Contents. Computational Fluid Dynamics. Fundamentals. Deformation of Materials

Contents R. Verfürth www.ruhr-uni-bochum.de/num1 Lecture Series / Bochum / Summer Term 2018 Variational Formulation of the Stokes Equations of the Stokes Equations Stationary Incompressible Navier-Stokes Equations Non-Stationary Incompressible Navier-Stokes Equations Compressible and Inviscid Problems References 1/ 221 2/ 221 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/ 221 4/ 221

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/ 221 6/ 221 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 f(x, t)dx dt V (t) = d f(φ(η, t), t)j(η, t)dη dt V ( = f(φ(η, t), t)j(η, t) t = V 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/ 221 8/ 221

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/ 221 10/ 221 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/ 221 12/ 221

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 13/ 221 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/ 221 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 + 1 2 ρ v 2, where ε is often identified with the temperature. σ = α ε. 15/ 221 16/ 221

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 + 1 2 ρ v 2 e = ρε + 1 2 ρ v 2 17/ 221 18/ 221 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 ε + ρv grad ε = λd(v) : D(v) + µ(div v) 2 p div v t + 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/ 221 20/ 221

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/ 221 22/ 221 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/ 221 24/ 221

Notations and Auxiliary Results Divergence Theorem Notations and Auxiliary Results Weak Derivative Divergence: d w i div w = x i i=1 Divergence Theorem: div wdx = w nds Γ The function u is said to be weakly differentiable w.r.t. x i with weak derivative w i, if every continuously differentiable function v with v = 0 on Γ satisfies w i vdx = u v dx. x i If u is weakly differentiable w.r.t. to all variables x 1,..., x d, we call u weakly differentiable and write u for the vector (w 1,..., w d ) of the weak derivatives. 25/ 221 26/ 221 Examples Notations and Auxiliary Results Every function which is continuously differentiable in the classical sense is weakly differentiable and its classical derivative coincides with the weak derivative. Every continuous piecewise differentiable function is weakly differentiable and its weak derivative is the piecewise classical derivative. 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 1 +... + 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} 27/ 221 28/ 221

Examples Notations and Auxiliary Results Every bounded function is in L 2 (). 1 v(x) = is not in x 2 +y L2 (B(0, 1)) (B(0, 1) circle with 2 radius 1 centred at the origin), since 1 v(x) 2 1 dx = 2π dr is not finite. r B(0,1) 0 Every continuously differentiable function is in H 1 (). A piecewise differentiable function is in H 1 (), if and only if it is globally continuous. Functions in H 1 () must not admit point values. v(x) = ln( ln( x 2 + y 2 ) ) is in H 1 (B(0, 1)) but has no finite value at the origin. 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 2 2 1 for all ϕ H 1 () c T,1 () diam() 1, c T,2 () diam() if is a simplex or parallelepiped 29/ 221 30/ 221 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 and in T are either disjoint or share a vertex or a complete edge or, if d = 3, a complete face. admissible not admissible 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 h T Remarks Notations and Auxiliary Results Curved boundaries can be approximated by piecewise straight lines or planes. The admissibility is necessary to ensure the continuity of the finite element functions and thus the inclusion of the finite element spaces in H 1 0 (). If the admissibility is violated, the continuity of the finite element functions must be enforced which leads to a more complicated implementation. Partitions can also consist of general quadrilaterals or hexahedra which leads to a more complicated implementation. 31/ 221 32/ 221

Notations and Auxiliary Results Finite Element Spaces Remarks Notations and Auxiliary Results span{x α 1 1... xα d d : α 1 +... + α d k} R k ( ) reference simplex = span{x α 1 1... xα d d : max{α 1,..., α d } k} reference cube R k () = { p F 1 : p R k } S k, 1 (T ) = {v : R : v R k () for all T } S k,0 (T ) = S k, 1 (T ) C() S k,0 0 (T ) = Sk,0 (T ) H 1 0 () = {v S k,0 (T ) : v = 0 on Γ} The global continuity ensures that S k,0 (T ) H 1 (). The polynomial degree k may vary from element to element; this leads to a more complicated implementation. 33/ 221 34/ 221 Notations and Auxiliary Results Notations and Auxiliary Results Element-wise Degrees of Freedom N,k k = 1 k = 2 k = 3 k = 4 Global Degrees of Freedom N T,k N T,k = N,k k = 1 T k = 2 The functions in S k,0 (T ) are uniquely defined by their values in N T,k thanks to the admissibility of T. 35/ 221 36/ 221

Notations and Auxiliary Results Nodal Basis Functions Properties Notations and Auxiliary Results The nodal basis function associated with a vertex z N T,k is uniquely defined by he conditions λ z,k S k,0 (T ), λ z,k (z) = 1, λ z,k (y) = 0 for all y N T,k \ {z}. {λ z,k : z N T,k } is a basis for S k,0 (T ). {λ z,k : z N T,k \ Γ} is a basis for S k,0 0 (T ). (Degrees of freedom on the boundary Γ are suppressed.) λ z,k vanishes outside the union of all elements that share the vertex z. The stiffness matrix is sparse. 37/ 221 38/ 221 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. 39/ 221 40/ 221

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 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. 41/ 221 42/ 221 Notations and Auxiliary Results Bubble Functions Notations and Auxiliary Results Inverse Estimates for the 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. 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. 43/ 221 44/ 221

Jumps Notations and Auxiliary Results Variational Formulation of the Stokes Equations Variational Formulation of the Stokes Equations n E : a unit vector perpendicular to a given face E J E (ϕ): jump of a given piece-wise continuous function across a given face E in the direction of n E J E (ϕ) depends on the orientation of n E but quantities of the form J E (n E ϕ) are independent thereof. A First Attempt Abstract Saddle-Point Problems Saddle-Point Formulation of the Stokes Equations 45/ 221 46/ 221 Variational 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 Variational Formulation of the Stokes Equations A First Attempt Corresponding 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. 47/ 221 48/ 221

Variational Formulation of the Stokes Equations A First Attempt The Space V (T ) = S 1,0 0 (T )d V Variational Formulation of the Stokes Equations A First Attempt The Space 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 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) 49/ 221 50/ 221 Variational Formulation of the Stokes Equations A First Attempt The Space V (T ) = S 1,0 0 (T )d V Variational Formulation of the Stokes Equations 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 ( 0 1 ) v T (x) = h ( 1 0 ) v T (x) v T = 0 in bottom left square Sweeping through squares yields: v T = 0 in Hence: S 1,0 0 (T )d V = {0} 51/ 221 52/ 221

Variational Formulation of the Stokes Equations A First Attempt Finite Element Subspaces of V 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. Variational Formulation of the Stokes Equations Saddle-Point Formulation of the Stokes Equations Another Variational Formulation of the Stokes Equations Multiply the momentum equation with v H 1 0 ()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 53/ 221 54/ 221 Variational Formulation of the Stokes Equations Saddle-Point Formulation of the Stokes Equations Corresponding Questions Variational Formulation of the Stokes Equations Saddle-Point Formulation of the Stokes Equations 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 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? 55/ 221 56/ 221

Variational Formulation of the Stokes Equations Saddle-Point Formulation of the Stokes Equations Properties of the Variational Problem The variational problem admits a unique solution. The velocity minimizes the energy 1 u 2 f u 2 under the constraint div u = 0. The pressure is the corresponding Lagrange multiplier. Velocity and pressure together are the unique saddle-point of the functional 1 u 2 p div u f u 2 Variational Formulation of the Stokes Equations Saddle-Point Formulation of the Stokes Equations Properties of the Discrete Problem ( ) A B The stiffness matrix has the block structure B T 0 with a symmetric positive definite matrix A and a rectangular matrix B. The discrete problem admits a unique solution if and only if B has maximal rank. The discrete problem has a unique solution and yields optimal error estimates if and only if the inf-sup condition p T div u T inf sup u T 1 p T β > 0 p T Y (T )\{0} u T X(T )\{0} is satisfied with β independent of T. 57/ 221 58/ 221 Variational Formulation of the Stokes Equations Saddle-Point Formulation of the Stokes Equations 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 rank 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. 59/ 221 Variational Formulation of the Stokes Equations 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 inf-sup condition is satisfied. 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 }. Choosing the polynomial degree of the pressure one less than the polynomial degree of the velocity is optimal. 60/ 221

Variational Formulation of the Stokes Equations Saddle-Point Formulation of the Stokes Equations Approximation of the Space V of the Stokes Equations of the Stokes Equations 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. A Second Attempt Stable Finite Element Pairs Petrov-Galerkin Methods Non-Conforming s Stream-Function Formulation 61/ 221 62/ 221 of the Stokes Equations A Second Attempt The P 1/P 0-Element 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. 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 () 1 +1 1 +1 +1 1 +1 1 +1 1 +1 +1 1 +1 1 1 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. 63/ 221 64/ 221

of the Stokes Equations A Second Attempt Proof of the Checker-Board Instability of the Stokes Equations A Second Attempt Conclusions div v T dx = v T n ij ds ij ij (ih,jh) = h { v T (ih, jh) ( ) 1 1 + vt ((i + 1)h, jh) ( ) 1 2 1 + 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 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. 65/ 221 66/ 221 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} 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 Y (T ) = S 0, 1 (T ) L 2 0 () Y (T ) = S 1,0 (T ) L 2 0 () 67/ 221 68/ 221

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 Y (T ) = S 1,0 (T ) L 2 0 () 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 () 69/ 221 70/ 221 of the Stokes Equations Stable Finite Element Pairs A Catalogue of Stable Elements 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 of the Stokes Equations Stable Finite Element Pairs Error Estimates There is a constant c which only depends on the shape parameter of T such that u u T 1 + p p T { } c inf u v T 1 + p q T. (v T,q T ) 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 1 + p p T c h k{ u k+1 + p k }. 71/ 221 72/ 221

of the Stokes Equations Petrov-Galerkin Methods Properties of the Mini Element 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. 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 B l B b D 1 b Bb T ) ( ) ( ul = p A straightforward calculation yields: ( Bb D 1 b Bb T ) i,j h 2 T f l B b D 1 b f b λ i λ j ) 73/ 221 74/ 221 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 J E (p T )J E (q T ). 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 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 J E (p T )J E (q T ) = δ h 2 f q T E E E T 75/ 221 76/ 221

of the Stokes Equations Petrov-Galerkin Methods Choice of Stabilization Parameters Set δ max = max{max δ, max δ E}, T E E min{ min δ, min δ E} T E E 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. 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 77/ 221 78/ 221 of the Stokes Equations Petrov-Galerkin Methods Error Estimates of the Stokes Equations Non-Conforming s The Basic Idea 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 1 + p p T ch k{ } u k+1 + p k. 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. 79/ 221 80/ 221

of the Stokes Equations Non-Conforming s The Crouzeix-Raviart Element (d = 2) of the Stokes Equations Non-Conforming s Properties of the Crouzeix-Raviart Element 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: 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. 81/ 221 82/ 221 of the Stokes Equations Non-Conforming s Drawbacks of the Crouzeix-Raviart Element Its accuracy deteriorates drastically in the presence of re-entrant corners. It has no higher order equivalent. It has no three-dimensional equivalent. 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 } 83/ 221 84/ 221

of the Stokes Equations Non-Conforming s Solution of the Discrete Problem of the Stokes Equations Non-Conforming s Computation of the Velocity 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 J E (p T ) T T 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 ). for all E E. 85/ 221 86/ 221 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. 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 87/ 221 88/ 221

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 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. 89/ 221 90/ 221 Summary of the Stokes Equations Summary standard Petrov-Galerkin non-conforming Bernardi-Raugel h 1 R k /R k 1 h k Crouzeix-Raviart h 1 mini element h 1 R k /R k h k Taylor-Hood h 2 R k /R k 1 (k 3) h k no parameters arbitrary pairs velocity-pressure decouple low order critical parameters critical non-convex domain critical not conservative not conservative fully conservative arbitrary dimension arbitrary dimension only two-dimensional Motivation Uzawa Type Algorithms Multigrid Algorithms Conjugate Gradient Type Algorithms 91/ 221 92/ 221

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 1+ 1 2d ) with the CG-algorithm with Gauß-Seidel preconditioning, O(N) with a multigrid (MG-) algorithm. 93/ 221 94/ 221 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 ). 95/ 221 96/ 221

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. 97/ 221 98/ 221 Uzawa Type Algorithms The Uzawa Algorithm 0. Given: an initial guess p 0, a tolerance ε > 0 and a relaxation parameter ω > 0. 1. Set i = 0. 2. 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 2. 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 ). 99/ 221 100/ 221

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. 101/ 221 Uzawa Type Algorithms The Improved Uzawa Algorithm 0. Given: an initial guess p 0 and a tolerance ε > 0. 1. 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 = 0. 2. 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 2. 102/ 221 Uzawa Type Algorithms Properties of the Improved Uzawa Algorithm Multigrid Algorithms The Basic Idea 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 0.5. 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. 103/ 221 104/ 221

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 105/ 221 106/ 221 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. 107/ 221 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 = 0. 3.2 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 1. 3.3 Update u k by u k u k + I k 1,k u k 1. 4. (Post-smoothing) Perform ν 2 steps of the iterative procedure u k u k + M k (f k L k u k ). 108/ 221

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. 0 1 2 0 1 1 2 0 0 1 1 2 0 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. 109/ 221 110/ 221 Multigrid Algorithms Smoothing for 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. Multigrid Algorithms Number of Operations 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. 111/ 221 112/ 221

Multigrid Algorithms Convergence Rate for Stokes Problem 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. 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). 113/ 221 114/ 221 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. 115/ 221 116/ 221

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 = 0. 2. 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 2. 117/ 221 118/ 221 Motivation Basic Ingredients A Posteriori Error Estimates for the Stokes Problem The Stokes Problem and its u H0 1()d, p L 2 0 () weak solution of the Stokes problem with no-slip boundary condition: 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 + 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 ) 119/ 221 120/ 221

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. 121/ 221 122/ 221 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. 123/ 221 124/ 221

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 2 + 1 } 1 h E J E (n E ( u T p T I)) 2 2 E. 2 E E, Then the error is bounded from above and from below by { u ut 2 1 + p p T 2} 1 2 c { ( η 2 R, + h 2 f f T 2 ) } 1 2 A Posteriori Error Estimates for the Stokes Problem Discussion of the A Posteriori Error Estimates I 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. T { } 1 η R, c u u T 2 1,ω + p p T 2 ω + h 2 f f T 2 2 ω 125/ 221 126/ 221 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 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. 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) 127/ 221 128/ 221

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 + J (n ( u T p T I)) v + q div u T. Set η N, = { u 2 1,T + p 2 T } 1 2. 129/ 221 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 }. 130/ 221 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. 131/ 221 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. 132/ 221

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. 133/ 221 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 3. 3. 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 2. 134/ 221 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. 135/ 221 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. 136/ 221

Mesh Refinement, Coarsening and Smoothing Regular Refinement Mesh Refinement, Coarsening and Smoothing Hanging Nodes Elements are subdivided by joining the midpoints of their edges. This preserves the shape parameter. 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. 137/ 221 138/ 221 Mesh Refinement, Coarsening and Smoothing Irregular Refinement Mesh Refinement, Coarsening and Smoothing Marked Edge Bisection Triangles Quadrilaterals 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. 139/ 221 140/ 221

Mesh Refinement, Coarsening and Smoothing Mesh Coarsening Mesh Refinement, Coarsening and Smoothing Going Back in the Grid Hierarchy 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. 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 = 0. 2. 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. 141/ 221 142/ 221 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 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. 143/ 221 144/ 221

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 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 with e i = h 1 2 Ei J Ei (n Ei u T ) 145/ 221 146/ 221 Stationary Incompressible Navier-Stokes Equations Stationary Incompressible Navier-Stokes Equations Variational Formulation A Posteriori Error Estimates Strong Form Stationary Incompressible Navier-Stokes Equations 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 multiply the momentum equation with a test function v H 1 0 ()d and integrate the result over. This is possible since u H 1 0 ()d implies (u )u H 1 () d. 147/ 221 148/ 221

Stationary Incompressible Navier-Stokes Equations 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 Stationary Incompressible Navier-Stokes Equations 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 ). 149/ 221 150/ 221 Stationary Incompressible Navier-Stokes Equations 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. Basic Idea Stationary Incompressible Navier-Stokes Equations 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 2 151/ 221 152/ 221

Stationary Incompressible Navier-Stokes Equations 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 Stationary Incompressible Navier-Stokes Equations 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, ) ). 153/ 221 154/ 221 Stationary Incompressible Navier-Stokes Equations 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. Stationary Incompressible Navier-Stokes Equations 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. 155/ 221 156/ 221

Stationary Incompressible Navier-Stokes Equations 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. 157/ 221 Conclusion Stationary Incompressible Navier-Stokes Equations 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). 158/ 221 Stationary Incompressible Navier-Stokes Equations Stationary Incompressible Navier-Stokes Equations 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 Drawbacks of the Upwind Scheme It does not fit well into the framework of variational methods. The discrete problem is no longer differentiable. y b x 159/ 221 160/ 221