Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 13B (Mathematical Biology)

For spatially homogeneous solutions, the bistable cubic reaction-diffusion equation reduces to $\dot{u}=f(u)=u(1-u)(u-r)$. Its equilibria are $0,r,1$, andThus $0$ and $1$ are asymptotically stable, while $r$ is unstable. Small spatial Fourier perturbations have growth rates $f^{\prime }(u_{\ast })-m{k}^2$, so diffusion also leaves the two outer uniform states linearly stable.

At $r=1/2$, write $F(u)=-u^2(1-u{)}^2/4$, so $F^{\prime }=f$. Multiplying the stationary equation $m{U}^{\prime \prime }+F^{\prime }(U)=0$ by $U^{\prime }$ gives $m(U^{\prime }{)}^2/2+F(U)=C$. The endpoint values and vanishing endpoint derivatives make $C=0$. For the decreasing connection,Separate variables and absorb the integration constant into a translation $x_0$. The logistic front of a bistable cubic equation isIt approaches $1$ on the left and $0$ on the right and satisfies the stationary differential equation directly.

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.