Let us try to find out the final configuration of this process.

For the static solutions, $$h = \pm  \frac{1}{2\alpha}\sqrt{\frac{4\alpha E - k}{2k}},
\label{staticsolution}$$ which stay at one of the two minima of the potential, the energy density $\mathcal H$ is given by $$\mathcal H = U + \frac{1}{2} E^2 = 
\frac k {4\alpha} \left( E - \frac{k}{8\alpha}\right)\;.$$ Especially, the energy density vanishes when $k=0$. Figure \[fig:flatEh\] shows the amplitude $h$ and the energy density $\mathcal H$ change as functions of $k$. Since the energy density is monotonically increasing in $k$, we expect that solutions with $k\neq 0$ are unstable and decay to the solution with $k=0$. Note that, although $h$ diverges as $k$ goes to $0$, the field strength $E$ vanishes in this limit. The constant electric field in the initial configuration is wiped out in the $k=0$ solution and the final configuration will be the trivial vacuum state with $E=0$.

We can directly check that the static solution (\[staticsolution\]) with $k\neq 0$ is unstable. The solution corresponds to the gauge field configuration, $A_1^{(0)}(t) = \frac{k}{4\alpha} t$ and $A_3^{(0)} + i A_4^{(0)} = \frac{1}{2\alpha}
\sqrt{\frac{4\alpha E - k}{ 2k}} e^{-ik x_2}$. Let us add a small perturbation $A^{(0)} \rightarrow 