Chapter 10. Function approximation Function approximation. The Lebesgue space L 2 (I)

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1 Capter 1 Function approximation We ave studied metods for computing solutions to algebraic equations in te form of real numbers or finite dimensional vectors of real numbers. In contrast, solutions to di erential equations are scalar or vector valued functions, wic only in simple special cases are analytical functions tat can be expressed by a closed matematical formula. Instead we use te idea to approximate general functions by linear combinations of a finite set of simple analytical functions, for example trigonometric functions, splines or polynomials, for wic attractive features are ortogonality and locality. We focus in particular on piecewise polynomials defined by te finite set of nodes of a mes, wic exibit bot near ortogonality and local support. 1.1 Function approximation Te Lebesgue space L 2 (I) Inner product spaces provide tools for approximation based on ortogonal projections on subspaces. We now introduce an inner product space for functions on te interval I =[a, b], te Lebesgue space L 2 (I), defined as te class of all square integrable functions f : I! R, L 2 (I) ={f : Z b a f(x) 2 dx < 1}. (1.1) Te vector space L 2 (I) isclosedundertebasicoperationsofpointwise addition and scalar multiplication, by te inequality, wic follows from Young s inequality. (a + b) 2 apple 2(a 2 + b 2 ), 8a, b, (1.2) 97

2 98 CHAPTER 1. FUNCTION APPROXIMATION Teorem 17 (Young s inequality). For a, b and >, Proof. apple (a b) 2 = a b 2 2ab. Te L 2 -inner product is defined by (f,g) =(f,g) L 2 (I) = ab apple 1 2 a2 + 2 b2 (1.3) Z b a f(x)g(x) dx, (1.4) wit te associated L 2 norm, Z b 1/2 kfk = kfk L 2 (I) =(f,f) 1/2 = f(x) dx 2, (1.5) for wic te Caucy-Scwarz inequality is satisfied, Approximation of functions in L 2 (I) a (f,g) applekfkkgk. (1.6) We seek to approximate a function f in a vector space V, by a linear combination of functions { j } n j=1 V,tatis f(x) f n (x) = j j (x), (1.7) wit j 2 R. Iflinearlyindependent,teset{ j } n j=1 spans a finite dimensional subspace S V, S = {f n 2 V : f n = j j (x), j 2 R}, (1.8) j=1 wit te set { j } n j=1 abasisfors. Forexample,inaFourier series te basis functions j are trigonometric functions, in a power series monomials. Te question is now ow to determine te coordinates j so tat f n (x) is a good approximation of f(x). One approac to te problem is to use te tecniques of ortogonal projections previously studied for vectors in R n, an alternative approac is interpolation, were j are cosen suc tat f n (x i )=f(x i ), in a set of nodes x i,fori =1,...,n. If we cannot evaluate te function f(x) inarbitrarypointsx, butonlyaveaccesstoasetof sampled data points {(x i,f i )} m i=1, witm n, we can formulate a least squares problem to determine te coordinates j tat minimize te error f(x i ) f i,inasuitablenorm. j=1

3 1.1. FUNCTION APPROXIMATION 99 L 2 projection Te L 2 projection Pf,ontotesubspaceS V,definedby(1.8),ofa function f 2 V,witV = L 2 (I), is te ortogonal projection of f on S, tat is, wic corresponds to, (f Pf,s)=, 8s 2 S, (1.9) j ( i, j) =(f, i), 8i =1,...,n. (1.1) j=1 By solving te matrix equation Ax = b, wita ij =( i, j), x j = j, and b i =(f, i), we obtain te L 2 projection as Pf(x) = j j (x). (1.11) j=1 We note tat if i(x) as local support, tat is i(x) 6= only for a subinterval of I, tentematrixa is sparse, and for { i } n i=1 an ortonormal basis, A is te identity matrix wit j =(f, j). Interpolation Te interpolant f 2 S, isdeterminedbyteconditiontat f(x i )=f(x i ), for n nodes {x i } n i=1. Tatis, f(x i )= f(x i )= j j (x i ), i =1,...,n, (1.12) j=1 wic corresponds to te matrix equation Ax = b, wita ij = j (x i ), x j = j,andb i = f(x i ). Te matrix A is an identity matrix under te condition tat j(x i )=1, for i = j, andzeroelse.wetenreferto{ i } n i=1 as a nodal basis, forwic j = f(x j ), and we can express te interpolant as f(x) = j j (x) = j=1 f(x j ) j (x). (1.13) j=1

4 1 CHAPTER 1. FUNCTION APPROXIMATION Regression If we cannot evaluate te function f(x) inarbitrarypoints,butonlyave access to a set of data points {(x i,f i )} m i=1, witm n, wecanformulate te least squares problem, min kf i f n (x i )k = min kf i f n2s { j } n j=1 j j (x i )k, i =1,...,m, (1.14) j=1 wic corresponds to minimization of te residual b Ax, wita ij = j (x i ), b i = f i,andx j = j,wicwecansolve,forexample,byformingtenormal equations, A T Ax = A T b. (1.15) 1.2 Piecewise polynomial approximation Polynomial spaces We introduce te vector space P q (I), defined by te set of polynomials p(x) = qx c i x i, x 2 I, (1.16) i= of at most order q on an interval I 2 R, wittebasisfunctionsx i and coordinates c i,andtebasicoperationsofpointwiseadditionandscalar multiplication, (p + r)(x) =p(x)+r(x), ( p)(x) = p(x), (1.17) for p, r 2P q (I) and 2 R. One basis for P q (I) istesetofmonomials {x i } q i=,anoteris{(x c)i } q i=,wicgivestepower series, p(x) = qx a i (x c) i = a + a 1 (x c) a q (x c) q, (1.18) i= for c 2 I, wit a Taylor series being an example of a power series, f(x) =f(y)+f (y)(x y)+ 1 2 f (y)(x y) (1.19)

5 1.2. PIECEWISE POLYNOMIAL APPROXIMATION 11 Langrange polynomials For a set of nodes {x i } q i=, we define te Lagrange polynomials { }q i=,by i(x) = (x x ) (x x i 1 )(x x i+1 ) (x x q ) (x i x ) (x i x i 1 )(x i x i+1 ) (x i x q ) = Y i6=j tat constitutes a basis for P q (I), and we note tat x x j, x i x j i(x j )= ij, (1.2) wit te Dirac delta function defined as ( 1, i = j ij =, i 6= j (1.21) so tat { } q i= is a nodal basis, wic we refer to as te Lagrange basis. We can express any p 2P q (I) as p(x) = qx p(x i ) i (x), (1.22) i=1 and by (1.13) we can define te polynomial interpolant q f 2P q (I), q f(x) = qx f(x i ) i (x), x 2 I, (1.23) i=1 for a continuous function f 2C(I). Piecewise polynomial spaces We now introduce piecewise polynomials defined over a partition of te interval I =[a, b], a = x <x 1 < <x m+1 = b, (1.24) for wic we let te mes T = {I i } denote te set of subintervals I j = (x i 1,x i )oflengt i = x i x i 1,wittemes function, (x) = i, for x 2 I i. (1.25) We define two vector spaces of piecewise polynomials, te discontinuous piecewise polynomials on I, definedby W (q) = {v : v Ii 2P q (I i ),i=1,...,m+1}, (1.26)

6 12 CHAPTER 1. FUNCTION APPROXIMATION and te continuous piecewise polynomials on I, definedby V (q) Te basis functions for W (q) basis, for example, i,(x) = = {v 2 W (q) : v 2C(I)}. (1.27) can be defined in terms of te Lagrange x i x = x i x x i x i 1 i (1.28) i,1(x) = x x i 1 x i x i 1 = x x i 1 i (1.29) defining te basis functions for W (1),by (, x 6= [x i 1,x i ], i,j(x) = i,j, x 2 [x i 1,x i ], (1.3) for i =1,...,m+1,andj =, 1. For V (q) we need to construct continuous basis functions, for example, 8 ><, x 6= [x i 1,x i+1 ], i(x) = i,1, x 2 [x i 1,x i ], (1.31) >: i+1,, x 2 [x i,x i+1 ], for V (1), wic we also refer to as at functions. φ i,1 (x) φ i (x) x =a x 1 x 2 x i-1 x i x i+1 x m x x m+1 =b x =a x 1 x 2 x i-1 x i x i+1 x m x x m+1 =b Figure 1.1: Illustration of a mes T = {I i }, wit subintervals I j = (x i 1,x i )oflengt i = x i x i 1, and i,1 (x) abasisfunctionforw (1) (left), and a basis function i (x) forv (1) (rigt).

7 1.2. PIECEWISE POLYNOMIAL APPROXIMATION 13 L 2 projection in V (1) Te L 2 projection of a function f 2 L 2 (I) ontotespaceofcontinuous piecewise linear polynomials V (1),onasubdivisionofteintervalI wit n nodes, is given by Pf(x) = j j (x), (1.32) wit te coordinates j determined by from te matrix equation j=1 Mx = b, (1.33) wit m ij =( j, i), x j = j,andb i =(f, i). Te matrix M is sparse, since m ij =for i j > 1, and for large n we need to use an iterative metod to solve (1.33). We compute te entries of te matrix M, referredtoas a mass matrix, fromtedefinitionoftebasisfunctions(1.31),starting wit te diagonal entries, Z xi Z xi m ii = ( i, i) = i (x) dx = i,1 (x) dx + i+1, (x) dx x i 1 x i Z xi (x x i 1) 2 Z xi+1 (x i+1 x) 2 = dx + dx x i 1 2 i x i 2 i+1 = 1 apple (x xi 1 ) 3 x i + 1 apple (xi+1 x) 3 x i+1 = i 2 i 3 x i 1 2 i i+1 3, and similarly we compute te o -diagonal entries, m ii+1 = ( i, i+1) = and = = = = = Z xi+1 i(x) i+1 (x) dx = Z xi+1 (x i+1 x) (x x i ) dx x i i+1 i+1 Z 1 xi+1 (x i+1 x x i+1 x i x 2 + xx i ) dx x apple i 1 xi+1 x 2 x 3 x i+1 x i x x2 x x i+1 i 2 x i 1 (x 3 i+1 3x 2 i+1x i +3x i+1 x 2 i x 3 i ) 2 i+1 2 i i i+1 (x i+1 x i ) 3 = i+1 6, m ii 1 =( i, i 1) = i(x) i x i x i 1 (x) dx =... = i 6. i+1,(x) i+1,1 (x) dx

8 14 CHAPTER 1. FUNCTION APPROXIMATION 1.3 Exercises Problem 34. Prove tat te sum of two functions f,g 2 L 2 (I) is a function in L 2 (I). Problem 35. Prove tat (1.1) follows from (1.9). Problem 36. Prove tat te L 2 projection Pf of a function f 2 L 2 (I), onto te subspace S L 2 (I), is te best approximation in S, in te sense tat kf Pfkapplekf sk, 8s 2 S. (1.34)

9 Capter 11 Boundary value problems in R Te boundary value problem in one variable is an ordinary di erential equation, for wic an initial condition is not enoug, instead we need to specify boundary conditions at eac end of te interval. Contrary to te initial value problem, te dependent variable does not represent time, but sould rater we tougt of as a spatial coordinate Te boundary value problem Te boundary value problem We consider te following boundary value problem, for wic we seek a function u(x) 2C 2 (, 1), suc tat u (x) =f(x), x 2 (, 1), (11.1) u() = u(1) =, (11.2) given a source term f(x), and boundary conditions at te endpoints of te interval I =[, 1]. We want to find an approximate solution to te boundary value problem in te form of a continuous piecewise polynomial tat satisfies te boundary conditions (11.2), tat is we seek U 2 V = {v 2 V (q) : v() = v(1) = }, (11.3) suc tat te error e = u U is small in some suitable norm k k. Te residual of te equation is defined as R(w) =w + f, (11.4) 15

10 16 CHAPTER 11. BOUNDARY VALUE PROBLEMS IN R wit R(u) =,foru = u(x) tesolutionofteboundaryvalueproblem. Our strategy is now to find an approximate solution U 2 V,suctat R(U) issmall. We ave two natural metods to find a solution U 2 V wit a minimal residual: (i) te least squares metod, wereweseektesolutionwitte minimal residual measured in te L 2 -norm, min kr(u)k, (11.5) U2V and (ii) Galerkin s metod, were we seek te solution for wic te residual is ortogonal te subspace V, (R(U),v)=, 8v 2 V. (11.6) Wit an approximation space consisting of piecewise polynomials, we refer to te metods as a least squares finite element metod, andagalerkin finite element metod. Wit a trigonometric approximation space we refer to Galerkin s metod as a spectral metod. Galerkin finite element metod Te finite element metod (FEM) based on (11.6) takes te form: find U 2 V, suc tat U (x)v(x) dx = f(x)v(x) dx, (11.7) for all test functions v 2 V. For (11.8) to be well defined, we need to be able to represent te second order derivative U,wicisnotobviousfor low order polynomials, suc as linear polynomials, or piecewise constants. To reduce tis constraint, we can use partial integration to move one derivative from te approximation U to te test function v, sotat U (x)v(x) dx = U (x)v (x) dx [U (x)v(x)] 1 = U (x)v (x) dx, since v 2 V,andtussatisfiesteboundaryconditions. Te Galerkin finite element metod now reads: find U 2 V,suctat, for all v 2 V. U (x)v (x) dx = f(x)v(x) dx, (11.8)

11 11.1. THE BOUNDARY VALUE PROBLEM 17 Te discrete problem We now let V be te space of continuous piecewise linear functions, tat satisfies te boundary conditions (11.2), tat is, U 2 V = {v 2 V (1) : v() = v(1) = }, (11.9) so tat we can write any function v 2 V as v(x) = v i i (x), (11.1) i=1 over a mes T wit n internal nodes x i,andv i = v(x i )since{ i } n i=1 is a nodal basis. We tus searc for an approximate solution U(x) = U j j (x), (11.11) j=1 wit U j = U(x j ). If we insert (11.1) and (11.11) into (11.8), we get U j j(x) i(x) dx = f(x) i (x) dx, i =1,...,n, (11.12) j=1 wic corresponds to te matrix equation Sx = b, (11.13) wit s ij =( j, i ), x j = U j,andb i =(f, i). Te matrix S is sparse, since s ij =for i j > 1, and for large n we need to use an iterative metod to solve (11.13). We compute te entries of te matrix S, referredtoasasti ness matrix, from te definition of te basis functions (1.31), starting wit te diagonal entries, s ii = ( i, = Z xi x i 1 Z xi Z xi+1 i) = ( i) 2 (x) dx = ( i,1) 2 (x) dx + x i 1 x i 2 xi dx + dx = 1 + 1, i i+1 i x i i+1 and similarly we compute te o -diagonal entries, s ii+1 = ( i, and i+1) = s ii 1 =( i, i(x) i+1(x) dx = i 1) = Z xi+1 x i 1 i+1 1 i+1 dx = i(x) i 1(x) dx =... = 1 i. ( i+1,) 2 (x) dx 1 i+1,

12 18 CHAPTER 11. BOUNDARY VALUE PROBLEMS IN R Te variational problem Galerkin s metod is based on te variational formulation, orweak form, of te boundary value problem, were we searc for solution in a vector space V,forwictevariationalformiswelldefined:findu 2 V,suctat u (x)v (x) dx = f(x)v(x) dx, (11.14) for all v 2 V. To construct an appropriate vector space V for (11.14) to be well defined, we need to extend L 2 spaces to include also derivatives, wic we refer to as Sobolev spaces. We introduce te vector space H 1 (, 1), defined by, H 1 (, 1) = {v 2 L 2 (, 1) : v 2 L 2 (, 1)}, (11.15) and te vector space tat also satisfies te boundary conditions (11.2), H 1 (, 1) = {v 2 H 1 (, 1) : v() = v(1) = }. (11.16) Te variational form (11.14) is now well defined for V = H 1 (, 1), since by Caucy-Scwarz inequality, and for f 2 L 2 (, 1). Optimality of Galerkin s metod u (x)v (x) dx appleku kkv k < 1 (11.17) f(x)v(x) dx applekfkkvk < 1, (11.18) Galerkin s metod (11.8) corresponds to searcing for an approximate solution in a finite dimensional subspace V V,forwic(11.1)issatisfied for all test functions v 2 V. Te Galerkin solution U is te best possible approximation in V,inte sense tat, ku Uk E appleku vk E, 8v 2 V, (11.19) wit te energy norm defined by 1/2 kwk E = w (x) dx 2. (11.2)

13 11.2. EXERCISES 19 Tus U 2 V represents a projection of u 2 V onto V,witrespectto te inner product defined on V, (v, w) E = v (x)w (x) dx, (11.21) wit kwk 2 E =(w, w) E. By subtracting (11.8) from (11.14), we expose te Galerkin ortogonality property, (u U, v) E =, 8v 2 V, (11.22) wic we can use to express te optimality of te approximation U. Teorem 18. Te Galerkin solution U 2 V is te optimal approximation of u wit respect to te energy norm, tat is Proof. For any v 2 V, ku Uk E appleku vk E, 8v 2 V. (11.23) ku Uk 2 E = (u U, u u ) E =(u U, u v) E +(u U, v u ) E = (u U, u v) E appleku Uk E ku vk E Exercises Problem 37. Derive te variational formulation and te finite element metod for te boundary value problem wit a(x) >, and c(x). (a(x)u (x)) + c(x)u(x) =f(x), x 2 (, 1), (11.24) u() = u(1) =, (11.25)

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15 Capter 12 Boundary value problems in R n 12.1 Di erential operators in R n Te gradient and Jacobian We recall te definition of te gradient of a scalar function f : R n! R, as grad f = T i in vector notation and index notation, respectively, wic we can interpret as te di erential operator @x n i, (12.2) acting on te function f. Te directional derivative r v f,intedirection of te vector v : R n! R n,isdefinedas r v f =(v r)f = j. (12.3) For a vector valued function F : R n! R m, we define te Jacobian J, n (rf 1 ) T J = F 6 = rf = = 4. 5 i, m n (rf m ) T j wit r v F =(v r)f = Jv = v j. (12.5) 111

16 112 CHAPTER 12. BOUNDARY VALUE PROBLEMS IN R n Divergence and rotation For F : R n! R n we define te divergence, and, for n =3,terotation, div F = r F n i, (12.6) rot F =curlf = r F, (12.7) were 2 3 e 1 e 2 e 3 r @ F 1 F 2 F wit e =(e 1,e 2,e 3 )testandardbasisinr 3. Te divergence can be understood in terms of te Gauss teorem, Z Z r Fdx= F nds, (12.8) wic relates te volume integral over a domain R 3,wittesurface integral over te boundary wit normal n. Similarly, te rotation can be interpreted in terms of te Kelvin-Stokes teorem, Z Z r F ds = F dr, (12.9) wic relates te surface integral of te rotation over a surface to te line integral over its witpositiveorientationdefinedbydr. Laplacian and Hessian We express te Laplacian f as, f = r 2 f = r T rf = r rf 2 1 and te Hessian H, 2 1 H = f = rr T 6 f = 2 n@x n@x n 3 f, 2 i 7 5 f. j

17 12.2. FUNCTION SPACES 113 Te vector Laplacian is defined as te and for m = n =3,weave Partial integration in R n F = r 2 F =( F 1,..., F n ), (12.12) F = r(r F ) r (r F ). (12.13) For a scalar function f : R n! R, and a vector valued function F : R n! R n,weavetefollowinggeneralizationofpartialintegrationoveradomain R n,referredtoasgreen s teorem, (rf,f) L 2 ( ) = (f,r F ) L 2 ( ) +(f,f n) L 2 ( ), (12.14) wit boundary and outward unit normal vector n = n(x) forx 2, were we use te notation, Z (v, w) L 2 ( ) = v wdx, (12.15) for two vector valued functions v, w, and Z (v, w) L 2 ( ) = v wds, (12.16) for te boundary integral. For two scalar valued functions te scalar product in te integrand is replaced by te usual multiplication. Wit F = rg, for g : R n! R a scalar function, Green s teorem gives, (rf,rg) L 2 ( ) = (f, g) L 2 ( ) +(f,rg n) L 2 ( ). (12.17) 12.2 Function spaces Te Lebesgue spaces L p ( ) Let be a domain in R n and let p be a positive real number, ten we define te Lebesgue space L p ( ) by L p ( ) = {f : kfk p < 1}, (12.18) wit te L p ( ) norm, Z 1/p kfk p = f(x) dx p, (12.19)

18 114 CHAPTER 12. BOUNDARY VALUE PROBLEMS IN R n were in te case of a vector valued function f : R n! R n, f(x) p = f 1 (x) p f n (x) p, (12.2) or a matrix valued function f : R n! R n n, f(x) p = f ij (x) p. (12.21) i,j=1 L p ( ) is a vector space, since (i) f 2 L p ( ) for any 2 R, and(ii) f + g 2 L p ( ) for f,g 2 L p ( ), by te inequality, (a + b) p apple 2 p 1 (a p + b p ), a, b, 1 apple p<1, (12.22) wic follows from te convexity of te function t 7! t p. L p ( ) is a Banac space, and L 2 ( ) is a Hilbert space wit te inner product (12.15) wic induces te L 2 ( )-norm. In te following we let it be implicitly understood tat (, ) =(, ) L 2 and k k= k k L 2 ( ). Sobolev spaces To construct appropriate vector spaces for variational formulations of partial di erential equations, we need to extend te spaces L 2 ( ) to include also derivatives. Te Sobolev space H 1 ( ) is defined by, and we define H 1 ( ) = {v 2 L 2 ( ) : rv 2 L 2 ( )}, (12.23) H 1 ( ) = {v 2 H 1 ( ) : v(x) =, 8x 2 }, (12.24) to be te space of functions in H 1 ( ) tat are zero on te boundary FEM for Poisson s equation Te Poisson equation We now consider te Poisson equation for an unknown function u : R n! R, u = f, x 2, (12.25) wit R n,andgivendataf : R n! R. For te equation to ave a unique solution we need to specify boundary conditions. We may prescribe Diriclet boundary conditions, u = g D, x 2, (12.26)

19 12.3. FEM FOR POISSON S EQUATION 115 Neumann boundary conditions, ru n = g N, x 2, (12.27) wit n = n(x) teoutwardunitnormalon N,oralinearcombinationof te two, wic we refer to as a Robin boundary condition. Homogeneous Diriclet boundary conditions We now state te variational formulation of Poisson equation wit omogeneous Diriclet boundary conditions, u = f, x 2, (12.28) u =, x 2, (12.29) wic we obtain by multiplication wit a test function v 2 V = H 1 ( ) and integration over. Using Green s teorem, we obtain te variational formulation: find u 2 V,suctat (ru, rv) =(f,v), (12.3) for all v 2 V,sinceteboundarytermvanisesastetestfunctionisan element of te vector space H 1 ( ). Homogeneous Neumann boundary conditions We now state te variational formulation of Poisson equation wit omogeneous Neumann boundary conditions, u = f, x 2, (12.31) ru n =, x 2, (12.32) wic we obtain by multiplication wit a test function v 2 V = H 1 ( ) and integration over. Using Green s teorem, we get te variational formulation: find u 2 V,suctat, (ru, rv) =(f,v), (12.33) for all v 2 V, since te boundary term vanises by te Neumann boundary condition. Tus te variational forms (12.3) and (12.33) are similar, wit te only di erence being te coice of test and trial spaces. However, it turns out tat te variational problem (12.33) as no unique solution, since for any solution u 2 V,alsov + C is a solution, wit C 2 R

20 116 CHAPTER 12. BOUNDARY VALUE PROBLEMS IN R n aconstant. Toensureauniquesolution,weneedanextraconditionfor te solution wic determines te arbitrary constant, for example, we may cange te approximation space to Z V = {v 2 H 1 ( ) : v(x) dx =}. (12.34) Non omogeneous boundary conditions Poisson equation wit non omogeneous boundary conditions takes te form, u = f, x 2, (12.35) u(x) =g D, x 2 D, (12.36) ru n = g N, x 2 N, (12.37) wit = D [ N. We obtain te variational formulation by by multiplication wit a test function v 2 V,wit V w = {v 2 H 1 ( ) : v(x) =w(x), x2 D }, (12.38) and integration over. Using Green s teorem, we get te variational formulation: find u 2 V gd, suc tat, (ru, rv) =(f,v)+(g N,v) L 2 ( N ). (12.39) for all v 2 V. Te Diriclet boundary condition is enforced troug te trial space, and is referred to as an essential boundary condition, wereas te Neumann boundary condition is enforced troug te variational form, referred to as a natural boundary condition. Galerkin finite element metod To compute approximate solutions to te Poisson equation, we can formulate a Galerkin metod based on te variational formulation of te equation, replacing te Sobolev space V wit a finite dimensional space V, constructed by a set of basis functions { i } M i=1, overamest,definedas acollectionofelements{k i } N i=1 and nodes {N i } M i=1. For te Poisson equation wit omogeneous Diriclet boundary conditions, te Galerkin element metod takes te form: Find U 2 V,suc tat, (ru, rv) =(f,v), v 2 V, (12.4)

21 12.4. LINEAR PARTIAL DIFFERENTIAL EQUATIONS 117 wit V H( ). 1 Te variational form (12.4) corresponds to a linear system of equations Ax = b, wita ij =( j, i), x j = U(N j ), and b i =(f, i), wit i (x) te basis function associated wit te node N i. For V apiecewisepolynomialspace,wereferto(12.4)asagalerkin finite element metod Linear partial di erential equations Te abstract problem We can express a general linear partial di erential equation as te abstract problem, A(u) =f, x 2, (12.41) wit boundary conditions, B(u) =g, x 2. (12.42) For a Hilbert space V,wecanderivetevariationalformulation: find u 2 V suc tat, a(u, v) =L(v), v 2 V, (12.43) wit a : V V! R a bilinear form, tatisafunctionwicislinearin bot arguments, and L : V! R a linear form, orlinear functional, wic is a linear function onto te scalar field R. Teorem 19 (Riesz representation teorem). For every linear functional L : V! R on te Hilbert space V, wit inner product (, ), tere exists a unique element u 2 V, suc tat Existence and uniqueness L(v) =(u, v), 8v 2 V. (12.44) We can prove te existence of unique solutions to te variational problem (12.43), under certain conditions. Assume te bilinear form a(, ) is symmetric, a(v, w) =a(w, v), 8v, w 2 V, (12.45) and coercive, orv-elliptic, a(v, v) c kvk V, v 2 V, (12.46)

22 118 CHAPTER 12. BOUNDARY VALUE PROBLEMS IN R n wit c >, and k k V te norm on V. A symmetric and elliptic bilinear form defines an inner product on V,wicinducesanormwicwerefer to as te energy norm, kwk E = a(w, w) 1/2. (12.47) But if te bilinear form is an inner product on V,byRieszrepresentation teorem tere exists a unique u 2 V, suc tat a(u, v) =L(v). (12.48) If te bilinear form is not symmetric, we still ave unique solution to (12.43) by te Lax-Milgram teorem, iftebilinearformiselliptic(12.46), and continuous, a(u, v) apple C 1 kuk V kvk V, C 1 >, (12.49) wit also te linear form continuous, A priori error estimation L(v) apple C 2 kvk V, C 2 >. (12.5) In a Galerkin metod we seek an approximation U 2 V suc tat a(u, v) =L(v), v 2 V, (12.51) wit V V afinitedimensionalsubspace,wicintecaseofafinite element metod is a piecewise polynomial space. Te Galerkin approximation is optimal in te energy norm, since by Galerkin ortogonality, we ave tat a(u U, v) =, v 2 V, (12.52) ku Uk 2 E = a(u U, u u )=a(u U, u v)+a(u U, v u ) so tat = a(u U, u v) appleku Uk E ku vk E, ku Uk E appleku vk E, v 2 V. (12.53) Specifically, by coosing v = u, we obtain te a priori error estimate kek E appleku uk E, (12.54) were te error e = u U is estimated in terms of an interpolation error.

23 12.5. EXERCISES 119 Aposteriorierrorestimation In contrast to an a priori error estimate wic is expressed in terms of te unknown exact solution u, ana posteriori error estimate is bounded in terms of te approximate solution U. We define a linear functional of te solution u, M(u) =(u, ), (12.55) wit te Riesz representer of te functional M( ), guaranteed to exist by Teorem 19. To estimate te error wit respect to M( ), we introduce an adjoint problem: find ' 2 V,suctat a(v, ') =M(v), 8v 2 V. (12.56) Te a posteriori error estimate ten follows, as M(u) M(U) =a(u, ') a(u, ') =L(') a(u, ') =r(u, '), (12.57) wit te weak residual r(u, ') =L(') a(u, '). (12.58) Wit U afiniteelementapproximationcomputedoveramest,we can split te integral over te elements K in T,sotat M(u) M(U) =r(u, ') = X X r K (U, ') = E K, (12.59) K2T K2T wit te error indicator E K = r K (U, '), (12.6) defined for eac element K. To approximate te error indicator we can compute an approximation to te adjoint problem, ', sotat E K r K (U, ). (12.61) Suc local error indicators can be used in an adaptive algoritm to identity wic elements K tat contribute te most to te global error Exercises Problem 38. Derive te variational formulation (12.39), and formulate te finite element metod.