Circular
Pursuit
Problems
Miguel Conner
[email protected]
Reed College
Statement of the Problem
Computer Solutions
Limit Cycles and Fixed Points
Herding Cats
"A dog at the center of a circular pond makes straight for
a duck which is swimming (counterclockwise) along the
edge of the pond. If the rate of swimming of the dog is
to the rate of swimming of the duck as k : 1, determine
the equation of the curve of pursuit and the discance the
dog swims to catch the duck [1]".
We’ll start by trying to compute the paths of the
dog and the duck numerically. Starting with a slow
dog (k < 1) at any initial position, we can let the
chase begin:
We can see why a limit cycle arises if we analyze
the fixed points of the system. Setting d /d✓ = 0
and dR/d✓ = 0 gives us that sin = k and cos =
R, which we can combine to say
p
R = 1 k2 .
Our dog is now running circles around Clops the cat.
Clops doesn’t like dogs so she runs the opposite direction. Does the dog manage to herd Clops?
We can answer this using the same methods as before. A phase diagram for k < 1 (fast dog) shows
that:
k<1
-π
Given that the radius of the circle is 1, we can solve
for the radius of the dog’s limit cycle using right
triangles:
duck
R
- π2
0
π
2
π
1.5
1.5
1
1
1 - k2
R
1 - k2
R
The dog seems to follow the duck for a while, but
then gets stuck in the same circular motion. If we
remove some of the initial path, we see that for different values of k (k = 0.3, k = 0.5, etc.) we get
different stable limit cycles of radius r = k!
0.75
Getting Differential Equations
The duck’s position is ~xduck = (cos ✓, sin ✓). Using
the equation for a line, y = mx + b, we can determine the slope of the line from the dog to the duck
and arrive to the relation:
0.3
y cos(✓ + ) + x sin(✓ + ) = sin .
dR
= sin
d✓
k
π
2
0.5
0.75
0.95
1.5
1 - k2
(1)
which (you might have guessed) turn out to be
impossible to solve in closed form. We’ll have to
resort to computational methods and techniques
learned in class to figure out what happens [2].
0
π
2
π
1
0.5
0.5
0
-π
- π2
0
π
2
π
- π2
0
π
2
k>1
π
ϕ
where we see that the stable fixed point
p (our limit
cycle) corresponds to the points R = 1 k 2 and
R = cos . For k > 1:
-π
- π2
k>1
0
π
2
π
1
1
0.5
0.5
0
-π
- π2
0
π
2
π
ϕ
For k = 1, the dog will approach the duck asymptotically, thus never truly catching the duck [3].
k>1
0
1
k<1
0
0
(2)
-π
- π2
Despite similar outcomes, the two values of k
translate to different types of escape paths. These
are illustrated in the two examples below.
0.5
-π
What about for a fast Fido (k > 1)? If the dog starts
from inside the pond it will always sneak up on the
duck from behind ( = ⇡/2). Starting the dog off
outside the pond adds the (unlikely) possibility of
frontal capture ( = ⇡/2).
π
ϕ
-0.5
-0.95
π
2
0
1
0.5
-0.75
0
the circular limit cycle fixed point is now unstable,
so Clops will always escape even if it takes a while.
At k > 1 (slow dog), Clops escapes easily.
R
0.3
-0.3
- π2
π
1 - k2
R
R.
Using some clever math tricks, we can turn the two
above equations into two differential equations
1
- π2
k<1
0
1
-0.5
0
ϕ
-0.3
With the same method we can find the line perpendicular to the dog-duck line and find a second relation:
d
cos
=
d✓
R
0
1.5
-0.95 -0.75
0.5
-π
-π
0.5
1
to show that r = k, as we saw before. We can also
verify stability by looking at the phase portrait of
the system for k < 1:
0.95
x cos(✓ + ) + y sin(✓ + ) = cos
ϕ
r
R
dog
0.5
we get a fixed point at R = 0, = ⇡/2, the stable case that corresponds to capture from behind,
and R = 0, = ⇡/2, the saddle point describing
frontal capture.
References
[1] Nahin, P.J. Chases and Escapes: The Mathematics of Pursuit and
Evasion. Princeton University Press, 2007.
[2] Strogatz, S.H. Nonlinear Dynamics and Chaos. Westview
Press, 2015.
[3] Davis, H. T. Introduction to Nonlinear Differential and Integral
Equations. Dover, 1962.