Paper 2, Section II, C

Differential Equations | Part IA, 2017

(a) Consider the system

dxdt=x(1x)xydydt=18y(4x1)\begin{aligned} &\frac{d x}{d t}=x(1-x)-x y \\ &\frac{d y}{d t}=\frac{1}{8} y(4 x-1) \end{aligned}

for x(t)0,y(t)0x(t) \geqslant 0, y(t) \geqslant 0. Find the critical points, determine their type and explain, with the help of a diagram, the behaviour of solutions for large positive times tt.

(b) Consider the system

dxdt=y+(1x2y2)xdydt=x+(1x2y2)y\begin{aligned} &\frac{d x}{d t}=y+\left(1-x^{2}-y^{2}\right) x \\ &\frac{d y}{d t}=-x+\left(1-x^{2}-y^{2}\right) y \end{aligned}

for (x(t),y(t))R2(x(t), y(t)) \in \mathbb{R}^{2}. Rewrite the system in polar coordinates by setting x(t)=x(t)= r(t)cosθ(t)r(t) \cos \theta(t) and y(t)=r(t)sinθ(t)y(t)=r(t) \sin \theta(t), and hence describe the behaviour of solutions for large positive and large negative times.

Typos? Please submit corrections to this page on GitHub.