Finite element discretization of Laplace and Poisson equations

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1 Finit lmnt discrtization of Laplac and Poisson quations Yashwanth Tummala Tutor: Prof S.Mittal 1

2 Outlin Finit Elmnt Mthod for 1D Introduction to Poisson s and Laplac s Equations Finit Elmnt Mthod for 2D-Discrtization tc Wak formulation Intrpolation functions Boundary Conditions Assmbly of lmnt quations Exampls Summary 2 Finit Elmnt Discrtization of Laplac and Poisson Equations

3 Finit Elmnt Mthod For 1D Domain rprsntd by simpl domain/finit lmnts Approximation functions constructd ovr lmnts 1D quation looks lik: Construct Finit Elmnt Msh Driv th lmnt quations Assmbl th quations Impos th Boundary conditions Find th solutions d du a + cu f = 0, 0<x<L dx dx 3 Finit Elmnt Discrtization of Laplac and Poisson Equations

4 Introduction 4 What ar Poisson s and Laplac s Equations? ( k u) = f( x, y) in domain Ω, whr is th gradint oprator In Cartsian coordinat systm r r = i + j x y and th Poisson's Equation taks th form: u u k k = f( x, y) x x y y Whn f ( x, y) is zro it bcoms a Laplac's Equation! Finit Elmnt Discrtization of Laplac and Poisson Equations

5 Introduction Som xampls: Hat Transfr-Conduction, Convction (Tmp. distribution) Elctrostatics- Scalar potntial Magntostatics- Magntic potntial Fluid mchanics, Strss Distribution tc. Problm Statmnt: -Find th solution u(x,y) of th scond ordr partial diffrntial quation namly th Poisson s Equation -Givn k, f(x,y) and spcifid boundary conditions in a domain 5 Finit Elmnt Discrtization of Laplac and Poisson Equations

6 Finit Elmnt Discrtization Discrtization/Msh Gnration: 2D gomtric shaps- triangl, rctangl, quadrilatral tc. Gnral ruls: Elmnts should charactriz govrning quations Numbr, shap, typ, ar as rquird (accuracy) Larg gradints ar adquatly modld ( dnsity of lmnts) Msh rfinmnts should vary gradually Ω Ω $n Γ 6 Finit Elmnt Discrtization of Laplac and Poisson Equations

7 Wak Form 7 For a typical lmnt Ω Finit lmnt modl ovr Ω Ω,, u u w k k f dxdy = 0 x x y x W also hav th following, u w u u w k = k wk x x x x x x u w u u w k = k wk y y y y y y From gradint (or divrgnc thorm) in componnt form: Ω Ω u u wk dxdy = w k nxds x x x u u wk dxdy = w k nyds y y y Γ Γ Finit Elmnt Discrtization of Laplac and Poisson Equations

8 Wak Form 8 Whr n$ = n $ i + n $ j and n$ = unit normal vctor Thus w gt th wak form of Poisson's Equation: Ω x w u w u u u k + k wf dxdy w n k + n k ds = 0 x y x x y y Γ x y u u W say qn = nx k + x ny k y y Finit Elmnt Modl: u is approximatd n u( x, y) uh( x, y) = u jψ j( x, y) j = 1 u x y u nod x y th h(, ) = valu of h at j ( j, j) n = no. of nods of th lmnt. Ψ j = Lagrang intrpolation function/shap function. Finit Elmnt Discrtization of Laplac and Poisson Equations

9 Finit Elmnt Modl 9 Chos n j=1 { Ψ1 Ψ2 Ψn} linarly indpndnt functions w=,,... Th wak form bcoms: Or Ψ Ψ k + k dxdy u fψ dxdy Ψ q ds = 0 Ψi j Ψi j j i i n x x y y Ω Ω Γ n j=1 quation th this i ca ku = f + Q ( i= 1,2,..., n) ij j i i Unknowns : u,u,...,u In 1 2 n { K } { uj } = { f } + { Q } n b rwrittn as: th matrix form th st of Equations is: nxn nx1 nx1 nx1 Finit Elmnt Discrtization of Laplac and Poisson Equations

10 Intrpolation Functions 10 For a triangular lmnt: u( x, y) = c + c x + c y = u Ψ + u Ψ + u Ψ h u ( x, y ) = u = c + c x + c y, h h h i { } { }{ } u ( x, y ) = u = c + c x + c y, or u = A c 1 {} { } {} u ( x, y ) = u = c + c x + c y, or c = A u Aftr a bit of math w gt : α = xy x j k y, β = y y, γ = ( x x ) k j i j k i j k 1 c1 = ( α1u1+ α2u2 + α3u3),... A=ara of th triangl 2A From th first Equation abov w gt: 1 Ψ i = ( αi + βi x+ γi y),( i = 1,2,3) 2A Finit Elmnt Discrtization of Laplac and Poisson Equations

11 Intrpolation Functions 11 { } W calculat K and f for linar triangular lmnt. Th K ij associatd lmnt cofficint matrix is: k = + 4A 1 fi = fa 3 An xampl : ( βi βj γi γ j ) a + b b a 1 k fab K = 0, f 1 2ab = a 0 a [ ] b b { } Finit Elmnt Discrtization of Laplac and Poisson Equations

12 Intrpolation Functions 12 For a rctangular lmnt: u ( x, y) = c + c x + c y + c xy h u ( x, y) = c + c x + c y + c xy h u = u (0,0) = c, u = u ( a,0) = c + c a 1 h 1 2 h 1 2 u = u ( a, b) = c + c a + c b + c ab, u = u (0, b) = c + c b 3 h h 1 3 Thus w gt th intrpolation functions as: x y x y Ψ 1 = 1 1, Ψ 2 = 1, a b a b x y x y Ψ 3 =, Ψ 4 = 1 or ab a b i + 1 x i + x y i + y Ψ 1 ( x, y) = ( 1) 1 1 a b Finit Elmnt Discrtization of Laplac and Poisson Equations

13 Intrpolation Functions Th lmnt cofficint matrix for linar rctangular lmnt: ( a + b ) a b ( a + b ) b a ( + ) 2 ( + ) ( ) ( ) ( + ) 2 2( + ) a b a b b a a b k K = 6ab a + b b 2a 2 a + b a 2b b a a b a b a b Whn a=b, w furthr simplify to gt : k K = Finit Elmnt Discrtization of Laplac and Poisson Equations

14 Assmbly of Elmnt Equations Basd on two principls: Continuity of primary variabls Equilibrium or balanc of scondary variabls (lik flux) Infrncs: Continuity of u at nods continuity along intrlmnt boundary. At th intrfac, flux should b qual in magnitud and opposit in sign. Rlation/Cofficints of th stiffnss matrix ar thus drivd, taking global and local nod numbring into considration. 14 Finit Elmnt Discrtization of Laplac and Poisson Equations

15 Assmbly of Elmnt Equations For bttr undrstanding: With two lmnts w hav u = U, u = u = U, u = u = U, u = U, u = U continuity of u ( n) ( n) q = q Equilibrium of scondary variabls So, th assmbld quations ar givn by: K11 K12 K U1 F K21 K22 + K11 K23 + K14 K12 K U F F K3 1 K32+ K41 K33 + K44 K42 K U 43 3 = F3 + F K21 K24 K22 K23 U4 F K 2 31 K34 K32 K 33 U 5 F 3 15 Finit Elmnt Discrtization of Laplac and Poisson Equations

16 Boundary Conditions Th boundary intgral is of th form: Q i = q n Ψ i ( s ) d s Γ Whn is it not ncssary to comput??? Equilibrium of intrnal flux In gnral th intgral ovr th boundary is: Q = Ψ ( sq ) ( sds ) + Ψ ( sq ) ( sds ) + Ψ ( sq ) ( sds ) i i n i n i n i = i1 + i2 + i3 Q Q Q Q Q 1 What is th contribution of sid 2-3 to What about th rctangular lmnts? How many non-zro trms will thy hav? 16 Finit Elmnt Discrtization of Laplac and Poisson Equations

17 Exampl Considr th problm u = f0 i = f0 in a x y squar rgion. Using FEM find u(x,y), u=0 on th ntir boundary, is uniformly distributd sourc. f 0 u u Along th diagonal : u n u = = 0 x u 1 u u = = 0 n 2 y x u n u = = 0 y 17 On th lins of symmtry, normal drivativ of u(x,y)=0 Finit Elmnt Discrtization of Laplac and Poisson Equations

18 Exampl 18 Solution by Linar Triangular Elmnts: Symmtry along diagonal-modl as shown Msh as 4 linar triangls - local and global nods Elmnt 1 = Typical Elmnt W gt th lmnt coff matrix and sourc vctor: 2 2 b b fab 0 K = b a b a { f } 1 2ab + = a a 1 whr a = b = A/2 = 0.5 Thus w hav : { } { } { } { } K = K = K = K, f = f = f = f Finit Elmnt Discrtization of Laplac and Poisson Equations

19 Exampl 19 So, finally : f0 K = 1 2 1, { f } 1 2 = Th assmbld systm of quations look lik: U U 2 3 Q2 + Q3 + Q U 3 f 3 Q Q2 + Q1 = + 3 U 4 Q U Q1 + Q3 + Q U Q From th boundary conditions w gt; U = U = U = Finit Elmnt Discrtization of Laplac and Poisson Equations Q

20 Exampl 20 By boundary condition and balancing of intrnal Flux : Q = Q = 0, Q = Q + Q + Q = Q3 = Q3 + Q2 + Q1 = 0 Th abov st of quations can b writtn as: K11 K12 K13 U1 f K21 K22 + K33 + K11 K23 + K32 U2 = f2 + f3 + f K K32 K23 K33 K22 K 11 U f3 + f2 + f1 Us in g numrical valus for K and f ( with f = 1), W gt : 1 ij U U 2 = U 3 3 Finit Elmnt Discrtization of Laplac and Poisson Equations

21 Exampl 21 Solution by Linar Rctangular Elmnts: fa 1 0 K =, { f } = Obsrv that U = U = U = U = U = Boundary conditions on scondary variabls ar: Q = 0, Q + Q = 0, Q + Q = Balancing at global nod 5 givs: Q + Q + Q + Q = By t aking f = 1, a=0.5, w can solv for U 0 i u = 0 x Finit Elmnt Discrtization of Laplac and Poisson Equations u=0 u=0 u = 0 y

22 Conduction k Conduction in 1D is givn by: Conduction in 2D is givn by: u(x,y)= T = Tmpratur x q n, k f y T k x = f ( x, y ) x x T T k x k y = f( x, y ) x x y y = ngativ of hat flux = thrmal conductivitis along x and y = intrnal hat gnration Considr hat transfr in two-dimnsional plan. Find Tmpratur distribution in th plan. 22 Finit Elmnt Discrtization of Laplac and Poisson Equations

23 Convction 23 For a convction boundary: Equation for nrgy tranfsfr: T T k n + k n + β ( T T ) = q x y x x y y n β = convctiv conductanc, T = Ambint tmpratur q = spcifid hat flux Th W ak Form: Ω n w T w T kx + ky wf dxdy w qn β T T x x y y Γ Finit Elmnt Modl: n j=1 j ( x y) T= T Ψ, j Finit Elmnt Discrtization of Laplac and Poisson Equations ( ) ds = 0

24 Conduction/Convction So, w finally gt : n j = 1 ( ) K + H T = F + P ij ij j i i, whr, Ψ Ψ i j Ψ Ψ i j K ij = k x + k y dxdy x x y y Ω F = f Ψ dxdy + q Ψ ds = f + Q i i n i i i Ω Γ H = β Ψ Ψ ds, P = Ψ T ds ij i j i i Γ Γ 24 Finit Elmnt Discrtization of Laplac and Poisson Equations

25 Summary Finit Elmnt Modl for 1D problms Poisson s and Laplac s Equations (Modl Equation) Finit Elmnt Discrtization Wak Form Finit Elmnt Modl and Algbraic quations Intrpolation Functions for Triangular lmnts Intrpolation Functions for Rctangular lmnts Exampls For th Intrpolation/Shap functions 25 Finit Elmnt Discrtization of Laplac and Poisson Equations

26 Summary Associatd Elmnt Cofficint Matrix Boundary Conditions and Intgrals Assmbly of lmnt Equations and th principls Exampl with a squar domain and triangular lmnts Exampl with a squar domain and rctangular lmnts Application of FEM to Conduction, Conduction & Convction tc. 26 Finit Elmnt Discrtization of Laplac and Poisson Equations

27 Rfrncs An Introduction to th Finit Elmnt Mthod- 3 rd Edition by J.N.REDDY Th Finit Elmnt Mthod- 4 th Edition by Zinkiwicz - THANK YOU 27 Finit Elmnt Discrtization of Laplac and Poisson Equations

AS 5850 Finite Element Analysis

AS 5850 Finite Element Analysis AS 5850 Finit Elmnt Analysis Two-Dimnsional Linar Elasticity Instructor Prof. IIT Madras Equations of Plan Elasticity - 1 displacmnt fild strain- displacmnt rlations (infinitsimal strain) in matrix form