computational methods for design lecture 4 – introduction to sensitivities john a. burns c enter...
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Computational Methods for Design Lecture 4 – Introduction to Sensitivities
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
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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
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System of Differential Equations
)()(
)(
)(
)(tvtv
mg
tv
tv
ty
dt
d
)()()( tvtvm
gtv
)()( tvty
0)0(
000,10)0(
v
y
)()(
)(),,),(()(
22
2
txtxm
g
txmgtxftx
dt
d
)(
)()(
2
1
tx
txtx
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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
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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),(
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Logistics Equation
)()(1
1)(2
1 txtxq
qtxdt
d
xx
qqqqpf
2121
11),,(
),),(()( 21 qqtxftxdt
d
tqexqx
xqqqtx
1020
0221 ),,(
),,( 211
qqtxq
),,( 212
qqtxq
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Computing Sensitivities
),,,,(),,,( 2122
2
mgxxfxx
mg
xmgxf
HOW DO WE COMPUTE THE SENSITIVITIES …
),,,(
),,,(),,,(
2
1
mgtx
mgtxmgtx
),,,(
),,,(),,,(
2
1
mgtx
mgtxmgtx
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SEM Example 1
5)0( ),()( xtqxtxdt
d
DIFFERENTIATE
qteqtxtx 5),()(
qtteqtxq
5),(
qtqtqt eteqett
qtxqt
qtsdt
d
555),(),(
qtdefine
teqtxq
qts 5),(),(
),(),(55),( qtxqtsqeteqqtsdt
d qtqt
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SEM Example 1
),(),( qtqxqtxdt
d
),(),(55),( qtxqtsqeteqqtsdt
d qtqt
),(),(),( qtxqtsqqtsdt
d
5)0( x
0)0( s
)()( tqxtxdt
d 5)0( x
)()()( txtsqtsdt
d 0)0( s
qttets 5)(
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SEM Method
SOLVE THE SYSTEM (DE) – (SE)
)(),( tsqtxq
(DE)
(SE)
)()( tqxtxdt
d 5)0( x
)()()( txtsqtsdt
d 0)0( s
1. WHY DO IT THIS WAY ?2. WE DERIVED (SE) BY USING THE KNOWN
SOLUTION …HOW DO WE FIND (SE) IN GENERAL?
3. HOW GENERAL IS THIS PROCESS?
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Derivation of SEN Eq
5)0( x)()( tqxtxdt
d
),()( qtxtx
),(),( qtqxqtxdt
d 5),0( qx
DIFFERENTIATE THE EQUATION WITH RESPECT TO q
),(),( qtqxq
qtxdt
d
q
),(),( qtxqtxq
q
),(),( qtxqdt
dqtx
dt
d
q),(),( qtxqtx
),( qts ),( qts
INTERCHANGE THE ORDER OF DIFFERENTIATION
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Derivation of SEN Eq
),(),(),( qtxqtsqqtsdt
d
0),0( qs
q
(DE)
(SE)
)()( tqxtxdt
d 5)0( x
)()()( txtsqtsdt
d 0)0( s
5),0( qx 05),0(
qqx
q
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SEM Method
(DE)
(SE)
)()( tqxtxdt
d 5)0( x
)()()( txtsqtsdt
d 0)0( s
)()(
)(
)(
)()(
12
1
2
1
txtqx
tqx
tx
tx
dt
dtx
dt
d
)()( ),()( 21 tstxtxtx
)(
)()(
2
1
tx
txtx
0
5)0(x
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Explicit Euler for SEQs
t0
(x0 ,s0)t
kt1t 2t 1kt
ihttth i 0 ,
),(1 qxfhxx kkk
R2
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Explicit Euler for SEQs
)),(()()(
)(
)(
)()(
12
1
2
1 qtxftxtqx
tqx
tx
tx
dt
dtx
dt
d
),(1 qxfhxx kkk
kkkxxq
xqh
x
x
x
x
12
1
2
1
12
1
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Explicit Euler for SEQs
)][][(][][
][][][
12112
1111
kkkk
kkk
xxqhxx
xqhxx
kkkxxq
xqh
x
x
x
x
12
1
2
1
12
1
SOLVE BOTH DE AND SE TOGETHER
HOW DOES IT WORK?
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MATLAB Code for SEM
Set q
Set x0 and s0
Set h
Time interval
Set ICs
Explicit Euler
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DE Solution x(t)
qteqtxtx 5),()(
05.h
01.h
1.h
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SE Solution s(t)
qtteqtsts 5),()(
05.h
01.h
1.h
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Special Structure of SE’s
(DE)
(SE)
)()( tqxtxdt
d 5)0( x
)()()( txtsqtsdt
d 0)0( s
(DE) )()( tqxtxdt
d 5)0( x
(SE) )()()( txtsqtsdt
d 0)0( s
FIRST: SOLVE (DE)qtetx 5)(
qte5
SECOND: SOLVE (SE)
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Logistics Equation
)()(1
1)(2
1 txtxq
qtxdt
d
xx
qqqqpf
2121
11),,(
),),(()( 21 qqtxftxdt
d
tqexqx
xqqqtx
1020
0221 ),,(
),,( 211
qqtxq
),,( 212
qqtxq
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SEQ for the Logistics Equation
DIFFERENTIATE THE EQUATION WITH RESPECT TO q1
),,(),,(1
1),,( 21212
121 qqtxqqtxq
qqqtxdt
d
1q
),,(),,(1
1),,( 21212
11
211
qqtxqqtxq
qqtxdt
d
q
),,(),,(1
1),,( 21212
121 qqtxqqtxq
qqqtxdt
d
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SEQ for the Logistics Equation
),,(),,(1
1),,( 21212
11
211
qqtxqqtxq
qqtxdt
d
q
221
2
1211
1
)],,([),,( qqtxq
qqqtxq
q
221
2
1
1211
1
)],,([),,( qqtxq
q
qqqtxq
q
),,(),,( 21211
1 qqtxqqtxq
q
2211
12
)],,([1
qqtxqqq
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SEQ for the Logistics Equation
),,(),,( 21211
1 qqtxqqtxq
q
),,(),,( 21211
1 qqtxqqtxq
q
22121
1211
2
)],,([),,()],,([21
qqtxqqtxq
qqtxqq
2211
12
)],,([1
qqtxqqq
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SEQ for the Logistics Equation
),,( 211
qqtxdt
d
q
),,(),,( 21211
1 qqtxqqtxq
q
),,( 211
qqtxqdt
d
),,(),,( 21211
1 qqtxqqtxq
q
INTERCHANGE THE ORDER OF DIFFERENTIATION
22121
1211
2
)],,([),,()],,([21
qqtxqqtxq
qqtxqq
22121
1211
2
)],,([),,()],,([21
qqtxqqtxq
qqtxqq
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SEQ for the Logistics Equation
),,( 211
qqtxqdt
d
),,(),,( 21211
1 qqtxqqtxq
q
),,()],,([2)],,([1
211
2112
212
qqtxq
qqtxqqqtxq
),,(),,( 211
211 qqtxq
qqtsdefine
),,( 211 qqts ),,( 211 qqts
),,( 211 qqts
),,( 211 qqtsdt
d ),,(),,( 212111 qqtxqqtsq
)),,()],,([2)],,(([1
2112112
212
qqtsqqtxqqqtxq
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SEQ for the Logistics Equation
),,( 211 qqtsdt
d ),,(),,( 212111 qqtxqqtsq
)),,()],,([2)],,(([1
2112112
212
qqtsqqtxqqqtxq
)(1 tsdt
d )()(11 txtsq
))()]([2)](([1
112
2
tstxqtxq
)(1 tsdt
d
2
21
21 )]([
1)()()]([
21 tx
qtxtstx
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SEQ for the Logistics Equation
NEED SENSITIVITY WITH RESPECT TO q2
)(1 tsdt
d
2
21
21 )]([
1)()()]([
21 tx
qtxtstx
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SEQ for the Logistics Equation 2
DIFFERENTIATE THE EQUATION WITH RESPECT TO q2
),,(),,(1
1),,( 21212
121 qqtxqqtxq
qqqtxdt
d
2q
),,(),,(1
1),,( 21212
12
212
qqtxqqtxq
qqtxdt
d
q
),,(),,(1
1),,( 21212
121 qqtxqqtxq
qqqtxdt
d
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SEQ for the Logistics Equation 2
),,(),,(1
1),,( 21212
12
212
qqtxqqtxq
qqtxdt
d
q
221
2
1211
2
)],,([),,( qqtxq
qqqtxq
q
221
2
1
2211
2
)],,([),,( qqtxq
q
qqqtxq
q
),,( 212
1 qqtxq
q
221
22
2212
21 )],,([
1)],,([
)(
1qqtx
qqqqtx
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SEQ for the Logistics Equation 2
),,( 212
1 qqtxq
q
),,()],,([21
)],,([)(
121
221
2
2212
21 qqtx
qqqtx
qqqtx
),,( 212
1 qqtxq
q
221
22
2212
21 )],,([
1)],,([
)(
1qqtx
qqqqtx
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SEQ for the Logistics Equation 2
),,( 212
1 qqtxq
q
2
2122
1 )],,([)(
1qqtx
),,()],,([2 212
212
1 qqtxq
qqtxq
q
),,( 212
qqtxdt
d
q
INTERCHANGE THE ORDER OF DIFFERENTIATION
),,(),,( 212
212 qqtxq
qqtsdefine
),,( 212
qqtxqdt
d ),,( 212 qqts ),,( 212 qqts
),,( 212 qqts
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SEQ for the Logistics Equation 2
)(2 tsdt
d)(21 tsq
2
22
1 )]([)(
1tx
qq )()]([2 2
2
1 tstxq
q
)(2 tsdt
d)()](
21[ 2
21 tstx
2
22
1 )]([)(
1tx
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SEQ’s for the Logistics Equation
FROM THE FIRST PARTIAL
)()(1
1)(2
1 txtxq
qtxdt
d
THE LOGISTICS EQUATION
0)0( xx
0)0(1 s
0)0(2 s
)(2 tsdt
d)()](
21[ 2
21 tstx
2
22
1 )]([)(
1tx
)(2 tsdt
d)()](
21[ 2
21 tstx
2
22
1 )]([)(
1tx
)(1 tsdt
d
2
21
21 )]([
1)()()]([
21 tx
qtxtstx
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SEQ’s for the Logistics Equation
)()(1
1)(2
1 txtxq
qtxdt
d
0)0( xx
FIRST: SOLVE (DE) )(tx
SECOND: SOLVE (SEs)
)(2 tsdt
d)()](
21[ 2
21 tstx
2
22
1 )]([)(
1tx
)(1 tsdt
d
2
21
21 )]([
1)()()]([
21 tx
qtxtstx
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Model Problem #1
dx)(w(x,q
(x),w)-w(x,)(dqd
1
0) ˆ
J qqq
SENSITIVITY
The sensitivity equation for s(x, q ) = q w(x , q) in the“physical” domain (q) = (0,q) is given by
Can be made “rigorous” by the method of mappings.MORE ABOUT THIS NEAR THE END
x) ,w(xq
),s(xs(x) 0
qq q
q ,
q x
)x(w)(s,)(s dxd | 00
q
x xsdxd,xw
dxdxs
dxd 0 ,0)(
2)(
83+)(
2
2q
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Typical Cost Function
WHERE w( x , q ) USUALLY SATISFIES A DIFFERENTIAL EQUATION AND q IS A PARAMETER (OR VECTOR OF PARAMETERS)
1
0
22
1 |)(ˆ) ,(|) ), ,((=) ( dxxwxwwFJ qqqq
1
0
]) ,([])(ˆ) ,([) ), ,((=) ( dxxwxwxww qdqd
dqd FJ q q q q q
1
0
)] ,( [),(ˆ) ,( ) ), ,( (=)] ( [ dxxwxwxww qdqd
dqd FJ q q q q q
THE CHAIN RULE PRODUCES
OR (Reality) USING NUMERICAL SOLUTIONS
hh h h
CONTINUOUSSENSITIVITY
DISCRETESENSITIVITY
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Computing Gradients
(I) BY FINITE DIFFERENCES
( )- ) (
)( qd
d JJJ
qq0 q0
q0
q
hh
h
TYPICAL APPROACHES TO COMPUTE
)(dqd J q
q =q0
h
(II) BY DISCRETE SENSITIVITIES
1
0
)] ,( [),(ˆ) ,( ) ), ,( (=)] ( [ dxxwxwxww qdqd
dqd FJ q0
q0q0q0 q0hh h h
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Computing Gradients
FINITE DIFFERENCES
• REQUIRES 2 NON-LINEAR SOLVES
• IF SHAPE IS A DESIGN VARIABLE, FD REQUIRES 2 MESH GENERATIONS
DISCRETE SENSITIVITIES
• REQUIRES THE EXISTENCE OF THE DISCRETE SENSITIVITY
• IF SHAPE IS A DESIGN VARIABLE, THE DISCRETE SENSITIVITY LEADS TO MESH DERIVATIVES COMPUTATIONS
WHAT IS THE “CONTINUOUS / HYBRID”SENSITIVITY EQUATION METHOD? --- SEM
1
0
)] ,( [),(ˆ) ,( )] ( [ dxxwxwxw qdqd J q0
q0q0hh h, k
APPROXIMATE
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A Sensitivity Equation Method
FOR q > 1 AND h=q/(N+1) CONSIDER (FORMAL)
h h hh h
NUMERICAL APPROXIMATION
x=0 x=1 x=q
x
w(x)
w h(x) = Finite Element Approximation
4 00 )(w,)(w qq ,
x,xwdxdxw
dxd 0 0
3)(
81+)(
22
DISCRETE STATE EQUATION
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A Sensitivity Equation Method
h h
h h hh h 4 00 )(w,)(w qq ,
x,xwdxdxw
dxd 0 0
3)(
81+)(
22
q
x)x(w)(s,)(s dx
d | 00q
q ,
x xsdxd,xw
dxdxs
dxd 0 ,0)(
2)(
83+)(
2
2q
h
IMPORTANT OBSERVATIONS The sensitivity equations are linear The sensitivity equation “solver” can be constructed
independently of the forward solver -- SENSE™ When done correctly “mesh gradients” are not required
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A Sensitivity Equation Method
FOR q > 1 AND k = q/(M+1) CONSIDER (FORMAL)
2nd NUMERICAL APPROXIMATION
x=0 x=1 x=q
x
s(x)= qw(x,q)
s h,k(x) = Finite Element Approximation ofh
h,k
) ,(),( xwq
xsh,k
q q
q ,
x xsdxd,xw
dxdxs
dxd 0 ,0)(
2)(
83+)(
2
2q
q
x)x(w)(s,)(s dx
d | 00q
h
h
h
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Convergence Issues
),,,( )] ,([),(ˆ) ,( )] ( [1
0
xdxxwxwxwdef
qdqd GJ
q qqhhhq k
h,k
THEOREM. The finite element scheme is asymptotically consistent.
0-)] ( [00
),,,x(lim dqd GJ qq
h
hh k
k
a trust region method should (might?) converge.
),,,(-)] ( [ xdqd GJ qqh
h kWhen the error is small, thenIDEA:
J. T. Borggaard and J. A. Burns, “A PDE Sensitivity Equation Method for Optimal Aerodynamic Design”, Journal of Computational Physics, Vol.136 (1997), 366-384.
R. G. Carter, “On the Global Convergence of Trust-Region Algorithms Using Inexact Gradient Information”, SIAM J. Num. Anal., Vol 28 (1991), 251-265.J. T. Borggaard, “The Sensitivity Equation Method for Optimal Design”, Ph.D. Thesis, Virginia Tech, Blacksburg, VA, 1995.
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Convergence Issues
N=16, M=32
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Convergence Issues
),,,(-)] ( [ xdqd GJ qqh
h h
THE CASE k = h is often used, but may not be “good enough”
NOT CONVERGENT
N=M=16 Tol = 0.00001 Tol = 0.0001 Total: 378.82Iter q Grad. Norm Step Time (secs) Cost Time Grad. Time
0 1.2000 4.3998E+00 -3.6427E-02 0.1231 1 1.1636 3.1583E+03 1.4051E-03 31.3210 31.2697 0.04782 1.1650 3.0910E+03 -1.4339E-03 36.2310 36.1798 0.04803 1.1635 5.8909E+02 7.8372E-03 46.1160 46.0075 0.10434 1.1714 5.0139E+03 -8.8462E-04 45.3550 45.3006 0.05115 1.1705 2.9396E+03 -1.5052E-03 43.6720 43.6208 0.04706 1.1690 1.7238E+04 2.5880E-04 42.5810 42.5301 0.04687 1.1693 2.5888E+03 1.7342E-03 46.3470 46.2965 0.04728 1.1710 4.6995E+04 -9.4732E-05 44.1900 44.1396 0.04689 1.1709 1.5743E+02 0.0000E+00 42.8790 42.8265 0.0485
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Timing Issues
THE CASE k = 2h offers flexibility and ),,,(-)] ( [ xdqd GJ qqh
h 2h
convergence. But, what about timings?
Approximately 96 .6% of cpu time spent in function evaluationsApproximately 02 .4% of cpu time spent in gradient evaluations
N=16, M=32 Tol = 0.00001 Tol = 0.0001 Total: 39.81Iter q Grad. Norm Step Time (secs) Cost Time Grad. Time
0 1.2000 4.8489E+00 -3.2414E-02 0.1968 1 1.1676 2.0720E+01 4.0347E-01 34.9270 34.8053 0.09112 1.5711 4.4544E+00 3.7808E-01 1.2613 1.1234 0.10753 1.9491 6.9846E-02 -1.2442E-02 0.9941 0.8714 0.09254 1.9367 1.5779E-02 2.7472E-03 0.4190 0.2907 0.09385 1.9394 3.3558E-03 -5.8723E-04 0.4095 0.2845 0.09436 1.9389 7.3235E-04 1.2801E-04 0.4525 0.3135 0.10837 1.9390 1.5892E-04 -2.7785E-05 0.8602 0.6327 0.19688 1.9390 3.0703E-05 0.0000E+00 0.2914 0.1451 0.1083
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Mathematics Impacts “Practically”
UNDERSTANDING THE PROPER MATHEMATICAL FRAMEWORK CAN BE EXPLOITED TO PRODUCE BETTER SCIENTIFIC COMPUTING TOOLS
A REAL JET ENGINE WITH 20 DESIGN VARIABLES PREVIOUS ENGINEERING DESIGN METHODOLOGY
REQUIRED 8400 CPU HRS ~ 1 YEAR USING A HYBRID SEM DEVELOPED AT VA TECH AS
IMPLEMENTED BY AEROSOFT IN SENSE™ REDUCED THE DESIGN CYCLE TIME FROM ...
8400 CPU HRS ~ 1 YEAR TO 480 CPU HRS ~ 3 WEEKS
NEW MATHEMATICS WAS THE ENABLING TECHNOLOGY
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Special Structure of SE’s
(DE)
(SE)
)()( tqxtxdt
d 5)0( x
)()()( txtsqtsdt
d 0)0( s
(DE) )()( tqxtxdt
d 5)0( x
(SE) )()()( txtsqtsdt
d 0)0( s
FIRST: SOLVE (DE)qtetx 5)(
qte5
SECOND: SOLVE (SE)
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END