 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/32a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

U (x) = \frac{x ^{2}}{( 1 + x ^{2} ) ^{2}}, E (x, y) = \frac{1}{2} y^{2} + U (x) .

The stated identity gives

U^{'} (x) = \frac{2 x ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{3}} .

Hence, along the system,

\dot{E} = y \overset{y}{˙} + U^{'} (x) \overset{x}{˙} = y [- U^{'} (x) - k y] + U^{'} (x) y = - k y^{2} \leq 0.

Near the origin,

E

is positive definite because

U (x) > 0

for

x \neq = 0

. The First Lyapunov theorem therefore gives stability.

Choose a sufficiently small compact invariant sublevel set

Ω_{c} = {(x, y) : E (x, y) \leq c}, 0 < c < \frac{1}{4} .

{\dot{E} = 0}

we have

y = 0

. A trajectory can remain there only if

\overset{y}{˙} = - U^{'} (x) = 0,

which occurs at

x = 0, \pm 1

c < 1/4 = U (\pm 1)

excludes the two latter points, so the largest invariant subset of

{\dot{E} = 0} \cap Ω_{c}

is the origin. LaSalle invariance principle now gives convergence to the origin. Thus, for every

k > 0

(0, 0) is asymptotically stable .

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