ordinary differential equations
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Ordinary Differential Equations. Jyun-Ming Chen. Review Euler’s method 2 nd order methods Midpoint Heun’s Runge-Kutta Method. Systems of ODE Stability Issue. Contents. Review. DE (Differential Equation) - PowerPoint PPT PresentationTRANSCRIPT
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Ordinary Differential Equations
Jyun-Ming Chen
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Contents
• Review• Euler’s method• 2nd order methods
– Midpoint
– Heun’s
• Runge-Kutta Method
• Systems of ODE• Stability Issue
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Review
• DE (Differential Equation)– An equation specifying the relations among the
rate change (derivatives) of variables
• ODE (Ordinary DE) vs. PDE (Partial DE)– The number of independent variables involved
)( 2
2
txxdt
xd ),( 1 yxf
y
f
x
f
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Review (cont)
• Solution of an equation:
• Solution of DE vs. Solution of Equation
0)1(1 :Solution
01)( 2
fx
xxf
• Geometrically,
f(x)
x
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Review (cont)
• Solution of an differential equation:
• Geometrically:
solutions validall are
,2,
is,That
:Solution
tt
ttt
exex
ecececx
xdt
dx
t
x
Need additional conditions to
specify a solution
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Review (cont)
• Order of an ODE– The highest derivative in the equation
• nth order ODE requires n conditions to specify the solution– IVP (initial value problem): All conditions
specified at the same (initial) point– BVP (boundary value problem): otherwise
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IVP VS. BVPRevisit Shooting Problem
t
g
y
(2) 5
(1) 10
10
)10(let
)(
212
1
2
2
ctcty
ctdt
dydt
yd
ggy
tfy
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8
IVP vs. BVP
ttty
c
c
ty
ty
55)(
5
0
5)0(
0)0(
:IVP
2
1
2
tty
ccy
cy
ty
ty
605
6010010500)10(
0)0(
100)10(
0)0(
:BVP
2
11
2
Physical meaning
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Review (cont)
• Linearity:– No product nor nonlinear functions of y and its
derivatives
• nth order linear ODE
)()()()()( 011
1
1 xbyxadx
dyxa
dx
ydxa
dx
ydxa
n
n
nn
n
n
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Focus of This Chapter
• Solve IVP of nth order ODE numerically
• e.g.,
0 ),( Solve
0)0(
1)0( conditions initial ,
2
2
xxy
yx
dx
yd
dxdy
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ODE (IVP)
• First order ODE (canonical form)
• Every nth order ODE can be converted to n first order ODEs in the following method:
0)0(),,( yxyyxfdx
dy
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)()()()()( 011
1
1 xbyxadx
dyxa
dx
ydxa
dx
ydxa
n
n
nn
n
n
1
11
2
22
3
12
1
)(
)(
)(
)()(
n
nn
n dx
yd
dx
dyxy
dx
yd
dx
dyxy
dx
dy
dx
dyxy
xyxy
1
11
2
22
3
12
1
)(
)(
)(
)()(
n
nn
n dx
yd
dx
dyxy
dx
yd
dx
dyxy
dx
dy
dx
dyxy
xyxy
)()()()()( 10211
1
32
21
xbyxayxayxadx
dyxa
ydx
dy
ydx
dy
ydx
dy
nnn
n
nn
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Example
xdx
dyx
dx
yd
dx
yd52 2
2
2
3
3
2
22
1
3
2
1
Let
dx
yddxdy
dxdy
dxdy
y
y
yy
21
32
22
33 52
ydx
dy
ydx
dy
xyxydx
dy
xyyx
y
y
yfy
y
y
y
y
52
)( ,Let
322
3
2
3
2
1
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End of Review
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xyxftytty
xyxfyxyxxy
xyxfy
yxfdx
dy
x
y
yxyyxfdx
dy
),()()(
),()()(
),(
),(
)(),,( 00
The Canonical Problem
dx
dy
x
yx
,0
This is Euler’s method
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Example
1)0(, yydx
dy
719.0081.081.01.0)81.0()2.0()3.0(
81.009.09.01.0)9.0()1.0()2.0(
9.01.011.0)1()0()1.0(
1.0 Choose
1)1,0(,),(
xx
xx
xx
t
fyyxf
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Example (cont)
1)0(
ln
:SolutionExact
x
x
ey
cyecy
cxy
dxy
dy
1.0
1
x
y
y=e–x
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Error Analysis(Geometric Interpretation)
xyxfyy
xyxfyyy
xyxfy
yxfdx
dy
x
y
yxyyxfdx
dy
),(
),(
),(
),(
)(),,(
0001
0001
00
If the true solution were a straight line, thenEuler is exact
Think in terms of Taylor’s
expansion
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Error Analysis(From Taylor’s Expansion)
2
62
2
32
),(),()(
)()()()()(x
dxd
xx
yxfxyxfxy
xyxyxxyxyxxy
Euler’sEuler’s truncation errorO(x2) per step
1st order method
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Cumulative Error
x
y
x = 0 x = T
Number of steps = T/xCumulative Err. = (T/x) O(x2) = O(x)
Remark:x Error But computation time
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Example (Euler’s)
xexyyydx
dy )(:exact 1, 0
)(9.0)1)(()(
1.0Let
)()()(
)(),(
xyxxyxxy
x
xxyxyxxy
xyyxf
,...729.0)3.0(,81.0)2.0(,9.0)1.0( yyy
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Methods to Improve Euler
Motivated by Geometric Interpretation
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Midpoint Method
xxyxfxyxxy
xyxfxyxyxx
xx
))(,()()(
))(,()()(
22
22
xxyxfxyxxy
xyxfxyxyxx
xx
))(,()()(
))(,()()(
22
22
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Example (Midpoint)
xxyxy
xxyxfxyxxy
xyxyxy
yxfxyxy
yyxf
x
xx
xx
xx
2
22
22
22
1)( )(
))(,()()(
1)()()(
),()()(
),(
)(905.0)1.0)(05.01)(()()1.0(
1.0Let
1)()()( 2
xyxyxyxy
x
xxyxyxxy x
xexyyydx
dy )(:exact 1, 0
,...741.0)3.0(,819.0)2.0(,905.0)1.0( yyy
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Heun’s Method
xxxyxxfyxfxyxxy
xxyxfxyxxy
))(,(),()()(
)(Predictor ))(,()()(0
21
0
xxxyxxfyxfxyxxy
xxyxfxyxxy
))(,(),()()(
)(Predictor ))(,()()(0
21
0
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Example (Heun’s)
)(905.01.09.011)()1.0(
1.0Let
1)()()(
)()( )(
))(,(),()()(
1)()()(
),()()(
),(
21
21
021
021
0
xyxyxy
x
xxxyxyxy
xxxyxyxy
xxxyxxfyxfxyxxy
xxyxxyxy
xyxfxyxxy
yyxf
xexyyydx
dy )(:exact 1, 0
,...741.0)3.0(,819.0)2.0(,905.0)1.0( yyy
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Remark
• Comparison of Euler, Heun, midpoint– 1st order: Euler
– 2nd order: Heun, midpoint
• “order”:• All are special cases of
RK (Runge-Kutta) methods
ExactEuler (error)
Midpoint (error)
y(0.1) 0.9050.9 (0.0
05)0.905
(0)
y(0.2) 0.8190.81 (0.
009)0.819
(0)
y(0.3) 0.7410.729 (0.
012)0.741
(0)
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RK Methods
)1 to from slope (estimate
function lincrementa :),,(
)( ),(
where
),,(
1
1
ii
hyx
hxyyxyy
hhyxyy
ii
iiii
iiii
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RK Methods (cont)
expansion sTaylor'match chosen to are ),,( Constants
constants:,
),(
),(
),(
),(
constant:'
11,111,1
22212123
11112
1
2211
ijii
iji
nnnniiin
ii
ii
ii
nn
qpa
qp
nkqnkqyhpxfk
hkqhkqyhpxfk
hkqyhpxfk
yxfk
sa
kakaka
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Taylor’s Expansion
fy
f
x
f
yy
f
x
f
dx
dy
y
f
dx
dx
x
fyxf
dx
d
'),(
)(][
)(][),()(
)()()()()(
32
32
32
2
2
2
hOfhfy
hOfhyxfxy
hOxyhxyxyhxy
hiiy
f
ixf
ii
hiiy
f
ixf
iii
hidx
diii
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RK 1st Order
method sEuler' ),(
1
expantion sTaylor' with Compare
),(
),(
1
1
1
11
111
hyxfyy
a
hyxfayy
yxfaka
n
iiii
iiii
ii
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RK 2nd Order
)(
)(][
)()()(),(
: Expand
22112
21221
211112111
211112
2
hOhfqahpahfahfay
hOhhkqhpfahkayy
hOhkqy
fhp
x
fyxfk
k
iyf
xf
iii
yf
xf
iii
iiii
hkakayy
hkqyhpxfk
yxfk
kakan
ii
ii
i
)(
),(
),(
2
22111
11112
11
2211
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RK 2nd Order (cont)
unknowns 4
eqns, 3
:
:
1 :
21
1122
21
122
21
qahf
pah
aahf
iyfxfi
21
1121
112 , ,01 qpaa
hyxfyhxfyy
hhkqyhpxfyhky
hkakayy
iiiiii
iiii
ii
)),(,(
),(
)(
21
21
1
11112
22111
11, , 11121
121
2 qpaa
),(),,(
)( )(
21
2121
22111
hfyhxfkyxfk
hkkyhkakayy
iiiii
iii
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RK 4th Order
• Mostly commonly used one
• Higher order … more evaluation, but less gain on accuracy
)(
)(
)(
)(
where
)22(
34
221
21
3
121
21
2
1
432161
1
hkh,yxfk
hkh,yxfk
hkh,yxfk
,yxfk
hkkkkyy
nn
nn
nn
nn
nn
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System of ODE
• Convert higher order ODE to 1st order ODEs
• All methods equally apply, in vector form
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Example (Mass-Spring-Damper System)
• Governing Equation
• After setting the initial conditions x(0) and x’(0), compute the position and velocity of the mass for any t > 0
0 kxxcxm
c
k
Initial Condition
x
m
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0)0(
1)0(:IC
)(10
then
,
2
1
212
21
122
21
x
x
cxkxm
x
xxkxcxxm
xxxx
Example (cont)
0)0(,),()()(
Euler Dim
yyxyxfxyxxy
n
0)0(,),()()(
Euler Dim
yyxyxfxyxxy
n
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set t=0.1
27.0
971.01.0
)19.0(99.0
19.0
19.0
99.01.0)2.0),2.0(()2.0()3.0(
19.0
99.01.0
)1.0(1
1.0
1.0
11.0)1.0),1.0(()1.0()2.0(
1.0
11.0
01
0
0
11.0)0.0),0.0(()0.0()1.0(
xfxx
xfxx
xfxx
Example (cont)
)(),(
211
2
2
1
cxkx
xtxf
x
xx
m
Assume m=1,c=1, k=1 (for ease of computation)
21
2
xx
xf
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Stability: Symptom
xexy
yydx
dy
10)( :solutionexact
1)0(,10
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Stability (cont)• Example Problem:
xexy
yydx
dy
)( :solutionexact
0,1)0(,
xexy
yydx
dy
)( :solutionexact
0,1)0(,
x
iiii
exy
h
h
hh
h
h
yhhyyy
)( :solutionexact
2limit stability
2
1 1or ,1 1
is,That
11 if unstable
11 if stable
)1(
:Euler
*
1
Conditionally stable
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Discussion
• Different algorithm different stability limit– Check Midpoint Method
• Different problem different stability limit– use the previous problem as benchmark
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Implicit Method (Backward Euler)
)(),(
)(),(
),()(
)1,(:Euler Backward
2111
21
1
hOyhyxfy
hOyhyxfy
yxfhOh
yy
ii
iiiii
iiii
iiii
)(),(
),()(
),1( :Euler
),(
21
1
hOhyxfyy
yxfhOh
yy
ii
yxf
iiii
iiii
dxdy
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Example
• Remark:– Always stable (for this problem)– Truncation error the same as Euler (only
improve the stability)
ihi
ii
iiiiii
yy
hOyyh
hOyhyyhyxfy
yyxf
11
1
21
211111
)()1(
)(),(
),(
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Linear System of ODE with Constant Coefficients
ii
ii
iii
yhCIy
hOyyhCI
hOhCyyy
Cyy
11
1
21
)(
)(
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Semi-Implicit Euler
• Not guaranteed to be stable, but usually is
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c
k
Initial Condition
x
m
0
:Motion ofEquation
kxxcxm
xmkxxcxmf
nn yhh
hyyy
xxx
xx
xx
xx
kcm
1
1
212
21
2
1
1
1
11
10
1Let
xc
force damping
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c
k
Initial Condition
x
m
0
:Motion ofEquation
2
2
kxxcxm
xmkxxcxmf
221
2
1
21
2
2212
21
2
1
21
1
21
10
)(
1Let
xx
x
hxh
hhyy
xy
f
yfy
xxx
xx
xx
xx
kcm
nn
2
force damping
xc
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Stiff Set of ODE
Stability limit
Use the change of variable
101)0(
121)0(
1000210002
2
2121
21
100021
ccccv
ccu
eczecy
zvuzvu
yvuyvu
zyvzyu
xx
101)0(
121)0(
1000210002
2
2121
21
100021
ccccv
ccu
eczecy
zvuzvu
yvuyvu
zyvzyu
xxGet the following solution:
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Adaptive Stepsize
• Solving ODE numerically … tracing the integral curve y(x)
• what’s wrong with uniform step size– Uniformly small: waste
effort
– Uniformly large: might miss details
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Step Doubling
• Idea:– Estimate the truncation error by taking each
step twice: one full step, two half steps– control the step size such that the estimated
error is not too big.
1(2h1)
2(h1)Desired h0
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52
Ex: RK 2nd Order
00
11
11
00
3
1
0
1
0
3312
432
431
432
431
error desired theproduce tostepsize the:
stepby error estimated the:
error)n truncatio(the 6
)()(2)()2(
)()(2)2(
)()2()2(
31
h
h
hhh
h
hchyy
hOhcyhOhcy
hOhcyhxy
hOhcyhxy
Overhead: # of f(x,y) evaluations24–2 = 6
Overhead: # of f(x,y) evaluations24–2 = 6
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53
Example
xxy
yyy
cos)( :Exact
1)0(,sin
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54
Ex (Semi-implicit Euler)
g
ty
h
hh
ty
ty
hty
hty
fh
hyfh
hyy
yy
ff
y
fJ
yygy
yyyygy
nnnnn
)(
0)(
)(
)(
)(
10
1
10
1
00
10
,
,
3)0(,0)0(,3)0(,0)0(,
22
2
1
2
1
1
1
21
21
212
21
hg
ty
ty
ty
hty
hty
hfyy
Euler
nnn
)(
)(
)(
)(
)( 2
2
1
2
1
1
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55
nnn hfyhJIyhJI
vu
ff
y
fJ
1
21
1999999
1998998
,
,