Why is the solution of the phi^6 potential not a soliton?

[Catria]

Homework Statement

Consider a theory with a $\phi^6$-scalar potential:

$\mathcal{L} = \frac{1}{2}(\partial_\mu\phi)^2-\phi^2(\phi^2-1)^2.$

Why is the solution to the equation of motion not a soliton?

Homework Equations

$\phi''=\frac{\partial V}{\partial\phi}$

The Attempt at a Solution

$\phi'\phi''=\phi'\frac{dV}{d\phi}\\<br /> \frac{d}{dx}\left(\frac{\phi'^2}{2}\right)=\frac{dV}{dx}\\<br /> \phi'=\pm \sqrt{2V}\\<br /> \phi'=\phi-\phi^3\\<br /> \Rightarrow\phi(x)= \frac{e^x}{\sqrt{e^{2x}-C_1}}-\frac{e^{-x}}{\sqrt{e^{-2x}-C_2}}$

Yet the solution on the last line is not a soliton. Why is that so?

 

[Catria]

More generally, one has

$\mathcal{L} = \frac{1}{2}(\partial_\mu\phi)^2-(\phi^2-\epsilon)(\phi^2-1)^2$

in which case there is no soliton, since the energy, defined as $E=\int_{-\infty}^{+\infty}dx [\frac{\phi'^2}{2}+V(\phi)]$ is such that

$\lim_{t\to +\infty}E(t) = -\infty$

for $\epsilon\neq 0$ and hence is topologically unstable.