FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS

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FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONSLONG CHENAs a model problem of general parabolic equations, we shall consider the followingheat equation and study corresponding finite element methods in Ω (0, T ), ut u fu 0on Ω (0, T ),(1) u(·, 0) u0 in Ω.Here u u(x, t) is a function of spatial variable x Ω Rn and time variable t (0, T ). The ending time T could be . The Laplace operator is taking with respectto the spatial variable. For the simplicity of exposition, we consider only homogenousDirichlet boundary condition and comment on the adaptation to Neumann and other typeof boundary conditions. Besides the boundary condition on Ω, we also need to assignthe function value at time t 0 which is called initial condition. For parabolic equations,the boundary Ω (0, T ) Ω {t 0} is called the parabolic boundary. Thereforethe initial condition can be also thought as a boundary condition of the space-time domainΩ (0, T ).1. VARIATIONAL FORMULATION AND ENERGY ESTIMATEWe multiply a test function v H01 (Ω) and apply the integration by part to obtain avariational formulation of the heat equation (1): given an f L2 (Ω) (0, T ], for anyt 0, find u(·, t) H01 (Ω), ut L2 (Ω) such that(2)(ut , v) a(u, v) (f, v), for all v H01 (Ω).where a(u, v) ( u, v) and (·, ·) denotes the L2 -inner product.We then refine the weak formulation (2). The right hand side could be generalized tof H 1 (Ω). Since map H01 (Ω) to H 1 (Ω), we can treat ut (·, t) H 1 (Ω) for afixed t. We thus introduce the Sobolev space for the time dependent functions!1/qZTLq (0, T ; W k,p (Ω)) : {u(x, t) kukLq (0,T ;W k,p (Ω)) : 0ku(·, t)kqk,p dt }.Our refined weak formulation will be: given f L2 (0, T ; H 1 (Ω)) and u0 H01 (Ω),find u L2 (0, T ; H01 (Ω)) and ut L2 (0, T ; H 1 (Ω)) such that hut , vi a(u, v) hf, vi, v H01 (Ω), and t (0, T ) a.e.(3)u(·, 0) u0where h·, ·i is the duality pair of H 1 (Ω) and H01 (Ω). We assume equation (3) is wellposed. For the existence and uniqueness of the solution [u, ut ], we refer to [3]. In mostplaces, we shall still use the formulation (2) and assume f, ut L2 (Ω) so that the dualitypair is realized by the L2 inner product (·, ·).1

2LONG CHENRemark 1.1. The topology for the time variable should also be treat in L2 sense. But in(2) and (3) we still pose the equation point-wise (almost everywhere) in time. In particular,one has to justify the point value u(·, 0) does make sense for an L2 type function whichcan be proved by the regularity theory of the heat equation. To easy the stability analysis, we treat t as a parameter and the function u u(x, t) asa mappingu : [0, T ] H01 (Ω),defined asu(t)(x) : u(x, t)(x Ω, 0 t T ).With a slight abuse of notation, we still use u(t) to denote the map. The norm ku(t)k orku(t)k1 is taken with respect to the spatial variable and thus becomes a function of time.We then introduce the operatorL : L2 (0, T ; H01 (Ω)) L2 (0, T ; H 1 (Ω)) L2 (Ω)as(Lu)(·, t) t u u in H 1 (Ω), for t (0, T ] a.e.(Lu)(·, 0) u(·, 0).Then the equation (3) can be written asLu [f, u0 ].Here we explicitly include the initial condition u0 . The spatial boundary condition is buildinto the space H01 (Ω). In most places, when it is clear from the context, we also use L forthe differential operator only.We shall prove several stability results of L which are known as energy estimates in [5].Theorem 1.2 (Energy estimates for the heat equation). Suppose [u, ut ] is the solution of(2) and ut L2 (0, T ; L2 (Ω)), then for t (0, T ] a.e.Z t(4)ku(t)k ku0 k kf (s)k ds0(5)ku(t)k2 tZ u(s) 21 ds ku0 k2 0(6) u(t) 21 Z0tZkf (s)k2 1 ds,0tkut (s)k2 ds u0 21 Ztkf (s)k2 ds.0Proof. The solution is defined via the action of all test functions. The art of the energyestimate is to choose an appropriate test function to extract desirable information.We first choose v u to obtain(ut , u) a(u, u) (f, u).We manipulate these three terms as:Z1 dd1 2(u )t kuk2 kuk kuk; (ut , u) 22dtdtΩ a(u, u) u 21 ;1 (f, u) kf kkuk or (f, u) kf k 1 u 1 (kf k2 1 u 21 ).2

FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS3The inequality (4) is an easy consequence of the following inequalitykukFromdkuk kf kkuk.dt1 d1kuk2 u 21 (kf k2 1 u 21 ),2 dt2we getdkuk2 u 21 kf k2 1 .dtIntegrating over (0, t), we obtain (5).The last energy estimate (6) can be proved similarly by choosing v ut and left as anexercise. From (5), we can obtain the stability of the operator LL : L2 (0, T ; H01 (Ω)) L2 (0, T ; H 1 (Ω)) L2 (Ω)askuk2L2 (0,T ;H 1 (Ω)) ku0 k2 kf k2L2 (0,T ;H 1 (Ω)) .0Since the equation is posed a.e for t, we could also obtain the maximum-norm estimate intime. For example, (4) can be formulated askukL (0,T ;L2 (Ω)) ku0 k kf kL1 (0,T ;L2 (Ω)) ,and (6) implieskukL (0,T ;H01 (Ω)) u0 1 kf kL2 (0,T ;L2 (Ω)) .Exercise 1.3. Prove the energy estimate(7)kuk2 e λt ku0 k2 Zte λ(t s)) kf k2 1 ds,0where λ λmin ( ) 0. The estimate (7) shows that the effect of the initial data isexponential decay.2. F INITE ELEMENT METHODS : SEMI - DISCRETIZATION IN SPACE2.1. Semi-discretization in space. Let {Th , h 0} be a quasi-uniform family of triangulations of Ω and Vh H01 (Ω) be a finite element space based on Th . The semidiscretized finite element method is: given f V0h (0, T ], u0,h Vh H01 (Ω), finduh L2 (0, T ; Vh ) such that ( t uh , vh ) a(uh , vh ) hf, vh i, vh Vh , t R .(8)uh (·, 0) u0,hThe scheme (8) is called semi-discretization since uh is still a continuous (indeed differential) function of t. The initial condition u0 is approximated by u0,h Vh .Take Vh as the linear finite element space as an example. We can expand uh PNi 1 ui (t)ϕi (x), where ϕi is the standard hat basis at the vertex xi for i 1, · · · , N ,the number of interior nodes, and the corresponding coefficient ui (t) now is a function oftime t. The solution uh can be computed by solving an ODE system(9)M u̇ Au f ,

4LONG CHENwhere u (u1 , · · · , uN ) is the coefficient vector, M , A are the mass matrix and thestiffness matrix, respectively, and f (f1 , · · · , fN ) with fi (f, ϕi ).When the linear finite element is used, one can use three vertices quadrature rule i.e.3Z1Xg(x) dx g(xi ) τ .3 i 1τThen the mass matrix becomes diagonal M diag(m1 , · · · , mN ). This is known as themass lumping. For 2-D uniform grids, mi h2 and A is the five point stencil discretizationof . Therefore (9) can be interpret as a rescaled finite difference discretization at eachvertex and the ODE system (9) can be solved efficiently by mature ODE solvers.2.2. Setting for the error analysis. We shall apply our abstract error analysis developedin Unified Error Analysis to estimate the error u uh in certain norms. The setting is R 1/2T X L2 (0, T ; H01 (Ω)), and kukX 0 u(t) 21 dt R 1/2T Y L2 (0, T ; H 1 (Ω), and kf kY 0 kf (t)k2 1 dt R 1/2T Xh L2 (0, T ; Vh ), and kuh kXh 0 uh (t) 21 dt R 1/2T Yh L2 (0, T ; V0h ), and kfh kYh 0 kfh (t)k2 1,h dt. Recall the dualnormhfh , vh i, for fh V0h .kfh k 1,h supvh Vh vh 1 Ih Rh (t) : H01 (Ω) Vh is the Ritz-Galerkin projection, i.e., Rh u Vh suchthat vh Vh .a(Rh u, vh ) a(u, vh ), Πh Qh (t) : H 1 (Ω) V0h is the projectionhQh f, vh i hf, vh i, vh Vh . Ph : Xh X is the natural inclusion L : X Y L2 (Ω) is Lu t u u, Lu(·, 0) u(·, 0), andLh L Xh : Xh Yh Vh .We summarize the setting in the following diagramLX Ry hY Q,y hLXh h Yhwhich is not commutative and the difference is the consistency error kQh Lu Lh Rh ukYh .The discrete equation we are solving isLh uh Qh f,uh (·, 0) u0,h .in V0h , t (0, T ] a.e.

FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS52.3. Stability. Adapt the proof of the energy estimate for L, we can obtain similar stabilityresults for Lh . The proof is almost identical and thus skipped here.Theorem 2.1 (Energy estimate for finite element discretization). Suppose uh satisfy Lh uh fh , uh (·, 0) u0,h , thenZ t(10)kfh (s)k dskuh (t)k ku0,h k 0Z tZ tkuh (t)k2 uh (s) 21 ds ku0,h k2 kfh (s)k2 1,h ds,(11)00Z tZ t uh (t) 21 (12)k t uh (s)k2 ds u0,h 21 kfh (s)k2 ds.00Note that in the energy estimate (11), the dual norm k · k 1 is replaced by a weaker onek · k 1,h since we can apply the inequalityhfh , uh i kfh k 1,h uh 1 .The weaker norm k · k 1,h can be estimated bykf k 1,h kf k 1 Ckf k.2.4. Consistency. Recall that the consistency error is kQh Lu Lh Rh ukYh . The choiceof Ih Rh simplifies the consistency error analysis.Lemma 2.2 (Error equation). For the semi-discretization, we have the error equationLh (uh Rh u) Qh (I Rh )ut , t 0, in V0h ,(13)(uh Rh u)(·, 0) u0,h Rh u0 .(14)Proof. Let A . By our definition of consistency, the error equation is: for t 0Lh (Rh u uh ) Lh Rh u Qh Lu t (Rh u Qh u) (ARh u Qh Au).The desired result then follows by noting that ARh Qh A in V0h and Qh Rh Rh . The error equation (13) holds in V0h which is a weak topology. The motivation to chooseIh Rh is that in this weak topologyhARh u, vh i ( Rh u, vh ) ( u, v) hAu, vh i hQh Au, vh i,or simply in operator formARh Qh A.This technique is firstly proposed by Wheeler [6].Apply the stability to the error equation, we obtain the following estimate on the discreteerror Rh u uh .Theorem 2.3 (Stability of discrete error). The solution uh of (8) satisfy the following errorestimateZ t(15)kRh u uh k ku0,h Rh u0 k kQh (I Rh )ut k ds0(16)kRh u uh k2 Zt (uh Rh u) 21 ds ku0,h Rh u0 k2 0(17) Rh u uh 21 ZtkQh (I Rh )ut k2 1,h ds,0Zt2k t (Rh u uh )k ds 0 u0,h Rh u0 21 Z0tkQh (I Rh )ut k2 ds.

6LONG CHENWe then estimate the two terms u0,h Rh u0 and Qh (I Rh )ut involved in these errorestimates. We use the linear element as an example since it is the most commonly choice.The first issue is on the choice of u0,h . An optimal one is obviously u0,h Rh u0 so thatno error coming from approximation of the initial condition. However, this choice requiresthe inversion of a stiffness matrix which is not cheap. A simple choice would be the nodalinterpolation, i.e. u0,h (xi ) u0 (xi ) or any other choice with optimal approximationpropertyku0,h Rh u0 k ku0 u0,h k ku0 Rh u0 k . h2 ku0 k2 ,(18)and similarly u0,h Rh u0 1 . hku0 k2 .According to (7) in Exercise 1.3, the effect of the initial boundary error will be exponentially decay to zero as t goes to infinity. So in practice, we can choose the simple nodalinterpolation.On the estimate of the second term, assume ut H 2 (Ω) and H 2 -regularity result holdfor Poisson equation (for example, the domain is smooth or convex), then(19)kQh (I Rh )ut k k(I Rh )ut k . h2 kut k2 .The negative norm can be bounded by the L2 -norm askQh (I Rh )ut k 1,h kQh (I Rh )ut k 1 CkQh (I Rh )ut k . h2 kut k2 .When using quadratic and above polynomial, we can prove a stronger estimate for thenegative norm and will be discussed in Section 2.6.2.5. Convergence. The convergence of the discrete error comes Rh u uh from the stability and consistency.Theorem 2.4 (Convergence of the discrete error). Suppose the solution u to (3) satisfyingut L2 (0, T ; H 2 (Ω)) and the H 2 -regularity holds for the Poisson equation. Let uh bethe solution of (8) with u0,h satisfying (18). We then have Z t(20) Rh u uh 1 kRh u uh k Ch2 ku0 k2 kut k2 ds .0To estimate the true error u uh , we need the approximation error estimate of theprojection Rh ; see Introduction to Finite Element Methodsh 1 ku Rh uk u Rh u 1 Chkuk2 .Theorem 2.5 (Convergence of the discretization error). Suppose the solution u to (3) satisfying ut L2 (0, T ; H 2 (Ω)). Then the solution uh of (8) with u0,h having optimalapproximation property (18) satisfy the following optimal order error estimate: Z t 12(21)h u uh 1 ku uh k Ch ku0 k2 kut k2 ds .02.6. *Superconvergence and error estimate in the maximum norm. The error eh Rh u uh satisfies the evolution equation (13) with eh (0) 0 with the chose u0,h Rh u0(22) t eh Ah eh τh ,where τh : Qh t (Rh u u). ThereforeZ teh (t) exp( Ah (t s))τh ds.0

FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS7Due to the smoothing effect of the semi-group e Ah t , we have the following estimate.Here we follow the work by Garcia-Archilla and Titi [4].Lemma 2.6 (Smoothing property of the heat kernel). For τh Vh , we haveZ texp( Ah (t s))Ah τh ds C log h max kτh k.(23)max0 t T0 t T0Proof. Let λmin and λmax be the minimal and maximal eigenvalue of Ah . Then it is easyto check λmax (t s) if (t s) λ 1max , λmax e Ah (t s) 1 1eAh (t s)if λmax (t s) λ 1min , λmin (t s)λmin eif (t s) λ 1.minNote that λmin O(1) and λmax Ch 2 . We getZ te Ah (t s) Ah τh ds C log h max kτh k.max0 t T0 t T0 Theorem 2.7 (Superconvergence in H 1 -norm). Suppose the solution u to (3) satisfyingut L (0, T ; H 2 (Ω)). Let uh be the solution of (8) with u0,h Rh u0 . Thenmax Rh u uh 1 C log h h2 max kut k2 .(24)0 t T0 t TWhen ut L2 (0, T ; H 2 (Ω)), then Rh u uh 1 Ch2(25) Zt 1/2.kut k22 ds0Proof. We multiply1/2Ahto (22) and apply Lemma 2.6 to getZ T1/21/2 eh (T ) 1 kAh eh (T )k e Ah (T s) Ah τh ds0TZe Ah (T s) Ah ds 0 1/2max kAh0 t Tτh k C log h h2 max kut k2 .0 t T 1/2In the last step, we have used the fact kAh τh k kτh k 1,h kτh k.To get (25), we use the energy estimate (17). Since the optimal convergent rate for u Rh u 1 or u uh 1 is only first order, thesecond order error estimate (24) and (25) are called superconvergence.To control the maximum norm, we use the discrete embedding result (for 2-D only)kRh u uh k C log h Rh u uh 1 ,and the error estimate of Rh in the maximum normku Rh uk C log h h2 kuk2, ,to obtain the following result.

8LONG CHENTheorem 2.8 (Maximum norm estimate for linear element in two dimensions). Supposethe solution u to (3) satisfying u L (0, T ; W 2, ) and ut L2 (0, T ; W 2, ) or ut L (0, T ; W 2, (Ω)). Let uh be the solution of (8) with u0,h Rh u0 . Then in twodimensions" Z t 1/2 #22(26)k(u uh )(t)k C log h h kuk2, kut k2 ds,0(27)2 2max k(u uh )(t)k C log h h max [ku(t)k2, kut (t)k2 ] .0 t T0 t TFor high order elements, we could get superconvergence in L2 norm. Let us define theorder of the polynomial as the degree plus 1, which is the optimal order when measuringthe approximation property in Lp norm. For example, the order of the linear polynomial is2. When the order of polynomial r is bigger than 3 (,i.e., quadratic and above polynomial),we can prove a stronger estimate for the negative normku Rh uk 1 Chr 1 kukr .(28)Using the technique in Lemma 2.6 and Theorem 2.7, we have the following estimate onthe L2 norm.Theorem 2.9 (Superconvergence in L2 -norm for high order elements). Suppose the solution u to (3) satisfying ut L (0, T ; H r (Ω)). Let uh be the solution of (8) withu0,h Rh u0 . Then, for r 3,max k(Rh u uh )(t)k C log h hr 1 max k(ut )(t)k2r .(29)0 t T20 t TrWhen ut L (0, T ; H (Ω)), then(30)r 1 ZtkRh u uh k Chkut k2r 1/2.ds0Proof. From (22), we haveZTe Ah (T s) τh dskeh (T )k 0Z 0T 1/2e Ah (T s) Ahds 1/2max kAh0 t Tτh k C log h kτh k 1 .In the second step, we have used the estimateZ T 1/2e Ah (T s) Ahds C log h ,0 1/2which can be proved by the estimate ke Ah (T s) Ah k (T s) 1To prove (30), we simply use the energy estimate (16) and (28. In 1-D, using the inverse inequality kRh u uh k h 1/2 kRh u uh k, we couldobtain the superconvergence in the maximum norm 1/2 Z tr 1/22kut kr ds.(31)kRh u uh k Ch0

FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS9Again using the inverse inequality kuh Rh uk Ch 1 kuh Rh uk, the superconvergence of L2 norm, and the maximum norm estimate of Ritz-Galerkin projection, forr 3, ku Rh uk Chr kukr, , we can improve the maximum norm error estimate.Theorem 2.10 (Maximum norm estimate for high order elements in two dimensions). Suppose the solution u to (3) satisfying u L (0, T ; W 2, ) and ut L2 (0, T ; W 2, ) orut L (0, T ; W 2, (Ω)). Let uh be the solution of (8) with u0,h Rh u0 . Then in twodimensions and for r 3" Z t 1/2 #r2k(u uh )(t)k Ch kukr, kut kr ds(32).03. F INITE ELEMENT METHODS : SEMI - DISCRETIZATION IN TIMEIn this section, we consider the semi-discretization in time. We first discretize the timeinterval (0, T ) into a uniform grid with size δt T /N and denoted by tn nδt forn 0, . . . N .3.1. Low order schemes. A continuous function in time will be interpolated into a vectorby f n : (I n f )(·, tn ) f (·, tn ). Recall that A : H01 H 1 . Below we list threelow order schemes in operator form. Forward Euler Method: u0 u0un un 1 Aun 1 f n 1 .δt Backward Euler Method: u0 u0un un 1 Aun f n .δt Crank-Nicolson Method: u0 u0un un 1 A(un un 1 )/2 f n 1/2 .δtNote that these equations hold in H 1 (Ω) sense. Taking Crank-Nicolson as an example,the equation reads as1 n1(33)(u un 1 , v) ( un un 1 , v) (f n 1/2 , v) for all v H01 .δt2We now study the stability of these schemes. We rewrite the Backward Euler method as(I δtA)un un 1 δtf n .Since A is SPD, λmin (I δtA) 1 and consequently, λmax ((I δtA) 1 ) 1. Thisimplies the L2 stability(34)kun k kun 1 k δtkf n k ku0 k nXδtkf k k.k 1The stability (35) is the discrete counter part of (4): discretize the integralRiemann sum.Similarly one can derive the L2 stability for the C-N scheme(35)kun k ku0 k nXk 1δtkf k 1/2 k.R tn0kf k ds by a

10LONG CHENThe integralR tn0kf k ds is approximated by the middle point rule.Remark 3.1. For C-N method, the right hand side can be also chosen as (f n f n 1 )/2.It corresponds to the trapezoid quadrature rule. For nonlinear problem A(u), it can beA((un un 1 )/2) or (A(un ) A(un 1 ))/2. Which one to chose is problem dependent.The energy estimate can be adapted to the semi-discretization in time easily. For example, we chose v (un un 1 )/2 in (33) to get1 n 2 1 n 1 2ku k kuk δt un 1/2 21 δt(f n 1/2 , un 1/2 ),22which implies the counter part of (5)kun k2 (36)nXδt uk 1/2 21 ku0 k2 k 1nXδtkf k 1/2 k2 1 .k 1Exercise 3.2. Study the stability of the forward and backward Euler method.We then study the convergence. We use C-N as a typical example. We apply the discreteoperator Ln to the error I n u unLn (I n u un ) Ln I n u I n Lu u(·, tn ) u(·, tn 1 ) t u(·, tn 1/2 ).δtThe consistency error isu(·, tn ) u(·, tn 1 ) t u(·, tn 1/2 ) C(δt)2 .δt(37)By the stability result, we then getkI n u un k Ctn δt2 .From the consistency error estimate (37), one can easily see the backward and forwardEuler methods are only first order in time.3.2. High order discretizations in time. We assume that U C(0, T ; H01 (Ω)) is a continuous piecewise q-th degree polynomial in time, that is, on the time interval Jn : (tn 1 , tn ),qXU Jn (x, t) (t tn 1 )j uj (x), uj (x) H01 (Ω).j 0We denote Pq (Jn ) as the set of such q-th degree polynomials on Jn , and define an operatorQnq 1 : C(0, T ; H01 (Ω)) Pq 1 (Jn ) satisfyingZZhQnq 1 u, pi dt hu, pi dt, p Pq 1 (Jn ).JnJnThen, the semi-discretization in time is to seek U (t) Pq (Jn ) such thatLn U (t) : t U (t) Qnq 1 AU (t) Qnq 1 f (t),(38)which is equivalent to the Petrov-Galerkin formulationZZZ(39)h t U, vi dt hAU, vi dt hf, vi dt,JnJn t Jn , v Pq 1 (Jn ).JnNote that the self-adjoint operator A is commute with the L2 projection, i.e., Qnq 1 A AQnq 1 .

FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS11The initial condition is given by U (0) u0 , and U (tn 1 ) is obtained from the problem(39) on the previous time interval Jn 1 for n 2. For example, if q 1, the solution U (t)is a piecewise linear function and the test function

FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS 3 The inequality (4) is an easy consequence of the following inequality kuk d dt kuk kfkkuk: From 1 2 d dt kuk2 juj2 1 1 2 (kfk2 1 juj 2 1); we get d dt kuk2 juj2 1 kfk 2 1: Integrating over (0;t), we obtain (5). The last energy estimate (6) can be proved similarly by choosing v u tand left .

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