=iv$, $v>0$, their imaginary parts are $-v/2$. A root cannot cross the real axis in $D_{+}$, since a real root would make $\mathrm{Re}\omega (k)=\mathrm{Re}\omega (\nu )=0$. Continuity therefore proves the claim throughout this connected domain. The branch points $\pm 2/\sqrt{3}$ are outside $\overline{D_{+}}$; interchanging the two root labels does not change the formula below.

Evaluate the global relation at each ${\nu }_j$, where its analyticity requirement holds:The left side is linear in ${\nu }_j$. Interpolate it to the argument $k$ with weightsThey satisfy $A_1+A_2=1$ and $A_1{\nu }_1+A_2{\nu }_2=k$. HenceFor the last coefficient, useadding $1-k^2$ and subtracting the interpolated $1-{\nu }_j^2$ gives $1-3k^2$.

The remaining solution-transform contribution is ${\int }_{\partial D_{+}}{e}^{ikx}{\sum }_j{A}_j\hat{q}({\nu }_j,t)dk$. Its integrand is holomorphic in $D_{+}$: the transformed arguments remain in the lower half-plane. The weights are bounded at infinity, and $e^{ikx}$ supplies decay on the upper closing arc for $x>0$. The Cauchy integral theorem therefore makes this integral zero. PutThe Dirichlet spectral representation for the linear dispersive Stokes equation is nowAll quantities in this expression are transforms of the prescribed initial and boundary values. Neither derivative trace remains.

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.