## Problem Set 5: Solutions

1. Use Lyapunov functions to describe the stability of the zero solution for
each of the

fol lowing systems .

This system looks like a stable linear
node

with an additional damping term, so we look for a Lyapunov function to prove

stability. The most basic choice is a quadratic function ,

This function is clearly positive definite , has level sets which are closed
orbits

around the origin, and is smooth. Thus, we can verify that

for (x, y) ≠ (0, 0). Thus, the origin is a global attractor.

This system looks quite similar to the previous

one - it is a a linear center plus a damping term . We would like the linear
terms

to cancel when we compute
, so we choose the quadratic function V (x, y) =

1/2(4x^{2} + y^{2}), which is positive de nite, has elliptical level sets around the
origin,

and is smooth. ComputeThis is

negative for x , y ∈(-π ,π )
- 0, so the origin is a local attractor.

2. We consider the system

Let us take the following Lyapunov function

where a, b > 0. V (x, y) is clearly positive definite and
its partial derivatives are all

continuous. Consider

We need to show that
(x, y) is negative definite in a neighborhood of (0, 0). In fact,

we will show that that
(x, y) is negative definite inside the ellipse V (x, y) = 1. Take

any (x, y) ≠ (0, 0) satisfying

This necessarily implies |x| < a, |y| < b and consequently
x - a < 0, y - b < 0, which

means
(x, y) < 0. Note also that in region (??) we have
(x, y) = 0 iff x = 0, y = 0,

so
(x, y) is negative definite there and all solutions starting in (??) approach
the

origin

3. If (and only if) we choose A,B > 0, we have that V is
C^{1}, positive definite, and,

since yf(y) > 0, defines closed orbits around the origin. Thus, to prove
asymptotic

stability, we just need to show that
< 0 in some neighborhood of the origin.

If we choose A = B = 1, the given condition that
(and, say, that α > 0 so

that the principal minors of the matrix are the correct signs ), we know that
this is a

negative definite form and is thus negative for all nonzero (x, f(y)) (which
correspond

to nonzero (x, y)).

4. The equation

can be rewritten as a system

This is a Lienard equation (see Theorem 11.2, page 402,
Jordan & Smith, 3rd edition).

We have f(x, y) = x^{2} + y^{2} - 1, g(x) = x, G(x) = 1/2x^{2} and all the as sumptions
of the

theorem are satisfied. Define the energy function

Note that E(0, 0) = 0, ε is
positive definite, continuous and increases monotonically

in every radial direction from the origin. The family of contours of ε
consists of circles

centered on the origin. Notice

is positive semi-definite inside the circle x ^{2} + y^{2} = 1,
so the critical point (0, 0) is

unstable. In fact, the linearization

yields the eigenvalues
, which means the origin is an unstable spiral. Note

further that
is negative outside the menti oned circle , except at y = 0, and is zero

on the circle. This suggests that the circle x^{2} + y^{2} = 1 is the only closed
curve in the

x , y plane that could potentially be a limit cycle. In fact, by direct
substitution we

verify that

is the periodic solution of (??) (whose existence is
asserted by Theorem 11.2). Given

the sign of
in a neighborhood of the cycle, we conclude it is stable.