Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 7A (Differential Equations)

Table of contents
- i 7A
  - Solution i
- ii 7A
  - Solution ii
- iii 7A
  - Solution iii
- iv 7A
  - Solution iv

i

Solution

Take

F = x^{2} + y^{2} .

Then

\dot{F} = 2 x \overset{x}{˙} + 2 y \overset{y}{˙} = 2 k (x^{2} + y^{2})^{2} = 2 k F^{2} .

With

F (0) = 1

, separation gives

- \frac{1}{F} = 2 k t - 1,

and hence

F (t) = \frac{1}{1 - 2 k t}

on its maximal interval of existence.

ii

Solution

The only equilibrium is

(0, 0)

. Its linearization is

(\overset{x}{˙} \overset{y}{˙}) = (0 - 1 10) (x y),

whose eigenvalues are

\pm i

, so the linearized system has a centre.

The exact radial identity from part (i) gives

\overset{r}{˙} = k r^{3} .

Thus the origin is asymptotically stable for

k < 0

, neutrally stable for

k = 0

, and unstable for

k > 0

. Moreover,

\dot{θ} = \frac{x y ˙ - y x ˙}{r ^{2}} = - 1,

so trajectories rotate clockwise.

iii

Solution

The radially symmetric planar dynamical system formulas give

f (r) = k r^{3}, g (θ) = - 1 .

With

r (0) = 1

and

θ (0) = 0

,

r (t) = \frac{1}{1 - 2 k t}, θ (t) = - t,

again on the maximal interval for which the denominator is positive.

iv

Solution

For the modified system,

F = x^{2} + y^{2}

satisfies

\dot{F} = 2 F - 4 F^{2} = 2 F (1 - 2 F) .

With

F (0) = 1

, the logistic solution is

x (t)^{2} + y (t)^{2} = \frac{1}{2 - e ^{- 2 t}} .

Consequently,

t \to \infty lim (x (t)^{2} + y (t)^{2}) = \frac{1}{2} .

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