chapter 3: modeling with first-order differential...

OverviewLinear Models

Nonlinear ModelsSummary

Chapter 3: Modeling with First-Order DifferentialEquations

王奕翔

Department of Electrical EngineeringNational Taiwan University

[email protected]

September 26, 2013

王奕翔 DE Lecture 3

[email protected]



1 Overview

2 Linear ModelsGrowth and DecayCooling and WarmingMixturesSeries Circuit

3 Nonlinear ModelsPopulation Dynamics and Logistic EquationChemical Reactions

4 Summary




Organization of Lectures in Chapter 2 and 3

��

��(2-1)

��(2-6)

Separable DE (2-2)

�DE(2-3)

Exact DE(2-4)

��(2-5)

Linear Models (3-1)

Nonlinear Models (3-2)




What is covered in this lecture

We will learn how to model a system by first-order differential equations,solve them by the methods that we have learned, and answer thequestions in which we are interested.

解應用題基本步驟:1 寫下描述系統變化的微分方程式：定義自變數和應變數

2 用學過的方法來解寫下的微分方程式：得到描述系統特性的函數

3 解決一開始待解的問題




Growth and DecayCooling and WarmingMixturesSeries Circuit

1 Overview



4 Summary





Population Growth

Thomas Malthus 人口論 (1798): 人口增加速率正比於人口總數。1 couple: 1 child per 10 months. 10 couples: 10 children per 10 months.This model can also be extended to other kinds of population, forexample, bacteria.

Example (培養皿中的細菌數)Initially (t = 0) there are P0 number of bacteria. At time t = 1 hour,there are 2P0 number of bacteria. If the rate of growth is proportional tothe total population, find the total population at time t, that is, P(t),and the time at which there are 4P0 number of bacteria.





Population Growth

A: First we write down the DE and initial conditions relating P(t) and t:

dPdt = kP, P(0) = P0, k > 0

which can be solved as follows.dPP = k dt, P ̸= 0 =⇒ ln |P| = kt + c 兩邊積分=⇒ ln P0 = c 代初始值 =⇒ P(t) = P0ekt, t ≥ 0.

To determine k, we plug in the other condition:P(1) = 2P0 = P0ek =⇒ k = ln 2 =⇒ P(t) = P02

t.Finally, the time it takes for the population to reach 4P0 is:log2 4 = 2 hours.





Radioactive Decay

For radioactive substances, the decay rate is proportional to theconcurrent total amount A(t). dA

dt = kA, A(0) = A0, k < 0.

Half-life: the time it takes for 1/2 of the atoms in a initial amount todegenerate. Example: the half-life of C-14 is 5730 years.

Example (碳14定年)A piece of fossil is found to contain 0.1% of the original amount of C-14.Estimate the age of the fossil given the half-life of C-14 is 5730 years.





Radioactive Decay

A: We already know the function A(t): A(t)/A0 = ekt, A(0) = A0.

Half-life = 5730 =⇒ 0.5 = exp(5730k) =⇒ k = − ln 25730 .

Now, 0.001 = ekt =⇒ t = − ln 1000k = 5730× 3× log2 10 ≈ 57104.





Growth and Decay

If the growth/decay rate is proportional to the current totalamount/population, we have

dXdt = kX, X(0) = X0

With experience, we know immediately that

XX0

= ekt.

To find k, we need to plug in another condition.





1 Overview



4 Summary





Cooling

I. Newton: 升(降)溫的速率與物體和周遭環境的溫差成正比

Example (Cooling of Hot Water)The initial temperature of a bottle of water is 100◦. At t = 10 minutes, itbecomes 40◦. At t = 20 minutes, it becomes 28◦. What is the roomtemperature?

A: According to Newton, we have

dTdt = k(T − Tm), T(0) = 100, k < 0.

We can solve it by first finding an integrating factor (exercise). Instead,here is a shortcut.





Cooling

Observe that d(T−Tm)dt = k(T − Tm), T(0)− Tm = 100− Tm. Hence,

we immediately have

T − Tm100− Tm

= ekt.

Plug in the two conditions

40− Tm100− Tm

= e10k,28− Tm100− Tm

= e20k

=⇒(

40− Tm100− Tm

)2

=28− Tm100− Tm

=⇒ Tm = 25.





1 Overview



4 Summary





Mixture of Two Fluids

Mixtures The mixing of two salt solutions of differing concentrations givesrise to a first-order differential equation for the amount of salt contained in the mix-ture. Let us suppose that a large mixing tank initially holds 300 gallons of brine (thatis, water in which a certain number of pounds of salt has been dissolved). Anotherbrine solution is pumped into the large tank at a rate of 3 gallons per minute; theconcentration of the salt in this inflow is 2 pounds per gallon. When the solution inthe tank is well stirred, it is pumped out at the same rate as the entering solution. SeeFigure 1.3.2. If A(t) denotes the amount of salt (measured in pounds) in the tank attime t, then the rate at which A(t) changes is a net rate:

. (7)

The input rate Rin at which salt enters the tank is the product of the inflow concentra-tion of salt and the inflow rate of fluid. Note that Rin is measured in pounds perminute:

Now, since the solution is being pumped out of the tank at the same rate that it ispumped in, the number of gallons of brine in the tank at time t is a constant 300 gal-lons. Hence the concentration of the salt in the tank as well as in the outflow isc(t) ! A(t)!300 lb/gal, so the output rate Rout of salt is

The net rate (7) then becomes

(8)

If rin and rout denote general input and output rates of the brine solutions,* thenthere are three possibilities: rin ! rout, rin " rout, and rin # rout. In the analysis lead-ing to (8) we have assumed that rin ! rout. In the latter two cases the number of gal-lons of brine in the tank is either increasing (rin " rout) or decreasing (rin # rout) atthe net rate rin $ rout. See Problems 10–12 in Exercises 1.3.

Draining a Tank In hydrodynamics, Torricelli’s law states that the speed v ofefflux of water though a sharp-edged hole at the bottom of a tank filled to a depth his the same as the speed that a body (in this case a drop of water) would acquire infalling freely from a height h—that is, , where g is the acceleration due togravity. This last expression comes from equating the kinetic energy with thepotential energy mgh and solving for v. Suppose a tank filled with water is allowed todrain through a hole under the influence of gravity. We would like to find the depth hof water remaining in the tank at time t. Consider the tank shown in Figure 1.3.3. Ifthe area of the hole is Ah (in ft2) and the speed of the water leaving the tank is

(in ft/s), then the volume of water leaving the tank per second is (in ft3/s). Thus if V(t) denotes the volume of water in the tank at time t, then

, (9)dVdt

! $Ah12gh

Ah12ghv ! 12gh

12mv2

v ! 12gh

dAdt

! 6 $A

100 or

dAdt

%1

100 A ! 6.

Rout ! ( lb/gal) " (3 gal/min) ! lb/min.A(t)––––300

A(t)––––100

concentrationof salt

in outflowoutput rate

of brineoutput rate

of salt

concentrationof salt

in inflowinput rateof brine

input rateof salt

Rin ! (2 lb/gal) " (3 gal/min) ! (6 lb/min).

dAdt

! "input rateof salt # $ "output rate

of salt # ! Rin $ Rout

24 ! CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

input rate of brine3 gal/min

output rate of brine3 gal/min

constant300 gal

FIGURE 1.3.2 Mixing tank

h

Aw

Ah

FIGURE 1.3.3 Draining tank

*Don’t confuse these symbols with Rin and Rout, which are input and output rates of salt.

92467_01_ch01_p01-034.qxd 2/10/12 12:53 PM Page 24

Example (Mixture of Salt Solutions)In the figure, the concentration of salt of the incoming flow is 2lb/gal. Initially there is 50 lb of salt in the tank. Find theconcentration of salt in the tank as time t → ∞.

A: Immediately we see that the final answer is 2 lb/gal. (Why?)

We should do the calculation if we also want to know how theconcentration changes with time. Let the total amount of saltat time t be A(t) lb.

Incoming rate of salt: Rin = 2× 3 = 6 lb/min;Outgoing: Rout = (current concentration)×3 = A

300×3 = A

100.

∴ dAdt = 6− A

100, A(0) = 50.





Mixture of Two Fluids

Solve dAdt = 6− A

100, A(0) = 50.

1 Derive Auxiliary DE: Let µ(t) := the integrating factor to be found.

d (µA)

dt = µdAdt + Adµ

dt = µ

{6− A

100

}+ Adµ

dt = 6µ+

{dµdt − µ

100

}A

2 Solve Auxiliary DE: dµdt =

µ

100. One solution: µ(t) = e t

100 .

3 Solve Original DE: Plug in µ(t) = e t100 and A(0) = 50:

et

100 A = 600et

100 − 550 =⇒ A(t) = 600− 550e−t100





1 Overview



4 Summary





LRC Series Circuits

where the minus sign indicates that V is decreasing. Note here that we are ignoringthe possibility of friction at the hole that might cause a reduction of the rate of flowthere. Now if the tank is such that the volume of water in it at time t can be writtenV(t) ! Awh, where Aw (in ft2) is the constant area of the upper surface of the water(see Figure 1.3.3), then dV!dt ! Aw dh!dt. Substituting this last expression into (9)gives us the desired differential equation for the height of the water at time t:

. (10)

It is interesting to note that (10) remains valid even when Aw is not constant. In thiscase we must express the upper surface area of the water as a function of h—that is,Aw ! A(h). See Problem 14 in Exercises 1.3.

Series Circuits Consider the single-loop LRC-series circuit shown in Fig-ure 1.3.4(a), containing an inductor, resistor, and capacitor. The current in a circuitafter a switch is closed is denoted by i(t); the charge on a capacitor at time t is de-noted by q(t). The letters L, R, and C are known as inductance, resistance, and capac-itance, respectively, and are generally constants. Now according to Kirchhoff’ssecond law, the impressed voltage E(t) on a closed loop must equal the sum of thevoltage drops in the loop. Figure 1.3.4(b) shows the symbols and the formulas for therespective voltage drops across an inductor, a capacitor, and a resistor. Since currenti(t) is related to charge q(t) on the capacitor by i ! dq!dt, adding the three voltages

inductor resistor capacitor

and equating the sum to the impressed voltage yields a second-order differentialequation

(11)

We will examine a differential equation analogous to (11) in great detail inSection 5.1.

Falling Bodies To construct a mathematical model of the motion of a bodymoving in a force field, one often starts with the laws of motion formulated by theEnglish mathematician Isaac Newton (1643–1727). Recall from elementary physicsthat Newton’s first law of motion states that a body either will remain at rest or willcontinue to move with a constant velocity unless acted on by an external force. Ineach case this is equivalent to saying that when the sum of the forces —that is, the net or resultant force—acting on the body is zero, then the accelerationa of the body is zero. Newton’s second law of motion indicates that when the netforce acting on a body is not zero, then the net force is proportional to its accelera-tion a or, more precisely, F ! ma, where m is the mass of the body.

Now suppose a rock is tossed upward from the roof of a building as illustratedin Figure 1.3.5. What is the position s(t) of the rock relative to the ground at time t?The acceleration of the rock is the second derivative d2s!dt2. If we assume that theupward direction is positive and that no force acts on the rock other than the force ofgravity, then Newton’s second law gives

. (12)

In other words, the net force is simply the weight F ! F1 ! "W of the rock near thesurface of the Earth. Recall that the magnitude of the weight is W ! mg, where m is

m d 2sdt2 ! "mg or

d 2sdt2 ! "g

F ! " Fk

L d 2qdt2 # R

dqdt

#1C

q ! E(t).

L didt

! L d 2qdt2 , iR ! R

dqdt

, and 1C

q

dhdt

! "Ah

Aw 12gh

1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ! 25

(a)

(b)

E(t)L

C

R

(a) LRC-series circuit

(b)

L

R

Inductorinductance L: henries (h)

voltage drop across: Ldidt

i

Capacitorcapacitance C: farads (f)

voltage drop across:1C

i

Resistorresistance R: ohms (!)voltage drop across: iR

i

q

C

FIGURE 1.3.4 Symbols, units, andvoltages. Current i(t) and charge q(t) aremeasured in amperes (A) and coulombs(C), respectively

groundbuilding

rock

s(t)s0

v0

FIGURE 1.3.5 Position of rockmeasured from ground level

92467_01_ch01_p01-034.qxd 2/10/12 12:53 PM Page 25


. (10)





(11)




. (12)


m d 2sdt2 ! "mg or

d 2sdt2 ! "g

F ! " Fk

L d 2qdt2 # R

dqdt

#1C

q ! E(t).

L didt


dqdt

, and 1C

q

dhdt

! "Ah

Aw 12gh


(a)

(b)

E(t)L

C

R


(b)

L

R



i



i


i

q

C


groundbuilding

rock

s(t)s0

v0


92467_01_ch01_p01-034.qxd 2/10/12 12:53 PM Page 25


. (10)





(11)




. (12)


m d 2sdt2 ! "mg or

d 2sdt2 ! "g

F ! " Fk

L d 2qdt2 # R

dqdt

#1C

q ! E(t).

L didt


dqdt

, and 1C

q

dhdt

! "Ah

Aw 12gh


(a)

(b)

E(t)L

C

R


(b)

L

R



i



i


i

q

C


groundbuilding

rock

s(t)s0

v0


92467_01_ch01_p01-034.qxd 2/10/12 12:53 PM Page 25 where the minus sign indicates that V is decreasing. Note here that we are ignoringthe possibility of friction at the hole that might cause a reduction of the rate of flowthere. Now if the tank is such that the volume of water in it at time t can be writtenV(t) ! Awh, where Aw (in ft2) is the constant area of the upper surface of the water(see Figure 1.3.3), then dV!dt ! Aw dh!dt. Substituting this last expression into (9)gives us the desired differential equation for the height of the water at time t:

. (10)





(11)




. (12)


m d 2sdt2 ! "mg or

d 2sdt2 ! "g

F ! " Fk

L d 2qdt2 # R

dqdt

#1C

q ! E(t).

L didt


dqdt

, and 1C

q

dhdt

! "Ah

Aw 12gh


(a)

(b)

E(t)L

C

R


(b)

L

R



i



i


i

q

C


groundbuilding

rock

s(t)s0

v0


92467_01_ch01_p01-034.qxd 2/10/12 12:53 PM Page 25

Current is the derivativeof charge:

i = dqdt

E(t) = total voltage drop:

E(t) = L d2qdt2

+ R dqdt

+qC





Example: LR Series Circuit

Example

Integrating the last equation and solving for A gives the general solution A(t) ! 600 " ce#t/100. When t ! 0, A ! 50, so we find that c ! #550. Thus theamount of salt in the tank at time t is given by

. (6)

The solution (6) was used to construct the table in Figure 3.1.5(b). Also, it can beseen from (6) and Figure 3.1.5(a) that A(t) : 600 as t : $. Of course, this is whatwe would intuitively expect; over a long time the number of pounds of salt in thesolution must be (300 gal)(2 lb/gal) ! 600 lb.

In Example 5 we assumed that the rate at which the solution was pumped in wasthe same as the rate at which the solution was pumped out. However, this need not bethe case; the mixed brine solution could be pumped out at a rate rout that is fasteror slower than the rate rin at which the other brine solution is pumped in. The nextexample illustrates the case when the mixture is pumped out at rate that is slowerthan the rate at which the brine solution is being pumped into the tank.

A(t) ! 600 # 550e#t/100

88 ! CHAPTER 3 MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS

t

A A = 600

500

(a)

t (min) A (lb)

50 266.41100 397.67150 477.27200 525.57300 572.62400 589.93

(b)

FIGURE 3.1.5 Pounds of salt in thetank in Example 5

FIGURE 3.1.7 LR-series circuit

EL

R

FIGURE 3.1.6 Graph of A(t) inExample 6

t

A

50

250

500

100

EXAMPLE 6 Example 5 Revisited

If the well-stirred solution in Example 5 is pumped out at a slower rate of, say, rout ! 2 gal/min, then liquid will accumulate in the tank at the rate of rin # rout ! (3 # 2) gal/min ! 1 gal/min. After t minutes,

(1 gal/min) . (t min) ! t gal

will accumulate, so the tank will contain 300 " t gallons of brine. The concentrationof the outflow is then c(t) ! A!(300 " t) lb/gal, and the output rate of salt is Rout !c(t) . rout, or

.

Hence equation (5) becomes

.

The integrating factor for the last equation is

and so after multiplying by the factor the equation is cast into the form

Integrating the last equation gives By applying theinitial condition and solving for A yields the solution A(t) ! 600 " 2t #(4.95 % 107)(300 " t)#2. As Figure 3.1.6 shows, not unexpectedly, salt builds up inthe tank over time, that is,

Series Circuits For a series circuit containing only a resistor and an inductor,Kirchhoff’s second law states that the sum of the voltage drop across the inductor(L(di!dt)) and the voltage drop across the resistor (iR) is the same as the impressedvoltage (E(t)) on the circuit. See Figure 3.1.7.

Thus we obtain the linear differential equation for the current i(t),

, (7)

where L and R are constants known as the inductance and the resistance, respectively.The current i(t) is also called the response of the system.

Ldidt

" Ri ! E(t)

A : $ as t : $.

A(0) ! 50(300 " t)2A ! 2(300 " t)3 " c.

ddt

[(300 " t)2 A] ! 6(300 " t)2.

e"2dt>(300" t) ! e2 ln(300" t) ! eln(300" t)2! (300 " t)2

dAdt

! 6 #2A

300 " t or

dAdt

"2

300 " tA ! 6

Rout ! # A300 " t

lb/gal$ ! (2 gal/min) !2A

300 " t lb/min

92467_03_ch03_p083-115.qxd 2/10/12 2:39 PM Page 88

Consider the LR circuit shown on the left,where E(t) = 10 volts, R = 10 ohms, L = 0.5henry, and initial current i(0) = 0.Find i(t).

A: Current i(t) satisfies the following DE:

Ldidt + Ri = E(t) =⇒ 1

2

didt + 10i = 10 =⇒ di

dt + 20i = 20





Example: LR Series Circuit

Solve didt + 20i = 20, i(0) = 0.

1 Derive Auxiliary DE: Let µ(t) := the integrating factor to be found.

d (µi)dt = µ

didt + i dµ

dt = 20µ {1− i}+ i dµdt = 20µ+

{dµdt − 20µ

}i

2 Solve Auxiliary DE: dµdt = 20µ. One solution: µ(t) = e20t.

3 Solve Original DE: Plug in µ(t) = e20t and i(0) = 0:

e20ti − 0 =

∫ t

0

20e20τdτ =⇒ e20ti(t) = e20t − 1 =⇒ i(t) = 1− e−20t




Population Dynamics and Logistic EquationChemical Reactions

1 Overview



4 Summary





Reality is not likely to be linear, usually

Thomas Malthus 人口論 (1798): 人口增加速率正比於人口總數。

dPdt = kP

Points that are missing:Limited resources: population ↑, competition ↑ (nonlinearity kicksin!)Emigration and immigrationDeath, etc.





1 Overview



4 Summary





Population Dynamics

Limited resources: population ↑, competition ↑.=⇒ the constant k may vary with P !

Hence, in general the population dynamics should be governed by

dPdt = Pf(P).





Logsitic Equation

dPdt = Pf(P).

Intuitively, the function f(P) should be a decreasing function of P.

Simplest case: f(P) = a − bP, for a, b > 0.

Definition (Logistic Equation)

dPdt = P(a − bP)





Solving Logistic Equation

Solve dPdt = P(a − bP).

A: We can solve this by separation of variables.

1

P(a − bP)dP = dt, P ̸= 0,ab

=⇒{1/aP +

b/aa − bP

}dP = dt

=⇒ 1

a ln |P| − 1

a ln |a − bP| = t + c′

=⇒ Pa − bP = ceat, why we pick 0 < P < a/b?

=⇒ P(t) = aceat

1 + bceat =ac

e−at + bc

Note: if P(0) = ab , P(t) = a

b for all t.王奕翔 DE Lecture 3




Solution Curves of Logistic EquationsThe solution is called Logistic Function: P(t) = ac

e−at+bc

Properties of Logistic Function

P(t): increasing function. Population saturation: ab .

P(t) → ab as t → ∞; P(t) → 0 as t → −∞.

Saddle point: P = a2b

The dashed line P ! a!2b shown in Figure 3.2.2 corresponds to the ordinate of apoint of inflection of the logistic curve. To show this, we differentiate (4) by theProduct Rule:

.

From calculus recall that the points where d2P!dt2 ! 0 are possible points of inflec-tion, but P ! 0 and P ! a!b can obviously be ruled out. Hence P ! a!2b is the onlypossible ordinate value at which the concavity of the graph can change. For 0 " P " a!2b it follows that P# $ 0, and a!2b " P " a!b implies that P# " 0.Thus, as we read from left to right, the graph changes from concave up to concavedown at the point corresponding to P ! a!2b. When the initial value satisfies 0 " P0 " a!2b, the graph of P(t) assumes the shape of an S, as we see inFigure 3.2.2(a). For a!2b " P0 " a!b the graph is still S-shaped, but the point ofinflection occurs at a negative value of t, as shown in Figure 3.2.2(b).

We have already seen equation (4) in (5) of Section 1.3 in the form dx!dt ! kx(n % 1 & x), k $ 0. This differential equation provides a reasonablemodel for describing the spread of an epidemic brought about initially by introduc-ing an infected individual into a static population. The solution x(t) represents thenumber of individuals infected with the disease at time t.

! 2b2P "P &ab#"P &

a2b#

! P(a & bP)(a & 2bP)

d 2Pdt2 ! P "&b

dPdt# % (a & bP)

dPdt

!dPdt

(a & 2bP)

3.2 NONLINEAR MODELS ! 97

P

P0

a/2b

a/b

t

P

P0 a/2b

a/b

t

(a)(a)

(b)

FIGURE 3.2.2 Logistic curves fordifferent initial conditions

(a)

t

x x = 1000

10

500

5

(a)

t (days) x (number infected)

4 50 (observed)5 1246 2767 5078 7359 882

10 953

(b)

FIGURE 3.2.3 Number of infectedstudents in Example 1

EXAMPLE 1 Logistic Growth

Suppose a student carrying a flu virus returns to an isolated college campus of 1000students. If it is assumed that the rate at which the virus spreads is proportional notonly to the number x of infected students but also to the number of students notinfected, determine the number of infected students after 6 days if it is furtherobserved that after 4 days x(4) ! 50.

SOLUTION Assuming that no one leaves the campus throughout the duration of thedisease, we must solve the initial-value problem

.

By making the identification a ! 1000k and b ! k, we have immediately from(5) that

.

Now, using the information x(4) ! 50, we determine k from

We find . Thus

.

Finally, .

Additional calculated values of x(t) are given in the table in Figure 3.2.3(b). Note thatthe number of infected students x(t) approaches 1000 as t increases.

x(6) !1000

1 % 999e&5.9436 ! 276 students

x(t) !1000

1 % 999e&0.9906t

&1000k ! 14 ln 19

999 ! &0.9906

50 !1000

1 % 999e&4000k.

x(t) !1000k

k % 999ke&1000kt !1000

1 % 999e&1000kt

dxdt

! kx(1000 & x), x(0) ! 1

92467_03_ch03_p083-115.qxd 2/10/12 2:39 PM Page 97

The dashed line P ! a!2b shown in Figure 3.2.2 corresponds to the ordinate of apoint of inflection of the logistic curve. To show this, we differentiate (4) by theProduct Rule:

.

From calculus recall that the points where d2P!dt2 ! 0 are possible points of inflec-tion, but P ! 0 and P ! a!b can obviously be ruled out. Hence P ! a!2b is the onlypossible ordinate value at which the concavity of the graph can change. For 0 " P " a!2b it follows that P# $ 0, and a!2b " P " a!b implies that P# " 0.Thus, as we read from left to right, the graph changes from concave up to concavedown at the point corresponding to P ! a!2b. When the initial value satisfies 0 " P0 " a!2b, the graph of P(t) assumes the shape of an S, as we see inFigure 3.2.2(a). For a!2b " P0 " a!b the graph is still S-shaped, but the point ofinflection occurs at a negative value of t, as shown in Figure 3.2.2(b).

We have already seen equation (4) in (5) of Section 1.3 in the form dx!dt ! kx(n % 1 & x), k $ 0. This differential equation provides a reasonablemodel for describing the spread of an epidemic brought about initially by introduc-ing an infected individual into a static population. The solution x(t) represents thenumber of individuals infected with the disease at time t.

! 2b2P "P &ab#"P &

a2b#

! P(a & bP)(a & 2bP)

d 2Pdt2 ! P "&b

dPdt# % (a & bP)

dPdt

!dPdt

(a & 2bP)

3.2 NONLINEAR MODELS ! 97

P

P0

a/2b

a/b

t

P

P0 a/2b

a/b

t

(a)(a)

(b)

FIGURE 3.2.2 Logistic curves fordifferent initial conditions

(a)

t

x x = 1000

10

500

5

(a)

t (days) x (number infected)

4 50 (observed)5 1246 2767 5078 7359 882

10 953

(b)

FIGURE 3.2.3 Number of infectedstudents in Example 1

EXAMPLE 1 Logistic Growth

Suppose a student carrying a flu virus returns to an isolated college campus of 1000students. If it is assumed that the rate at which the virus spreads is proportional notonly to the number x of infected students but also to the number of students notinfected, determine the number of infected students after 6 days if it is furtherobserved that after 4 days x(4) ! 50.

SOLUTION Assuming that no one leaves the campus throughout the duration of thedisease, we must solve the initial-value problem

.

By making the identification a ! 1000k and b ! k, we have immediately from(5) that

.

Now, using the information x(4) ! 50, we determine k from

We find . Thus

.

Finally, .

Additional calculated values of x(t) are given in the table in Figure 3.2.3(b). Note thatthe number of infected students x(t) approaches 1000 as t increases.

x(6) !1000

1 % 999e&5.9436 ! 276 students

x(t) !1000

1 % 999e&0.9906t

&1000k ! 14 ln 19

999 ! &0.9906

50 !1000

1 % 999e&4000k.

x(t) !1000k

k % 999ke&1000kt !1000

1 % 999e&1000kt

dxdt

! kx(1000 & x), x(0) ! 1

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Variants of Logistic Equations

dPdt = P(a − bP)± h immigration or emigrationdPdt = P(a − bP) + cekP 西瓜偎大邊

dPdt = P(a − b ln P) Gompertz DE

ExerciseDerive the population saturation point in each of the above models.





1 Overview



4 Summary





Chemical Reactions

Chemical A and B react and produce C.大一普化：反應速率正比於反應物濃度的乘積

Initial concentration: [A] = a, [B] = b.X(t): concentration of C at time t.To produce one unit of C we need λa unit of A and λb unit of B.

Then, we have the following DE governing the rate of reaction:

dXdt = k(a − λaX)(b − λbX).

How to solve? Separation of variables!




1 Overview



4 Summary




Short Recap

Growth and decay

Series circuit

Population dynamics

Logistic equation

Chemical reaction




Self-Practice Exercises

3-1: 3, 19, 27, 33, 35, 39, 43, 45

3-2: 5, 7, 9, 11, 15, 19, 21


chapter 3: modeling with first-order differential...

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