The solution of a nonlinear solitary wave equation can be written as $$\phi(x) = \pm v \tanh\left[\frac{m}{ \sqrt 2} (x-x_0)\right]$$ where the plus and minus refer to the kink and antikink solutions respectively and $x_0$ is a constant of integration.
I don't understand the following claim
The energy density is localized near $x = x_0$ and goes to zero exponentially fast for $|x- x_0 | > 1/m$.
Can anyone explain this to me with a mathematical argument and graphical picture?
Please let me know if you need more information. Thanks in advance.
EDIT:
The illustration looks like:
Can we do it by Mathematica? Source: R. Rajamaran, Solitons and Instantons (North-Holland,Amsterdam, 1987) page no 21.
The energy functional for $\phi^4$ theory is given by:
$$E(\phi) = \int_{-\infty}^{\infty} \varepsilon(\phi,x) dx = \int_{-\infty}^{\infty} \left[\frac12 |\nabla \phi(x)|^2 + \frac{m^2}{4v^2} (\phi(x)^2 - v^2)^2 \right] dx$$
If one substitute the kink or anti-kink solution to the energy density $\varepsilon(\phi,x)$, one get $$\varepsilon(\phi,x) = \frac{m^2v^2}{2 \cosh(\frac{m}{\sqrt{2}}(x-x_0))^4}$$
Since $\cosh(x) \sim \frac12 e^{|x|}$ for large $x$, we find $$\varepsilon(\phi,x) \sim 8m^2v^2 e^{-2\sqrt{2}m |x-x_0|}\quad\text{ as }\quad x \to \pm \infty$$
To understand the exponential behavior for large $x$, let us look at the equation of motion of $\phi$ derived from varying the energy functional:
$$\frac{\delta E(\phi)}{\delta \phi} = 0 \quad\implies\quad -\frac{d^2\phi(x)}{dx^2} + \frac{m^2}{v^2}\phi(x)(\phi(x)^2 - v^2) = 0\tag{*} $$
For concreteness, let us look as the case where $\phi(x)$ is the kink solution and $x$ is large and positive. We know $\phi(x) \to v$ as $x \to +\infty$. This suggest us rewrite $\phi(x)$ as $v + \eta(x)$ and expand $(*)$ in powers of $\eta(x)$, we obtain: $$-\frac{d^2\eta(x)}{dx^2} + 2 m^2 \eta(x) + O(\eta(x)^2) = 0$$
This implies for large positive $x$, $\eta(x)$ falls off exponentially as $e^{-\sqrt{2}m x}$. If we do the same thing to the energy density, we find:
$$\varepsilon(\phi,x) \sim \frac12 |\nabla \eta(x)|^2 + m^2 \eta(x)^2 + O(\eta(x)^3)$$
Since the leading dependence of $\eta(x)$ in the energy density is quadratic, $\varepsilon(\phi,x)$ falls off twice as fast as $\phi(x)$, i.e. exponentially as $e^{-2\sqrt{2}m x}$. The is what we have obtained before by an explicit computation of $\varepsilon(\phi,x)$.