Overview
Linear Models
Nonlinear Models
Summary
Chapter 3: Modeling with First-Order Differential
Equations
王奕翔
Department of Electrical Engineering
National Taiwan University
ihwang@ntu.edu.tw
September 26, 2013
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Organization of Lectures in Chapter 2 and 3
Separable
DE (2-2)
(2-1)
(2-6)
DE
(2-3)
Exact DE
(2-4)
(2-5)
王奕翔
DE Lecture 3
Linear
Models (3-1)
Nonlinear
Models (3-2)
Overview
Linear Models
Nonlinear Models
Summary
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 the
questions 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
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
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, for
example, 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 to
the total population, find the total population at time t, that is, P(t),
and the time at which there are 4P0 number of bacteria.
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
Population Growth
A: First we write down the DE and initial conditions relating P(t) and t:
dP
= kP, P(0) = P0 , k > 0
dt
which can be solved as follows.
dP
= k dt, P ̸= 0 =⇒ ln |P| = kt + c 兩邊積分
P
=⇒ ln P0 = c 代初始值 =⇒ P(t) = P0 ekt , t ≥ 0.
To determine k, we plug in the other condition:
P(1) = 2P0 = P0 ek =⇒ k = ln 2 =⇒ P(t) = P0 2t .
Finally, the time it takes for the population to reach 4P0 is:
log2 4 = 2 hours.
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
Radioactive Decay
For radioactive substances, the decay rate is proportional to the
dA
concurrent total amount A(t).
= kA, A(0) = A0 , k < 0.
dt
Half-life: the time it takes for 1/2 of the atoms in a initial amount to
degenerate. 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.
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
Radioactive Decay
A: We already know the function A(t): A(t)/A0 = ekt , A(0) = A0 .
ln 2
Half-life = 5730 =⇒ 0.5 = exp(5730k) =⇒ k = − 5730
.
Now, 0.001 = ekt =⇒ t =
− ln 1000
k
王奕翔
= 5730 × 3 × log2 10 ≈ 57104.
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
Growth and Decay
If the growth/decay rate is proportional to the current total
amount/population, we have
dX
= kX, X(0) = X0
dt
With experience, we know immediately that
X
= ekt .
X0
To find k, we need to plug in another condition.
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
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, it
becomes 40◦ . At t = 20 minutes, it becomes 28◦ . What is the room
temperature?
A: According to Newton, we have
dT
= k(T − Tm ), T(0) = 100, k < 0.
dt
We can solve it by first finding an integrating factor (exercise). Instead,
here is a shortcut.
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
Cooling
m)
Observe that d(T−T
= k(T − Tm ), T(0) − Tm = 100 − Tm . Hence,
dt
we immediately have
T − Tm
= ekt .
100 − Tm
Plug in the two conditions
40 − Tm
28 − Tm
= e10k ,
= e20k
100 − Tm
100 − Tm
(
)2
40 − Tm
28 − Tm
=⇒
=
100 − Tm
100 − Tm
=⇒ Tm = 25.
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
01_ch01_p01-034.qxd
2/10/12
12:53 PM
Overview
Linear Models
Nonlinear Models
Page 24 Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
Mixture of Two Fluids
24
●
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
input rate of brine
3 gal/min
Example
(Mixture
of Salt
Mixtures
The mixing
of twoSolutions)
salt solutions of differing concentrations gives
rise to a first-order differential equation for the amount of salt contained in the mix-
Inture.
theLetfigure,
the
saltholds
of the
incoming
us suppose
thatconcentration
a large mixing tank of
initially
300 gallons
of brineflow
(that is 2
is, water Initially
in which a there
certain number
of salt
dissolved).
lb/gal.
is 50 oflbpounds
of salt
in has
thebeen
tank.
FindAnother
the
brine solution is pumped into the large tank at a rate of 3 gallons per minute; the
concentration
ofsalt
salt
in inflow
the tank
as time
t →When
∞. the solution in
concentration of the
in this
is 2 pounds
per gallon.
constant
300 gal
the tank is well stirred, it is pumped out at the same rate as the entering solution. See
A:Figure
Immediately
we seethethat
the
final
answer
is 2 lb/gal.
(Why?)
1.3.2. If A(t) denotes
amount
of salt
(measured
in pounds)
in the tank
at
time t, then the rate at which A(t) changes is a net rate:
We should dodAthe input
calculation
if we also want to know how the
rate
output rate
(7)salt
$ time.
Rin total
$ Rout. amount of
!"
#
"
concentrationdtchanges
with
Let# !
the
of salt
of salt
atThe
time
t
be
A(t)
lb.
input rate R in at which salt enters the tank is the product of the inflow concentraoutput rate of brine
3 gal/min
FIGURE 1.3.2 Mixing tank
tion of salt and the inflow rate of fluid. Note that R in is measured in pounds per
Incoming
rate of salt: Rin = 2 × 3 = 6 lb/min;
minute:
concentration concentration) × 3 =
Outgoing: Rout = (current
of salt
in inflow
dA
dt
input rate
of brine
A
100
input rate
of salt
A
300
×3 =
=6−
, A(0) = 50.
R∴
in ! (2 lb/gal) " (3 gal/min) ! (6 lb/min).
Now, since the solution is being pumped out of the tank at the same rate that it is
pumped in, the number of gallons of brine in the tank at time t is a constant 300 gallons. Hence the concentration of the salt in the tank as well as in the outflow is
王奕翔
DE Lecture 3
A
.
100
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
Mixture of Two Fluids
Solve
dA
A
=6−
, A(0) = 50.
dt
100
1
Derive Auxiliary DE: Let µ(t) := the integrating factor to be found.
}
}
{
{
d (µA)
dA
dµ
A
dµ
dµ
µ
=µ
+A
=µ 6−
+A
= 6µ +
−
A
dt
dt
dt
100
dt
dt
100
2
Solve Auxiliary DE:
3
Solve Original DE: Plug in µ(t) = e 100 and A(0) = 50:
t
dµ
µ
=
. One solution: µ(t) = e 100 .
dt
100
t
t
−t
t
e 100 A = 600e 100 − 550 =⇒ A(t) = 600 − 550e 100
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
p01-034.qxd
E(t)
L
2/10/12
12:53 PM
R
LRC SeriesC Circuits
(a) LRC-series
(a) circuit
Inductor
inductance L: henries (h)
di
voltage drop across: L
dt
L
R
E(t)
i
CL
(a) LRC-series
(a) circuit
Resistor
resistance R: ohms (Ω)
Inductordrop across: iR
voltage
inductance L: henries (h)
di
voltage drop across: L
dt
i
R
i
L
where the minus sign indic
Growth and Decay
the possibility
of Warming
friction at the
that might
cause aatr
Cooling and
thehole
possibility
of friction
Itthat
is interesting
toof
note
that
Mixtures (h)
L
L:
henries
there.
Now
if
the
tank
is
such
the
volume
water
Series
Circuit
there.
Now
if
the
tank
is sui
E(t)
di R
2case we must express the up
voltage
L
V(t)drop
! Aacross:
h,
where
A
(in
ft
)
is
the
constant
area
of
thef
w
w
V(t) ! A w h, where A w (in
dt
A!
A(h). SeeSubstituting
Problem 14
w!
(see Figure 1.3.3), then dV!dt(see
AFigure
w dh!dt.
1.3.3), then dV!
gives C
us the desired differential
equation
the height
o
gives
us the for
desired
differen
Series Circuits Con
i
(a) LRC-series
(a)
circuit
L
ure dh
1.3.4(a),Acontaining
an i
! " h AS
12gh
.
1.3 DIFFERENTIAL EQUATIONS
MATHEMA
afterdta switch
Awis closed is d
Current
the
derivative
notedisby
q(t).
The letters L,
Inductor
Resistor
It
is
interesting
to
note
that
(10)
remains
validtoeven
when
of
charge:
It
is
interesting
note
that
itance,
respectively,
and
ar
where
the
minus
sign
indicates
that
V
is
decreasing.
Not
inductanceR:L:ohms
henries
resistance
(Ω)(h)
case
weacross:
must express
the upper
surface
area
ofimpressed
the
water
difriction
case
we
must
express
the
up
second
law,
the
voltage
drop
iR
the
possibility
of
at
the
hole
that
might
cause
a
r
dq
voltage drop across: L
iA(h).
= 1.3.
A
Problem
involtage
Exercises
w ! A(h).
Athat
drops
in Problem
the
Fii
there.
Now See
if thedt
tank is 14
such
volume
of loop.
water14
w ! the
dtSee
voltage
drops
acr
V(t) ! A w h, where A w (in ft 2)respective
is the constant
area
of the
toSubstituting
charge
q(t)
ii
(see Series
Figure
1.3.3),
then Consider
dV!dti(t)
! is
A
dh!dt.
Circuits
the
single-loop
LRC-se
wrelated
Series
Circuits
Con
R
=equation
totalresistor,
voltage
drop:
L the desired
gives
us
differential
for
the
height
ure
1.3.4(a),
containing
anE(t)
inductor,
and
capacit
ure
1.3.4(a),
containing
an oi
inducto
after a switch is closed is denoted
i(t);
the
charge
ond
2
after aby
switch
is
closed
is
d q
q
ARh dq +asdi
=dh
LCby
+known
Capacitor
noted by q(t). The letters L,E(t)
R,noted
and
are
inducta
2
q(t).
The
letters
L
!
!
"
12gh
.
dt
dt
C L,L
Resistor
dtand
capacitance
farads (f)
dt respectively,
Aw
itance, C:respectively,
and are itance,
generally
constants.
Nowar
resistance R: ohms (Ω)
1
q
voltage
drop
across:
second
law,
the
impressed
voltage
E(t)
on
a
closed
loop
second
law, the
voltage drop across: C
iR
andremains
equating
theimpressed
sum when
to th
It is interesting
that
(10)
valid
even
voltage
drops into
thenote
loop.
Figure
1.3.4(b)
shows
the
symb
voltage
drops
in
the
loop.
Fi
equation
case
we
must
express
the
upper
surface
area
of
the
water
respective voltage drops across
an inductor,
a capacitor,
respective
voltage
drops acra
Aw !
A(h). See
14 on
in Exercises
1.3. i ! dq!d
ii
i(t)
is related
to Problem
charge q(t)
theiscapacitor
i(t)
related toby
charge q(t)
C
R
Overview
Inductor
Linear Models
Nonlinear
Models
Page
25
inductance
Summary
resistor
inducto
Series
Circuits inductor
Consider the single-loop
LRC-se
(b)
di an inductor,
d 2q
王奕翔
DE Lecture
3
ure
1.3.4(a),
containing
resistor, dq
and di
capacit
and so after multiplying
by the factor the equation is cast into the fo
Growth and Decay
Overview
Linear Models
Nonlinear Models
Summary
250
Cooling and Warming
Mixtures
Series Circuit
d
(300 " t)2 A ! 6(300 " t)2.
dt
[
]
Integrating the last equation gives (300 " t) A ! 2(300 " t)
Example: 50LR Series
t Circuit
100
FIGURE 3.1.6 Graph of A(t) in
Example 6
Example
L
E
R
FIGURE 3.1.7 LR-series circuit
2
3
" c.
initial condition A(0) ! 50 and solving for A yields the solution A(t
(4.95 % 10 7)(300 " t) #2. As Figure 3.1.6 shows, not unexpectedly,
the tank over time, that is, A : $ as t : $.
Series Circuits For a series circuit containing only a resistor
Kirchhoff’s second law states that the sum of the voltage drop acr
(L(di!dt)) and the voltage drop across the resistor (iR) is the same a
Consider
the LR circuit shown on the left,
voltage (E(t)) on the circuit. See Figure 3.1.7.
where
E(t)
= 10 the
volts,
= 10 ohms,
L=
Thus
we obtain
linearRdifferential
equation
for0.5
the current i(
henry, and initial current i(0)di = 0.
L " Ri ! E(t),
Find i(t).
dt
where L and R are constants known as the inductance and the resistan
The current i(t) is also called the response of the system.
A: Current i(t) satisfies the following DE:
L
di
1 di
di
+ Ri = E(t) =⇒
+ 10i = 10 =⇒
+ 20i = 20
dt
2 dt
dt
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
Example: LR Series Circuit
Solve
di
+ 20i = 20, i(0) = 0.
dt
1
Derive Auxiliary DE: Let µ(t) := the integrating factor to be found.
{
}
d (µi)
di
dµ
dµ
dµ
=µ +i
= 20µ {1 − i} + i
= 20µ +
− 20µ i
dt
dt
dt
dt
dt
2
Solve Auxiliary DE:
3
dµ
= 20µ. One solution: µ(t) = e20t .
dt
Solve Original DE: Plug in µ(t) = e20t and i(0) = 0:
∫ t
e20t i − 0 =
20e20τ dτ =⇒ e20t i(t) = e20t − 1 =⇒ i(t) = 1 − e−20t
0
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Population Dynamics and Logistic Equation
Chemical Reactions
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Population Dynamics and Logistic Equation
Chemical Reactions
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。
dP
= kP
dt
Points that are missing:
Limited resources: population ↑, competition ↑ (nonlinearity kicks
in!)
Emigration and immigration
Death, etc.
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Population Dynamics and Logistic Equation
Chemical Reactions
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Population Dynamics and Logistic Equation
Chemical Reactions
Population Dynamics
Limited resources: population ↑, competition ↑.
=⇒ the constant k may vary with P !
Hence, in general the population dynamics should be governed by
dP
= Pf(P).
dt
王奕翔
DE Lecture 3
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Nonlinear Models
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Population Dynamics and Logistic Equation
Chemical Reactions
Logsitic Equation
dP
= Pf(P).
dt
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)
dP
= P(a − bP)
dt
王奕翔
DE Lecture 3
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Nonlinear Models
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Chemical Reactions
Solving Logistic Equation
dP
= P(a − bP).
dt
A: We can solve this by separation of variables.
Solve
1
a
dP = dt, P ̸= 0,
P(a − bP)
b
}
{
b/a
1/a
+
dP = dt
=⇒
P
a − bP
1
1
=⇒ ln |P| − ln |a − bP| = t + c′
a
a
P
= ceat , why we pick 0 < P < a/b?
=⇒
a − bP
aceat
ac
= −at
=⇒ P(t) =
at
1 + bce
e
+ bc
Note: if P(0) = ab , P(t) =
a
b
for all t.
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
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Chemical Reactions
Solution Curves of Logistic Equations
P
The solution is called Logistic Function: P(t) =
92467_03_ch03_p083-115.qxd
2/10/12 2:39
Properties of Logistic
Function
PM
Page 97
P
a
2b
a
b.
a/b
P0
t
王奕翔
The dashed l
point of infle
Product Rule:
d 2P
dt2
t
3.2
(a)
a/2b
(a)
a/2b
P0
P(t): increasing function. Population saturation:
P(t) → ab as t → ∞; P(t) → 0 as t → −∞.
Saddle point: P =
a/b
ac
e−at +bc
From calculu
tion, but P !
ordi
The dashed Pline P ! a !2b a/b
shown inpossible
Figure 3.2.2
0 " PTo"show
a!2
point of inflection of the logistic curve.
Thus, as we r
Product Rule:
P0
down dP
at thed
d 2P
dPa/2b
! P &b
% (a &0 bP)
" P0 "!a !2
2
dt
dt
dt
d
Figure 3.2.2(a
P
inflection!
occ
We have
t
dx!dt ! !
kx(n
2b
model for des
(b)
DEFrom
Lecture calculus
3
ing and 2infecte
recall that the points where
P!dt 2
"
#
Overview
Linear Models
Nonlinear Models
Summary
Population Dynamics and Logistic Equation
Chemical Reactions
Variants of Logistic Equations
dP
= P(a − bP) ± h
dt
dP
= P(a − bP) + cekP
dt
dP
= P(a − b ln P)
dt
immigration or emigration
西瓜偎大邊
Gompertz DE
Exercise
Derive the population saturation point in each of the above models.
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Population Dynamics and Logistic Equation
Chemical Reactions
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
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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:
dX
= k(a − λa X)(b − λb X).
dt
How to solve? Separation of variables!
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
1 Overview
2 Linear Models
Growth and Decay
Cooling and Warming
Mixtures
Series Circuit
3 Nonlinear Models
Population Dynamics and Logistic Equation
Chemical Reactions
4 Summary
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Short Recap
Growth and decay
Series circuit
Population dynamics
Logistic equation
Chemical reaction
王奕翔
DE Lecture 3
Overview
Linear Models
Nonlinear Models
Summary
Self-Practice Exercises
3-1: 3, 19, 27, 33, 35, 39, 43, 45
3-2: 5, 7, 9, 11, 15, 19, 21
王奕翔
DE Lecture 3