Circular Pursuit ProblemsMiguel Conner mconner@reed.edu Reed College

Statement of the Problem"A dog at the center of a circular pond makes straight fora duck which is swimming (counterclockwise) along theedge of the pond. If the rate of swimming of the dog isto the rate of swimming of the duck as k : 1, determinethe equation of the curve of pursuit and the discance thedog swims to catch the duck [1]".

ᶰ R

ᶚ dog

duck

Getting Differential EquationsThe duck’s position is ~xduck = (cos ✓, sin ✓). Usingthe equation for a line, y = mx + b, we can deter-mine the slope of the line from the dog to the duckand arrive to the relation:

�y cos(✓ + �) + x sin(✓ + �) = sin�.

With the same method we can find the line perpen-dicular to the dog-duck line and find a second re-lation:

x cos(✓ + �) + y sin(✓ + �) = cos��R.

Using some clever math tricks, we can turn the twoabove equations into two differential equations

d�

d✓

=

cos�

R

� 1 (1)

dR

d✓

= sin�� k (2)

which (you might have guessed) turn out to beimpossible to solve in closed form. We’ll have toresort to computational methods and techniqueslearned in class to figure out what happens [2].

Computer SolutionsWe’ll start by trying to compute the paths of thedog and the duck numerically. Starting with a slowdog (k < 1) at any initial position, we can let thechase begin:

The dog seems to follow the duck for a while, butthen gets stuck in the same circular motion. If weremove some of the initial path, we see that for dif-ferent values of k (k = 0.3, k = 0.5, etc.) we getdifferent stable limit cycles of radius r = k!

0.3 0.5 0.75 0.95-0.3-0.5-0.75-0.95

0.3

0.5

0.75

0.95

-0.3

-0.5

-0.75

-0.95

What about for a fast Fido (k > 1)? If the dog startsfrom inside the pond it will always sneak up on theduck from behind (� = ⇡/2). Starting the dog offoutside the pond adds the (unlikely) possibility offrontal capture (� = �⇡/2).

For k = 1, the dog will approach the duck asymp-totically, thus never truly catching the duck [3].

Limit Cycles and Fixed PointsWe can see why a limit cycle arises if we analyzethe 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

R =

p1� k

2.

Given that the radius of the circle is 1, we can solvefor the radius of the dog’s limit cycle using righttriangles:

1rR

ϕ

to show that r = k, as we saw before. We can alsoverify stability by looking at the phase portrait ofthe system for k < 1:

- π2

π2 π-π 0

1

1.5

0.5

1- k2

0

- π2

π2 π-π 0

1

1.5

0.5

1- k2

0

ϕ

R

k < 1

where we see that the stable fixed point (our limitcycle) corresponds to the points R =

p1� k

2 andR = cos�. For k > 1:

- π2

π2 π-π 0

1

0.5

0

- π2

π2 π-π 0

1

0.5

0

ϕ

Rk > 1

we get a fixed point at R = 0, � = ⇡/2, the sta-ble case that corresponds to capture from behind,and R = 0, � = �⇡/2, the saddle point describingfrontal capture.

Herding CatsOur dog is now running circles around Clops the cat.Clops doesn’t like dogs so she runs the opposite direc-tion. Does the dog manage to herd Clops?We can answer this using the same methods as be-fore. A phase diagram for k < 1 (fast dog) showsthat:

- π2

π2 π-π 0

1

1.5

0.5

1- k2

0

- π2

π2 π-π 0

1

1.5

0.5

1- k2

0

ϕ

R

k < 1

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.

- π2

π2 π-π 0

1

0.5

0

- π2

π2 π-π 0

1

0.5

0

ϕ

R

k > 1

Despite similar outcomes, the two values of k

translate to different types of escape paths. Theseare illustrated in the two examples below.

k < 1 k > 1

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.