Variational Formulation of Boundary Value Problems

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1 Wht is α =3 Dα u? Chpter 4 Vritionl Formultion of Boundry Vlue Problems 4.1 Elements of Function Spces Spce of Continuous Functions N is set of non-negtive integers. 1 An n-tuple α =α 1,, α n N n is clled multi-index. 2 The length of α is 3 0 = 0,, 0. Set D α := / x 1 α1 / x n αn. α := α j. Exmple Assume n =3, α =α 1, α 2, α 3 N 3, ux 1,x 2,x 3 :R 3 R. 29 j=1 α = 3 = 3 α j = 3. j=1 = α =3, 0, 0, 0, 3, 0, 0, 0, 3, 2, 1, 0, 2, 0, 1, 0, 2, 1, = α =3 1, 2, 0, 1, 0, 2, 0, 1, 2, 1, 1, 1. D α u = 3 u x u x u x u 3 x 2 1 x + 3 u 2 x 2 1 x + 3 u 3 x 2 2 x u x 1 x u 2 x 1 x u 3 x 2 x u +. x 1 x 2 x 3 This sort of list cn get very long. Hence D α is useful nottion. Definition Let be n open set in R n. Let k N. Define spces C k, C k nd C by C k := {u : R D α u is continuous in for ll α k}, C k := {u : R D α u is continuous in for ll α k}, C := {u : R D α u is continuous in for ll α N n }, where is the closure of. If is bounded, =, where is the boundry of. We denote C =C 0 nd C =C 0. Exmple Set I := 0, 1 nd ux := 1/x 2 x I. Then clerly for ll k 0 u C k I. However, in I = [0, 1] u is not continuous t 0. Thus, u/ CI. Definition For bounded open set R n, k N nd u C k, the norm u C k is defined by u C k := sup D α ux. α k x Exmple Let I = 0, 1, ux := x, u CI. Then, sup x I ux =1. Definition For n open set R n nd u C the support of u denoted by supportu R n is defined by supportu := the closure of {x ux 0}. Remrk. The support of u is the smllest closed subset of such tht u = 0 in \supportu. 30

2 Exmple Let 0=x 0 <x 1 < <x n =1be ptition of [0, 1]. Define φ j x : [0, 1] R by x x j 1 x x j 1,x j, h φ j x := x j+1 x x x j,x j+1, h 0 elsewhere. Then we see φ j CI nd supportφ j =[x j 1,x j+1 ]. 2 Define wx :R n R by 1 wx := e 1 x 2 x < 1, 0 otherwise. Then we see w C R n nd supportw = {x R n x 1}. Definition Define C k 0 Ck by C k 0 := {u C k supportu is bounded subset of } Spces of Integrble Functions Definition Let denote n open subset of R n nd ssume 1 p<. We define spce of integrble functions L p by { } L p := v : R vx p < +. The spce L p is Bnch spce with norm L p defined by 1/p v L p = vx p. Especilly the spce L 2 is Hilbert spce with inner product, L 2 defined by u, v L 2 := uxvx nd norm L 2 defined by u L 2 := u, v L We hve Minkowski s inequlity s follows. For u, v L p, 1 p< u + v L p u L p + v L p. We lso hve Hölder s inequlity. For u L p nd v L q, 1 p, q < with 1/p +1/q =1 uxvx u L p v L q. Now, ny two integrble functions re equl if they re equl lmost everywhere, tht is, they re equl except on set of zero mesure. Strictly speking, L p consists of equivlent clsses of functions. Exmple Let u, v : 1, 1 R be { 1 x 0, 1, ux = 0 x 1, 0], { 1 x [0, 1, vx = 0 x 1, 0, The functions u nd v re equl lmost everywhere, since the set {0} where u0 v0 hs zero mesure in the intervl 1, 1. So u nd v re equl s integrble functions in 1, 1. Suppose tht u C k, where is n open set of R n. Let v C0. Then we see by integrtion by prts D α uxvx = 1 α uxd α vx, where α k. Definition A function u : R is loclly integrble if u L 1 U for every bounded open set U such tht U. Definition Suppose u : R is loclly integrble nd there is loclly integrble function w α : R such tht w α xφx = 1 α uxd α φx for ll φ C 0. Then the wek derivtive of u of order α denoted by D α u is defined by D α u = w α. 32

3 Note tht t most only one w α stisfies so the wek derivtive of u is welldefined. Indeed, the following DuBois-Rymond lemm shows such w α is unique. Lemm DuBois-Rymond Suppose is n open set in R n nd w : R is loclly integrble. If wxφx =0 for ll φ C0, then wx =0for.e x. 0 x 1, 1+x 1 x 0, ux := 1 x 0 x 1, 0 1 x. y x We will used for both clssicl nd wek derivtives. Exmple Let = R. Set ux = 1 x +, x, where Thus, { x x > 0, x + := 0 x 0. Clerly we see tht u is loclly integrble, u C nd u/ C 1. However, it my hve wek derivtive. Tke ny φ C0 nd α =1. Then, where 1 α uxd α φx = uxφ x 1 = 1 x φ x = 1 + xφ x 1 xφ x = 1 φx + 1φx 1 0 = wxφx, 0 x< 1, 1 1 <x<0, wx := 1 0 <x<1, 0 1 < x. Here we do not worry bout the points x = 1, 0, 1, since they hve zero mesure. Thus, u hs its wek derivtive Du = w. Definition Let k be non-negtive integer nd p [0,. W k,p defined by The spce W k,p := {u L p D α u L p α k} 33 34

4 is clled Sobolev spce. It is Bnch spce with the norm u W k,p := 1/p D α u p L p. α k Especilly, when p = 2, we denote H k s W k,2. It is Hilbert spce with the inner product u, v H k := D α u, D α v L 2. α k Of specil interest re H 1 nd H 2. If =, b, R, we see tht u, v H 1 = u, v L 2 + Du, Dv L 2 b b = uxvx + DuxDvx. u, v H 2 = u, v L 2 + Du, Dv L 2 + D 2 u, D 2 v L 2 b b b = uxvx + DuxDvx + D 2 uxd 2 vx. Remrk. 1 By using Hölder s inequlity, we cn prove Cuchy-Schwrz inequlity for the inner product of H k s follows. u, v H k D α u, D α v L 2 α k D α u L 2 D α v L 2 α k D α u 2 D L 2 α v 2 L 2 α k α k = u H k v H k. 2 Let =, b R nd u H 1. Then u C. In higher spce dimensions this sttement is no longer true. 4.2 One Dimensionl Problem: Dirichlet condition Find u : R such tht { d pdu +qu = f, x, ux =0, x. BVP Specifying the vlue of u t boundry points is sid to be Dirichlet boundry condition. Now the methodology is 1 multiply the eqution by test function, integrte by prts nd use boundry conditions ppropritely, 2 identify V,, nd l, 3 verify, if possible, the ssumptions of Lx-Milgrm. = Unique existence to the vritionl formultion of the BVP. Let φ : R be sufficiently smooth. We will cll φ our test function. Let us follow the methodology. 1 [ = fxφx = px dux φx d pxdux ] x=1 + x=0 px dux φx+qxuxφx dφx + qxuxφx. We wnt to eliminte the term [pxdux/φx] x=1 x=0, so we suppose tht the test function φ stisfies the sme Dirichlet conditions s u, i.e, φ0 = φ1 = 0. Then we hve tht px dux dφx + qxuxφx = fxφx for ny test function φ. We wnt u, v to be from the sme spce. For the term uφ to mke sense, we need u, φ L2. For the derivtives du/, dφ/ to mke sense, we tke this further, so u, φ H 1. Let = 0, 1, p,q C nd f L 2. Note tht = {x =0} { x = 1}. We consider the following problem Let us choose V := { φ H 1 v0 = v1 = 0 }, 36

5 where H 1 = { φ L 2 Dφ L 2 }. We equip the inner product, V :=, H1. Let us define u, v := pdudv + quv, lv := fv. Moreover, ssume tht px p 0 > 0, qx q 0 > 0 for ll x. 3 We will verify the ssumptions of Lx-Milgrm s theorem. i For φ V nd f L 2, we see by Cuchy-Schwrz inequlity tht lφ = fφ f L 2 φ L 2 f L 2 φ 2 L 2 + Dφ 2 L 2 = c l φ V, where we hve set c l := f L 2. Thus, l : V R is bounded. Clerly l is liner, i.e, lαφ + βψ =αlφ+βlψ for ny φ, ψ V nd α, β R. ii Obviously, : V V R is biliner. Moreover, is bounded. Indeed, φ, ψ pdφdψ + qφψ mx px DφDψ + mx qx φψ x x mx px Dφ L 2 Dψ L 2 + mx qx φ L 2 ψ L 2 x x C Dφ L 2 Dψ L 2 + φ L 2 ψ L 2 C Dφ 2 L 2 + φ 2 Dψ L 2 2 L 2 + ψ 2 L 2 = C φ H 1 ψ H 1, where we hve set C := mx{mx x 37 px, mx qx }. x The biliner form, is coercive, since for ll φ V φ, φ = p Dφ 2 + q φ 2 p 0 Dφ 2 + q 0 φ 2 = Ĉ φ 2 V, where we hve set Ĉ := min{p 0,q 0 }. We cn now pply Lx-Milgrm s theorem to see tht there uniquely exists solution to the following problem P Find u V such tht for ny φ V. pdudφ + quφ = fφ, Remrk. For V = H0 1 we cn use the norm V defined by φ 2 V = Dφ 2, since the following Poincre inequlity holds:- there exists C>0such tht φ 2 C Dφ 2 for φ V. By using this inequlity we cn prove the unique existence of the solution solving P with q 0 in the sme wy s bove. 4.3 One Dimensionl Problem: Neumnn condition Let = 0, 1, px,qx C nd fx L 2. We consider the following problem. Find u : R such tht d p du + qu = f, x, NBVP d ux =0, x. 38

6 Specifying the vlue of du/ t boundry points is sid to be Neumnn boundry condition. We ssume the sme conditions for p, q s before, i.e, px p 0 > 0, qx q 0 > 0 in. Let us derive the vritionl form. Tke sufficient smooth test function φ, multiply NBVP by φ nd integrte. fxφx = d px dux [ = px dux ] x=1 φx + px dux x=0 = px dux dφx + qxuxφx. φx+qxuxφx dφx + qxuxφx We hve eliminted the term [pxdux/φx] x=1 x=0 by tking into ccount the Neumnn boundry conditions dux/ = 0 for x =0, 1. Let us choose the functionl spce V := H 1 in this cse nd define u, v := pdudv + quv, lv := fv. The corresponding vritionl problem is tht:- P Find u V such tht for ny φ V. pdudφ + quφ = fφ, Agin the liner form l :V R nd the biliner form, :V V R stisfy the ssumptions of Lx-Milgrm s theorem, hence the problem P hs the unique solution. Remrk. Consider NBVP with q 0. Then we see fx = d pxdux ] x=1 =0. x=0 [ = px dux 39 Thus, we need to ssume fx = 0 s comptibility condition in this cse. In order to prove the unique existence of the solution, we need to modify the functionl spce. Let us define Hm 1 by Hm 1 := {φ H 1 φ =0}, nd equip the sme inner product s H 1. Agin Poincre s inequlity is vilble for this spce Hm, 1 i.e, for ll φ Hm 1 Dφ 2 C φ 2, where C > 0 is constnt. By using this inequlity we cn prove the unique existence of the solution solving P with q 0 nd V = Hm 1 in the sme wy s bove on the ssumption fx = One Dimensionl Problem: Robin/Newton Condition Let = 0, 1, px,qx C, fx L 2, δ,g 0,g 1 R be constnts. We consider the following problem. Find u : R such tht d p du + qu = f, x, p d u0 + δu0 = g 0, RNBVP p d u1 + δu1 = g 1. Let us derive the vritionl form. Tke sufficiently smooth φ. This kind of boundry condition is sid to be Robin/Newton boundry condition. We ssume the sme conditions for p, q s before, i.e, px p 0 > 0, qx q 0 > 0 in nd tht δ 0. fxφx = d px dux [ = px dux ] x=1 φx + px dux x=0 = g 1 + δu1φ1 g 0 δu0φ0 + px dux dφx + qxuxφx, 40 φx+qxuxφx dφx + qxuxφx

7 which is equl to px dux dφx + qxuxφx + δu1φ1 + δu0φ0 = fxφx g 1 φ1 + g 0 φ0. This suggests tht we should define,, l nd the functionl spce V s following. u, v := pdudv + quv + δu1v1 + δu0v0, lv := fv g 1 v1 + g 0 v0, V := H 1. As usul we need to show tht the biliner forms, l re bounded nd is coercive to estblish the unique solvbility of RNBV P. Let us ssume the following inequlity holds true for while. φx C φ H for ll φ H 1. Then it is esy to see tht,,l re bounded. Now φ, φ minp 0,q 0 φ 2 V + δφ1 2 + φ0 2 minp 0,q 0 φ 2 V = α φ 2 V, where α = minp 0,q 0. Hence the biliner form becomes coercive nd Lx- Milgrm s theorem ssures the unique existence of the solution. 4.5 One Dimensionl H 1 inequlities Here we derive some inequlities in the one dimensionl cse =, b. We define function spce H 1 by ɛ0 H 1 ɛ0 := {φ H1 φ =0}. The following inequlity is one exmple of Poincré-Friedrichs inequlity. 41 Proposition For ll φ H 1 ɛ0, φ L b Dφ L 2. Proof. We cn write tht for x b x φx = Dφηdη. Then we see tht b φ 2 L 2 = φx 2 b x 2 = Dφηdη b x x 1 2 Dφη 2 dη b x = x Dφη 2 dη b b x Dφη 2 dη = 1 2 b 2 Dφ 2 L 2,b. Exmple We cn pply this inequlity to prove the unique solvbility of the Dirichlet problem with functionl spce V, biliner form, nd liner functionl l defined by V := {φ H 1 0, 1 φ0 = φ1 = 0}, 1 u, v := pdudv for u, v V, 0 1 lu := fu for u V. 0 We only check tht the biliner form is bounded nd coercive. We see tht u, v sup px Du L 2 Dv L 2 sup px u H 1 v H 1 x 0,1 x 0,1 42

8 nd v, v p 0 Dv 2 L 2 = p 0 2 Dv 2 L 2 + p 0 2 Dv 2 L 2 p 0 2 α v 2 V, where α := p 0 min1, 2/b 2 /2. Dv 2 L v 2 L 2 b 2 Proposition The following Agmon s inequlity holds. For ll φ H 1 ɛ0 Proof. which gives the inequlity. Noting tht this Agmon s inequlity yields or for ny φ H 1 ɛ0. mx φx 2 2 φ L 2 φ H 1. x x φx 2 dφη 2 = dη dη x =2 φηdφηdη x 2 b 2 2 φ L 2 φ H 1, 1/2 x φη 2 dη 1/2 Dφη 2 dη 1/2 b φη 2 dη Dφη 2 dη φ L 2 b mx x [,b] φx, 1/2 mx φx 2 2 b mx φx φ H x x [,b] 1, mx φx 2 b φ H 1 x Wek Solutions to Elliptic Problems The simplest elliptic eqution is Lplces eqution: u =0, 4.2 where := n j=1 2 is the Lplce opertor. A generl second order elliptic x 2 j eqution is: given bounded open set R n find u such tht: x j ij x u + b i x u + cxu = fx x, 4.3 where clssiclly i j C 1, i, j =1,..., n; b i C,i =1,..., n; c C; f C. For the eqution to be elliptic we require ij xξ i ξ j C ξi 2 ξ =ξ 1,..., ξ n R n, 4.4 where C >0 is independent of x, ξ. Condition 4.4 is clled uniform ellipticity. The eqution is usully supplemented with boundry conditions - Dirichlet, Neumnn, Robin, or mixed Dirichlet/Neumnn boundry. In the cse of the homogeneous Dirichlet problem u = 0 on u is sid to be clssicl solution provided u C 2 C. Elliptic theory tells us tht there exists unique clssicl solution provided ij,b i, c, f nd re sufficiently smooth. However we re only intersted in problems where the dt is not smooth, for exmple f = sign1/2 x, = 1, 1. This problem cn t hve u C 2 becuse u hs jump discontinuity re x = 1/2. With the help of functionl nlysis the existence/uniqueness theory for wek, vritionl solutions turn out to be esy nd is good for FEM. 4.7 Vritionl Formultion of Elliptic Eqution: Neumnn Condition Let be bounded domin in R n with smooth boundry. Let p, q C such tht px p 0 > 0, qx q 0 > 0 x, 44

9 nd f L 2. Find u : R such tht { p u+qu = f, x, u n =0, x, NBVP where n is the unit outwrd norml to the boundry. Note tht p u = xi = p u p 2 u x 2 + p u i = p u + p u, u = u n. n So we hve second order PDE. In one dimensionl problem, in order to derive the vritionl formultion we used integrtion by prts. Let us revise some formule relted to the integrtion by prts. Nottion: v v =,..., v T x 1 x 1 v = 2 v = v A = A i D 2 v ij = 2 v x j TrD 2 v= v. Theorem Divergence theorem. Let A : R n be C 1 vector field. The following equlity holds. A = A nds. 45 Remrk. Suppose A = fe i with the coordinte vector e i = 0,, 1,, 0 T, i.e, the jth component is {e i } j = δ i,j. Then we see A = f. So by the Divergence theorem f = fn i ds. In one dimensionl cse where =, b, = {, b}, Divergence theorem becomes b f = fb f. x Let us derive the integrtion by prts formul. Proposition Integrtion by prts. For A C 1 ; R n, g C 1, A g = ga nds g A. Proof. By Divergence theorem we see tht Ag = A ngds. Alterntively, Ag =g A + A g. By combining these equlity we get the desired formul. For exmple, if A = u nd g = v, we hve by integrtion by prts formul tht u v = v u nds v u. Noting tht u n = u nd u = u, n 46

10 where is Lplcin, we obtin v u = v u n ds u v. Looking t our boundry vlue problem p u+qu = f, we hve tht p uv = p u v p u n vds. Let v be sufficiently smooth test function. Multiply NBVP by v nd integrte using Divergence theorem. fv = p u+quv = p u v p u n vds + quv. Since u/ n = 0 on, we do not need to plce restriction on the test function v. So if u solve BVP, then for ny sufficiently smooth function v. p u v + quv = fv, Now to use Lx-Milgrm, we hve to set up V,, nd l. In order for the two inner products on the left hnd side to mke sense, we tke V = H 1, u, v = p u v + quv, lv = fv, for ll u, v V. Note tht V is rel Hilbert spce with the norm 1/2 v V = v H 1 = v 2 + v 2 nd obviously, :V V R is biliner nd l :V R is liner. Moreover 47 we observe v, v = p v 2 + qv 2 p 0 v 2 + q 0 v 2 min{p 0,q 0 } v 2 H 1 = α v 2 H 1, where we hve put α := min{p 0,q 0 }. Thus, is coercive. v, w = p v w + qvw p u w + qvw C v 2 + v 2 1/2 w 2 + w 2 1/2 1/2 1/2 C v 2 + v 2 w 2 + w 2 = C v H 1 w H 1. Therefore,, is bounded. Finlly let us check the boundedness of l. lv = fv f v 1/2 1/2 f 2 v 2 = f L 2 v L 2 f H 1 v H 1 = C l v H 1. Thus, l is bounded. Now we cn pply Lx-Milgrm to prove tht there exists unique solution u V to the following problem P. P Find u V such tht for ll v V. u, v =lv 48

11 4.8 Vritionl Formultion of Elliptic Eqution: Dirichlet Problem On the sme ssumptions on, p, q, f, we consider the following problem. Find u : R such tht { p u+qu = f, x, u =0, x, DBVP Let us derive the vritionl form of DBVP s in the section 2.5. Multiply DBVP by sufficiently smooth test function v nd integrte. Then we see fv = p u v + quv p u n v. Since we hve u 0 on, we hve to force our test function v to stisfy the sme condition; v 0 on. Then we obtin p u v + quv = fv for ny sufficient smooth function v with v 0 on. Set V := {v H 1 v = 0 on } = H 1 0. Note tht V is rel Hilbert spce with the inner product v, w V := v, w L 2 + v, w L 2 nd v V = v H 1. As before we define v, w := p v w + qvw, lv = fv. The sme rgument s the previous section shows tht, is coersive nd bounded bi-liner form nd l is bounded liner functionl on V. Therefore Lx- Milgrm s theorem tells us tht there uniquely exists solution to the vritionl problem of DBVP Inhomogeneous Boundry Condition Let V be Hilbert spce nd, be biliner coercive form on V V, let l be liner, let V 0 be closed subspce of V nd g V. Set V g = {v V : v = v 0 +g, v 0 V 0 } nd consider the problem: find u V g such tht u, v =lv v V 0. We cn show tht there exists unique solution. Let u 0 = u g. Then the problem becomes: find u 0 V 0 such tht u 0 =lv g, v v V For the finite element method it becomes: find u h such tht: M 1 u h = g 0 φ 0 + α j φ j + g 1 φ M. 4.6 j=1 this mens tht A is the sme s the homogeneous cse but b now hs contributions from g 0,g Second Order Elliptic Problems Consider the problem x j ij x u + b i x u + cxu = fx x 4.7 with u = 0 on. Multiply by test funciton nd integrte by prts in the second order term using the divergence theorem. The result is the wek vritionl form of the BVP: find u V such tht u, v =lv v V where V = H0 1 nd w, v := lv := ij x w v x j + b i x w vx+ cxwx, fxvx = f, v. We seek to pply the Lx-Milgrm theorem. Recll v, w H 1 0 = vw + v w = v, w + v, w. We hve three conditions to check to stisfy the theorem. 50

12 1 Is l bounded liner functionl? Clerly lαv + βw =f, αv + βw = αf, v+ βf, w = αlv+ βlw so l is liner functionl on V nd lv = fxvx f L 2 v L 2 f L 2 v H 1 0 where we hve used the Cuchy-Schwrtz inequlity nd thus l is bounded. 2 Is, bounded? Assume tht ij L, b i L, c L re ll bounded for ll i, j nd tht f L 2. Then w, v ij w xi v xj + b i w xi v + cwv mx ijx w xi v xj + mx b ix w xi v + mx cx w v x x x c w xi v xj + w xi v + w v c w xi v xj + w xi v + w v c w V v V + w V v V + w V v V = c 1 w V v V where c = mx{mx ij mx ij x, mx i mx b i x, mx cx } nd c 1 = cn 2 + n Is, coercive? The crucil ssumption is tht the ij stisfies the ellipticity ssumption ij xξ i ξ j c ξi 2 ξ 1,..., ξ n R n, x, 4.8 i.e. for ll x we must hve ξ T Axξ ĉ ξ 2 = ˆξ T ξ We lso ssume tht cx 1 2 b i x 0 x Then v, v = ij xv xi v xj + b i xv xi v + cxvx 2 c v 2 + b xi i x v2 /2 + cv 2. The middle integrl here is 1 2 b v2, which fter integrtion by prts equls 1 2 v2 b so tht v, v c v 2 + v 2 cx 1 xi 2 bx c v 2 xi = c v 2 Note tht we need b L for this to work. We wish to show tht Recll the Poincre-Friedrichs inequlities Hence v, v c 0 v 2 V = c 0 v + v v 2 c v 2 v H 1 0. v, v c v 2 c v c 1 2 v, v+1 2 v, v c 2 v 2 + c v 2c c 0 v 2 + v Remrks on the Lx-Milgrm Result 1. Uniqueness: by our usul methods this follows from the linerity of l, the bilinerity of, nd the coercivity of,. 52

13 2. Stbility estimte: we know tht c 0 u 2 V u, u =lu c 2 u V so we cn deduce tht the solution to our BVP stisfies u H 1 1 c 0 f L Continuity with repsect to l. Consider the two problems Then Choosing v = u 1 u 2 : u 1 V s.t. u 1,v=l 1 v v V u 2 V s.t. u 2,v=l 2 v v V. u 1 u 2 =l 1 v l 2 v =ˆlv c 0 u 1 u 2 2 V = ˆlu 1 u 2 l 1 l 2 V u 1 u 2 u 1 u 2 V l 1 l 2 V c 0 In terms of our originl elliptic bvp s we hve tht u 1 u 2 H 1 1 c 0 f 1 f 2 L If l is the zero element of V i.e. lv =0 v V then 0 = u, u u V = 0 by coercivity nd u = Inhomogeneous Boundry Conditions Consider the elliptic problem p u+qu = f x 4.15 u = g x 4.16 where is bounded open subset of R 2. We ssume tht the dt p, q, f, g re sufficiently smooth nd tht p M px p 0 > 0 q M qx q 0 > 0 53 x x Let v be test function. Multiply by v nd integrte: 0= v p u+ quv fv = I 1 + I 2 + I 3 Nov choosing ϕ = v, f = p u in: ϕf =ϕ f + ϕ f vp u =v p u+ v p u I 1 = vp u v p u = p v u vp u ν Choosing v = 0 on we hve I 1 = p v u. Thus 0= p v u + quv fv v H Set V 0 = H 1 0,u, v = p v u + quv, lv = fv. Note tht u V 0. However g H 1 so u g H 1 nd u g H 1 0 =V 0, i.e u V g := {w V = H 1 :w = g + v, v V 0 }. Thus our vritionl problem P is to find u V g such tht u, v =lv v V 0. Observe tht V 0 is liner spce but V g = g + V 0 is n ffine spce. We cn t pply Lx-Milgrm directly. Consider u = u g V 0 : so u V 0 solves u + g, v =u, v =lv v V 0 u,v=lv g, v =: l v v V 0 Now we just need to check Lx-Milgrm for this problem. Clerly, is biliner nd symmetric. Coercivity: v, v = p v 2 + qv 2 p 0 v 2 + q 0 v 2 minp 0,q 0 v 2 + v 2 c 0 v H 1. Boundedness: using the Cuchy-Schwrtz inequlity we hve w, v = p w v + qwv p M w v + q M w v mxp M,q M w v + w v c w H 1 v H 1. 54

14 Clerly l is liner nd l v = lv g, v lv + g, v f v + c g H 1 v H 1 L v H 1. Thus there exists unique u nd we conclude therefore tht there exists unique u = u + g. The biliner form is symmetric so there is n energy nd ssocited minimistion problem: Jv = 1 v, v lv 2 Find u V g s.t. Ju Jv v V g. Exercise: Prove tht these two problems re equivlent Finite Element Method in 2D Tke to be polygon. Let T h be tringultion of, T h = {κ} nd set h κ = dim κ the length of the longest side, h = mx dim κ. We ssume tht T h <. Any tringles in T h must intersect long complete edge, t vertex of not t ll. Note tht ny liner function on R 2 is of the form vx, y =+bx+cy nd is defined by three prmeters. Thus ny function v h V h := {η C :v h κ is liner} 4.18 is uniquely determined by it vlues t the vertices of the tringultion: φ i x j =δ ij i, j =1,..., N h,x j is tringle vertex The support of the bsis functions is locl, so A will gin be sprse. Exmple Find u H 1 such tht p u v + quv = fv v H A finite element method pplied to this yields the problem: find u h H 1 such tht p u h v h + qu h v h = fv h v h V h