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Mathematical Modeling

2301675

Dr. Kitiporn Plaimas

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Chapter 6

Differential Equations

Outlines

• Setting up a model with a differential equation

• Solving a differential equation

• Slope fields

• Models and its applications

• Equilibrium solution and stability

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Setting up a model with a

differential equation• Sometimes we do not know a function, but we have

information about its rate of change, or its derivative.

• Then we may be able to write a new type of equation, called a

differential equation, from which we can get information about

the original function.

• For example, we may use what we know about the derivative

of a population function (its rate of change) to predict the

population in the future.

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MARINE HARVESTING

• The effect of fishing on a fish population.

• Suppose that, left alone, a fish population increases at a

continuous rate of 20% per year. Suppose that fish are also being

harvested (caught) by fishermen at a constant rate of 10 million

fish per year.

• How does the fish population change over time?

• Notice that we have been given information about the rate of

change, or derivative, of the fish population. Combined with

information about the initial population, we can use this to

predict the population in the future.

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• We know that

• Suppose the fish population, in millions, is P and its derivative

is dP/dt, where t is time in years. If left alone, the fish

population increases at a continuous rate of 20% per year,

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• Since the rate of change of the fish population is dP/dt

• This is a differential equation that models how the fish

population changes. The unknown quantity in the equation is

the function giving P in terms of t.

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MATHEMATICAL MODELING USING DIFFERENTIAL EQUATIONS - Applied Calculus 5th Edition

NET WORTH OF A COMPANY

A company earns revenue (income) and also makes payroll

payments.

• Assume that revenue is earned continuously, that payroll payments

are made continuously, and that the only factors affecting net

worth are revenue and payroll.

• The company's revenue is earned at a continuous annual rate of

5% times its net worth. At the same time, the company's payroll

obligations are paid out at a constant rate of 200 million dollars a

year.

Write a differential equation to model the net worth of the

company, W, in millions of dollars, as a function of time, t, in years.

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POLLUTION IN A LAKE

• If clean water flows into a polluted lake and a stream takes

water out, the level of pollution in the lake will decrease

(assuming no new pollutants are added).

• Example 1: The quantity of pollutant in the lake decreases at a

rate proportional to the quantity present.

• Write a differential equation to model the quantity of pollutant in

the lake.

• Is the constant of proportionality positive or negative?

• Use the differential equation to explain why the graph of the

quantity of pollutant against time is decreasing and concave up

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Quantity of pollutant in a lake

Since no new pollutants are being added to the lake, the quantity Q is

decreasing over time, so dQ/dt is negative. Thus, the constant of

proportionality k is negative.

Notice that the rate at which pollutants leave a lake is proportional to the

quantity of pollutants in the lake. This model works for any contaminants

flowing in or out of a fluid system with complete mixing.

THE QUANTITY OF A DRUG IN THE BODY

• Example 2: A patient having major surgery is given the

antibiotic vancomycin intravenously at a rate of 85 mg per

hour.

• The rate at which the drug is excreted from the body is

proportional to the quantity present, with proportionality

constant 0.1 if time is in hours.

(a) Write a differential equation for the quantity, Q in mg, of

vancomycin in the body after t hours.

(b) When Q = 100 mg, is Q increasing or decreasing?

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So

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2301675 Mathematical Modeling


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Solutions of Differential Equations

• A solution to a differential equation is any function that

satisfies the differential equation.

• Check whether or not a function is a solution to a differential

equation.

• Techniques to solve a differential equation.

• Solve the Differential Equation Numerically

• Formula or function solution

• Visualization of a solution. 23

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Solve the Differential Equation Numerically

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MARINE HARVESTING

Suppose at time t = 0, the fish population is 60 million.

Find approximate values for P(t) for t = 1, 2, 3,4, 5.

the change in P is approximately dP/dt · Δt when Δt is small.

Formula or function solution

• Separation of variables

• Exact equations

• Integrating factor

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Visualization of a solution

• How to visualize a differential equation and its solutions. 2

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Visualizing the slope of y

Solutions

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Models and its applications

• Exponential growths and decays

• Growths with the differential equation

• Logistics growths

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EXPONENTIAL GROWTH AND DECAY

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Applications

• Population growth

• Continuously compounded interest

• Pollution in the Great Lakes

• Quantity of a drug in the body

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Population growth

• Consider the population P of a region where there is no

immigration or emigration. The rate at which the population is

growing is often proportional to the size of the population.

This means larger populations grow faster

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Continuously compounded interest

• Imagine interest being accrued at a rate proportional to the

balance at that moment. Thus, the larger the balance, the

faster interest is earned and the faster the balance grows.

• Example: A bank account earns interest continuously at a rate

of 5% of the current balance per year. Assume that the initial

deposit is $1000 and that no other deposits or withdrawals

are made.

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Pollution in the Great Lakes

• In the 1960s pollution in the Great Lakes became an issue of

public concern. We will set up a model for how long it would

take the lakes to flush themselves clean, assuming no further

pollutants were being dumped in the lake.

• Let Q be the total quantity of pollutant in a lake of volume V at

time t.

• Suppose that clean water is flowing into the lake at a constant rate

r and that water flows out at the same rate.

• Assume that the pollutant is evenly spread throughout the lake,

and that the clean water coming into the lake immediately mixes

with the rest of the water.

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Setting Up a Differential Equation for the Pollution

where the negative sign represents the fact that Q is decreasing.

• At time t, the concentration of pollutants is Q/V and water

containing this concentration is leaving at rate r. Thus,

• So, the differential equation is

and its general solution is

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Pollutant in lake versus time

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Lakes

Volume and outflow in Great Lakes

How long will it take for 90% of the pollution to

be removed from Lake Erie?

For Lake Erie, r/V = 175/460 = 0.38

When 90% of the pollution has been removed,

10% remains

Solving for t gives

For 99% to be removed?

Quantity of a drug in the body

• The rate at which a drug leaves a patient's body is proportional to the quantity of the drug left in the body.

Let Q represent the quantity of drug left, then

• The negative sign indicates the quantity of drug in the body is decreasing.

• The solution to this differential equation is � � �� ; the

quantity decreases exponentially. The constant k depends on the drug and Q is the amount of drug in the body at time zero. Sometimes physicians convey information about the relative decay rate with a half life, which is the time it takes for Q to decrease by a factor of 1/2.

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Example

Valproic acid is a drug used to control epilepsy; its halflife in the

human body is about 15 hours.

(a) Use the half-life to find the constant k in the differential

equation dQ/dt = −kQ, where Q represents the quantity of

drug in the body t hours after the drug is administered.

(b) At what time will 10% of the original dose remain?

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(a) (b)

Growths with the differential equation

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• A patient is given the drug warfarin, an anticoagulant,

intravenously at the rate of 0.5 mg/hour. Warfarin is

metabolized and leaves the body at the rate of about 2% per

hour.

• A differential equation for the quantity, Q (in mg), of warfarin

in the body after t hours is given by

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in the body for different initial values of Q?

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• If Q is small, then 0.02Q is also small and the rate the drug is

excreted is less than the rate at which the drug is entering the body.

Since the rate in is greater than the rate out, the rate of change is

positive and the quantity of drug in the body is increasing.

• If Q is large enough that 0.02Q is greater than 0.5, then 0.5 − 0.02Q

is negative, so dQ/dt is negative and the quantity is decreasing.

• For small Q, the quantity will increase until the rate in equals the

rate out. For large Q, the quantity will decrease until the rate in

equals the rate out.

What is the value of Q at which the rate in

exactly matches the rate out?

What is the value of Q at which the rate in exactly

matches the rate out?

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• If the amount of warfarin in the body is initially 25 mg, then

the amount being excreted exactly matches the amount being

added. The quantity of drug Q will stay constant at 25 mg.

Notice also that when Q = 25, the derivative dQ/dt is zero.

• If the initial quantity is 25, then the solution is the horizontal

line Q = 25. This solution is called an equilibrium solution.

The slope field for this differential equation

• Solution curves drawn for Q = 20, Q = 25, and Q = 30.

• In each case, the quantity of drug in the body is approaching the

equilibrium solution of 25 mg.

• The solution curve with Q = 30 should remind you of an exponential

decay function. It is, in fact, an exponential decay function that has

been shifted up 25 units.

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Company’s revenue

A company's revenue is earned at a continuous annual rate of 5% of its

net worth. At the same time, the company's payroll obligations are

paid out at a constant rate of 200 million dollars a year.

• The differential equation governing the net worth, W (in millions of

dollars), of this company in year t is given by

• The general solution is

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• To find C we use the initial condition that W = W0 when t = 0.

• Then, the solution is

• If W0 = 4000, then W = 4000, the equilibrium solution.

• If W0 = 5000, then W = 4000 + 1000e0.05t.

• If W0 = 3000, then W = 4000 − 1000e0.05t.

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Equilibrium solutions and stability

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EQUILIBRIUM SOLUTIONS

• An equilibrium solution is constant for all values of the independent variable. The graph is a horizontal line. Equilibrium solutions can be identified by setting the derivative of the function to zero.

• An equilibrium solution is stable if a small change in the initial conditions gives a solution which tends toward the equilibrium as the independent variable tends to positive infinity.

• An equilibrium solution is unstable if a small change in the initial conditions gives a solution curve which veers away from the equilibrium as the independent variable tends to positive infinity.

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• Because Q always gets closer and closer to the equilibrium value of 25 as t → ∞, we call Q = 25 a stable equilibrium for Q.


Company’s revenue

• This equilibrium solution is called unstable because if W starts near, but not equal to, 4000, the net worth W moves further away from 4000 as t → ∞.

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THE LOGISTIC MODEL

• A population in a confined space grows proportionally to the

product of the current population, P, and the difference

between the carrying capacity, L, and the current population.

• The rate of change of P is proportional to the product of P and

L − P, so

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This is called a logistic differential equation.

Logistic growth curve

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• The derivative dP/dt is the product of k and P and L − P, so

when P is small, the derivative dP/dt is small and the opulation

grows slowly.

• As P increases, the derivative dP/dt increases and the

population grows more rapidly. However, as P approaches the

carrying capacity L, the term L − P is small, and dP/dt is again

small and the population grows more slowly.

NEWTON'S LAW OF HEATING AND COOLING

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NEWTON'S LAW OF HEATING AND COOLING

• Newton proposed that the temperature of a hot object decreases at

a rate proportional to the difference between its temperature and

that of its surroundings. Similarly, a cold object heats up at a rate

proportional to the temperature difference between the object and

its surroundings.

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Example: the temperature of two cups of coffeeagainst time, one starting at a highertemperature than the other, but both tendingtoward room temperature in the long run.

• Let H be the temperature at time t of a cup of coffee in a 70°F

room.

• Newton's Law says that the rate of change of H is proportional

to the temperature difference between the coffee and the

room:

• The rate of change of temperature is dH/dt. The temperature

difference between the coffee and the room is (H − 70), so

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k = 1

Example

• The body of a murder victim is found at noon in a room with a

constant temperature of 20°C. At noon the temperature of the

body is 35°C; two hours later the temperature of the body is

33°C.

(a) Find the temperature, H, of the body as a function of t, the

time in hours since it was found.

(b) Graph H against t. What happens to the temperature in the

long run?

(c) At the time of the murder, the victim's body had the normal

body temperature, 37°C. When did the murder occur?

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MODELING THE INTERACTION OF

TWO POPULATIONS

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MODELING THE INTERACTION OF

TWO POPULATIONS

• Predator-prey model (Lotka-Volterra equations)

• Two species competing for food (Competition)

• Two species helping each other (symbiosis)

• Others

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A PREDATOR-PREY MODEL: ROBINS AND WORMS

• We model a predator-prey system using what are called the

Lotka-Volterra equations.

Example: robins are the predators and worms the prey

• Suppose there are r thousand robins and w million worms.

If there were no robins, the worms would increase

exponentially according to the equation

If there were no worms, the robins would have no food and

so their population would decrease according to the equation

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the effect of the two populations on

one another2

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How exactly do the two populations interact?

Interaction of two populations

• Let's assume the effect of one population on the other is

proportional to the number of “encounters.”

• An encounter is when a robin eats a worm.

• The number of encounters is likely to be proportional to the

product of the populations because if one population is held

fixed, the number of encounters should be directly proportional

to the other population.

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Analysis of this system

• Let's look at the specific example with a = b = c = k = 1:

• To see the growth of the populations, we want graphs of r and

w against t. However, it is easier to obtain a graph of r against

w first. If we plot a point (w, r) representing the number of

worms and robins at any moment, then, as the populations

change, the point moves.

• The wr-plane on which the point moves is called the phase

plane and the path of the point is called the phase trajectory.

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Phase plane and trajectory

• To find the phase trajectory, we need a differential equation

relating w and r directly. We have the two differential

equations

• Thinking of r as a function of w and w as a function of t, the

chain rule gives

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Slope field of this differential equation in the

phase plane

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The Slope Field and Equilibrium Points

• What solutions of this equation look like from the slope field.

• At the point (1, 1) there is no slope drawn because dr/dw is undefined there since the rates of change of both populations with respect to time are zero:

• In terms of worms and robins, this means that if at some moment w = 1 and r = 1 (that is, there are 1 million worms and 1 thousand robins), then w and r remain constant forever. The point w = 1, r = 1 is therefore an equilibrium solution.

• The origin is also an equilibrium point, since if w = 0 and r = 0, then w and r remain constant. The slope field suggests that there are no other equilibrium points.

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Equilibrium Points

• By solving this

Only w = 0, r = 0 and w = 1, r = 1 as solutions.

Thus, (0, 0) and (1, 1) are equilibrium points

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Trajectories in the wr-Phrase Plane

• Let's look at the trajectories in the phase plane.

• A point on a curve represents a pair of populations (w, r) existing

at the same time t (though t is not shown on the graph).

• A short time later, the pair of populations is represented by a

nearby point. As time passes, the point traces out a trajectory. It

can be shown that the trajectory is a closed curve.

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Solution curve is closed

In which direction does the point move on the

trajectory?

• Look at the original pair of differential equations. They tell us

how w and r change with time.

• Imagine that we are at the point P0 in Figure 9.41,

where w = 2.2 and r = 1; then

Therefore, r is increasing, so the point is moving in the direction

shown by the arrow.

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Populations change over time

• Suppose that at time t = 0, there are 2.2 million worms and 1

thousand robins. Describe how the robin and worm populations

change over time.

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Lynxes and Hares

• A predator-prey system for which there are long-term data is

the Canadian lynx and the hare.

• Both animals were of interest to fur trappers and the records

of the Hudson Bay Company shed some light on their

populations through much of the 20 century.

• These records show that both populations oscillated up and

down, quite regularly, with a period of about ten years. This is

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OTHER FORMS OF SPECIES INTERACTION

• Other types of interactions between two species, such as

competition and symbiosis.

• Try to describe the interactions between two populations x

and y modeled by the following systems of differential

equations.

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Competitive interactions

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If we ignore the interaction terms with xy, we have dx/dt = 0.2x and

dy/dt = 0.6y, so both populations grow exponentially.

Since both interaction terms are negative, each species inhibits the

other's growth, such as when deer and elk compete for food.

Symbiotic interactions

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If we ignore the interaction terms, the populations of both species

decrease exponentially.

However, both interaction terms are positive, meaning each species

benefits from the other, so the relationship is symbiotic.

An example is the pollination of plants by insects.

One-side benefit interaction

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Ignoring interaction, x grows and y decays.

But the interaction term means that y benefits from x, in

the way birds that build nests benefit from trees.

Predator-prey interaction

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Without the interaction, x grows and y decays.

The interaction terms show that y hurts x while x benefits y.

This is a predator-prey model where y is the predator and x is the prey.

MODELING THE SPREAD OF A DISEASE

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The spread of a disease

• Differential equations can be used to predict when an

outbreak of a disease becomes so severe that it is called an

epidemic and to decide what level of vaccination is necessary

to prevent an epidemic.

• Example: FLU IN A BRITISH BOARDING SCHOOL

In January 1978, 763 students returned to a boys' boarding

school after their winter vacation. A week later, one boy

developed the flu, followed immediately by two more.

By the end of the month, nearly half the boys were sick. Most

of the school had been affected by the time the epidemic was

over in mid-February.

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THE S-I-R MODEL

• We apply one of the most commonly used models for an

epidemic, called the S-I-R model, to the boarding school flu

example. Imagine the population of the school divided into

three groups:

• S = the number of susceptibles, the people who are not yet sick

but who could become sick

• I = the number of infecteds, the people who are currently sick

• R = the number of recovered, or removed, the people who have

been sick and can no longer infect others or be reinfected.

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• In this model, the number of susceptibles decreases with

time, as people become infected. We assume that the rate

people become infected is proportional to the number of

contacts between susceptible and infected people.

• We expect the number of contacts between the two groups to

be proportional to both S and I. (If S doubles, we expect the

number of contacts to double; similarly, if I doubles, we expect

the number of contacts to double.) Thus, we assume that the

number of contacts is proportional to the product, SI. In other

words, we assume that for some constant a > 0,

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• The number of infecteds is changing in two ways:

• newly sick people are added to the infected group and others are

removed.

• The newly sick people are exactly those people leaving the

susceptible group and so accrue at a rate of aSI (with a positive

sign this time).

• People leave the infected group either because they recover (or

die), or because they are physically removed from the rest of the

group and can no longer infect others.

• We assume that people are removed at a rate proportional to

the number sick, or bI, where b is a positive constant. Thus,

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• Assuming that those who have recovered from the disease are

no longer susceptible, the recovered group increases at the

rate of bI, so

• We are assuming that having the flu confers immunity on a

person, that is, that the person cannot get the flu again.

• We can use the fact that the total population S+I+R is not

changing. Thus, once we know S and I, we can calculate R. So

we restrict our attention to the two equations

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The Constants a and b

• The constant a measures how infectious the disease is—that

is, how quickly it is transmitted from the infecteds to the

susceptibles.

• In the case of the flu, we know from medical accounts that the

epidemic started with one sick boy, with two more becoming

sick about a day later.

• Thus, when I = 1 and S = 762, we have dS/dt ≈ −2, enabling us

to roughly approximate a:

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• The constant b represents the rate at which infected people

are removed from the infected population.

• In this case of the flu, boys were generally taken to the

infirmary within one or two days of becoming sick.

• Assuming half the infected population was removed each day,

we take b ≈ 0.5. Thus,

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The Phrase Plane

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Trajectory

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The value of I first increases, then decreases to zero. This peak value of I

occurs when S ≈ 200. We can determine exactly when the peak value

occurs by solvingS = 192

What does the SI-Phase Plane Tell Us?

• For this example, the value S =192 is called a threshold population. If

S is around or below 192, there is no epidemic. If S is significantly

greater than 192, an epidemic occurs.

• The phase diagram makes clear that the maximum value of I is about

300, which is the maximum number infected at any one time.

• In addition, the point at which the trajectory crosses the S-axis

represents the time when the epidemic has passed (since I = 0).

Thus, the S-intercept shows how many boys never get the flu and,

hence, how many do get sick.

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THRESHOLD VALUE

• For the general SIR model, we have the following result:

• If S , the initial number of susceptibles, is above b/a, there is an epidemic; if S is below b/a, there is no epidemic.

• How Many People Should Be Vaccinated?

• To answer this, we can think of vaccination as removing people from the S category (without increasing I), which amounts to moving the initial point on the trajectory to the left, parallel to the S-axis. To avoid an epidemic, the initial value of S should be around or below the threshold value.

• Therefore, the boarding-school epidemic would have been avoided if all but 192 students had been vaccinated.

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Graphs of S and I Against t

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References

• MATHEMATICAL MODELING USING DIFFERENTIAL EQUATIONS

- Applied Calculus 5th Edition

• D.D. Mooney and R.J. Swift, A Course in Mathematical

Modeling, The Mathematical Association of America, 1999.

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