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# MATH542 MATH542

SDSU

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This 6 page Class Notes was uploaded by Camryn Rogahn on Tuesday October 20, 2015. The Class Notes belongs to MATH542 at San Diego State University taught by P.Blomgren in Fall. Since its upload, it has received 19 views. For similar materials see /class/225268/math542-san-diego-state-university in Mathematics (M) at San Diego State University.

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Date Created: 10/20/15

Outline Numerical Solutions to Differential Equations 0 The Finite Element Method 0 Recap 0 Looking for Solutions a Stiffness Matrix and Load Vector 0 Identifying a Linear System 0 Properties of the Stiffness Matrix 9 Error Estimation Lecture Notes 22 The Finite Element Method Peter Blomgren blomgrenpeter gmail com Department of Mathematics and Statistics 9 BUiiding the TOOIbOX D IS t G computamifl39cie ei ee center Error COWOI San Diego State University 9 Teaser FEM for Partial Differential Equations 53quot D39ego CA 92182397720 0 Initial Example The Poisson Equation I MPtequot i39iUSSdSUEdu 0 Analytical Tools Vector Calculus O Variational Formulation Spring 2009 O Minimization Formulation m 0 The Road Ahead In Some Other Class m The Finite Element Method 7 124 The Finite Element Method 7 224 Quick Recap llll Quick Recap lllll Last time we formulated the Galerkin variational Vh and Ritz inner PI OdUCtSI We defined the tWO for nOW identical inner minimization Mh methods which will give us the finite element PrOdUCtSi solutions to the differential equation D aLlvledx7 uvuvdx I I Ll f Dirichlet Boundary Conditions Fth df39 dth f t39 I The Function Space Vh Given the basis functions 4b the tent ur er we 8 me 8 energy unc Iona functions we can write Fu 3307 V 7 f7 V N V h V V n x Given these definitions we formulated the Galerkin and Ritz problems m 5393 The Finite Element Method 7 324 The Finite Elemehl Method 424 Quick Recap Galerkin s Method Variational Approach Looking for the Solution If uh E Vh is a solution to Vh then in particular 8Ul77 f7 j7 j17277N Also we can write uh in terms of the basis functions Vh Find uh E Vh so that au1v1fvh Vvh E Vh N UhX ZEMKXL 51 1109 Ritz Method Minimization Approach j1 Therefore we can rewrite the above equation Mh Find uh E V so that FUh S FVh Vvh E Vh N gradingbum 1172N m 1 m The Finite Element Method 7 524 The Finite Element Method 7 624 The Stiffness Matrix and Load Vector Computing the Stiffness Matrix F The elements of the stiffness matrix can be computed rom N First we notice that if the basis is the piecewise linear 39 a 1 741507 J 1727 39 39 39 7 N tent functionsquot then if if 7jl gt 1 then and Q are IT non overlapping which means a f O we identify the vectors 5 152 5NT and The Diagonal Entries j 1727 7 N s i T I i I b 7 b1 b2 7bN where b 7 f and the matrix A 9 1 X1 1 1 1 Where Aij a iv j a j j W dX h dX T T 1 We can now write our problem as X171 1 X 11 1 1 Ag B The Super and Sub Diagonal Entries j 23 N X1 71 1 1 For historical reasons Structural Mechanics A IS known as the a j717 j a j 41514 77 dX 77 stiffness matrix and b as the load vector Xrl hj hj hj 39 m 5393 The Finite Element Method 7 324 The Finite Element Method 7 724 Illustration Overlapping and Nonoverlapping Tent Functions The Stiffness Matrix is Symmetric Positive Definite SPD A is symmetric since au7 v av7 u and with VX nj X we get N Z Tiia li 193le a N N Zm li ETDb v1X2 dX 2 O 1 j1 09 03 07 06 I J1 05 M 39i39 Equality holds if and only if VX E O which implies VX E O by 03 i the boundary conditions 2 Fact A matrix A is SPD if 01 H 1 3 A XTAX gt 07 VX E Rn0 Fact An SPD matrix has positive eigenvalues ii is non m singular The Finite Element Method 7 924 The Finite Element Method 7 1024 Estimating the Error for the Model Problem The Special Case hj h7 j 1727 7 N We are now going to look at the error u 7 uh where Ll is the exact solution of the differential equation and uh is the solution to If we equi partition the interval we get the linear system 71 2 0 71 bN 0 If we divide by h the we recover the standard finite difference method for the problem where the right hand side X1 b 3 fx gti dx h Xil over the interval XI719941 h is a weighted mean of fx The Finite Element Method wDh l the finite element problem Vh Since V C V 8W7 VIC f7 Vhi VVh E Vii 3047 VIZ f7 Vhi VVh E Vii Vvh E V 8U i thi VA 07 That means that the error is orthogonal to the function space V as measured by ao7 0 The Finite Element Method 7 1124 5353 7 1224 Definition and Tools Cauchy39s Inequality Proof Cauchy39s Inequality Given v and W define the renormalized functions V and W Def39n39tlon The L239n0rm Which means 1 Now 1 A A A A A A 0 g lliviwll2aiviw iviw 7 2 7 llWll2 alwtw VO W dX avV2av 17vavT W 2 I 23V717v Hence Theorem Cauchy39s Inequality lalvi S 1 80 W S J Removing the linear normalization factors give laV7 W S llVllllWll m D as The Finite Element Method 7 1324 The Finite Element Method 7 1424 Error Control Theorem Error Control Proof Proof Let vh E Vh be arbitrary and set Wh uh 7 vh Then Wh E Vh and we Theorem get For any V E V we have llu Uh ll2 3u Uh 7 u Uhl 3u bl7 WL lluuhll lawnii 3U Uh7 h l Wh u Uh 7 u Vhl So measured in the L2 norm of the derivative also known as the Cauchyvs S uhHHU Vhy H1 seminorm the solution uh to the discrete problem is closer to the solution u to the original continuous ODE than any other DlVldlng through by blll glVeS function in the function space Vh llu Uhl ll S llu Vhl ll 3 5 The Finite Element Method 7 1524 The Finite Element Method 7 1624 Error Control Applying the Theorem FEM for PDEs the Poisson Equation From the theorem we can get a quantitative estimate upper bound for the error 7 uhH by estimating 7 VhH where V E V is a suitably chosen function Let V be the linear interpolant of u ie vhxj then luX ViXl S hmaxxe01 lu Xl 1 lUX VhXl S h maxx 01 lU Xl 2 The first result and the theorem gives 7 lth Hu uh H Xrglg llu Xl Since u 7 uh0 O we can integrate and get lluX 7 LlhXll S h max lLlXl can be made sharper h2 xE01 The Finite Element Method 539 7 1724 Consider the following boundary value problem 7Auf uO inQ on F3Q where Q is a bounded domain in R2 X1X2 X1 6 R7X2 E R F is the boundary of Q fX1X2 a given function and 32 2 Au 7 7 LIX1 X2 u X1 X2 u X1 X2 2 2 7 x1x1 7 xzxz 7 3X1 3X2 Physical problems Our simple 1 D model problems from last lecture carry over to this 2 D model heat distribution in a plate the displacement of an elastic membrane fixed at the boundary under transverse load etc The Finite Element Method 7 1324 Vector Calculus Review or Crash Course Vector Calculus Review or Crash Course We need to generalize integration by partsquot to higher dimensions We start with the divergence theorem Vo di o ds Q r where G L117 u2T and s 5 VOU7TXl 3U2 3X2 n1 n2T is the unit length one outward normal to F d denotes area integration and d5 integration along the boundary cf in 1 D dX 27b dX 1 d5 50 bl 27b The Finite Element Method 7 1924 If we apply the divergence theorem to the vector functions 11 vw0 and g O7 vw we find 3 3 7Vw V7W di vwnds7 139 12 m 3X7quot 3X7quot r Let Vv denote the gradient 3v 3v T Vv 7 7 3X1 3X2 and use the result above to write Green s Formula 7 8v 8W 8v 8W fQ Vvo dex f9 871671 872672 dx 2 2 fl vg7gn1 vg7gn2 d5 7 9 v E 721 dx f vginv d5 7 f0 vAwdi 5353 The Finite Element Method 7 2024 Vector Calculus Review or Crash Course llllll The normal derivative 3w 3w 3w V F 7 in in W 0 3n 3X1 1 3X2 2 is the derivative in the outward normal direction to the boundary F With this notation we are ready to state the variational formulation corresponding to the differential equation Variational Formulation V Find u E V so that au7 v f7 v VV 6 V where au7 v Vu ond 0 f7 v fv d 0 v v 6 C9 V VX1 and VX2 are piecewise continuous in 9 v O on F Note We have skightly changed the notation of the 3070 inner product This is very common in the PDE framework The ao7 o inner product generically involves integrating the products of m derivatives of the two arguments over the domain The Finite Element Method 7 2124 The Finite Element Method 7 2224 Minimization Formulation Next u is a solution to V if and only if it is also a solution to the following minimization problem M Find u E V such that Fu g Fv VV 6 V Constructing a Finite Element Method for this problem where Fv is the potential energy 1 FM 5am v e f v nl 5393 The Finite Element Method 7 2324 The Finite Element Method 7 2424

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