Compatible Discretization Schemes

Compatible Discretization Schemes Marc Gerritsma 1 1 Collaborators Artur Palha, Guido Oud, Jasper Kreeft & Mick Bouman Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 1 / 81

What if... Why mimetic methods Aims of the course What are we going to do? 1. Give a more geometric interpretation to differential equations, Geometry ; 2. Make a distinction between topological relations and metric relations, Topological vs. Metric ; 3. Use higher order approximations, Why Higher Order ; 4. General curvilinear domains; 5. Advection equation and the Poisson equation. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 2 / 81

Compatible schemes Introduction What if... Why mimetic methods A = A T A = A T Ax = b Discrete Physical model Continuous physical model Discretization Ω C x Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 3 / 81

Compatible schemes Introduction What if... Why mimetic methods A = A T A = A T Ax = b Continuous physical model Ω C x Discrete Physical model Compatible Discretization Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 3 / 81

Simple Poisson equation What if... Why mimetic methods Provocative introduction Consider the good old Poisson equation p = f and rewrite this as a first order system Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 4 / 81

Simple Poisson equation What if... Why mimetic methods Provocative introduction Consider the good old Poisson equation p = f and rewrite this as a first order system divu = f gradp = u Exact discretization? Discrete formulation Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 4 / 81

Simple Poisson equation What if... Why mimetic methods Provocative introduction Consider the good old Poisson equation p = f and rewrite this as a first order system divu = f gradp = u Exact discretization? Yes Yes Discrete formulation Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 4 / 81

Simple Poisson equation What if... Why mimetic methods Provocative introduction Consider the good old Poisson equation p = f and rewrite this as a first order system Exact discretization? Discrete formulation divu = f Yes u i,j u i 1,j + v i,j v i,j 1 = f ij gradp = u Yes u i,j = p i,j p i 1,j v i,j = p i,j p i,j 1 If we can discretize both equations without error, then it is tempting to conclude that we can solve the Poisson equation without error... Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 4 / 81

Simple Poisson equation What if... Why mimetic methods Provocative introduction Consider the good old Poisson equation p = f and rewrite this as a first order system Exact discretization? Discrete formulation divu = f Yes u i,j u i 1,j + v i,j v i,j 1 = f ij gradp = u Yes u i,j = p i,j p i 1,j v i,j = p i,j p i,j 1 If we can discretize both equations without error, then it is tempting to conclude that we can solve the Poisson equation without error... Reason: the u in the first equation is not the same u as in the second equation, so the error is introduced in equating these to vectors. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 4 / 81

Discrete vs. continuous modeling What if... Why mimetic methods Physical theories can be decomposed into global relations and local relations [Tonti 1973, Matiussi 2002]. The global relations are metric-free and the representation in discrete form is exact. The matrix coefficient in these relations are 1, 0 and 1, so no round-off error. Topological relations remain unchanged on moving meshes or non-uniform meshes. The global relations and associated unknowns are represented as co-chains on dual cell-complexes. Local relations provide a coupling between the dual cell-complexes. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 5 / 81

What if... Why mimetic methods Outline of the course We will look in this course at the following topics: Expressing physical quantities as differential forms; Discrete analysis with chains and co-chains; The relation between differential forms and co-chains: histopolation; The Poisson equation; Convection equation. In the exercises on Friday afternoon we will play with higher order mimetic schemes for the Poisson equation and convection equation on arbitrary grids. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 6 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors I We are all acquainted with vectors in R N : Take two points in R N and connect them by an arrow. But this construction does not make sense in non-euclidean space. How do you define, for instance, a vector on a sphere in R 3? We can resolve this as follows: Suppose we have a (local) coordinate system, which assigns to every point a set of numbers (x 1,..., x n ), then we can define the curve t (x 1 (t),..., x n (t)), such that for t = 0 (x 1 (0),..., x n (0)) = (x 1..., xn ). t = 0 (x 1...,x n ) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 7 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors I We are all acquainted with vectors in R N : Take two points in R N and connect them by an arrow. But this construction does not make sense in non-euclidean space. How do you define, for instance, a vector on a sphere in R 3? We can resolve this as follows: Suppose we have a (local) coordinate system, which assigns to every point a set of numbers (x 1,..., x n ), then we can define the curve t (x 1 (t),..., x n (t)), such that for t = 0 (x 1 (0),..., x n (0)) = (x 1..., xn ). With the curve parametrized by t, we can now define the the tangent vector along the curve: (ẋ 1 (0),..., ẋ n (0)). t = 0 (x 1...,x n ) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 7 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors I We are all acquainted with vectors in R N : Take two points in R N and connect them by an arrow. But this construction does not make sense in non-euclidean space. How do you define, for instance, a vector on a sphere in R 3? We can resolve this as follows: Suppose we have a (local) coordinate system, which assigns to every point a set of numbers (x 1,..., x n ), then we can define the curve t (x 1 (t),..., x n (t)), such that for t = 0 (x 1 (0),..., x n (0)) = (x 1..., xn ). With the curve parametrized by t, we can now define the the tangent vector along the curve: (ẋ 1 (0),..., ẋ n (0)). Of course, we may consider multiple curves passing through (x 1..., xn ), each with their own tangent vectors. t = 0 (x 1...,x n ) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 7 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors II t = 0 (x 1...,x n ) The collection of tangent curves through x form a vector space T x M, the so-called tangent space at x M. Note that if v, w T x M and α, β R then αv + βw T x M So M itself is in general not a vector space, but its tangent space at x is. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 8 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors III t = 0 (x 1...,x n ) Suppose that we had chosen a different coordinate system to label (locally) the points near x, say (y 1,..., y n ) and the same curve is now given in terms of these new coordinate by (y 1 (t),..., y n (t)), with (y 1 (0),..., y n (0)) = (x 1,..., xn ), then the tangent vector is given by (ẏ 1 (0),..., ẏ n (0)). The components (ẏ 1 (0),..., ẏ n (0)) are not the same as (ẋ 1 (0),..., ẋ n (0)), but they describe exactly the same tangent vector! Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 9 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors III t = 0 (x 1...,x n ) Suppose that we had chosen a different coordinate system to label (locally) the points near x, say (y 1,..., y n ) and the same curve is now given in terms of these new coordinate by (y 1 (t),..., y n (t)), with (y 1 (0),..., y n (0)) = (x 1,..., xn ), then the tangent vector is given by (ẏ 1 (0),..., ẏ n (0)). The components (ẏ 1 (0),..., ẏ n (0)) are not the same as (ẋ 1 (0),..., ẋ n (0)), but they describe exactly the same tangent vector! Let y 1 = y 1 (x 1,..., x n ),..., y n = y n (x 1,..., x n ), then dy i n ( y i ) dx j dt = t=0 x j=1 j x=x dt t=0 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 9 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors IV Let y 1 = y 1 (x 1,..., x n ),..., y n = y n (x 1,..., x n ), then dy i n ( y i ) dx j dt = t=0 x j=1 j x=x dt t=0 The tangent vector is the same in each coordinate system but its representation in terms of components depends on the coordinate system Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 10 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors IV Let y 1 = y 1 (x 1,..., x n ),..., y n = y n (x 1,..., x n ), then dy i n ( y i ) dx j dt = t=0 x j=1 j x=x dt t=0 The tangent vector is the same in each coordinate system but its representation in terms of components depends on the coordinate system x 2 = 2 x 2 = 1 x 2 = 0 x 2 = 1 x 1 = 0 x 1 = 1 x 1 = 2 x 1 = 3 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 10 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors IV Let y 1 = y 1 (x 1,..., x n ),..., y n = y n (x 1,..., x n ), then dy i n ( y i ) dx j dt = t=0 x j=1 j x=x dt t=0 The tangent vector is the same in each coordinate system but its representation in terms of components depends on the coordinate system x 1 = 3 x 1 = 2 x 1 = 1 x 1 = 0 x 1 = 5 x 1 = 4 x 2 = 1 x 2 = 0 x 2 = 1 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 10 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors as differential operators I We have just calculated the coefficients of the tangent vector, but coefficients only make sense with respect to a basis. Based on the transformation rules for vector fields we are tempted to write a vector as v x=x = ẋ 1 (0) x 1 + + ẋn (0) x n Suppose we have another coordinate system, y i = y i (x j ), then we have v x=x = n n i=1 j=1 x j (0) yi x j y i Although it might seems unfamiliar to use partial derivatives as basis functions, it is very easy to work with. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 11 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors as differential operators II Natural basis vectors and partial derivatives behave exactly the same under coordinate transformations. Mathematically, there is no need to distinguish between them and we shall therefore write vectors as v x=x = ẋ 1 (0) x 1 + + ẋn (0) x n This is only true when we use coordinate basis vectors. If we normalize the basis vectors we use this nice property. We can now also define how a vector acts on smooth functions f [ v(f) = v 1 x 1 + + ] vn x n (f) := v 1 f f + + vn x1 x n Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 12 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Vectors as differential operators II Natural basis vectors and partial derivatives behave exactly the same under coordinate transformations. Mathematically, there is no need to distinguish between them and we shall therefore write vectors as v x=x = ẋ 1 (0) x 1 + + ẋn (0) x n This is only true when we use coordinate basis vectors. If we normalize the basis vectors we use this nice property. We can now also define how a vector acts on smooth functions f [ v(f) = v 1 x 1 + + ] vn x n (f) := v 1 f f + + vn x1 x n Using the transformation rules for vectors, we can show that this construction is independent of the local coordinates: n v(f) x := v j f n x x j=1 j = f n ( x j ) n x j=1 j y i=1 i vy i = v i f y y i=1 i = v(f)y Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 12 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Linear functionals & Dual space I Let E be a real vector space, i.e., for all v, w E and a, b R we have that av + bw E, then we can define the dual space, E, consisting of all linear functionals defined on E. Definition: A linear functional, α, on E is a real-valued linear functional α, α : E R. Thus α (av + bw) = aα(v) + bα(w) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 13 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Linear functionals & Dual space I Let E be a real vector space, i.e., for all v, w E and a, b R we have that av + bw E, then we can define the dual space, E, consisting of all linear functionals defined on E. Definition: A linear functional, α, on E is a real-valued linear functional α, α : E R. Thus α (av + bw) = aα(v) + bα(w) Definition: The collection of all linear functionals α on a vector space E form a new vector space E, the dual space to E, under the operations (aα + bβ) (v) := aα(v) + bβ(v), a, b R α, β E, v E Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 13 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Linear functionals & Dual space II Consider a vector v = v 1 e 1 + + v n e n 1, the due to linearity α(v) can be written as α(v) = α ( v 1 e 1 + + v n e n ) = v 1 α(e 1 ) + + v n α(e n). So if we know how α operates on the basis vectors e i, we know how α acts on all vectors. Let us define the dual basis, e i, by e i (e j ) = 1 if i = j 0 if i j 1 We write ei, because the basis does not necessarily has to be a coordinate basis Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 14 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Linear functionals & Dual space II Consider a vector v = v 1 e 1 + + v n e n 1, the due to linearity α(v) can be written as α(v) = α ( v 1 e 1 + + v n e n ) = v 1 α(e 1 ) + + v n α(e n). So if we know how α operates on the basis vectors e i, we know how α acts on all vectors. Let us define the dual basis, e i, by e i (e j ) = 1 if i = j 0 if i j e i (v) = e i ( v 1 e 1 + + v n e n ) = v 1 e i (e 1 ) + + v n e i (e n) = v i So e i is the linear functional that reads off the i th component of each vector v. 1 We write ei, because the basis does not necessarily has to be a coordinate basis Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 14 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Linear functionals & Dual space III So now we have e i (v) = e i ( v 1 e 1 + + v n e n ) = v 1 e i (e 1 ) + + v n e i (e n) = v i α(v) = = n α ( v j ) n e j = v j α(e j ) j=1 j=1 n n α(e j )e j (v) = α(e j )e j (v) j=1 j=1 v E Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 15 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Linear functionals & Dual space III So now we have e i (v) = e i ( v 1 e 1 + + v n e n ) = v 1 e i (e 1 ) + + v n e i (e n) = v i α(v) = = n α ( v j ) n e j = v j α(e j ) j=1 j=1 n n α(e j )e j (v) = α(e j )e j (v) j=1 j=1 v E This equality implies that n n α = α(e j )e j = α j e j j=1 j=1 So the functionals e j form a basis for E. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 15 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Linear functionals & Dual space IV Remarks: Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 16 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Linear functionals & Dual space IV Remarks: The linear functionals acting on vectors are generally called covariant vectors, co-vectors or 1-forms. Co-vectors can be associated with vectors, but this association depends on the metric and since we want to avoid as much as possible metric-dependent constructions, we treat co-vectors as separate entities. In classical tensor analysis there is the operation of lowering and raising the indices by means of the metric tensor g. This is precisely the conversion from vector to co-vector. This operation suggests that vectors and co-vectors are different ways to write the same mathematical entity, which is not true. Co-vectors behave differently from vectors and in the remainder of the course we prefer to work with differential forms Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 16 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The differential of a function I Definition: Let f be a function on a manifold M, then the differential, df, at the point p M is defined by df(v) p := v p(f) R It is easy to show that df is a linear functional which acts on vectors and therefore df is a 1-form. We saw that v p(f) is independent of any basis, therefore this defintion of df is coordinate-free. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 17 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The differential of a function I Definition: Let f be a function on a manifold M, then the differential, df, at the point p M is defined by df(v) p := v p(f) R It is easy to show that df is a linear functional which acts on vectors and therefore df is a 1-form. We saw that v p(f) is independent of any basis, therefore this defintion of df is coordinate-free. If we write v in terms of a coordinate basis, we have n df v j n x j=1 j = v j (p) f x j=1 j (p) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 17 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The differential of a function I Definition: Let f be a function on a manifold M, then the differential, df, at the point p M is defined by df(v) p := v p(f) R It is easy to show that df is a linear functional which acts on vectors and therefore df is a 1-form. We saw that v p(f) is independent of any basis, therefore this defintion of df is coordinate-free. If we write v in terms of a coordinate basis, we have n df v j n x j=1 j = v j (p) f x j=1 j (p) If we take the spacial function f(x) = x i this gives n dx i v j n x j=1 j = v j xi x j=1 j = vi Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 17 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The differential of a function II If we take the spacial function f(x) = x i this gives n dx i v j n x j=1 j = v j xi x j=1 j = vi so for each i dx i reads off the i th component of the vector v, i.e. for a coordinate basis we have e i = dx i So, we can express every 1-form as n n α = α j e j = α j dx j j=1 j=1 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 18 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator / x and dx Major conceptual change: We have seen / x and dx before, where in each case there was a limiting process involved: dx was interpreted as the infinitesimal small distance in the x-direction between two points and / x measures the change in the x-direction when we move an infinitesimal distance in the x-direction. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 19 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator / x and dx Major conceptual change: We have seen / x and dx before, where in each case there was a limiting process involved: dx was interpreted as the infinitesimal small distance in the x-direction between two points and / x measures the change in the x-direction when we move an infinitesimal distance in the x-direction. In the new interpretation / x and dx are basis vectors for vectors and 1-forms, respectively, which can be defined without having to made any recourse to a limiting process! Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 19 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The pullback operator I Let φ : M n V r which maps an r-dimensional manifold onto a r-dimensional space, i.e., φ(x 1,..., x n ) = (y 1,..., y r ), then a vector v in M is mapped onto a vector φ (v) in V n φ (v M ) = φ v j (x k ) n r x j=1 j = v j (φ(x k )) j=1 i=1 y i x j y i = v V Here you see why it is so convenient to associate the basis vectors with / x. The transformation boils down to the application of the chain-rule. Definition: Let φ : M n V r be a smooth map of manifolds and let φ(x) = y. Let φ be the push-forward of φ. The pullback φ is a linear transformation that maps co-vectors defined on V onto co-vectors defined on M. If β is a co-vector defined in V and v a vector in M then φ (β)(v) := β (φ v) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 20 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The pullback operator II Definition: Let φ : M n V r be a smooth map of manifolds and let φ(x) = y. Let φ be the push-forward of φ. The pullback φ is a linear transformation that maps co-vectors defined on V onto co-vectors defined on M. If β is a co-vector defined in V and v a vector in M then φ (β)(v) := β (φ v) Example Let β be a 1-form defined in y V by β = b i dy i, then φ (β) = b i y i x j dxj So the transformation of 1-forms is (again) obtained through application of the chain rule. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 21 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The pullback operator III M V Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 22 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The pullback operator III M x φ V y Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 22 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The pullback operator III T x M T y V M x V y φ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 22 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The pullback operator III T x M φ T y V M x V y φ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 22 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The pullback operator III T xm R T y V T x M φ T y V M x V y φ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 22 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The pullback operator III T xm R φ T y V T x M φ T y V M x V y φ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 22 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator 2-forms I 0-forms are functions defined on M; they assign a number ( R) to every point x M. 1-forms are functionals defined on T M; they assign a number to every vector associated with the x M. Higher order differential forms can be constructed from 1-forms by means of the wedge product: Let α and β be two 1-forms and v and w be two vectors, then (α β) (v, w) = α(v) β(w) α(w) β(v) = α(v) β(v) Note that the wedge product of twp 1-forms is bilinear operators, i.e. α(w) β(w) (α β) (av 1 + bv 2, w) = a (α β) (v 1, w) + b (α β) (v 2, w) and (α β) (v, aw 1 + bw 1 ) = a (α β) (v, w 1 ) + b (α β) (v, w 2 ) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 23 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator 2-forms II Furthermore, we have that (aα 1 + bα 2 ) β = aα 1 β + bα 2 β and α (aβ 1 + bβ 2 ) = aα β 1 + bα β 2 The wedge product of two 1-forms is skew-symmetric, i.e. (α β) (v, w) = α(v) α(w) β(v) β(w) = α(w) α(v) β(w) β(v) = (α β) (w, v) and α β = β α Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 24 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator 2-forms: Example I Consider the 2 vectors in R 3 given by v = x + y + 3 z and w = 3 x + 3 y z then (dx dy) (v, w) = dx(v) dx(w) dy(v) dy(w) = v 1 w 1 v 2 w 2 = 1 3 1 3 = 0 (dy dz) (v, w) = dy(v) dy(w) dz(v) dz(w) = v 2 w 2 v 3 w 4 = 1 3 3 1 = 10 and (dz dx) (v, w) = dz(v) dx(v) dz(w) dx(w) = v 3 w 3 v 1 w 1 = 3 1 1 3 = 10 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 25 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator 2-forms: Example II z w v x y Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 26 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator 2-forms: Example II z w v x y (dx dy)(v,w) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 26 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator 2-forms: Example II z w v x y (dx dy)(v,w) So dx dy assigns to two vectors the number which corresponds to the area spanned by the two vectors projected onto the xy-plane. Similarly, dy dz and dz dx give the projected areas onto the yz- and xz-plane, respectively. Note that (dx dy)(v, w) = (dy dx)(v, w) = (dx dy)(w, v) = (dy dx)(w, v) The change of sign indicates that dx dy is the oriented projected area. If we swap the vectors or the 1-forms we reverse the orientation and consequently the value assigned to the 2-form. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 26 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The flux vector A well known 2-form in physics is the flux vector given by f (2) (x, y, z) = f x(x, y, z) dy dz + f y(x, y, z) dz dx + f z(x, y, z) dx dy In conventional physical modeling the flux vector is defined with respect to the normal to a surface. The two signs which correspond to the orientation are in conventional physical modeling distinguished by outward unit normal and inward unit normal. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 27 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Flux in different coordinates I f (2) (x, y, z) = f x(x, y, z) dy dz + f y(x, y, z) dz dx + f z(x, y, z) dx dy Consider a transformation (or mapping), i.c., transformation to spherical coordinates: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ dx = x x x dr + dθ + r θ φ = dy = y y y dr + dθ + r θ φ dz = z z z dr + dθ + r θ φ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 28 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Flux in different coordinates I f (2) (x, y, z) = f x(x, y, z) dy dz + f y(x, y, z) dz dx + f z(x, y, z) dx dy Consider a transformation (or mapping), i.c., transformation to spherical coordinates: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ = dx = x x x dr + dθ + = sin θ cos φ dr + r cos θ cos φ dθ r sin θ sin φ dφ r θ φ dy = y y y dr + dθ + r θ φ dz = z z z dr + dθ + r θ φ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 28 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Flux in different coordinates I f (2) (x, y, z) = f x(x, y, z) dy dz + f y(x, y, z) dz dx + f z(x, y, z) dx dy Consider a transformation (or mapping), i.c., transformation to spherical coordinates: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ = dx = x x x dr + dθ + = sin θ cos φ dr + r cos θ cos φ dθ r sin θ sin φ dφ r θ φ dy = y y y dr + dθ + = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ r θ φ dz = z z z dr + dθ + r θ φ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 28 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Flux in different coordinates I f (2) (x, y, z) = f x(x, y, z) dy dz + f y(x, y, z) dz dx + f z(x, y, z) dx dy Consider a transformation (or mapping), i.c., transformation to spherical coordinates: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ = dx = x x x dr + dθ + = sin θ cos φ dr + r cos θ cos φ dθ r sin θ sin φ dφ r θ φ dy = y y y dr + dθ + = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ r θ φ dz = z z z dr + dθ + = cos θ dr r sin θ dθ r θ φ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 28 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Flux in different coordinates I dx = sin θ cos φ dr + r cos θ cos φ dθ r sin θ sin φ dφ dy = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ dz = cos θ dr r sin θ dθ = dy dz = (sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ) (cos θ dr r sin θ dθ) = cos θ sin θ sin φ dr dr r sin 2 θ sin φ dr dθ +r cos 2 θ sin φ dθ dr r 2 cos θ sin θ sin φ dθ dθ +r cos θ sin θ cos φ dφ dr r 2 sin 2 cos φ dφ dθ = r sin θ dr dθ + r 2 sin 2 θ cos φ dθ dφ + 1 r sin 2θ cos φ dφ dr 2 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 29 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Flux in different coordinates I dx = sin θ cos φ dr + r cos θ cos φ dθ r sin θ sin φ dφ dy = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ dz = cos θ dr r sin θ dθ dz dx = r cos φ dr dθ + r 2 sin 2 θ sin φ dθ dφ + 1 r sin 2θ sin φ dφ dr 2 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 29 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Flux in different coordinates I dx = sin θ cos φ dr + r cos θ cos φ dθ r sin θ sin φ dφ dy = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ dz = cos θ dr r sin θ dθ dx dy = r 2 sin 2θ dθ dφ r sin 2 θ dφ dr Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 29 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Flux in different coordinates I dx = sin θ cos φ dr + r cos θ cos φ dθ r sin θ sin φ dφ dy = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ dz = cos θ dr r sin θ dθ f (2) (x, y, z) = f x(x, y, z) dy dz + f y(x, y, z) dz dx + f z(x, y, z) dx dy = f (2) = [ f x r sin φ + f y r cos φ] dr dθ + [ fx r 2 sin 2 θ cos φ + f y r 2 sin 2 θ sin φ + f z r 2 sin 2θ ] dθ dφ + [ f x 1 ] 1 sin 2θ cos φ + fy 2 2 sin 2θ sin φ fz sin2 θ dφ dr Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 29 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Flux in different coordinates I dx = sin θ cos φ dr + r cos θ cos φ dθ r sin θ sin φ dφ dy = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ dz = cos θ dr r sin θ dθ Although the equations look awful, the recipe to derive them is quite easy and so is the implementation in a solver. Either provide the mapping and the partial derivatives at specified points (to be determined later) or simply provide the mapping only and determine the derivatives by means of interpolation. We were given the map Φ : (r, θ, φ) (x, y, z) and we calculated Φ f 2 (x, y, z) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 29 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator 3-form Let α, β and δ be 1-forms and u, v and w be vectors, then a 3-form is defined by (α β δ) (u, v, w) = α(u) α(v) α(w) β(u) β(v) β(w) δ(u) δ(v) δ(w) Note that (dx dy dz) (u, v, w) = u 1 v 1 w 1 u 2 v 2 w 2 u 3 v 3 w 3 So, dx dy dz assigns to three vectors the volume of the parallelepiped spanned by those vectors A well-know example of a physical quantity described by a 3-form is density: ρ dx dy dz. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 30 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The exterior derivative I The exterior derivative, d, is an operator which converts k-forms in (k + 1)-forms, which satisfies the following properties: 1. For smooth 0-forms, df is the differential of f defined by df(v) = v(f) 2. d ( dα (k)) = 0 (k+2) 3. Let α be a p-form and β a q-form, then d (α β) = dα β + ( 1) p α dβ. As a consequence of this definition we have that the exterior derivative n-form in a n-dimensional space vanishes. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 31 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The exterior derivative I The exterior derivative, d, is an operator which converts k-forms in (k + 1)-forms, which satisfies the following properties: 1. For smooth 0-forms, df is the differential of f defined by df(v) = v(f) 2. d ( dα (k)) = 0 (k+2) 3. Let α be a p-form and β a q-form, then d (α β) = dα β + ( 1) p α dβ. As a consequence of this definition we have that the exterior derivative n-form in a n-dimensional space vanishes. The importance of the exterior derivative is that it can be represented exactly in a discrete setting. Furthermore, the exterior derivative is coordinate-free and metric free and therefore has the same form and properties in all coordinate systems. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 31 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Example exterior derivative 0-form Let ω be the 1-form given by then the exterior derivative gives ω 0 (x, y, z) = f(x, y, z) dω 0 (x, y, z) = f f f dx + dy + x y z dz We can associate the vector gradf = (f x, f y, f z) T with the exterior derivative of ω 0 = f. This is denoted by gradf = dω 0 or dω 0 = gradf. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 32 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Example exterior derivative 1-form Let ω be the 1-form given by ω 1 (x, y, z) = P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz then the exterior derivative gives dω 1 (x, y, z) = P P P dx dx + dy dx + dz dx + x y z Q Q Q dx dy + dy dy + dz dy + x y z R R R dx dz + dy dz + dz dz x y z = ( R y Q z ) dy dz + ( P z R x where we used dx dy = dy dx, dy dz = dz dy, etc. and dx dx = dy dy = dz dz = 0. ) ( Q dz dx + x P ) dx dy y Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 33 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Example exterior derivative 2-form Let ω be the 1-form given by ω 2 (x, y, z) = A(x, y, z) dydz + B(x, y, z) dzdx + C(x, y, z) dxdy then the exterior derivative gives dω 2 (x, y, z) = A A dx dy dz + x y B B dx dz dx + x y C C dx dx dy + x y = ( A x + B y + C z A dy dy dz + dz dy dz + z B dy dz dx + dz dz dx + z C dy dx dy + dz dx dy z ) dx dy dz where we used dx dy = dy dx, dy dz = dz dy, etc. and dx dx = dy dy = dz dz = 0. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 34 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Differential forms 2 Let M R n a differentiable manifold, then k-form ω k Λ k : a rank-k, anti-symmetric, tensor field over M Example diff. form ω k : T xm T }{{ xm R, } k copies ω k (..., v i,..., v j,... ) = ω k (..., v j,..., v i,... ) Wedge product: Let ω k Λ k and ω l Λ l then : Λ k Λ l : Λ k+l Exterior derivative: d : Λ k Λ k+1. If n = 3 we have for k = 0 d grad, k = 1 d curl and for k = 2 d div. Since d d 0 this means curl grad 0 and div curl 0. Exact sequence (De Rham complex) R Λ 0 (Ω) d Λ 1 (Ω) d Λ 2 (Ω) d Λ 3 (Ω) d 0 2 Cartan [1], Spivak [3], Flanders [2] Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 35 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Differential forms Stokes Theorem: Let Ω k+1 be a k + 1-dimensional manifold and ω Λ k then ω k = Ω k+1 Ω k+1 dω k Example generalized Stokes Two ways to interpret Stokes Theorem: ω k = Ω k+1 Ω k+1 dω k ω k, Ω k+1 = dω k, Ω k+1 Either take the exterior derivative and establish the Stokes Theorem; Take the Stokes Theorem to define the exterior derivative. In the latter interpretation the exterior derivative is the adjoint of the boundary operator. This is the interpretation that we will use in the discrete setting. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 36 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The interior product The exterior derivative was a mapping from k-forms onto (k + 1)-forms. The interior product ι v is a mapping from (k + 1)-forms to k-forms, defined by: Given a vector v, then ( ι vω (k+1)) (w 1,..., w k ) = ω (k+1) (v, w 1,..., w k ) The interior product of a 0-form is 0 by definition. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 37 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Convection The Lie-derivative I The Lie-derivative of a k-form, L vα (k) measures the change of the k-form in the direction of the vector v. This derivative can be defined by Cartan s magic formula in terms of the interior product and the exterior derivative Consider R 3 and v = (v 1, v 2, v 3 ), then L vα (k) = d ι v α (k) + ι v dα (k) L vα (1) = [ ( α1 v 1 + α 2 v 2 + α 3 v 3) ( x 1 + v 2 α1 x 2 α ) ( 2 x 1 v 3 α3 x 1 α ) ] 1 x 3, dx 1 + [ ( α1 v 1 + α 2 v 2 + α 3 v 3) ( x 2 + v 3 α2 x 3 α ) ( 3 x 2 v 1 α1 x 2 α ) ] 2 x 1 dx 2 + [ ( α1 v 1 + α 2 v 2 + α 3 v 3) ( x 3 + v 1 α3 x 1 α ) ( 1 x 3 v 2 α2 x 3 α ) ] 3 x 2 dx 3 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 38 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Convection The Lie-derivative I The Lie-derivative of a k-form, L vα (k) measures the change of the k-form in the direction of the vector v. This derivative can be defined by Cartan s magic formula in terms of the interior product and the exterior derivative Consider R 3 and v = (v 1, v 2, v 3 ), then [ ( v 1 α1 x 1 + α 2 [ ( v 2 α1 x 1 + α 2 x 2 + α 3 x 3 x 2 + α 3 x 3 [ ( v 3 α1 x 1 + α 2 x 2 + α 3 x 3 L vα (k) = d ι v α (k) + ι v dα (k) ) + ) + L vα (2) = x 2 ( v 2 α 1 v 1 α 2 ) ( v 3 x 3 α 2 v 2 ) α 3 ) + ( v 1 x 1 α 3 v 3 ) α 1 ] ( v 3 x 3 α 1 v 1 ) α 3 dx 2 dx 3 + ( v 1 x 1 α 2 v 2 ) α 1 ( v 2 x 2 α 3 v 3 ) α 2 ] dx 3 dx 1 + ] dx 1 dx 2 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 38 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Convection The Lie-derivative II If we restrict ourselves to convection of a scalar quantity then we have two forms L vα (0) = v 1 α α α + v2 + v3 x1 x2 x 3 ( ( v L vα (3) 1 α ) = x 1 + ( v 2 α ) x 2 + ( v 3 α ) ) x 3 dx 1 dx 2 dx 3 Or written in our conventional vector analysis notation L vα (0) = v α L vα (3) = (αv) convection form conservation form Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 39 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator Convection The Lie-derivative II If we restrict ourselves to convection of a scalar quantity then we have two forms L vα (0) = v 1 α α α + v2 + v3 x1 x2 x 3 ( ( v L vα (3) 1 α ) = x 1 + ( v 2 α ) x 2 + ( v 3 α ) ) x 3 dx 1 dx 2 dx 3 Or written in our conventional vector analysis notation L vα (0) = v α L vα (3) = (αv) convection form conservation form This is not just writing the equation in the most convenient form! Both convection terms are equally valid, but they refer to different types of variables! Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 39 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The Hodge operator Exact sequence (De Rham complex) R Λ 0 (Ω) d Λ 1 (Ω) which for n = 3 can be recognized as d Λ 2 (Ω) d Λ 3 (Ω) d 0 R H P grad H L curl H S div H V 0 How do we represent then the Laplace operator? Applying d twice yields exactly zero due to the exactness of the De Rham complex and R(grad) / D(div). R Λ 0 d Λ 1 d Λ 2 d Λ 3 d 0 0 Λ 3 d Λ 2 d Λ 1 d Λ 0 R Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 40 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The Hodge operator Exact sequence (De Rham complex) R Λ 0 (Ω) d Λ 1 (Ω) which for n = 3 can be recognized as d Λ 2 (Ω) d Λ 3 (Ω) d 0 R H P grad H L curl H S div H V 0 How do we represent then the Laplace operator? Applying d twice yields exactly zero due to the exactness of the De Rham complex and R(grad) / D(div). R H p grad H L curl H S div H V 0 0 HṼ div H S curl H L grad H P R Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 40 / 81

Vectors Differential forms The exterior derivative The Lie-derivative Hodge- operator The Hodge operator For the 2D case the De Rham complex is given by R Λ 0 d Λ 1 d Λ 2 Λ 2 Λ 1 d d Λ0 R correspond, in the proxies to the following exact sequences: R H 1 H(curl) L 2 L 2 H(div) H 1 R Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 41 / 81

Chains Introduction Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex A p-chain is a formal sum of oriented p-cells. 0-cells are points, 1-cells are curves, 2-cells surfaces and 3-cells represent volumes. The collection of all p-chains is called a cell complex. P 13 L 22 P 14 L 23 P 15 L 24 P 16 L 18 S 7 L S 8 19 L S 9 20 L 21 L 15 P 10 L 16 P 11 L 17 P 9 P 12 L 11 S 4 L S 5 12 L S 6 13 L 14 P 5 P 8 L 8 P 6 L 9 P 7 L 10 L 4 S 1 L 5 S 2 L 6 S 3 L 7 P 1 L 1 P 2 L 2 P 3 L 3 P 4 Cell complex K and labeling of the 0-cells, 1-cells and 2-cells Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 42 / 81

Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex Chains A p-chain is a formal sum of oriented p-cells. C (p) = N p cell i=1 m i σ (p) i If σ is a p-cell, then its boundary constitutes a (p 1)-chain denoted by σ The boundary of the boundary is empty for every p-cell σ 0 σ σ σ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 43 / 81

Cochains Introduction Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex With every p-cell we can associate a value from a field F. Assume F = R, σ p : σ p R By doing so, we can associate with every p-chain a so-called co-chain C p = N p cell i=1 α i σ i,(p) Duality pairing between co-chains and chains satisfies all the properties of geometric integration C p, C p R Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 44 / 81

Cochains Introduction Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex With every p-cell we can associate a value from a field F. Assume F = R, σ p : σ p R By doing so, we can associate with every p-chain a so-called co-chain C p = N p cell i=1 α i σ i,(p) Duality pairing between co-chains and chains satisfies all the properties of geometric integration C p, C p R Note that both chains and co-chains are metric free. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 44 / 81

Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex The coboundary operator Let C p+1 be a (p + 1)-chain, then its boundary C p+1 is a p-chain, then formally we can write C p, C p+1 = δc p, C p+1 where δ is the formal adjoint of the boundary operator. This adjoint relation is the discrete analogue of the generalized Stokes theorem. Remember that the Stokes Theorem was given by ω k, Ω k+1 = dω k, Ω k+1 δ is a discrete version of the exterior derivative and is called the coboundary operator δ : C p C p+1 Note that δδc p 0 just like ddω k 0, since δδc p, C p+2 = δc p, C p+2 = C p, C p+2 0 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 45 / 81

Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex The action of the coboundary operator I The action of the coboundary operator in pictures 1 1 5 6 5 5 7 2 a) Action of δ on 0-chain 7 4 4 4 4 7 2 2 1 b) Action of δ on 1-chain 1 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 46 / 81

Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex The action of the coboundary operator II Example illustrating the property δδc (0) = 0 (2) 4 7 9 given a 0-cochain c (0) = (12, 3, 5,9, 7,4) 12 3 5 δ 4 +7 7 +9 δc (0) = ( 9,2,4, 2,4, 3, 8) +3 +2 +4 12 7 3 +9 = 5 8 +4 +4 9 +2 12 +3 3 +5 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 47 / 81

Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex The action of the coboundary operator II Example illustrating the property δδc (0) = 0 (2) 4 7 9 given a 0-cochain c (0) = (12, 3, 5,9, 7,4) 12 3 5 δ 4 +7 7 +9 δc (0) = ( 9,2,4, 2,4, 3, 8) +3 +2 +4 12 7 3 +9 = 5 8 +4 +4 9 +2 12 +3 3 +5 δ δδc (0) = (0,0) 3 +8 +4 9 2 4 +4 +2 = 0 0 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 47 / 81

Incidence matrices Introduction Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex The coboundary operator can be represented by a so-called incidence matrix. Consider for instance the cell-complex as shown in the Figure. P 4 L 3 P 3 L 4 S1 L 2 P 1 P 2 L 1 K-complex Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 48 / 81

Incidence matrices Introduction Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex The coboundary operator can be represented by a so-called incidence matrix. Consider for instance the cell-complex as shown in the Figure. P 4 L 3 P 3 E 1,0 relates 0-cohains to 1-cochains, i.e discrete gradient operator E 1,0 = 1 1 0 0 0 1 1 0 0 0 1 1 1 0 0 1 L 4 S1 L 2 P 1 P 2 L 1 K-complex Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 48 / 81

Incidence matrices Introduction Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex The coboundary operator can be represented by a so-called incidence matrix. Consider for instance the cell-complex as shown in the Figure. P 4 L 3 P 3 E 2,1 relates 0-cochains to 1-cochains, i.e discrete curl operator E 2,1 = ( 1 1 1 1 ) L 4 S1 L 2 P 1 P 2 L 1 K-complex Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 48 / 81

Incidence matrices Introduction Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex The coboundary operator can be represented by a so-called incidence matrix. Consider for instance the cell-complex as shown in the Figure. P 4 L 3 P 3 Note that E 2,1 E 1,0 = ( 1 1 1 1 ) ( 0 0 0 0 ) 1 1 0 0 0 1 1 0 0 0 1 1 1 0 0 1 = L 4 S1 L 2 This is matrix expression for δδ 0. P 1 P 2 L 1 K-complex Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 48 / 81

Dual cell complex Introduction Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex In the first slide we mentioned polar vectors and twisted vectors, or the association of k-forms with (n k)-forms. We also saw this coupling between k-forms with (n k)-forms. R Λ 0 d Λ 1 d Λ 2 d Λ 3 d 0 0 Λ 3 d Λ 2 d Λ 1 d Λ 0 R Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 49 / 81

Dual cell complex Introduction Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex This can be accommodated in the discrete case by introducing a dual cell complex. The dual cell complex is constructed such that every p-cell on the primal complex corresponds to (n p)-cell on the dual grid. P 4 = S 4 L 3 = L 3 L 4 = L 4 S 1 = P 1 P 3 = S 3 L 2 = L 2 The orientation of the dual p-cells is determined by the orientation of the n-dimensional space in which the complex is embedded. P 1 = S 1 P 2 = S 2 L 1 = L 1 Dual cell complex Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 49 / 81

Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex Incidence matrices on dual cell complex The incidence matrix Ẽ1,0 relates 0-cochains on the dual grid to 1-cochains on the dual grid Ẽ 1,0 = 1 1 1 1 = ( E 2,1) T P 4 = S 4 L 3 = L 3 L 4 = L 4 S 1 = P 1 P 3 = S 3 L 2 = L 2 This matrix corresponds to the divergence operator on the primal grid P 1 = S 1 P 2 = S 2 L 1 = L 1 Dual cell complex Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 50 / 81

Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex Incidence matrices on dual cell complex The incidence matrix Ẽ2,1 relates 1-cochains on the dual grid to 2-cochains on the dual grid Ẽ 2,1 = In general: 1 0 0 1 1 1 0 0 0 1 1 0 0 0 1 1 = ( E 1,0) T P 4 = S 4 L 3 = L 3 L 4 = L 4 S 1 = P 1 P 3 = S 3 L 2 = L 2 Ẽ n+1 p,n p = ( E p,p 1) T P 1 = S 1 P 2 = S 2 L 1 = L 1 For n = 3 this gives the symmetry relations div = grad T and curl = curl T. Dual cell complex Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 50 / 81

Again the Poisson equation Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex Remember that we started with the good old Poisson equation p = f rewritten as a first order system divu = f δũ 1 = f 2 gradp = u δp 0 = u 1 So we see that both first order equations can be discretized exactly and without any reference to metric on a cell complex. The first equations can be exactly represented on the dual grid, while second equation is represented on the primal grid. 3 In general we have ũ d 1 where d =dimω. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 51 / 81

Again the Poisson equation Chains and co-chains The coboundary operator The Incidence matrix Dual cell complex Remember that we started with the good old Poisson equation p = f rewritten as a first order system divu = f δũ 1 = f 2 gradp = u δp 0 = u 1 So we see that both first order equations can be discretized exactly and without any reference to metric on a cell complex. The first equations can be exactly represented on the dual grid, while second equation is represented on the primal grid. The somewhat provocative statement on the first slide that u u now becomes clear: u 1 is internally oriented with the line segment, whereas ũ 1 is externally oriented (it is the normal velocity) 3. 3 In general we have ũ d 1 where d =dimω. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 51 / 81

Interpolation operator I Spectral Element mesh(es) Consider the dual spectral element mesh consisting of a Gauss-Lobatto-Legendre (GLL) points and Extended Gauss-Legendre (EGL) points. Dual elements, N = 4 1 0 1 1 0 1 Dual cell complex: GLL-grid (black) and EGL-grid (red) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 52 / 81

Interpolation - edge interpolation I Interpolation operator I Spectral Element mesh(es) We just saw, that we can write the gradient equation and the divergence equation exactly, without reference to any coordinate system or metric. Now we need to establish a relation ũ 1 and u 1. This connection needs approximation and depends on the metric. This can no longer be done at a purely algebraic level. We need to interpolate our 1-cochains to continuous 1-forms, because only as differential forms this connection can be established. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 53 / 81

Interpolation - edge interpolation I Interpolation operator I Spectral Element mesh(es) We just saw, that we can write the gradient equation and the divergence equation exactly, without reference to any coordinate system or metric. Now we need to establish a relation ũ 1 and u 1. This connection needs approximation and depends on the metric. This can no longer be done at a purely algebraic level. We need to interpolate our 1-cochains to continuous 1-forms, because only as differential forms this connection can be established. You are probably familiar with polynomial interpolation: Given a collection of points on an interval, determine the polynomial such that the value of the polynomial equals the value of a prescribed function in the points. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 53 / 81

Interpolation - edge interpolation II Interpolation operator I Spectral Element mesh(es) N=1 N=3 N=7 1 1 1 0 0 0 1 1 0 1 1 1 0 1 1 1 0 1 N=15 N=31 2 N=63 1 1 1 0 0 0 1 1 0 1 1 1 0 1 1 2 1 0 1 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 54 / 81

Interpolation - edge interpolation III Interpolation operator I Spectral Element mesh(es) Instead of requiring that the nodal values of the function and the interpolant agree we could impose that the integrals of the functions and the interpolant agree between the points. If f(x) is an integrable function and P (x) is a polynomial then we impose xi xi f(x) dx = P (x) dx x i 1 x i 1 Note that f(x) dx and P (x) dx are 1-forms, so we actually impose that the 1-forms agree on each interval. This will be called edge interpolation or histopolation. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 55 / 81

Interpolation - edge interpolation IV Interpolation operator I Spectral Element mesh(es) 1.5 1.5 1 1 0.5 0.5 0 0 0.5 0.5 1 1 1.5 1 0.5 0 0.5 1 1.5 1 0.5 0 0.5 1 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 56 / 81

Nodal interpolation Introduction Interpolation operator I Spectral Element mesh(es) The construction of such an interpolating polynomial can be most conveniently done by means of Lagrange polynomials, h i (ξ), through the points ξ i. The Lagrange polynomial has the property 1 if i = j h i (ξ j ) =. (1) 0 if i j Then the interpolation of a function f(ξ) in the points ξ i is given by N N (I(f)) (ξ) = f(ξ j )h j (ξ) := f j h j (ξ) j=0 Here I is the interpolation operator. It takes values at points, 0-cochains and maps it onto a function, a 0-form. So nodal interpolation maps 0-cochains onto 0-forms. j=0 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 57 / 81

Nodal interpolation Introduction Interpolation operator I Spectral Element mesh(es) The construction of such an interpolating polynomial can be most conveniently done by means of Lagrange polynomials, h i (ξ), through the points ξ i. The Lagrange polynomial has the property 1 if i = j h i (ξ j ) =. (1) 0 if i j Then the interpolation of a function f(ξ) in the points ξ i is given by N N (I(f)) (ξ) = f(ξ j )h j (ξ) := f j h j (ξ) j=0 Here I is the interpolation operator. It takes values at points, 0-cochains and maps it onto a function, a 0-form. So nodal interpolation maps 0-cochains onto 0-forms. Evaluation of a function, i.e., a 0-form in a node ξ i is a map from 0-forms to 0-cochains (Rf) (ξ i ) = f(ξ) = f(ξ i ) := f i ξ i So nodal integration maps 0-forms onto 0-cochains. j=0 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 57 / 81

Edge interpolation I Introduction Interpolation operator I Spectral Element mesh(es) Analogous to the Lagrange functions, we use for histopolation the so-called edge polynomials, e i (ξ), which have the property that ξk 1 if i = k e i (x) dx =. (2) ξ k 1 0 if i k If we can defined such edge polynomials, we can find the polynomial which satisfies these requirements by setting P (N 1) (ξ) = N f i e i (ξ). (3) It can be shown, that the edge polynomials can be derived from the Lagrange polynomials in the following way: Let h i (ξ) of order N be the Lagrange polynomials through the nodes ξ 0 < ξ 1 < < ξ N 1 < ξ N, then the edge polynomials, e i (ξ), of order N 1 are given by i=1 i 1 e i (ξ) = Note that the edge functions are 0-forms k=0 dh k dξ. (4) Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 58 / 81

Edge interpolation II Introduction Interpolation operator I Spectral Element mesh(es) Edge interpolation is a mapping from 1-cochains to 1-forms Integration of a 1-form over a line segment (ξ i 1, ξ i ) is a mapping from 1-forms to 1-cochains. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 59 / 81

Edge interpolation III Introduction Interpolation operator I Spectral Element mesh(es) Edge function e3(ξ) 2.5 2 1.5 1 0.5 ξ 0 ξ 1 ξ 2 ξ 3 ξ 4 ξ 5 ξ 6 0 0.5 1 1.5 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 ξ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 60 / 81

Interpolation operator I Spectral Element mesh(es) Connection between differential forms and cochains The two equations divu = f and gradp = u are global relations and they can be represented in terms of cochains. The constitutive equation u 1 = ũ 1 is a local relations which can only be satisfied in terms of differential forms. The operator which provides this connection is the discrete analogue of the Hodge -operator, but this operator is not defined in algebraic topology. The recipe to resolve this problem is to interpolate a differential form from the cochains and then apply the Hodge operator at the continuous level. So the two important operators are: the Interpolation Operator I and the Reduction Operator R. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 61 / 81

Reduction operator R Introduction Interpolation operator I Spectral Element mesh(es) The Reduction Operator R is naturally given by R : Λ p C p, σ p = R(ω p ) := ω p σ p Note that this operation requires that the p-cell σ p is no longer considered as topological element, but now acquires metrical properties. It is quite straightforward to show that Rd = δr so external derivative and coboundary operator commute with R. In principle, no error is introduced by the application of R. Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 62 / 81

Interpolation operator I Interpolation operator I Spectral Element mesh(es) The Interpolation Operator I must satisfy two basic conditions RI = Id i.e. I is a right-inverse of R and IR = Id + O(h s ) i.e. I is an approximate left-inverse of R. Furthermore, we require that di = Iδ Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 63 / 81

Interpolation operator I Spectral Element mesh(es) Consider the dual spectral element mesh consisting of a Gauss-Lobatto-Legendre (GLL) points and Extended Gauss-Legendre (EGL) points. Rehuel Lobatto 1797-1866 Carl Friedrich Gauß1777-1855 Marc Gerritsma Burgers Course CFD I, 25-29 January 2010 64 / 81