= Solution
A <travelling wave> $u(x,t)=U(x-vt)$ obeys $mU''+vU'+f(U)=0$, so the mechanical potential in the requested form is $V=F$, up to an additive constant. Here
$$
F(U)=-\frac14U^2(1-U)^2+\frac16(r-\tfrac12)U^2(2U-3).
$$
Its 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 $r<1/2$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-2-front-potentials.png]
{title=Bistable front potentials with equal and unequal endpoint maximum heights}
Multiply the wave equation by $U'$ and integrate along the <heteroclinic orbit>. The endpoint derivatives vanish, and $U(-\infty)=1$, $U(\infty)=0$, so
$$
0=\frac m2[(U')^2]_{-\infty}^{\infty}+v\int_{-\infty}^{\infty}(U')^2dz+F(0)-F(1).
$$
Consequently
$$
\boxed{v\int_{-\infty}^{\infty}(U')^2dz=\frac{1-2r}{12},\qquad \operatorname{sgn}v=\operatorname{sgn}(1-2r).}
$$
The 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.
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