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

<br /> \phi&#039;\phi&#039;&#039;=\phi&#039;\frac{dV}{d\phi}\\<br /> \frac{d}{dx}\left(\frac{\phi&#039;^2}{2}\right)=\frac{dV}{dx}\\<br /> \phi&#039;=\pm \sqrt{2V}\\<br /> \phi&#039;=\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&#039;^2}{2}+V(\phi)] is such that

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

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