Post on 11-Jan-2016
Computational Methods for Design Lecture 3 – Elementary Differential
Equations
John A. Burns
Center for Optimal Design And Control
Interdisciplinary Center for Applied MathematicsVirginia Polytechnic Institute and State University
Blacksburg, Virginia 24061-0531
A Short Course in Applied Mathematics
2 February 2004 – 7 February 2004
N∞M∞T Series Two Course
Canisius College, Buffalo, NY
Today’s Topics
Lecture 3 – Elementary Differential Equations A Review of the Basics Equilibrium Stability Dependence on Parameters: Sensitivity Numerical Methods
A Falling Object
( ) ( )F t ma t“Newton’s Second Law”
. y(t)
( ) ( ) ( ) ( ) ( )g dampmy t F t F t mg y t y t
)()()( tytym
gty
)()()( tvtvm
gtv
)()( tvty
0)0(
000,10)0(
v
y
{
{
AIR RESISTANCE
Height: y(t)
)(ty
Velocity: v(t)=y’(t)
)()( tytv
System of Differential Equations
)()(
)(
)(
)(tvtv
mg
tv
tv
ty
dt
d
)()()( tvtvm
gtv
)()( tvty
0)0(
000,10)0(
v
y
)()(
)(
)(
)()(
22
2
2
1
txtxm
g
tx
tx
tx
dt
dtx
dt
d
)(
)()(
2
1
tx
txtx
State Space
)()(
)(
)(
)()(
22
2
2
1
txtxm
g
tx
tx
tx
dt
dtx
dt
d
))(),((
))(),((
)(
)(
)(
)(
212
211
2
1
2
1
txtxf
txtxf
tx
txf
tx
tx
dt
d
2
0
0
2
1
)0(
)0(R
v
y
x
x
STATE SPACE
System of Differential Equations
2
2
1
)(
)()( R
tx
txtx
2
2
121 )(
)()(),()( R
tx
txftxtxftxf
)()( txftxdt
d
THE PHYSICS – BIOLOGY – CHEMISTRY IS FINDING 21, xxf
SELECTION OF THE “CORRECT STATE SPACE” IS A COMBINATION OF PHYSICS – BIOLOGY –
CHEMISTRY AND MATHEMATICS
Parameters
),,,,(),,,( 2122
2
mgxxfxx
mg
xmgxf
IN REAL PROBLEMS THERE ARE PARAMETERS
SOLUTIONS DEPEND ON THESE PARAMETERS
),,,( mgtx
WE WILL BE INTERESTED IN COMPUTINGSENSITIVITIES WITH RESPECT TO THESE PARAMETERS
),,,(
mgtx
MORE LATER
Logistics Equation
10)0( Rpp
)()(1
1)( 0 tptpK
rtpdt
d
LE
),),(()()(1
1)( 00 KrtpftptpK
rtpdt
d
ppK
rKrpf
11),,( 00
Analytical Solution
0 5 10 150
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
K
trepKp
KpKrtp
000
00 ),,(
Initial p0: 1 < p0 < 20,000
0 5 10 15 20 250
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2x 10
4
500,9 ,9163.0 Kr
1000 p
10 p
000,150 p
100 p
000,200 p
7500 p
15000 pK
Equilibrium States
)()(1
1)( 0 tptpK
rtpdt
d
LE
EQUILIBRIUM STATES ARE CONSTANT SOLUTIONS
constant a )( eptp
Kpp ee or 0
0)( tp
0)( tp ee ppK
r
110 0
0)( tp
0)( ee ptp Kptp ee )(
Equilibrium States
0 5 10 15 20 250
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2x 10
4
Kpp ee or 0
K
0
UNSTABLE STABLE
A Falling Object
)()()( tvtvm
gtv
)()( tvty {
. y(t)
)()(0 tvtvm
g
)(0 tv
0
0
)(
)(
tv
ty
0 0
0gNO EQUILIBRIUM STATES
Terminal Velocity2 /
2 /
1( )
1
t g mmg mgt
t g m
ev t
e
220 ft/sec 150 m/hr
( ) ( )v t y t
A Falling Object
)(ty)()( tytv
terter vvm
g
0 /gmvter
)()()( tytym
gty (0) 10,000 (0) 0y y
Systems of DEs
)()(
)(
)(
)(
)(
)(
22
2
2
1
2
1
txtxm
g
tx
tx
txf
tx
tx
dt
d
n
nn
n
n
nn
R
txtxtxf
txtxtxf
txtxtxf
tx
tx
tx
f
tx
tx
tx
dt
d
))(),...(),((
))(),...(),((
))(),...(),((
)(
)(
)(
)(
)(
)(
21
212
211
2
1
2
1
MORE EQUATIONS
Epidemic Models SIR Models (Kermak – McKendrick, 1927)
Susceptible – Infected – Recovered/Removed
( ) ( ) ( )dS t S t I t
dt
( ) ( ) ( ) ( )dI t S t I t I t
dt
( ) ( )dR t I t
dt
( ) ( ) ( ) constantS t I t R t N
Epidemic Models SIR Models (Kermak – McKendrick, 1927)
Susceptible – Infected – Recovered/Removed
)()()( tItStSdt
d
)()()()( tItItStIdt
d
)()( tItRdt
d
)()(
)()(
)()(
3
2
1
tRtx
tItx
tStx
),,(
),,(
),,(
3213
3212
3211
2
221
21
3
2
1
xxxf
xxxf
xxxf
x
xxx
xx
x
x
x
f
)()()( 211 txtxtxdt
d
)()()()( 2212 txtxtxtxdt
d
)()( 23 txtxdt
d
Systems of DEs
n
nn
n
n
nn
R
txtxtxf
txtxtxf
txtxtxf
tx
tx
tx
f
tx
tx
tx
dt
d
))(),...(),((
))(),...(),((
))(),...(),((
)(
)(
)(
)(
)(
)(
21
212
211
2
1
2
1
)()( txftxdt
d
n
n
Rx
x
x
x
2
1
nn RRf :)(
Initial Value Problems
)()( txftxdt
d nRxtx 00 )(
MOST OF THE TIME WE FORGET THE ARROW
)()( txftxdt
d nRxtx 00 )(
AND f CAN DEPEND ON TIME t AND PARAMETERS q
qtxtftxdt
d),(,)( nRxtx 00 )(
Basic Results
A solution to the ordinary differential equation (Σ) is adifferentiable function
)(,)( txtftxdt
d(Σ)
nRbatx ),(:)(
defined on a connected interval (a,b) such that x(t)
satisfies (Σ) for all t (a,b).
2)()( txtxdt
d
ttx
1
1)(
1 ,1
1)(
t
ttxl
tt
txr 1 ,1
1)(
TWO SOLUTIONS
Solutions
-3 -2 -1 0 1 2 3 4-20
-15
-10
-5
0
5
10
15
20
1 ,1
1)(
t
ttxl
tt
txr 1 ,1
1)(
Initial Condition
-3 -2 -1 0 1 2 3 4-20
-15
-10
-5
0
5
10
15
20
1)0( x
1 ,1
1)(
t
ttxl
Basic Theorems
Theorem 1. Let f: Rn ---> Rn be a continuous function on a domain D Rn, and x0 D. Then there exists at least one solution to the initial value problem (IVP).
)(,)( txtftxdt
d nRxtx 00 )((IVP)
),...,(
),...,(
),...,(
),...,,(
21
212
211
21
nn
n
n
n
xxxf
xxxf
xxxf
xxxtf
),...,,( 21 nij
xxxtfx
TO GET UNIQUENESS WE NEED MORE
Basic Theorems
Theorem 2. If there is an open rectangle about (t0, x0) such that
is continuous at all points (t, x) , then there a unique solution to the initial value problem (IVP).
),...,,( 21 nij
xxxtfx
nR
t
x0
t0
)(tx
SIR Model
)()()( 211 txtxtxdt
d
)()()()( 2212 txtxtxtxdt
d
)()( 23 txtxdt
d
2
221
21
3213
3212
3211
),,(
),,(
),,(
x
xxx
xx
xxxf
xxxf
xxxf
21211 ),,,( xxxxf
12122
),,,( xxxfx
22121
),,,( xxxfx
0),,,( 2123
xxfx
12112
),,,( xxxfx
22111
),,,( xxxfx
0),,,( 2113
xxfx
221212 ),,,( xxxxxf
SIR Model
00
0
0
),,( 12
12
3,...13,..1
321
xx
xx
xxxfx
ji
ij
ALL ENTRIES ARE CONTINUOUS FOR ALL
),,,,( 321 xxx
Theorem 1. IS OK
A Falling Object
),,,,(
),,,,(),,,,(
212
211
22
2
21
mgxxf
mgxxfxx
mg
xmgxxf
2211 ),,,,( xmgxxf
0 ,)(
0 ,)(),,,,(
22
2
22
2
212
xxm
g
xxm
gmgxxf
0),,,,( 21
2111
xx
mgxxfx
1),,,,( 22
2112
xx
mgxxfx
NO PROBLEM SO FAR
A Falling Object
0 ,)(
0 ,)(),,,,(
22
2
22
2
212
xxm
g
xxm
gmgxxf
0),,,,( 2121
mgxxfx
0 ,2
0 ,2
),,,,(
22
22
2122 xx
m
xxmmgxxf
x
AGAIN … CONTINUOUS FOR ALL ),,,,( 21 mgxx
Theorem 1. IS OK
Parameter Dependence
n
mnn
mn
mn
mn R
qqqxxxf
qqqxxxf
qqqxxxf
qqqxxxf
),...,,,...,(
),...,,,...,(
),...,,,...,(
),...,,,...,(
2121
21212
21211
2121
nn RRqf :),(
),,,,(
),,,,(),,,,(
212
211
22
2
21
mgxxf
mgxxfxx
mg
xmgxxf
FOR THE FALLING OBJECT …
Examples: n=m=1
5)0( ),()( xtqxtxdt
d
qxqxf ),(
qqxfx
),(
CONTINUOUS EVERYWHEREqteqtxtx 5),()(
UNIQUE SOLUTION
1)0( ,)(
)(
xqt
txtx
dt
d
qt
xqxtf
),,(
qtqxf
x
1
),(
CONTINUOUS WHEN 0 qt
22 /)(),( qqqtqtxq
qqtqtxtx /)(),()(
UNIQUE SOLUTION
qtteqtxq
5),(
Logistic Equation
2
2121
1),,( x
qxqqqxf x
qqqqxf
x 2121
2),,(
)()(1
1)( 0 tptpk
rtpdt
d
0)0( pp
)()(1
1)(2
1 txtxq
qtxdt
d
0)0( xx
kqrqtptx 101 , ),()(
tqexqx
xqqqtx
1020
0221 ),,(
),,( 211
qqtxq
),,( 212
qqtxq
Numerical Methods
))(,()()( 1
kk
kk
txtft
txtx
FORWARD EULER
)(,)( txtftxdt
d nRxtx 00 )((IVP)
))(,()()( 1 kkkk txttftxtx
t0
x0
t
kt1t 2t 1kt
Explicit Euler
t0
x0
t
kt1t 2t 1kt
1))(,()()( 0001 xtxttftxtxdefine
2),()( 1112 xxttfxtxdefine
))(,()()( 1112 txttftxtx
1),()( 1 k
define
kk xxttfxtx kk
Explicit Euler
t0
x0
t
kt1t 2t 1kt
ihttth i 0 ,
),(1 kkkk xtfhxx
Example 1
tetx 21)( 0)0( ),()( xxtqxtx
dt
d
qtextx 0)(
Explicit Euler
h=.2
h=.01
h=.1
tetx 21)(
k
kkk
qxhx
xtfhxx
k
k
),(1
qtextx 0)( 0)0( ),()( xxtqxtxdt
d
Example 2
02 )0( ,)()( xxtxqtx
dt
d
ttx
1
1)(
Example 2
h=.2
h=.1
02 )0( ,)()( xxtxqtx
dt
d
]1[
][
),(2
1
kk
k
k
qxhx
xqhx
xtfhxx
k
kkk
Typical MATLAB m files
Eeuler_1.m Eeuler_2.m
Simple Example 3
101)( ,1
101)( ,)),((
tx
txtxf(t)x
10 )x(
0
10 ),10(101
100 ,1
tt
ttx(t)
10
1
10
Simple Example 3
10 x),(1 kkk xfhxx
hfhxfhxx 1),1(1),( 01 0
,epsh IF
10
1
PROBLEM ISFINITE PRECISION
ARITHMETIC
MESH REFINEMENTMAKES THE PROBLEM
WORSE
11... 11 hxxx kk
101 ,1
101 ,),(
x
xxf ),1(f
hx 11 1