norms and spaces definition: the space of all square integrable funcions defined in the domain is a...
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Norms and spaces
Definition:
)(2L The space of all square integrable funcions defined in the domain
dxf 2 is a finite number not infinity
L2 norm of f
2/12
)(2 dxff
L
Example)(2 L
fcompute
)0,0(
)2,2(
2),( yxyxf
)(2Lf
2
0
2
0
222
)()(
2dxdyyxf
L
5
1442
0
423
8 24 dyyy5
12
)(2
Lf
Definition:
)(1H The function and its first derivatives are square integrable
)}(,,:{ 2 Luuuu yx
Example
)0,0(
)2,2(
2),( yxyxf
5
1442
0
2
0
222
)()(
2
dxdyyxfL
4)1(2
0
2
0
22
)(2
dxdyfLx
3
642
0
2
0
22
)()2(
2
dxdyyf
Ly
Def:
2
)(
2
)()(222
LyLxL
vvv
2)(
2)()( 22
1 LvLvH
v
3
642
)(
2
)()(4
222
LyLxLfff
Def:
3
644
3
28)(1
Hf
Norms and spaces
)(1 Hf
Example
)0,0(
)2,2(
2),( yxyxf
3
282
0
2
0
222
)()(
2
dxdyyxfL
4)1(2
0
2
0
22
)(2
dxdyfLx
3
642
0
2
0
22
)()2(
2
dxdyyf
Ly
Def:
2
)(
2
)()(222
LyLxL
vvv
3
642
)(
2
)()(4
222
LyLxLfff
Norms and spaces
)(2||||||
Lvv
Energy norm
Example
xyxyxv 3),( 2
Compute the energy norm of the function
Norms and spaces
Example
xyxyxv 3),( 2
Which of the following belongs to
Definition: )(10H }on 0),(:{ 1 uHuu
)(10 H
)0,0(
)2,2(
)2()2(),( yyxxyxu
Triangle inequality
Norms and spaces
Cauchy-Schwartz inequality
baba (Real numbers)
)()()( 222
LLLvuvu
Triangle inequality
22
LLvuuv
Cauchy-Schwartz inequality
|||||| |||||| vuuv
KKPv
CvV
K
h trianglesallfor )(
),(
1
0
the space of all continuous piecewise linear polynomials
} 0:{0, onvVvV hh
the space of all continuous piecewise linear polynomials which vanishes on the boundary
Definition:
Definition:
Norms and spaces
Relations:
)( 100, HVh
)( 1 HVh
81 Kh
)0,0( )0,4(
)4,0( )4,4(
1
2 3
45
1K
2K3K
4K
5K6K
7K
)2,0(
)4,2( )4,3(
82 Kh
43 Kh
44 Kh
87 Kh
56 Kh85 Kh
the length of the longest edge on K
Kh
Mesh size h
local mesh size
Example:
KKhh max
mesh size
},,max{ 71 KK hhh Example:
4h
on 0 in
ufu
A Priori Error Estimates
0,
0,
)(),(
such that Find
hhhhh
hh
VvvFvua
Vu
u hu
)( )(),(
such that )( Find10
10
HvvFvua
Hu
? huue in what sense the error e becomes small
vuvua ),(
A Priori Error Estimates
? huue
)( 10
Hvfvvu0, hhhhh Vvfvvu
0,
0,
)(),(
such that Find
hhhhh
hh
VvvFvua
Vu
)( )(),(
such that )( Find10
10
HvvFvua
Hu
variational formulation finite element method
0, allfor ,0)( hhhh Vvvuu
Galerkin Orthogonality
The finite element approximation uh, satisfies the orthogonality
Theorem 1 (Galerkin Orthogonality).
Proof: from the variational formulation we have:
0, )( hhhhh VvvFvu
0, )( hhhh VvvFvu
subtract
0, 0)( hhhh Vvvuu
)(100, HVh
)( 10
Hvfvvu
0, hhhhh Vvfvvu
variational formulation
finite element method
0, allfor |||,||| |||||| hhhh Vvvuuu
Best Approximation
The finite element approximation uh, satisfies
Theorem 2 (Best Approximation).
Proof: have weany for 0,hh Vv
)()( |||||| 2hhh uuuuuu
)()( hhhh uvvuuu
)()()()( hhhhh uvuuvuuu
)()( hh vuuu
|||||| |||||| hh vuuu
This shows that the finite element solution uh is the closest of all functions in Vhto the exact solution u when measuring distance using the energy norm.
K
2
)(
2222
||||||KLKh uDChuu
Energy Norm Error
The finite element approximation uh, satisfies the a priori error estimate
Theorem 3 (error depends on meshsize).
Proof:
K
2
)(
222
KLK uDCh
This shows that how the error depends on the mesh size.
with a constant C independent of Kh
Start from the best approximation
|||||| |||||| vuuu h
choosing v= πu22 ||| ||| |||||| uuuu h
K
2
)(2) ( KL
uu
the gradient of the error tends to zero as the maximum mesh size h tend to zero.
Remark
K
2
)(
2222
||||||KLKh uDChuu
K
2
)(
222
C KL
uDh
2
)(
222
C
L
uDh
)(
22
||||||
Lh uDChuu
)()( 22
LLvCv
Error Depends on hFor any function Theorem (Poincare Inequality).
These constants C’s are different
Remark
)(
22
||||||
Lh uDChuu
|||||| vC
)(10 Hv
||||||)(2 hLh uuCuu
)(
22
L
uDCh
)(
2
)( 22
LLh uDChu-u
L2 error is h2we expect that the L2 error to be h2 and not h
)(
22
)( 22
LLh uDChuu
The finite element approximation uh, satisfies the a priori error estimate
Theorem 4 (The L2-error)
with a constant C independent of h
The proof uses a well-known technique called Nitsche’s trick,
on 0 in
ufu
on 0
in
e
Dual problem or adjoint problemMain problem
Remark:
Continuous Piecewise Linear Interpolation
Definition:
we define its continuous piecewise linear interpolant by
)(0 CuLet
n
kkkNuu
1
)( hVu
approximates by taking on the same values in the nodes Ni.u u
u
u
uDChuu 2)(
L2 error is h2
)(
22
)( 22
LLh uDChuu
The finite element approximation uh, satisfies the a priori error estimate
Theorem 4 (The L2-error)
with a constant C independent of h
on 0
in
e
Proof: let ϕ be the solution of the dual problem
Multiplying by e and integrating using Green’s formula
ee2
ene
e
)(
e
Cauchy-Schwartz inequality
)(2
ee
)( e
2 Dche
Che
ChuDCh 2
eChuDCh 2
uDeCh 22
Dividing by e∥∥
uDeChe 222
uDChe 22
)(
22
)( 22
LLh uDChuu
Error norms
)(
22
)( 22
LLh uDChuu
The finite element approximation uh, satisfies the a priori error estimate
Theorem (L2 Norm)
with a constant C independent of h
The finite element approximation uh, satisfies the a priori error estimate
Theorem (Energy Norm)
with a constant C independent of Kh
)(
22
||||||
Lh uDChuu
Calcuate the L2 error
K
2
)(
2
)( 22 KLhLh uuuu
K
2)( K
huu
K
2 ||),(),( Kyxuyxu h
K
22 |||||| |||||| Khh uuuu
K
yy
K
xx uhuuhu 2
K
2 )()(
||),(),(||),(),( 2
K
2 KyxuhyxuKyxuhyxu yyxx
Calcuate the L2 error
on 0 in
ufu
)1,1()1,1(
)2(2),( 22 yxyxf
)1)(1(),( 22 yxyxu
Exact solution
)(
22
)( 22
LLh uDChuu
2.01 h 1.02 h
11 Chuu h
22 Chuu h
1
2
1
2 h
h
uu
uu
h
h
1
2
1
2 h
h
uu
uu
h
h
1
2
1
2 ln / lnh
h
e
e
function [l2error] = compute_error(p, t, uh, u, u_x, u_y)% calculates the error nt = size(t, 2); % number of triangles
np = size(p,2); t1 = t(1,:); t2 = t(2,:); t3 = t(3,:); x1 = p(1, t1); x2 = p(1, t2); x3 = p(1, t3); y1 = p(2, t1); y2 = p(2, t2); y3 = p(2, t3); xc = (x1 + x2 + x3)/3; % x-coord of element midpoints
yc = (y1 + y2 + y3)/3; % y-coord of element midpoints
exact = feval(u, xc, yc); % exact sol at the midpoints
uhc = pdeintrp(p, t, uh); % FE sol at the midpoints
l2error2 =0;for K = 1:nt loc2glob = t(1:3,K); x = p(1,loc2glob); y = p(2,loc2glob); area = polyarea(x,y); loc_er = (exact(loc2glob) - uhc(loc2glob)).^2; l2error2 = l2error2 + sum(loc_er)*areaendl2error = sqrt(l2error2); [exact' , uhc', abs(exact-uhc)']
function [p,e,t,uh] = solveE(hmax)% Poisson's equation on a square [-1,1]X[-1,1] is solved% and the resulting finite element solution is stored in % the vector uh.% g = 'squareg'; % domainb = 'squareb1'; % Dirichlet dataf = '-2*((x.^2 - 1) + (y.^2 - 1))'; % right hand side [p, e, t] = initmesh(g,'Hmax',hmax); % triangulationuh = assempde(b, p, e, t, 1, 0, f); % solve pde
u = inline('(x.^2 - 1).*(y.^2 - 1)', 'x', 'y'); % exact solu_x = inline('2*x.*(y.^2 - 1)', 'x', 'y'); % u_x exact u_y = inline('2*y.*(x.^2 - 1)', 'x', 'y'); % u_y exact h1=0.2;[p1,e1,t1,uh1] = solveE(h1);[l2error1] = compute_error(p1, t1, uh1, u, u_x, u_y) h2=0.1;[p2,e2,t2,uh2] = solveE(h2);[l2error2] = compute_error(p2, t2, uh2, u, u_x, u_y) log(l2error1/l2error2)/log(h1/h2)
Calcuate the energy norm of the error
function [energy_error] = energy_error(p, t, uh, u, u_x, u_y)
)(
22
||||||
Lh uDChuu
a) Write matlab file to compute the energy norm of the error
b) Write matlab file to compute the energy norm of the error for two different meshes with meshsizes h1 and h2 and verify the rate of convergence
Exercise