6570 notes - chapters 1-3

Mathematical Modeling - Math 6570

Dr. John Travis

Mississippi College

Discuss the nuclear arms race model created in Chapter 1. Students should pick a problem at the end of each section for homework.

CHAPTER TWO

Basics and Setting Up Framework

Modeling Procedure: pg 31

Through observation, identify primary factors and make simplifications
Conjecture tentative relationships among the factors
Apply mathematical analysis to model
Validate the model by checking to see if it meets the real-world.

Modeling Approaches: pg 32

Using "real-world" trials (Simulation or Experiments)
Identifying factors involved (Graphical or Symbolic)

Scientific Method: page 38

General observations
Formulate a hypothesis
Determine method to test hypothesis
Gather data
Test hypothesis using the data
Confirm or deny hypothesis

Modeling Process: page 35

Identify the problem
Make assumptions
Solve/Interpret the model
Verify the model....does it address the problem, make sense and correspond to reality?
Implement the model
Maintain the model

Remark: A major difference between the modeling process and the scientific method is the goal. The modeling process is designed to gather evidence to corroborate a model, not to confirm or deny it since we already know that it is most probably a simplification of the real problem. We are only trying to test its reasonableness.

A model is said to be robust when its conclusions do not depend on the precise satisfaction of the assumptions; otherwise, it is a fragile model.
The sensitivity of the model is a measure of the model's robustness or fragility.

Sources of Errors:

Formulation - using incorrect assumptions or leaving out important facts
Truncation - taking an exact process and approximating it
Round-off - computer arithmetic is very seldom exact
Convergence - taking a convergent process and stopping it after a finite number of steps
Measurement - imprecise data

Measuring Errors:

Absolute Error
Relative Error

Class Discussion Problems: page 40-41

HOMEWORK: page 41 # 1-8

Review the "illustrative examples" on pages 41-47

HOMEWORK: Pick one of the projects on page 48

CHAPTER THREE

Modeling with Discrete Dynamical Systems

Remark: Things that are dynamical undergo change. In modeling change, we utilize the following paradigm:

future value = present value + change

or, by solving

future value - present value = change

Notice, we have change is a difference in value.

Measurement time frames:

Discrete - behavior takes place over finite time intervals. Leads to Difference Equations
Continuous - - behavior takes place continually. Leads to Differential Equations

Example: Modeling the Arms Race

Suppose we have two countries, X and Y, which are engaged in a nuclear arms race.

Strategies:

Friendly Strategy: To survive a msssive first strike and inflict unacceptable damage on the enemy. Implies targeting of population and industrial centers. Assumed when determining its own missile force.
Enemy Strategy: To conduct a massive first strike to destroy the friendly missile force. Implies targeting missile sites. Presumed by each country for its enemy.

We will suppose each country adopts these strategies.

Create variables:

x = number of missiles possessed by country X
y = number of missiles possessed by country Y
We will consider y = f(x) ( or x = g(y) ) to be the minimum number of missiles needed for country Y to accomplish its strategies when country X has x missiles (or Y has y missiles). We desire to see how a change in one country's number affects the other country's number.

Creating y = f(x):

If x=0, then y₀ missiles are needed by the Friendly Strategy.
For x>0, y > y₀.
We will assume that each missile from X can destroy no more than one missile from Y. Hence, y < y₀+x
Result: y₀ < y < y₀ + x
Let 0 < s < 1 be the percentage of Y's missiles which survive the Enemy Strategy.
Relative sizes of missile force:

Case 1: x<y. Then, y-x of Y's missiles are not even targeted and hence will not be destroyed. Also, s x of Y's missiles will survive the attack. After all this, Y will apply the Friendly Strategy and needs y₀ missiles to remain. Therefore, y₀ = number of missiles remaining = (y-x) + s x, or

y = y₀ + (1-s) x.

Case 2: y=x. Since s x of Y's missiles survive the attack and Y needs y₀ afterwards, then y₀ = x s = y s.

y = y₀/s.

Case 3: y < x < 2y. So, x-y of Y's missiles will be targeted twice and s²(x-y) will survive both rounds. The rest, y-(x-y) = 2y-x, will be target once and s(2y-x) of these will survive. Therefore, to apply the Friendly Strategy, Country Y will need for y₀ missiles to survive both attacks and so

y₀ = s²(x-y) + s(2y-x), or y = (y₀ + x(s-s²)) / (2s-s²)

Case 4: x=2y. In the same manner as in case 3, we get

y₀ = s²y, or y = y₀ / s².

Continue in this manner and get a decreasing slope, piecewise-linear graph similar to Figure 1.7, pg 9. A continuous model for this graph (generalizing from cases 2 and 4) would be the function y = y₀ / s ^x/y. This curve is strictly increasing (from earlier) and concave down. Indeed, using logarithms,

ln(y) = ln(y₀) - x ln(s) / y, or

y ln(y) = y ln(y₀) - x ln(s), or by differenting w.r.t. x,

y ( 1/y + ln(y) ) = y ' ln(y₀) - ln(s), or

y = ln(s) / ( 1/y + ln(y) - ln(y₀) ), or

y = ln(s) (y/y) / ( 1/y + ln(y) - ln(y₀) )², or

y = ln(s) / (-1 + ln(y₀/y) ),

which is always negative since ln(s)<0, and ln(y₀/y) < 0.

Note, the case for x=g(y) is symmetrical since X is employing the same strategy. See Figure 1.8, pg 10.

Therefore, since the curves are increasing and strictly concave down or up, respectively, then there is only one point of intersection, that is a place where both countries are satisfied with their strategies.

Dynamic Modeling: Consider the development of an arms race. Use the following situation:

n = number of periods that have passed
x_n = number of weapons possessed by X at period n
y_n = number of weapons possessed by Y at period n
s_x = success rate of X's missiles
s_y = success rate of Y's missiles

Result: A Discrete Dynamical System for the Arms Race:

y_n+1 = y₀ + s_x x_n
x_n+1 = x₀ + s_y y_n

with initial values x₀ and y₀ and parameters s_x and s_y given.

Differences: Given a sequence of numbers A = { a₁, a₂, a₃, ... }

the nth first forward difference is given by D⁺ a_n = a_n+1 - a_n
the nth first backward difference is given by D^- a_n = a_n - a_n-1
the nth first centered difference is given by D^o a_n = a_n+1 - a_n-1

Unless we otherwise specify, we will utilize forward differences and write Da_n = D⁺ a_n.

Example: Interest

Let

P_n = present value at time n,
r = interest rate per period.
Interest earned per period = present value * rate = P_n r.

Using Future value = Present Value + Interest Earned, yields the difference equation

P_n+1 = P_n + r P_n = P_n (1+r), or

P₀ = Initial deposit
DP_n = r P_n

Example: A Home Mortgage

Notice, in this case the amount is changed each month not only by interest but also by payments
Let

B_n = balance owed at time n,
R = period payment amount,
r = interest rate per period.

Using Future balance = Present balance + Interest owed - Periodic payment yields

B_n+1 = B_n + r B_n - R = B_n (1+r) - R, or

B₀ = Initial amount borrowed
DB_n = r B_n - R

HOMEWORK: page 58 #2, 4, 6

Remark: Using change = future value - present value, an effective way to model change is using

Da_n = f(a₀, a₁, ... , a_n) for some sequence { a_n }.

Example: Population growth
Let

P_n = population at time n,
b = birth rate per period,
d = death rate per period,
r = growth rate per period = b-d.

Notice, k>0 implies increasing population, k<0 implies decreasing population. Hence, we get

P_n+1 = P_n + r P_n = P_n (1+r), or

P₀ = Initial population
DP_n = r P_n

This is exactly the same difference equation as that for Interest.

Example: Population growth with limited resources
Let

M = carrying capacity = maximum sustainable population

Then, the above model could be modified so that the growth rate is dependent upong population size relative to the carrying capacity to yield:

P_n+1 = P_n + {r (M - P_n)/M} P_n , or

P₀ = Initial population
DP_n ={r (M - P_n)/M} P_n

Multiplying this one out yields a difference equation which is nonlinear.

Numerical Solution: Develope a reasonable initial value and iterate the difference equation (actually using the equation that preceeded the equation) for as many values as desired. This can be compared to experimentally created test values to check for the model's validity. (See yeast example on page 63, Figure 3.8)

Example: Population growth with another competing species

Let

C_n = competing species population at time n
r_p = growth rate for species P
r_c = growth rate for species C

With unlimited resources, the species won't compete and so will undergo change as before yielding

P₀ and C₀ = Initial populations
DP_n = r_p P_n
DC_n = r_c C_n

With limited resources, the species will compete and will tend to slow down the growth rate depending upon the size of the other population. One such model:

P_n+1 = P_n + {r_p (M - C_n)/M } P_n, and similarly

C_n+1 = C_n + {r_c (M - P_n)/M } C_n, or

P₀ and C₀ = Initial populations
DP_n ={r_p (M - C_n)/M} P_n
DC_n ={r_c (M - P_n)/M} C_n

Example: Population growth with a varying prey species (Predator-prey)

As above, let C_n represent the prey species and P_n the predator species and assume the predator only eats prey. Further assume the following:

In the absence of predators, the prey grow unconstrained assuming r_c > 0
In the absence of prey, the predators have zero birth date and so r_p < 0
The presence of a large number of prey should increase the growth rate of predators
The presence of a large number of predators should decrease the growth rate of prey

The product P_nC_n is large only if both P_n and C_n are large. So, a possible model reflecting these assumptions is:

P_n+1 = P_n + r_p P_n + k₁ P_nC_n, and similarly
C_n+1 = C_n + r_c C_n - k₂ P_nC_n, or

P₀ and C₀ = Initial populations
DP_n = r_p P_n + k₁ P_nC_n
DC_n = r_c C_n - k₂ P_nC_n

HOMEWORK: page 67 #1, 4, 6 (discuss the solution for various initial whale populations), 8, 11

Numerical Solutions for Dynamical Systems

Start with an initial value and iterate a sufficient number of subsequent values to determine the patterns involved. Then, redo the problem to see the effect of small changes in the data and any necessary parameters. Generally, we can "solve" difference equations by simply iterating the equation for a sufficiently large number of terms. Long-term behavior can often be predicted by noting any trends in these solutions.

Defn: An equilibrium value is a value for which a dynamical system remains constant. This equilibrium is stable (or attracting) if there is a number e such that

|a₀ - a| < e => a_n -> a This equilibrium is unstable (or repelling) if there is a number e such that 0<|a₀ - a| < e => |a_n - a| > e for some but not necessarily all values of n.

Linear Dynamical Systems:

General System 1: Consider the sytem given by

a₀ = given
a_n+1 = r a_n

Consider the parameter r: What if:

r=0? Then a_n=0 for all n.
r>0? Then a_n is the same sign for all n
r<0? Then a_n oscillates in sign as n increases.
r=1? Then a_n=constant for all n...ie, an equilibrium value.
0<r<1? Then a_n->0 as n increases
r>1? Then |a_n|->oo as n increases

Are there any other equilibrium values? If so, then c = a_n = a_n+1 = r a_n = r c implies c=0 is an equilibrium and if r=1, then c = anything is also an equilibruim value.

General System 2: Consider the system given by

a₀ = given
a_n+1 = r a_n + b

Theorem 3.2: For General System 2 with nonzero b, if r=1, no equilibrium exists and the solution of the difference equation is a 2-cycle. Otherwise, the system has an equilibrium of a=b / (1-r). This equilibrium is stable provided |r| < 1 and unstable provided |r|>1.

Pf: Certainly, if r=1 and b is nonzero, there is no equilibrium.
Indeed, if so, there would exist a value a = a_n for all n which would imply a = a + b or 0 = b.
If r is not equal to 1, an equilibrium exists <=> a = ra+b, for some a
<=> a=b / (1-r), as desired.
Notice |a₁ - a| = |ra₀ + b - b / (1-r)| = |r| |a₀ - a|.
Similarly, |a₂ - a| = |ra₁ + b - b / (1-r)| = ... = |r|² |a₀ - a|.
By induction, we get
|a_n - a| = |r|ⁿ |a₀ - a|. Therefore, if |r|<1, this converges to zero and we have a stable system.
If |r|>1, this diverges and we have a unstable system.

Nonlinear Systems: These arise when the difference equation uses powers of a_n. These are in general very difficult to solve and numerical methods for solving them must be utilized.

HOMEWORK: page 80 #2, 4, 8