2 Multi-Dimensional Variational Principles y s n u = α θ pu = β n Ω Figure 3..: Two-dimensional region wit and normal vector n x were

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1 Capter 3 Multi-Dimensional Variational Principles 3. Galerkin's Metod and Extremal Principles Te construction of Galerkin formulations presented in Capters and 2 for one-dimensional problems readily extends to iger dimensions. Following our prior developments, we'll focus on te model two-dimensional self-adjoint diusion problem L[u] =;(p(x y)u x ) x ; (p(x y)u y ) y + q(x y)u = f(x y) (x y) 2 (3..a) were < 2 wit (Figure 3..) and p(x y) > 0, q(x y) 0, (x y) 2. Essential boundary conditions u(x y) =(x y) (x y) E (3..b) are prescribed on te E and natural boundary y) p(x y) = pru n := p(u x cos + u y sin ) =(x y) (x y) (3..c) are prescribed on te remaining N Te angle is te angle between te x-axis and te outward normal n (Figure 3..). Te Galerkin form of (3..) is obtained by multiplying (3..a) by a test function v and integrating over to obtain v[;(pu x ) x ; (pu y ) y + qu ; f]dxdy =0: (3..2) In order to integrate te second derivative terms by parts in two and tree dimensions, we use Green's teorem or te divergence teorem radxdy = Z a nds

2 2 Multi-Dimensional Variational Principles y s n u = α θ pu = β n Ω Figure 3..: Two-dimensional region wit and normal vector n x were s is a coordinate a =[a a 2 ] T,and : (3..3b) In order to use tis result in te present circumstances, let us introduce vector notation (pu x ) x +(pu y ) y := r(pru) and use te \product rule" for te divergence and gradient operators r (vpru) =(rv) (pru) +vr (pru): (3..3c) Tus, ;vr(pru)dxdy = [(rv) (pru) ;r(vpru)]dxdy: Now apply te divergence teorem (3..3) to te second term to obtain ;vr(pru)dxdy = rv prudxdy ; Z vpru Tus, (3..2) becomes [rv pru + v(qu ; f)]dxdy ; Z vpu n ds =0

3 3.. Galerkin's Metod and Extremal Principles 3 were (3..c) was used to simplify te surface integral. Te integrals in (3..4) must exist and, wit u and v of te same class and p and q smoot, tis implies (u 2 x + u 2 y + u 2 )dxdy exists. Tis is te two-dimensional Sobolev space H. Drawing upon our experiences in one dimension, we expect u 2 H E, were functions in H E are in H and satisfy te Diriclet boundary conditions (3..b) on E. Likewise, we expect v 2 H 0, wic denotes tat v = 0 E. Tus, te variation v sould vanis were te trial function u is prescribed. Let us extend te one-dimensional notation as well. Tus, te L 2 inner product is (v f) := vf dxdy (3..5a) and te strain energy is A(v u) :=(rv pru)+(v qu) = [p(v x u x + v y u y )+qvu]dxdy: (3..5b) We also introduce a boundary L 2 inner product as <v w>= Z vwds: (3..5c) Te boundary integral may be restricted N since v =0on@ E. Wit tis nomenclature, te variational problem (3..4) may bestated as: nd u 2 H E A(v u) =(v f)+ <v > 8v 2 H 0: (3..6) Te Neumann boundary condition (3..c) was used to replace pu n in te boundary inner product. Te variational problem (3..6) as te same form as te one-dimensional problem (2.3.3). Indeed, te teory and extremal principles developed in Capter 2 apply to multi-dimensional problems of tis form. Teorem 3... Te function w 2 H E tat minimizes is te one tat satises (3..6), and conversely. I[w] =A(w w) ; 2(w f) ; 2 <w >: (3..7) Proof. Te proof is similar to tat of Teorem 2.2. and appears as Problem at te end of tis section.

4 4 Multi-Dimensional Variational Principles Corollary 3... Smoot functions u 2 H E satisfying (3..6) or minimizing (3..7) also satisfy (3..). Proof. Again, te proof is left as an exercise. Example 3... Suppose tat te Neumann boundary conditions (3..c) are canged to Robin boundary conditions pu n + u = (x y) N : (3..8a) Very little canges in te variational statement of te problem (3..a,b), (3..8). Instead of replacing pu n by in te boundary inner product (3..5c), we replace it by ; u. Tus, te Galerkin form of te problem is: nd u 2 H E satisfying A(v u) =(v f)+ <v ; u > 8v 2 H 0: (3..8b) Example Variational principles for nonlinear problems and vector systems of partial dierential equations are constructed in te same manner as for te linear scalar problems (3..). As an example, consider a tin elastic seet occupying a twodimensional region. As sown in Figure 3..2, te Cartesian components (u u 2 ) of te displacement vector vanis on te E of of te and te components of te traction are prescribed as (S S 2 ) on te remaining N Te equations of equilibrium for suc a problem are (cf., e.g., [6], 2 = 0 (3..9a) = 0 (x y) 2 (3..9b) were ij, i j = 2, are te components of te two-dimensional symmetric stress tensor (matrix). Te stress components are related to te displacement components by Hooke's law = E ; 2 ) (3..0a) 22 = E ; ) (3..0b) 2 = E 2( + ) ) (3..0c)

5 3.. Galerkin's Metod and Extremal Principles 5 y s n u = 0, u = 0 2 θ Ω S 2 S Figure 3..2: Two-dimensional elastic seet occupying te region. Displacement components (u u 2 )vanis E and traction components (S S 2 ) are prescribed N. x were E and are constants called Young's modulus and Poisson's ratio, respectively. Te displacement and traction boundary conditions are u (x y) =0 u 2 (x y) =0 (x y) E (3..a) n + n 2 2 = S n 2 + n 2 22 = S 2 (x y) N (3..b) were n = [n n 2 ] T = [cos sin ] T is te unit outward normal vector (Figure 3..2). Following te one-dimensional formulations, te Galerkin form of tis problem is obtained by multiplying (3..9a) and (3..9b) by test functions v and v 2, respectively, integrated over, and using te divergence teorem. Wit u and u 2 being components of a displacement eld, te functions v and v 2 are referred to as components of te virtual displacement eld. We use (3..9a) to illustrate te process tus, multiplying by v and integrating over, we nd 2 ]dxdy Te tree stress components are dependent on te two displacement components and are typically replaced by tese using (3..0). Were tis done, te variational principle

6 6 Multi-Dimensional Variational Principles would involve second derivatives of u and u 2. Hence, wewould want to use te divergence teorem to obtain a symmetric variational form and reduce te continuity requirements on u and u 2. We'll do tis, but omit te explicit substitution of (3..0) to simplify te presentation. Tus, we regard and 2 as components of a two-vector, we use te divergence teorem (3..3) to @y 2]dxdy = v [n + n 2 2 ]ds: Selecting v 2 H 0 implies tat te boundary integral vanises E. Tis and te subsequent use of te natural boundary condition (3..b) @y 2]dxdy = v S ds 8v 2 H 0: (3..2a) Similar treatment of (3..9b) gives ]dxdy = v 2 S 2 ds 8v 2 2 H 0: (3..2b) Equations (3..2a) and (3..2b) may be combined and written in a vector form. Letting u =[u u 2 ] T, etc., we add (3..2a) and (3..2b) to obtain te Galerkin problem: nd u 2 H 0 suc tat A(v u) =< v S > 8v 2 H0 (3..3a) were A(v u) ) 2]dxdy < v S >= (3..3b) (v S + v 2 S 2 )ds: (3..3c) Wen a vector function belongs to H, we mean tat eac of its components is in H. Te spaces H E and H 0 are identical since te displacement is trivial E. Te solution of (3..3) also satises te following minimum problem. Teorem Among all functions w =[w w 2 ] T 2 HE te solution u =[u u 2 ] T of (3..3) is te one tat minimizes I[w] = E 2( ; 2 ) f( )2 )2

7 3.. Galerkin's Metod and Extremal Principles 7 and conversely. + ( ; ) )2 gdxdy ; (w S + w 2 S 2 )ds Proof. Te proof is similar to tat of Teorem Te stress components ij, i j = 2, ave been eliminated in favor of te displacements using (3..0). Let us conclude tis section wit a brief summary. A solution of te dierential problem, e.g., (3..), is called a \classical" or \strong" solution. Te function u 2 H 2 B, were functions in H 2 ave nite values of [(u xx ) 2 +(u xy ) 2 +(u yy ) 2 +(u x ) 2 +(u y ) 2 + u 2 ]dxdy and functions in H 2 B also satisfy all prescribed boundary conditions, e.g., (3..b,c). Solutions of a Galerkin problem suc as (3..6) are called \weak" solutions. Tey may be elements of a larger class of functions tan strong solutions since te igorder derivatives are missing from te variational statement of te problem. For te second-order dierential equations tat we ave been studying, te variational form (e.g., (3..6)) only contains rst derivatives and u 2 H E. Functions in H ave nite values of [(u x ) 2 +(u y ) 2 + u 2 ]dxdy: and functions in HE also satisfy te prescribed essential (Diriclet) boundary condition (3..b). Test functions v are not varied were essential data is prescribed and are elements of H0. Tey satisfy trivial versions of te essential boundary conditions. Wile essential boundary conditions constrain te trial and test spaces, natural (Neumann or Robin) boundary conditions alter te variational statement of te problem. As wit (3..6) and (3..3), inomogeneous conditions add boundary inner product terms to te variational statement. Smoot solutions of te Galerkin problem satisfy te original partial dierential equation(s) and natural boundary conditions, and conversely. Galerkin problems arising from self-adjoint dierential equations also satisfy extremal problems. In tis case, approximate solutions found by Galerkin's metod are best in te sense of (2.6.5), i.e., in te sense of minimizing te strain energy of te error.

8 8 Multi-Dimensional Variational Principles Problems. Prove Teorem 3.. and its Corollary. 2. Prove Teorem 3..2 and aslo sow tat smoot solutions of (3..3) satisfy te dierential system (3..9) - (3..). 3. Consider an innite solid medium of material M containing an innite number of periodically spaced circular cylindrical bers made of material F. Te bers are arranged in a square array wit centers two units apart in te x and y directions (Figure 3..3). Te radius of eac ber is a (< ). Te aim of tis problem is to nd a Galerkin problem tat can be used to determine te eective conductivity of te composite medium. Because of embedded symmetries, it suces to solve a y M F a r θ x Figure 3..3: Composite medium consisting of a regular array of circular cylindrical bers embedded in in a matrix (left). Quadrant of a Periodicity cell used to solve tis problem (rigt). problem on one quarter of a periodicity cell as sown on te rigt of Figure Te governing dierential equations and boundary conditions for te temperature

9 3.. Galerkin's Metod and Extremal Principles 9 (or potential, etc.) u(x y) witin tis quadrant are r(pru) =0 (x y) 2 F [ M u x (0 y)=u x ( y)=0 0 y u(x 0) = 0 u(x ) = 0 x u 2 C 0 pu r 2 C 0 (x y) 2 x 2 + y 2 = a 2 : (3..4) Te subscripts F and M are used to indicate te regions and properties of te ber and matrix, respectively. Tus, letting we ave and :=f(x y)j 0 x 0 y g F := f(r )j 0 r a 0 =2g M := ; F : Te conductivity p of te ber and matrix will generally be dierent and, ence, p will jump at r = a. If necessary, we can write pf if x p(x y) = 2 + y 2 <a 2 : p M if x 2 + y 2 >a 2 Altoug te conductivities are discontinuous, te last boundary condition conrms tat te temperature u and ux pu r are continuous at r = a. 3.. Following te steps leading to (3..6), sow tat te Galerkin form of tis problem consists of determining u 2 H E as te solution of F [M p(u x v x + u y v y )dxdy =0 8v 2 H 0: Dene te spaces H E and H 0 for tis problem. Te Galerkin problem appears to be te same as it would for a omogeneous medium. Tere is no indication of te continuity conditions at r = a Sow tat te function w 2 H E tat minimizes I[w] = F [M p(w 2 x + w 2 y)dxdy is te solution u of te Galerkin problem, and conversely. Again, tere is little evidence tat te problem involves an inomogeneous medium.

10 0 Multi-Dimensional Variational Principles 3.2 Function Spaces and Approximation Let us try to formalize some of te considerations tat were raised about te properties of function spaces and teir smootness requirements. Consider a Galerkin problem in te form of (3..6). Using Galerkin's metod, we nd approximate solutions by solving (3..6) in a nite-dimensional subspace S N of H. Selecting a basis f j g N j= for S N, we consider approximations U 2 SE N of u in te form U(x y) = NX j= c j j (x y): (3.2.) Wit approximations V 2 S N 0 of of v aving a similar form, we determine U as te solution A(V U) =(V f)+ <V > 8V 2 S N 0 : (3.2.2) (Nontrivial essential boundary conditions introduce dierences between S N E and S N 0 and we ave not explicitly identied tese dierences in (3.2.2).) We've mentioned te criticality of knowing te minimum smootness requirements of an approximating space S N. Smoot (e.g. C ) approximations are dicult to construct on nonuniform two- and tree-dimensional meses. We ave already seen tat smootness requirements of te solutions of partial dierential equations are usually expressed in terms of Sobolev spaces, so let us dene tese spaces and examine some of teir properties. First, let's review some preliminaries from linear algebra and functional analysis. Denition V is a linear space if. u v 2Vten u + v 2V, 2. u 2V ten u 2V, for all constants, and 3. u v 2Vten u + v 2V, for all constants,. Denition A(u v) isabilinear form on VV if, for u v w 2Vand all constants and,. A(u v) 2<,and 2. A(u v) is linear in eac argument tus, A(u v + w)=a(u v)+a(u w) A(u + v w)=a(u w)+a(v w):

11 3.2. Function Spaces and Approximation Denition An inner product A(u v) is a bilinear form on VVtat. is symmetric in te sense tat A(u v) =A(v u), 8u v 2V, and 2. A(u u) > 0, u 6= 0andA(0 0)=0,8u 2V. Denition Te norm kk A associated wit te inner product A(u v) is kuk A = p A(u u) (3.2.3) and it satises. kuk A > 0, u 6= 0,k0k A =0, 2. ku + vk A kuk A + kvk A, and 3. kuk A = jjkuk A, for all constants. Te integrals involved in te norms and inner products are Lebesgue integrals rater tan te customary Riemann integrals. Functions tat are Riemann integrable are also Lebesgue integrable but not conversely. We ave neiter time nor space to delve into Lebesgue integration nor will it be necessary for most of our discussions. It is, owever, elpful wen seeking understanding of te continuity requirements of te various function spaces. So, we'll make a few brief remarks and refer tose seeking more information to texts on functional analysis [3, 4, 5]. Wit Lebesgue integration, te concept of te lengt of a subinterval is replaced by te measure of an arbitrary point set. Certain sets are so sparse as to ave measure zero. An example is te set of rational numbers on [0 ]. Indeed, all countably innite sets ave measure zero. If a function u 2 V possesses a given property except on a set of measure zero ten it is said to ave tat property almost everywere. A relevant property is te notion of an equivalence class. Two functions u v 2Vbelong to te same equivalence class if ku ; vk A =0: Wit Lebesgue integration, two functions in te same equivalence class are equal almost everywere. Tus, if we are given a function u 2 V and cange its values on a set of measure zero to obtain a function v, tenu and v belong to te same equivalence class. We need one more concept, te notion of completeness. A Caucy sequence fu n g n= 2 V is one were lim ku m ; u n k A =0: m n!

12 2 Multi-Dimensional Variational Principles If fu n g n= converges in kk A to a function u 2 V ten it is a Caucy sequence. Tus, using te triangular inequality, lim ku m ; u n k A lim fku m ; uk A + ku ; u n k A g =0: m n! m n! AspaceV were te converse is true, i.e., were all Caucy sequences fu n g n= converge in kk A to functions u 2V, issaidtobecomplete. Denition A complete linear space V wit inner product A(u v) and corresponding norm kuk A, u v 2V is called a Hilbert space. Let's list some relevant Hilbert spaces for use wit variational formulations of boundary value problems. We'll present teir denitions in two space dimensions. Teir extension to one and tree dimensions is obvious. Denition Te space L 2 () consists of functions satisfying L 2 () := fuj u 2 dxdy < g: (3.2.4a) It as te inner product (u v) = uvdxdy (3.2.4b) and norm kuk 0 = p (u u): (3.2.4c) Denition Te Sobolev space H k consists of functions u wic belong to L 2 wit teir rst k 0 derivatives. Te space as X te inner product and norm (u v) k := (D u D v) (3.2.5a) jjk kuk k = p (u u) k (3.2.5b) were wit and 2 non-negative integers, and =[ 2 ] T jj = + 2 (3.2.5c) D u 2 : (3.2.5d)

13 3.2. Function Spaces and Approximation 3 In particular, te space H as te inner product and norm (u v) =(u v)+(u x v x )+(u y v y )= (uv + u x v x + u y v y )dxdy (3.2.6a) kuk = 2 4 Likewise, functions u 2 H 2 ave nite values of Example kuk 2 2 = (u 2 + u 2 x + u 2 y)dxdy 3 5 =2 [u 2 xx + u 2 xy + u 2 yy + u 2 x + u 2 y + u 2 ]dxdy: : (3.2.6b) We ave been studying second-order dierential equations of te form (3..) and seeking weak solutions u 2 H and U 2 S N H of (3..6) and (3.2.2), respectively. Letusverify tat H is te correct space, at least in one dimension. Tus, consider a basis of te familiar piecewise-linear at functions on a uniform mes wit spacing ==N j (x) = 8 < : (x ; x j; )= if x j; x<x j (x j+ ; x)= if x j x<x j+ 0 oterwise : (3.2.7) Since S N H, j and 0 j must be in L 2, j = 2 ::: N. Consider C approximations of j (x) and 0 j(x) obtained by \rounding corners" in O(=n)-neigboroods of te nodes x j;, x j, x j+ as sown in Figure3.2.. A possible smoot approximation of 0 j(x) is 0 j(x) 0 j n(x) = 2 [tan n(x ; x j+) + tan n(x ; x j;) ; 2 tan n(x ; x j) ]: A smoot approximation j n of j is obtained by integration as j n (x) = cos n((x ; xj+)=) 2n ln cos n((x ; x j; )=) : cos 2 n((x ; x j )=) Clearly, j n and 0 j n are elements of L 2. Te \rounding" disappears as n!and Z lim n! 0 [ 0 j n(x)] 2 dx 2(=) 2 =2=: Te explicit calculations are somewat involved and will not be sown. seems clear tat te limiting function 0 j 2 L 2 and, ence, j 2 S N for xed. However, it

14 4 Multi-Dimensional Variational Principles Figure 3.2.: Smoot version of a piecewise linear at function (3.2.7) (top), its rst derivative (center), and te square of its rst derivative (bottom). Results are sown wit x j; = ;, x j =0,x j+ =( = ), and n = 0. Example Consider te piecewise-constant basis function on a uniform mes j (x) = if xj; x<x j 0 oterwise : (3.2.8) A smoot version of tis function and its rst derivative are sown in Figure and may be written as j n (x) = 2 [tan n(x ; x j;) ; tan n(x ; x j) ] 0 j n(x) = n n(x ; x j; ) n(x ; x 2 [sec2 ; sec 2 j ) ]: As n!, j n approaces a square pulse owever, 0 j n is proportional to te combination of delta functions 0 j n(x) / (x ; x j; ) ; (x ; x j ):

15 3.2. Function Spaces and Approximation 5 Tus, we anticipate problems since delta functions are not elements of L 2. 0 j n(x) Squaring [ 0 j n(x)] 2 =( n n(x ; x 2 )2 [sec 4 j; ) n(x ; x ;2sec 2 j; ) n(x ; x sec 2 j ) n(x ; x +sec 4 j ) ]: As sown in Figure 3.2.2, te function secn(x ; x j )= is largest at x j and decays exponentially fast from x j tus, te center term in te above expression is exponentially small relative to te rst and tird terms. Neglecting it yields [ 0 j n(x)] 2 ( n 2 )2 [sec 4 n(x ; x j; ) n(x ; x + sec 4 j ) ]: Tus, Z [ 0 j n(x)] 2 dx n 0 2 [tan n(x ; x j;) n(x ; x (2 + sec 2 j; ) ) + tan n(x ; x j) n(x ; x (2 + sec 2 j ) )] 0: Tis is unbounded as n! ence, 0 j(x) =2 L 2 and j (x) =2 H Figure 3.2.2: Smoot version of a piecewise constant function (3.2.8) (left) and its rst derivative (rigt). Results are sown wit x j; =0,x j =( = ), and n = 20. Altoug te previous examples lack rigor, we may conclude tat a basis of continuous functions will belong to H in one dimension. More generally, u 2 H k implies tat u 2 C k; in one dimension. Te situation is not as simple in two and tree dimensions. Te Sobolev space H k is te completion wit respect to te norm (3.2.5) of C k functions wose rst k partial derivatives are elements of L 2. Tus, for example, u 2 H implies tat u, u x, and u y are all elements of L 2. Tis is not sucient to ensure tat u is continuous in two and tree dimensions. Typically, if@ is smoot ten u 2 H k implies tat u 2 C s ( were s is te largest integer less tan (k ; d=2) in d dimensions [, 2]. In two and tree dimensions, tis condition implies tat u 2 C k;2. Problems

16 6 Multi-Dimensional Variational Principles. Assuming tat p(x y) > 0 and q(x y) 0, (x y) 2, nd any oter conditions tat must be satised for te strain energy A(v u) = [p(v x u x + v y u y )+qvu]dxdy to be an inner product and norm, i.e., to satisfy Denitions and Construct avariational problem for te fourt-order biarmonic equation (pu) =f(x y) (x y) 2 were u = u xx + u yy and p(x y) > 0 is smoot. Assume tat u satises te essential boundary conditions u(x y) =0 u n (x y) =0 (x y) were n is a unit outward normal vector To wat function space sould te weak solution of te variational problem belong? 3.3 Overview of te Finite Element Metod Let us conclude tis capter wit a brief summary of te key steps in constructing a niteelement solution in two or tree dimensions. Altoug not necessary, we will continue to focus on (3..) as a model.. Construct a variational form of te problem. Generally, we will use Galerkin's metod to construct a variational problem. As described, tis involves multiplying te dierential equation be a suitable test function and using te divergence teorem to get a symmetric formulation. Te trial function u 2 H E and, ence, satises any prescribed essential boundary conditions. Te test function v 2 H 0 and, ence, vanises were essential boundary conditions are prescribed. Any prescribed Neumann or Robin boundary conditions are used to alter te variational problem as, e.g., wit(3..6) or (3..8b), respectively. Nontrivial essential boundary conditions introduce dierences in te spaces H E and H0. Furtermore, te nite element subspace SE N cannot satisfy non-polynomial boundary conditions. One way of overcoming tis is to transform te dierential equation to one aving trivial essential boundary conditions (cf. Problem at te end of tis section). Tis approac is dicult to use wen te boundary data is discontinuous or wen te problem is nonlinear. It is more important for teoretical tan for practical reasons.

17 3.3. Overview of te Finite Element Metod 7 Te usual approac for andling nontrivial Diriclet data is to interpolate it by te nite element trial function. Tus, consider approximations in te usual form U(x y) = NX j= c j j (x y) (3.3.) owever, we include basis functions k for mes entities (vertices, edges) k tat are E. Te coecients c k associated wit tese nodes are not varied during te solution process but, rater, are selected to interpolate te boundary data. Tus, wit a Lagrangian basis were k (x j y j )= k j, we ave U(x k y k )=(x k y k )=c k (x k y k ) E : Te interpolation is more dicult wit ierarcical functions, but it is manageable (cf. Section 4.4). Wewillave to appraise te eect of tis interpolation on solution accuracy. Altoug te spaces S N E and S N 0 dier, te stiness and mass matrices can be made symmetric for self-adjoint linear problems (cf. Section 5.5). A tird metod of satisfying essential boundary conditions is given as Problem 2 at te end of tis section. 2. Discretize te domain. Divide into nite elements aving simple sapes, suc as triangles or quadrilaterals in two dimensions and tetraedra and exaedra in tree dimensions. Tis nontrivial task generally introduces errors Tus, te problem is typically solved on a polygonal region ~ dened by te nite element mes (Figure 3.3.) rater tan on. Suc errors may be reduced by using nite elements wit curved sides and/or faces (cf. Capter 4). Te relativeadvantages of using fewer curved elements or a larger number of smaller straigt-sided or planar-faced elements will ave to be determined. 3. Generate te element stiness and mass matrices and element load vector. Piecewise polynomial approximations U 2 S N E of u and V 2 S N 0 of v are cosen. Te approximating spaces S N E and S N 0 are supposed to be subspaces of H E and H 0, respectively owever, tis may not be te case because of errors introduced in approximating te essential boundary conditions and/or te domain. Tese eects will also ave to be appraised (cf. Section 7.3). Coosing a basis for S N, we write U and V in te form of (3.3.). Te variational problem is written as a sum of contributions over te elements and te element stiness and mass matrices and load vectors are generated. For te model problem (3..) tis would involve solving XN e= [A e (V U) ; (V f) e ; <V > e ]=0 8V 2 S N 0 (3.3.2a)

18 8 Multi-Dimensional Variational Principles 5 6 s n u = α 4 7 θ pu + γu = β n U 7 y K = x l = K, l e e 8 Figure 3.3.: Two-dimensional domain aving E N wit unit normal n discretized by triangular nite elements. Scematic representation of te assembly of te element stiness matrix K e and element load vector l e into te global stiness matrix K and load vector l. were A e (V U) = (V x pu x + V y pu y + VqU)dxdy (3.3.2b) e

19 3.3. Overview of te Finite Element Metod 9 (V f) e = e V fdxdy (3.3.2c) <V > e = Z V ds ~ N e is te domain occupied by element e,andn is te number of elements in te mes. Te boundary integral (3.3.2d) is zero unless a portion e coincides wit te boundary of te nite element ~. Galerkin formulations for self-adjoint problems suc as (3..6) lead to minimum problems in te sense of Teorem 3... Tus, te nite element solution is te best solution in S N in te sense of minimizing te strain energy of te error A(u ; U u ; U). Te strain energy of te error is ortogonal to all functions V in SE N for tree-vectors. as illustrated in Figure H E 0 u U S E N Figure 3.3.2: Subspace SE N of H E solution of Galerkin's metod. illustrating te \best" approximation property of te 4. Assemble te global stiness and mass matrices and load vector. Te element stiness and mass matrices and load vectors tat result from evaluating (3.3.2b-d) are added directly into global stiness and mass matrices and a load vector. As depicted in Figure 3.3., te indices assigned to unknowns associated wit mes entities (vertices as sown) determine te correct positions of te elemental matrices and vectors in te global stiness and mass matrices and load vector.

20 20 Multi-Dimensional Variational Principles 5. Solve te algebraic system. For linear problems, te assembly of (3.3.2) gives rise to a system of te form d T [(K + M)c ; l] =0 (3.3.3a) were K and M are te global stiness and mass matrices, l is te global load vector, c T =[c c 2 ::: c N ] T (3.3.3b) and d T =[d d 2 ::: d N ] T : (3.3.3c) Since (3.3.3a) must be satised for all coices of d, we must ave (K + M)c = l: (3.3.4) For te model problem (3..), K + M will be sparse and positive denite. Wit proper treatment of te boundary conditions, it will also be symmetric (cf. Capter 5). Eac step in te nite element solution will be examined in greater detail. Basis construction is described in Capter 4, mes generation and assembly appear in Capter 5, error analysis is discussed in Capter 7, and linear algebraic solution strategies are presented in Capter. Problems. By introducing te transformation ^u = u ; sow tat (3..) can be canged to a problem wit omogeneous essential boundary conditions. Tus, we can seek ^u 2 H Anoter metod of treating essential boundary conditions is to remove tem by using a \penalty function." Penalty metods are rarely used for tis purpose, but tey are important for oter reasons. Tis problem will introduce te concept and reinforce te material of Section 3.. Consider te variational statement (3..6) as an example, and modify it by including te essential boundary conditions A(v u) =(v f)+ <v N +<v ; 8v 2 H : Here is a penalty parameter and subscripts on te boundary integral indicate teir domain. No boundary conditions are applied and te problem is solved for u and v ranging over te wole of H.

21 3.3. Overview of te Finite Element Metod 2 Sow tat smoot solutions of tis variational problem satisfy te dierential equation (3..a) as well as te natural boundary conditions (3..c) and u = (x y) 2 E: Te penalty parameter must be selected large enoug for tis natural boundary condition to approximate te prescribed essential condition (3..b). Tis can be tricky. If selected too large, it will introduce ill-conditioning into te resulting algebraic system.

22 22 Multi-Dimensional Variational Principles

23 Bibliograpy [] R.A. Adams. Sobolev Spaces. Academic Press, New York, 975. [2] O. Axelsson and V.A. Barker. Finite Element Solution of Boundary Value Problems. Academic Press, Orlando, 984. [3] C. Geoman and G. Pedrick. First Course in Functional Analysis. Prentice-Hall, Englewood Clis, 965. [4] P.R. Halmos. Measure Teory. Springer-Verlag, New York, 99. [5] J.T. Oden and L.F. Demkowicz. Applied Functional Analysis. CRC Press, Boca Raton, 996. [6] R. Wait and A.R. Mitcell. Te Finite Element Analysis and Applications. Jon Wiley and Sons, Cicester,