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
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
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
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
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),(
Logistics Equation
)()(1
1)(2
1 txtxq
qtxdt
d
xx
qqqqpf
2121
11),,(
),),(()( 21 qqtxftxdt
d
tqexqx
xqqqtx
1020
0221 ),,(
),,( 211
qqtxq
),,( 212
qqtxq
Computing Sensitivities
),,,,(),,,( 2122
2
mgxxfxx
mg
xmgxf
HOW DO WE COMPUTE THE SENSITIVITIES …
),,,(
),,,(),,,(
2
1
mgtx
mgtxmgtx
),,,(
),,,(),,,(
2
1
mgtx
mgtxmgtx
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
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)(
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?
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
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
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
Explicit Euler for SEQs
t0
(x0 ,s0)t
kt1t 2t 1kt
ihttth i 0 ,
),(1 qxfhxx kkk
R2
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
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?
MATLAB Code for SEM
Set q
Set x0 and s0
Set h
Time interval
Set ICs
Explicit Euler
DE Solution x(t)
qteqtxtx 5),()(
05.h
01.h
1.h
SE Solution s(t)
qtteqtsts 5),()(
05.h
01.h
1.h
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)
Logistics Equation
)()(1
1)(2
1 txtxq
qtxdt
d
xx
qqqqpf
2121
11),,(
),),(()( 21 qqtxftxdt
d
tqexqx
xqqqtx
1020
0221 ),,(
),,( 211
qqtxq
),,( 212
qqtxq
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
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
SEQ for the Logistics Equation
),,(),,( 21211
1 qqtxqqtxq
q
),,(),,( 21211
1 qqtxqqtxq
q
22121
1211
2
)],,([),,()],,([21
qqtxqqtxq
qqtxqq
2211
12
)],,([1
qqtxqqq
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
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
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
SEQ for the Logistics Equation
NEED SENSITIVITY WITH RESPECT TO q2
)(1 tsdt
d
2
21
21 )]([
1)()()]([
21 tx
qtxtstx
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
Convergence Issues
N=16, M=32
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
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
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
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)
END
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