Finite Element Methods for Parabolic Equations

FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS

LONG CHEN

As a model problem of general parabolic equations, we shall consider the following heat equation and study corresponding ﬁnite element methods   ut − ∆u = f in Ω × (0,T ), (1) u = 0 on ∂Ω × (0,T ),  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 respect to the spatial variable. For the simplicity of exposition, we consider only homogenous Dirichlet boundary condition and comment on the adaptation to Neumann and other type of boundary conditions. Besides the boundary condition on ∂Ω, we also need to assign the 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. Therefore the initial condition can be also thought as a boundary condition of the space-time domain Ω × (0,T ).

1. VARIATIONAL FORMULATION AND ENERGY ESTIMATE 1 We multiply a test function v ∈ H0 (Ω) and apply the integration by part to obtain a variational formulation of the heat equation (1): given an f ∈ L2(Ω) × (0,T ], for any 1 2 t > 0, find u(·, t) ∈ H0 (Ω), ut ∈ L (Ω) such that 1 (2) (ut, v) + a(u, v) = (f, v), for all v ∈ H0 (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 to −1 1 −1 −1 f ∈ H (Ω). Since ∆ map H0 (Ω) to H (Ω), we can treat ut(·, t) ∈ H (Ω) for a fixed t. We thus introduce the Sobolev space for the time dependent functions

1/q Z T ! q k,p q L (0,T ; W (Ω)) := {u(x, t) | kukLq (0,T ;W k,p(Ω)) := ku(·, t)kk,p dt < ∞}. 0

2 −1 1 Our reﬁned weak formulation will be: given f ∈ L (0,T ; H (Ω)) and u0 ∈ H0 (Ω), 2 1 2 −1 ﬁnd u ∈ L (0,T ; H0 (Ω)) and ut ∈ L (0,T ; H (Ω)) such that hu , vi + a(u, v) = hf, vi, ∀v ∈ H1(Ω), and t ∈ (0,T ) a.e. (3) t 0 u(·, 0) = u0

−1 1 where h·, ·i is the duality pair of H (Ω) and H0 (Ω). We assume equation (3) is well posed. For the existence and uniqueness of the solution [u, ut], we refer to [3]. In most 2 places, we shall still use the formulation (2) and assume f, ut ∈ L (Ω) so that the duality pair is realized by the L2 inner product (·, ·). 1 2 LONG CHEN

Remark 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 which can 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) as a mapping 1 u : [0,T ] → H0 (Ω), deﬁned as u(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 or ku(t)k1 is taken with respect to the spatial variable and thus becomes a function of time. We then introduce the operator 2 1 2 −1 2 L : L (0,T ; H0 (Ω)) → L (0,T ; H (Ω)) × L (Ω) as −1 (Lu)(·, t) = ∂tu − ∆u in H (Ω), for t ∈ (0,T ] a.e. (Lu)(·, 0) = u(·, 0). Then the equation (3) can be written as

Lu = [f, u0].

Here we explicitly include the initial condition u0. The spatial boundary condition is build 1 into the space H0 (Ω). In most places, when it is clear from the context, we also use L for the 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 2 (2) and ut ∈ L (0,T ; L (Ω)), then for t ∈ (0,T ] a.e. Z t (4) ku(t)k ≤ ku0k + kf(s)k ds 0 Z t Z t 2 2 2 2 (5) ku(t)k + |u(s)|1 ds ≤ ku0k + kf(s)k−1 ds, 0 0 Z t Z t 2 2 2 2 (6) |u(t)|1 + kut(s)k ds ≤ |u0|1 + kf(s)k ds. 0 0 Proof. The solution is deﬁned via the action of all test functions. The art of the energy estimate 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: Z 1 2 1 d 2 d • (ut, u) = (u )t = kuk = kuk kuk; Ω 2 2 dt dt 2 • a(u, u) = |u|1; 1 •| (f, u)| ≤ kfkkuk or |(f, u)| ≤ kfk |u| ≤ (kfk2 + |u|2). −1 1 2 −1 1 FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS 3

The inequality (4) is an easy consequence of the following inequality d kuk kuk ≤ kfkkuk. dt From 1 d 1 kuk2 + |u|2 ≤ (kfk2 + |u|2), 2 dt 1 2 −1 1 we get d kuk2 + |u|2 ≤ kfk2 . dt 1 −1 Integrating over (0, t), we obtain (5). The last energy estimate (6) can be proved similarly by choosing v = ut and left as an exercise. From (5), we can obtain the stability of the operator L 2 1 2 −1 2 L : L (0,T ; H0 (Ω)) → L (0,T ; H (Ω)) × L (Ω) as 2 2 2 kuk 2 1 ≤ ku0k + kfk 2 −1 . L (0,T ;H0 (Ω)) L (0,T ;H (Ω)) Since the equation is posed a.e for t, we could also obtain the maximum-norm estimate in time. For example, (4) can be formulated as

kukL∞(0,T ;L2(Ω)) ≤ ku0k + kfkL1(0,T ;L2(Ω)), and (6) implies

kuk ∞ 1 ≤ |u | + kfk 2 2 . L (0,T ;H0 (Ω)) 0 1 L (0,T ;L (Ω))

Exercise 1.3. Prove the energy estimate Z t 2 −λt 2 −λ(t−s)) 2 (7) kuk ≤ e ku0k + e kfk−1 ds, 0 where λ = λmin(−∆) > 0. The estimate (7) shows that the effect of the initial data is exponential decay.

2. FINITEELEMENTMETHODS: SEMI-DISCRETIZATION IN SPACE

2.1. Semi-discretization in space. Let {Th, h → 0} be a quasi-uniform family of tri- 1 angulations of Ω and Vh ⊂ H0 (Ω) be a finite element space based on Th. The semi- 0 1 discretized finite element method is: given f ∈ Vh × (0,T ], u0,h ∈ Vh ⊂ H0 (Ω), find 2 uh ∈ L (0,T ; Vh) such that (∂ u , v ) + a(u , v ) = hf, v i, ∀v ∈ , t ∈ +. (8) t h h h h h h Vh R uh(·, 0) = u0,h

The 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 = PN i=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 of time t. The solution uh can be computed by solving an ODE system (9) Mu˙ + Au = f, 4 LONG CHEN where u = (u1, ··· , uN )| is the coefficient vector, M, A are the mass matrix and the stiffness 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.

3 Z 1 X g(x) dx ≈ g(x )|τ|. 3 i τ i=1

Then the mass matrix becomes diagonal M = diag(m1, ··· , mN ). This is known as the 2 mass lumping. For 2-D uniform grids, mi = h and A is the five point stencil discretization of −∆. Therefore (9) can be interpret as a rescaled finite difference discretization at each vertex 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 developed in Uniﬁed Error Analysis to estimate the error u − uh in certain norms. The setting is 1/2 2 1 R T 2 • X = L (0,T ; H0 (Ω)), and kukX = 0 |u(t)|1 dt 1/2 2 −1 R T 2 • Y = L (0,T ; H (Ω), and kfkY = 0 kf(t)k−1 dt 1/2 2 R T 2 • Xh = L (0,T ; Vh), and kuhkXh = 0 |uh(t)|1 dt 1/2 2 0 R T 2 • Yh = L (0,T ; Vh), and kfhkYh = 0 kfh(t)k−1,h dt . Recall the dual norm

hfh, vhi 0 kfhk−1,h = sup , for fh ∈ Vh. vh∈Vh |vh|1 1 • Ih = Rh(t): H0 (Ω) → Vh is the Ritz-Galerkin projection, i.e., Rhu ∈ Vh such that

a(Rhu, vh) = a(u, vh), ∀vh ∈ Vh. −1 0 • Πh = Qh(t): H (Ω) → Vh is the projection

hQhf, vhi = hf, vhi, ∀vh ∈ Vh.

• Ph : Xh → X is the natural inclusion 2 •L : X → Y × L (Ω) is Lu = ∂tu − ∆u, Lu(·, 0) = u(·, 0), and

Lh = L|Xh : Xh → Yh × Vh. We summarize the setting in the following diagram

L X −−−−→ Y     , yRh yQh

Lh Xh −−−−→ Yh which is not commutative and the difference is the consistency error kQhLu−LhRhukYh . The discrete equation we are solving is

0 Lhuh = Qhf, in Vh, ∀t ∈ (0,T ] a.e.

uh(·, 0) = u0,h. FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS 5

2.3. Stability. Adapt the proof of the energy estimate for L, we can obtain similar stability results for Lh. The proof is almost identical and thus skipped here.

Theorem 2.1 (Energy estimate for ﬁnite element discretization). Suppose uh satisfy Lhuh = fh, uh(·, 0) = u0,h, then Z t (10) kuh(t)k ≤ ku0,hk + kfh(s)k ds 0 Z t Z t 2 2 2 2 (11) kuh(t)k + |uh(s)|1 ds ≤ ku0,hk + kfh(s)k−1,h ds, 0 0 Z t Z t 2 2 2 2 (12) |uh(t)|1 + k∂tuh(s)k ds ≤ |u0,h|1 + kfh(s)k ds. 0 0

Note that in the energy estimate (11), the dual norm k · k−1 is replaced by a weaker one k · k−1,h since we can apply the inequality

hfh, uhi ≤ kfhk−1,h|uh|1.

The weaker norm k · k−1,h can be estimated by

kfk−1,h ≤ kfk−1 ≤ Ckfk.

2.4. Consistency. Recall that the consistency error is kQhLu − LhRhukYh . The choice of Ih = Rh simpliﬁes the consistency error analysis. Lemma 2.2 (Error equation). For the semi-discretization, we have the error equation 0 (13) Lh(uh − Rhu) = Qh(I − Rh)ut, t > 0, in Vh,

(14) (uh − Rhu)(·, 0) = u0,h − Rhu0. Proof. Let A = −∆. By our deﬁnition of consistency, the error equation is: for t > 0

Lh(Rhu − uh) = LhRhu − QhLu = ∂t(Rhu − Qhu) + (ARhu − QhAu). 0 The desired result then follows by noting that ARh = QhA in Vh and QhRh = Rh. 0 The error equation (13) holds in Vh which is a weak topology. The motivation to choose Ih = Rh is that in this weak topology

hARhu, vhi = (∇Rhu, ∇vh) = (∇u, ∇v) = hAu, vhi = hQhAu, vhi, or simply in operator form ARh = QhA. This technique is ﬁrstly proposed by Wheeler [6]. Apply the stability to the error equation, we obtain the following estimate on the discrete error Rhu − uh.

Theorem 2.3 (Stability of discrete error). The solution uh of (8) satisfy the following error estimate Z t (15) kRhu − uhk ≤ ku0,h − Rhu0k + kQh(I − Rh)utk ds 0 (16) Z t Z t 2 2 2 2 kRhu − uhk + |(uh − Rhu)|1 ds ≤ ku0,h − Rhu0k + kQh(I − Rh)utk−1,h ds, 0 0 Z t Z t 2 2 2 2 (17) |Rhu−uh|1+ k∂t(Rhu−uh)k ds ≤ |u0,h−Rhu0|1+ kQh(I−Rh)utk ds. 0 0 6 LONG CHEN

We then estimate the two terms u0,h − Rhu0 and Qh(I − Rh)ut involved in these error estimates. 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 = Rhu0 so that no error coming from approximation of the initial condition. However, this choice requires the inversion of a stiffness matrix which is not cheap. A simple choice would be the nodal interpolation, i.e. u0,h(xi) = u0(xi) or any other choice with optimal approximation property 2 (18) ku0,h − Rhu0k ≤ ku0 − u0,hk + ku0 − Rhu0k . h ku0k2, and similarly |u0,h − Rhu0|1 . hku0k2. According to (7) in Exercise 1.3, the effect of the initial boundary error will be exponen- tially decay to zero as t goes to inﬁnity. So in practice, we can choose the simple nodal interpolation. 2 2 On the estimate of the second term, assume ut ∈ H (Ω) and H -regularity result hold for Poisson equation (for example, the domain is smooth or convex), then 2 (19) kQh(I − Rh)utk ≤ k(I − Rh)utk . h kutk2. The negative norm can be bounded by the L2-norm as 2 kQh(I − Rh)utk−1,h ≤ kQh(I − Rh)utk−1 ≤ CkQh(I − Rh)utk . h kutk2. When using quadratic and above polynomial, we can prove a stronger estimate for the negative norm and will be discussed in Section 2.6.

2.5. Convergence. The convergence of the discrete error comes Rhu − uh from the stability and consistency. Theorem 2.4 (Convergence of the discrete error). Suppose the solution u to (3) satisfying 2 2 2 ut ∈ L (0,T ; H (Ω)) and the H -regularity holds for the Poisson equation. Let uh be the solution of (8) with u0,h satisfying (18). We then have Z t 2 (20) |Rhu − uh|1 + kRhu − uhk ≤ Ch ku0k2 + kutk2 ds . 0

To estimate the true error u − uh, we need the approximation error estimate of the projection Rh; see Introduction to Finite Element Methods −1 h ku − Rhuk + |u − Rhu|1 ≤ Chkuk2. Theorem 2.5 (Convergence of the discretization error). Suppose the solution u to (3) sat- 2 2 isfying ut ∈ L (0,T ; H (Ω)). Then the solution uh of (8) with u0,h having optimal approximation property (18) satisfy the following optimal order error estimate: Z t −1 2 (21) h |u − uh|1 + ku − uhk ≤ Ch ku0k2 + kutk2 ds . 0

2.6. *Superconvergence and error estimate in the maximum norm. The error eh = Rhu−uh satisﬁes the evolution equation (13) with eh(0) = 0 with the chose u0,h = Rhu0

(22) ∂teh + Aheh = τh, where τh := Qh∂t(Rhu − u). Therefore Z t eh(t) = exp(−Ah(t − s))τhds. 0 FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS 7

Due to the smoothing effect of the semi-group e−Aht, 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 have Z t

(23) max exp(−Ah(t − s))Ahτhds ≤ C| log h| max kτhk. 0≤t≤T 0 0≤t≤T

Proof. Let λmin and λmax be the minimal and maximal eigenvalue of Ah. Then it is easy to check  λ e−λmax(t−s) if (t − s) ≤ λ−1 ,  max max e−Ah(t−s)A ≤ −1 −1 −1 h (t − s) if λmax ≤ (t − s) ≤ λmin,  −λmin(t−s) −1 λmine if (t − s) ≥ λmin. −2 Note that λmin = O(1) and λmax ≤ Ch . We get Z t −Ah(t−s) max e Ahτhds ≤ C| log h| max kτhk. 0≤t≤T 0 0≤t≤T Theorem 2.7 (Superconvergence in H1-norm). Suppose the solution u to (3) satisfying ∞ 2 ut ∈ L (0,T ; H (Ω)). Let uh be the solution of (8) with u0,h = Rhu0. Then 2 (24) max |Rhu − uh|1 ≤ C| log h| h max kutk2. 0≤t≤T 0≤t≤T 2 2 When ut ∈ L (0,T ; H (Ω)), then Z t 1/2 2 2 (25) |Rhu − uh|1 ≤ Ch kutk2 ds . 0 1/2 Proof. We multiply Ah to (22) and apply Lemma 2.6 to get Z T 1/2 1/2 −Ah(T −s) |eh(T )|1 = kAh eh(T )k = e Ah τhds 0 Z T −1/2 −Ah(T −s) ≤ e Ah ds max kAh τhk 0 0≤t≤T 2 ≤ C| log h|h max kutk2. 0≤t≤T

−1/2 In the last step, we have used the fact kAh τhk = kτhk−1,h ≤ kτhk. To get (25), we use the energy estimate (17).

Since the optimal convergent rate for |u − Rhu|1 or |u − uh|1 is only ﬁrst order, the second order error estimate (24) and (25) are called superconvergence. To control the maximum norm, we use the discrete embedding result (for 2-D only)

kRhu − uhk∞ ≤ C| log h||Rhu − uh|1, and the error estimate of Rh in the maximum norm 2 ku − Rhuk∞ ≤ C| log h|h kuk2,∞, to obtain the following result. 8 LONG CHEN

Theorem 2.8 (Maximum norm estimate for linear element in two dimensions). Suppose ∞ 2,∞ 2 2,∞ the solution u to (3) satisfying u ∈ L (0,T ; W ) and ut ∈ L (0,T ; W ) or ut ∈ ∞ 2,∞ L (0,T ; W (Ω)). Let uh be the solution of (8) with u0,h = Rhu0. Then in two dimensions " Z t 1/2# 2 2 (26) k(u − uh)(t)k∞ ≤ C| log h|h kuk2,∞ + kutk2 ds , 0 2 2 (27) max k(u − uh)(t)k∞ ≤ C| log h| h max [ku(t)k2,∞ + kut(t)k2] . 0≤t≤T 0≤t≤T For high order elements, we could get superconvergence in L2 norm. Let us deﬁne the order of the polynomial as the degree plus 1, which is the optimal order when measuring the approximation property in Lp norm. For example, the order of the linear polynomial is 2. 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 norm r+1 (28) ku − Rhuk−1 ≤ Ch kukr. Using the technique in Lemma 2.6 and Theorem 2.7, we have the following estimate on the L2 norm. Theorem 2.9 (Superconvergence in L2-norm for high order elements). Suppose the so- ∞ r lution u to (3) satisfying ut ∈ L (0,T ; H (Ω)). Let uh be the solution of (8) with u0,h = Rhu0. Then, for r ≥ 3, r+1 2 (29) max k(Rhu − uh)(t)k ≤ C| log h|h max k(ut)(t)kr. 0≤t≤T 0≤t≤T 2 r When ut ∈ L (0,T ; H (Ω)), then Z t 1/2 r+1 2 (30) kRhu − uhk ≤ Ch kutkr ds . 0 Proof. From (22), we have

Z T −Ah(T −s) keh(T )k = e τhds 0 Z T −1/2 −1/2 −Ah(T −s) ≤ e Ah ds max kAh τhk 0 0≤t≤T

≤ C| log h|kτhk−1. In the second step, we have used the estimate Z T −1/2 −Ah(T −s) e Ah ds ≤ C| log h|, 0

−Ah(T −s) −1/2 −1 which can be proved by the estimate ke Ah k ≤ (T − s) To prove (30), we simply use the energy estimate (16) and (28. −1/2 In 1-D, using the inverse inequality kRhu − uhk∞ ≤ h kRhu − uhk, we could obtain the superconvergence in the maximum norm Z t 1/2 r+1/2 2 (31) kRhu − uhk∞ ≤ Ch kutkr ds . 0 FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS 9

−1 Again using the inverse inequality kuh − Rhuk∞ ≤ Ch kuh − Rhuk, the superconvergence of L2 norm, and the maximum norm estimate of Ritz-Galerkin projection, for r r ≥ 3, ku − Rhuk∞ ≤ Ch kukr,∞, we can improve the maximum norm error estimate. Theorem 2.10 (Maximum norm estimate for high order elements in two dimensions). Sup- ∞ 2,∞ 2 2,∞ pose the solution u to (3) satisfying u ∈ L (0,T ; W ) and ut ∈ L (0,T ; W ) or ∞ 2,∞ ut ∈ L (0,T ; W (Ω)). Let uh be the solution of (8) with u0,h = Rhu0. Then in two dimensions and for r ≥ 3 " Z t 1/2# r 2 (32) k(u − uh)(t)k∞ ≤ Ch kukr,∞ + kutkr ds . 0

3. FINITEELEMENTMETHODS: SEMI-DISCRETIZATION IN TIME In this section, we consider the semi-discretization in time. We ﬁrst discretize the time interval (0,T ) into a uniform grid with size δt = T/N and denoted by tn = nδt for n = 0,...N. 3.1. Low order schemes. A continuous function in time will be interpolated into a vector n n n n 1 −1 by f := (I f)(·, t ) = f(·, t ). Recall that A = −∆ : H0 → H . Below we list three low order schemes in operator form. 0 • Forward Euler Method: u = u0 un − un−1 + Aun−1 = f n−1. δt 0 • Backward Euler Method: u = u0 un − un−1 + Aun = f n. δt 0 • Crank-Nicolson Method: u = u0 un − un−1 + A(un + un−1)/2 = f n−1/2. δt Note that these equations hold in H−1(Ω) sense. Taking Crank-Nicolson as an example, the equation reads as 1 1 (33) (un − un−1, v) + (∇un + ∇un−1, ∇v) = (f n−1/2, v) for all v ∈ H1. δt 2 0 We now study the stability of these schemes. We rewrite the Backward Euler method as (I + δtA)un = un−1 + δtf n. −1 Since A is SPD, λmin(I + δtA) ≥ 1 and consequently, λmax((I + δtA) ) ≤ 1. This implies the L2 stability n X (34) kunk ≤ kun−1k + δtkf nk ≤ ku0k + δtkf kk. k=1

R tn The stability (35) is the discrete counter part of (4): discretize the integral 0 kfk ds by a Riemann sum. Similarly one can derive the L2 stability for the C-N scheme n X (35) kunk ≤ ku0k + δtkf k−1/2k. k=1 10 LONG CHEN

R tn The integral 0 kfk 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 be A((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 get 1 1 kunk2 − kun−1k2 + δt|un−1/2|2 = δt(f n−1/2, un−1/2), 2 2 1 which implies the counter part of (5) n n n 2 X k−1/2 2 0 2 X k−1/2 2 (36) ku k + δt|u |1 ≤ ku k + δtkf k−1. k=1 k=1 Exercise 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 discrete operator Ln to the error Inu − un u(·, tn) − u(·, tn−1) Ln(Inu − un) = LnInu − InLu = − ∂ u(·, t ). δt t n−1/2 The consistency error is n n−1 u(·, t ) − u(·, t ) 2 (37) − ∂tu(·, tn−1/2) ≤ C(δt) . δt By the stability result, we then get n n 2 kI u − u k ≤ Ctnδt . From the consistency error estimate (37), one can easily see the backward and forward Euler methods are only ﬁrst order in time.

1 3.2. High order discretizations in time. We assume that U ∈ C(0,T ; H0 (Ω)) is a continuous piecewise q-th degree polynomial in time, that is, on the time interval Jn := (tn−1, tn), q X n−1 j 1 U|Jn (x, t) = (t − t ) uj(x), uj(x) ∈ H0 (Ω). j=0

We denote Pq(Jn) as the set of such q-th degree polynomials on Jn, and deﬁne an operator n 1 Qq−1 : C(0,T ; H0 (Ω)) → Pq−1(Jn) satisfying Z Z n hQq−1u, pi dt = hu, pi dt, ∀p ∈ Pq−1(Jn). Jn Jn

Then, the semi-discretization in time is to seek U(t) ∈ Pq(Jn) such that n n n (38) L U(t) := ∂tU(t) + Qq−1AU(t) = Qq−1f(t), ∀t ∈ Jn, which is equivalent to the Petrov-Galerkin formulation Z Z Z (39) h∂tU, vi dt + hAU, vi dt = hf, vi dt, ∀v ∈ Pq−1(Jn). Jn Jn Jn 2 n Note that the self-adjoint operator A is commute with the L projection, i.e., Qq−1A = n AQq−1. FINITE ELEMENT METHODS FOR PARABOLIC EQUATIONS 11

n−1 The initial condition is given by U(0) = u0, and U(t ) 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 v is piecewise constant in time. More exactly, un − un−1 ∂ U| = . t Jn δt n−1/2 n n Thus, if we use the midpoint t to compute Qq−1U(t) and Qq−1f(t) in the problem (38), then we arrive at the Crank-Nicolson method. Moreover, using left and right end points yields the Forward and Backward methods, respectively. For high order discretiza- n tion, inside one time interval Jn, one needs to solve a mass matrix equation to get u . The naive basis for polynomial {(t − tn−1)j, j = 0, . . . , q} is not friendly to the implementa- tion. Instead one can use quadrature points, e.g., Gauss-Legendre points or Radau points; see [2]. 3.3. Stability. We shall adapt the energy estimate to the discretization. Let us choose v = Qq−1U. Then, we control the following terms: Z Z n 1 n 2 n−1 2 • h∂tU, Qq−1Ui dt = h∂tU, Ui dt = kU(t )k − kU(t )k ; Jn Jn 2 Z Z Z n n n n n 2 • hQq−1AU, Qq−1Ui dt = hAQq−1U, Qq−1Ui dt = |Qq−1U|1 dt; Jn Jn Jn Z Z Z n 1 2 1 n 2 • hf, Qq−1Ui dt ≤ kfk−1dt + |Qq−1U|1 dt. Jn 2 Jn 2 Jn We then obtain the energy estimate as follows.

Theorem 3.3 (Energy estimate for semi-discretization in time). Suppose U satisfies ∂tU + n n Qq−1AU = Qq−1f, for n ≥ 1, then Z t Z t 2 n 2 2 2 (40) kU(t)k + |Qq−1U|1 ds ≤ kU(0)k + kfk−1 ds. 0 0 n 1 3.4. Consistency. Let I : C(0,T ; H0 (Ω)) → Pq(Jn) be the operator of the Lagrange interpolation such that Inu(tn−1) = u(tn−1), Inu(tn) = u(tn), and Z Z n hI u, pi dt = hu, pi dt, ∀p ∈ Pq−2(Jn). Jn Jn Note that we use moments instead of point values to define the Lagrange interpolation. n It is straight forward to verify that I u = u if u ∈ Pq(Jn), and Z 1/2 n 2 (q+1) ku − I uk dt . (δt) kukHq+1(tn−1,tn;L2(Ω)). Jn From the definition of the consistency error, we have n n n n n n n L (U − I u) = (Qq−1∂tu − ∂tI u) + (Qq−1Au − Qq−1AI u). For the first time, the integration by part in terms of t provides Z Z n n n hQq−1∂tu − ∂tI u, vi dt = h∂tu − ∂tI u, vi dt Jn Jn n Z n t n = hu − I u, vi tn−1 − hu − I u, ∂tvi dt Jn = 0, 12 LONG CHEN

n for every v ∈ Pq−1(Jn). The ﬁrst identity holds because ∂tI u ∈ Pq−1(Jn) so that n n n ∂tI u = Qq−1∂tI u. The third one comes from the deﬁnition of the interpolation operator In. Hence, we have the error equation n n n n n L (U − I u) = Qq−1Au − Qq−1AI u =: τn. And Z Z n 2 n n 2 kτ k dt = kQq−1(I − I )Auk dt Jn Jn Z ≤ k(I − In)Auk2 dt Jn 2(q+1) 2 . (δt) kAukHq+1(tn−1,tn;L2(Ω)). n n n 1 3.5. Convergence. Let e = U − I u. Clearly, e ∈ Pq(Jn) and e (0) = U(0) − 1 (I u)(0) = u0 − u(0) = 0. Then, from the consistency result (40), we get N Z N 2 X n 2 2(q+1) 2 ke (T )k ≤ kτ k dt . (δt) kAukHq+1(0,T ;L2(Ω)). n=1 Jn Therefore, we obtain the following convergence rate

Theorem 3.4 (Energy estimate for semi-discretization in time). Suppose U satisﬁes ∂tU + n n Qq−1AU = Qq−1f, for n ≥ 1, then for any t ∈ (0,T ) (q+1) ku(t) − U(t)k . (δt) kAukHq+1(0,t;L2(Ω)). Reﬁned analysis including a posteriori error analysis and superconvergence at nodal points can be found in [1].

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