two-point boundary value problem de weak form discrete form linear system 1 2 3 4
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Two-Point Boundary Value Problem
DEWeak Form
Discrete FormLinear System
0)1()0(
')''(
uu
fcubuau
1)( )(),(
such that )( Find10
10
HLua
Hu
2
h
hh
SLua
Su
)(),(
such that FindbAU 34
spaces
)(2L
The space of all square integrable funcions defined in the domain
Definition:
dxffL 22 :)(
Definition:
)(',:)( 21 LvvvH
The function and its first derivative are square integrable
Remark: Both spaces are Hilbert spaces. R2 is also a Hilbert space
reals are and ,:2 yx
y
xvvR with inner product 21212
2
1
1 ,),( yyxxvuy
x
y
x
R2 is also a Hilbert space
Triangle inequality
)(2 LTriangle inequality:
)(, 2 Lgf)()()( 222
LLLgfgf
)()()( 111
HHHwvwv
Triangle inequality: )(1 H
)(, 1 Hwv
wvwv
Triangle inequality: 2R2, Rwv
2/12
221
2/122
21
2/12
22
2
11
wwvv
wvwv
Cauchy-Schwarz inequality
)(2 LCauchy-Schwarz inequality:
)(, 2 Lgf )()( 22
,
LL
gfgf
Cauchy-Schwarz inequality: (integral form)
2/11
0
22/11
0
21
0)()()()(
dxxgdxxfdxxgxf
Example: verify CS-inequalityxexgxxf )( ,3)(
Cauchy-Schwarz inequality:2, Rwv wvwv , 2/12
221
2/122
21
2121
wwvv
wwvv
3.09572/13 2 e
Cauchy-Schwarz inequality
Is this true?:
??, gf 11
',' gfgf
Cauchy-Schwarz inequality: (integral form)
2/11
0
22/11
0
21
0)()()()(
dxxgdxxfdxxgxf
Bilinear form
Definition:
A bilinear form on V is a function : V × V → R, which is linear in each argument separately
)or ( 12 HL
Example:
1
0)(')('),( )1 dxxvxuvua
1
0)('')(''),( )2 dxxvxuvua
1
0'''),( )3 dxuvvuvua
the bilinear form a( ・ , ・ ) on V is bounded if there is a constant M such that.
Definition:
1
11 Hw,v , v wM|a(w, v)|
a(w, v). a(w, u) v) u a(w,
a(v,w), a(u,w) v,w) u a(
R , V and u,v,w
),( aThe bilinear form is said to be symmetric if
a(w, v) = a(v,w), ∀v,w ∈ V,
Definition:),( a
Example: prove that a is bounded bilinear form on
1
0)(')('),( dxxvxuvua
1H
Bilinear form
the bilinear form a( ・ , ・ ) on is coercive if there is a constant α > 0 such that.
Definition:
12
1 Hv , v|a(v, v)|
Example: prove that a is coercive on
1
0)(')('),( dxxvxuvua
1H
1H
Linear functional
A linear functional L : V → R is said to be bounded
Definition:
Vv, vc|L(v)|V
Example:
is the smallest constant cL
Remark:V
vL|v|L )(
xxf
, dxxvxfvL
)( where
)()()(0
0
1HV
Lax-Milgram lemma
Lax-Milgram lemma
L
),( a bounded coercive bilinear form on
bounded linear functional on
VV V
there exists a unique vector u ∈ V such that (2) is satisfied
V Hilbert space
Then
VvvLvua
Vu
)(),(
such that Find
L
),( a bilinear form on
linear functional on VV
V
V Hilbert space
whereConsider:
2 )or ( 12 HL
DEWeak Form
0)1()0(
')''(
uu
fcubuau
1)( )(),(
such that )( Find10
10
HLua
Hu
2
dxfL
dxcubuauua
)(
)'''(),(
Example: 2)( xxa
0)( xb
1)( xc
33)( 2 xxxf
0)1()0(
33)'')2(( 2
uu
xxuux
1
)( )(),(
such that )( Find10
10
HLua
Hu
2
1
0
33
1
0)2(
)()(
)''(),(
2 dxL
dxuuua
xx
x
Show that there exist a unique solution for (2)
Lax-Milgram lemma
Example: 2)( xxa
0)( xb
1)( xc
1
)( )(),(
such that )( Find10
10
HLua
Hu
2
Show that there exist a unique solution for (2)
Lax-Milgram lemma
solution: In order to show that there exist a unique solution for (2), we need to satisfy all the conditions of Lax-Milgram lemma
bounded )(
in coercive ),(
bounded ),(
L
a
a proof
Later we will do another proof for a symmetric a
33)( 2 xxxf
0)1()0(
33)'')2(( 2
uu
xxuux
1
0
33
1
0)2(
)()(
)''(),(
2 dxL
dxuuua
xx
x
Poincare’s inequality (HW) 1
010 Hvvv
Show that:10010
'2' Hvvvv
Example: 2)( xxa
0)( xb
1)( xc
44)( 2 xxxf
0)1()0(
44)'')2(( 2
uu
xxuux
1
)( )(),(
such that )( Find10
10
HLua
Hu
2
1
0
44
1
0)2(
)()(
)''(),(
2 dxL
dxuuua
xx
x
Show that there exist a unique solution for (3)
Lax-Milgram lemma
hh
hh
SLua
Su
)(),(
such that Find
3
Thm: A finite dimensional subspace of a Hilbert space is Hilbert
Hilbert V dim-finite V0 V Hilbert 0V
solution: In order to show that there exist a unique solution for (3), we need to satisfy all the conditions of Lax-Milgram lemma
bounded )(
in coercive ),(
bounded ),(
L
a
a
Example: 2)( xxa
0)( xb
1)( xc
44)( 2 xxxf
)( )(),(
such that )( Find10
10
HLua
Hu
2
Stability
Setting ϕ = u in (2) and using (coercive) and (Poincare), we find
Poincare’s inequality
1010
Hvvv
)(),( uLuua
1000),( ufufuf
2
1u
01
fu
continuous dependence of solutions with respect to perturbations of data
A problem that satisfies the three conditions is said to be well posed
1)existence of solutions,2)uniqueness of solutions,3)stability
Solution bounded by the data of the problem
Definintion:
Small change in the data produce small chang in the solution
Stability:
Linear System of Equations
bAU hh
hh
SLua
Su
)(),(
such that Find34
1
1
M
jjjh Uu
),( ijij aa ),( ii fb
(4) has a unique solution iff that the matrix A is invertible ( non-singular )
Remark:Under what condition that(4) has solution
Remark:
A is symmetric and positive definite
Definition:
An nxn matrix A is symmetric and positive definite if
0, 0 VRVAVV nT
Example: show that A is SPD
51
13A
Linear System of Equations
bAU hh
hh
SLua
Su
)(),(
such that Find34
1
1
M
jjjh Uu
),( ijij aa ),( ii fb
(4) has a unique solution iff that the matrix A is invertible ( non-singular )
Remark:Under what condition that(4) has solution
Remark:
A is symmetric and positive definite
Proof:
1
111 )()(,
M
iii
TM xVxvVVV
),( vva
1
1
1
1
)(,)(M
jjj
M
iii xVxVa
1
1
1
1
)(),(M
ijiji
M
j
xxaVV
1
1
1
1
M
ijiji
M
j
VaV
AVV Tcoercively 2
1),( vvvaAVV T 2
0'v
0)0(,0'0'02
0 vCvvvAVV T