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Consider a scalar field $\phi(t, x, y, z)$ obeying the waves equation with an Higgs-like potential (the "mexican hat"): $$\tag{1} \mathcal{V}(\phi) = \frac{\lambda}{4} (\phi^2 - \phi_0^2)^2, $$ where $\lambda > 0$ and $\phi_0$ are constants. $\phi_0$ is the value of the "true vacuum" field, which can be positive or negative. The waves equation of motion of the self-interacting field is this (I'm using units such that $c \equiv 1$): $$\tag{2} \frac{\partial^2 \phi}{\partial t^2} - \frac{\partial^2 \phi}{\partial x^2} - \frac{\partial^2 \phi}{\partial y^2} - \frac{\partial^2 \phi}{\partial z^2} + \lambda (\phi^2 - \phi_0^2) \, \phi = 0. $$ To solve the waves equation (2), we need a consistent set of initial conditions and boundary conditions. For the initial conditions, we usually give the field values at time $t = 0$ and its rate of change: \begin{align} \phi(0, x, y, z) &= \mathcal{F}(x, y, z), \tag{3} \\[2ex] \frac{\partial \phi}{\partial t} \, \bigg|_{t = 0} &= \mathcal{G}(x, y, z). \tag{4} \end{align} As far as I know, there is no constraint on the values of $\phi(0, x, y, z)$, in the classical theory of relativistic fields. The function $\mathcal{F}$ can be anything.

But is there any constraint on the values of the field derivative? Can the function $\mathcal{G}$ be completely arbitrary too in relativity? (this time derivative is not the same as a real velocity, which of course is constrained by causality: $v < 1$).

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