nonconforming finite element approximation of time-dependent maxwell's equations in debye...

Nonconforming Finite Element Approximation ofTime-Dependent Maxwell’s Equations in DebyeMediumDongyang Shi, Changhui YaoSchool of Mathematics and Statistics, Zhengzhou University, Zhengzhou 450001,China

Received 31 May 2013; accepted 17 October 2013Published online 17 March 2014 in Wiley Online Library (wileyonlinelibrary.com).DOI 10.1002/num.21843

In this article, a new nonconforming mixed finite element method (FEM) for approximating the Maxwell’sequations with Debye medium in three-dimension are developed. By employing traditional variational for-mula, without adding “stability” or “penalty” terms, we show that the discrete scheme is robust. With thehelp of the element’s typical properties, interpolation and derivative transfer skills, the convergence analysisis presented and error estimates for semidiscrete and leap-frog fully-discrete schemes are obtained, respec-tively. Numerical example shows the validity of the proposed method. © 2014 Wiley Periodicals, Inc. NumerMethods Partial Differential Eq 30: 1654–1673, 2014

Keywords: Debye medium; error estimates; Maxwell’s equations; nonconforming mixed FEMs; semi-discrete and fully-discrete scheme

I. INTRODUCTION

Debye medium is one of basic physical concepts when one investigates dielectric in electro-magnetic theory and materials science [1]. It is a kind of isotropic dispersive medium, and itspermittivity and conductivity are functions of frequency. With the help of the polarization anddielectric relaxation, one can establish phenomenological theory in Debye medium. That is tosay, in the process of polarization, microscopic particles complex energy exchange actions canbe taken into the following dielectric time parameters. Therefore, numerical studies of Maxwell’sequations in Debye medium have attracted considerable attention.

Since early 1990s, a lot of investigation has been devoted to the applications of time-domainfinite difference (FDTD) methods for dispersive medium, including Debye medium, see [1, 2],and references therein. But few works can be explored about time-domain finite element methods(TDFEMs) untill 2001 by Jiao and Jin [3]. Later, the studies on TDFEM bloom up, see J.C. Li’swork [4–7].

Correspondence to: Dongyang Shi, School of Mathematics and Statistics, Zhengzhou University, Zhengzhou 450001,China (e-mail: [email protected])Contract grant sponsor: P. R. China NSFC; contract grant numbers: 11271340, 11101384, and 11101381

© 2014 Wiley Periodicals, Inc.

NONCONFORMING FINITE ELEMENT APPROXIMATION 1655

In [5], Li considered three popular dispersives medium models (the isotropic cold plasmamedium one-pole Debye medium, and two-pole Lorentz medium) of time-dependent Maxwell’sequations in a bounded three-dimensional (3D) domain by Nédélec’s conforming FE, and provedthe optimal order error estimates. In [8], Li and Lin obtained the global superconvergence bysemidiscrete mixed FEMs. In [9,10], they developed a leap-frog mixed finite element scheme forsolving Maxwell’s equations. In [11], the interior penalty discontinuous Galerkin (DG) methodsfor the time-dependent Maxwell’s equations in cold plasma were set up. However, the aboveresearches are only concentrated on Nédélec’s conforming FEMs.

For nonconforming FEMs used to electromgnetic systems, series of works can be found in Dou-glas [12], Brenner [13–15], Shi [16–18], and references therein. In [12], the authors constructeda new Crouzeix–Raviart type nonconforming mixed cubic finite element in 3D and presented theclassical error estimates with absorbing boundary conditions. Unfortunately, there is no numericalexperiment to verify their theoretical analysis.

The first constructive theoretical and numerical analysis for Maxwell’s equations can be foundin Brenner’s work [13–15], in which the Crouzeix–Raviart triangular nonconforming FE wasanalyzed and applied to 2D curl − curl system. At the same time, they found that the traditionalweak formula can not lead to a convergence scheme even if the mesh is refined. Therefore, thetraditional weak formula was modified in [13–15] by adding penalty terms involving the tangentand normal jumps. The error estimates and numerical experiments showed that the modified formworked well. The so-called crucial difference is that the piecewise H(curl)

⋂H(div) seminorm,

unlike the piecewise H 1 seminorm, is too weak to control the jumps even if the mesh is refined.Hence, the two terms involving the jumps must be included in the discretization so as to controlthe consistency error.

In [16], three Crouzeix–Raviart-type rectangular nonconforming FE spaces were utilized totime-dependent Maxwell’s system in (2D) by a FE scheme as well as a mixed FE scheme withtraditional variational formulas. In [17], the new nonconforming FE spaces were establishedon anisotropic meshes by mixed FE formulations in 2D and 3D, respectively, and derived theanisotropic error estimations in L2 norm and H(curl) norm. In [18], the convergence analysisof QuasiWilson nonconforming FE to Maxwell’s equations for arbitrary quasiuniform quadrilat-eral meshes was studied. The subsequent work can also be found in Qiao [19], where a pair ofnew mixed nonconforming FE spaces were set up, and applied to the time-harmonic Maxwell’sequations. At the same time, the analysis of extrapolation and superconvergence were presented,respectively. But, all the above studies are based on perfect electric conductor. Until now, there isno numerical example to show that the rectangular or cubic Crouziex–Raviart-type nonconformingfinite element can result in a convergence scheme for electromagnetic system.

The objective of this article is to analyze and develop a new nonconforming mixed FEM toapproximate the Maxwell system in Debye medium. We construct a novel nonconforming finiteelement space, and use the degrees of freedom of electric field defined by the centroids of faces,which is in accord with Yee’s scheme [20] implicitly. From the experiment results, we find thatthe new FE spaces can lead to robust discrete scheme only by employing classical variational for-mula without adding “stability” term. Furthermore, the semidiscrete and leap-frog fully discreteschemes are investigated and the optimal order error convergent rate is obtained, respectively.

This rest of this work is organized as follows: In Section 2, we introduce the one-pole Maxwell’sequations in Debye medium. In Section 3, we present the new nonconforming mixed finite ele-ment spaces and give some typical properties. In Section 4, a semidiscrete scheme is proposedand the optimal order error estimate is obtained. In Section 5, a full-discrete scheme is establishedand optimal order error estimate is derived. In Section 6, numerical results are provided to verifythe validity of the theoretical analysis.

Numerical Methods for Partial Differential Equations DOI 10.1002/num

1656 SHI AND YAO

II. MAXWELL’S EQUATIONS IN ONE-POLE DEBYE MEDIUM

For the medium model, the governing equations are described as follows[4]

ε0ε∞ �Et − ∇ × �H + (εs − ε∞)ε0

t0�E − 1

t0�P = �f (X, t), (1)

μ0 �Ht = −∇ × �E, (2)

1

(εs − ε∞)ε0

�Pt + 1

(εs − ε∞)ε0t0�P = 1

t0�E, (3)

where �E is the electric field, �H is the magnetic field, f is the source term, ε0 is the permittivity offree space, μ0 is the permeability of free space, �P is the polarization vector, ε∞ is the permittivityat infinite frequency, εs(>ε∞) is the permittivity at zero frequency, and t0 is the relaxation time,X = (x, y, z).

Assume that the perfect conducting boundary condition is

n × �E = 0, ∂� × (0, T ) (4)

and the initial conditions are

�E(X, 0) = �E0(X), �H(X, 0) = �H0(X), �P(X, 0) = �P0(X), ∀X ∈ �, (5)

where �E0, �H0, and �P0 are the given functions.Let H0(curl, �) = {ϕ ∈ [L2(�)]3 : ∇ × ϕ ∈ [L2(�)]3, n × φ|∂� = 0}. The weak for-

mulation for Eq. (1)–(4) reads as: find �E ∈ C(0, T ; H0(curl; �)) ∩C1(0, T ; (L2(�))3), �H ∈C1(0, T ; (L2(�))3), and �P ∈ C1(0, T ;

(L2(�))3) such that ∀�φ ∈ H0(curl; �), �ψ and �φ ∈ (L2(�))3

ε0ε∞( �Et , �φ) − ( �H , ∇ × �φ) + (εs − ε∞)ε0

t0( �E, �φ) − 1

t0( �P , �φ) = ( �f , �φ), (6)

μ0( �Ht , �ψ) + (∇ × �E, �ψ) = 0, (7)

1

(εs − ε∞)ε0( �Pt ,

�φ) + 1

(εs − ε∞)ε0t0( �P , �

φ) − 1

t0( �E, �

φ) = 0, (8)

with the initial conditions (5).The proof of existence and uniqueness of solution to (6)–(8) can be found [21].

III. NONCONFORMING MIXED FINITE ELEMENT SPACES

Let � be a bounded cubic domain in R3, and n the unit outer normal vector of the correspondingface. For each 0 < h < 1, let T h be a regular partition. For K ∈ T h, let 2hx , 2hy and 2hz be thelength, height, and width of element K , respectively, hK = max {hx , hy , hz}, h = maxK{hK}, and(xK , yK , zK) be the center coordinates of K . Let K = [−1, 1]3 be the reference element. Thenthere is an affine mapping FK : K → K

x = xK + hxx, y = yK + hyy, z = zK + hzz.



Define R = Rx × Ry × Rz and S = Sx × Sy × Sz on K as⎧⎪⎨⎪⎩Rx = Span{1, y, z, (y2 − 5

3 y4) − (z2 − 5

3 z4)},

Ry = Span{1, z, x, (z2 − 53 z

4) − (x2 − 53 x

4)},Rz = Span{1, x, y, (x2 − 5

3 x4) − (y2 − 5

3 y4)}.

(9)

and ⎧⎪⎨⎪⎩Sx = Span{1, y − 10

3 y3, z − 103 z3},

Sy = Span{1, z − 103 z3, x − 10

3 x3},Sz = Span{1, x − 10

3 x3, y − 103 y3},

(10)

respectively. The associated interpolation operators are � : φ ∈ [H 2(K)]3 → �φ ∈ R(K) andQ : ψ = (ψx , ψy , ψz) ∈ [L2(K)]3 → Qψ ∈ S(K), respectively, which satisfy

1

|Fi |∫

Fi

(�ϕ − −→ϕ )ds = 0, i = 1, . . . , 6. (11)

and ∫K

(Qψl − ψl)dxdydz = 0,∫

K

curl(Qψl − ψl)dxdydz = 0, l = x, y, z. (12)

The 2D curl is defined as

curlψx =(

∂ψx

∂z, −∂ψx

∂y

), curlψy =

(∂ψy

∂x, −∂ψy

∂z

), curlψz =

(∂ψz

∂y, −∂ψz

∂x

).

Obviously, ∇ × R = S.Let �h : [H 2(�)]3 → V h and Qh : [L2(�)]3 → Wh be the corresponding interpolation

operators with respect to � and Q, respectively.Define the nonconforming mixed finite element spaces

Uh = {�ϕ ∈ [L2(�)]3 : �ϕ|K ∈ R, ∀K ∈ T h}, Uh0 =

{�ϕ ∈ Uh,

∫F

n × �ϕdf = 0, F ⊂ ∂�

},

Wh = { �ψ ∈ [L2(�)]3 : �ψ |K ∈ S, ∀K ∈ T h}.

Then the operator Qh satisfies

( �ψ − Qh�ψ , �χ) = 0, ∀ �χ ∈ Wh. (13)

Due to ∇ × Uh ⊆ Wh, we also have

( �ψ − Qh�ψ , ∇ × �ϕ)h = 0, ∀�ϕ ∈ Uh, where (·, ·)h =

∑K

(·, ·)K . (14)

Therefore, we can get the following results from interpolation theory [21].


1658 SHI AND YAO

Lemma 3.1. Let T h be quasi-regularity, then there exist constants Ci , i = 1, 2, 3, independentof h such that

‖�ϕ − �h �ϕ‖0 ≤ C1hi‖�ϕ‖i , i = 1, 2, ‖ �ψ − Qh

�ψ‖0 ≤ C2h‖ �ψ‖1. (15)

‖∇ × ( �ϕ − �h �ϕ)‖0 ≤ C3h‖∇ × �ϕ‖1. (16)

Lemma 3.2. Assume that �H ∈ H 2(�), then for ∀�φ = (φ1, φ2, φ3) ∈ Uh0 , there exists a constant

C4 independent of h such that

∑K

∫∂K

�Hn × �φds ≤ C4h| �H |2‖ �φ‖0. (17)

Proof. ∀Fj ⊂ ∂K , let φij = 1|Fj |

∫Fj

φids, i = 1, 2, 3, j = 1, . . . , 6. By use of the definition

of Uh0 , we have

∑K

[(∫F3

−∫

F1

)(H2φ33 − H3φ23)dydz +

(∫F2

−∫

F4

)(H1φ32 − H3φ12)dxdz

+((∫

F6

−∫

F5

))(H1φ26 − H2φ16)dxdy

]= 0.

Therefore,

∑K

∫∂K

�Hn × �φds =∑K

[(∫F3

−∫

F1

)(H2φ3 − H3φ2)dydz

+(∫

F2

−∫

F4

)(H1φ3 − H3φ1)dxdz +

((∫F6

−∫

F5

))(H1φ2 − H2φ1)dxdy

],

=∑K

[(∫F3

−∫

F1

)(H2(φ3 − φ33) − H3(φ2 − φ23))dydz

+(∫

F2

−∫

F4

)(H1(φ3 − φ32) − H3(φ1 − φ12))dxdz

+((∫

F6

−∫

F5

))(H1(φ2 − φ26) − H2(φ1 − φ16))dxdy

]�

∑K

(M1 + M2 + M3)

Note that ∂φi

∂x, ∂φi

∂yand ∂φi

∂z(i = 1, 2, 3) are independent of x, y and z, respectively, with the

same argument of [17], we have

|Mi | ≤ 8

3h2| �H |2,K | �φ|1,K , i = 1, 2, 3. (18)

Then employing the inverse inequality yields the desired result.



Remark 1. Lemma 3.2 is very crucial for all of the following error estimates in this article.Because the proposed element satisfies a typical property, that is, for φ = (φ1, φ2, φ3) ∈ Uh

0 , φ1, φ2

and φ3 are only independent of the variable x, y and z, respectively, which can guarantee divφ = 0.However, Lemma 3.2 is not valid to the Crouzeix–Raviart nonconforming FE discussed in [13–15],for this element does not have the above propertiy. Thus, the authors therein did not present aconvergent scheme by traditional variational formula and two “stability” terms have to be addedso as to make the scheme become robust.

IV. THE SEMI-DISCRETE SCHEME

In this section, we develop a semidiscrete scheme and present the corresponding error estimate.The weak formulation for (6)–(8) reads as: find ( �Eh, �Hh, �Ph) ∈ Uh

0 ×Wh ×Uh ,∀�φ ∈ Uh0 , �ψ ∈

Wh, �φ ∈ Uh such that

ε0ε∞( �Eht , �φh)h − ( �Hh, ∇ × �φh)h + (εs − ε∞)ε0

t0( �Eh, �φh)h − 1

t0( �Ph, �φh)h = ( �f , �φh), (19)

μ0( �Hht , �ψh)h + (∇ × �Eh, �ψh)h = 0, (20)

1

(εs − ε∞)ε0( �Pht ,

�φh)h + 1

(εs − ε∞)ε0t0( �Ph, �

φh)h − 1

t0( �Eh, �

φh)h = 0. (21)

with the initial conditions

�Eh(0) = �h�E0, �Hh(0) = Qh

�H0, �Ph(0) = �h�P0.

Theorem 4.1. Let ( �E, �H , �P) and ( �Eh, �Hh, �Ph) be the solutions of (6)–(8) and (19)–(21), respec-tively. Assume that �E ∈ C1(0, T ; (H 2(�))3), �H ∈ C0(0, T ; (H 2(�))3), �P ∈ C1(0, T ; (H 1(�))3).Then there exists a constant C > 0 independent of h, such that

ε0ε∞‖ �E − �Eh‖20 + μ0‖ �H − �Hh‖2

0 + 1

(εs − ε∞)ε0‖ �P − �Ph‖2

0 ≤ Ch2. (22)

Proof. Subtracting (19)–(21) from (6)–(8) with �φ = �φh ∈ Uh0 , �ψ = �ψh ∈ Wh, �

φ = �φh ∈ Uh,

respectively, we obtain the error equations

ε0ε∞( �Et − �Eht , φh)h − ( �H − �Hh, ∇ × �φh)h + (εs − ε∞)ε0

t0( �E − �Eh, �φh)h

− 1

t0( �P − �Ph, �φh)h − 〈 �H , n × �φh〉h = 0, ∀�φh ∈ Uh

0 , (23)

μ0( �Ht − �Hht , �ψh)h + (∇ × ( �E − �Eh), �ψh)h = 0, ∀ �ψh ∈ Wh, (24)

1

(εs − ε∞)ε0( �Pt − �Pht ,

�φh)h + 1

(εs − ε∞)ε0t0( �P − �Ph, �

φh)h

− 1

t0( �E − �Eh, �

φh)h = 0, ∀ �φh ∈ Uh. (25)


1660 SHI AND YAO

Letting �ξ(t) = (�h�E − �Eh)(t), �η(t) = (Qh

�H − �Hh)(t),�ξ(t) = (�h

�P − �Ph)(t), choosing�φh = �ξ , �ψh = �η, �

φ = �ξ in (19)–(21), and rearranging the terms lead to

ε0ε∞(�ξt , �ξ)h − (�η, ∇ × �ξ)h + (εs − ε∞)ε0

t0(�ξ , �ξ)h = 1

t0(�ξ , �ξ)h + ε0ε∞(�h

�Et − �Et , �ξ)h

− (Qh�H − �H , ∇ × �ξ)h + (εs − ε∞)ε0

t0(�h

�E − �E, �ξ)h

− 1

t0(�h

�P − �P , �ξ)h − 〈 �H , n × �ξ〉h, (26)

μ0(�ηt , �η)h + (∇ × �ξ , �η)h = μ0(Qh�Ht − �Ht , �η)h + (∇ × (�h

�E − �E), �η)h, (27)

1

(εs − ε∞)ε0(�ξt ,

�ξ)h + 1

(εs − ε∞)ε0t0(�ξ , �ξ)h = 1

t0(�ξ , �ξ)h

+ 1

(εs − ε∞)ε0(�h

�Pt − �Pt ,�ξ)h + 1

(εs − ε∞)ε0t0(�h

�P − �P , �ξ)h − 1

t0(�h

�E − �E, �ξ)h. (28)

Adding (26)–(28) together, we have

1

2

d

dt

(ε0ε∞‖�ξ‖2

0 + μ0‖�η‖20 + 1

(εs − ε∞)ε0‖�ξ‖2

0

)

+ 1

(εs − ε∞)ε0t0‖�ξ‖2

0 + (εs − ε∞)ε0

t0‖�ξ‖2

0 =11∑i=1

Erri . (29)

Now we will estimate Erri (i = 1, 2, . . . , 11) one by one.First, for Err3, Err6, using ∇ × Wh = Uh, we have

Err3 = −(Qh�H − �H , ∇ × �ξ)h = 0, Err6 = μ0(Qh

�Ht − �Ht , �η)h = 0.

Second, by use of the inequality |ab| ≤ δa2 + 14δ

b2, ∀δ > 0, similar to [4], the other terms in(29) can be estimated as:

|Err1| ≤ 2

t0

(δ1

(εs − ε∞)ε0‖�ξ‖2

0 + (εs − ε∞)ε0

4δ1‖�ξ‖2

0

),

|Err2| ≤ ε0ε∞δ2‖�ξ‖20 + ε0ε∞

C1h2

4δ2‖ �Et‖2

1,

|Err4| ≤ (εs − ε∞)ε0δ4

t0‖�ξ‖2

0 + (εs − ε∞)ε0

4δ4t0C1h

2‖ �E‖21,

|Err5| ≤ ε0ε∞δ5

t0‖�ξ‖2

0 + 1

4ε0ε∞δ5t0C1h

2‖ �P ‖21,



|Err7| ≤ δ7μ0‖�η‖20 + 1

4δ7μ0C2h

2‖ �E‖22,

|Err8| ≤ δ8

(εs − ε∞)ε0‖�ξ‖2

0 + 1

4δ8(εs − ε∞)ε0C1h

2‖ �Pt‖21,

|Err9| ≤ δ9

(εs − ε∞)ε0t0‖�ξ‖2

0 + 1

4δ9(εs − ε∞)ε0t0C1h

2‖ �P ‖21,

|Err10| ≤ δ10

(εs − ε∞)ε0t0‖�ξ‖2

0 + (εs − ε∞)ε0

4δ10t0C1h

2‖ �E‖21.

Finally, by Lemma 3.2, we have

|Err11| = |〈 �H , n × �ξ〉h| ≤ C4h| �H |2‖�ξ‖0 ≤ ε0ε∞δ11‖�ξ‖20 + 1

4ε0ε∞δ11C4h

2‖ �H‖22.

Thus, choosing the proper parameters such that 12δ1

+ δ4 ≤ 1, 2δ1 + δ9 + δ10 ≤ t20 , and

C5 = max{δ2, δ5t0

, δ7, δ8, δ11}, the following important inequality can be provided

d

dt

(ε0ε∞‖�ξ‖2

0 + μ0‖�η‖20 + 1

(εs − ε∞)ε0‖�ξ‖2

0

)≤ C5

(ε0ε∞‖�ξ‖2

0 + μ0‖�η‖20 + 1

(εs − ε∞)ε0‖�ξ‖2

0

)+ C1h

2 ε0ε∞4δ2

‖ �Et‖21

+(

C1(εs − ε∞)ε0

4δ4t0+ C1(εs − ε∞)ε0

4δ10t0+ C2

4δ7μ0

)h2‖ �E‖2

2 + C4

4ε0ε∞δ11h2‖ �H‖2

2

+ C1

4δ8(εs − ε∞)ε0h2‖ �Pt‖2

1 +(

1

4δ5t0ε0ε∞+ 1

4δ9(εs − ε∞)ε0t0

)C1h

2‖ �P ‖21.

Then integrating the above inequality with respect to t , and using the facts �ξ(0) = 0, �η(0) =0, �ξ(0) = 0, we obtain

ε0ε∞‖�ξ‖20 + μ0‖�η‖2

0 + 1

(εs − ε∞)ε0‖�ξ‖2

0 ≤ C5

∫ t

0

(ε0ε∞‖�ξ‖2

0 + μ0‖�η‖20 + 1

(εs − ε∞)ε0‖�ξ‖2

0

)dt

+ C6h2

∫ t

0(‖ �Et‖2

1 + ‖ �E‖22 + ‖ �H‖2

2 + ‖ �Pt‖21 + ‖ �P ‖2

1)dt , (30)

where

C6 = max

{C1ε0ε∞

δ2,

(C1(εs − ε∞)ε0

4δ4t0+ C1(εs − ε∞)ε0

4δ10t0+ C2

4δ7μ0

),

C4

4ε0ε∞δ11,

C11

4δ8(εs − ε∞)ε0, C1

(1

4δ5t0ε0ε∞+ 1

4δ9(εs − ε∞)ε0t0

)}.

Employing Grownwall inequality, yields

ε0ε∞‖�ξ‖20 + μ0‖�η‖2

0 + 1

(εs − ε∞)ε0‖�ξ‖2

0 ≤ Ch2

∫ t

0(‖ �Et‖2

1 + ‖ �E‖22 + ‖ �H‖2

2 + ‖ �Pt‖21 + ‖ �P ‖2

1)dt ,

(31)

which together with Lemma 3.1 and triangle inequality yield the desired result.


1662 SHI AND YAO

V. THE FULLY-DISCRETE SCHEME

In this section, we will develop a fully-discrete scheme, and discuss its stability and optimal errorestimates.

To construct a fully-discrete scheme, we divide the time interval [0, T ] into M uniform subin-tervals with points 0 = t0 < t1 < · · · < tM = T , where tk = kτ . Let Ik = [tk−1, tk], 1 ≤ k ≤ M ,and denote uk = u(·, kτ),

δτuk = uk − uk−1/τ , uk− 1

2 = (uk + uk−1)/2.

Then, the fully-discrete mixed finite element scheme for (6)–(8) can be formulated as: For all

k = 1, . . . , M , find ( �Ekh, �Hk

h , �P kh ) ∈ Uh

0 × Wh × Uh such that for all �φ ∈ Uh0 , �ψ ∈ Wh, �

φ ∈ Uh

there hold

ε0ε∞(δτ�Ek

h, �φ) − ( �Hk− 12

h , ∇ × �φ)h + (εs − ε∞)ε0

t0( �Ek

h, �φ) − 1

t0( �P k− 1

2h , �φ) = ( �f , �φ), (32)

μ0(δτ�Hk+ 1

2h , �ψ) + (∇ × �Ek

h, �ψ)h = 0, (33)

1

(εs − ε∞)ε0(δτ

�P k+ 12

h , �φ) + 1

(εs − ε∞)ε0t0( �P k+ 1

2h , �

φ) − 1

t0( �Ek−1

h , �φ) = 0. (34)

Similar to the proof in [22], a unique solution ( �Ekh, �Hk

h , �P kh ) to (32)–(34) at each time step can

be guaranteed.

Theorem 5.1. Let ( �Enh , �Hn+ 1

2h , �P n+ 1

2h ) be the solution of (32)–(34), cv = 1√

μ0ε0the speed of light

in vacuum, c1 the constant from the inverse estimate

‖∇ × �ψh‖0 ≤ c1h−1‖ �ψh‖0, �ψh ∈ Vh,

then for n ≥ 1, if the time step τ satisfies

τ ≤ min

( √ε∞h√2c1cv

,√

ε∞2(εs − ε∞)

t0, 1

), (35)

we have the following discrete stability:

ε0ε∞‖ �Enh‖2

0 + μ0‖ �Hn+ 12

h ‖20 + 1

(εs − ε∞)ε0‖ �P n+ 1

2h ‖2

0

≤ 3

(ε0ε∞‖ �E0

h‖20 + μ0‖ �H 1

2h ‖2

0 + 1

(εs − ε∞)ε0‖ �P 1

2h ‖2

0

). (36)



Proof. Choosing �φ = τ( �Ekh + �Ek−1

h ), �ψ = τ( �Hk− 12

h + �Hk+ 12

h ) and �φ = τ( �P k− 1

2h + �P k+ 1

2h ) for

(32)–(34), respectively, and then adding them together, we have

ε0ε∞(‖ �Ekh‖2

0 − ‖ �Ek−1h ‖2

0) − τ( �Hk− 12

h , ∇ × ( �Ekh + �Ek−1

h )) + (εs − ε∞)ε0τ

2t0‖ �Ek

h + �Ek−1h ‖2

0

− τ

t0( �P k− 1

2h , �Ek

h + �Ek−1h ) + μ0(‖ �Hk+ 1

2h ‖2

0 − ‖ �Hk− 12

h ‖20) + τ(∇ × �Ek

h, �Hk+ 12

h + �Hk− 12

h )

− 1

(εs − ε∞)ε0(‖ �P k+ 1

2h ‖2

0 − ‖ �P k− 12

h ‖20) + τ

2(εs − ε∞)ε0t0‖ �P k+ 1

2h + �P k− 1

2h ‖2

0

− τ

t0( �Ek−1

h , �P k+ 12

h + �P k− 12

h ) = 0.

Note that

τ [( �Hk− 12

h , ∇ × ( �Ekh + �Ek−1

h )) − (∇ × �Ekh, �Hk+ 1

2h + �Hk− 1

2h )]

= τ [( �Hk− 12

h , ∇ × �Ek−1h ) − (∇ × �Ek

h, �Hk+ 12

h )], (37)

and

− τ

[(εs − ε∞)ε0

2t0‖ �Ek

h + �Ek−1h ‖2

0 − 1

t0( �P k+ 1

2h + �P k− 1

2h , �Ek

h + �Ek−1h )

+ 1

2(εs − ε∞)ε0t0‖ �P k+ 1

2h + �P k− 1

2h ‖2

0

]

= −τ

2

∥∥∥∥∥√

(εs − ε∞)ε0

t0( �Ek

h + �Ek−1h ) − 1√

(εs − ε∞)ε0t0( �P k+ 1

2h + �P k− 1

2h )

∥∥∥∥∥2

0

, (38)

we obtain

ε0ε∞(‖ �Ekh‖2

0 − ‖ �Ek−1h ‖2

0) + μ0(‖ �Hk+ 12

h ‖20 − ‖ �Hk− 1

2h ‖2

0) + 1

(εsε∞)ε0(‖ �P k+ 1

2h ‖2

0 − ‖ �P k− 12

h ‖20)

≤ τ [( �Hk− 12

h , ∇ × �Ek−1h ) − (∇ × �Ek

h, �Hk+ 12

h )] + τ

t0( �P k− 1

2h , �Ek

h + �Ek−1h )

+ τ

t0( �Ek−1

h , �P k+ 12

h + �P k− 12

h ) − τ

t0( �P k+ 1

2h + �P k− 1

2h , �Ek

h + �Ek−1h ) (39)

≤ τ [( �Hk− 12

h , ∇ × �Ek−1h ) − ( �Hk+ 1

2h , ∇ × �Ek

h)]

+ τ

t0( �Ek−1

h , �P k+ 12

h + �P k− 12

h ) − τ

t0( �P k+ 1

2h , �Ek

h + �Ek−1h )

= τ [( �Hk− 12

h , ∇ × �Ek−1h ) − ( �Hk+ 1

2h , ∇ × �Ek

h)] + τ

t0( �Ek−1

h , �P k− 12

h ) − τ

t0( �Ek

h, �P k+ 12

h ). (40)


1664 SHI AND YAO

Then summing (40) from k = 1 to n , we have

ε0ε∞(‖ �Enh‖2

0 − ‖ �E0h‖2

0) + μ0(‖ �Hn+ 12

h ‖20 − ‖ �H 1

2h ‖2

0) + 1

(εsε∞)ε0(‖ �P n+ 1

2h ‖2

0 − ‖ �P 12

h ‖20)

≤ τ( �H 12

h , ∇ × �E0h) − τ( �Hn+ 1

2h , ∇ × �En

h) + τ

t0( �E0

h, �P 12

h ) − τ

t0( �En

h , �P n+ 12

h ). (41)

By the inverse estimate and Cauchy–Schwartz inequality, we get

τ( �Hn+ 12

h , ∇ × �Enh) ≤ τ · c1h

−1‖ �Enh‖0‖ �Hn+ 1

2h ‖0

= τ · c1h−1 · cv

√ε0‖ �En

h‖0 · √μ0‖ �Hn+ 1

2h ‖0

≤ 1

4ε0ε∞‖ �En

h‖20 +

(c1cvτ

h

)2 μ0

ε∞‖ �Hn+ 1

2h ‖2

0. (42)

Similarly,

τ

t0( �En

h , �P n+ 12

h ) ≤ 1

4ε0ε∞‖ �En

h‖20 + εs − ε∞

ε∞t20

· τ 2

(εs − ε∞)ε0

∥∥∥ �P n+ 12

h

∥∥∥2

0. (43)

τ( �H 12

h , ∇ × �E0h) ≤ 1

4ε0ε∞‖ �E0

h‖20 +

(c1cvτ

h

)2 μ0

ε∞

∥∥∥ �H 12

h

∥∥∥2

0, (44)

τ

t0( �E0

h, �P 12

h ) ≤ 1

4ε0ε∞‖ �E0

h‖20 + εs − ε∞

ε∞t20

· τ 2

(εs − ε∞)ε0

∥∥∥ �P 12

h

∥∥∥2

0. (45)

Substituting (42)–(45) into (41), and using (35), we can complete the proof.

The following lemma is useful for the error estimates of full-discrete scheme which can befound in [5, 7, 9–11, 22].

Lemma 5.1. Let uj = u(·, jτ), we have

i.

∥∥∥∥∥∥uk − 1

τ

∫ tk+ 1

2

tk− 1

2

u(s)ds

∥∥∥∥∥∥2

0

≤ τ 3

4

∫ tk+ 1

2

tk− 1

2

‖utt (s)‖20ds, ∀u ∈ (L2(�))3,

ii.

∥∥∥∥1

2(uk + uk+1) − 1

τ

∫ tk+1

tk

u(s)ds

∥∥∥∥2

0

≤ τ 3

4

∫ tk+1

tk

‖utt (s)‖20ds, ∀u ∈ (L2(�))3,

iii.∥∥δτu

k∥∥2

0≤ 1

τ

∫ tk

tk−1

‖ut(t)‖20dt , ∀u ∈ (L2(�))3.

Theorem 5.2. Let ( �En, �Hn, �P n) and ( �Enh , �Hn+ 1

2h , �P n+ 1

2h ) be the solutions of (6)–(8) and (32)–

(34) at time tn = nτ , respectively. Then, there exists a constant C > 0, independent of τ and h ,such that

max1≤n≤M

(‖ �En − �Enh‖h + ‖ �Hn+ 1

2 − �Hn+ 12

h ‖h + ‖ �P n+ 12 − �P n+ 1

2h ‖h) ≤ C(h + τ 2). (46)



Proof. Let �ξ kh = �h

�Ek − �Ekh, �ηk

h = Qh�Hk − �Hk

h , �ξ kh = �h

�P k − �P kh . Integrating (6)–(8) over

Ik = [tk−1, tk] , subtracting from (32)-(34), respectively. Then for all �φh ∈ Uh0 , �ψh ∈ Wh, �

φh ∈ Uh,we have

i. ε0ε∞(δτ�ξ kh , �φh)h − (�ηk− 1

2h , ∇ × �φh)h

= ε0ε∞(δτ (�h�Ek − �Ek), �φh)h −

(Qh

�Hk− 12 − 1

τ

∫Ik

�H(s)ds, ∇ × �φh

)h

+⟨

1

τ

∫Ik

�Hds, n × �φh

⟩h

+ (εs − ε∞)ε0

t0

(�Ek

h − 1

τ

∫Ik

�E(s)ds, �φh

)h

+ 1

t0

(1

τ

∫Ik

�P(s)ds − �P k− 12

h , �φh

)h

,

ii. μ0(δτ �ηk+ 12

h , �ψh)h + (∇ × �ξ kh , �ψh)h

= μ0(δτ (Qh�Hk+ 1

2 − �Hk+ 12 ), �ψh)h +

⎛⎝∇ × (�h�Ek − 1

τ

∫Ik+ 1

2

�E(s)ds), �ψh

⎞⎠h

,

iii.1

(εs − ε∞)ε0(δτ

�ξ

k+ 12

h , �φh)h = 1

(εs − ε∞)ε0(δτ (�h

�P k+ 12 − �P k+ 1

2 ), �φh)

+ 1

(εs − ε∞)ε0t0

⎛⎝ �P k+ 12

h − 1

τ

∫Ik+ 1

2

�P(s)ds

⎞⎠ , �φh)h + 1

t0

⎛⎝1

τ

∫Ik+ 1

2

�E(s)ds − �Ek−1h , �

φh

⎞⎠h

.

Let �φh = τ(�ξ kh + �ξ k−1

h ), �ψh = τ(�ηk+ 12

h + �ηk− 12

h ), �φh = τ(

�ξ

k+ 12

h + �ξ

k− 12

h ), and note that

(∇ × �ξ kh , �ηk+ 1

2h + �ηk− 1

2h )h − (�ηk− 1

2h , ∇ × (�ξ k

h + �ξ k−1h ))h = (∇ × �ξ k

h , �ηk+ 12

h )h − (∇ × �ξ k−1h , �ηk− 1

2h )h,

we have

ε0ε∞(‖�ξ k0 ‖2

0 − ‖�ξ k−1h ‖2

0) + μ0(‖�ηk+ 12

h ‖20 − ‖�ηk− 1

2h ‖2

0) + 1

(εs − ε∞)ε0(‖�

ξk+ 1

2h ‖2

0 − ‖�ξ

k− 12

h ‖20)

= τ [(∇ × �ξ k−1h , �ηk− 1

2h )h − (∇ × �ξ k

h , �ηk+ 12

h )h] + τε0ε∞(δτ (�h�Ek − �Ek), �ξ k

h + �ξ k−1h )h

− τ

(�Hk− 1

2 − 1

τ

∫Ik

�H(s)ds, ∇ × (�ξ kh + �ξ k−1

h )

)h

+ τ(εs − ε∞)ε0

t0

(�Ek

h − �h�Ek + �h

�Ek − �Ek + �Ek − 1

τ

∫Ik

�E(s)ds, �ξ kh + �ξ k−1

h

)h

+ τ

t0

(1

τ

∫Ik

�P(s)ds − �P k− 12 + �P k− 1

2 − �h�P k− 1

2 + �ξ

k− 12

h , �ξ kh + �ξ k−1

h

)h

+ τ

⎛⎝∇ × (�h�Ek − �Ek + �Ek − 1

τ

∫Ik+ 1

2

�E(s)ds), �ηk+ 12

h + �ηk− 12

h

⎞⎠h


1666 SHI AND YAO

+ τ

(εs − ε∞)ε0(δτ (�h

�P k+ 12 − �P k+ 1

2 ), �ξ k+ 12

h + �ξ

k− 12

h )h

+ τ

(εs − ε∞)ε0t0

⎛⎝ �P k+ 12

h − �h�P k+ 1

2 + �h�P k+ 1

2 − �P k+ 12 + �P k+ 1

2

− 1

τ

∫Ik+ 1

2

�P(s)ds, �ξ k+ 12

h + �ξ

k− 12

h

⎞⎠h

+ τ

t0

⎛⎝1

τ

∫Ik+ 1

2

�E(s)ds − �Ek−1 + �Ek−1 − �h�Ek−1 + �ξ k−1

h , �ξ k+ 12

h + �ξ

k− 12

h

⎞⎠h

+ τ

⟨1

τ

∫Ik

�Hds, n × (�ξ kh + �ξ k−1

h )

⟩h

, (47)

It is easy to check that

�Ekh − �h

�Ek = −1

2(�ξ k

h + �ξ k−1h ), �P k+ 1

2h − �h

�P k+ 12 = −1

2(�ξ

k+ 12

h + �ξ

k− 12

h ), (48)

and

(εs − ε∞)ε0

2t0‖�ξ k

h + �ξ k−1h ‖2

0 − 1

t0(�ξ

k− 12

h + �ξ

k+ 12

h , �ξ kh + �ξ k−1

h ) + 1

2(εs − ε∞)ε0t0‖�ξ

k− 12

h + �ξ

k+ 12

h ‖20

= 1

2

∥∥∥∥∥√

(εs − ε∞)ε0

t0(�ξ k

h + �ξ k−1h ) − 1√

(εs − ε∞)ε0t0(�ξ

k− 12

h + �ξ

k+ 12

h )

∥∥∥∥∥2

0

. (49)

Summing up (49) from k = 1 to n, we have

ε0ε∞(‖�ξnh ‖2

0 − ‖�ξ 0h‖2

0) + μ0(‖�ηn+ 12

h ‖20 − ‖�η 1

2h ‖2

0) + 1

(εs − ε∞)ε0(‖�

ξn+ 1

2h ‖2

0 − ‖�ξ

12

h ‖20)

= τ(∇ × �ξ 0h , �η 1

2h )h − τ(∇ × �ξn

h , �ηn+ 12

h )h + τε0ε∞n∑

k=1

(δτ (�h�Ek − �Ek), �ξ k

h + �ξ k−1h )h

− τ

n∑k=1

(�Hk− 1

2 − 1

τ

∫Ik

�H(s)ds, ∇ × (�ξ kh + �ξ k−1

h )

)h

+ τ(εs − ε∞)ε0

t0

n∑k=1

(�h

�Ek − �Ek + �Ek − 1

τ

∫Ik

�E(s)ds, ξ kh + ξ k−1

h

)h

+ τ

t0

n∑k=1

(1

τ

∫Ik

�P(s)ds − �P k− 12 + �P k− 1

2 − �h�P k− 1

2 − �ξ

k+ 12

h , �ξ kh + �ξ k−1

h

)h

+ τ

n∑k=1

⎛⎝∇ ×⎛⎝�h

�Ek − �Ek + �Ek − 1

τ

∫Ik+ 1

2

�E(s)ds

⎞⎠ , �ηk+ 12

h + �ηk− 12

h

⎞⎠h



+ τ

(εs − ε∞)ε0

n∑k=1

(δτ (�h

�P k+ 12 − �P k+ 1

2 ), �ξ k+ 12

h + �ξ

k− 12

h

)h

+ τ

(εs − ε∞)ε0t0

n∑k=1

⎛⎝�h�P k+ 1

2k − �P k+ 1

2k + �P k+ 1

2k − 1

τ

∫Ik+ 1

2

�P(s)ds, �ξ k+ 12

h + �ξ

k− 12

h

⎞⎠h

+ τ

t0

n∑k=1

⎛⎝1

τ

∫Ik+ 1

2

�E(s)ds − �Ek + �Ek − �h�Ek + �ξ k−1

h , �ξ k+ 12

h + �ξ

k− 12

h

⎞⎠h

+ τ

n∑k=1

⟨1

τ

∫Ik

�Hds, n × (�ξ kh + �ξ k−1

h )

⟩h

�11∑i=1

Zi . (50)

Now, we will estimate each term Zi(i = 1, . . . , 11).First, by inverse inequality, the fist two terms can be estimated by

Z1 = τ(∇ × �ξ 0h , �η 1

2h )h ≤ τc1h

−1‖�ξ 0h‖0 · ‖�η 1

2h ‖0

= τc1h−1cν

√ε0‖�ξ 0

h‖0 · √μ0‖�η 1

2h ‖0 ≤ δ1ε0ε∞‖�ξ 0

h‖20 + 1

4δ1

(c1cντ√

ε∞h

)2

μ0‖�η 12h ‖2

0.

Z2 = τ(∇ × �ξnh , �ηn+ 1

2h )h ≤ δ2ε0ε∞‖�ξn

h ‖20 + 1

4δ2

(c1cντ√

ε∞h

)2

μ0‖�ηn+ 12

h ‖20.

Second, using Lemma 3.1, Lemma 5.1 and Green’s formula, we obtain

Z3 = τε0ε∞n∑

k=1

(δτ (�h�Ek − �Ek), �ξ k

h + �ξ k−1h )h

≤ τε0ε∞n∑

k=1

[δ3

∥∥∥�ξ kh + �ξ k−1

h

∥∥∥2

0+ 1

4δ3τ

∫ tk

tk−1

∥∥∥(�h�Ek − �Ek)t

∥∥∥2

0dt

]

≤ τε0ε∞δ3(‖�ξnh ‖2

0 + ‖�ξ 0h‖2

0) + 2τε0ε∞δ3

n−1∑k=1

‖�ξ kh‖2

0 + ε0ε∞T

4δ3C1h

2‖ �Et‖21,

Z4 = τ

n∑k=1

(�Hk− 1

2 − 1

τ

∫Ik

�H(s)ds, ∇ × (�ξ kh + �ξ k−1

h )

)h

= τ

n∑k=1

(∇ × �Hk− 1

2 − 1

τ

∫Ik

∇ × �H(s)ds, (�ξ kh + �ξ k−1

h )

)h

+ τ

n∑k=1

⟨�Hk− 1

2 − 1

τ

∫Ik

�H(s)ds, n × (�ξ kh + �ξ k−1

h )

⟩h

≤ 2τδ4

n∑k=1

(‖�ξ kh‖2

0 + ‖�ξ k−1h ‖2

0) + τ

4δ4

n∑k=1

∥∥∥∥∇ × �Hk− 12 − 1

τ

∫Ik

∇ × �H(s)ds

∥∥∥∥2

0


1668 SHI AND YAO

+ τC4h

n∑k=1

∥∥∥∥ �Hk− 12 − 1

τ

∫Ik

�H(s)ds

∥∥∥∥2

‖�ξ kh + �ξ k−1

h ‖0

≤ 2τδ4(‖�ξnh ‖2

0 + ‖�ξ 0h‖2

0) + 4τδ4

n−1∑k=1

‖�ξ kh‖2

0 + τ 4T

16δ4‖∇ × �Htt‖2

0

+ τC4h2

n∑k=1

τ 3

4δ4

∫ tk

tk−1

‖ �Htt (s)‖20ds + τδ4

n∑k=1

‖�ξ kh + �ξ k−1

h ‖0

≤ 6τδ4(‖�ξnh ‖2

0 + ‖�ξ 00 ‖2

h) + 8τδ4

n−1∑k=1

‖�ξ kh‖2

0 + τ 4T

16δ4‖∇ × �Htt‖2

0 + T τ 4

4δ4C4h

2‖ �Htt‖20,

Z5 = τ(εs − ε∞)ε0

t0

n∑k=1

(�h

�Ek − �Ek + �Ek − 1

τ

∫Ik

�E(s)ds, �ξ kh + �ξ k−1

h

)h

≤ τ(εs − ε∞)ε0

t0

n∑k=1

2δ5(‖�ξ kh‖2

0 + ‖�ξ k−1h ‖2

0)

+ τ(εs − ε∞)ε0

t0

n∑k=1

1

2δ5

(∥∥∥∥1

2(�h

�Ek + �h�Ek−1 )

− 1

2( �Ek + �Ek−1)

∥∥∥∥2

0

+∥∥∥∥1

2( �Ek + �Ek−1) − 1

τ

∫Ik

�E(s)ds

∥∥∥∥2

0

)

≤ 2τ(εs − ε∞)ε0

t0δ5(‖�ξn

h ‖20 + ‖�ξ 0

h‖20) + 4τδ5

(εs − ε∞)ε0

t0

n−1∑k=1

‖�ξ kh‖2

0

+ τ(εs − ε∞)ε0T

2t0δ5C1h

2‖ �E‖21 + τ 4(εs − ε∞)ε0T

16t0δ5‖ �Ett‖2

0.

With the similar arguments the above, we have

Z6 = τ

t0

n∑k=1

(1

τ

∫Ik

�P(s)ds − �P k− 12 + �P k− 1

2 − �h�P k− 1

2 − �ξ

k+ 12

h , �ξ kh + �ξ k−1

h

)h

≤ 2δ6τ

t0(‖�ξn

h ‖20 + ‖�ξ 0

h‖20) + 4δ6

τ

t0

n−1∑k=1

‖�ξ kh‖2

0 + τ

2δ6t0

n∑k=1

‖�ξ

k+ 12

h ‖20

+ τ 4

8δ6t0

∫ tn

t0

‖ �Ptt (s)‖20ds + τT

2δ6t0C1h

2‖ �P ‖21,

Z7 = τ

n∑k=1

⎛⎝∇ ×⎛⎝�h

�Ek − �Ek + �Ek − 1

τ

∫Ik+ 1

2

�E(s)ds

⎞⎠ , �ηk+ 12

h + �ηk− 12

h

⎞⎠h

≤ 2τδ7(‖�ηn+ 12

h ‖20 + ‖�η 1

2h ‖2

0) + 4τδ7

n−1∑k=1

‖�ηk+ 12

h ‖20 + C2T

δ7h2‖ �E‖2

2 + T τ 4

8δ7‖∇ × �Htt‖2

0,



Z8 = τ

(εs − ε∞)ε0

n∑k=1

(δτ (�h�P k+ 1

2 − �P k+ 12 ), �ξ k+ 1

2h + �

ξk− 1

2h )h

≤ τ

(εs − ε∞)ε02δ8(‖�

ξn+ 1

2h ‖2

0 + ‖�ξ

12

h ‖20) + τ

(εs − ε∞)ε04δ8

n−1∑k=1

‖�ξ

k+ 12

h ‖20

+ C1T

4δ8(εs − ε∞)ε0h2‖ �Pt‖2

1,

Z9 = τ

(εs − ε∞)ε0t0

n∑k=1

⎛⎝�h�P k+ 1

2k − �P k+ 1

2k + �P k+ 1

2k − 1

τ

∫Ik+ 1

2

�P(s)ds, �ξ k+ 12

h + �ξ

k− 12

h

⎞⎠h

≤ τ

(εs − ε∞)ε0t02δ9(‖�

ξn+ 1

2h ‖2

0 + ‖�ξ

12

h ‖20) + τ

(εs − ε∞)ε0t04δ9

n−1∑k=1

‖�ξ

k+ 12

h ‖20

+ C1T τ

4δ9(εs − ε∞)ε0t0h2‖ �P ‖2

1 + τ 4T

16δ9(εs − ε∞)ε0t0‖ �Ptt‖2

0,

Z10 = τ

t0

n∑k=1

⎛⎝1

τ

∫Ik+ 1

2

�E(s)ds − �Ek−1 + �Ek−1 − �h�Ek−1 + �ξ k−1

h , �ξ k+ 12

h + �ξ

k− 12

h

⎞⎠h

≤ 2δ10τ

t0(‖�

ξn+ 1

2h ‖2

0 + ‖�ξ

12

h ‖20) + τ

t04δ10

n−1∑k=1

‖�ξ

k+ 12

h ‖20

+ τ 4T

8δ10t0‖ �Ett‖2

0 + C1T τ

2δ10t0h2‖ �E‖2

1 + τ

2δ10t0

n∑k=1

‖�ξ k−1h ‖2

0,

Z11 = τ

n∑k=1

⟨1

τ

∫Ik

�Hds, n × (�ξ kh + �ξ k−1

h )

⟩h

≤ τ

n∑k=1

⟨ �H , n × (�ξ kh + �ξ k−1

h )⟩h

≤ 2δ11τ(‖�ξnh ‖2

0 + ‖�ξ 0h‖2

0) + 4δ11τ

n−1∑k=1

‖�ξ kh‖2

0 + τT

4δ11C4h

2| �H |22.

Now, we chose δi , (i = 1, 2, . . . , 11) satisfying

δ2ε0 + τε0ε∞δ3 + 4δ4τ + 2τ(εs − ε∞)ε0

t0δ5 + 2δ6τ

t0+ 2δ11τ ≤ 1

2ε0ε∞,

(c1cvτ

h

)2 μ0

4δ2ε∞+ 2τδ7 ≤ 1

2μ0,

τ

2δ6t0+ 2δ8τ

(εs − ε∞)ε0+ 2δ9τ

(εs − ε∞)ε0t0+ 2δ10τ

t0≤ 1

2(εs − ε∞)ε0,


1670 SHI AND YAO

and let

C7 = max

{δ1ε0 + τε0ε∞δ3 + 4δ4τ + 2τ(εs − ε∞)ε0

t0δ5 + 2δ6τ

t0+ 2δ11τ + ε0ε∞,

(c1cvτ

h

)2 μ0

4δ2ε∞+ 2τδ7 + μ0,

2δ8τ

(εs − ε∞)ε0+ 2δ9τ

(εs − ε∞)ε0t0+ 2δ10τ

t0+ 1

(εs − ε∞)ε0

},

C8 = max

{(2τε0ε∞δ3 + 2τδ4 + 4τδ5

(εs − ε∞)ε0

t0+ 4δ6τ

t0,

4δ7τ

μ0,

(τ

2δ6t0+ 4τδ8

(εs − ε∞)ε0+ 4τδ9

(εs − ε∞)ε0t0+ 4τδ10

t0

)· (εs − ε∞)ε0

}.

Then, substituting the estimates of Z1 − Z11 into (50), we have

ε0ε∞‖�ξnh ‖2

0 + μ0‖�ηn+ 12

h ‖20 + 1

(εs − ε∞)ε0‖�ξ

n+ 12

h ‖20

≤ C7(‖�ξ 0h‖2

0 + ‖�η 12h ‖2

0 + ‖�ξ

12

h ‖20) +

[(ε0ε∞4δ3

C1T ‖ �Et‖22 + C4τ

4T

4δ4‖ �Htt‖2

0

+(

C1(εs − ε∞)ε0T

2t0δ5+ C2T

δ7+ C1T τ

2δ10t0

)‖ �E‖2

1 +(

τT

2δ6t0+ T τ

4δ9(εs − ε∞)ε0t0

)C1‖ �P ‖2

1

+ C1T

4δ8(εs − ε∞)ε0‖ �Pt‖2

1 + C4T τ

4δ11‖ �H‖2

2

)h2 + τ 4

((T

16δ4+ T

8δ7

)‖ �Htt‖2

1

+(

(εs − ε∞)ε0T

16δ5t0+ T

8δ10t0

)‖ �Ett‖2

0 +(

T

8δ6t0+ T

16δ9(εs − ε∞)ε0t0

)‖ �Ptt‖2

0

)]

+ C8

n−1∑k=1

(ε0ε∞‖�ξ k

h‖20 + μ0‖�ηk+ 1

2h ‖2

0 + 1

(εs − ε∞)ε0‖�ξ

k+ 12

h ‖20

),

which implies


0 + μ0‖�ηn+ 12

h ‖20 + 1

(εs − ε∞)ε0‖�ξ

n+ 12

h ‖20 ≤ C7(‖�ξ 0

h‖20 + ‖�η 1

2h ‖2

0 + ‖�ξ

12

h ‖20) + C9(h

2 + τ 4)

+ C8

n−1∑k=1

(ε0ε∞‖�ξ k

h‖20 + μ0‖�ηk+ 1

2h ‖2

0 + 1

(εs − ε∞)ε0‖�ξ

k+ 12

h ‖20

), (51)

where

C9 = max

{(ε0ε∞4δ3

C1T ‖ �Et‖21 + C4τ

4T

4δ4‖ �Htt‖2

0

+(

C1(εs − ε∞)ε0T

2t0δ5+ C2T

δ7+ C1T τ

2δ10t0

)‖ �E‖2

2 +(

C1τT

2δ6t0+ C1T τ

4δ9(εs − ε∞)ε0t0

)‖ �P ‖2

1



FIG. 1. The left figure shows the error curve at t = 0.4 and the right figure shows at t = 1 by nonconform-ing FE and Nédélec’s FE, respectively. [Color figures can’ be viewed in the online issue, which is availableat wileyonlinelibraray.com.]

+ C1T

4δ8(εs − ε∞)ε0‖ �Pt‖2

1 + C4T τ

4δ11‖ �H‖2

2

),

(T

16δ4+ T

8δ7

)‖ �Htt‖2

1

+(

(εs − ε∞)ε0T

16δ5t0+ T

8δ10t0

)‖ �Ett‖2

0 +(

T

8δ6t0+ T

16δ9(εs − ε∞)ε0t0

)‖ �Ptt‖2

0

}.

So by the discrete Grownwall inequality, we get


0 + μ0‖�ηn+ 12

h ‖20 + 1

(εs − ε∞)ε0‖�ξ

n+ 12

h ‖20 ≤ C(‖�ξ 0

h‖20 + ‖�η 1

2h ‖2

0 + ‖�ξ

12

h ‖20) + C(h + τ 2)2,

(52)

which together with (48) and triangle inequality yield the desired result.

VI. NUMERICAL EXPERIMENTS

To confirm our theoretical analysis and compare with Nédélec’s FE, we compute the electricfield E and magnetic field H by fully-discrete scheme (32)–(34). We assume that � = [0, 1]3 ,I = [0, 1]. All the physical parameters are taken as ε0 = 1, ε∞ = 1, εs = 2, t0 = 1, μ0 = 1. Then,we can construct the analytical solutions of (1)–(3) as

�E =⎛⎝E1

E2E3

⎞⎠ =⎛⎝A cos(π ∗ x) sin(π ∗ y) sin(π ∗ z)

B sin(π ∗ x) cos(π ∗ y) sin(π ∗ z)

C sin(π ∗ x) sin(π ∗ y) cos(π ∗ z)

⎞⎠ e−t cos t ,

�H =⎛⎝H1

H2H3

⎞⎠ =⎛⎝ π(C − B) sin(π ∗ x) cos(π ∗ y) cos(π ∗ z)

π(A − C) cos(π ∗ x) sin(π ∗ y) cos(π ∗ z)

π(B − A) cos(π ∗ x) cos(π ∗ y) sin(π ∗ z)

⎞⎠ e−t (cos t − sin t)/2,

�P =⎛⎝ P 1

P 2P 3

⎞⎠ =⎛⎝ A cos(π ∗ x) sin(π ∗ y) sin(π ∗ z)

B sin(π ∗ x) cos(π ∗ y) sin(π ∗ z)

C sin(π ∗ x) sin(π ∗ y) cos(π ∗ z)

⎞⎠ e−t sin t ,


1672 SHI AND YAO

TABLE I. Errors at t = 0.4.

N × N × N Non(Err) order Ned(Err) order

4 × 4 × 4 0.2712 – 0.3218 –8 × 8 × 8 0.1371 0.9837 0.1627 0.983916 × 16 × 16 0.0687 0.9985 0.0817 0.994732 × 32 × 32 0.0343 0.9998 0.0409 0.9977

TABLE II. Errors at t = 1.

N × N × N Non(Err) order Ned(Err) order

4 × 4 × 4 0.1545 – 0.1853 –8 × 8 × 8 0.0777 0.9921 0.0961 0.946416 × 16 × 16 0.0389 0.9974 0.0483 0.992132 × 32 × 32 0.0195 0.9992 0.0242 0.9984

where A = 1, B = 13 , C = − 4

3 . It is easy to check that the source term is

�f (X, t) = ε0ε∞∂ �E∂t

− ∇ × �H + (εs − ε∞)ε0

t0�E − 1

t0�P .

The errors of our nonconforming FE and Nédélec’s FE at time t = 0.4 and 1 with τ = 0.001are listed in the following Tables I and II, respectively, where N is the number of partition inx, y, z−direction. The total degrees of freedom(Dofs) for Nédélec’s FE can be calculated by for-mula 3N 3 + 6N 2 + 3N and the total Dofs for the proposed nonconforming FE can be calculatedby the formula 6N 3 + 6N 2. The error curves in Fig. 1 are given. Let

Err = max1≤n≤M

(‖ �En − �Enh‖0 + ‖ �Hn+ 1

2 − �Hn+ 12

h ‖0 + ‖ �P n+ 12 − �P n+ 1

2h ‖0).

Non(Err) and Ned(Err) denote the errors of our proposed nonconforming FE and Nédélec’sFE, respectively.

From Tables I and II and Fig. 1, we may see that our proposed element indeed has a goodperformance as that of Nédélec’s element.

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