and $g-1$ be in $\h{1}(\R)$. Define the *sliding metric* on $\h{1}$, $\dist$, $$\dist(f,g) = \inf_{y} \norm{f(\cdot+y)-g}_\h{1}
\label{eq:distance}$$

Given $\f-1\in C^{1}([0,T):H^{1}(\mathbb{R}))$, $T>0$, we say that $\f$ is *orbitally stable*, if for all $\eps>0$, there exists $\delta>0$ such that if $$\norm{\f(t=0)- \psi(t=0)}_\h{1}\leq \delta$$ with $\psi-1 \in C^{1}([0,T):H^{1}(\mathbb{R}))$ $$\dist(\f(t),\psi(t))<\ep$$ for $t< T$.

A [*solitary traveling wave*]{} is a solution of the form $\f=\f_c(x-ct)$, where $\f_c(x)$ asymptotes to a constant, say $\f\equiv1$, as $x\to\pm\infty$. Thus, solitary waves, $\f_c$, of , for the case of $n+m=0$ satisfy $$\label{eq:solitarywave1}
-c\dx \f_c +\dx(\f_c^n) + c \dx\paren{\f_c^n \dx(\f_c^n \dx\f_c) }=0$$ After one integration, and using the boundary condition at $\infty$, $$\label{eq:solitarywave2}
-c\f_c +\f_c^n + c \f_c^n \dx(\f_c^n\dx\f_c) +c-1=0$$ In general, there is no closed form expression for $\f_c$ as a function of $x$. As previously noted, for $c>n>1$, a solution in excess of the reference state at $\infty$ exists, [@Nakayama92]. $\f_c(x)$ can be shown to be exponentially decaying as $x\to\pm\infty$ and analytic in a strip about the real axis in the complex plane, see [@Simpson07b] for more details. Two such waves are pictured in Figure \[fig:solwaves\].

![Two solitary waves for $n=2$. Note that as the speed parameter, $c$, increases, the waves become both taller *and* broader.[]{data-label="fig:solwaves"}](solwave-fig){width="4in"}

It is 