plane. The solution is also translationally invariant in the transverse directions. It is then natural to look for a non-linear solution with the same set of symmetries, and we choose the following ansatz, $$\label{E:FlatAnsatz}
\begin{split}
A_0&=0\;,\qquad A_2=g(t)\\
A_1&=A_1(t)\\
A_3+ i A_4 &= h(t) e^{-i k x_2}\;.
\end{split}$$ We denote the time coordinate by $x^0$ or $t$ interchangeably. The equation of motion for the function $g(t)$ sets $g(t)=0$. The remaining equations of motion become $$\label{E:EOMflat5D}
\begin{split}
&\ddot{A}_1(t) -4 \alpha k h(t) \dot h(t)=0\\
&\ddot h(t) + k^2 h(t) + 4\alpha k \dot A_1(t) h(t) =0\;.
\end{split}$$ Note that the momentum $\Pi_1$ conjugate to $A_1$ is given by $$\label{E:Pi1Flat}
\Pi_1 = -\dot A_1(t) +2\alpha k h(t)^2 \;.$$ The first equation of can be written as $\partial \Pi_1/\partial t =0$ and solved by $\Pi_1=constant$. The integration constant is fixed as $\Pi_1 = E$ by the initial configuration where $h=0$ and $\Phi_1= -\dot A_1 =E$.

From and , $$\ddot h(t) + k^2 h(t) - 4\alpha k E h(t) + 8 \alpha^2 k^2 h^3(t)=0\;.$$ This can be integrated and yields $$\frac 1 2 \dot h(t)^2 - \frac 1 2 k (4\alpha E - k) h(t)^2 + 2\alpha^2 k^2 h(t)^4 = constant.$$ This is equal to 