Boundary condition for solitons in 1+1 dimensions to have finite energy

Question

Suppose a classical field configuration of a real scalar field $\phi(x,t)$, in $1+1$ dimensions, has the energy $$E[\phi]=\int\limits_{-\infty}^{+\infty} dx\, \left[\frac{1}{2}\left(\frac{\partial\phi}{\partial t}\right)^2+\frac{1}{2}\left(\frac{\partial \phi}{\partial x} \right)^2+a\phi^2+b\phi^4 \right]$$ where the potential is given by $$V(\phi)=a\phi^2+b\phi^4.$$

For the energy $E[\phi]$ to be finite, one necessarily requires $\phi\rightarrow 0$, as $x\rightarrow\pm\infty$. But in addition, shouldn't we also necessarily require that both $\frac{\partial\phi}{\partial t}$ and $\frac{\partial \phi}{\partial x}$ must vanish as $x\rightarrow \pm\infty$?

However, if I'm not mistaken, to find soliton solutions (which have finite energy), one imposes only the boundary condition $\phi\rightarrow 0$, as $x\rightarrow\pm\infty$. Does it mean that if this is satisfied, the vanishing boundary condition on the derivatives of the field are also automatically satisfied?

EDIT: I know that for arbitrary functions $f(x)$ this is not true i.e., if $f(x)\rightarrow 0$ as $x\rightarrow \pm \infty$, $f^\prime(x)$ need not vanish as $x\rightarrow \pm \infty$. But $\phi(x,t)$ are not arbitrary functions in the sense that they are solutions of Euler-Lagrange equations. Therefore, it may be possible that the condition $\phi\rightarrow 0$, as $x\rightarrow\pm\infty$ is sufficient.

Answer 1

In order to the energy be finite it is necessary that the energy density asymptotically vanishes. Notice that this is achieved if the scalar field asymptotically approaches a constant value.

Now recall that a soliton is not only a finite energy solution of the equations of motions but also a stable one. For topological solitons this requires that the vacuum manifold be degenerate. In your example this is only possible if $a<0$ (the potential is not positive definite!). Then as you can see the potential vanishes for $\phi=\pm\sqrt{-a/b}$. When the scalar field interpolates this two values, i.e., $$\phi(t,x\rightarrow -\infty)=-\sqrt{-a/b},\quad \phi(t,x\rightarrow +\infty)=+\sqrt{-a/b},$$ then we have a topologically stable solution. The plots bellow shows these features for the kink in $1+1$. On the left a degenerate potential. On top right a scalar field interpolating two different vacua and on the bottom right the energy density as function of the position. Note that this solution cannot be deformed to the vacuum solution (either $\phi(t,x)=-v$ or $\phi(t,x)=+v$) since it will cost an infinite amount of energy. That is what gives the stability of this solution.

Answer 2

It seems that You're wrong. You can look for static configurations $\varphi = \varphi (x)$. Then the expression for the energy is reduced to $$ E[\varphi] = \int \limits_{-\infty}^{\infty} dx \frac{1}{2}\left(\partial_{x}\varphi \mp \sqrt{2V(\varphi(x))} \right)^{2} \pm \int \limits_{\varphi(x = -\infty)}^{\varphi (x = \infty)}\sqrt{2V(\varphi)}d\varphi, $$ where $$ V(\varphi) = a\varphi^{2} + b\varphi^{4} $$ If we require the energy to be finite, we have $$ \partial_{x}\varphi \to \pm \sqrt{2V(\varphi(x))} \ \ \text{at } \ \ x \to \infty , \quad V(\varphi(x)) \leqslant \infty $$