MA 342N
Assignment 2
Due 16 March 2016
Id:
342N-s2016-2.m4,v 1.3 2016/05/13 14:01:35 john Exp john
1. For the following system
u′ = v + uw,
v ′ = u + vw,
w ′ = w + uv
(a) find the linearisation at the equilibrium (0, 0, 0),
Solution:
u′ = v
v′ = u w′ = w
or, in matrix form,





u
0 1 0
u′
 

 ′
 v  = 1 0 0 v .
w
0 0 1
w′
(b) find the eigenvectors and eigenvalues of the coefficient matrix,
Solution:
Eigenvalues λ1 = 1, λ2 = −1, and λ3 = 1 with eigenvectors
 
1
 
v1 =  1  ,
0


1


, v2 =  −1  ,
0
 
0
v3 =
 
0.
1
(c) find the dimensions of the stable manifold, unstable manifold and
centre manifold,
Solution:
The dimension of the stable manifold is equal to the number of
eigenvalues with negative real part, counted with multiplicity, so 1
1
Id:
342N-s2016-2.m4,v 1.3 2016/05/13 14:01:35 john Exp john 2
in this case. The dimension of the unstable manifold is equal to the
number of eigenvalues with positive real part, so 2. The dimension
of any centre manifold is equal to the number of purely imaginary
eigenvalues, which is 0 here.
(d) find the lowest order for which there is a resonance,
Solution: There are six resonances of order three, namely
λ1 = 2λ1 + λ2 = λ1 + λ2 + λ3 = λ2 + 2λ3,
and
λ3 = 2λ1 + λ2 = λ1 + λ2 + λ3 = λ2 + 2λ3,
but none of lower order.
(e) find a linear change of variable which diagonalises the coefficient
matrix,
Solution: The change of variable
p = u + v,
q = u − v,
r=w
converts the system into
p′ = p + pr,
q ′ = −q + qr,
1
1
r ′ = r + p2 − q 2 ,
4
4
which has a coefficient matrix


1 0 0


 0 −1 0  .
0 0 1
(f) remove the quadratic part of the non-linearity via a quadratic
change of variable.
Solution: It is easier to do this in the p, q, r coordinates than in
the u, v, w coordinates. The change of variable we need is
p̃ = p − pr,
q̃ = q − qr,
1
1
r̃ = r − p2 − q 2 .
4
12
As explained in lecture, the coefficient of yj yk in ỹi is just the
coefficient of yj yk in yi′ divided by λi − λj − λk . This change of
variable works because
1
1
p̃′ = p′ − p′ r − pr ′ = p + pr − pr − pr 2 − pr + p3 + pq 2 = p̃ + · · ·
4
4
Id:
342N-s2016-2.m4,v 1.3 2016/05/13 14:01:35 john Exp john 3
1
1
q̃ ′ = q ′ −q ′ r −qr ′ = −q + qr + qr −qr 2 −qr − p2 q + q 3 = −q̃ + · · ·
4
4
1
1
1
1
1
1
1
1
r̃ ′ = r ′ − pp′ − qq ′ = r+ p2 − q 2 − p2 − p2 r+ q 2 − q 2 r = r̃+· · ·
2
6
4
4
2
2
6
6
where the omitted terms are all of order at least three. If you
want a change of variables in the original coordinates then
ũ =
p̃ + q̃
p − q + pr − qr
=
= v + vw,
2
2
p − q + pr − qr
p̃ − q̃
=
= u + uw,
2
2
1
1
1
1
1
w̃ = r̃ = r − p2 − q 2 = w − u2 − uv − v 2
4
12
3
3
3
ṽ =
works.
2. (a) Find the equilibria of the equation
y ′ = y 3 − zy 2 + 2y 2 − zy + z 2 − 2z.
Hint: The cubic factors into a quadratic times a linear factor.
Solution: The equation is
y ′ = (y 2 − z)(y − z + 2).
We have an equilibrium where either factor is zero, so there is√
an
equilibrium√at η = z −2 for all values of z and equilibria at η = z
and η = − z for non-negative z.
(b) What are the dimensions of the stable, unstable and centre manifolds at these equilibria?
Solution: These are the numbers of eigenvalues with negative real
part, positive real part and zero real part, respectively. There’s
only one eigenvalue here, since the coefficient matrix
∂F
(η) = ( 3η 2 − 2zη + 4η − z )
∂y
is 1×1, and this eigenvalue is just 3η 2 −2zη +4η −z. For η = z −2
this is positive for z < 1, negative for 1 < z < 4 and positive again
√
for z > 4. It is, of course, zero at z = 1 and z = 4. For η = z
it is positive for 0 < z < 4 and
√ negative for z > 4. It is zero at
z = 0 and z = 4. For η = − z it is negative for 0 < z < 1 and
positive for z > 1. It is zero at z = 0 and z = 1.
Id:
342N-s2016-2.m4,v 1.3 2016/05/13 14:01:35 john Exp john 4
So we have a one-dimensional centre manifold when (z, η) is one of
(0, 0), (1, −1) or (4, 2). We have a one-dimensional stable manifold
on the part of the line η = z − 2 between (1, −1) and (4, 2) and
on the parts of the parabola η 2 = z between (4, 2) and (0, 0) and
between (1, −1) and infinity. We have a one dimensional unstable
manifold on the parts of the line η = z − 2 between infinity and
(1, −1) and between (4, 2) and infinity and also on the parts of
the parabola between infinity and (4, 2) and between (0, 0) and
(1, −1).
(c) Draw a diagram of the stable and unstable equilibria in the z-η
plane.
Solution:
In this diagram the stable equilibria are shown in green and the
unstable in red.
(d) There are three bifurcation points, z = 0, z = 1 and z = 4. Are
these transcritical, saddle-node, Hopf or pitchfork bifurcations?
Solution: From the diagram it’s clear that z = 0 is a saddle-node
bifurcation and the other two are transcritical.