Solution of the exam in TMA4212 Monday 23rd May 2013 Time: 9:00 13:00

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1 Norwegi Uiversity of Sciece d Techology Deprtmet of Mthemticl Scieces Cotct durig the exm: Ele Celledoi, tlf Pge 1 of 7 of the exm i TMA4212 Mody 23rd My 2013 Time: 9:00 13:00 Allowed ids: Approved simple pocket clcultor. All writte d hdwritte mteril from the course. Lerig outcome L6 d L3 hve bee tested through the project work 1. Problem 1 Lerig outcome L2, L7 Cosider the differetil equtio (p(x)u ) + r(x)u = f(x), for x (, b), with p C 1 [, b], p(x) c 0 > 0 for ll x [, b], r C 0 [, b], r(x) 0 for ll x [, b] d f L 2 [, b], subject to the boudry coditios p()u () + α u() = A, p(b)u (b) + β u(b) = B, where α d β re positive rel umbers d A d B re rel umbers. ) Show tht the wek formultio of the boudry vlue problem is where d fid A(u, v) = u H 1 (, b), such tht A(u, v) = l(v) for ll v H 1 (, b), [p(x)u (x)v (x) + r(x)u(x)v(x)] dx + αu()v() + βu(b)v(b), l(v) = f, v + Av() + Bv(b). We multiply the boudry vlue problem o both sides by test fuctio v H 1 (, b) d itegrte betwee d b to obti [ (p(x)u ) v + r(x)uv] dx = fv dx, 1 The lerig outcome is published o the course webpge d o the officil descriptio of the course.

2 TMA4212 Numericl PDEs Pge 2 of 7 d itegrtig by prts we get ( pu )v b [ (p(x)u )v r(x)uv] dx = fv dx, [(p(x)u )v + r(x)uv] dx + [p()u ()v() p(b)u (b)v(b)] = usig the boudry coditios d rerrgig the terms we get [(p(x)u )v + r(x)uv] dx + αu()v() + βu(b)v(b) = fv dx, fv dx + Av() + Bv(b), ledig to the proposed defiitio of A(u, v) (left hd side) d l(v) (right hd side). b) Costruct fiite elemet pproximtio of the boudry vlue problem bsed o this wek formultio usig piecewise lier fiite elemet bsis fuctios o the subdivisio = x 0 < x 1 < < x 1 < x = b o the itervl [, b]. Show tht the fiite elemet method gives rise to set of + 1 simulteous lier equtios with + 1 ukows u i = u h (x i ), i = 0, 1,...,. Show tht this lier system hs uique solutio. Commet o the structure of the mtrix M R (+1) (+1) of the lier system: Is M symmetric? Is M positive deiite? Is M tridigol? Expli your swers. We cosider + 1 piecewise lier bsis fuctios φ 0,..., φ such tht φ j (x j ) = 1, j = 0,..., : x x j 1, x [x h j 1, x j ), x φ j (x) = j+1 x, x [x h j, x j+1 ], j = 1,..., 1 0 otherwise, d φ 0 = { x1 x h, x [x 0, x 1 ), 0 otherwise, The fiite elemet spce is defied by φ = { x x 1 h, x [x 1, x ], 0 otherwise. S h := {w h H 1 w h = w j φ j }, j=0 d the Glerki method is Fid u h S h s.t. A(u h, v h ) = l(v h ), v h S h.

3 TMA4212 Numericl PDEs Pge 3 of 7 Therefore u h c be writte s u h = j=0 u jφ j, d oticig tht it is ecessry d sufficiet to impose A(u h, φ i ) = l(φ i ), i = 0,..., to determie u h, usig the bilierity of A d the lierity of l we get d lettig we get the lier system A(φ j, φ i )u j = l(φ i ), j=0 i = 0,..., M i,j := A(φ j, φ i ), b i := l(φ i ), Mu = b, with M the fiite elemet mtrix ( + 1) ( + 1). M is symmetric becuse M i,j = A(φ j, φ i ) = A(φ i, φ j ) = M j,i. M is positive defiite, i fct cosider the vector v R +1 d v 0 the v T Mv = v i M i,j v i = v i A(φ j, φ i )v i i j i j d usig the bilierity of the mp we get v T Mv = A(v, v), where v = j=0 v jφ j H 1. Further we hve v T Mv = A(v, v) pv v dx > c 0 (v ) 2 dx + αv() 2 + βv(b) 2, d sice v 0 (lmost everywhere) d v H 1, d x v (x) dx = v(x) v() the v is ot zero (lmost everywhere) uless v(x) = v() 0 for lmost every x so either or v(x) = v() lmost everywhere i [, b] d (v ) 2 dx > 0, α v() 2 > 0, we c coclude tht v T Mv > 0

4 TMA4212 Numericl PDEs Pge 4 of 7 d M is positive defiite. To uderstd if M is tridigol we must cosider the etries M i,j = A(φ j, φ i ), i j > 1, the result follows fter showig tht φ i or φ j re ot simulteously differet from zero i y of the subitervls [x l, x l+1 ] for ll l whe i j > 1, d the sme is true for their derivtives. Problem 2 Lerig outcome L5, L7 Cosider Lplce equtio i 2D o squre domi Ω = [0, 1] [0, 1], with Dirichlet boudry coditios, u xx + u yy = 0, u = g, o Ω. Use the 5-poits formul to discretize the equtio o the grid h = 1 M+1, x i = i h, i = 1,..., M, d y j = j h, j = 1,..., M d obti lier system of lgebric equtios A u = b, where u is vector whose compoets re the umericl pproximtios of u o the grid of the discretiztio u i,j u(x i, y j ), ordered s follows: u := [u 1,1, u 2,1,..., u M,1, u 1,2,..., u M,2,..., u 1,M,..., u M,M ] T. Cosider the use of the cojugte grdiet method to solve this lier system. Recll the followig formul for the eigevlues of the discrete Lplci: µ j,l = 2 h (cos (πjh) 1) + 2 (cos (πlh) 1), j = 1,..., M, l = 1,..., M. 2 h2 ) Write out A explicitly. Show tht A is positive defiite mtrix whose eigevlues lie i itervl [, b] with πh pproximtely. b 2 A is M 2 M 2 with M blocks B M M log the digol ech of which is of the form 4 1 B = h

5 TMA4212 Numericl PDEs Pge 5 of 7 log the sub d super digol there re M 1, M M idetity blocks scled with 1 h 2. The eigevlues of A re µ j,l oe fids mi( µ j,l ) = mx j,l j d mi( µ j,l ) = 4 j,l h (1 cos(πh)) = 4 π 2 h 2 2 h (1 cos(πjh)) = mx h2 j = 2π 2, 8 h 2 si(π/2jh)2 = 8 h 2 si(π/2mh)2 = b 8 h 2, b πh 2. b) Assume you re give rbitrry iitil guess x 0. Use the Theorem for the covergece of the cojugte grdiet method to estimte the umber of itertios K ecessry to obti reltive error below the tolerce ε, tht is Recll tht for vector v, e K A e 0 A ε. v A := v T Av. Fid estimte for K ssumig M + 1 = 100 d ε = We use the theorem bout the covergece of the cojugte grdiet method i chpter 14.5 i the book by Strikwerd d we hve e K A 2 ( b ) K e 0 A ( b + ) = 2(1 b )K K (1 + = 2(1 πh 2 )K b )K (1 + πh d to fid K we impose the coditio 2 )K Tkig logrithms we get πh (1 2 2 )K = ε. (1 + πh )K 2 K = log ε 2 (log(1 πh πh ) log(1 + )). 2 2 Isertig the prescribed vlues for ε d M + 1 we obti the estimte K = 536.

6 TMA4212 Numericl PDEs Pge 6 of 7 Problem 3 Lerig outcome L1, L4, L7 Cosider the lier PDE u t + u xxx = 0, u(, t) = u(, t) = 0, x R, t 0. Cosider the itervl [ L, L] with L > 0 sufficietly lrge, cosider the grid x m = L+hm, h = 2L/M, m = 0,..., M. Discretise with cetrl fiite differeces i spce d the Bckwrd Euler method i time (bckwrd differeces i time), let u(x 0, t) = u(x M, t) = 0. Use the followig cetrl differeces pproximtio of the third derivtive u xxx (xm,t) = u(x m+3, t) 3u(x m+1, t) + 3u(x m 1, t) u(x m 3, t) 8h 3 + O(h 2 ). Show tht the obtied method is Vo Neum stble. Usig cetrl differeces i spce d bckwrd Euler i time we obti ledig to U +1 m = U m + k 8h 3 ( U +1 m 3 3U +1 m 1 + 3U +1 m+1 U +1 m+3), U +1 m + k 8h 3 ( U +1 m 3 + 3U +1 m 1 3U +1 m+1 + U +1 m+3) = U m, by ssumig U m = ξ e iβxm, d i = 1, d substitutig i the method we obti ξ = iα, with α = k (si(3βh) 3 si(βh)). 8h3 Filly we obti ξ ξ = ξ 2 = iα 1 iα = α 1, 2 implyig Vo Neum stbility. Problem 4 Lerig outcome L1, L4, L7 Let r d α be positive rel umbers. Show tht if [w 1,..., w M ] T stisfies rw m 1 + (1 + 2r + α)w m rw m+1 = v m, 1 m M 1, w 0 = w M = 0, the mx w m mx v m. 0 m M 1 m M 1

7 TMA4212 Numericl PDEs Pge 7 of 7 Use this to show tht Bckwrd Euler coverges for rbitrry r = k/h 2 o the problem u t = u xx u, u(x, 0) = f(x), u(0, t) = g 0 (t), u(1, t) = g 1 (t). Hit: You will hve to tke w m = e +1 m where e m := Um u(x m, t m ) is the error, d Um is the umericl pproximtio give by the Bckwrd Euler method. We hve d the d the (1 + 2r + α)w m = rw m 1 + rw m+1 + v m (1 + 2r + α) mx w m 2r mx w m + mx v m, 1 m M 1 m M 1 m M 1 Cosiderig Bckwrd Euler we hve mx w m mx v m. 1 m M 1 m M 1 ru +1 m 1 + (1 + 2r + k)u +1 m ru +1 m+1 = U m d the error e m := U m u(x m, t ) stisfies the equtio re +1 m 1 + (1 + 2r + k)e +1 m re +1 m+1 = e m kτ m where τm = (O(k) + O(h 2 )). Usig the property prove t the begiig of the exercise, with w m = e +1 m d v m = e m kτm mx 1 m M e+1 m mx 1 m M e m + k mx τ m k m j=1 mx τ m j k mx τ m T mx τ m m m, m, d sice the method is cosistet (i fct for Bckwrd Euler mx m, τ m = O(k)+O(h 2 )) so we get mx m e m T (O(k) + O(h 2 )), m d the lst iequlity implies covergece.