Spring 2017 MATH 63CM
Solutions to Homework Assignment 6
1. We assume without loss of generality that an 6= 0.
Let us define n variables by u1 = y, u2 = y 0 , ..., un = y (n−1) . It is easy to
check that these satisfy the first-order linear system u0 (t) = Au(t) where


0
1
0
...
0
 0
0
1
...
0 



 ..
..
.
.

.
.
0
0 
A= .


 ..
..
..
 .
.
.
1 
− aan0 − aan1 − aan2 . . . − an−1
an
We see then that
χA (λ) = det(λI − A) =
1
(an λn + an−1 λn−1 + ... + a1 λ + a0 )
an
The Jordan normal form of A will consist of Jordan blocks with eigenvalues
the distinct roots λ1 , ..., λm of the polynomial an λn + an−1 λn−1 + ... + a1 λ +
a0 . Therefore the first coordinate of u(t), which is y(t), will generally be
expressible in the form
m
X
y(t) =
pj (t)eλj t
j=1
where pj (t) is a polynomial of degree less than the multiplicity νj of the root
λj in an λn + an−1 λn−1 + ... + a1 λ + a0 .
2. We give pictures of the phase curves and the vector fields in each case.
For parts (b) and (c), the phase curves lie in two-dimensional planes, so we
give pictures of the phase curves in these planes instead.
(a)
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(b)
(c)
3. We know that A, B are similar to the rotation matrices
cos β − sin β
cos α − sin α
, Rβ =
Rα =
sin α cos α
sin β cos β
respectively. Then the two systems are topologically equivalent if and only
if u0 (t) = Rα u(t) and u0 (t) = Rβ u(t) are.
The flow of the former is φt (u) = eRα t u = Rαt u and of the latter ψt (u) =
Rβt u. Suppose that the two systems are topologically equivalent via the
homeomorphism f : R2 → R2 . Then we have
f (φt (u)) = ψt (f (u)) ⇔ f (Rαt u) = Rβt f (u)
Then for t = 2π
α , we obtain f (u) = R2πβ/α f (u) for all u and thus R2πβ/α =
β
α
I ⇔ α ∈ Z. By the same argument, setting t = 2π
β , we get β ∈ Z. We
conclude that we must have α = ±β. Both are possible, since we can easily
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check that taking f = Id when α = β or f to be the reflection along the
x-axis when α = −β we indeed get topological equivalences.
4. Suppose that f gives a topological equivalence between the two systems.
Let φt be the flow of u0 (t) = Au(t) and ψt the flow of u0 (t) = Bu(t). Then
f (φt (u)) = ψt (f (u)) ⇔ f (eAt u) = eBt f (u). In particular, for u = 0 we
get f (0) = eBt f (0) ⇒ f (0) = 0 (the eigenvalues of eBt are eλt , where λ
is an eigenvalue of B, and therefore cannot be equal to 1). Then if u is
in the stable subspace V− ∼
= R of the first system we have eAt u → 0 as
Bt
t → +∞ and thus e f (u) → f (0) = 0 as t → +∞, i.e. f (u) lies in the
stable subspace W− ∼
= R2 of the second system. The same argument works
for the inverse of f , which in particular yields a continuous, injective map
g
W− ∼
= R2 → R ∼
= V− .
We show that such a map cannot exist. Consider the x-axis L inside R2 .
Then g|L is a continuous, injective map from R to R. In particular, it has
to be monotonic and its image must be an open interval around 0. The
same argument for the y-axis L0 shows that g|L0 has image an open interval
around 0. This implies that g(L) and g(L0 ) intersect in more than one point,
violating the injectivity of g. We thus have a contradiction.
5. Let us take A = 0 and G(x1 , ..., xn ) = (x21 , ..., x2n ). These clearly satisfy
the two requirements. We see that any solution to u0 (t) = Au(t) + G(u(t))
ai
satisfies ui (t) = 1−a
where ai is the i-th coordinate of the initial condition
it
u0 . For any > 0, taking u0 = 2 e1 which satisfies ||u0 || < , we get
u(t) = 2 1−t e1 ⇒ ||u(t)|| = 2 1−t → +∞ as t → 2 .
6. (a) This is algebraic manipulation using the formulas x(t) = r(t) cos θ(t)
and y(t) = r(t) sin θ(t). For example, dropping the t from the notation for
brevity, we have r2 = x2 + y 2 and hence
r0 r = x0 x + y 0 y = x2 − r(x2 + xy) + x2 y + y 2 + r(xy − y 2 ) − x2 y = r2 (1 − r)
Dividing by r, we get r0 (t) = r(t)(1 − r(t)). This covers the case r(t) = 0
for some t too, since the origin is an equilibrium point for the system. We
can derive the other formula analogously.
(b) In polar coordinates we need to show equivalently that r(t) → 1 and
θ(t) → 2nπ as t → +∞ for some integer n.
We can solve the first equation for r(t) by using separation of variables to
r(t)
r(0)
obtain |r(t)−1|
= |r(0)−1|
et , which implies that r(t) → 1 as t → +∞ assuming
r(0) 6= 0.
If R(t) is an anti-derivative for r(t), then we can use separation of variables
3
2 θ
again together with the double-angle
formula 1 − cos θ = 2 sin ( 2 ) to solve
the second equation and get cot θ(t)
= −R(t) + C for some constant C.
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It is clear that since r(t) → 1, R(t) → +∞ as t → +∞, which implies that
θ(t) → 2nπ as desired.
(c) Note that if the initial condition is (r0 , θ0 ), where for example we can
take r0 < 1, θ0 > 0 both close to 1 and 0 respectively, then since both
r(t), θ(t) are increasing by the equations of our system, we will have
θ0 (t) = r(t)(1 − cos θ(t)) ≥ r0 (1 − cos θ0 ) > 0
as long as θ(t) < π for example.
This clearly implies that for some time t0 we will have θ(t0 ) = π2 and hence
the solution will move “far” from (1, 0). Therefore (1, 0) is not a stable
equilibrium point.
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