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A posteriori error estimates Part II Complementary estimates Tom a s Vejchodsk y vejchod@math.cas.cz Institute of Mathematics, Academy of Sciences Zitn a 25, 115 67 Praha 1 Czech Republic E M H A A T T I C M S f

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A posteriori error estimates

Part II – Complementary estimates Tom´ aˇ s Vejchodsk´ y vejchod@math.cas.cz

Institute of Mathematics, Academy of Sciences ˇ Zitn´ a 25, 115 67 Praha 1 Czech Republic

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M A T H E M A T I C S

A c a d e m y

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May 29, 2012, Technical University of Ostrava

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Outline

◮ Toy problem ◮ Derivation of complementary estimates ◮ Two options

◮ (A) Error majorant ◮ (B) Dual finite elements

◮ Energy minimization ◮ Method of hypercircle ◮ Numerical examples ◮ Conclusions

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Toy problem

Classical formulation: −∆u = f in Ω, u = 0 on ∂Ω Weak formulation: V = H1

0(Ω)

u ∈ V : a(u, v) = F(v) ∀v ∈ V Notation:

◮ a(u, v) = (∇u, ∇v) ◮ F(v) = (f , v) ◮ (ϕ, ψ) =

  • Ω

ϕψ dx

◮ Error: e = u − uh ◮ Energy norm: |

| |e| | |2 = a(e, e) = (∇e, ∇e) = ∇e2

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Derivation

Divergence thm.: (div y, v) + (y, ∇v) = 0 ∀y ∈ H(div, Ω), v ∈ V Friedrichs’ inequality: v0 ≤ CF| | |v| | | ∀v ∈ V Theorem: Let uh ∈ V be arbitrary then | | |u − uh| | | ≤ η(uh, y) η(uh, y) = CF f + div y0 + y − ∇uh0 ∀y ∈ H(div, Ω) Proof: v ∈ V a(u − uh, v) = (f , v) − (∇uh, ∇v) = (f + div y, v) + (y − ∇uh, ∇v) ≤ f + div y0 v0 + y − ∇uh0 ∇v0 ≤ (CF f + div y0 + y − ∇uh0) | | |v| | | Set v = u − uh.

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Derivation

Divergence thm.: (div y, v) + (y, ∇v) = 0 ∀y ∈ H(div, Ω), v ∈ V Friedrichs’ inequality: v0 ≤ CF| | |v| | | ∀v ∈ V Theorem: Let uh ∈ V be arbitrary then | | |u − uh| | | ≤ η(uh, y) η(uh, y) = CF f + div y0 + y − ∇uh0 ∀y ∈ H(div, Ω) Lemma: Let u ∈ V be the exact solution. Then | | |u − uh| | | = η(uh, ∇u).

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Two options

(A) Error majorant | | |u − uh| | | ≤η(uh, y) η(uh, y) = CF f + div y0 + y − ∇uh0 ∀y ∈ H(div, Ω) [S. Repin et al., 2000–] (B) Dual finite elements | | |u − uh| | | ≤ η(uh, y)

  • η(uh, y) = y − ∇uh0

∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} [J. Haslinger, I. Hlav´ aˇ cek, M. Kˇ r´ ıˇ zek, 1970s–80s]

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(A) Error majorant

| | |u − uh| | | ≤ η(uh, y) = CF f + div y0 + y − ∇uh0 ∀y ∈ H(div, Ω) | | |u − uh| | |2 ≤ η2(uh, y, β) =

  • 1 + 1

β

  • C 2

F f + div y2 0 +(1 + β) y − ∇uh2

∀β > 0 Proof: (A + B)2 ≤

  • 1 + 1

β

  • A2 + (1 + β)B2

∀β > 0 Equality for β = A/B.

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(A) Error majorant

| | |u − uh| | | ≤ η(uh, y) = CF f + div y0 + y − ∇uh0 ∀y ∈ H(div, Ω) | | |u − uh| | |2 ≤ η2(uh, y, β) =

  • 1 + 1

β

  • C 2

F f + div y2 0 +(1 + β) y − ∇uh2

∀β > 0 Notation: W = H(div, Ω) Complementary problem (equivalent formulations): (i) Find y ∈ W : η(uh, y) ≤ η(uh, w) ∀w ∈ W (ii) Find y ∈ W and β > 0 : η2(uh, y, β) ≤ η2(uh, w, β) ∀w ∈ W, β > 0 If β > 0 fixed: (iii) Find y ∈ W : η2(uh, y, β) ≤ η2(uh, w, β) ∀w ∈ W (iv) Find y ∈ W : (div y, div w) + β C 2

F

(y, w) = β C 2

F

(∇uh, w) − (f , div w) ∀w ∈ W

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(A) Error majorant

| | |u − uh| | | ≤ η(uh, y) = CF f + div y0 + y − ∇uh0 ∀y ∈ H(div, Ω) | | |u − uh| | |2 ≤ η2(uh, y, β) =

  • 1 + 1

β

  • C 2

F f + div y2 0 +(1 + β) y − ∇uh2

∀β > 0 Notation: W = H(div, Ω) Practical implementation:

◮ Wh ⊂ W,

dim Wh < ∞ e.g. Raviart-Thomas elements of degree p: Wp

h = {wh ∈ H(div, Ω) : wh|K ∈ Pp(K) ∀K ∈ Th} ◮ Set values for β and CF ◮ Find yh ∈ Wh :

(div yh, div wh) + β C 2

F

(yh, wh) = β C 2

F

(∇uh, wh) − (f , div wh) ∀wh ∈ Wh

◮ Compute η(uh, yh)

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Friedrichs’ constant CF

v0 ≤ CF| | |v| | | ∀v ∈ V (a) Analytical estimate (Mikhlin, 1986): V = H1

0(Ω)

CF ≤ 1 π 1 |a1| + · · · + 1 |ad| −1/2 , Ω ⊂ a1 × · · · × ad,

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Friedrichs’ constant CF

v0 ≤ CF| | |v| | | ∀v ∈ V (b) Numerical upper bound: CF = sup

v∈V

v0 | | |v| | | ⇔ λ1 = inf

v∈V

| | |v| | |2 v2 , C 2

F = 1/λ1

Eigenvalue problem: ui ∈ V : a(ui, v) = λi(ui, v) ∀v ∈ V Galerkin approxim.: uh

i ∈ Vh : a(uh i , vh) = λh i (uh i , vh)

∀vh ∈ Vh Vh ⊂ V ⇒ λh

1 = inf vh∈Vh

| | |vh| | |2 vh2 ⇒ λ1 ≤ λh

1

⇒ 1/λh

1 ≤ C 2 F

Sigillito, Kuttler (1970s): λ

h 1 ≤ λ1

⇒ C 2

F ≤ 1/λ h 1

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 ∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0)

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 ∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (i) ⇔ (ii) y − ∇uh2

0 ≤ w − ∇uh2

y2

0 − 2(y, ∇uh) + ∇uh2 0 ≤ w2 0 − 2(w, ∇uh) + ∇uh2

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 ∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (i) ⇔ (ii) y − ∇uh2

0 ≤ w − ∇uh2

y2

0 − 2(y, ∇uh)

≤ w2

0 − 2(w, ∇uh)

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 ∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (i) ⇔ (ii) y − ∇uh2

0 ≤ w − ∇uh2

y2

0 − 2(f , uh)

≤ w2

0 − 2(f , uh)

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 ∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (i) ⇔ (ii) y − ∇uh2

0 ≤ w − ∇uh2

y2 ≤ w2

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 ∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (ii) ⇒ (iii) J(t) =

  • y + tw0

2

0,

J(t) has minimum at t = 0 0 = J′(0) = lim

t→0

  • y + tw0

2

0 − y2

t

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 ∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (ii) ⇒ (iii) J(t) =

  • y + tw0

2

0,

J(t) has minimum at t = 0 0 = J′(0) = lim

t→0

y2

0 + 2t(y, w0) + t2

w0 2

0 − y2

t = 2(y, w0)

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 ∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) Lemma 1: (i) ⇔ (ii) ⇔ (iii) Proof: (iii) ⇒ (ii) w ∈ Q(f ), ∃w0 ∈ Q(0) : w = y + w0, (y, w) = y2 0 ≤ w − y2

0 = w2 0 − 2(y, w) + y2 0 = w2 0 − y2

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 ∀y ∈ Q(f ) Q(f ) = {y ∈ H(div, Ω) : f + div y = 0} Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) Lemma 2: y = ∇u ∈ Q(f ) is the unique solution of (i)–(iii) Proof: If y1 ∈ Q(f ) and y2 ∈ Q(f ) satisfy (iii): ⇒ y2 − y1 ∈ Q(0) and (y2 − y1, w0) = 0 ⇒ y2 − y12

0 = 0

⇒ y2 = y1

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(B) Dual finite elements

| | |u − uh| | | ≤ η(uh, y) = y − ∇uh0 Q(f ) = {y ∈ H(div, Ω) : f + div y = 0}

y = ∇u yh ∇uh ∇V Q(f) [L2(Ω)]d

Complementary problem: (i) Find y ∈ Q(f ) : η(uh, y) ≤ η(uh, w) ∀w ∈ Q(f ) (ii) Find y ∈ Q(f ) :

1 2 y2 0 ≤ 1 2 w2

∀w ∈ Q(f ) (iii) Find y ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) Lemma 3:

  • η2(u, yh) +

η2(uh, y) = η2(uh, yh) ∀uh ∈ V , yh ∈ Q(f ) yh − y2

0 + ∇u − ∇uh2 0 = yh − ∇uh2

Proof: yh − ∇u + ∇u − ∇uh2

0 = yh − ∇u2 0 + ∇u − ∇uh2

(yh − ∇u, ∇u − ∇uh) = 0

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(B) Dual finite elements

Practical implementation: d = 2, Ω simply connected

◮ q(x1, x2) = −

x1 f (s, x2) ds, 0 ⊤ ⇒ − div q = f

◮ Q(f ) = q + Q(0) = q + curl H1(Ω)

curl = (∂2, −∂1)⊤

◮ Complementary problem:

(iii) y = q + curl z ∈ Q(f ) : (y, w0) = 0 ∀w0 ∈ Q(0) (iv) z ∈ H1(Ω) : (curl z, curl v) = −(q, curl v) ∀v ∈ H1(Ω) (v) z ∈ H1(Ω) : (∇z, ∇v) = −(q, curl v) ∀v ∈ H1(Ω)

◮ Galerkin approximation:

zh ∈ Zh ⊂ H1(Ω) : (∇zh, ∇vh) = −(q, curl vh) ∀vh ∈ Zh

◮ yh = q + curl zh ◮ Compute

η(uh, yh)

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Energy minimization

a(u, v) = (∇u, ∇v), F(v) = (f , v), a∗(y, w) = (y, w) Primal problem: u ∈ V : a(u, v) = F(v) ∀v ∈ V u ∈ V : J(u) = min

v∈V J(v),

J(v) = 1 2a(v, v) − F(v) Complementary problem: y ∈ Q(f ) : a∗(y, w0) = 0 ∀w0 ∈ Q(0) y ∈ Q(f ) : J∗(y) = min

w∈Q(f ) J∗(w),

J∗(w) = 1 2a∗(w, w) Complementarity of energies: J(u) + J∗(y) = −1 2a(u, u) + 1 2a∗(∇u, ∇u) = 0

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Method of hypercircle

y = ∇u yh ∇uh Guh ∇V Q(f) [L2(Ω)]d

Theorem: If

◮ u ∈ V is primal solution ◮ uh ∈ V , yh ∈ Q(f ) arbitrary ◮ Guh = (yh + ∇uh)/2

Then ∇u − Guh0 = 1 2 η(uh, yh). Proof: 4 ∇u − Guh2

0 = ∇u − yh + ∇u − ∇uh2

= ∇u − yh2

0 + ∇u − ∇uh2 0 = ∇uh − yh2

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Numerical illustration of hypercircle

−u′′(x) = 2 − 12x + 12x2 in (0, 1), u(0) = u(1) = 0 u(x) = x2(1 − x)2

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.01 0.02 0.03 0.04 0.05 0.06 0.07 Primal solution

x u(x) uh(x)

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Numerical illustration of hypercircle

−u′′(x) = 2 − 12x + 12x2 in (0, 1), u(0) = u(1) = 0 yh(x) = u′(x) = 2x(2x − 1)(x − 1) Guh(x) = (yh(x) + u′

h)/2

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 −0.2 −0.1 0.1 0.2 FEM approximation

x u′(x) u′

h(x)

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Numerical illustration of hypercircle

−u′′(x) = 2 − 12x + 12x2 in (0, 1), u(0) = u(1) = 0 yh(x) = u′(x) = 2x(2x − 1)(x − 1) Guh(x) = (yh(x) + u′

h)/2

∇u − Guh0 uh0 . = 9.3 %

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 −0.2 −0.1 0.1 0.2 Hypercircle approximation

x u′(x) Guh(x)

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Example 1

−∆u = f in (0, 2) × (0, 1) u = 0 on ΓD n⊤∇u = 0 on ΓN

ΓD ΓN

f = 5π2 16 u u = sin πx1 4 sin πx2 2

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Example 1

−∆u = f in (0, 2) × (0, 1) u = 0 on ΓD n⊤∇u = 0 on ΓN

ΓD ΓN

f = 5π2 16 u u = sin πx1 4 sin πx2 2 CF = 4 √ 5π . = 0.5694 C low

F

= 0.5693 C up

F = 0.6004

10 10

1

10

2

10

3

10

4

0.2 0.4 0.6 0.8 1 Friedrichs’ constant − Example 1 Number of elements Upper bound p=1 Upper bound p=2 Exact value Lower bound

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Example 1

−∆u = f in (0, 2) × (0, 1) u = 0 on ΓD n⊤∇u = 0 on ΓN

ΓD ΓN

Lower bound: reference solution Upper bound: error majorant

10 10

1

10

2

10

3

10

4

10

−1

10 Error bounds − Example 1 Number of elements Upper bound p=1 Upper bound p=2 Lower bound

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Example 2

−∆u = f in (0, 2) × (0, 1) u = 0 on ΓD n⊤∇u = 0 on ΓN

ΓD ΓN

f = 5π2 16 sin πx1 4 sin πx2 2

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Example 2

−∆u = f in (0, 2) × (0, 1) u = 0 on ΓD n⊤∇u = 0 on ΓN

ΓD ΓN

f = 5π2 16 sin πx1 4 sin πx2 2 CF = ? C low

F

= 0.7750 C up

F = 0.8712

10 10

1

10

2

10

3

10

4

0.25 0.5 0.75 1 1.25 1.5 Friedrichs’ constant − Example 2 Number of elements Upper bound p=1 Upper bound p=2 Lower bound

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Example 2

−∆u = f in (0, 2) × (0, 1) u = 0 on ΓD n⊤∇u = 0 on ΓN

ΓD ΓN

Lower bound: reference solution Upper bound: error majorant

10 10

1

10

2

10

3

10

4

10

−1

10 Error bounds − Example 2 Number of elements Upper bound p=1 Upper bound p=2 Lower bound

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Conclusions

◮ Guaranteed upper bounds

◮ if exact arithmetic ◮ if exact quadrature

◮ Arbitrary numerical method for uh ∈ V

◮ solving complementary problem is expensive

◮ Particular numerical method for uh ∈ V

◮ postprocessing of ∇uh

⇒ fast yh

◮ Total error

◮ including quadrature and algebraic errors

◮ Technical difficulties

◮ Friedrichs’ constant ◮ handling of Q(f )

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History

2000– S. Repin (S. Korotov, J. Valdman, S. Sauter, M. Frolov, . . . )

  • M. Vohral´

ık (R. Fuˇ c´ ık, I. Cheddadi, M.I. Prieto, . . . ) 1976– I. Hlav´ aˇ cek (J. Haslinger, M. Kˇ r´ ıˇ zek, J. Vacek, J. Weisz, . . . ) 1971 J.P. Aubin and H.G. Burchard 1957 J.L. Synge

Books:

  • P. Neittaanm¨

aki, S. Repin, Reliable methods for computer simulation, error control and a posteriori estimates, Elsevier, Amsterdam, 2004.

  • S. Repin, A posteriori estimates for partial differential

equations, de Gruyter, Berlin, 2008.

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References

  • T. Vejchodsk´

y: Complementary error bounds for elliptic systems and applications, in press Appl. Math. Comput., 2011. (Preprint 232.)

  • T. Vejchodsk´

y: Complementarity - the way towards guaranteed error estimates, in: Programs and Algorithms of Numerical Mathematics 15, Institute of Mathematics, Prague, 2010, pp. 205–220. (Preprint 231.)

  • M. Ainsworth, T. Vejchodsk´

y: Fully computable robust a posteriori error bounds for singularly perturbed reaction-diffusion problems, Numer. Math. 119 (2011) 219–243. (Preprint 208.)

  • T. Vejchodsk´

y: Complementarity based a posteriori error estimates and their properties, in press Math. Comput. Simulation, 2011. (Preprint 190.)

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Thank you for your attention

Part II – Complementary estimates Tom´ aˇ s Vejchodsk´ y vejchod@math.cas.cz

Institute of Mathematics, Academy of Sciences ˇ Zitn´ a 25, 115 67 Praha 1 Czech Republic

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A c a d e m y

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May 29, 2012, Technical University of Ostrava