Paper 2, Section I, B

Mathematical Biology
Part II, 2016

(a) The populations of two competing species satisfy

dN1dt=N1[b1λ(N1+N2)]dN2dt=N2[b2λ(N1+N2)]\begin{aligned} \frac{d N_{1}}{d t} &=N_{1}\left[b_{1}-\lambda\left(N_{1}+N_{2}\right)\right] \\ \frac{d N_{2}}{d t} &=N_{2}\left[b_{2}-\lambda\left(N_{1}+N_{2}\right)\right] \end{aligned}

where b1>b2>0b_{1}>b_{2}>0 and λ>0\lambda>0. Sketch the phase diagram (limiting attention to N1,N20)\left.N_{1}, N_{2} \geqslant 0\right).

The relative abundance of species 1 is defined by U=N1/(N1+N2)U=N_{1} /\left(N_{1}+N_{2}\right). Show that

dUdt=AU(1U)\frac{d U}{d t}=A U(1-U)

where AA is a constant that should be determined.

(b) Consider the spatial system

ut=u(1u)+D2ux2\frac{\partial u}{\partial t}=u(1-u)+D \frac{\partial^{2} u}{\partial x^{2}}

and consider a travelling-wave solution of the form u(x,t)=f(xct)u(x, t)=f(x-c t) representing one species (u=1)(u=1) invading territory previously occupied by another species (u=0)(u=0). By linearising near the front of the invasion, show that the wave speed is given by c=2Dc=2 \sqrt{D}.

[You may assume that the solution to the full nonlinear system will settle to the slowest possible linear wave speed.]