= Solution
The <Hamiltonian> <functional> is
$$
H[\psi,\psi^*]=\int\left[\frac{\hbar^2}{2m}|\nabla\psi|^2+V_{\rm ext}|\psi|^2+\frac U2|\psi|^4\right]d^3x.
$$
Treating the field and its <complex conjugate> as independent variables, and integrating the <gradient> term by parts with periodic, decaying or fixed-field boundary conditions, gives
$$
\frac{\delta H}{\delta\psi^*}=-\frac{\hbar^2}{2m}\nabla^2\psi+V_{\rm ext}\psi+U|\psi|^2\psi.
$$
Thus $i\hbar\psi_t=\delta H/\delta\psi^*$ is the <Gross–Pitaevskii equation>. The factor $1/2$ in the interaction energy avoids counting each pair twice.
At prescribed particle number, minimize $H$ subject to $N=\int|\psi|^2d^3x$. Introducing the <chemical potential> as a <Lagrange multiplier>, the variation of $H-\mu N$ vanishes at a ground-state profile $\psi_0$, proving
$$
\boxed{\left[-\frac{\hbar^2}{2m}\nabla^2+V_{\rm ext}+U|\psi_0|^2\right]\psi_0=\mu\psi_0.}
$$
The corresponding time-dependent field is $\psi(\mathbf x,t)=e^{-i\mu t/\hbar}\psi_0(\mathbf x)$. A solution of this stationary equation is only a candidate <ground state>; the <ground state> is the minimizer among all fields satisfying the normalization. For the uniform repulsive gas in a periodic volume, the <gradient> energy is nonnegative and $\int n^2d^3x\ge N^2/\mathcal V$, so the constant-density, constant-phase state is indeed the minimum. The uniform-background solutions below assume $U>0$ and a positive background density.
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