A Finite Element Primer

Transcription

1 A Finite Element Primer David J. Silvester Scool of Matematics, University of Mancester Version.3 updated 4 October Contents A Model Diffusion Problem x. Domain x. Continuous Function x.3 Normed Vector Space x.4 Square Integrable Function x.5 Inner Product Space x.6 Caucy Scwarz Inequality x.7 Sobolev Space x.8 Weak Derivative Galerkin Approximation x.9 Caucy Sequence x. Complete Space Finite Element Galerkin Approximation Tis is a summary of finite element teory for a diffusion problem in one dimension. It provides te matematical foundation for Capter of our reference book Finite Elements and Fast Iterative Solvers wit Applications in Incompressible Fluid Dynamics, see ttp:// i

2 A Model Diffusion Problem. A Model Diffusion Problem Te problem we consider erein is a two point boundary value problem. A formal statement is: given a real function f C (,) (see Definition x. below), we seek a function u C (,) C [,] (see below) satisfying u (x) = f(x) for < x < u() = ; u() =. Te term u = d u dx represents diffusion, and f is called te source term. A sufficiently smoot function u satisfying (D) is called a strong (or, in te case of continuous function f, a classical ) solution. Problem. (f = ) Tis is a model for te temperature in a wire wit te ends kept in ice. Tere is a current flowing in te wire wic generates eat. Solving (D) gives te parabolic ump u(x) = (x x ). } (D). Problem. Problem. temperature.5..5 deflection..5.5 x.5 x Some basic definitions will need to be added if our statement of (D) is to make sense. Definition x. (Domain) A domain is a bounded open set; for example, Ω = (,), wic identifes were a differential equation is defined. Te closure of te set, denoted Ω, includes all te points on te boundary of te domain; for example, Ω = [,].

3 A Model Diffusion Problem Definition x. (Continuous function) A real function f is mapping wic assigns a unique real number to every point in a domain: f : Ω R. C (Ω) is te set of all continuous functions defined on Ω. C k (Ω) is te set of all continuous functions wose kt derivatives are also continuous over Ω. C (Ω) is te set of all functions u C (Ω) suc tat u can be extended to a continuous function on Ω. Te standard way of categorizing spaces of functions is to use te notion of a norm. Tis is made explicit in te following definition. Definition x.3 (Normed vector space) A normed vector space V, as (or more formally is equipped wit ) a mapping : V R wic satisfies four axioms: u u V; (were means for all ) u = u = ; (were means if and only if ) 3 αu = α u, α R and u V; 4 u+v u + v u,v V. Note tat, if te second axiom is relaxed to te weaker condition u = u = ten V is only equipped wit a semi norm. A normed vector space tat is complete (see Definition x. below) is called a Banac Space. Two examples of normed vector spaces are given below. Example x.3. Suppose tat V = R, tat is, all vectors u = Valid norms are [ ux u y ]. u = u x + u y ; l norm u = (u x +u y) / ; l norm u = max{ u x, u y }. l norm Example x.3. Suppose tat V = C (Ω). A valid norm is u,ω = max u(x). L norm. x Ω Returning to (D), te source function f(x) may well be roug, f C (Ω). An example is given below.

4 A Model Diffusion Problem 3 Problem. (f(x) = H(x /); wereh(x)iste unitstep function wic switces from to at te origin) f(x) = x = x = f(x) = Tisisamodelforte deflectionofasimplysupportedelasticbeamsubject to a discontinuous load. Solving te differential equation over te two intervals and imposing continuity of te solution and te first derivative at te interface point x = / gives te generalized solution sown in te figure on page : { x u(x) = x x < x x. We will see later tat an appropriate place to look for u is inside te space of square integrable functions. Definition x.4 (Square integrable function) L (Ω) is te vector space of square integrable functions defined on Ω : u L (Ω) if and only if u <. Functions tat are not continuous in [,] may still be square integrable. We give two examples below. Example x.4. Consider f = x /4. f dx = Ω x / dx = ence f < so tat f L (Ω). { x < Example x.4. Consider f = x. f dx = = / / f dx+ dx+ / / f dx dx /

5 A Model Diffusion Problem 4 ence f < so tat f L (Ω). L (Ω) is a Banac space. Example x.3.3 Suppose tat V = L (Ω) wit Ω = (,). A valid norm is ( / u Ω = u dx). L norm L (Ω) is pretty special it is also equipped wit an inner product. Definition x.5 (Inner product space) An inner product space V, as a mapping (, ) : V V R wic satisfies four axioms: ➊ (u,w) = (w,u) u,w V; ➋ (u,u) u V; ➌ (u,u) = u = ; ➍ (αu+βv,w) = α(u,w)+β(v,w) α,β R; u,v,w V. Example x.5. Suppose tat V = R. A valid inner product is given by (u,w) = u x w x +u y w y = u w Example x.5. Suppose tat V = L (Ω), wit Ω = (,). A valid inner product is given by (u,w) = uw. A complete inner product space like L (Ω) is called a Hilbert Space. Note tat an inner product space is also a normed space. Tere is a natural (or energy ) norm u = (u,u). Inner products and norms are related by te Caucy Scwarz inequality. Definition x.6 (Caucy Scwarz inequality) (u,v) u v u,v V. (C S)

6 A Model Diffusion Problem 5 Example x.6. Suppose tat V = R. We ave te discrete version of C S: u w u w (u x +u y) / (w x +w y) /. Example x.6. Suppose tat V = L (Ω), wit Ω = (,). We ave uw ( ) /( ) / uw u w. Returning to Problem., we now address te question of were (in wic function space) do we look for te generalized solution u wen te function f is square integrable but not continuous? Te answer to tis is in a Sobolev space. Definition x.7 (Sobolev space) For a positive index k, te Sobolev space H k (,) is te set of functions v : (,) R suc tat v and all (weak-) derivatives up to and including k are square integrable: u H k (,) u <, (u ) <,..., (u (k) ) <. Note tat H k (,) defines a Hilbert space wit inner product and norm (u,w) k = uw+ ( u k = u + u w + (u ) u w (u (k) ) ) /. u (k) w (k) Returning to Problem., te appropriate solution space turns out to be } H (,) = {u L (,), u L (,); u() =, u() =. u H (,) essential b.c. s Tis is a big surprise: Tere are no second derivatives in te definition of H! First derivatives need not be continuous!

7 A Model Diffusion Problem 6 To elp understand were H (,) comes from, it is useful to reformulate (D) as a minimization problem. Specifically, given f L (,), we look for a minimizing function u H (,) satisfying F(u) F(v), v H (,), (M) were F : H (,) R is te so-called energy functional F(v) = (v ) fv, and H (,) is te associated minimizing set. Note tat bot rigt and side terms are finite, ( ) v H (,) ( ) fv ( (v ) < )/( )/ f v f L (Ω) v H (,) using C S fv <. Note also tat te definition of (M) and te construction ( ) suggests tat even rouger load data f may be allowable. Problem.3 (f(x) = δ(x /); were δ(x) is te Dirac delta function wic is zero at all points apart from te origin) x = / Tis gives a model for te deflection of a simply supported elastic beam subject to a point load. Te solution is called a fundamental solution or a Green s function. In tis case, since v H (,) v C (,), we ave tat fv = v(/) <, so tat (M) is well defined. (Note tat tis statement is only true for domains in R in iger dimensions te Dirac delta function is not admissible as load data.)

8 A Model Diffusion Problem 7 Returning to (M), we can compute u by solving te following variational formulation : given f L (,) find u H(,) suc tat u v = fv v H (,). (V) A solution to (V) (or, equivalently a solution to (M)) is called a weak solution. Te relationsip between (D), (M) and (V) is explored in te following tree teorems. Teorem. ((D) (V)) If u solves (D) ten it solves (V). Proof. Let u satisfy (D). Since continuous functions are square integrable ten u L (,) and u L (,). Furtermore since u() = = u() from te statement of (D), we ave tat u H(,). To sow (V), let v H (,), multiply (D) by v and integrate over Ω: Using integrating by parts gives were u v = u v = fv. u v [u v], [u (x)v(x)] = u ()v() u ()v(), and since v H (,) we ave v() = v() = so tat te boundary term is zero. Tus we ave tat u satisfies as required. u v = fv v H (,) Te above proof sows us ow to construct a weak formulation from a classical formulation. We now use te properties of an inner product in Definition x.5 to sow tat (V) as a unique solution. Teorem. A solution to (V) is unique. Proof. Let V = H(,) and assume tat tere are two weak solutions u (x) V, u (x) V suc tat (u,v ) = (f,v) v V, (a,b) = (u,v ) = (f,v) v V. ab;

9 A Model Diffusion Problem 8 Subtracting and ten using ➍ gives We now define w = u u, so tat (u,v ) (u,v ) = v V, (u u,v ) = v V. (w,v )= v V. ( ) Our aim is to sow tat w = in (,) (so tat u = u.) To do tis we note tat w V (by te definition of a vector space) and set v = w in ( ). Using ➌ ten gives (w,w ) = w =. ( ) From wic we migt deduce tat w is constant. Finally, if we use te fact tat functions in V are continuous over [,], and are zero at te end points, we can see tat w = as required. Te ole in te above argument is tat two square integrable functions wic are identical in [,] except at a finite set of points are equivalent to eac oter. (Tey cannot be distinguised from eac oter in te sense of taking teir L norm.) Tus, a more precise statement of ( ) is tat w = almost everwere. Tus a more rigorous way of establising uniqueness is to use te famous Poincaré Friedric inequality. Lemma.3 (Poincaré Friedric) If w H (,) ten Proof. See below. w (w ). (P F) Tus, starting from ( ) and using P F gives w (w ) = and we deduce tat w = almost everywere in (,), so tat tere is a unique solution to (V) in te L sense. Proof. (of P F) Suppose w H (,), ten w(x) = w()+ x w (ξ)dξ.

10 A Model Diffusion Problem 9 Tus, since w() = we ave x w (x) = w dξ ( x )( x dξ ( )( dξ }{{ } = ) (w ) dξ ) (w ) dξ using C S because x. Hence w (x) (w ) dξ, and integrating over (,) gives { } { } w (w ) dξ dx = (w ) dx } {{ } R + = as required. Teorem.4 ((V) (M)) If u solves (V) ten u solves (M) and vice versa. Proof. (I) (V) (M) Let u H (,) be te solution of (V), tat is, (u,v ) = (f,v) v H (,). Suppose v H (,), and define w = v u H (,), ten using te symmetry ➊ and linearity ➍ of te inner product gives F(v) = F(u+w) = ((u+w),(u+w) ) (f,u+w) = (u,u ) (f,u)+ (u,w ) + (w,u ) (f,w)+ (w,w ) (u,w ) = (u,u ) (f,u)+ (w,w )+(u,w ) (f,w) = = F(u)+ (w,w ). Finally, using ➋, we ave tat F(v) F(u) as required.

11 A Model Diffusion Problem (II) (V) (M). Let u H (,) be te solution of (M), tat is F(u) F(v) v H (,). Tus, given a function v H (,) and ε R, u + εv H (,), so tat F(u) F(u+εv). We now define g(ε) := F(u+εv). Tis function is is minimised wen ε =, so tat we ave tat dg =. Now, ε= dε g(ε) = ((u+εv),(u+εv) ) (f,u+εv) = ε (v,v )+ε(u,v ) ε(f,v)+ (u,u ) (f,u) dg and so, = ε(v,v )+(u,v ) (f,v). dε Finally, setting dg = we see tat u solves (V). ε= dε In summary, we ave tat (D) (V) (M). Te converse implication (D) (V) is not true unless u H (,) is smoot enoug to ensure tat u C (,). In tis special case, we ave tat (D) (V). Returning to (M) and te space H (,), a very important property of L (Ω) is te concept of a weak derivative. Definition x.8 (Weak derivative) u L (Ω) possesses a weak derivative u L (Ω) satisfying (φ, u) = (φ,u) φ C (Ω), (W D) were C (Ω) is te space of infinitely differentiable functions wic are zero outside Ω. Example x.8. Consider te function u = x wit Ω = (,). x = x =

12 Galerkin Approximation Tis function is not differentiable in te classical sense. However, starting from te rigt and side of W D, and integrating by parts gives ( ) dφ dx,u = = = = x dφ dx ( x) dφ dx x dφ dx d dx ( x)φ+ d dx (x)φ [ ] [ ] ( x)φ (x)φ φ φ() φ( ) { } d dx ( x) + φ() φ() Tus te weak derivative is a step function, {, < x < u =, < x <. Note tat te value of u is not defined at te origin.. Galerkin Approximation φ { d dx (x) } = (φ, u). We now introduce a finite dimensional subspace V k H (,). Tis is associated wit a set of basis functions V k = span{φ (x),φ (x),...,φ k (x)} so tat every element of V k, say u k, can be uniquely written as u k (x) = k α j φ j (x), α j R. j= To compute te Galerkin approximation, we pose te variational problem (V) over V k. Tat is, we seek u k V k suc tat (u k,v k ) = (f,v k) v k V k. (V )

13 Galerkin Approximation Equivalently, since {φ i } k i= are a basis set, we ave tat ( (u k,φ i) = (f,φ i ) i =,,...,k k α j φ j ),φ i = (f,φ i ) j= k j= ( α j φ j,φ ) i = (f,φ i ). Tis can be written in matrix form as Ax = f (V ) wit A ij = ( φ j i),φ, xj = α j, and f i = (f,φ i ), i,j =,...,k. (V ) is called te Galerkin system, A is called te stiffness matrix, f is te load vector and u k = k j= α jφ j is te Galerkin solution. Teorem. Te stiffness matrix is symmetric and positive definite. Proof. Symmetry follows from ➊. To establis positive definiteness we consider te quadratic form and use ➍: x T Ax = k k j= i= α ja ji α i = k k j= i= α ( j φ j,φ i ( k ) = j= α jφ j, k i= α iφ i = (u k,u k ). Tus from ➋ we see tat A is at least semi-definite. Definiteness follows from te fact tat x T Ax = if and only if u k =. But since u k H (,) ten u k = implies tat u k =. Finally, since {φ i } k i= are a basis set, we ave tat u k = implies tat x =. Teorem. implies tat A is nonsingular. Tis means tat te solution x (and ence u k ) exists and is unique. An alternative approac, te so-called Rayleig Ritz metod, is obtained by posing te minimization problem (M) over te finite dimensional subspace V k. Tat is, we seek u k V k suc tat ) αi F(u k ) F(v k ) v k V k. (M ) Doing tis leads to te matrix system (V ) so tat te Ritz solution and te Galerkin solution are one and te same. Te beauty of Galerkin s metod is te best approximation property.

14 Galerkin Approximation 3 Teorem. (Best approximation) If u is te solution of (V) and u k is te Galerkin solution, ten u u k u v k v k V k. (B A) Proof. Te functions u k and u satisfy te following u k V k ; (u k,v k) = (f,v k ) v k V k But since V k V we ave tat u V; (u,v ) = (f,v) v V. (u,v k ) = (f,v k) v k V k. Subtracting equations and using ➍ gives (u,v k) (u k,v k) = v k V k (u u k,v k ) = v k V k (G O) Tis means tat te error u u k is ortogonal to te subspace V k a property known as Galerkin ortogonality. To establis te best approximation property we start wit te left and side of B A and use Galerkin ortogonality as follows: u u k = (u u k,u u k ( ) ) = (u u k,u ) u u k,u k u k V k } {{ } { = }} G O { = (u u k,u ) (u u k,v k) v k V k = (u u k,u v k) u u k u v k. using C S Hence, dividing by u u k > as required. u u k u v k v k V k, An important observation ere is tat we ave a natural norm to measure errors wic is inerited from te minimimization problem (M). If u u k = ten B A olds trivially.

15 Galerkin Approximation 4 Example x.3.4 Suppose V = H (,). A valid norm is v E = (v,v ) / = v. Tis is called te energy norm. A tecnical issue tat arises ere is tat te best approximation property does not automatically imply tat te Galerkin metod converges in te sense tat u u k E as k. For convergence, we really need to introduce te concept of a complete space tat was postponed earlier. Definition x.9 (Caucy sequence) A sequence (v (k) ) V is called a Caucy sequence in a normed space V if for any ε >, tere exists a positive integer k (ε) suc tat v (l) v (m) V < ε l,m k. A Caucy sequence is convergent, so te only issue is weter te limit of te sequence is in te correct space. Tis motivates te following definition. Definition x. (Complete space) AnormedspaceV iscompleteifitcontainstelimitsofallcaucysequences in V. Tat is, if (v (k) ) is a Caucy sequence in V, ten tere exists ξ V suc tat lim k v(k) ξ V =. We write tis informally as lim k v (k) = ξ. Example x.. Te space H(,) is complete wit respect to te energy norm E. Te upsot is tat te Galerkin approximation is guaranteed to converge to te weak solution in te limit k. We now introduce a simple-minded Galerkin approximation based on global polynomials. Tat is, we coose V k = span {,x,x,...,x k }. To ensure tat V k V, te function u k = j α jx j must satisfy two conditions: (I) u k H (,); (II) u k () = = u k ().

16 Galerkin Approximation 5 Te first condition is no problem, u k C (,)! To satisfy te second condition we need to modify te basis set to te following, V k = span { x(x ),x (x ),x 3 (x )...,x k (x ) }. Problem. (f = ) Consider V = span{ x(x ),x } (x ) φ Ten constructing te matrix system (V ), we ave tat Tis gives ( 3 A = A = A = 6 f = f = 6 5 )( α α (φ ) = φ φ = (φ ) = φ = φ = ) So te Galerkin solution is = φ ( d dx (x x) ) = 3 d dx (x x) d dx (x3 x ) = 6 = A ( d dx (x3 x ) ) = 5 x x = 6 x 3 x =. ( 6 ) ( α α u (x) = φ +φ = (x x ) = u(x). ) = ( Te fact tat te Galerkin approximation agrees wit te exact solution is to be expected given te best approximation property and noting tat u V. Problem. (f(x) = H(x /)) Consider V as above. In tis case, ). f = f = φ = φ = x 3 x = 5 9.

17 Finite Element Galerkin Approximation 6 Tis gives te Galerkin system ( )( α α ) = ( 5 9 Note tat te Galerkin solution is not exact in tis instance. Te big problem wit global polynomial approximation is tat te Galerkin matrix becomes increasingly ill conditioned as k is increased. Computationally, it beaves like a Hilbert matrix and so reliable computation for k > is not possible. For tis reason, piecewise polynomial basis functions are used in practice instead of global polynomial functions. 3. Finite Element Galerkin Approximation A piecewise polynomial approximation space can be constructed in four steps. Step(i) Subdivision of Ω into elements. For Ω = [,] te elements are intervals as illustrated below. n x = x = Step (ii) Piecewise approximation of u using a low-order polynomial (e.g. linear): u u e u n x x In general, a linear function u e (x) is defined by its values at two distinct points x x l e (x) = u e (x) = (x x ) (x x ) l e (x) ) u e (x )+ (x x ) (x x ) l e (x). u e (x ). l i (x) are called nodal basis functions and satisfy te interpolation conditions linear over e if x = x if x = x ; l e (x) = linear over e if x = x if x = x.

18 Finite Element Galerkin Approximation 7 Step (iii) Satisfaction of te smootness requirement (I). Tis is done by carefully positioning te nodes, so tat x and x are at te end points of te interval. u k (x ) u k (x ) u k (x 3 ) u k (x n ) u k (x ) u k (x n ) x = x = 3 n x x x x 3 x n x n Concatenatingte element functions u e (x) givesaglobalfunction u k (x) u k (x) = n u e (x). e = Tusu k (x) is defined byk = n internalvalues, andte twoboundary values u k (x ) = u k () and u k (x n ) = u k (). We can ten define N i (x), te so called global basis function, so tat N i (x) = if x = x i if x = x j (j i) linear in every element e, N i (x) x i 3 x i x i x i x i+ x i+ and write u k (x) = α N (x)+α N (x)+...+α n N n (x). Note tat u k (x i ) = α i, so tat te unknowns are te function values at te nodes.

19 Finite Element Galerkin Approximation 8 Step (iv) Satisfaction of te essential boundary condition requirement (II). Tis is easy we simply remove te basis functions N (x) and N n (x) from te basis set. Te modified Galerkin approximation is u k (x) = α N (x)+...+α n N n (x), and is associated wit te approximation space Problem 3. (f(x) = ) Consider five equal lengt elements V n = span{n,...,n n } x = x = 5 x = 5 x 3 = 3 5 x 4 = 4 x 5 = 5 so tat V 6 = span{n i } 5 i=. For computational convenience te Galerkin system coefficients A ij = N j N i, b i = may be computed element by element. Tat is, A ij = 5 k= xk x k N j N i xk dls dl t x k dx dx, b i = 5 k= N i, xk x k N i xk x k l t were s and t are local indices referring to te associated element basis functions illustrated below. l (x) N k (x) x k k x k x k+,

20 Finite Element Galerkin Approximation 9 N k (x) l (x) x k x k k x k In particular, in element k tere are two nodal basis functions so tat l (x) = (x x k ) (x k x k ), l (x) = (x x k) (x k x k ), dl dx =, dl dx =, wit = x k x k = /5. Tis generates a element contribution matrix xk xk (l ) l l A k x k x k = xk xk l l (l = ) x k x k and a ( ) element contribution vector b k = xk x k l xk x k l =.

21 Finite Element Galerkin Approximation Summing over te element contributions gives te stiffness matrix and te load vector, A = + + A A... + A 5 b =

22 Finite Element Galerkin Approximation Tus, te assembled Galerkin system is A x b α α α α 3 α 4 α 5 =. Note tat tis system is singular (and inconsistent!). All te columns sum to zero, so tat A =. Tis problem arises because we ave not yet imposed te essentialboundary conditions. To do tis we simply need to removen (x) and N 5 (x) from te basis set, tat is, delete te first and last row and column from te system. Doing tis gives te nonsingular reduced system α α =. α 3 α 4 A x b Setting = /5, and solving gives α = α 4 =.8; α = α 3 =.; so te Galerkin finite element solution is u 4(x) =.8N (x)+.n (x)+.n 3 (x)+.8n 4 (x). It is illustrated below. Note tat u 4(x i ) = u(x i ) so te finite element solution is exact at te nodes! It is not exact at any point in between te nodes owever.

23 Finite Element Galerkin Approximation 8 x x x 3 x 4 Note also tat te generic equation x i x i x i+ = corresponds to a centered finite difference approximation to u =. To complete te discussion we would like to sow tat te finite element solution converges to te weak solution in te limit. Teorem 3. (Convergence in te energy norm) If u is te solution of (V) and u k is te finite element solution based on linear approximation ten u u k E f were is te lengt of te longest element in te subdivision (wic does not ave to be uniform.) Proof. We now formally introduce te linear interpolant, u V k of te exact solution, so tat u(x i ) = u (x i ) i =,,,...,n. Note tat we cannot assume tat u k = u in general. Introducing e(x) = u(x) u (x), we see tat e V k and tat e(x i ) =, i =,,,...,n.

24 Finite Element Galerkin Approximation 3 We can now bound te element interpolation error using standard tools from approximation teory xi xi (e ) (x i x i ) (e ) x i x i = (x i x i ) xi xi x i (u ) x i (u ) were = max i x i x i. Summingoverteintervalstengivesteestimate Finally, using B A (e ) (u ). (u u k ) (u u ) u = f <. Tus lim u k = u in te energy norm. To get an error estimate in L we use a very clever duality argument. Teorem 3. (Aubin Nitsce) If u is te solution of (V) and u k is te finite element solution based on linear approximation ten u u k f. Proof. Let w be te solution of te dual problem Ten we ave w = u u k x (,); w() = = w(). u u k = (u u k,u u k ) = (u u k, w ) = (u u k,w ) (since w() = w() = ) = (u u k,w ) (u u k,(w ) ), w V k, G O were w is te interpolant of w in V k. Hence u u k = (u u k,(w w ) ) u u k (w w ) C S w = u u k.

25 Finite Element Galerkin Approximation 4 Hence, assuming tat u u k >, we ave tat as required. u u k u u k f, A more complete discussion of tese issues can be found in Capters and 4 of te following reference book. Endre Süli & David Mayers, An Introduction to Numerical Analysis, Cambridge University Press, 3.