Substitute the <Focusing Schrödinger lens ansatz> into
$$
i u_t+\Delta u+|u|^2u=0.
$$
The imaginary terms proportional to $P_\eta+y\cdot\nabla P_\eta$ cancel because $b=-\lambda\lambda_t$. After multiplying the remaining real equation by $\lambda^3$, one obtains
$$
\Delta P_\eta+P_\eta^3
-\lambda^2\gamma_tP_\eta
+\frac{\lambda^2b_t+b^2}{4}|y|^2P_\eta=0.
$$
The dynamical system gives $\lambda^2b_t+b^2=-\eta$. Comparison with the equation for $P_\eta$ therefore requires
$$
\gamma_\eta'(t)=\frac1{\lambda_\eta(t)^2}
=\frac1{t^2+\eta}.
$$
Thus, up to an arbitrary constant phase,
$$
\boxed{\gamma_\eta(t)
=\gamma_\eta(-1)
+\frac1{\sqrt\eta}
\left(\arctan\frac t{\sqrt\eta}
+\arctan\frac1{\sqrt\eta}\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 154 / 3 / 4 / Solution

{c}

If $P_\eta$ solves the <trapped focusing cubic ground-state equation>, the ansatz
$$
u(t,x)=\lambda(t)^{-1}P_\eta(x/\lambda(t))
\exp\left(-\frac{i}{4}b(t)|x/\lambda(t)|^2+i\gamma(t)\right)
$$
solves the two-dimensional <mass-critical focusing nonlinear Schrödinger equation> when
$$
b=-\lambda\lambda_t,
\qquad
\lambda^2b_t+b^2=-\eta,
\qquad
\gamma_t=\lambda^{-2}.
$$

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