Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 70 / 1

Use the half-line Fourier transform and finite-time boundary transformsThe half-line Fourier transform is analytic for $\mathrm{Im}k<0$ under spatial decay. All spectral integrals below have their usual oscillatory, or vanishing Gaussian damping, interpretation until absolute convergence is established.

The free Schrodinger equation has the local divergence identityIntegrating in $x,t$ gives the global relation for the half-line free Schrodinger equationLet $D_{+}=\{\mathrm{Re}k>0,\mathrm{Im}k>0\}$. Orient its boundary from $i\infty$ down to $0$, and then from $0$ to $+\infty$, so that $D_{+}$ lies on the left. Fourier inversion followed by contour integration in the second quadrant yieldsThis is a complex spectral representation involving the initial trace and both boundary traces. The contour deformation works because each boundary-time integrand contains $e^{-i{k}^2(t-s)}$ with $s\le t$, which decays in the second quadrant, as well as $e^{ikx}$ for $x>0$. This fixes both the quadrant and the orientation; changing either without changing the signs would produce an incorrect representation.

The unknown Neumann boundary condition can be removed by the Fokas method. Evaluate the global relation for the half-line free Schrodinger equation at $-k$:Consequently $k{G}_0-i{G}_1=2k{G}_0-{\hat{q}}_0(-k)+e^{i{k}^{2}t}\hat{q}(-k,t)$. In the representation from part (a), the last term has integral ${\int }_{\partial D_{+}}{e}^{ikx}\hat{q}(-k,t)dk=0$, by the Cauchy integral theorem and Jordan lemma. It is analytic in the upper half-plane, and the exponential decays on the closing first-quadrant arc. Hence an expression involving only the given data isOne may replace $t$ by $T$ in the upper limit defining $G_0$: the contribution of boundary times $s>t$ closes to zero in $D_{+}$, since $e^{-i{k}^2(t-s)}$ then decays there. Using $t$ makes causality transparent.

For an explicit proof of uniform convergence, it is useful to apply boundary lifting before inversion. Set $h(x)=e^{-x}$, $c=g_0(0)$, $w_0(x)=q_0(x)-ch(x)$, andThe compatibility condition gives $w_0(0)=0$. The Fourier sine transform of $w=q-g_{0}h$ satisfies $S_t+i{k}^{2}S=Hf$, with initial value $S_0$. The integrating factor therefore gives the equivalent representationThis last integral is absolutely and uniformly convergent for $x\ge 0$, $0\le t\le T$ under concrete sufficient hypotheses $w_0,w_0^{\prime },w_0^{\prime \prime }\in L^1$, decay of the boundary terms, and $g_0\in C^2[0,T]$. Indeed, two integrations by parts give $|S_0(k)|\le \Vert w_0^{\prime \prime }{\Vert }_1/k^2$ for $k\ge 1$. Another integration by parts, this time in $s$, givesThus the initial term is uniformly $O(k^{-2})$ and the forcing term uniformly $O(k^{-3})$. For $0<k<1$, use $|S_0(k)|\le \Vert w_0{\Vert }_1$ and the bounded time integral. An integrable majorant proves the claimed uniform convergence and permits evaluation at both boundaries.

At $t=0$, Fourier sine inversion gives $q(x,0)=c{e}^{-x}+w_0(x)=q_0(x)$. At $x=0$, the integral vanishes, giving $q(0,t)=g_0(t)$, including the compatible corner. To verify the equation, note that $h^{\prime \prime }=h$ and the transformed equation impliesAdding the lifted part gives $i{q}_t+q_{xx}=0$. Under the stated smoothness, this holds classically in the interior; differentiated spectral integrals can first be Gaussian-regularized, or read in the sine-transform $L^2$ sense and then identified with the smooth solution. Uniform convergence of $q$ itself does not require claiming uniform convergence of every differentiated integral at the corner.

Splitting the free Schrodinger equation into real and imaginary parts givesDifferentiating the first relation in time and using the second yields the Euler-Bernoulli beam equation $u_{tt}+u_{xxxx}=0$. This is the Schrodinger factorization of the elastic beam equation.

Assume the initial velocity has an integrable first spatial moment, as allowed by sufficient decay, and defineThen $v_0^{\prime \prime }=-u_1$ and $v_0$ decays at infinity. To encode the second boundary datum, defineThe Dirichlet boundary condition for the resulting free Schrodinger equation is compatible at the corner, since $q_0(0)=g_0(0)$. Insert these explicit $q_0,g_0$ into the data-only complex integral in part (b), with ${\hat{q}}_0$ and $G_0$ defined as in part (a). The required displacement is the real part of that integral. Equivalently, the uniformly convergent lifted integral in part (b) may be used with the same complex data.

The Schrodinger factorization of the elastic beam equation verifies every condition: $u(x,0)=u_0(x)$, $u_t(x,0)=-v_0^{\prime \prime }(x)=u_1(x)$, $u(0,t)={\tilde{u}}_0(t)$, andThe corner requirements on $u_0^{\prime \prime }$ and ${\tilde{u}}_0^{\prime }$ ensure consistency of these derivative traces; the natural interpretation of the last printed compatibility is ${\tilde{u}}_0^{\prime }(0)=u_1(0)$. If its prime were instead imposed for every $t$, that would simply be an extra restriction on the data, and the same construction would still solve them.