Contents. Computational Fluid Dynamics. Fundamentals. Deformation of Materials

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1 Contents R. Verfürth Fakultät für Mathematik Ruhr-Universität Bochum Lecture Series / Milan / March 2010 of the Stokes Equations Non-Stationary Incompressible Navier-Stokes Equations Compressible and Inviscid Problems References 1/ 300 2/ 300 Modelization Deformation of Materials Modelization Notations and Auxiliary Results Notation: R d : domain initially occupied by a material moving under the influence of interior and exterior forces η : initial position of an arbitrary particle x = Φ(η, t): position of particle η at time t > 0 (t) = Φ(, t): domain occupied by the material at time t > 0 Basic assumptions: Φ(, t) : (t) is an orientation preserving diffeomorphism for all t > 0. Φ(, 0) is the identity. 3/ 300 4/ 300

2 Modelization Lagrange and Euler Representation Velocity Modelization Lagrange representation: Fix η and look at the trajectory t Φ(η, t). η is called Lagrange coordinate. The Langrange coordinate system moves with the fluid. Euler representation: Fix the point x and look at the trajectory t Φ(, t) 1 (x) which passes through x. x is called Euler coordinate. The Euler coordinate system is fixed. Velocity of the movement at the point x = Φ(η, t) is v(x, t) = Φ(η, t). t 5/ 300 6/ 300 Properties Modelization DΦ = ( Φ i η j ) 1 i,j d Jacobi matrix of Φ, J = det DΦ Jacobi determinant of Φ, A ij co-factors of DΦ (1 i, j d): t J = i,j = i,j = i,k (DΦ) ij J t (DΦ) ij ( 1) i+j A ij v i η j Jδ i,k v i x k = ( 1) i+j 2 A ij Φ i t η i,j j = ( 1) i+j A ij Φ k v i η j x k i,j,k = J div v Modelization Transport Theorem d dt = d dt = = V V (t) V f(x, t)dx f(φ(η, t), t)j(η, t)dη ( f(φ(η, t), t)j(η, t) t V (t) + f(φ(η, t), t) v(φ(η, t), t)j(η, t) ) + f(φ(η, t), t) div v(φ(η, t), t)j(η, t) dη ( t f(x, t) + div[ f(x, t)v(x, t) ]) dx 7/ 300 8/ 300

3 Modelization Conservation of Mass ρ denotes the density of the material. ρdx is the total mass of a control volume. V (t) Total mass is conserved: 0 = d ρdx = dt V (t) V (t) ( t ρ + div[ ρv ]) dx. This holds for every control volume, hence: t ρ + div[ ρv ] = 0. Modelization Conservation of Momentum V (t) ρvdx is the total momentum of a control volume. Its temporal change is d [ ] [ ] ρvdx = ρv + div ρv v dt V (t) V (t)( ) dx. t This is in equilibrium with exterior and interior forces. Exterior forces are given by ρfdx. V (t) 9/ / 300 Modelization Interior Forces Modelization Cauchy Theorem Basic assumptions: Interior forces act via the surface of a volume V (t). Interior forces only depend on the normal direction of the surface of the volume. Interior forces are additive and continuous. The previous assumptions imply: There is a tensor field T : R d d such that the interior forces are given by T nds. V (t) T is such that the divergence theorem of Gauß holds T nds = div Tdx. V (t) V (t) 11/ / 300

4 Modelization Conservation of Momentum (ctd.) The conservation of momentum and the Cauchy theorem imply: ( ) ( ) t (ρv) + div(ρv v) = ρf + div T. V (t) V (t) This holds for every control volume, hence: (ρv) + div(ρv v) = ρf + div T. t Modelization Conservation of Energy V (t) edx is the total energy of a control volume. Its temporal change is in equilibrium with the internal energy and the energy of exterior and interior forces. Exterior forces contribute ρf vdx. V (t) Interior forces give n T vds = div [ T v ] dx. V (t) V (t) The Cauchy theorem implies that the internal energy is of the form n σds = div σdx. V (t) V (t) Hence, conservation of energy implies t e + div(ev) = ρf v + div(t v) + div σ. 14/ / 300 Modelization Constitutive Laws Basic assumptions: T only depends on the gradient of the velocity. The dependence on the velocity gradient is linear. T is symmetric. (Due to the Cauchy theorem this is a consequence of the conservation of angular momentum.) In the absence of internal friction, T is diagonal and proportional to the pressure, i.e. all interior forces act in normal direction. The total energy e is the sum of internal and kinetic energy. σ is proportional to the variation of the internal energy. Modelization Consequences of the Constitutive Laws Above assumptions imply: T = 2λD(v) + µ(div v) I pi, where D(v) = 1 2 ( v + vt ) is the deformation tensor, λ, µ are the dynamic viscosities, p is the pressure, I is the unit tensor. e = ρε ρ v 2, where ε is often identified with the temperature. σ = α ε. 15/ / 300

5 Modelization Compressible Navier-Stokes Equations in Conservative Form t ρ + div(ρv) = 0 (ρv) + div(ρv v) = ρf + 2λ div D(v) t + µ grad div v grad p e + div(ev) = ρf v + 2λ div[d(v) v] t + µ div[div v v] div(pv) + α ε p = p(ρ, ε) Modelization Euler Equations Inviscid flows, i.e. λ = µ = 0: t ρ + div(ρv) = 0 (ρv) + div(ρv v + pi) = ρf t e + div(ev + pv) = ρf v + α ε t p = p(ρ, ε) e = ρε ρ v 2 e = ρε ρ v 2 17/ / 300 Modelization Compressible Navier-Stokes Equations in Non-Conservative Form Insert first equation in second one and first and second equation in third one: t ρ + div(ρv) = 0 ρ[ v + (v )v] = ρf + λ v + (λ + µ) grad div v grad p t ρ[ t ε + ρv grad ε] = λd(v) : D(v) + µ(div v)2 p div v + α ε p = p(ρ, ε) Modelization Non-Stationary Incompressible Navier-Stokes Equations Assume that the density ρ is constant, replace p by p ρ, denote by ν = λ ρ the kinematic viscosity, drop the energy equation: div v = 0 v + (v )v = f + ν v grad p t 19/ / 300

6 Modelization Reynolds Number Introduce a reference length L, a reference time T, a reference velocity U, a reference pressure P, and a reference force F and new variables and quantities by x = Ly, t = T τ, v = Uu, p = P q, f = F g. Choose T, F and P such that T = L U, F = νu L 2 Then div u = 0 u + Re(u )u = f + u grad q, t and P L νu = 1. Modelization Stationary Incompressible Navier-Stokes Equations Assume that the flow is stationary: div v = 0 ν v + (v )v + grad p = f where Re = LU ν is the dimensionless Reynolds number. 21/ / 300 Modelization Stokes Equations Modelization Boundary Conditions Around 1827, Pierre Louis Marie Henri Navier suggested the general boundary condition Linearize at velocity v = 0: div v = 0 v + grad p = f λ n v n + (1 λ n )n T n = 0 λ t [v (v n)n] + (1 λ t )[T n (n T n)n] = 0 with parameters λ n, λ t [0, 1] depending on the actual flow-problem. A particular case is the slip boundary condition v n = 0, T n (n T n)n = 0. Around 1845, Sir George Gabriel Stokes suggested the no-slip boundary condition v = 0. 23/ / 300

7 Notations and Auxiliary Results Sobolev Spaces and Norms L 2 () Lebesgue space with norm ϕ = ϕ = { ϕ2 } 1 2 H k () = {ϕ L 2 () : D α ϕ L 2 () α α d k}, k 1, Sobolev spaces with semi-norm { } 1 ϕ k, = ϕ k = α1+...+αd=k Dα ϕ 2 2 and norm ϕ k, = ϕ k = { k l=0 ϕ 2 l } 1 2 Norms of vector- or tensor-valued functions are defined component-wise. H 1 0 () = {ϕ H1 () : ϕ = 0 on Γ = } V = {v H 1 0 ()d : div v = 0} L 2 0 () = {ϕ L2 () : ϕ = 0} Notations and Auxiliary Results Poincaré, Friedrichs and Trace Inequalities Poincaré inequality: ϕ c P diam() ϕ 1 for all ϕ H 1 () L 2 0 () c P = 1 π if is convex. Friedrichs inequality: ϕ c F diam() ϕ 1 for all ϕ H0 1() { } 1 Trace inequality: ϕ Γ c T,1 () ϕ 2 + c T,2 () ϕ for all ϕ H 1 () c T,1 () diam() 1, c T,2 () diam() if is a simplex or parallelepiped 25/ / 300 Notations and Auxiliary Results Finite Element Meshes T Γ is the union of all elements in T. Affine equivalence: Each T is either a triangle or a parallelogram, if d = 2, or a tetrahedron or a parallelepiped, if d = 3. Admissibility: Any two elements in T are either disjoint or share a vertex or a complete edge or if d = 3 a complete face. Shape-regularity: For every element, the ratio of its diameter h to the diameter ρ of the largest ball inscribed into is bounded independently of. Mesh-size: h = h T = max T h Notations and Auxiliary Results Finite Element Spaces span{x α xα d d : α α d k} if is a triangle or a tetrahedron R k () = span{x α xα d d : max{α 1,..., α d } k} if is a parallelogram or a parallelepiped S k, 1 (T ) = {ϕ : R : ϕ R k () T } S k,0 (T ) = S k, 1 (T ) C() S k,0 0 (T ) = Sk,0 (T ) H 1 0 () = {ϕ S k,0 (T ) : ϕ = 0 on Γ} 27/ / 300

8 Notations and Auxiliary Results Approximation Properties Notations and Auxiliary Results Vertices and Faces inf ϕ ϕ T ch k+1 ϕ k+1 ϕ T S k, 1 (T ) ϕ H k+1 (), k N inf ϕ ϕ T j ch k+1 j ϕ k+1 ϕ T S k,0 (T ) inf ϕ T S k,0 0 (T ) ϕ ϕ T j ch k+1 j ϕ k+1 ϕ H k+1 (), j {0, 1}, k N ϕ H k+1 () H 1 0 (), j {0, 1}, k N N : set of all element vertices E: set of all (d 1)-dimensional element faces A subscript, or Γ to N or E indicates that only those vertices or faces are considered that are contained in the respective set. 29/ / 300 Patches Notations and Auxiliary Results ω = E E ω = N N ω E = E E ω E = N E N ω z = z N Notations and Auxiliary Results Nodal Shape Functions λ z denotes the nodal shape function associated with the vertex z. It is uniquely defined by the conditions λ z S 1,0 (T ), λ z (z) = 1, λ z (y) = 0 y N \ {z}. ω z is the support of λ z. 31/ / 300

9 Notations and Auxiliary Results A Quasi-Interpolation Operator Define the quasi-interpolation operator R T : L 1 () S 1,0 0 (T ) by R T ϕ = z N λ z ϕ z with ϕ z = ω z ϕdx ω z dx. It has the following local approximation properties for all ϕ H 1 0 () ϕ R T ϕ c A1 h ϕ 1, ω ϕ R T ϕ c A2 h 1 2 ϕ 1, ω. Notations and Auxiliary Results Proof of the Local Approximation Properties The Poincaré inequality implies for every vertex z ϕ ϕ z ωz c z diam(ω z ) ϕ 1,ωz. The trace inequality yields for all faces E of all elements ϕ E c 1 h 1 2 ϕ + c 2 h 1 2 ϕ 1,. The properties of the nodal shape functions imply ϕ R T ϕ ϕ ϕ z + ϕ z. z N z N,Γ The first term is bounded using the Poincaré inequality, the second one using ϕ H 1 0 (). 33/ / 300 Notations and Auxiliary Results Bubble Functions Define element and face bubble functions by ψ = α λ z, ψ E = α E λ z. z N z N E The weights α and α E are determined by the conditions max x ψ (x) = 1, max ψ E(x) = 1. x E is the support of ψ ; ω E is the support of ψ E. Notations and Auxiliary Results Inverse Estimates for the Bubble Functions For all elements, all faces E and all polynomials ϕ the following inverse estimates are valid c I1,k ϕ ψ 1 2 ϕ, (ψ ϕ) c I2,k h 1 ϕ, c I3,k ϕ E ψ 1 2 E ϕ E, (ψ E ϕ) ωe c I4,k h 1 2 E ϕ E, ψ E ϕ ωe c I5,k h 1 2 E ϕ E. 35/ / 300

10 Notations and Auxiliary Results Proof of the Inverse Estimates Jumps Notations and Auxiliary Results Transform the left hand-sides to the reference simplex or cube. Take into account that the left-hand sides define semi-norms. Invoke the equivalence of norms on finite dimensional spaces to prove the corresponding estimates on the reference element. Transform the right-hand sides back to the current element or face. n E : a unit vector perpendicular to a given face E [ϕ] E : jump of a given piece-wise continuous function across a given face E in the direction of n E [ϕ] E depends on the orientation of n E but quantities of the form [n E ϕ] E are independent thereof. 37/ / 300 A First Attempt Abstract Saddle-Point Problems Saddle-Point Formulation of the Stokes Equations A First Attempt A Variational Formulation of the Stokes Equations Stokes equations with no-slip boundary condition u + grad p = f in, div u = 0 in, u = 0 on Γ Multiply momentum equation with v V = {w H 1 0 ()d : div w = 0}, integrate over and use integration by parts. Resulting variational formulation: Find u V such that for all v V u : v = f v 39/ / 300

11 A First Attempt Corresponding A First Attempt The Space V (T ) = S 1,0 0 (T )d V Find u T V (T ) V such that for all v T V (T ) u T : v T = f v T Advantage: The discrete problem is symmetric positive definite. Disadvantage: The discrete problem gives no information on the pressure. Candidate for lowest order discretization: V (T ) = S 1,0 0 (T )d V. = (0, 1) 2 T Courant triangulation consisting of 2N 2 isosceles right-angled triangles with short sides of length h = N 1 v T S 1,0 0 (T )d V arbitrary 41/ / 300 A First Attempt The Space V (T ) = S 1,0 0 (T )d V A First Attempt The Space V (T ) = S 1,0 0 (T )d V 0 = div v T on 0 = div v T = 0 = n v T n v T = 2h 1 2 ( 1 1 ) vt (x) + h ( 1 0 ) v T (x) = h ( 0 1 ) v T (x) 0 = div v T on 0 = div v T = 0 = n v T n v T = 2h 1 2 ( 1 1 ) vt (x) + h ( 0 1 ) v T (x) = h ( 1 0 ) v T (x) 43/ / 300

12 A First Attempt The Space V (T ) = S 1,0 0 (T )d V A First Attempt Finite Element Subspaces of V v T = 0 in bottom left square Sweeping through squares yields: v T = 0 in Hence: S 1,0 0 (T )d V = {0} In order to obtain a non-trivial space S k,0 0 (T )d V, the polynomial degree k must be at least 5. Despite the high polynomial degree, the approximation properties of S k,0 0 (T )d V are rather poor. 45/ / 300 A First Attempt Another Variational Formulation of the Stokes Equations Multiply the momentum equation with v H0 1()d, integrate over and use integration by parts. Multiply the continuity equation with q L 2 0 () and integrate over. Resulting variational formulation: Find u H0 1()d and p L 2 0 () such that for all v H0 1()d and q L 2 0 () u : v p div v = f v q div u = 0 A First Attempt Corresponding Choose finite element spaces X(T ) H 1 0 ()d and Y (T ) L 2 0 (). Find u T X(T ) and p T Y (T ) such that for all v T X(T ) and q T Y (T ) u T : v T p T div v T = f v q T div u T = 0 47/ / 300

13 Questions A First Attempt Abstract Saddle-Point Problems The Setting Does the variational problem admit a unique solution? Does the discrete problem admit a unique solution? What is the quality of the approximation? What are good choices for the discrete spaces? X and Y are Hilbert spaces with norms X and Y. a : X X R and b : X Y R are continuous bilinear forms. l : X R and χ : Y R are continuous linear functionals. Problem: Find u X and λ Y such that for all v X and µ Y a(u, v) + b(v, λ) = l, v b(u, µ) = χ, µ (S) 49/ / 300 Abstract Saddle-Point Problems Auxiliary Operators and Spaces Define continuous linear operators A : X X, B : X Y, B : Y X by setting for all u, v X, λ Y Set Au, v = a(u, v), Bu, λ = b(u, λ), B λ, v = b(u, λ). V = ker B, V = {g X : g, v = 0 v V }, V = {u X : (u, v) X = 0 v V }. Define the continuous linear operator π : X V by πf, v = f, v v V. Abstract Saddle-Point Problems The Inf-Sup Condition The following conditions are equivalent: 1. There is a constant β > 0 such that (inf-sup condition) inf sup λ Y \{0} u X\{0} b(u, λ) u X λ Y β. 2. B is an isomorphism of Y onto V and B λ X β λ Y for all λ Y. 3. B is an isomorphism of V onto Y and Bu Y β u X for all u X. 51/ / 300

14 Abstract Saddle-Point Problems Motivation of the Inf-Sup Condition Assume that X = R n, Y = R m with m < n and b(u, λ) = λ T Bu with a rectangular matrix B R m n. Then the following conditions are equivalent: B has maximal rang m. The rows of B are linearly independent. λ T Bu = 0 for all u R n implies λ = 0. inf λ sup u λ T Bu u λ > 0. The linear system B T λ = 0 only admits the trivial solution. For every f R m there is a unique u R n which is orthogonal to ker B and which satisfies Bu = f. Abstract Saddle-Point Problems Proof of the Equivalences Condition (1), the definition of B and X imply B λ X = b(u, λ) sup β λ Y. u X\{0} u X Hence, B is injective and its range is closed. The closed graph theorem then proves (2) This is a consequence of the above equality From the definitions of V and V one concludes that V and (V ) are isometric. Hence, B is an isomorphism of V onto Y if and only if B is an isomorphism of (Y ) Y onto (V ) V and both isomorphisms have the same norm. 53/ / 300 Abstract Saddle-Point Problems Well-Posedness of Problem (S) Abstract Saddle-Point Problems Proof of Problem (S) admits a unique solution for every right-hand side if and only if (i) πa is an isomorphism of V onto V and (ii) b satisfies the inf-sup condition. If problem (S) is well-posed, its solution satisfies u X + λ Y c { l X + χ Y }. The constant c grows with β 1. Due to (ii) there is a unique u 0 V with Bu 0 = χ and u 0 X 1 β χ Y. Due to (i) there is a unique w V with πaw = π(l Au 0 ) and w X (πa) 1 L(V,V ) l Au 0 X. u = u 0 + w satisfies π(l Au) = 0 whence l Au V. Due to (ii) there is a unique λ Y with B λ = l Au and λ Y 1 β l Au X. u, λ solve (S) and satisfy the stability estimate. 55/ / 300

15 Proof of Abstract Saddle-Point Problems Problem (S) with l = 0 and arbitrary χ Y admits a unique solution. Hence Y = range B. The open mapping theorem proves that B is an isomorphism and thus establishes (ii). Consider a u V with πau = 0. Due to (ii), there is a unique λ Y with B λ = Au. Thus u, λ solve (S) with homogeneous right-hand side. Hence, u = 0 and πa is injective. Due to the Hahn-Banach theorem, for every g V, there is an l X with πl = g. Problem (S) admits a unique solution u, λ for the right-hand side l, χ = 0. Hence, there is a u X with πau = g and πa is surjective. The open mapping theorem proves that πa is an isomorphism and thus establishes (i). Abstract Saddle-Point Problems Coercive Forms a Assume that a is symmetric and coercive on X, i.e. there is an α > 0 such that a(u, u) α u 2 X holds for all u X. Then: πa is an isomorphism and (πa) 1 L(V,V ) 1 α. Problem (S) is well-posed if and only if the form b satisfies the inf-sup condition. The solution of problem (S) is the unique saddle-point of the functional L(u, λ) = 1 2a(u, u) + b(u, λ) l, u χ, λ. The solution u of (S) minimizes the functional J(u) = 1 2a(u, u) l, u under the constraint b(u, µ) = χ, µ for all µ Y. 57/ / 300 Abstract Saddle-Point Problems of Saddle-Point Problems Replace X and Y by finite dimensional subspaces X n and Y n. Resulting discrete problem: Find u n X n and λ n Y n such that for all v n X n and µ n Y n a(u n, v n ) + b(v n, λ n ) = l, v n (S n ) b(u n, µ n ) = χ, µ n Abstract Saddle-Point Problems Well-Posedness of Problem (S n ) Assume for simplicity that the form a is coercive on X. Then: Problem (S n ) is well posed if and only if the form b satisfies the discrete inf-sup condition inf sup λ n Y n\{0} u n X n\{0} b(u n, λ n ) u n X λ n Y β n > 0. If problem (S n ) is well-posed, its solution satisfies u n X + λ n Y c { l X + χ Y }. The constant c grows with β 1 n. 59/ / 300

16 Abstract Saddle-Point Problems Error Estimates Assume that a is coercive and that b satisfies both the inf-sup condition and the discrete inf-sup condition. Denote by u, λ the unique solution of problem (S) and by u n, λ n the unique solution of problem (S n ). Then there is a constant c which grows with β 1 n u u n X + λ λ n Y { } c inf u v n X + inf λ µ n Y v n X n µ n Y n such that Abstract Saddle-Point Problems Proof of the Error Estimates For any v n X n, µ n Y n define l X, χ Y by l, v = a(u v n, v) + b(v, λ µ n ), χ, µ = b(u v n, µ). Then l X + χ Y c { u v n X + λ µ n Y }. Subtracting problems (S) and (S n ) gives for every w n X n and ρ n Y n l, w n = a(u n v n, w n ) + b(w n, λ n µ n ), χ, ρ n = b(u n v n, ρ n ). The stability estimate for problem (S n ) and the triangle inequality now prove the error estimate. 61/ / 300 Abstract Saddle-Point Problems A Duality Argument H is another Hilbert space with norm H such that X is dense in H with continuous injection. For every g H denote by u g, λ g the solution of problem (S) with l = g and χ = 0. Then u and u n satisfy the error estimate u u n H c { u u n X + λ λ n Y } 1 { sup inf u v n X + g H\{0} g H v n X n } inf λ µ n Y. µ n Y n Proof Abstract Saddle-Point Problems The density of X in H implies u u n H = (g, u u n ) H sup. g H\{0} g H Subtracting (S) and (S n ) and using the definition of u g, λ g yields for every v n X n, µ n Y n (g, u u n ) H = a(u u n, u g ) + b(u u n, λ g ) + b(u g, λ λ n ) }{{} =0 = a(u u n, u g v n ) + b(u g v n, λ λ n ) + b(u u n, λ g µ n ) + a(u u n, v n ) + b(v n, λ λ n ) + b(u u }{{} n, µ n ). }{{} =0 =0 63/ / 300

17 Saddle-Point Formulation of the Stokes Equations Saddle-Point Formulation of the Stokes Equations The saddle-point formulation of the Stokes equations fits into the abstract framework with: X = H 1 0 () d, Y = L 2 0(), H = L 2 () d a(u, v) = u : v, b(u, p) = p div u l, v = f v, χ = 0 The bilinear form a is coercive on X. Hence, we only have to ascertain the inf-sup condition p div u inf sup u 1 p β > 0. p L 2 0 ()\{0} u H0 1()d \{0} A Proof in R 2 Saddle-Point Formulation of the Stokes Equations Assume that R 2 is either convex or has a C 2 boundary. Choose an arbitrary p L 2 0 (). Set v = ϕ where ϕ H 2 () L 2 0 () is the unique weak solution of the Neumann problem ϕ = p in, ϕ = 0 on Γ. n Set w = ( ψ x 2, ψ x 1 ) where ψ H 2 () is the unique weak solution of the biharmonic equation 2 ψ = 0 in, ψ = 0 on Γ, ψ n = v t on Γ. Set u = v + w. Then div u = p and u 1 c p. 65/ / 300 Saddle-Point Formulation of the Stokes Equations A Proof by Duvaut, Lions and Nečas. 1 st Step p c() { p 1 + p 1 } Set X() = {p H 1 () : p H 1 () d } equipped with p = p 1 + p 1. The definition H 1 () and the open mapping theorem imply that it suffices to prove the inclusion X() L 2 (). Due to the characterization of Sobolev spaces by Fourier transforms, the inclusion holds for R d. Using suitable reflections shows that the inclusion also holds for C -functions on R d 1 R +. The Hahn-Banach theorem implies that C (R d 1 R + ) is dense in X(R d 1 R + ). Combining the previous results with suitable partitions of unity establishes the inclusion for all Lipschitz domains. 67/ 300 Saddle-Point Formulation of the Stokes Equations A Proof by Duvaut, Lions and Nečas. 2 nd Step p c() p 1 Assume the contrary. Then there is a sequence (p n ) in L 2 0 () with p n = 1 and p n 1 1 n for all n. Since H0 1() is compactly embedded in L2 (), the latter space is compactly embedded in H 1 (). Hence, there is subsequence (p nk ) such that p nk p strongly in H 1, p nk p weakly in L 2 and p nk v p v for all C vector-fields v. This proves p = 0 and, since p n L 2 0 (), p = 0. This contradicts the estimate on the previous slide. 68/ 300

18 Saddle-Point Formulation of the Stokes Equations A Proof by Duvaut, Lions and Nečas. 3 rd Step The inf-sup condition is fulfilled. The operator grad : L 2 0 () H 1 () d is injective and continuous. The previous result implies that range(grad) is a closed subspace of H 1 () d. The open mapping theorem implies that grad is an isomorphism of L 2 0 () onto range(grad). The closed range theorem implies that range(grad) = ker(div) = V. Due to the abstract results, this proves the inf-sup condition. 69/ 300 Saddle-Point Formulation of the Stokes Equations A Proof by Bogovskii Assume that the domain is the union of a finite number of (eventually overlapping) subdomains which are star-shaped with respect to an inscribed ball. Then the inf-sup condition holds. A suitable additive decomposition of the pressure and velocity shows that it suffices to establish the inf-sup condition for a single subdomain. Consider a subdomain ω which is star-shaped with respect to an open ball with ω. Choose a C -function ϕ with support in and ϕ = 1. Properties of singular integrals imply that u(x) = p(y) x y x y d ϕ ( y + t x y ) t d 1 dtdy x y ω x y satisfies div u = p in ω and u 1,ω c diam(ω) p ω. 70/ 300 Saddle-Point Formulation of the Stokes Equations A Regularity Result Assume that the boundary Γ is of class C m+2 and that f H m () d. Then the weak solution of the Stokes problem satisfies: u H m+2 () d H0 1 () d, p H m+1 () L 2 0(), u m+2 + p m+1 c() f m. If is a convex polyhedron, the above regularity result holds with m = 0. Saddle-Point Formulation of the Stokes Equations Finite Element The finite element discretization of the Stokes equations fits into the abstract framework with X n = X(T ), Y n = Y (T ) The bilinear form a is coercive on X. Hence, we only have to ascertain the discrete inf-sup condition p T div u T inf sup p T Y (T )\{0} u T 1 p T β T > 0. u T X(T )\{0} In order to obtain optimal error estimates, the discretization must be uniformly stable, i.e. β T β > 0 for all T. 71/ / 300

19 Saddle-Point Formulation of the Stokes Equations Resulting Error Estimates Assume: u H k+1 () d H0 1 () d, p H k () L 2 0(). The discretization is uniformly stable. S k,0 (T ) d X(T ). S k 1,0 (T ) L 2 0() Y (T ) or S k 1, 1 (T ) L 2 0() Y (T ). Then: u u T 1 + p p T ch k{ u k+1 + p k }. If in addition is a convex polyhedron, then: u u T ch k+1{ u k+1 + p k }. Saddle-Point Formulation of the Stokes Equations Approximation of the Space V The space V = {v H0 1()d : div v = 0} is approximated by { } V (T ) = v T X(T ) : p T div v T = 0 p T Y (T ). For almost all discretizations used in practice V (T ) is not contained in V. In this sense, all these discretizations are non-conforming and not fully conservative. 73/ / 300 of the Stokes Equations of the Stokes Equations of the Stokes Equations A Second Attempt The P 1/P 0-Element A Second Attempt Stable Finite Element Pairs Petrov-Galerkin Methods Non-Conforming s Stream-Function Formulation T is a triangulation of a two-dimensional domain. X(T ) = S 1,0 0 (T )d, Y (T ) = S 0, 1 (T ) L 2 0 () Every solution u T X(T ), p T Y (T ) of every discrete Stokes problem satisfies: div ut is element-wise constant and div u T = 0 for every T. Hence, div ut = 0. Our first attempt yields u T = 0. Hence, this pair of finite element spaces is not stable and not suited for the discretization of the Stokes problem. 75/ / 300

20 of the Stokes Equations A Second Attempt The Q1/Q0-Element T is a partition of the unit square = (0, 1) 2 into N 2 squares with sides of length h = N 1 where N 2 is even. X(T ) = S 1,0 0 (T )d, Y (T ) = S 0, 1 (T ) L 2 0 () Denote by ij the square with bottom left corner (ih, jh). p T Y (T ) is the pressure with p T ij = ( 1) i+j (checker-board mode) Then p T div v T = 0 for every v T X(T ) (checker-board instability). Hence, this pair of finite element spaces is not stable and not suited for the discretization of the Stokes problem. of the Stokes Equations A Second Attempt Proof of the Checker-Board Instability div v T dx = v T n ij ds ij ij (ih,jh) = h { v T (ih, jh) ( ) vt ((i + 1)h, jh) ( ) v T ((i + 1)h, (j + 1)h) ( 1 1 ) + v T (ih, (j + 1)h) ( ) } 1 1 p T div v T dx = ( 1) i+j div v T dx = 0 i,j ij 77/ / 300 Conclusions of the Stokes Equations A Second Attempt The velocity space must contain enough degrees of freedom in order to balance element-wise the gradient of the pressure, face-wise the jump of the pressure. of the Stokes Equations Stable Finite Element Pairs An Auxiliary Result Define a mesh-dependent norm on S k, 1 (T ) by { ϕ 1,T = h 2 ϕ 2 + } 1 h E [ϕ] E 2 2 E. E E T Assume that: S 1,0 0 (T )d X(T ), Y (T ) S k, 1 (T ) for some k, There is a constant β > 0 independent of T such that p T div u T inf sup p T Y (T )\{0} u T 1 p T β. 1,T u T X(T )\{0} Then the pair X(T ), Y (T ) is uniformly stable. 79/ / 300

21 of the Stokes Equations Stable Finite Element Pairs Proof of the Auxiliary Result. 1 st Step Choose a pressure p T Y (T ) with p T = 1. Due to the well-posedness of the Stokes problem, there is a velocity u H0 1(T )d with u 1 = 1 and p T div u β. R T u satisfies R T u 1 c 1 u 1 = c 1, p T div(r T u) = p T div u + p T div(r T u u) β + p T div(r T u u). of the Stokes Equations Stable Finite Element Pairs Proof of the Auxiliary Result. 2 nd Step Integration by parts and the properties of R T imply p T div(r T u u) = p T (u R T u) + [p T ] E (R T u u) n E T E E E c 2 p T 1,T u 1. The last two estimates yield p T div u T sup u T X(T )\{0} u T 1 1 c 1 { β c2 p T 1,T }. 81/ / 300 of the Stokes Equations Stable Finite Element Pairs Proof of the Auxiliary Result. 3 rd Step The previous estimate and the third assumption imply p T div u T sup u T X(T )\{0} u T 1 max { β pt 1,T, 1 { } } β c2 p T 1,T c 1 min z 0 max { βz, 1 c 1 { β c2 z }} = β β c 1 β + c2. of the Stokes Equations Stable Finite Element Pairs The Bernardi-Raugel Element The following pair of finite element spaces is uniformly stable: T is any affine equivalent partition of a two or three dimensional domain. X(T ) = S 1,0 0 (T )d span{ψ E n E : E E} Y (T ) = S 0, 1 (T ) L 2 0 () 83/ / 300

22 of the Stokes Equations Stable Finite Element Pairs Proof of the Stability of the Bernardi-Raugel Element For p T Y (T ) set u T = E E h E [p T ] E n E. Integration by parts element-wise and the properties of the bubble-functions imply p T div u T = [p T ] E u T n E β p T 2 1,T E E E of the Stokes Equations Stable Finite Element Pairs The Mini Element of Brezzi-Fortin The following pair of finite element spaces is uniformly stable: T is any simplicial partition of a two or three dimensional domain. X(T ) = S 1,0 0 (T )d span{ψ : T } d and u T 1 c p T 1,T. Y (T ) = S 1,0 (T ) L 2 0 () Hence, the auxiliary result proves the stability. 85/ / 300 of the Stokes Equations Stable Finite Element Pairs Proof of the Stability of the Mini Element For p T Y (T ) set u T = h 2 ψ p T. T Integration by parts element-wise and the properties of the bubble-functions imply p T div u T = p T u T β p T 2 1,T T of the Stokes Equations Stable Finite Element Pairs The Hood-Taylor Element The following pair of finite element spaces is uniformly stable: T is any simplicial partition of a two or three dimensional domain. X(T ) = S 2,0 0 (T )d and u T 1 c p T 1,T. Hence, the auxiliary result proves the stability. Y (T ) = S 1,0 (T ) L 2 0 () 87/ / 300

23 of the Stokes Equations Stable Finite Element Pairs The Modified Hood-Taylor Element The following pair of finite element spaces is uniformly stable: T is any simplicial partition of a two or three dimensional domain. T /2 is obtained from T by uniform refinement connecting the midpoints of edges. X(T ) = S 1,0 0 (T /2)d Y (T ) = S 1,0 (T ) L 2 0 () of the Stokes Equations Stable Finite Element Pairs Proof of the Stability of the Hood-Taylor and Modified Hood-Taylor Elements For every p T Y (T ) there is a u T X(T ) such that u T coincides with the tangential derivative of p T at the midpoints of edges. Bercovier and Pironneau proved in 1979 that with this choice of u T the third condition of the auxiliary result is fulfilled. Hence, the auxiliary result proves the stability. 89/ / 300 of the Stokes Equations Stable Finite Element Pairs A Catalogue of Stable Elements of the Stokes Equations Petrov-Galerkin Methods Properties of the Mini Element The previous arguments can be modified to prove that the following pairs of spaces are uniformly stable on any affine equivalent partition in R d, d 2: X(T ) = S k,0 0 (T )d span{ϕψ E n E : E E, ϕ R k 1 (E)} span{ρψ : T, ρ R k 2 ()} d, Y (T ) = S k 1, 1 (T ) L 2 0 (), k 2 X(T ) = S k+d 1,0 0 (T ) d, Y (T ) = S k 1, 1 (T ) L 2 0 (), k 2 X(T ) = S k,0 0 (T )d, Y (T ) = S k 1,0 (T ) L 2 0 (), k 3 ψ ψ = 0 for all ϕ ψ = ϕ ψ = for all ϕ S 1,0 (T ), T ϕψ = 0 Hence, the bubble part of the velocity of the mini element can be eliminated by static condensation. The resulting system only incorporates linear velocities and pressures. 91/ / 300

24 of the Stokes Equations Petrov-Galerkin Methods The Mini Element with Static Condensation Original system: A l 0 B T l u l f l 0 D b Bb T u b = f b B l B b 0 p 0 System with static condensation: ( Al Bl T ) ( ) ( ul f B l B b D 1 b Bb T = l p B b D 1 b A straightforward calculation yields: ( Bb D 1 b Bb T ) i,j h 2 T f b λ i λ j ) of the Stokes Equations Petrov-Galerkin Methods Idea of Petrov-Galerkin Methods Try to obtain control on the pressure by adding element-wise terms of the form δ h 2 p T q T, face-wise terms of the form δe h E [p T ] E [q T ] E. The form of the scaling parameters is motivated by the Mini element and the request that element and face contributions should be of comparable size. The resulting problem should be coercive. Contrary to penalty methods, the additional terms should be consistent with the variational problem, i.e. they should vanish for the weak solution of the Stokes problem. Pressure-jumps are no problem. Test the momentum equation element-wise with δ h 2 q T. E 93/ / 300 of the Stokes Equations Petrov-Galerkin Methods General Form of Petrov-Galerkin Methods Find u T X(T ), p T Y (T ) such that for all v T X(T ), q T Y (T ) u T : v T p T div v T = f v T q T div u T + δ h 2 ( u T + p T ) q T T + δ E h E T ] E [q T ] E = E E T E[p δ h 2 f q T T of the Stokes Equations Petrov-Galerkin Methods Choice of Stabilization Parameters Set δ max = max{max δ, max δ E }, T E E T min{ min δ, min δ E } T E E T δ min = min δ T if pressures are discontinuous, if pressures are continuous. A reasonable choice of the stabilization parameters then is determined by the condition δ max δ min. 95/ / 300

25 of the Stokes Equations Petrov-Galerkin Methods Choice of Spaces Optimal with respect to error estimates versus degrees of freedom: X(T ) = S k,0 0 (T )d { S k 1,0 (T ) L 2 0 Y (T ) = () S k 1, 1 (T ) L 2 0 () Equal order interpolation: X(T ) = S k,0 0 (T )d { S k,0 (T ) L 2 0 Y (T ) = () S k, 1 (T ) L 2 0 () continuous pressure discontinuous pressure continuous pressure discontinuous pressure of the Stokes Equations Petrov-Galerkin Methods Mesh-Dependent Norms and (Bi-)Linear Forms (u T, p T ) 1,T = { } 1 u T p T 2 + p T 2 2 1,T B T ((u T, p T ), (v T, q T )) = u T : v T p T div v T + q T div u T + δ h 2 ( u T + p T ) q T T + δ E h E [p T ] E [q T ] E E E E l T ((v T, q T )) = f v T + δ h 2 f q T T 97/ / 300 of the Stokes Equations Petrov-Galerkin Methods Stability of the Petrov-Galerkin Assume that δ min > 0 and δ max < δ 0 where δ 0 only depends on the shape parameter of T. Then there is a constant γ > 0 which does not depend on T such that inf sup (u T,p T ) (v T,q T ) B T ((u T, p T ), (v T, q T )) (u T, p T ) 1,T (v T, q T ) 1,T γ. of the Stokes Equations Petrov-Galerkin Methods Proof of the Stability Inverse estimates imply that B T ((u T, p T ), (u T, p T )) 1 2 δ min { (u T, p T ) 2 1,T p T 2} Due to the well-posedness of the Stokes problem, there is a velocity v H0 1(T )d with v 1 = p T and p T div v β p T 2. The properties of R T imply that B T ((u T, p T ), (R T v, 0)) ( β2) p T 2 β 2 (u T, p T ) 2 1,T, (R T v, 0) 1,T c p T. Taking the maximum of both estimates proves the stability. 99/ / 300

26 of the Stokes Equations Petrov-Galerkin Methods Error Estimates There is a constant c δ max γ 1 such that (u u T, p p T ) 1,T { c inf (u v T, p q T ) 2 1,T (v T,q T ) + T h 2 (u v T ) 2 If u H k+1 () d H0 1()d, p H k () L 2 0 (), S k,0 0 (T )d X(T ) and S k 1, 1 (T ) L 2 0 () Y (T ) or S k 1,0 (T ) L 2 0 () Y (T ) then (u u T, p p T ) 1,T c h k{ u k+1 + p k }. } 1 2. of the Stokes Equations Petrov-Galerkin Methods Proof of the Error Estimates The stability and the definitions of B T and l T yield (u u T, p p T ) 1,T (u v T, p q T ) 1,T + (v T u T, q T p T ) 1,T (v T u T, q T p T ) 1,T 1 B T ((v T u T, q T p T ), (w T, r T )) sup γ (w T,r T ) (w T, r T ) 1,T B T ((v T u T, q T p T ), (w T, r T )) = B T ((v T u, q T p), (w T, r T )) c { (u v T, p q T ) 2 1,T + T (w T, r T ) 1,T h 2 (u v T ) 2 } / / 300 of the Stokes Equations Non-Conforming s The Basic Idea of the Stokes Equations Non-Conforming s The Crouzeix-Raviart Element (d = 2) We want a fully conservative discretization, i.e. the discrete solution has to satisfy div u T = 0. As a trade-off, we are willing to relax the conformity condition X(T ) H 1 0 ()d. T a triangulation X(T ) = {v T : v T R 1 () 2, v T is continuous a midpoints of edges, vanishes at midpoints of boundary edges} v T Y (T ) = S 0, 1 (T ) L 2 0 () All integrals are taken element-wise. Degrees of freedom: 103/ / 300

27 of the Stokes Equations Non-Conforming s Properties of the Crouzeix-Raviart Element of the Stokes Equations Non-Conforming s Drawbacks of the Crouzeix-Raviart Element The Crouzeix-Raviart discretization admits a unique solution u T, p T. The discretization is fully conservative, i.e. the continuity equation div u T = 0 is satisfied element-wise. If is convex, the following error estimates hold { } 1 u u T 2 2 1, + p p T ch f, T u u T ch 2 f. Its accuracy deteriorates drastically in the presence of re-entrant corners. It has no higher order equivalent. It has no three-dimensional equivalent. 105/ / 300 of the Stokes Equations Non-Conforming s Construction of a Solenoidal Bases Denote by ϕ E S 1, 1 (T ) the function which takes the value 1 at the midpoint of E and vanishes at all other midpoints of edges. Set w E = ϕ E t E where t E is a unit vector tangential to E. Set w x = E E x 1 E ϕ En E,x. Then V (T ) = { u T X(T ) : div u T = 0 } = span { w x, w E : x N, E E } of the Stokes Equations Non-Conforming s Solution of the Discrete Problem The velocity u T V (T ) is determined by the conditions u T : v T = f v T T T for all v T V (T ). The pressure p T is determined by the conditions f n E ϕ E u T : ( ϕ E n E ) = E [p T ] E T T for all E E. 107/ / 300

28 of the Stokes Equations Non-Conforming s Computation of the Velocity The problem for the velocity is symmetric positive definite, corresponds to a Morley element discretization of the biharmonic equation, has condition number O(h 4 ). of the Stokes Equations Non-Conforming s Computation of the Pressure Set F =, M =. Choose an element T with an edge on the boundary. Set p T = 0 on. Add to M. While M do: Choose an element M. For all elements which share an edge with and which are not contained in F do: On set p T equal to the value of p T on plus the jump across the common edge. If is not contained in M, add it to M. Remove from M and add it to F. Compute the average of p T and subtract it from p T on every element. 109/ / 300 of the Stokes Equations Stream-Function Formulation The curl Operators (d = 2) curl ϕ = ( ) ϕ x 2 ϕ x 1 curl v = v 1 x 2 v 2 x 1 curl(curl ϕ) = ϕ curl(curl v) = v + (div v) curl( ϕ) = 0 div u = 0 if and only if there is a stream-function ψ with ψ = 0 on Γ and u = curl ψ in of the Stokes Equations Stream-Function Formulation Stream-Function Formulation of the Two-Dimensional Stokes Equations Taking the curl of the momentum equation proves: u is a solution of the two-dimensional Stokes equations if and only if u = curl ψ and ψ solves the biharmonic equation 2 ψ = curl f ψ = 0 ψ n = 0 in on Γ on Γ 111/ / 300

29 of the Stokes Equations Stream-Function Formulation Drawbacks of the Stream-Function Formulation It is restricted to two dimensions. It gives no information on the pressure. A conforming discretization of the biharmonic equation requires C 1 -elements. Low order non-conforming discretizations of the biharmonic equation are equivalent to the Crouzeix-Raviart discretization. Mixed formulations of the biharmonic equation using the vorticity ω = curl u as additional variable are at least as difficult to discretize as the original Stokes problem. Motivation Uzawa Type Algorithms Multigrid Algorithms Subspace Decomposition Methods Conjugate Gradient Type Algorithms 113/ / 300 Motivation Direct Solvers Typically require O(N 2 1 d ) storage for a discrete problem with N unknowns. Typically require O(N 3 2 d ) operations. Yield the exact solution of the discrete problem up to rounding errors. Yield an approximation for the differential equation with an O(h α ) = O(N α d ) error (typically: α {1, 2}). Motivation Iterative Solvers Typically require O(N) storage. Typically require O(N) operations per iteration. Their convergence rate deteriorates with an increasing condition number of the discrete problem which typically is O(h 2 ) = O(N 2 d ). In order to reduce an initial error by a factor 0.1 one typically needs the following numbers of operations: O(N 1+ 2 d ) with the Gauß-Seidel algorithm, O(N 1+ 1 d ) with the conjugate gradient (CG-) algorithm, O(N d ) with the CG-algorithm with Gauß-Seidel preconditioning, O(N) with a multigrid (MG-) algorithm. 115/ / 300

30 Motivation Nested Grids Often one has to solve a sequence of discrete problems L k u k = f k corresponding to increasingly more accurate discretizations. Typically there is a natural interpolation operator I k 1,k which maps functions associated with the (k 1)-st discrete problem into those corresponding to the k-th discrete problem. Then the interpolate of any reasonable approximate solution of the (k 1)-st discrete problem is a good initial guess for any iterative solver applied to the k-th discrete problem. Often it suffices to reduce the initial error by a factor 0.1. Motivation Nested Iteration Compute ũ 0 = u 0 = L 1 0 f 0. For k = 1,... compute an approximate solution ũ k for u k = L 1 k f k by applying m k iterations of an iterative solver for the problem L k u k = f k with starting value I k 1,k ũ k 1. m k is implicitly determined by the stopping criterion f k L k ũ k ε f k L k (I k 1,k ũ k 1 ). 117/ / 300 Uzawa Type Algorithms Structure of Discrete Stokes Problems Discrete Stokes problems have the form with: δ = 0 for mixed methods, 0 < δ 1 for Petrov-Galerkin methods, ( A B B T δc ) ( u p ) = ( f δg ) a square, symmetric, positive definite n u n u matrix A with condition of O(h 2 ), a rectangular n u n p matrix B, a square, symmetric, positive definite n p n p matrix C with condition of O(1), a vector f of dimension n u discretizing the exterior force, a vector g of dimension n p which equals 0 for mixed methods. Consequences Uzawa Type Algorithms ( ) The stiffness matrix A B B T δc is symmetric but indefinite, i.e. it has positive and negative real eigenvalues. Hence, standard iterative methods such as the Gauß-Seidel and CG-algorithms fail. 119/ / 300

31 Uzawa Type Algorithms The Uzawa Algorithm 0. Given: an initial guess p 0, a tolerance ε > 0 and a relaxation parameter ω > Set i = Apply a few Gauß-Seidel iterations to the linear system 3. If Au = f Bp i and denote the result by u i+1. Compute p i+1 = p i + ω{b T u i+1 δg δcp i }. Au i+1 + Bp i+1 f + B T u i+1 δcp i+1 δg ε return u i+1 and p i+1 as approximate solution; stop. Otherwise increase i by 1 and go to step / 300 Uzawa Type Algorithms Properties of the Uzawa Algorithm ω (1, 2), typically ω = 1.5. Typically v = 1 n u v v and q = 1 n p q q. The problem Au = f Bp i is a discrete version of d Poisson equations for the components of the velocity field. The Uzawa algorithm falls into the class of pressure correction schemes. The convergence rate of the Uzawa algorithm is 1 O(h 2 ). 122/ 300 Uzawa Type Algorithms Uzawa Type Algorithms Idea for an Improvement of the Uzawa Algorithm The problem ( ) A B B T δc ( u p ) = ( ) f δg is equivalent to u = A 1 (f Bp) and B T A 1 (f Bp) δcp = δg. The matrix B T A 1 B + δc is symmetric, positive definite and has a condition of O(1). Hence, a standard CG-algorithm can be applied to the pressure problem and has a uniform convergence rate independently of any mesh-size. The evaluation of A 1 g corresponds to the solution of d discrete Poisson equations Au = g for the components of u. The discrete Poisson problems can efficiently be solved with a MG-algorithm. Properties of B T A 1 B Identify vectors with corresponding finite element functions. u = A 1 Bp satisfies: u : v = p div v for all v, u 1 d p, u 1 β p (inf-sup condition). q = B T u = B T A 1 Bp satisfies: qr = r div u for all r, q d u 1 d p, β 2 p 2 u 2 1 = p div u = qp q p. 123/ / 300

32 Uzawa Type Algorithms The Improved Uzawa Algorithm 0. Given: an initial guess p 0 and a tolerance ε > Apply a MG-algorithm with starting value zero and tolerance ε to Av = f Bp 0 and denote the result by u 0. Compute r 0 = B T u 0 δg δcp 0, d 0 = r 0, γ 0 = r 0 r 0. Set u 0 = 0 and i = If γ i < ε 2 compute p = p 0 + p i, apply a MG-algorithm with starting value zero and tolerance ε to Av = f Bp and denote the result by u, stop. 3. Apply a MG-algorithm with starting value u i and tolerance ε to Av = Bd i and denote the result by u i+1. Compute s i = B T u i+1 + δcd i, α i = γ i d i s i, p i+1 = p i + α i d i, r i+1 = r i α i s i, γ i+1 = r i+1 r i+1, β i = γ i+1 γ i, d i+1 = r i+1 + β i d i. Increase i by 1 and go to step / 300 Uzawa Type Algorithms Properties of the Improved Uzawa Algorithm It is a nested iteration with MG-iterations in the inner loops. Typically 2 to 4 MG-iterations suffice in the inner loops. It requires O(N) operations per iteration. Its convergence rate is uniformly less than 1 for all meshes. It yields an approximate solution with error less than ε with O(N ln ε) operations. Numerical experiments yield convergence rates less than / 300 Uzawa Type Algorithms Convergence Rate of the Improved Uzawa Algorithm Denote by Mv the result of the MG-algorithm applied to a problem with right-hand side v. The improved Uzawa algorithm then corresponds to a CG-algorithm applied to the problem B T M(f Bp) δcp = δg. Properties of the MG-algorithm imply that M is symmetric, M satisfies Mv A 1 v ε A 1 v for all v. Hence, (1 ε)p T B T A 1 Bp p T B T MBp (1 + ε)p T B T A 1 Bp for all p. Thus, B T MB is symmetric, positive definite and has a condition of O(1) uniformly for all meshes. 127/ 300 Multigrid Algorithms The Basic Idea Classical iterative methods such as the Gauß-Seidel algorithm quickly reduce highly oscillatory error components. Classical iterative methods such as the Gauß-Seidel algorithm are very poor in reducing slowly oscillatory error components. Slowly oscillating error components can well be resolved on coarser meshes with fewer unknowns. 128/ 300

33 Multigrid Algorithms The Basic Two-Grid Algorithm Perform several steps of a classical iterative method on the current grid. Correct the current approximation as follows: Compute the current residual. Restrict the residual to the next coarser grid. Exactly solve the resulting problem on the coarse grid. Prolongate the coarse-grid solution to the next finer grid. Perform several steps of a classical iterative method on the current grid. Multigrid Algorithms Schematic Form Multigrid Two-Grid G R G G R R E E G P G P G P 129/ / 300 Multigrid Algorithms Basic Ingredients A sequence T k of increasingly refined meshes with associated discrete problems L k u k = f k. A smoothing operator M k, which should be easy to evaluate and which at the same time should give a reasonable approximation to L 1 k. A restriction operator R k,k 1, which maps functions on a fine mesh T k to the next coarser mesh T k 1. A prolongation operator I k 1,k, which maps functions from a coarse mesh T k 1 to the next finer mesh T k. 131/ 300 Multigrid Algorithms The Multigrid Algorithm 0. Given: the actual level k, parameters µ, ν 1, and ν 2, the matrix L k, the right-hand side f k, an initial guess u k. Sought: improved approximate solution u k. 1. If k = 0 compute u 0 = L 1 0 f 0; stop. 2. (Pre-smoothing) Perform ν 1 steps of the iterative procedure u k u k + M k (f k L k u k ). 3. (Coarse grid correction) 3.1 Compute f k 1 = R k,k 1 (f k L k u k ) and set u k 1 = Perform µ iterations of the MG-algorithm with parameters k 1, µ, ν 1, ν 2, L k 1, f k 1, u k 1 and denote the result by u k Update u k by u k u k + I k 1,k u k (Post-smoothing) Perform ν 2 steps of the iterative procedure u k u k + M k (f k L k u k ). 132/ 300

34 Multigrid Algorithms Typical Choices of Parameters µ = 1 V-cycle or µ = 2 W-cycle ν 1 = ν 2 = ν or ν 1 = ν, ν 2 = 0 or ν 1 = 0, ν 2 = ν 1 ν 4. Multigrid Algorithms Prolongation and Restriction The prolongation is typically determined by the natural inclusion of the finite element spaces, i.e. a finite element function corresponding to a coarse mesh is expressed in terms of the finite element bases functions corresponding to the fine mesh The restriction is typically determined by inserting finite element bases functions corresponding to the coarse mesh in the variational form of the discrete problem corresponding to the fine mesh. 133/ / 300 Multigrid Algorithms Smoothing (Positive Definite Problems) Gauß-Seidel iteration SSOR iteration: Perform a forward Gauß-Seidel sweep with over-relaxation as pre-smoothing. Perform a backward Gauß-Seidel sweep with over-relaxation as post-smoothing. ILU smoothing: Perform an incomplete lower upper decomposition of L k by suppressing all fill-in. The result is an approximate decomposition Lk U k L k. Compute vk = M k u k by solving the system L k U k v k = u k. 135/ 300 Multigrid Algorithms Smoothing (Stokes Problem) Squared Jacobi iteration: ( Mk = 1 A h 2 ) ω 2 k B h 2 k BT h 4 k δc The factors h 2 k and h 4 k compensate the different order of differentiation for the velocity and pressure. Vanka smoothers: Similarly to the Gauß-Seidel iteration, simultaneously adjust all degrees of freedom for the velocity and pressure corresponding to an element or to a patch of elements while fixing the remaining degrees of freedom. Patches typically consist of two elements sharing a common face or the elements sharing a given vertex. 136/ 300

35 Multigrid Algorithms Number of Operations Multigrid Algorithms Convergence Rate (Positive Definite Problems) Assume that one smoothing step requires O(N k ) operations, the prolongation requires O(N k ) operations, the restriction requires O(Nk ) operations, µ 2, Nk > µn k 1, then one iteration of the multigrid algorithm requires O(N k ) operations. The convergence rate is uniformly less than 1 for all meshes. The convergence rate is bounded by c+ν 1 +ν 2 with a constant which only depends on the shape parameter of the meshes. Numerical experiments yield convergence rates less than 0.1. c 137/ / 300 Multigrid Algorithms Convergence Rate (Stokes Problem) Multigrid Algorithms Techniques for Proving the Convergence of Multigrid Algorithms The convergence rate is uniformly less than 1 for all meshes. c ν1 +ν 2 The convergence rate is bounded by with a constant which only depends on the shape parameter of the meshes. Numerical experiments yield convergence rates less than 0.5. Methods of linear algebra and discrete Fourier analysis (Hackbusch) Spectral decomposition and scales of discrete Sobolev spaces (Bank-DuPont and Braess-Hackbusch) Subspace decomposition methods (Bramble-Pasciak-Xu and Wang) 139/ / 300

36 Multigrid Algorithms Convergence Proof à la Hackbusch The iteration matrix of the smoother is N k = I M k L k. The iteration matrix of the two-grid algorithm is Ŝ k = N ν 2 k (I I k 1,kL 1 k 1 R k,k 1)N ν 1 k. The iteration matrix of the multigrid algorithm is S k = Ŝk + N ν 2 k I k 1,kS µ k 1 L 1 k 1 R k,k 1N ν 1 k. Prove the smoothing property L k Nk ν η(ν)h α k with η(ν) 0 for ν and α 0. Prove the approximation property L 1 k I k 1,k L 1 k 1 R k,k 1 ch α. The smoothing and approximation property imply Ŝk cη(ν 1 + ν 2 ). If µ 2 a perturbation argument yields S k 2cη(ν 1 + ν 2 ). Multigrid Algorithms Establishing the Smoothing and Approximation Property The proof of the smoothing property is usually based on a spectral decomposition of L k and N k. The proof of the approximation property is usually based on arguments used in the proof of a priori error estimates. The crucial point is to correctly link both techniques. 141/ / 300 Multigrid Algorithms Convergence Proof à la Braess-Hackbusch Assume that L k is symmetric, positive definite and set v k = (v k, L k v k ) 1 2. Denote by Q k = I k 1,k L 1 k 1 R k,k 1L k the Ritz projection. Denote by J k the iteration matrix of the Jacobi iteration and set v k = J 1 2 k v k and ρ(v k ) = v k 2. v k 2 Prove that J ν k v k ρ ν v k with ρ = ρ(jk νv k), vk Q k v k min{1, c 1 ρ(v k )} v k. Then the convergence rate δ k of the multigrid algorithm with µ = 1{ and ν 1 = ν 2 = ν and Jacobi smoothing satisfies δ k max ρ 2ν[ δ k 1 + (1 δ k 1 ) min{1, c(1 ρ)} ]}. 0 ρ 1 By induction this proves δ k c c+2ν. 143/ 300 The Setting Subspace Decomposition Methods V a finite dimensional Hilbert space with inner product (, ) V i, 1 i N, subspaces of V with i V i = V The decomposition usually neither is direct nor orthogonal. Q i : V V i orthogonal projection w.r.t. to (, ) A : v V a symmetric, positive definite operator A i : V i V i the restriction of A to V i R i : V i V i an easy-to-evaluate, symmetric, positive definite approximation to A 1 i 144/ 300

37 Subspace Decomposition Methods The Subspace Decomposition Algorithm Examples Subspace Decomposition Methods Given an initial guess u 0 V. For n = 0, 1,... and j = 1,..., N compute u n+ j N = u n+ j 1 N + R j Q j (f Au n+ j 1 ). N V = R N and V i = span{e i } corresponds to the classical Gauß-Seidel algorithm. V = S k,0 0 (T N), V i = S k,0 0 (T i), R 0 = A 1 0 and R i = 1 ω i I with uniformly or locally refined nested meshes T i corresponds to the multigrid algorithm with Jacobi smoothing and ν 1 = 1, ν 2 = / / 300 Subspace Decomposition Methods Convergence Rate Set v = (Av, v) 1 2. Set λ = min i λ max (R i A i ) and Λ = max i λ max (R i A i ). Assume that Λ < 2. Assume that there are two constants 0 and 1 such that { N v i 2} v for all v = i v i, i=1 1 i,j N { N (Av i, w j ) 1 v i 2} 1 { N 2 w j 2} 1 2 for all v i V i, w j V j. i=1 Then the convergence rate of the subspace decomposition [ algorithm w.r.t. is less than 1 ( 2 Λ 1)( ) ] 1 λ 2 2 Λ 0 1. j=1 Subspace Decomposition Methods Proof of the Bound for the Convergence Rate Set T i = R i Q i A, E 0 = I, E j = (I T j )... (I T 1 ). For every j this gives E j + E j 1 = T j E j 1 and E j 1 v 2 E j v 2 = (AT j E j 1 v, (2I T j )E j 1 v) (2 Λ)(AT j E j 1 v, E j 1 v). Summation yields v 2 E N v 2 (2 Λ) j (AT je j 1 v, E j 1 v). The first assumption yields for v = i v i v 2 = { } 1 i (A(R ia i ) 1 T i v, v) λ 1 0 v i T iv 2 2. The second assumption implies i T iv 2 Λ j (AT je j 1 v, E j 1 v). Combining all estimates yields the bound for the convergence rate. 147/ / 300

38 Subspace Decomposition Methods Verification of the Assumptions The condition Λ < 2 can be satisfied by a suitable scaling of R i. Since V = i V i the mapping V 1... V N (v 1,..., v N ) i v i V is surjective. The open mapping theorem therefore proves the first assumption. The crucial point is to obtain an explicit bound for 0 which does not depend on N. This requires deep results concerning the characterization of Sobolev spaces as approximation spaces. Due to the Cauchy-Schwarz inequality, the second assumption is always satisfied with 1 N. If the subspaces satisfy a strengthened Cauchy-Schwarz inequality (Av i, w j ) γ i j v i w j with γ < 1, the second assumption is satisfied with γ. 149/ 300 Conjugate Gradient Type Algorithms CG-Type Algorithms for Non-Symmetric and Indefinite Systems of Equations The classical CG-algorithm breaks down for non-symmetric or indefinite systems of equations. A naive remedy is to apply the CG-algorithm to the system L T Lu = L T f of the normal equations. This approach cannot be recommended since passing to the normal system squares the condition number. The following variants of the CG-algorithm are particularly adapted to non-symmetric and indefinite problems: the stabilized bi-conjugate gradient algorithm (Bi-CG-stab in short), the generalized minimal residual method (GMRES in short). 150/ 300 Conjugate Gradient Type Algorithms The Idea of the Bi-CG-stab Algorithm The algorithm tries to simultaneously solve the original problem Lu = f and its adjoint problem L T v = f. For both problems it performs a simultaneous three-term recursion similar to the CG-iteration. It incorporates particular devices to detect possible break-downs and to restart the iteration before breaking down. It can be combined with preconditioning. Possible methods for preconditioning are: the SSOR-iteration applied to the symmetric part of L, incomplete factorizations of L as used in the context of ILU-smoothing. Conjugate Gradient Type Algorithms The Idea of the GMRES Algorithm It performs a three-term recursion to build increasingly larger rylov spaces n = span{u, Lu,..., L n 1 u}. For every rylov space n it approximately solves the minimization problem v n = argmin w n Lw f using a QR-method. 151/ / 300

39 Motivation Drawbacks of A Priori Error Estimates Motivation A Posteriori Error Estimates for the Stokes Problem Mesh Refinement, Coarsening and Smoothing They only yield information on the asymptotic behaviour of the error. They require regularity properties of the solution which often are not realistic. They give no information on the actual size of the error. They are not able to detect local singularities arising from re-entrant corners or boundary or interior layers which deteriorate the overall accuracy of the discretization. 153/ / 300 Motivation Goal of A Posteriori Error Estimation and Adaptivity We want to obtain explicit information about the error of the discretization and its spatial (and temporal) distribution. The information should a posteriori be extracted from the computed numerical solution and the given data of the problem. The cost for obtaining this information should be far less than for the computation of the numerical solution. We want to obtain a numerical solution with a prescribed tolerance using a (nearly) minimal number of grid-points. To this end we need reliable upper and lower bounds for the true error in a user-specified norm. Motivation General Adaptive Algorithm 0. Given: The data of a partial differential equation and a tolerance ε. Sought: A numerical solution with an error less than ε. 1. Construct an initial coarse mesh T 0 representing sufficiently well the geometry and data of the problem; set k = Solve the discrete problem on T k. 3. For every element in T k compute an a posteriori error indicator. 4. If the estimated global error is less than ε then stop. Otherwise decide which elements have to be refined or coarsened and construct the next mesh T k+1. Replace k by k + 1 and return to step / / 300

40 Motivation Basic Ingredients A Posteriori Error Estimates for the Stokes Problem The Stokes Problem and its An error indicator which furnishes the a posteriori error estimate. A refinement strategy which determines which elements have to be refined or coarsened and how this has to be done. u H0 1()d, p L 2 0 () weak solution of the Stokes problem with no-slip boundary condition: u + grad p = f div u = 0 u = 0 in in on Γ u T X(T ), p T Y (T ) solution of a mixed or Petrov-Galerkin discretization of the Stokes problem Assume that S 1,0 0 (T )d X(T ) 157/ / 300 Residual A Posteriori Error Estimates for the Stokes Problem Define two residuals R m H 1 () d and R c L 2 () associated with the momentum and continuity equation by R m, v = f v u T : v + p T div v R c, q = q div u T Then the error u u T, p p T solves the Stokes problem (u u T ) : v (p p T ) div v = R m, v q div(u u T ) = R c, q A Posteriori Error Estimates for the Stokes Problem Equivalence of Error and Residual The well-posedness of the saddle-point formulation of the Stokes problem implies 1 c { Rm 1 + R c } u u T 1 + p p T c { R m 1 + R c }. c and c depend on the space dimension d. c in addition depends on the constant in the inf-sup condition for the Stokes problem. The above equivalence holds for every discretization be it stable or not. 159/ / 300

41 A Posteriori Error Estimates for the Stokes Problem Evaluation of R c The definition of R c implies R c = div u T. Hence, R c can be evaluated easily and is a measure for the lacking incompressibility of u T. A Posteriori Error Estimates for the Stokes Problem Evaluation of R m 1 The explicit evaluation of R m 1 would require the solution of an infinite dimensional variational problem which is as expensive as the solution of the original Stokes problem. Hence, we must obtain estimates for R m 1 which at the same time are as sharp as possible and easy to evaluate. Main tools for achieving this goal are: properties of the discrete problem, the Galerkin orthogonality of Rm, an L 2 -representation of R m, approximation properties of the quasi-interpolation operator R T, inverse estimates for the bubble functions. 161/ / 300 A Posteriori Error Estimates for the Stokes Problem The Discrete Problem Reviewed u T, p T satisfies for every v T X(T ), q T Y (T ) 0 = l T ((v T, q T )) B T ((u T, p T ), (v T, q T )) = f v T u T : v T + p T div v T }{{} + T T E E δ h 2 δ h 2 δ E h E E = R m,v T f q T ( u T + p T ) q T [p T ] E [q T ] E q T div u T A Posteriori Error Estimates for the Stokes Problem Galerkin Orthogonality of R m The form of the discrete problem and the assumption S 1,0 0 (T )d X(T ) imply the Galerkin orthogonality R m, v T = 0 v T S 1,0 0 (T )d. 163/ / 300

42 A Posteriori Error Estimates for the Stokes Problem A Posteriori Error Estimates for the Stokes Problem L 2 -Representation of R m Integration by parts element-wise yields for every v H0 1()d the L 2 -representation R m, v = ( ) f + ut p T v T [ ne ( u T p T I) ] E v. E E E Upper Bounds for R m 1 The Galerkin orthogonality, the L 2 -representation and the approximation properties of R T imply R m, v = R m, v R T v c A1 h f + u T p T v 1, ω T c A2 he [n E ( u T p T I)] E E v 1, ωe E E The Cauchy-Schwarz inequality therefore yields { R m 1 c h 2 f + u T p T 2 T + E E h E [n E ( u T p T I)] E 2 E } / / 300 A Posteriori Error Estimates for the Stokes Problem Lower Bounds for R m 1 Denote be f T any piece-wise polynomial approximation of f. Inserting the functions ψ (f T + u T p T ) and ψ E [n E ( u T p T I)] E in the definition of R m and using the inverse estimates for the bubble functions proves h f T + u T p T c { R m 1, + h f f T } h 1 2 E [n E ( u T p T I)] E E c { R m 1,ωE + h E f f T ωe } A Posteriori Error Estimates for the Stokes Problem Residual A Posteriori Error Estimates Define the residual a posteriori error indicator η R, by { η R, = h 2 f T + u T p T 2 + div u T } 1 h E [n E ( u T p T I)] E 2 2 E. 2 E E, Then the error is bounded from above and from below by { u ut p p T 2} 1 2 c { ( η 2 R, + h 2 f f T 2 ) } 1 2 T { } 1 η R, c u u T 2 1,ω + p p T 2 ω + h 2 f f T 2 2 ω 167/ / 300

43 A Posteriori Error Estimates for the Stokes Problem Discussion of the A Posteriori Error Estimates I A Posteriori Error Estimates for the Stokes Problem Discussion of the A Posteriori Error Estimates II The first term in η R, is related to the residual of u T, p T with respect to the strong form of the momentum equation. The constants c and c depend on the shape parameter of T. The constant c in addition depends on the polynomial degrees of u T, p T, and f T. The second term in η R, is related to the residual of u T with respect to the strong form of the continuity equation. The third term in η R, is related to the boundary operator which canonically links the strong and weak form of the momentum equation. The third term in η R, is crucial for low order discretizations. The different scalings of the three terms in η R, take into account the different order of the differential operators. 169/ / 300 A Posteriori Error Estimates for the Stokes Problem Discussion of the A Posteriori Error Estimates III The upper bound is global. This is due to the fact that it is based on the norm of the inverse of the Stokes operator which is a global operator. (local force global flow) The lower bound is local. This is due to the fact that it is based on the norm of the Stokes operator itself which is a local operator. (local flow local force) 171/ 300 A Posteriori Error Estimates for the Stokes Problem Auxiliary Discrete Stokes Problems With every element T associate a patch T T containing, finite element spaces X(T ), Y (T ) on T. Find u X(T ), p Y (T ) such that for all v X(T ), q Y (T ) u : v p div v + q div u = {f + u T p T } v + [n ( u T p T I)] v + q div u T. Set η N, = { u 2 1,T + p 2 T } / 300

44 A Posteriori Error Estimates for the Stokes Problem Choice of Patches and Spaces Patches typically consist of: the element itself: T =, all elements sharing a face with : T = ω, all elements sharing a vertex with : T = ω. The spaces X(T ), Y (T ) typically consist of finite element functions of a sufficiently high degree, e.g. X(T ) = span{ψ v, ψ E w : v R kt ( ) d, w R ke (E ) d, T, E E T,}, Y (T ) = span{ψ q : q R ku 1( ), T }. with k T = max{k u + d, k p 1}, k E = max{k u 1, k p }. A Posteriori Error Estimates for the Stokes Problem Comparison of the Error Indicators Both indicators yield global upper and local lower bounds for the error. Each indicator can be bounded from above and from below by the other one. Both indicators well predict the spatial distribution of the error. Both indicators are well suited for adaptive mesh refinement. The evaluation of the residual indicator is less expensive. The indicator based on the auxiliary Stokes problems more precisely predicts the size of the error. 173/ / 300 Overview Mesh Refinement, Coarsening and Smoothing The mesh refinement requires two key-ingredients: a marking strategy that decides which elements should be refined, refinement rules which determine the actual subdivision of a single element. To maintain the admissibility of the partitions, i.e. to avoid hanging nodes, the refinement process proceeds in two stages: Firstly refine all those elements that are marked due to a too large value of η (regular refinement). Secondly refine additional elements in order to eliminate the hanging nodes which are eventually created during the first stage (irregular refinement). The mesh refinement may eventually be combined with mesh coarsening and mesh smoothing. 175/ 300 Mesh Refinement, Coarsening and Smoothing Maximum Strategy for Marking 0. Given: A partition T, error estimates η for the elements T, and a threshold θ (0, 1). Sought: A subset T of marked elements that should be refined. 1. Compute η T,max = max T η. 2. If η θη T,max mark by putting it into T. 176/ 300

45 Mesh Refinement, Coarsening and Smoothing Equilibration Strategy for Marking (Bulk Chasing or Dörfler Marking) 0. Given: A partition T, error estimates η for the elements T, and a threshold θ (0, 1). Sought: A subset T of marked elements that should be refined. 1. Compute Θ T = η. 2 Set Σ T = 0 and T =. T 2. If Σ T θθ T return T ; stop. Otherwise go to step Compute η T,max = max T \ T η. 4. For all elements T \ T check whether η = η T,max. If this is the case, mark by putting it into T and add η 2 to Σ T. Otherwise skip. When all elements have been checked, return to step / 300 Mesh Refinement, Coarsening and Smoothing Comparison of the Marking Strategies The maximum strategy is cheaper. At the end of the equilibration strategy the set T satisfies η 2 θ η. 2 T T Convergence proofs for adaptive finite element methods are often based on this property. 178/ 300 Mesh Refinement, Coarsening and Smoothing Ensuring a Sufficient Refinement Sometimes very few elements have an extremely large estimated error, whereas the remaining ones split into the vast majority with an extremely small estimated error and a third group of medium size consisting of elements with an estimated error of medium size. Then the marking strategies only refine the elements of the first group. This deteriorates the performance of the adaptive algorithm. This can be avoided by the following modification: Given a small percentage ε, first mark the ε% elements with largest estimated error for refinement and then apply the marking strategies to the remaining elements. 179/ 300 Mesh Refinement, Coarsening and Smoothing Regular Refinement Elements are subdivided by joining the midpoints of their edges. This preserves the shape parameter. 180/ 300

46 Mesh Refinement, Coarsening and Smoothing Hanging Nodes Mesh Refinement, Coarsening and Smoothing Irregular Refinement Hanging nodes destroy the admissibility of the partition. Therefore either the continuity of the finite element spaces must be enforced at hanging nodes or an additional irregular refinement must be performed. Enforcing the continuity at hanging nodes may counteract the refinement. Triangles Quadrilaterals 181/ / 300 Mesh Refinement, Coarsening and Smoothing Marked Edge Bisection Mesh Refinement, Coarsening and Smoothing Mesh Coarsening The first mesh is constructed such that the longest edge of an element is also the longest edge of its neighbour. The longest edges in the first mesh are marked. An element is refined by joining the midpoint of its marked edge with the vertex opposite to this edge (bisection). When bisecting the edge of an element, its two remaining edges become the marked edges of the resulting triangles. The coarsening of meshes is needed to ensure the optimality of the adaptive process, i.e. to obtain a given accuracy with a minimal amount of unknowns, to resolve moving singularities. The basic idea is to cluster elements with too small an error. This is achieved by either by going back in the grid hierarchy or by removing resolvable vertices. 183/ / 300

47 Mesh Refinement, Coarsening and Smoothing Going Back in the Grid Hierarchy 0. Given: A hierarchy T 0,..., T k of adaptively refined partitions, error indicators η for the elements of T k, and parameters 1 m k and n > m. Sought: A new partition T k m+n. 1. For every element T k m set η = For every element T k determine its ancestor T k m and add η 2 to η2. 3. Successively apply the maximum or equilibration strategy n times with η as error indicator. In this process, equally distribute η over the descendants of once an element is subdivided. Mesh Refinement, Coarsening and Smoothing Resolvable Vertices An element of the current partition T has refinement level l if it is obtained by subdividing l times an element of the coarsest partition. Given a triangle of the current partition T which is obtained by bisecting a parent triangle, the vertex of which is not a vertex of is called the refinement vertex of. A vertex z N of the current partition T and the corresponding patch ω z are called resolvable if z is the refinement vertex of all elements contained in ωz, all elements contained in ωz have the same refinement level. resolvable vertex non-resolvable vertex 185/ / 300 Mesh Refinement, Coarsening and Smoothing Removing Resolvable Vertices 0. Given: A partition T, error indicators η for all elements of T, and parameters 0 < θ 1 < θ 2 < 1. Sought: Subsets T c and T r of elements that should be coarsened and refined, respectively. 1. Set T c =, T r = and compute η T,max = max T η. 2. For all T check whether η θ 2 η T,max. If this is the case, put into T r. 3. For all vertices z N check whether z is resolvable. If this is the case and if max ω z η θ 1 η T,max, put all elements contained in ω z into T c. Mesh Refinement, Coarsening and Smoothing Mesh Smoothing Improve the quality of a given partition T by moving its vertices while retaining the adjacency of the elements. The quality is measured by a a quality function q such that a larger value of q indicates a better quality. The quality is improved by sweeping through the vertices with a Gauß-Seidel type smoothing procedure: For every vertex z in T, fix the vertices of ω z and find a new vertex z inside ω z such that min q( ) > min q(). ω ω z z 187/ / 300

48 Mesh Refinement, Coarsening and Smoothing Quality Functions Based on geometric criteria: q() = Based on interpolation: 4 3µ 2 () µ 1 (E 0 ) 2 + µ 1 (E 1 ) 2 + µ 1 (E 2 ) 2 q() = (u Q u L ) 2 with linear and quadratic interpolants of u Based on an error indicator: 2 2 q() = e i ψ Ei i=0 Stationary Incompressible Navier-Stokes Equations Variational Formulation A Posteriori Error Estimates with e i = h 1 2 Ei [n Ei u T ] Ei 189/ / 300 Strong Form Variational Formulation Stationary incompressible Navier-Stokes equations in dimensionless form with no-slip boundary condition u + Re(u )u + grad p = f div u = 0 u = 0 in in on Γ For the variational formulation, we want to multiply the momentum equation with a test function v H 1 0 ()d and integrate the result over. This is possible if u H 1 0 ()d implies (u )u H 1 () d. Variational Formulation Properties of the Non-Linear Term Hölder s inequality yields for v H0 1()d, u, w L 4 () d [ ] (u )v w u L 4 () v 1 w L 4 (). Since H0 1() is continuously embedded in L4 () for d 4 (compactly for d 3), this proves that [ H 1 0 () d] 3 [ ] (u, v, w) (u )v w is a continuous trilinear form. Integration by parts yields for all u, v, w H0 1()d with div u = 0 [ ] [ ] [ (u )v w = (u )w v, (u )v ] v = / / 300

49 Variational Formulation Variational Form Find u H0 1()d, p L 2 0 () such that for all v H1 0 ()d, q L 2 0 () u : v p div v + Re[(u )u] v = f v q div u = 0 Equivalent form: Find u V such that for all v V u : v + Re[(u )u] v = f v Variational Formulation Fixed-Point Formulation Denote by T : H 1 () d V the Stokes operator which associates with g H 1 () d the weak solution T g = v of the Stokes problem with right-hand side g, i.e. v : w = g w w V. Then the variational formulation of the Navier-Stokes equations is equivalent to u = T ( f Re(u )u ). 193/ / 300 Variational Formulation Properties of the Variational Problem Every solution satisfies the a priori bound u 1 c F diam() f, where c F is the constant in the Friedrichs inequality. If Re 2 d 1 2 solution. [ cf diam() ] 3 d 2 f < 1, there is at most one For every Reynolds number Re there exists at least one solution. Every solution has the same regularity as the solution of the corresponding Stokes problem. Every solution belongs to a differentiable branch Re u Re of solutions which has at most a countable number of turning or bifurcation points. Variational Formulation A Priori Bound Every solution u of the variational problem satisfies u 2 1 = u : u [ ] = u : u + Re (u )u u = f u f u f c F diam() u / / 300

50 Uniqueness Variational Formulation The difference v = u 1 u 2 of any two solutions satisfies v 2 [ ] [ ] 1 = Re (u1 )u 1 v + Re (u2 )u 2 v [ = Re (u1 )v ] [ ] v Re (v )u2 v [ ] = Re (v )u2 v Re v 2 L 4 () u 2 1. Combined with the a priori bound, the estimate v L 4 () 2 d 1 4 v 1 d 4 v d 4 1 and the Friedrichs inequality this proves the uniqueness. 197/ 300 Existence Variational Formulation V is separable, i.e. there is as sequence of nested finite dimensional subspaces V m such that m V m is dense in V. Denote by T m the Stokes operator corresponding to V m. The properties of the non-linear term and Schauder s fixed-point theorem imply that for every m there is a u m V m with u m = T m ( f Re(um )u m ). The arguments used to prove the a priori bound imply that the sequence u m is bounded in H 1. The compact embedding of H 1 in L 4 and the properties of the non-linear term imply that the u m converge weakly to an u H 1 0 ()d which solves the variational problem with V replaced by anyone of the V m. The density of m V m in V proves that u solves the variational problem. 198/ 300 Regularity Variational Formulation The regularity is proved by a bootstrap-argument using the fixed-point equation u = T ( f Re(u )u ) and the regularity of the Stokes problem: u H 1 = u L 6 = (u )u L 5 3 = (u )u H 3 10 = u H = u L = (u )u L 2 = u H 2 199/ 300 Variational Formulation Branches of Solutions The fixed-point formulation implies that there are no isolated solutions and that every solution depends in a differentiable way on Re. Differentiation of the fixed-point equation yields that v, the derivative w.r.t. Re of any solution u, satisfies v = Re T ( (u )v + (v )u ) T ( (u )u ). The compact embedding of H 1 in L 4 implies that the operator w T ( (u )w + (w )u ) is compact. The statement concerning limit and bifurcation points therefore follows from properties of the spectra of compact operators, the Fredholm alternative and the a priori bound. 200/ 300

51 Basic Idea Replace H0 1()d, L 2 0 () by a pair X(T ), Y (T ) of finite element spaces which is uniformly stable for the Stokes problem. Denote by V (T ) the corresponding approximation of V. Since V (T ) V the anti-symmetry of the non-linear term is lost. To recover the anti-symmetry replace the non-linear term by Ñ(u, v, w) = 1 [ ] 1 [ ] (u )v w (u )w v. 2 2 Discrete Problem Find u T X(T ), p T Y (T ) such that for all v T X(T ), q T Y (T ) u T : v T p T div v T + Re Ñ(u T, u T, v T ) = f v T q T div u T = 0 Equivalent form: Find u T V (T ) such that for all v T V (T ) u T : v T + Re Ñ(u T, u T, v T ) = f v T 201/ / 300 Fixed-Point Formulation of the Discrete Problem Denote by T T : H 1 () d V (T ) the discrete Stokes operator which associates with g H 1 () d the weak solution T T g = v T of the Stokes problem with right-hand side g, i.e. v T : w T = g w T w T V (T ). Then the discrete problem is equivalent to u T = T T ( f Re Ñ(u T, u T, ) ). Properties of the Discrete Problem Every solution satisfies the a priori bound u T 1 c F diam() f. [ If Re 2 d 1 2 cf diam() ] 3 d 2 f < 1, there is at most one solution. For every Reynolds number Re there exists at least one solution. Every solution belongs to a differentiable branch Re u T,Re of solutions which has at most a finite number of turning or bifurcation points. 203/ / 300

52 Error Estimates Assume that: Λ (0, ) is a compact, non empty interval. Λ Re ure is a regular branch of solutions of the Navier-Stokes equations. ure H k+1 () d, p Re H k () for all Re Λ with k 1. S k,0 0 (T )d X(T ) and S k 1, 1 (T ) L 2 0() Y (T ) or S max{k 1,1},0 (T ) L 2 0() Y (T ). Then there is a maximal mesh-size h 0 = h 0 (Λ, f, u Re ) > 0 such that for every partition T with h T h 0 the discrete problem has a solution u T,Re X(T ), p T,Re Y (T ) with u Re u T,Re 1 + p Re p T,Re ch k T sup u Re 2 k+1. Re Λ u Re u T,Re ch T u Re u T,Re 1 if is convex. Proof of the Error Estimates. The Basic Steps The continuous and discrete problem can be written as F (Re, u Re ) = 0 and F T (Re, u T,Re ) = 0. F T evaluated at the H 1 -projection of u Re is small. The derivative of F T evaluated at the H 1 -projection of u Re is close to the derivative of F evaluated at u Re. The derivative of F T evaluated at the H 1 -projection of u Re is invertible. The derivative of F T is Lipschitz-continuous. The discrete problem has a solution in a neighbourhood of the H 1 -projection of u Re. Compare the obtained solution with the solution of the Stokes problem with right-hand side f Re (u Re )u Re. 205/ / 300 Proof of the Error Estimates. 1 st Step Auxiliary quantities Define G, G C (H0 1()d, H 1 () d ) by [ ] G(v), w = (v )v w, G(v), w = Ñ(v, v, w). Define F, F T C ((0, ) H 1 0 ()d, H 1 0 ()d ) by F (Re, v) = v + Re T G(v) T f, F T (Re, v) = v + Re T T G(v) TT f. Note that F (Re, u Re ) = 0 and F T (Re, u T,Re ) = 0. Denote by û T,Re the H 1 -projection of u Re on V (T ) and set Proof of the Error Estimates. 2 nd Step ε T (Re) ch k T sup u Re 2 k+1 Re Λ ε T (Re) = F T (Re, û T,Re ) F (Re, u Re ) 1 u Re û T,Re 1 + Re T T ( G(û T,Re ) G(u Re )) 1 + Re (T T T )G(u Re ) 1 + (T T T )f 1 u Re û T,Re 1 ch k T u Re k+1 T T ( G(û T,Re ) G(u Re )) 1 G(û T,Re ) G(u Re ) 1 c u Re 1 u Re û T,Re 1 (T T T )G(u Re ) 1 ch k T G(u Re) k 1 ch k T u Re 2 k+1 (T T T )f 1 ch k T f k 1 ch T u Re k+1 ε T (Re) = F T (Re, û T,Re ) / / 300

53 Proof of the Error Estimates. 3 rd Step Proof of the Error Estimates. 4 th Step D v F T (Re, û T,Re ) D v F (Re, u Re ) ch u Re 2 D v F T (Re, û T,Re )w D v F (Re, u Re )w = Re T T (D G(û T,Re )w D G(u Re )w) + Re (T T T )D G(u Re )w T T (D G(û T,Re )w D G(u Re )w) 1 D G(û T,Re )w D G(u Re )w 1 ch T u Re 2 w 1 (T T T )D G(u Re )w 1 ch T D G(u Re )w ch T u Re 2 w 1 D v F T (Re, û T,Re ) 1 2 D v F (Re, u Re ) 1 for sufficiently small h T A 1 2 implies that (I A) is invertible and satisfies (I A) 1 2 D v F T (Re, û T,Re ) = D v F (Re, u Re ) [D v F (Re, u Re ) D v F T (Re, û T,Re )] = D v F (Re, u Re )[I D v F (Re, u Re ) 1 (D v F (Re, u Re ) D v F T (Re, û T,Re )] 209/ / 300 Proof of the Error Estimates. 5 th Step D v F T (Re, v 1 ) D v F T (Re, v 2 ) cre v 1 v 2 1 D v F T (Re, v 1 )w D v F T (Re, v 2 )w = Re T T (D G(v 1 )w D G(v 2 )w) T T (D G(v 1 )w D G(v 2 )w) 1 D G(v 1 )w D G(v 2 )w 1 c v 1 v 2 1 w 1 Proof of the Error Estimates. 6 th Step The discrete problem has a solution u T,Re with û T,Re u T,Re 1 4 D v F (Re, u Re ) 1 ε T (Re) Thanks to the fourth step we can define a mapping Φ by Φ(v) = v D v F T (Re, û T,Re ) 1 F T (Re, v). Then u T,Re is a solution of the discrete problem if and only if it is a fixed-point of Φ. The fourth and fifth step imply that Φ is a contraction on the ball B in H 1 with centre û T,Re and radius 4 D v F (Re, u Re ) 1 ε T (Re). 211/ / 300

54 Proof of the Error Estimates. 7 th Step Proof of the error estimate The sixth step implies that u Re u T,Re 1 ( D v F (Re, u Re ) 1 ) ε T (Re). Set ũ T,Re = T T (f G(u Re )) and denote by p T,Re the corresponding pressure. The stability of the discretization implies u T,Re ũ T,Re 1 + p T,Re p T,Re cre G(u Re ) G(u T,Re ) 1 cre ( u Re 1 + u T,Re 1 ) ure u T,Re 1. The error estimates for the Stokes problem with right-hand side f Re (u Re )u Re yield ũ T,Re u Re 1 + p T,Re p Re ch k T u Re k / 300 A Warning Example Consider the two-point boundary value problem u + Re uu = 0 in ( 1, 1) with boundary conditions u( 1) = 1, u(1) = 1. The solution is u(x) = tanh(α Rex) tanh(α Re ) where the parameter α Re is determined by the relation 2α Re tanh(α Re ) = Re. The solution exhibits a strong interior layer at x = 0. Explicitly solving the difference equations shows that: central differences are unstable, one-sided differences with a constant orientation on the whole interval are unstable, one-sided differences with their orientation depending on the sign of u are stable. 214/ 300 Conclusion We must stabilize the convective derivative. This can be achieved by upwind schemes or adding an artificial consistent viscosity in the direction of the streamlines (streamline diffusion method or SDFEM in short). An Upwind Scheme Approximate the integral involving the convective derivative by a one-point quadrature rule T )u T ] v T [(u [(u T (x ) )u T (x )] v T (x ). T Replace the convective derivative by an up-wind difference (u T (x ) )u T (x ) u T (x ) x y (u T (x ) u T (y )). Replace u T (y ) by I T u T (y ), the linear interpolate of u T in the vertices of the face of which contains y. a y b x 215/ / 300