Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 (Introduction to Non-Linear Analysis)

For $\psi (x)=e^{-|x{|}^2/2}$ in two dimensions,Thus the Laplacian satisfies $\Delta \psi =(|x{|}^2-2)\psi$, and the isotropic harmonic-oscillator operator gives

Split the integral at radius $R$. The Holder inequality on the ball and the spatial moment outside it giveTaking $R={\epsilon }^{-1/2}$ proves the stated estimate with $C_{\epsilon }=\sqrt{\pi }{\epsilon }^{-1/2}$. Choose $\epsilon <1/\omega$. If $X=\Vert v{\Vert }_4^2$, thenThe final two terms form a quadratic polynomial bounded below, so ${\inf }_{\Sigma }J>-\infty$.

Use the Gaussian from part 1. Since $(-\Delta +|x{|}^2)\psi =2\psi$,The quadratic coefficient is negative because $\omega >2$. For every sufficiently small nonzero $a$, the negative quadratic term dominates the quartic term, so $J(a\psi )<0$ and therefore ${\inf }_{\Sigma }J<0$.

Let $(v_n)$ be a minimizing sequence. The coercive lower bound from part 2 makes it bounded in the harmonic-oscillator energy space and in $L^4$. After taking a subsequence, $v_n$ converges weakly in both spaces. The compact embedding of the harmonic-oscillator energy space gives strong convergence in $L^2$, while weak lower semicontinuity of the gradient, moment, and $L^4$ terms yieldsThus $v$ attains the infimum. Replacing $v$ by $|v|$ does not increase the gradient norm, so a minimizer may be chosen nonnegative. It is nonzero because the infimum is negative whereas $J(0)=0$.

Taking the first variation against a smooth compactly supported function gives the Euler-Lagrange equationwhich is the Schrödinger trapped defocusing stationary equation.

Put $y=x/\lambda (t)$ andDifferentiate the scaling and quadratic phase. After multiplying the nonlinear Schrödinger equation by ${\lambda }^3{e}^{ib|y{|}^2/4-i\gamma }$, the terms proportional to $y\cdot \nabla v+v$, $|y{|}^{2}v$, and $v$ vanish respectively whenThe remaining expression iswhich vanishes by part 4. Hence the Schrödinger expanding lens ansatz for the mass-critical equation solves the defocusing cubic equation.

The Riccati equation $b_s+b^2=-4$ with $b(0)=0$ has solutionSince ${\lambda }_s/\lambda =-b$, the initial condition $\lambda (0)=1$ gives $\lambda (s)=\sec (2s)$. ConsequentlyFinally ${\gamma }_s=-\omega$. In physical time the solution is therefore

The scaling in part 5 givesThus $u\in L_{t,x}^4([-1,\infty )\times {\mathbb{R}}^2)$. For $T<T^{\prime }$, the dual Strichartz estimate for the free Schrödinger equation giveswhich tends to zero as $T,T^{\prime }\to \infty$. The integrals in the question are therefore Cauchy in $L^2$.

The Duhamel principle in the interaction representation defines an $L^2$ limit $u_{\infty }$ and expresses $S(-t)u(t)-u_{\infty }$ as the tail integral from $t$ to infinity. The same estimate sends that tail to zero, proving the scattering from a finite Strichartz norm conclusion

For a smooth compactly supported $u$, expand the nonnegative squareSince $\nabla \cdot (x/|x{|}^2)=|x{|}^{-2}$ in three dimensions, integration by parts turns the cross term into $-\frac{1}{2}\int |u{|}^2/|x{|}^2$. HenceDensity extends this Hardy inequality in Euclidean space to $H^1({\mathbb{R}}^3)$.

Differentiate the energy and use the chain rule:After integration by parts, the middle term is $-\int (\Delta u)u_t$. Thereforeby the defocusing semilinear wave equation. Thus the total energy is conserved.

SetDifferentiate, substitute $u_{tt}=\Delta u-u|u{|}^{p-1}$, and integrate every second derivative off $u$. The mixed first-derivative terms cancel because of the correction $(\Delta \psi )u/2$. For radial $u$ and radial $\psi$, the Hessian term is ${\psi }^{\prime \prime }|\nabla u{|}^2$. The potential term uses $u|u{|}^{p-1}\nabla u=\nabla (|u{|}^{p+1})/(p+1)$. One obtains the Morawetz identity for the defocusing wave equation

The displayed estimate requires the standard choice $\psi (r)=r$; read literally, the printed ${\psi }^{\prime }(r)=r$ would give $\psi =r^2/2$, $-{\Delta }^2\psi =0$, and would not imply the claimed estimate. For $\psi =|x|$, the distributional bilaplacian of the radial coordinate in three dimensions isThe delta term is nonnegative. The Morawetz action is bounded by $CE$ using Cauchy-Schwarz and the Hardy inequality in Euclidean space. Integrating the identity from $0$ to $T$ and discarding the delta term givesuniformly in $T$, proving the Morawetz estimate for the defocusing wave equation.

A finite-energy stationary solution would make the nonnegative spatial integral on the left constant in time. Its integral over $[0,\infty )$ can be finite only when that spatial integral is zero, so the stationary solution is $u=0$.

For a radial function in three dimensions,Set $w=ru$. Multiplying the wave equation by $r$ gives the radial reduction of the three-dimensional wave equationwith the regularity boundary condition $w(t,0)=0$.

Write $q=w_t+w_r$. The equation in part 5 saysDifferentiate $J_{\psi }$, replace $q_t$ by $q_r-|w{|}^{p-1}w/r^{p-1}$, and integrate the $q{q}_r$ and $|w{|}^{p-1}w{w}_r$ terms by parts. This gives the outgoing-energy identity for a radial defocusing wave

Since $w=ru$,and radial integration satisfies $dx=4\pi r^{2}dr$. Multiplying the identity from part 6 by $4\pi$ therefore giveswhich is the modified Morawetz identity.

Choose the constant weight $\psi =1$. The first term on the right vanishes and the second isThe functional on the left is bounded by the conserved energy using the Hardy inequality in Euclidean space. Integrating in time gives another proof of the Morawetz estimate for the defocusing wave equation.