 Solution

A travelling wave $u(x,t)=U(x-vt)$ obeys $m{U}^{\prime \prime }+v{U}^{\prime }+f(U)=0$, so the mechanical potential in the requested form is $V=F$, up to an additive constant. HereIts stationary points are maxima at $0,1$ and a minimum at $r$. Since $F(0)=0$ and $F(1)=(1-2r)/12$, the two maxima have equal height at $r=1/2$; the maximum at $1$ is lower for $r>1/2$ and higher for $r<1/2$.

Multiply the wave equation by $U^{\prime }$ and integrate along the heteroclinic orbit. The endpoint derivatives vanish, and $U(-\infty )=1$, $U(\infty )=0$, soConsequentlyThe integral is positive for a nonconstant front. In fact substitution of the same logistic shape from part (b) gives the exact velocity $v=\sqrt{m/2}(1-2r)$. Thus the $u=1$ phase invades rightwards when $r<1/2$, the zero phase invades leftwards when $r>1/2$, and the balanced front is stationary.