Put $y=x/\lambda(t)$ and
$$
u(t,x)=\lambda^{-1}v(y)e^{-ib|y|^2/4+i\gamma}.
$$
Differentiate the scaling and quadratic phase. After multiplying the nonlinear Schrödinger equation by $\lambda^3e^{ib|y|^2/4-i\gamma}$, the terms proportional to $y\cdot\nabla v+v$, $|y|^2v$, and $v$ vanish respectively when
$$
\frac{ds}{dt}=\lambda^{-2},
\qquad
b=-\lambda^{-1}\frac{d\lambda}{ds},
\qquad
\frac{db}{ds}+b^2=-4,
\qquad
\frac{d\gamma}{ds}=-\omega.
$$
The remaining expression is
$$
-\Delta v+|y|^2v-\omega v+v^3,
$$
which vanishes by part 4. Hence the <Schrödinger expanding lens ansatz for the mass-critical equation> solves the defocusing cubic equation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 154 / 1 / 5 / Solution

{c}

If $v$ solves the trapped stationary equation, the rescaled quadratic-phase ansatz
$$
u(t,x)=\lambda(t)^{-1}v(x/\lambda(t))
\exp\!\left(-\frac{i}{4}b(t)|x/\lambda(t)|^2+i\gamma(t)\right)
$$
solves the free mass-critical equation when the scaling, chirp, and phase parameters satisfy the associated lens-transform system.


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