Dynamical Systems: Lecture 2 - University of...

Naima Hammoud

Feb 22, 2017

Dynamical Systems: Lecture 2

Recall: the logistic equation

N = rN

✓1� N

K

◆N

N

Strogatz,NonlinearDynamicsandChaos(1994)

Recall: the logistic equation

N = rN

✓1� N

K

◆

fixed points

N = 0

rN

✓1� N

K

◆= 0

rN = 0

=) N = 0

✓1� N

K

◆= 0

=) N = K

N

N


Tumor Growth

N

N

˙N = �aN log(bN)


Tumor Growth

˙N = �aN log(bN)

N

N


Tumor Growth

˙N = �aN log(bN)

fixed points

N = 0

�aN log(bN) = 0

�aN = 0

! N = 0

log(bN) = 0

! bN = 1

! N = 1/b

N

N


Tumor Growth

˙N = �aN log(bN)

fixed points

N = 0

�aN log(bN) = 0

�aN = 0

! N = 0

log(bN) = 0

! bN = 1

! N = 1/b

N

N1/b


Synchronization

Synchronization of Fireflies

• Consider a system with one firefly and one flashlight

• Start flashing the light at a constant frequency (not too far from the firefly’s natural frequency), e.g. if a firefly fires every 0.8 seconds, then maybe the light can flash every second.

• The firefly will modify its frequency by speeding up or slowing down so as to match the flashlight

• If the stimulus is too slow or too fast, entrainment won’t occurStrogatz,NonlinearDynamicsandChaos(1994)


Model

✓(t)

✓ = 0

phase of the firefly’s flashing rhythm

the instant when a flash is emitted by the firefly

If there are no outside stimuli, the firefly flashes at a constant frequency w

✓ = !



• Suppose there’s a periodic stimulus (a flashlight), whose phase f satisfies

• If stimulus is ahead in the cycle, firefly speeds up. If stimulus is behind, firefly slows down

⇥ = ⌦

phase of the flashlight’s flashing rhythm

the instant when a flash is emitted by the stimulus

⇥(t)

⇥ = 0



✓ = ! +A sin(⇥� ✓)

sin(⇥� ✓) > 0 : if ⇥ ahead, then 0 < ⇥� ✓ < ⇡,

then firefly speeds up

sin(⇥� ✓) < 0 : if ⇥ behind, then ⇡ < ⇥� ✓ < 2⇡,

then firefly slows down



✓ = ! +A sin(⇥� ✓)

Let � = ⇥� ✓

=) � = ⇥� ✓

� = ⌦� ! �A sin�

✓ = !

⇥ = ⌦



✓ = ! +A sin(⇥� ✓)

Let � = ⇥� ✓

=) � = ⇥� ✓

� = ⌦� ! �A sin�

✓ = !

⇥ = ⌦

Change of variables(you’re not expected to know how to make

such variable changes)

redefine time to be At

define µ = ⌦�!A



✓ = ! +A sin(⇥� ✓)

Let � = ⇥� ✓

=) � = ⇥� ✓

� = ⌦� ! �A sin�

✓ = !

⇥ = ⌦

Change of variables(you’re not expected to know how to make

such variable changes) redefine time to be At

define µ = ⌦�!A

� = µ� sin�


µ =⌦� !

A

If µ is small, then frequencies are close to one another and we expect entrainment

If µ is large, we don’t expect entrainment

� = µ� sin�



µ =⌦� !

A



�

�

+

�

µ = 0

� = µ� sin�


µ =⌦� !

A



�

�

+

�

� = µ� sin�

µ is small


� = µ� sin� µ =⌦� !

A



�

�

+ +

µ is large


Fixed points when

� = µ� sin�

� = 0

µ� sin� = 0

µ = sin�


We plot sinf and µseparately and find their intersections, which will be the fixed points

Fixed points when

� = µ� sin�

� = 0

µ� sin� = 0

µ = sin�


�

sin�


Fixed points when

� = µ� sin�

� = 0

µ� sin� = 0

µ = sin�


�

sin�

µ We plot sinf and µseparately and find their intersections, which will be the fixed points

Fixed points when

� = µ� sin�

� = 0

µ� sin� = 0

µ = sin�


�

sin�

µ+

�+

� > 0 when µ > sin�

� < 0 when µ < sin�


Fixed points when

� = µ� sin�

� = 0

µ� sin� = 0

µ = sin�


�

sin�

+

�

µ�




Fixed points when

� = µ� sin�

� = 0

µ� sin� = 0

µ = sin�


�

sin�

�

µ

�




Fixed points when

� = µ� sin�

� = 0

µ� sin� = 0

µ = sin�


�

sin�µ



+ + We plot sinf and µseparately and find their intersections, which will be the fixed points

Fixed points when

� = µ� sin�

� = 0

µ� sin� = 0

µ = sin�

Two-dimensional Systems

Example: spring-mass system

Transform into a system of coupled equation

Fixed points:

x+ !

2x = 0

(x = y

y = �!

2x

x = y = 0 ! x = y = 0

Two-dimensional Systems

Example: spring-mass system

Transform into a system of coupled equation

Fixed points:

x+ !

2x = 0

(x = y

y = �!

2x

x = y = 0 ! x = y = 0

vector field: (x, y) = (y,�!

2x)

take specific points:

y = 0 ! (0,�!

2x)

x = 0 ! (y, 0)

x

y

Love affairs

Romeo is in love with Juliet, but Juliet is a fickle lover: the more Romeo loves her, the more she wants to run away and hide. But when Romeo gets discouraged and backs off, Juliet begins to find him strangely attractive! Romeo on the other hand tends to echo her: he warms up when she loves him, and grows cold when she hates him.

Let R(t) be Romeo’s love/hate for Julietand J(t) be Juliet’s love/hate for Romeo


Love affairs(R = aJ

J = �bRa, b > 0

fixed points: R = J = 0 ! R = J = 0

vector field: (R, J = 0) = (aJ,�bR)


Love Affairs: general case(R = aR+ bJ

J = cR+ dJ

a > 0, b > 0

a < 0, b > 0

Romeo gets excited by Juliet’s love for him, and is further spurred by his own affection

Romeo gets excited by Juliet’s love for him, but he is turned off by his own affection

eager

cautious


Love affairs: Cautious Lovers

(R = aR+ bJ

J = bR+ aJ

a < 0, b > 0

measure ofcautiousness

measure ofresponsiveness




(R = aR+ bJ

J = bR+ aJ

a < 0, b > 0



a = �2, b = 1

R

J




(R = aR+ bJ

J = bR+ aJ

a < 0, b > 0




a = �1, b = 2

R

J


Do opposites attract?(R = aR+ bJ

J = �bR� aJ

fixed points:

R = J = 0 ! R = J = 0



J = �bR� aJ

fixed points:

a = 2, b = 1

R

JR = J = 0 ! R = J = 0



J = �bR� aJ

fixed points:

a = 1, b = 2

R

JR = J = 0 ! R = J = 0



J = �bR� aJ

fixed points:

R = J = 0 ! R = J = 0

a = 2, b = �1

R

J


Competition: Rabbits versus Sheep

Suppose both species are competing for the same food resources (say grass), and that the amount is limited. Ignore predators, seasonal changes, and other sources of food.

1. Each species would grow to its carrying capacity in the absence of the other species. So we can use the logistic model for each, and maybe assign rabbits a higher growth rate?

2. When rabbits and sheep encounter each other, trouble starts. Conflict occurs at a rate proportional to the size of each population. We assume conflicts reduce the growth rate for each species, with rabbits taking the bigger hit.


Rabbits versus Sheep

Recall the logistic equation:

Let R(t) be the population of rabbitsand S(t) be the population of sheep

N = rN

✓1� N

K

◆

(R = R(3�R� 2S)

S = S(2� S �R)





N = rN

✓1� N

K

◆

(R = R(3�R� 2S)

S = S(2� S �R)

fixed points: R = S = 0

R(3�R� 2S) = 0

S(2� S �R) = 0

four fixed points:

(0, 0), (0, 2), (3, 0), (1, 1)Strogatz,NonlinearDynamicsandChaos(1994)




N = rN

✓1� N

K

◆

(R = R(3�R� 2S)

S = S(2� S �R)

four fixed points:

(0, 0), (0, 2), (3, 0), (1, 1)

R

S



(R = R(3�R� S)

S = S(2� S �R)

(R = R(3� 2R� S)

S = S(2� S �R)

(R = R(3� 2R� 2S)

S = S(2� S �R)



(R = R(3�R� S)

S = S(2� S �R)

(R = R(3� 2R� S)

S = S(2� S �R)

(R = R(3� 2R� 2S)

S = S(2� S �R)

R

S

R

S

R

S


Dynamical Systems: Lecture 2 - University of...

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