= Solution
For real $h\in\mathcal D(\mathbb R^d)$, differentiation under the integral gives
$$
\left.\frac d{dt}E(Q+th)\right|_{t=0}
=\int\nabla Q\mathbin{\cdot}\nabla h-\int Q^{1+4/d}h.
$$
The ground-state equation $\Delta Q-Q+Q^{1+4/d}=0$ and <integration by parts> reduce this to
$$
\left.\frac d{dt}E(Q+th)\right|_{t=0}=-\int Qh.
$$
In particular, along the amplitude direction $h=Q$ the derivative is $-\|Q\|_2^2<0$. Hence $u_0=(1+\delta)Q$ has negative energy for every sufficiently small $\delta>0$, while
$$
\|u_0\|_2=(1+\delta)\|Q\|_2<\|Q\|_2+\epsilon
$$
when $\delta$ is chosen small enough. The ground state has finite variance, so <negative-energy blowup for the mass-critical focusing nonlinear Schrödinger equation> shows that the corresponding solution blows up in finite time.
Back to article page
