Discretization error estimation

Discretization error estimation

Witold Cecot e-mail: plcecot@cyf-kr.edu.pl

Jerzy Pamin e-mail: jpamin@L5.pk.edu.pl

This lecture was prepared within project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction”, co-financed by the European Union within the European Social Fund

Discretization error

Errors committed

I Modelling error

I Discretization error (of FEM approximation)

I Solution error

Methods of discretization error estimation

I hierarchical (Runge)

I explicit residual (implicit not considered here)

I based on averaging (Zienkiewicz-Zhu)

I interpolation error analysis (not considered here)

FE approximation using linear functions

Example problem

Solve BVP using 4 linear elements

−u ⁰⁰ + u = f , f = x ³ − 6x ² + 12 , x ∈ (0, 5) bcs: u(0) = 0 , u(5) = 5

Analytical solution:

u ^analit = x ³ − 6x ² + 6x

Approximate solution u _h (h - element size)

x 0 1 2 4 5

u _h 0 0.938 -4.797 9.153 5

Error measure: e ^def = u − u h

4

−4

−8

x

1 2 3 4

u

This lecture was prepared within project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction”, co-financed by the European Union within the European Social Fund

Hierarchical method e ^H ≈ u _h/2 − u _h

u _h u _h/2

x 0 0.5 1 1.5 2 3 4 4.5 5

u h 0 0.469 0.938 -1.930 -4.797 -6.975 -9.153 -2.077 5 u h/2 0 1.647 1.000 -1.179 -4.138 -9.324 -8.299 -3.543 5 u h/2 −u h 0 1.178 0.062 0.751 0.660 -2.349 0.855 -1.467 0 η ^H _i = q R x _i+1

x _i (u _h/2 − u h ) ² dx → η ₁ ^H = 0.69, η ₂ ^H = 0.59, η ^H ₃ = 1.70, η ^H ₄ = 0.79

||e ^H || ² ≈ ||u h/2 − u h || ² = R 5

0 (u h/2 − u h ) ² dx → ||e ^H || ≈ 2.08

||u h/2 || ² = R 5

0 (u h/2 ) ² dx → ||u h/2 || = 12.35 → _||u ^{||e H ||}

h/2 || ≈ 17%

Error estimation based on residuum (explicit)

Residuum of differential equation

−u ⁰⁰ + u = f, f = x ³ − 6x ² + 12 → R(x) = f − (−u ⁰⁰ _h + u _h ) Residuum provides bound on error

||e|| ≤ C||R||

In 2D (J - jump of 1st derivative)

||e|| ² ≤ C(h ² ||R|| ² + h||J || ² ) Error indicator in 1D element i η ^R _i = h i

q R x i+1

x _i R ² dx, u ⁰⁰ _h = 0 → R = x ³ − 6x ² + 12 − u h

Substitute interpolation for u h , e.g.

η ^R ₁ = 1 q

R 1

0 {x ³ − 6x ² + 12 − [0(x − 1) + 0.938x]} ² dx = 9.94 Compute relative error norm

η ^R ₁

||f || ≈ 33%

and compare solution quality in elements

This lecture was prepared within project ”The development of the didactic potential of Cracow University of Technology in the range of modern construction”, co-financed by the European Union within the European Social Fund

Error estimation based on smoothing e ^S = ˜ u ⁰ _h − u ⁰ _h

x 4

1 8

−2

−5

3 4

2 1 u

13

˜ u

Smoothed solution derivative ˜ u ⁰ _h (through points determined at element edges)

h

h

y d

d

d

− d

y = _h ^{h 2}

1 +h 2 (d 1 − d 2 )

˜

u ⁰ = d 2 + y = d 1 h ₂

h ₁ +h ₂ + d 2 h ₁ h ₁ +h ₂

If h 1 = h 2 then ˜ u ⁰ = ^{d 1 +d} ₂ ²

˜

u ⁰ 1 = 4.2 (extrapolation to node 1),

˜

u ⁰ ₂ = −2.4, ˜ u ⁰ ₃ = −4.5, ˜ u ⁰ ₄ = 8.7, ˜ u ⁰ ₅ = 19.6 η i ^S = q

R x _i+1

x _i (˜ u ⁰ _h − u ⁰ _h ) ² dx η 1 ^S =

q R 1

0 {[4.2(1 − x) + (−2.4)x] − 0.94} ² dx = 1.93 η 2 ^S = 2.35, η ^S 3 = 8.09, η 4 ^S = 3.14

||e ^S || ² = R 5

0 (˜ u ⁰ _h − u ⁰ _h ) ² dx = P

i (η _i ^S ) ² → ^||e _{|| ˜ u} ^S 0 ^||

h || ≈ 57%