9. approximation theory (spectral methods) curve...
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1
9. Approximation Theory (Spectral Methods)
§ Curve fitting
GIVEN: , , 1,2, ,i ix y i N
Find: a function which best approxomates
the unknown function
(not necessary passing the data points)
y f x
Approximation Theory :
STEP2: optimize the fitting in some way, i.e., minimize the error term
STEP1: choose a proper funtion from ; , , ,
with some adjustable parameters , , ,
y f x a b c
a b c
2
1
; , , , , , ,N
i ii
E y f x a b c E a b c
i.e. looking for a set of a,b,c, such that
0
0
0
EaEbEc
2
Example: Given , and find a linear curve fitting.
Assume .i ix y
y ax b
To minimize the error , , we look for , such that E a b a b
Define the derivation as
2
1
,N
i ii
E y ax b E a b
1
1
0 2
0 2 1
N
i i ii
N
i ii
Ey ax b x
a
Ey ax b
b
2
1 1 1
0N N N
i i i ii i i
x y a x b x
1 1 1
0 1N N N
i ii i i
y a x b
1 1 12
2
1 1
N N N
i i i ii i i
N N
i ii i
N x y x ya
N x x
2
1 1 1 12
2
1 1
N N N N
i i i i ii i i i
N N
i ii i
x y x y xb
N x x
3
§ Approximation Theory
GIVEN: some function and some domain a,bf x
STEP1: select a proper set of linearly independent
funtions i x
which is expected to be convergent
to the exact function as f x n
0
STEP2: approximate the funtion from of the form
n
i ii
y f x
f x S x c x
STEP3: adjust the values of parameters
so that the resulting approximation is the best.ic
S x
Definition (linearly Independence)
0 1 2 n
0 0 1 1 n
0 1
The set of funtions , , , , is said
to be linearly independent on , if whenever
0 for and all , ,
then 0.n i
n
x x x x
a b
c x c x c x c s R x a b
c c c
4
Definition (Orthogonal)
if function weight therespect to with interval on the
functionsofset orthogonalan betosaidis,,,, 210
xωa,bn
ijii
b
a ji ji
jidxxxx
if0
if0
. of lsubintervaany on 0but
allfor 0 where
a,bxω
a,bxxω
Definition (Ortho-Normal):: orthogonal with i=1
n
iii xcxSxf
0
• Best approximation: define the -weighted error term as follows
b
adxxSxfxE 2
b
a
n
iii dxxcxfx
2
0
The best approximation has a set of {ci} which minimizes the error E.
njcE
j
,,2,1,0 allfor 0
~ (n+1) equations solve (n+1) unknowns {ci} ~
5
Solution:
0
2nb
i i jai
x f x c x x dx
b
a
n
i
b
a jiij dxxxxcdxxxfx0
2
b
a j
n
i
b
a jii dxxxfxdxxxxc 0
~ (n+1) equations (j=0,1,2,…,n) for (n+1) unknowns ~
(solutions exists as long as {i} is L.I.)
2
0
0nb
i iaij j
Ex f x c x dx
c c
With orthogonality (not necessary), we have
b
a j
n
iijii dxxxfxc
0
b
j j jac x f x x dx
0 if
0 if
b
i j i ijai
i jx x x dx
i j
b
j j jac x f x x dx
b
a jj
b
a j dxxxxdxxxfx
6
§ Commonly used functions {i}
nkkxx
nkkxx
x
xwba
nk
k
,,2,1forsin)(
,,2,1forcos)(
1
and1and,,choose
0
dxkxdxkx 22 cossin
nkkxxk ,,2,1,0for,1exp
(i) Fourier series (orthogonal)
2
1
choose , 1,1 and 1 1 and
( ) cos cos for 0,1,2, ,k k
a b w x x
x T x k x k n
0for
1for2/1
1
1
22
k
kdxxxTk
(ii) Chebychev polynomials (orthogonal)
choose , 1,1 and 1 and
( ) for 0,1,2, ,k k
a b w x
x L x k n
1 2
1
2
2 1kL x dxk
(iii) Legendre polynomials (orthogonal)
7
§ Application to ODEs ~ Rayleigh-Ritz method
xyy :ODEorder -2ndlinear aConsider
1 0where 0,1 , , 0,1
and 0 0 and 0 for 0 1
sufficient conditions for a unique solution
p x C q f C
p x q x x
for 0 1 and 0 1 0
d dyp x q x y f x
dx dx
x y y
The unique solution is the function which
1,02C
1
0
22
2 dxxuxfxuxqdxdu
xpxuI
PRINCIPLE OF VARIATION
satisfies BCs
minimizes the integral
8
§ Approximation Theory ~ ONE IDEA
0 select a set of
each of which satisfies the given BCs:
0 1 0
n
k k
k k
LI x
0
0
Define the set of functions spaned by
n
n k k
n
n n k kk
x
y x y x c x
xsoultuion exact theeapproximat to
fromfunction aselect
yn
2The exact solution 0 1 and B.C.sy x C ,
n
n
kkkn xcxyxy
0
011
000
0
0
n
kkkn
n
kkkn
cy
cy
2The approximation 0 1 and B.C.sn ny x C ,
9
nCn as1,02
Which is the best? The one having a minimum value of ny I
nxyyn as:eConvergenc
0
n
n n k kk
I y x I c x
SOLUTION::
1
0
22
2 dxxuxfxuxqdxdu
xpxuI
2 2
1
00 0 0
2n n n
k k k k k kk k k
p x c x q x c x f x c x dx
MINIMIAZATION: 0, for 0,1, 2, ,
j
Ij n
c
10
jc
I
0
1
00 0
2 2 2n n
j k k j k k jk k
p x x c x q x x c x f x x dx
1 1
0 00
2 2n
k j k j k jk
c p x x x q x x x dx f x x dx
0
Thus we obtain for 0,1, 2, ,n
jk k jk
a c , jb n
~ Solving ODE becomes solving matrix (algebraic) equations ~
2 2
1
00 0 0
2n n n
k k k k k kk k kj
p x c x q x c x f x c x dxc
kj jka a jb
txfuLtu
, :Given
§ Application to PDE ~ spectral methods, finite element methods, etc.
2Laplacian e.g. operatior, spartial some is where LL
term.sourceexternalknown a is x,tf
bxaxgu ,x,0 :I.C.
0fortx, :FIND tu
11
STEP1: find a set of LI and
the associated weight function
j x
w x
§ Global approximation theory:
fuLtu
:PDE into formula above theSubstitute :STEP2
txfxcLxdt
dc n
jjj
n
jj
j ,0
?
0
0
P.S. The coefficients is a function of
,
n
,
ow.j
n
n j jj
u x t
c
u x t c t x
t
for all x
§ Global approximation theory:
txfxcLxdt
dc n
jjj
n
jj
j ,0
?
0
0
Let the space spaned by . nk
n j jx
nLHS
?
nRHS
not necessary!
12
• Several ways to obtain an approximation:
0 0
,n n
jj j j
j j
dcx L c x f x t
dt
(i) best approximation: look for a set of which minimizesjc t
at which nodes ofset a choose :methodsn Collocatio (ii) n
0iix
0 0
,n n
jj j j
j ji i i
dcL c f t
dtx x x
(iii) Galerkin approximation
0
0
,
nj
j
b
ia
b b
i
j
n
j j ia aj
x x dx
x x d
dcx
dt
L c x dx tx xfx x
§ Galerkin approximation – linear L
dxtxfxxdxxLxxc
dxxxxdt
dc
b
a i
b
a ji
n
jj
n
j
b
a jij
,0
0
define if orthonorma
,
lb
ij i ja
b
ij
ij
i ja
b
i ia
a x x x dx
b x x L x dx
r t x x f x t dx
ω
trtcbdt
dca i
n
jjij
n
j
jij
00
~ PDE is reduced to a system of coupled ODEs ~
~ go time-marching ~
13
trtcbdt
dci
n
jjij
j 0
~ decoupled on LHS ~
special case: -orthonormal : j ij ijx a δ
§ Initial Conditions, i.e. cj(0)
0
,0 ,0 0n
n j jj
u x g x u x c x
n
jjj xcxg
0
0 minimize :ionapproximatbest
0
Collocation: 0n
i j j ij
g x c x
dxxxxcdxxgxxb
a ji
n
jj
b
a i
0
0 :Galerkin
e.g. Bu(x,t) = 0 for some linear spatial operator B
§ Boundary Conditions:
0 0
, 0n n
n j j j jj j
Bu x t B c t x c t B x
i.e. 0jB x
(a) select with the addional restriction
that each .
j
j
x
x BCs
(b) (Tau method) Discard as may ODEs as the number
of BCs and impose
0
, 0n
n j jj
Bu x t c t B x
14
§ Example: thermal boundary layer
2
2
:equation GoverningxT
DtT
iTtxT 0, :I.C.
sTtxT 0,1 :B.C.s
define ( , ) i
s i
T Tx t
T T
2
2
:equation GoverningxT
DtT
iTtxT 0, :I.C.
sTtxT 0,1 :B.C.s
2
2D
t x
,0 0x
1, 1t
15
1near developtoexpectedislayer bounday A thermal x
1near pointsgriddenser better x
Expect the solution is symmetric.
12
choose Chebychev polynomials of even degrees
cos 2 cosjT x j x
§ collocation + methods
collocation grid points: choose the locations of peaks of Chebychev,
xTx jj 2 choose
20
, ,n
n j jj
x t x t c t T x
ni
nin
xTx ni ,2,1,04
2cos ofpeak of locations 4
16
2
2D
t x
n
j
jj
n
jj
j
dx
xTdcDxT
dt
dc
02
22?
02
)()(
20
, ,n
n j jj
x t x t c t T x
2
22 2
0 0
( )( )
n nj j
j i j ij j
dc d T xT x D c t x
dt dx
discareded1,2,1,0for nini
20 0
B.C.: 1, 1 1n n
j j jj j
t c t T c t
§ collocation + methods
• Initial Conditions
20
,0 0 0
for 0 1 2 1
n
i j j ij
x c T x
i , , , ,n
0
1,0 1 0n
jj
c
20
, ,n
n j jj
x t x t c t T x
~ (n+1) equations for (n+1) unknowns ~
17
§ Galerkin method+{j(x)} B.C.s
11 and 1 Since 20 jTxT
1 1
1, 1 1 1 0 1n n
n j j jj j
t c t c t
1
2 01
, , 1
1
n
n j jj
n
j jj
x t x t c t x
c t T x T x
2 0Let . Thus 1 0j j jx T x T x
B.C.s
n
j
jj
n
jj
j
dx
xdtcDx
dt
dc
12
2?
1
)()(
• Use orthogonal property:
1 1
2
2 2
21 1
1 1
( ( ))( )
1 1
( ) in n
jj jj
j
i
j
d x dxdc xD c
xxdx dx
xt
dt x
n
jjij
n
j
jij nitcbD
dt
dca
11
,1,2,for
• Substitute into governing equation:2
2D
t x
1
, 1n
n j jj
x t c t x
18
1
1
20202 1 dxxxTxTxTxT ji
1
1
220202022 1 dxxTTTTTTT jiji
1 2 22 2 01
1i jT T T x dx
1 if2
3
if
ji
ji
1
21
( ) ( ) , , 1, 2, ,
1
i jij
x xa dx i j n
x
1
1
2022
2
02 1 dxxxTxTdx
dxTxT ji
2
1 2 22 0 21
1ji
d TT T x dx
dx
21 2
21
( )( ) 1j
ij i
d xb x x dx
dx
2 1
22 2 22
0
jj
j k kk
d T xP x T x
dx
1 1 2
2 2 0 210
1j
ij k i k kk
b T T T T x dx
19
Initial Conditions: ,0 0x
1 1
2 21 11
( ) ( )( ,0) 1 ( )0
1 1
ni ji
jj
x xx xdx c dx
x x
1 2
211
1 ( )0 for 1,2, ,
1
ni
ij jj
T xdx a c i n
x
1
, 1n
n j jj
x t c t x
1
,0 1 0n
n j jj
x c x
1
divide the interested domain
into several subdomins K
kk
a,b
♣ Local approximation theory
0
0P.S. is different from one subdomain to anoth
for all ,
.
,
er
nk k kn j j k
j
nkj j
x
u x t u x t c t x x
0
0
i.e. require sets of L.I. functions
and will have sets of corresponding coefficients
nkj j
nkj j
K x
K c t
20
§ Finite Element Methods ~ piecewise approximation
)boundary with (domain :problemTest
2221
( ) ( , ) ( , ) ( , ) ( , ) ( , )2
u uI u p x y q x y r x y u f x y u x y dxdy
x y
ofvaluetheminimizesandBCs satisfieswhich
function theis unique) and exists (ifsolution exact The
::VARIATION OF PRINCIPLE2 C
onyxgyxu
yxfyxuyxryu
yxqyx
uyxp
x
,,
,,,,,
1
STEP1: Divide the interested domain into several parts, i.e.
Each subdomain is called an "element".
K
kk
k
0
STEP2: within each subdomain , choose a proper set of , j
nk
kj
x y
k0
for ,,,
x,yyxcyxuyxun
j
kkk
jjn
0
P.S. , are different from one element to another in general.j
nk
jx y
21
01 1
Put , , ,K K n
k k kn j j
jk k
u x y u x y c x y
1
Thusk
Kkj
k
I u I c
STEP3: Look for the values of 0 1 and 0 1
such that , minizes
kj
kn
c , j , , ,n k , , ,K
u x y I u
2221
( ) ( , ) ( , ) ( , ) ( , ) ( , )2
u uI u p x y q x y r x y u f x y u x y dxdy
x y
2
0
1( , )
2m
mnjm
jmji
p x y c dxdyc x
)(minimized required 0
10alland10 allfor Therefore
micuI
,K,,m,n,,i
K
k
n
j
kjk
jmi
k xcyxp
c 1
2
0
,21
2
0
1,
2m
mnjm
jmji
p x y cc x
0 0
( , )m
mnjm
jj
mnjm
jmji
p x y c dcx
x yc
dx
0
m mnj i
ijj x x
22
0
0 ( , ) ( , ) ( , )k
k kk knj jk k ki i
j j ij
c p x y q x y r x y dxdyx x y y
( , ) ( , )k
kif x y x y dxdy
)(minimized required 0 m
icuI
0 0
0
, ,
, ,
m
m mm mn nj jm mi i
j jj j
nm m m mj j i i
j
p x y c q x y cx x y y
r x y c f x y dxdy
)variable(change :ntrearangemeAfter km
, , ,k
k kk kj jk k ki i
ij i j
kji
a p x y q x y r x y dxdyx x y y
a
, ,k
k ki ib f x y x y dxdy
0
or ~ elemental matrix equation
for 1 2
nk k k k k kij j i
j
a c b A c b
k , , ,K
• Ak is singular! i.e. Each element matrix equation is underminate.
► too many degrees of freedom!
23
then,alsoand,,if:trequiremen Continuity21 kkyx
1 1 1
0
, ,n
k k kn j j
j
u x y u c x y
2 2 2
0
, ,n
k k kn j j
j
u x y u c x y
needst requiremen Thus 0-C
1 1 2 2
0 0
, ,n n
k k k kj j j j
j j
c x y c x y
tsrequiremen the with
redonebeshouldproblemon minimizati The0-C
STEP1: within element , choose 1 points
denoted by , for 0 1 2
k
k ki i
n
x y i , , , ,n
,k k ki j j ijx y
n
ill
kl
ki
kl
ki
kl
klk
i
yyxx
yyxxyx
022
22
,
§ Example: use collocation points
yxki , s,polynomial Lagrangian ldimentiona-say two
function,interpointan chooseelement,each within :STEP2
and approximate the desired function as
k
n
i
ki
ki
kn yxyxcuyxu
,whenever,,,0
24
Notice:
iikj
n
j
kj
kiii
kn yxcuyxu ,,
0
ij
n
j
kj
ki cu
0
kic
0
, ,n
k k kn j j
j
u x y u x y
k
n
i
ki
ki
kn yxyxcuyxu
,whenever,,,0
§ example – triangular elements (linear interpolation)
1 2
3
11u
12u
13u
21u
22u
23u
31u
33u
32u
0 1 1 1 2 2 2 3 3 31 2 3 1 2 3 1 2 3If no constraint: d.o.f.s , , , , , , , ,C u u u u u u u u u
0with constraint: d.o.f.s = , , , ,a b c d eC - u u u u u
a
b c
d
e
11
1 22 1
0 2 32 1
32
1 2 33 3 3
:
a
b
c
d
e
u u
u u u
C u u u
u u
u u u u
25
: d.o.f. 11uua
1 0
,k
K nk kj j
k ja
u x yu
0
auuI
1
1 1
0
,n
j jja
u x yu
1
1 11
01
,n
j jj
u x yu
11
13
113
12
112
11
1110 buauaua
32d.o.f. :du u 3 3 3 3 3 3 3
21 1 22 2 23 3 20 a u a u a u b
1 2
1 1 2 2
0 0
, ,n n
j j j jj jb
u x y u x yu
0
buuI
: d.o.f. 21
12 uuub
1 2
1 1 2 21 2
0 02 1
, ,n n
j j j jj j
u x y u x yu u
21
23
213
22
212
21
211
12
13
123
12
122
11
1210 buauauabuauaua
26
2 3
2 2 3 3
0 0
, ,n n
j j j jj jc
u x y u x yu
0
cuuI
: d.o.f. 31
22 uuuc
2 3
2 2 3 32 3
0 02 1
, ,n n
j j j jj j
u x y u x yu u
31
33
313
32
312
31
311
22
23
223
22
222
21
2210 buauauabuauaua
321
cu
0
e
I u
u
1 2 33 3 3d.o.f. :eu u u u
1 2
3
1 1 2 21 2
0 03 3
3 33
03
, ,
,
n n
j j j jj j
n
j jj
u x y u x yu u
u x yu
1 1 1 1 1 1 1 2 2 2 2 2 2 231 1 32 2 33 3 3 31 1 32 2 33 3 3
3 3 3 3 3 3 331 1 32 2 33 3 3
0 a u a u a u b a u a u a u b
a u a u a u b
27
1 1 111 12 13
1 1 2 2 1 221 22 11 12 23 13
2 2 3 3 2 321 22 11 12 23 13
0 0
0
0
0
a a a
a a a a a a
a a a a a a
3 3 321 22 23
1 1 2 2 3 3 1 2 331 32 31 32 31 32 33 33 33
0
a a a
a a a a a a a a a
11
1 22 1
2 32 1
32
1 2 33 3 3
a
b
c
d
e
u u
u u u
u u u
u u
u u u u
11
1 22 1
2 32 1
32
1 2 33 3 3
b
b b
b b
b
b b b
• Global matrix equation Au=b b c d ea
1 1 111 12 13
1 1 121 22 23
1 1 131 32 33
0 0
0 0
0 0 0 0 0
0 0 0 0 0
0 0
a a a u
a a a
a a a
11
12
13
0
0
u
u
2 2 2 211 12 13 1
2 2 2 221 22 23 2
22 2 2331 32 33
0 0 0 0 0 0
0 0
0 0
00 0 0 0 0
0 0
a a a u
a a a u
ua a a
33
32
31
333
332
331
323
322
321
313
312
311
0
0
0 0
0 0
0 0
0 0 0 0 0
0 0 0 0 0
u
u
u
aaa
aaa
aaa
11
12
13
0
0
b
b
b
21
22
23
0
0
b
b
b
31
32
33
0
0
b
b
b
stiffness summation
28
• local matrix equation
1 1 1 1 1 1 1 111 12 13 1 11 12 13 11 1 1 1 1 1 1 121 22 23 2 21 22 23 21 1 1 1 1 1 1 131 32 32 3 31 32 32 3
1:a
b
e
a a a u a a a u b
k a a a u a a a u b
a a a u a a a u b
2 2 2 2 2 2 2 211 12 13 1 11 12 13 12 2 2 2 2 2 2 221 22 23 2 21 22 23 22 2 2 2 2 2 2 231 32 32 3 31 32 32 3
2 :b
c
e
a a a u a a a u b
k a a a u a a a u b
a a a u a a a u b
33
32
31
332
332
331
323
322
321
313
312
311
33
32
31
332
332
331
323
322
321
313
312
311
:3
b
b
b
u
u
u
aaa
aaa
aaa
u
u
u
aaa
aaa
aaa
k
e
d
c
§ Example – 2D Poisson Solver
tyxfyP
xP
P ,,2
2
2
22
29
• Let
K
k
N
i
N
j
kj
ki
kij yhxhPtyxP
1 0 0
',,
K
k
N
i
N
j
kj
ki
kij yhxhftyxf
1 0 0
',,
N
imm
ki
km
kik
i xx
xxxh
0
N
jmm
kj
km
kjk
j yy
yyyh
0
nodeson Intergrati Quadrature Gauss are and where00
N
j
kj
N
iki yx
Numerical Analysis• 挑戰人類結合數學及物理知識能力
• 激發人類的創造力與邏輯思考
• 數位化儀器的秘密武器
• 提供經濟的設計分析工具
• 數值誤差 – truncation error & rounding error
• 數值現象 – numerical diffusion & dispersion
• 穩定性、準確性、效率
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Happy Summer Vacation!