Lecture Note
Math 216 Lecture 27
-
University:
University of Michigan -
Course:
MATH 216 | Introduction to Differential Equations Academic year:
2019
-
Views:
416
Pages:
19
Author:
Calvin S
Math 216: Lecture 27
Math 216: Lecture 27 Cycles
We consider systems of autonomous equations
x 0 (t) = f x(t), y (t)
y 0 (t) = g x(t), y (t)
(1)
Definition
A solution x(t), y (t) is called periodic with period T > 0 if
x(t + T ) = x(t)
and y (t + T ) = y (t)
for all
t
(in particular, a periodic solution is defined for all t). Note that an
equilibrium solution
x(t) = x0
and y (t) = t0
is periodic with period T for any T > 0. A periodic solution that
is not an equilibrium, is called a cycle.
Math 216: Lecture 27 Cycles
Example
Consider the system
x 0 (t) = −y (t)
y 0 (t) = x(t).
It has one equilibrium solution x(t) = 0, y (t) = 0 and many cycles
x(t) = C cos t,
y (t) = C sin t
for
that are periodic with period T = 2π.
Math 216: Lecture 27
C 6= 0 Cycles
Exercise: Consider a system
x 0 (t) = ax(t) + by (t)
y 0 (t) = cx(t) + dy (t).
Show that cycles exist if and only if
a+d =0
and ad − bc > 0.
Here is a more interesting example.
Example (Lotka - Volterra equations)
Consider the system
x 0 (t) = x(t) 1 − y (t)
y 0 (t) = y (t) x(t) − 1
This is a particular case of Lotka - Volterra system of equations,
after A.J. Lotka (1880 – 1949) and V. Volterra (1860 – 1940),
modeling predator-prey interactions.
Math 216: Lecture 27 Cycles
Example (Continued)
Exercise: If x(t) and y (t) represent populations, which one is
predator and which one is prey?
As in Lecture 14, we deduce the equation for the trajectories (we
are interested in the case of x, y > 0).
dy
y (x − 1)
1−y
x −1
=
=⇒
dy =
dx
dx
x(1 − y )
y
x
Z
Z
1−y
x −1
=⇒
dy =
dx =⇒ ln y − y + C1 = x − ln x + C2
y
x
=⇒ ln x + ln y − x − y = C
for some constant C .
Math 216: Lecture 27 Cycles
Example (Finished)
Trajectories turn out to be cycles, here is a picture.
Exercise: Find the equilibrium solution with x, y > 0 and show
that we must have C ≤ −2. Is the equilibrium solution
asymptotically stable, stable, or unstable?
Exercise: Show that the curve ln x + ln y − x − y = C , where
x, y > 0 is bounded.
Math 216: Lecture 27 Trapped trajectories
Definition
Let R be a connected bounded region in the (x, y )-plane. We say
that R traps trajectories of (1), if any trajectory of (1) originating
in R, remains in R.
R
Math 216: Lecture 27 Trapped trajectories
Poincaré - Bendixson Theorem: If a region R traps trajectories of
(1) and
∂f
∂g
∂g
∂f
,
,
and
∂x
∂y
∂x
∂y
exist and continuous in R, then there is a periodic trajectory in R.
After H. Poincaré (1854 – 1912) and I. Bendixson (1861 – 1935).
In particular, if R traps trajectories and there are no equilibrium
solutions in R, then R must contain a cycle.
A useful tool: the dot product. For vectors
a
b
~a = 1
and ~b = 1 ,
a2
b2
we define their dot product by
~a · ~b = a1 b1 + a2 b2 .
If ~a · ~b > 0 then the angle between ~a and ~b is acute and if ~a · ~b < 0
then the angle between ~a and ~b is obtuse.
Math 216: Lecture 27 Trapped trajectories
Example
Consider the system
x 0 (t) = −y (t) + x(t) 1 − x 2 (t) − y 2 (t)
y 0 (t) = x(t) + y (t) 1 − x 2 (t) − y 2 (t) .
First, let us find equilibrium solutions. We must have
−y + x 1 − x 2 − y 2 = 0
=⇒ x 2 + y 2 1 − x 2 − y 2 = 0,
2
2
x +y 1−x −y
= 0
from which we conclude that the only equilibrium is x(t) = 0 and
y (t) = 0.
Math 216: Lecture 27 Trapped trajectories
Example (Continued)
Next, we claim that the ring defined by the inequalities
1
≤ x2 + y2 ≤ 2
2
traps trajectories.
y
(x,y)
(x,y)
0
x
Math 216: Lecture 27 Trapped trajectories
Example (Continued)
x
For a point (x, y ) on the outer boundary, the normal vector is
,
y
the direction of the field is
0
x
−y − x
=
y0
x −y
and
x
−y − x
·
= x(−y − x) + y (x − y ) = −x 2 − y 2 < 0,
y
x −y
so if a trajectory ever reaches the outer boundary, it will have to
“bounce back”.
Math 216: Lecture 27 Trapped trajectories
Example (Continued)
Similarly, at apoint (x, y ) on the inner boundary, the normal
−x
vector is
, the direction of the field is
−y
−y + 12 x
x + 12 y
and
1
1
−x
−y + 12 x
·
= − x 2 − y 2 < 0,
1
−y
x + 2y
2
2
so if a trajectory ever reaches the inner boundary, it will also have
to “bounce back”.
Math 216: Lecture 27 Trapped trajectories
Example (Continued)
Here is the picture of the direction field.
Math 216: Lecture 27 Trapped trajectories
Example (Finished)
It follows now that the ring
1
≤ x2 + y2 ≤ 2
2
contains a cycle. Indeed, it is easy to see that
x(t) = cos t,
y (t) = sin t
is a solution of the system.
Exercise: Consider a slightly perturbed system
x 0 (t) = −y (t) + x(t) 1 − 0.99x 2 (t) − 1.01y 2 (t)
y 0 (t) = x(t) + y (t) 1 − 1.01x 2 (t) − 0.99y 2 (t)
Math 216: Lecture 27 Trapped trajectories
Show that the same region R
1
≤ x2 + y2 ≤ 2
2
traps trajectories of the new system. Show that there are no
equilibrium solutions in R and conclude that R must contain a
cycle.
Exercise: Consider the system of equations
x 0 (t) = −x(t) + y (t) + x 2 (t)y (t)
y 0 (t) =
1 − y (t) − x 2 (t)y (t).
Show that the region pictured below traps trajectories of the
system.
y
1
0
(1,1)
1
slope=−1
x
y=
1+x 2
x
Math 216: Lecture 27 Trapped trajectories
The boundary of the region is constructed as follows: we start at
(0, 0), walk vertically to (0, 1), walk horizontally to (1, 1), walk
with slope −1 to the intersection with the curve y = x/(1 + x 2 ),
walk vertically till the intersection with the x-axis, and walk
horizontally back to the origin.
Exercise: Let us go back to the example of Lotka - Volterra
equations. Argue that every region of the type
ln x + ln y − x − y > C ,
where
x, y > 0
where C < −2 is a constant, traps trajectories.
Math 216: Lecture 27 Cycles and equilibrium solutions
Suppose that in system (1) the functions f (x, y ) and g (x, y ) are
continuous and suppose that the system has a cycle. Then inside
the cycle we must have an equilibrium solution (we don’t prove
that).
A connected region R is called simply connected if it contains no
holes.
R
simply
connected
R
not
simply connected
Math 216: Lecture 27 When there are no cycles
Another useful result: suppose that in a simply connected region
R, we have
∂f
∂g
∂g
∂f
+
> 0 everywhere or
+
< 0 everywhere.
∂x
∂y
∂x
∂y
Then R contains no cycles.
Indeed, if there was a cycle C of period T , then
I
f (x, y )dy − g (x, y ) dx
C
Z T
=
f x(t), y (t) y 0 (t) − g x(t), y (t) x 0 (t) dt
0
Z T
=
f x(t), y (t) g x(t), y (t) − g x(t), y (t) f x(t), y (t) dt = 0.
0
On the other hand, by the Green formula
I
ZZ
∂f
∂g
f (x, y )dy − g (x, y ) dx =
+
dxdy 6= 0.
∂x
∂y
C
R
Math 216: Lecture 27 When there are no cycles
Exercise: Show that the system
x 0 (t) = −5y (t) + x 3 (t)
y 0 (t) = 3x(t) + y 3 (t)
contains no cycles.
Math 216: Lecture 27
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