Applied Math / Dynamical Systems Workshop Lecture Noteswtang/files/Workshop2013.pdf · Applied Math...

Applied Math / Dynamical Systems Workshop

Lecture Notes

W. Tang

School of Mathematical & Statistical SciencesArizona State University

May. 21st-23rd, 2013

Tang (ASU) AMDS Workshop Notes May. 21st-23rd, 2013 1 / 46

Foreword: Mathematical tools are often used to obtain qualitative andquantitative information of the evolution (dynamics) of physical systems.These tools are of utmost importance in studies on applied mathematics,science and engineering. In this workshop, we discuss how differentmathematical tools, typically studied in several college level math courses,are related and used to understand physical systems. We use simple butillustrative examples, built from physically motivated (word) questions, andintroduce necessary concepts that are new but useful in our approach. Wehope that this workshop will serve as a proxy for you to understand thevarious concepts discussed in college level math courses, and help youestablish abstract understanding of mathematics on applied problems.This workshop is no substitute for the actual math courses, but isthe starting point to understand the connections. Please askquestions if you feel puzzled!

Some of the material in this workshop can be found in “NonlinearDynamics And Chaos: With Applications To Physics, Biology, Chemistry,And Engineering”, Steven Strogatz, Westview Press, 2001

May 21st

Lecture 1: Review of important concepts

Continuous functions

Take the infinitesimal limit

Slope: rate of change of the function df /dx = f ′ = lim∆x→0 ∆f /∆x

If f (t), df /dt = f denotes the instantaneous change of f over time

Link to real world: in many problems you can only describef , f , f ′, f ′′, etc

Calculus Example

Consider the Newton’s law of motion

dx(t)/dt = x = v , dv(t)/dt = v = a, F = ma = mx

Finding F? trivial. Given F , find a, then v , then x , can be difficult

Message: We can link quantities of physical interest to mathematicalequations

We’ll discuss how to solve for a subset of these equations later

Calculus Example

Taylor series expansionInstantaneous change of a function — focus on ‘local’ behavior

Taylor series expansion gives local information to certain orderExpanding a function f near a point x0 or (x0, y0)

f (x) ≈ f (x0) + f ′|x=x0(x − x0) +1

2f ′′|x=x0(x − x0)2 + · · ·

f (x , y) ≈ f (x0, y0) + fx |x=x0,y=y0(x − x0) + fy |x=x0,y=y0(y − y0)

2[fxx |x=x0,y=y0(x − x0)2 + fyy |x=x0,y=y0(y − y0)2

+ 2fxy |x=x0,y=y0(x − x0)(y − y0)] + · · ·Local approximation as plane, quadratic surface, etc.Are the following functions differentiable? Where do you see trouble?

Taylor series expansionInstantaneous change of a function — focus on ‘local’ behaviorTaylor series expansion gives local information to certain order

Expanding a function f near a point x0 or (x0, y0)

f (x) ≈ f (x0) + f ′|x=x0(x − x0) +1

2f ′′|x=x0(x − x0)2 + · · ·

Taylor series expansionInstantaneous change of a function — focus on ‘local’ behaviorTaylor series expansion gives local information to certain orderExpanding a function f near a point x0 or (x0, y0)

f (x) ≈ f (x0) + f ′|x=x0(x − x0) +1

2f ′′|x=x0(x − x0)2 + · · ·

+ 2fxy |x=x0,y=y0(x − x0)(y − y0)] + · · ·

Local approximation as plane, quadratic surface, etc.Are the following functions differentiable? Where do you see trouble?

f (x) ≈ f (x0) + f ′|x=x0(x − x0) +1

2f ′′|x=x0(x − x0)2 + · · ·

+ 2fxy |x=x0,y=y0(x − x0)(y − y0)] + · · ·Local approximation as plane, quadratic surface, etc.

Are the following functions differentiable? Where do you see trouble?

f (x) ≈ f (x0) + f ′|x=x0(x − x0) +1

2f ′′|x=x0(x − x0)2 + · · ·

Euler’s formula

Euler’s formula is given as

e ix = cos(x) + i sin(x)

This comes from exponentiating a purely complex number

e ix = 1 + ix +(ix)2

3!+ · · ·+ (ix)n

n!+ · · ·

= 1− x2

4!· · ·+ i

(x − x3

5!· · ·)

= cos(x) + i sin(x)

Hence we can also write

cos(x) =e ix + e−ix

2, sin(x) =

e ix − e−ix

Why important? Often analyze oscillatory systems (season, day-night)

Extension

cosh(x) =ex + ex

2, sinh(x) =

ex − ex

Euler’s formula

e ix = 1 + ix +(ix)2

3!+ · · ·+ (ix)n

n!+ · · ·

= 1− x2

4!· · ·+ i

(x − x3

5!· · ·)

= cos(x) + i sin(x)

2, sin(x) =

e ix − e−ix

Extension

cosh(x) =ex + ex

2, sinh(x) =

ex − ex

Euler’s formula

e ix = 1 + ix +(ix)2

3!+ · · ·+ (ix)n

n!+ · · ·

= 1− x2

4!· · ·+ i

(x − x3

5!· · ·)

= cos(x) + i sin(x)

2, sin(x) =

e ix − e−ix

Extension

cosh(x) =ex + ex

2, sinh(x) =

ex − ex

Euler’s formula

e ix = 1 + ix +(ix)2

3!+ · · ·+ (ix)n

n!+ · · ·

= 1− x2

4!· · ·+ i

(x − x3

5!· · ·)

= cos(x) + i sin(x)

2, sin(x) =

e ix − e−ix

Extension

cosh(x) =ex + ex

2, sinh(x) =

ex − ex

Euler’s formula

e ix = 1 + ix +(ix)2

3!+ · · ·+ (ix)n

n!+ · · ·

= 1− x2

4!· · ·+ i

(x − x3

5!· · ·)

= cos(x) + i sin(x)

2, sin(x) =

e ix − e−ix

Extension

cosh(x) =ex + ex

2, sinh(x) =

ex − ex

Differential equations

We already see one example of differential equation mx = F (x , t)

For unknown function x , usually in the form

F (x , x ′, x ′′, · · · , t) = 0

Example

x ′ = t, x ′ = x + t, x ′′ = x2 + sin(t), · · ·

Can be VERY difficult with nonlinearity and with partial derivatives

We will learn to solve some very simple ones coming from physicalproblems

Focus on what each term means (so you feel how to construct amodel and interpret a solution)

F (x , x ′, x ′′, · · · , t) = 0

Example

x ′ = t, x ′ = x + t, x ′′ = x2 + sin(t), · · ·

F (x , x ′, x ′′, · · · , t) = 0

Example

x ′ = t, x ′ = x + t, x ′′ = x2 + sin(t), · · ·

F (x , x ′, x ′′, · · · , t) = 0

Example

x ′ = t, x ′ = x + t, x ′′ = x2 + sin(t), · · ·

F (x , x ′, x ′′, · · · , t) = 0

Example

x ′ = t, x ′ = x + t, x ′′ = x2 + sin(t), · · ·

F (x , x ′, x ′′, · · · , t) = 0

Example

x ′ = t, x ′ = x + t, x ′′ = x2 + sin(t), · · ·

Matrix Algebra

We can usually write a system of equations into matrix form(a11 a12

a21 a22

(a11 a12

a21 a22

Example: intersection of lines / 2D vector fields

Properties of matrix leads to understandings of the system

We’ll learn some important concept in matrix (linear) algebra

Matrix Algebra

a21 a22

(a11 a12

a21 a22

)Example: intersection of lines / 2D vector fields

Matrix Algebra

a21 a22

(a11 a12

a21 a22

Matrix Algebra

a21 a22

(a11 a12

a21 a22

Numerical methods

Without knowing analytic solutions, we can solve for differentialequations numerically

Solving for x = x , x(0) = 1

Idea: Starting from x(0) = 1, estimate the slope x , then advance ...

Pseudo code: (specify dx , say, 0.1)

x(0) = 1, x(0) = x(0) = 1, x(dx) = x(0) + dx × x(0) = 1.1

x(dx) = 1.1, x(dx) = 1.1, x(2dx) = x(dx) + dx × x(dx) = 1.21

· · ·x [(n − 1)dx ] = xn−1, x(n − 1) = xn−1, x(ndx) = xn−1 + dx × xn−1

· · ·

Some caveat: accuracy, stability

Numerical methods

x(0) = 1, x(0) = x(0) = 1, x(dx) = x(0) + dx × x(0) = 1.1

· · ·

Numerical methods

x(0) = 1, x(0) = x(0) = 1, x(dx) = x(0) + dx × x(0) = 1.1

· · ·

Numerical methods

x(0) = 1, x(0) = x(0) = 1, x(dx) = x(0) + dx × x(0) = 1.1

· · ·

Numerical methods

x(0) = 1, x(0) = x(0) = 1, x(dx) = x(0) + dx × x(0) = 1.1

· · ·

May 21st

Lecture 2: Examples in 1-D

Fishery in Tempe Town Lake

Consider fish population in TTL. Suppose there’s no fishing, foodresource is abundant, no lag in maturation, what can we write downas an equation to describe the change in fish population?

Now consider there is limitation of food resource, so growth isactually slowing down due to increasing population, what equationcan we write down?

The logistic equationf = kf (1− f /C ),

where k is the growth rate at no competition, C the carrying capacityindicating availability of resources

Qualitative understanding: Is there anything you can say about thepopulation without solving for the differential equations?

Quantitative analysis (ODE)

Let’s consider the simpler part df /dt = kf . Physical meaning?

Separation of variables

df /f = kdt∫df /f =

∫kdt + c0

ln|f | = kt + c0

f = ec0ekt

The actual logistic equation?

Separation of variable, partial fraction, integrate∫[1

C − f]df = kt + c0,

f =Cc0e

1 + c0ekt

∫kdt + c0

ln|f | = kt + c0

f = ec0ekt

C − f]df = kt + c0,

f =Cc0e

1 + c0ekt

∫kdt + c0

ln|f | = kt + c0

f = ec0ekt

C − f]df = kt + c0,

f =Cc0e

1 + c0ekt

∫kdt + c0

ln|f | = kt + c0

f = ec0ekt

C − f]df = kt + c0,

f =Cc0e

1 + c0ekt

Solution behavior

Solution of the logistic equation subject to different initial conditions:

Solution behavior

Solution of the logistic equation subject to different initial conditions:

Qualitative analysis (Dynamical System)

How to describe behavior?

Phase portrait

Long term behavior? Fixed points? Stability?

Phase portrait

Consider fishery

How shall we control fishery?

Concept: maintain stable population

Interpret each curve

Consider fishery

Numerics

Let’s try solve the example equations with numerical methods

SynchronizationFirefly sync:

dΘ/dt = Ω + K1 sin(θ −Θ)

dθ/dt = ω + K2 sin(Θ− θ)

Subtract the first equation from the second by writing φ = Θ− θ:

dt= Ω− ω − (K1 + K2) sinφ

If |(Ω− ω)/(K1 + K2)| ≤ 1, φ? = arcsin( Ω−ωK1+K2

)Two equilibrium points, one stable, corresponding to phase lock

Two equilibrium points, one stable, corresponding to phase lock

Exercise for the day

Suppose the population equation for fish in TTL is

f = f 2 − f 4,

how should we regulate fishery?Hint: look for maximum of f = f 2 − f 4 − p, when the functionreaches max at f = 0, stable structure with max profit

Use numerical methods to solve for the above equation, with severalchoices of fishery control pHint: you should see that at maximum fishing rate solution convergefrom above fixed point and diverge from below

The moth population dynamics in a forest obeys the logistic equation.With the introduction of one species of wood pecker, the mothdecreases by the predation function p(f ) = Af /(B + f ). Discuss thedynamics.Hint: look for fixed points and discuss stability based on the touchingof the curves

f = f 2 − f 4,

May 22nd

Lecture 3: Linear 2D dynamics

Population dynamics

Let’s consider population migration(T n+1

(0.3 0.80.7 0.2

)with T 0 = 50000,P0 = 50000. Use the computer to check the longterm behavior of T n+1,Pn+1.

There is a convergence to some T ∗,P∗.

The above equation then becomes(T ∗

(0.3 0.80.7 0.2

)(T ∗

)But how to find them?

Population dynamics

(0.3 0.80.7 0.2

)(T ∗

Population dynamics

(0.3 0.80.7 0.2

)(T ∗

Eigenvalues and eigenvectors

Let vn = [T n,Pn]T , vn+1 = k[T n,Pn]T (because of alignment)

(0.3 0.80.7 0.2

So (A− kI )vn = [0, 0]T , or(0.3− k 0.8

0.7 0.2− k

)But we need [T n,Pn] to be non-trivial

Hence |(A− kI )| = 0 (explain determinant)

For this case k1 = 1, v1 = [8, 7]T , k2 = −0.5, v1 = [−1, 1]T

(0.3 0.80.7 0.2

)So (A− kI )vn = [0, 0]T , or(

0.3− k 0.80.7 0.2− k

(0.3 0.80.7 0.2

)So (A− kI )vn = [0, 0]T , or(

0.3− k 0.80.7 0.2− k

(0.3 0.80.7 0.2

)So (A− kI )vn = [0, 0]T , or(

0.3− k 0.80.7 0.2− k

Romeo-Juliet

A surreal love affair with a twisted story

Verbal statement: Romeo loves Juliet when she loves him, he backsoff when she doesn’t love him; Juliet is protective when Romeo lovesher, but will approach him when he walks away.

Mathematical model:

dt= aJ,

dt= −bR, a, b > 0

Differentiate left eq. w.r.t. t and use the right eq. we get

dt2+ abR = 0

Close analogy with the spring-mass problem

Indeed, many physically motivated problems have analogy, so westudy a class of problems

Romeo-Juliet

Mathematical model:

dt= aJ,

dt= −bR, a, b > 0

dt2+ abR = 0

Romeo-Juliet

Mathematical model:

dt= aJ,

dt= −bR, a, b > 0

dt2+ abR = 0

Romeo-Juliet

Mathematical model:

dt= aJ,

dt= −bR, a, b > 0

dt2+ abR = 0

Romeo-Juliet

Mathematical model:

dt= aJ,

dt= −bR, a, b > 0

dt2+ abR = 0

Romeo-Juliet

Mathematical model:

dt= aJ,

dt= −bR, a, b > 0

dt2+ abR = 0

Analytic solutions

From physics, solutions are sine and cosine functions

Let R = c1 sin(kt) + c2 cos(kt), plugging into the differentialequation, k =

c1, c2 are the initial conditions that prescribes the individual love witheach other

The problem has two degrees of freedom

Physical interpretation: Romeo and Juliet will NEVER be together,unless initially together

What’s worse: unstable — or, marginally stable

Let’s handle this with complex exponentials

Let R = cekt