MA 2327
Assignment 9
Due never
Id:
2327-f2015-9.m4,v 1.2 2016/03/21 11:04:19 john Exp john
1. For which of the following systems is (0, 0) a stable equibrium? For
which is it strictly stable?
(a)
x′ (t) = −x(t) + 2y(t),
y ′(t) = 2x(t) − y(t).
(b)
x′ (t) = −2x(t) + y(t),
y ′(t) = x(t) − 2y(t).
(c)
x′ (t) = −y(t),
y ′ (t) = x(t).
(d)
x′ (t) = y(t),
y ′ (t) = 0.
Solution: These are all linear autonomous systems, so it suffices to
consider eigenvalues of the corresponding matrix, and possibly its associated eigenspaces.
1
Id:
2327-f2015-9.m4,v 1.2 2016/03/21 11:04:19 john Exp john 2
(a) The eigenvalues are 1 and −3. Since one of these has positive real
part the equilibrium is neither stable nor strictly stable.
(b) The eigenvalues are −1 and −3. Both have negative real part, so
the equilibrium is both stable and strictly stable.
(c) The eigenvalues are i and −i. Neither has positive real part and
both are of multiplicity 1, so the equilibrium is stable. It is not
strictly stable, because they do not have negative real part.
(d) The only eigenvalue is 0, with algebraic multiplicity 2. Its eigenspace
is however only of dimension 1, so the Jordan blocks are not all
simple. In this case that’s trivial to see because the matrix is already in Jordan normal form. So the equilibrium is neither stable
nor strictly stable.
2. Consider the system
dx/dt = −y + a(x2 + y 2 )x,
dy/dt = x + a(x2 + y 2)y.
depending on a parameter a. Its linearisation at the equilibrium (0, 0)
is
dx/dt = −y,
dy/dt = x,
whose stability you examined in the previous question. In particular, it
is independent of a. For which values of a is (0, 0) a stable equilibrium
of the original system? For which is it strictly stable?
Hint: The system can be solved explicitly via a change to polar coordinates x = r cos θ, y = r sin θ.
Solution: In polar coordinates,
dr
x dx y dy
=
+
= ar 3 ,
dt
r dt
r dt
y dx
x dθ
dθ
=− 2
+ 2
= 1.
dt
r dt
r dt
The solution is
r0
r(t) = q
,
1 − ar02 (t − t0 )/2
θ(t) = θ0 + t − t0 .
Id:
2327-f2015-9.m4,v 1.2 2016/03/21 11:04:19 john Exp john 3
In the original coordinates this is
x0 cos(t − t0 ) + y0 sin(t − t0 )
,
x(t) = q
1 − a(x20 + y0 )2 (t − t0 )/2
y0 cos(t − t0 ) − x0 sin(t − t0 )
y(t) = q
.
1 − a(x20 + y0 )2 (t − t0 )/2
If a > 0 then the solution exists only for
t < t0 +
2
2
= t0 +
2
2
ar0
a(x0 + y02 )
and at least one of x(t) or y(t) blows up as t tends to its upper limit.
It follows that (0, 0) is neither stable nor strictly stable.
If a < 0 then
k(x(t), y(t)) − (0, 0)k < ǫ
for all t ≥ t0 if
k(x0 , y0) − (0, 0)k < δ
where
δ = ǫ.
This follows immediately from the fact that r is decreasing, since
k(x(t), y(t)) − (0, 0)k = r(t)
and
k(x0 , y0 ) − (0, 0)k = r0 = r(t0 ).
That establishes stability. For strict stability we need that
lim (x(t), y(t)) = (0, 0).
t→+∞
This can be proved directly from the solution formula above or by
noting the V (x, y) = x2 + y 2 is a strict Lyapunov function.
Finally, if a = 0 then the system is just
dx/dt = −y + a(x2 + y 2 )x,
dy/dt = x + a(x2 + y 2)y,
and we saw in the previous problem that this is stable, but not strictly
stable.