Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 1

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