M2AA1 Diffferential Equations: Problem Sheet 9
1. Consider x0 = (3 − y − x)x, y 0 = (x − 1 − y)y. This is a predator-prey model in which
one has species x has one constant source of food, and species y has limited growth.
(a) Draw the regions where (x0 > 0, y 0 > 0), (x0 < 0, y 0 > 0), (x0 > 0, y 0 < 0) and
(x0 < 0, y 0 < 0).
Answer: in this case the lines 3 − y − x = 0 and x − 1 − y = 0 intersect in (2, 1).
The arrows are as suggested in the figure below.
Notice that solutions which start on the x-axis remains forever on the x axis,
and similarly a solution which starts not the y axis remains forever on the y
axis.
(b) What are the singular points for this differential equation and determine the
linearisation at these points.
If (3−y −x)x = 0 then either x = 0 or 3−y −x = 0. If x = 0 and (x−1−y)y = 0
then y = −1 or y = 0. The point (0, −1) is outside the region we are interested
in, so this gives only (0, 0). If (x−1−y)y = 0 then either y = 0 or x−1−y = 0. If
y = 0 then (3−y−x)x = 0 gives either (0, 0) or (3, 0). If x−1−y = 0, x−1−y = 0
then (x, y) = (2, 1).
The linearisation at (0, 0) is x0 = 3x, y 0 = −y, which is a saddle (with eigenvalues
3, −1). In (x, y) the linear part of the r.h.s. is
3 − y − 2x
−x
.
y
x − 1 − 2y
At (3, 0) this gives
−3 −3
0
2
.
which has eigenvalues −3, 2 and so corresponds to a saddle. At (2, 1) this gives
−2 −2
.
1 −1
To compute the eigenvalues, we compute
−2 − λ
−2
det
= λ2 + 3λ + 4 = 0.
1
−1 − λ
√
This gives eigenvalues (−3 ± −7)/2, so a sink with complex eigenvalues (which
means spiralling).
1
(c) Show that there exists R so that any solution which starts in {(x, y); 0 < x <
R1 , 0 < y < R2 } remains in this rectangle forever. Conclude that orbits do
exists for all t ≥ 0.
Answer: You should take R1 , R2 so that the top right corner of the rectangle
goes through the line M and so that R1 > 3. Then along the top side of R the
solutions point downwards, and on the right hand side of R the solutions point
to the left.
Since for any initial condition x, y ≥ 0 one can find such a rectangle R, we
obtain that no solution ever can tend to infinity. It follows that solutions exist
for all time.
Can we say more? Let’s try whether we can use the function V (x, y) =
11.2 Predator/Prey Systems
245
(x − 2 log x) + A(y − log y). Provided we take A > 0, this function has a unique
and predators to increase at the same time. If the prey population is above
minimum
ata (2,
1)the
solack
essentially
measures
the distance to (2, 1). Then
its limiting value, it must decrease.
After
while
of prey causes
the
predator population to begin to decrease (when the solution crosses M ). After
that point the prey population can never increase past a/λ, and so the predator
population continues to decrease. If the solution crosses L, the prey population
increases again (but not past a/λ), while the predators continue to die off. In
the limit the predators disappear and the prey population stabilizes at a/λ.
Suppose now that L and M cross at a point Z = (x0 , y0 ) in the quadrant Q;
of course, Z is an equilibrium. The linearization of the vector
field at Z is
2
= ẋ(1 − 2/x) + Aẏ(1 − 1/y)
x−2
y−1
= ((2 − x) + (1 − y))x
+ A((x − 2) + (1 − y))y
x
y
2 + A(x − 2)(y − 1)
= −(x
−
2)
+
(1
−
y)(x
−
2)
−
A(y
−
1)
"
V̇
X! =
!
−λx0
dy0
−bx0
−µy0
X.
If we take A = 1 then this gives
The characteristic polynomial has trace given by −λx0 − µy0 < 0 and determinant (bd + λµ)x0 y0 > 0. From the trace-determinant plane of Chapter 4,
2
we see that Z has eigenvalues that are either both negative or both complex
with negative real parts. Hence Z is asymptotically stable.
Note that, in addition to the equilibria at Z and (0, 0), there is still an
equilibrium at (a/λ, 0). Linearization shows that this equilibrium is a saddle;
its stable curve lies on the x -axis. See Figure 11.7.
It is not easy to determine the basin of Z , nor do we know whether there
are any limit cycles. Nevertheless we can obtain some information. The line
L meets the x -axis at (a/λ, 0) and the y -axis at (0, a/b). Let " be a rectangle
whose corners are (0, 0), (p, 0), (0, q), and (p, q) with p > a/λ, q > a/b, and
the point (p, q) lying in M . Every solution at a boundary point of " either
−(x − 2) − 1(y − 1)2 ≤ 0.
and this is only = 0 when x = 2 and y = 1. Hence (2, 1) is asymptotically stable
and in fact that every initial condition for which x, y > 0 is attracted to this
singularity.
q
Γ
M
L
p
Figure 11.7 The nullclines and
phase portrait for a predator/prey
system with limited growth when
the nullclines do meet in Q.
2
2. Consider the system
x0 = sin x(−0.1 cos x − cos y)
y = sin y(cos x − 0.1 cos y)
on [0, π] × [0, π]. Determine the singular points of this differential equation, linearise
at each of these singular points and determine the local behaviour at each of these
singular points. Show that all solutions emanating from the source at (π/2, π/2) have
ω-limit sets equal to the square bounded by x = 0, π and y = 0, π. (Hint; can you
think of a Lyapounov function?) Draw the phase portrait of the solutions.
Answer: Assume that (x, y) is a singular point.
(a) If sin x = 0 then sin y = 0, i.e. at (kπ, lπ) because t if sin x = 0 then (cos x −
0.1 cos y) 6= 0 for all y.
(b) Similarly, sin y = 0 implies sin x = 0 because when sin y = 0 then (−0.1 cos x −
cos y) 6= 0 for all x.
(c) So the only other singular point is when (cos x − 0.1 cos y) = 0 = (−0.1 cos x −
cos y). This corresponds to cos x = cos y = 0, i.e. (x, y) = (π/2, π/2) + (kπ, lπ).
If (x, y) is chosen so that x = kπ then the x-coordinate of φt (x, y) is kπ for all t ≥ 0.
This is true because x0 = 0 along the line x = kπ. One says that the line x = kπ is
invariant. Similarly, the line y = lπ is invariant, because along this line y 0 = 0.
Using the Taylor expansion of sin and cos we obtain that linearisation at (0, 0) is
x0 = x(−0.1 − 1), y 0 = y(1 − 0.1). So (0, 0) is a saddle point. In the same way we get
(kπ, lπ) is a saddle-point for each k, l.
3 ), the linearisation at (x, y) = (π/2, π/2)
Since
= −(x − π/2)
+ O(|x − π/2|
cos(x)
x − π/2
0.1 1
x0
. So the eigenvalues there are 0.1 ± i and so
=
is
y − π/2
−1 0.1
y0
(π/2, π/2) is a spiralling source.
To determine the global dynamics, take U = sin x sin y. Then U > 0 on the square
(0, π) × (0, π) and U = 0 on its boundary. Note that
U̇
= cos x sin y · ẋ + sin x cos y · ẏ
= cos x sin y[sin x(−0, 1 cos x − cos y)] + sin x cos y[sin y(cos x − 0.1 cos y)]
= −0.1 sin x cos2 x sin y − 0.1 sin x sin y cos2 y.
Hence U̇ < 0 and therefore
solutions217
go to the zero set of U .
10.1 Limit Sets
(o, o)
(0, 0)
Figure 10.2 The ω-limit set of any
solution emanating from the source at
(π /2, π /2) is the square bounded by the
four equilibria and the heteroclinic
solutions.
solutions connecting these equilibria in the order listed. See Figure 10.2. There
is also a spiral source at (π /2, π /2). All solutions emanating from this source
accumulate on the four heteroclinic solutions connecting the equilibria (see
Exercise 4 at the end of this chapter). Hence the ω-limit set of any point on
!
these solutions is the square bounded by x = 0, π and y = 0, π .
3
3. Consider the system
x0 = −y − x4 /4 − x2 /2 + y 2 /2 (x3 − x)
y 0 = (x3 − x) − x4 /4 − x2 /2 + y 2 /2 y.
(i) Find the singularities of this system. (ii) Determine the local behaviour near these
singularities; (iii) Show that all non-stationary solutions have ω-limit sets consisting
of one of the homoclinic solutions plus a saddle.
Answer: Write r = x4 /4 − x2 /2 + y 2 /2 . Then one has a singularity if −y − r(x3 −
x) = 0 and (x3 − x) − ry = 0. Substituting the 2nd equation in the first one gives
(x3 − x)[1 + r2 ] = 0. Hence y = 0 and x3 − x = 0. So singular points are (0, 0) and
(±1, 0).
Linearising at (0, 0) gives
x0 = −y
y 0 = −x
1
1
. This is a sadand
which has eigenvalues ±1, with eigenvectors
1
−1
dle and its stable and unstable manifold are tangent to these eigenvectors at (0, 0).
(Sometimes the stable and unstable manifold of a saddle point are called separatrices.
0 x−1
0.5 −1
x
.
=
The linear part of the r.h.s. at (1, 0) is equal to
y
2 1/4
y0
So the eigenvalues are roots of λ2 − (3/4)λ + 1/8 + 2 = 0, i.e. λ1,2 = ((3/4) ±
p
9/16 − 17/2)/2. This means that (1, 0) is a spiralling source.
Considering the nulclines, one gets the impression that phase portrait is as follows
10.6 Applications
of Poincaré-Bendixson
227 way around in this figure):
(note
that the arrow
are drawn the wrong
Figure 10.9 A pair of
homoclinic solutions in the
ω-limit set.
How to prove this and get information about the global dynamics? Perhaps we
can try again something inspired by the idea of a Lyapounov function? Let’s try
(x, there
y) =arex4three
/4 −equilibria:
x2 /2 + aty 2(0,
/2.
setand
X := {(x, y); V (x, y) = 0} corresponds to a
A computation showsVthat
0), The
(−1, 0),
(1, 0). The origin is a saddle, while the other two equilibria are sources. The
phase portrait of this system is shown in Figure 10.9. Note that solutions far
from the origin tend to accumulate on the origin and a pair of homoclinic
solutions, each of which leaves and then returns to the origin. Solutions emanating from either source have ω-limit set that consists of just one homoclinic
4
!
solution and (0, 0). See Exercise 6 for proofs of these facts.
10.6 Applications of Poincaré-Bendixson
The Poincaré-Bendixson theorem essentially determines all of the possible
limiting behaviors of a planar flow. We give a number of corollaries of this
important theorem in this section.
A limit cycle is a closed orbit γ such that γ ⊂ ω(X ) or γ ⊂ α(X ) for some
X # ∈ γ . In the first case γ is called an ω-limit cycle; in the second case, an
α-limit cycle. We deal only with ω-limit sets in this section; the case of α-limit
√
figure ∞ meeting the x-axis at (0, 0) and (± 2, 0) and V > 0 is outside the figure
and < 0 in one of the two regions bounded by this figure. Then
V̇
= x3 ẋ − xẋ + y ẏ = x3 (−y − V · (x3 − x)) − x(−y − V · (x3 − x)) + y((x3 − x) − V · y)
= −V · ((x − x3 )2 + y 2 )
So V V̇ < 0 unless (x, y) is a singularity or belongs to the zero set of V . So as in the
proof of the Lyapounov theorem one obtains that φt (x, y) → X := {(x, y); V (x, y) =
0} whenever (x, y) 6= (0, 0), (±1, 0).
Note that if (x, y) ∈ X then φt (x, y) ∈ X for t ∈ R (because V̇ = 0). Since the only
singularity in X is (0, 0) the set X consists of an orbit. Since this orbit converges to
(0, 0) it turns out that W s (0, 0) = X and also W u (0, 0) = X.
4. Let ẋ = xP (x, y) and ẏ = yQ(x, y) and assume P, Q < 0. Show that (0, 0) is
attracting.
Answer: Take V (x, y) = x2 + y 2 . Then V̇ = 2xẋ + 2y ẏ = 2x2 P (x, y) + 2y 2 Q(x, y) < 0
except when (x, y) = (0, 0).
5. Consider the equation
x0 = x2 − y 2
y 0 = x2 + y 2
(a) Show that there is an invariant straight line containing (0, 0)
(b) Show that there are no periodic orbits.
(c) Sketch the phase portrait.
Answer: (a) If x2 − y 2 = 0 then x = ±y and this implies x2 + y 2 > 0 unless x = y = 0.
So the only singularity is (0, 0).
If (x, kx) then x0 = (1 − k 2 )x2 and y 0 = (1 + k 2 )x2 . If k = (1 + k 2 )/(1 − k 2 ) then
(x0 , y 0 ) points in the direction of the line (x, kx) and so the line L = {(x, kx); x ∈ R
is invariant. Looking at the graph of g(k) = (1 + k 2 )/(1 − k 2 ) this has precisely one
fixed point g(k) = k and this is the required invariant straight line containing (0, 0).
Clearly this invariant straight line consists of an orbit of the system.
Answer (b) If there was a periodic orbit, then this periodic orbit would have to
surround a critical point, see the last theorem in the notes. This is impossible since
it would then intersect the invariant straight line.
A simpler answer to (b): y 0 > 0 (unless (x, y) = (0, 0) and so y is strictly increasing
which is impossible along a periodic orbit.
5
6. Consider y 00 + (2y 2 + 3[y 0 ]2 − 1)y 0 + y = 0, which, when writing z = y 0 is equivalent
to the system
z 0 = −y − (2y 2 + 3z 2 − 1)z
y0 = z
(a) Show that if (y, z) starts in the annulus 1/3 < y 2 + z 2 < 1/2 then it stays in
this annulus.
(b) Show that the system has a periodic solution.
Answer: write V (y, z) = y 2 + z 2 . Then V̇ = 2y ẏ + 2z ż = 2yz + 2z(−y − (2y 2 + 3z 2 −
1)z) = −2z 2 (2y 2 +3z 2 −1). So V̇ > 0 if y 2 +z 2 < 1/3 and V̇ < 0 if y 2 +z 2 > 1/2. So it
follows that if (y, z) is in the annulus, then it stays in this annulus. (That V̇ > 0 along
the inner boundary of the annulus means that orbit starting on the inner boundary
flow towards A, and similarly V̇ < 0 along the outer boundary means that hat orbit
starting on the outer boundary flow towards A.)
Since this annulus does not contain a singular point, the Poincaré-Bendixson theorem
implies that the orbit must spiral to a periodic orbit.
7. Consider
x0 = x + 2y
y 0 = −2x − y − (1/2)x3
(a) Determine the singularities of this differential equation
(b) Show that the solutions of this are on level sets of V (x, y) = x2 +xy+y 2 +(1/8)x4 .
(c) Sketch the level sets of V .
Answer: y = −(1/2)x and 2x − (1/2)x + (1/2)x3 = 0. So(1/2)(3x + x3 ) = 0
1
2
. So the
which implies (x, y) = (0, 0). Linearizing in this point gives
−2 −1
determinant is 3 and the trace is 0. So λ1 + λ2 = 0 and λ1 λ2 = 3. This implies
that the eigenvalues are pure imaginary. So (0, 0) is not hyperbolic, and we can’t say
anything about the local behaviour from the linear part. Let’s try the function V :
= 2xẋ + ẋy + xẏ + 2y ẏ + (1/2)x3 ẋ
= 2x(x + 2y) + (x + 2y)y + x(−2x − y − (1/2)x3 )+
+2y(−2x − y − (1/2)x3 ) + (1/2)x3 (x + 2y)
= 0.
V̇
So solutions stay on level sets of V .
4
3
2
1
0
−1
−2
6
−3
−4
−4
−3
−2
−1
0
1
2
3
4
8. Consider
x0 = µx − y − (x + (3/2)y)(x2 + y 2 )
y 0 = x + µy + ((3/2)x − y)(x2 + y 2 ).
(a) What are singular points of this system and compute the eigenvalues of these
points.
To determine the singularities, multiply the first equation by x and 2nd one by
y and then add them. This gives µ(x2 + y 2 ) − (x2 + y 2 )2 = 0. This means (0, 0)
or x2 + y 2 = µ. Now substitute this into the equations gives
0 = µx − y − (x + (3/2)y)µ = (−1 − (3/2)µ)y
0 = x + µy + ((3/2)x − y)µ = (1 + (3/2)µ)x.
So x = y = 0. At this point, the linearisation of the system is
x0 = µx − y
y 0 = x + µy.
(As
at (0, 0), just ignore higher order points.) The eigenvalues of
we linearise
µ −1
are λ = µ + ±i. This is hyperbolic when µ 6= 0 and non-hyperbolic
1 µ
when µ = 0.
(b) Show that this system in polar coordinates becomes
r0 = µr − r3
θ0 = 1 + (3/2)r2 .
Answer:
ṙ cos θ − r sin θθ̇ = ẋ = µr cos θ − r sin θ − (r cos θ + (3/2)r sin θ)r2
ṙ sin θ + r cos θθ̇ = ẏ = r cos θ + µr sin θ + ((3/2)r cos θ − r sin θ)r2
Hence, we get
ṙ = µr − r3
θ̇0 = 1 + (3/2)r2 .
√
(c) Show that a periodic orbit emerges when µ > 0 with radius µ. One calls this
a Hopf bifurcation.
ṙ = 0 gives r = 0 or r2 = µ. The latter set corresponds to a circle and since
ṙ = 0 along this circle, orbits stay on this circle. Since θ̇ = 1 + (3/2)µ along this
circle, this circle does not contain singular points if µ is close to zero, and the
circle is a periodic orbit of the differential equation.
7