Error estimates of the Crank-Nicolson Galerkin method for the time ...

arXiv:1703.02274v1 [math.NA] 7 Mar 2017

IMA Journal of Numerical Analysis (2017) Page 1 of 27 doi:10.1093/imanum/drnxxx

Error estimates of the Crank-Nicolson Galerkin method for the time-dependent Maxwell-Schrödinger equations under the Lorentz gauge C HUPENG M A†, L IQUN C AO‡ Institute of Computational Mathematics and Scientific/Engineering Computing, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, China AND

YANPING L IN§ Department of Applied Mathematics, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong, China [Received on 8 March 2017] In this paper we study the numerical method and the convergence for solving the time-dependent Maxwell-Schrödinger equations under the Lorentz gauge. An alternating Crank-Nicolson finite element method for solving the problem is presented and the optimal error estimate for the numerical algorithm is obtained by a mathematical inductive method. Numerical examples are then carried out to confirm the theoretical results. Keywords: error estimates; Crank-Nicolson; Galerkin method; Maxwell-Schrödinger.

1. Introduction Light-matter interaction at nanoscale is a central topic in the study of optical properties of nanophotonic systems, for example, metallic nanostructures and quantum dots. In view of practical numerical simulation, a semiclassic model is often used for modelling light-matter interaction. The basic idea is to use the classical Maxwell’s equations for the electromagnetic field and the Schrödinger equation for the matter. In this paper, we study the following Maxwell-Schrödinger coupled system, which describes the

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interaction between an electron and its self-consistent generated and external electromagnetic fields.  1 ∂ ψ (x,t)  2  i¯h = [i¯h∇ + qA(x,t)] + qφ (x,t) + V0 ψ (x,t),    ∂t 2m    (x,t) ∈ Ω × (0, T ),        − ∂ ∇ · ε A(x,t) − ∇ · ε ∇φ (x,t) = q|ψ (x,t)|2 , (x,t) ∈ Ω × (0, T ),    ∂t ∂ 2 A(x,t) ∂ (∇φ (x,t)) (1.1) ε + ∇ × µ −1 ∇ × A(x,t) + ε = Jq (x,t),   2  ∂ t ∂ t    (x,t) ∈ Ω × (0, T ),      2  |q| iq¯h ∗   ψ ∇ψ − ψ ∇ψ ∗ − Jq = − |ψ |2 A,    2m m   ψ , φ , A subject to the appropriate initial and boundary conditions,

where Ω ⊂ Rd , d > 2 is a bounded Lipschitz polygonal convex domain in R2 (or a bounded Lipschitz polyhedron convex domain in R3 ), ψ ∗ denotes the complex conjugate of ψ , ε and µ respectively denote the electric permittivity and the magnetic permeability of the material and V0 is the constant potential energy. It is well-known that the solutions of the Maxwell-Schrödinger equations (1.1) lack uniqueness. In fact, for any function χ : Ω × (0, T ) → R, if (ψ , φ , A) is a solution of (1.1), then (exp(iχ )ψ , φ − ∂t χ , A+ ∇χ ) is also a solution of (1.1). To obtain mathematically well-posed equations, some extra constraint, commonly known as gauge choice, is often enforced on the solutions of the Maxwell-Schrödinger equations. The most common gauges are listed below. (i) The Lorentz gauge ∂φ = 0. ∇·A+ ∂t (ii) The Coulomb gauge ∇ · A = 0. (iii) The temporal gauge

φ = 0. For simplicity, we employ the atomic units, i.e. h¯ = m = q = 1, and assume that ε = µ = 1 without loss of generality. In this paper, we consider the time-dependent Maxwell-Schrödinger equations under the Lorentz gauge as follows:  ∂ψ 1  + (i∇ + A)2 ψ + V0ψ + φ ψ = 0, (x,t) ∈ Ω × (0, T ), −i    ∂ t 2   2  A ∂ i   + ∇ × (∇ × A) − ∇(∇ · A) + ψ ∗ ∇ψ − ψ ∇ψ ∗ ∂ t2 2 (1.2)   +|ψ |2 A = 0, (x,t) ∈ Ω × (0, T ),     2    ∂ φ − ∆ φ = |ψ |2 , (x,t) ∈ Ω × (0, T ). ∂ t2 The boundary conditions are

ψ (x,t) = 0, φ (x,t) = 0, A(x,t) × n = 0, ∇ · A(x,t) = 0,

(x,t) ∈ ∂ Ω × (0, T ),

(1.3)

¨ CRANK-NICOLSON GALERKIN METHOD FOR MAXWELL-SCHRODINGER EQUATIONS

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and the initial conditions are

ψ (x, 0) = ψ0 (x),

φ (x, 0) = φ0 (x),

A(x, 0) = A0 (x),

φt (x, 0) = φ1 (x),

At (x, 0) = A1 (x),

(1.4)

where φt and At denote the derivative of φ and A with respect to the time t respectively, n = (n1 , n2 , n3 ) is the outward unit normal to the boundary ∂ Ω . The gauge choice and the equations (1.2) impose the following constraints on the initial datas: ∇ · A0 + φ1 = 0,

∇ · A1 + ∆ φ0 + |ψ0 |2 = 0.

(1.5)

R EMARK 1.1 The boundary condition ψ (x) = 0 on ∂ Ω implies that the particle is confined in a whole domain Ω . The boundary condition A(x,t) × n = 0 and φ (x,t) = 0 on ∂ Ω are direct results of the perfect conductive boundary (PEC). The boundary condition ∇ · A = 0 on ∂ Ω can be deduced from the boundary condition of φ and the gauge choice. As for the determination of the boundary conditions for the vector potential A and the scalar potential φ , we refer to Chew (2014). The local and global well-posedness of solutions on all of R3 for the time-dependent MaxwellSchrödinger equations (1.1) have been studied in, for example, Ginibre & Velo (2003), Guo et al. (1995), Nakamitsu & Tsutsumi (1986), Nakamura & Wada (2005), Nakamura & Wada (2007), Shimomura (2003) and Wada (2012). To the best of our knowledge, the existence and uniqueness of the solution for the Maxwell-Schrödinger equations in a bounded domain seem to be open. There are a number of results for the numerical methods for the coupled Maxwell-Schrödinger equations. We recall some interesting studies. Sui et al. (2007) proposed the finite-difference time-domain (FDTD) method to solve the Maxwell-Schrödinger equations in the simulation of electron tunneling problem. Pierantoni et al. (2008) studied a carbon nanotube between two metallic electrodes by solving the Maxwell-Schrödinger equations with the transmission line matrix (TLM)/FDTD hybrid method. Ahmed & Li (2012) used the FDTD method for the Maxwell-Schrödinger system to simulate plasmonics nanodevices. However, in the numerical studies listed above, the EM fields are all described by the Maxwell’s equations involving electric fields E and magnetic fields H, instead of the A-φ formulation. Recently, Ryu (2015) applied the FDTD scheme to discretize the Maxwell-Schrödinger equations (1.1) under the Lorentz gauge and to simulate the interaction between a single electron in an artificial atom and an incoming electromagnetic field. For more results on this topic, we refer to Ohnuki et al. (2013), Sato & Yabana (2014), Turati & Hao (2012) and the references therein. There are few results on the finite element method (FEM) and its convergence analysis of the Maxwell-Schrödinger equations (1.1). In this paper we will present an alternating Crank-Nicolson finite element method for solving the problem (1.2)-(1.4), i.e. the finite element method in space and the Crank-Nicolson scheme in time. Then we will derive the optimal error estimates for the proposed method. Our work is motivated by Gao et al. (2014) in which Gao and his collaborators proposed a linearized Crank-Nicolson Galerkin method for the time-dependent Ginzburg-Landau equations and derived an optimal error estimate via a mathematical inductive method under the assumption that h and ∆ t are sufficiently small. Here h and ∆ t are the spatial mesh size and the time step, respectively. Compared to the time-dependent Ginzburg-Landau model, the error analysis of numerical schemes for the time-dependent Maxwell-Schrödinger system is much more difficult. The main difficulties and tricky parts in this paper are the estimates of the current term Jq and the error analysis for the wave function ψ . In particular we derive the energy-norm error estimates for the Schrödinger’s equation. The remainder of this paper is organized as follows. In section 2, we introduce some notation and propose a decoupled alternating Crank-Nicolson scheme with the Galerkin finite element approximation

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for the Maxwell-Schrödinger equations (1.2)-(1.4). The proof of the main theorem (see Theorem 2.1) in this paper will be given in section 3. Finally, some numerical tests are carried out to validate the theoretical results in this paper. Throughout the paper the Einstein summation convention on repeated indices is adopted. By C we shall denote a positive constant independent of the mesh size h and the time step ∆ t without distinction. 2. An alternating Crank-Nicolson Galerkin finite element scheme In this section, we present a numerical scheme for the Maxwell-Schrödinger equations (1.2)-(1.4) using Galerkin finite element method in space and a decoupled alternating Crank-Nicolson scheme in time. To start with, here and afterwards, we assume that Ω is a bounded Lipschitz polygonal convex domain in R2 (or a bounded Lipschitz polyhedron convex domain in R3 ). We introduce the following notation. Let W s,p (Ω ) denote the conventional Sobolev spaces of the real-valued functions. As usual, W s,2 (Ω ) and W0s,2 (Ω ) are denoted by H s (Ω ) and H0s (Ω ) respectively. We use W s,p (Ω ) = {u + iv | u, v ∈ W s,p (Ω )} and H s (Ω ) = {u + iv | u, v ∈ H s (Ω )} with calligraphic letters for Sobolev spaces of the complex-valued functions, respectively. Furthermore, let Ws,p (Ω ) = [W s,p (Ω )]d and Hs (Ω ) = [H s (Ω )]d with bold faced letters be Sobolev spaces of the vector-valued functions with d components (d=2, 3). L2 inner-products in H s (Ω ), H s (Ω ) and Hs (Ω ) are denoted by (·, ·) without ambiguity. In particular, we introduce the following subspace of H1 (Ω ): Ht1 (Ω ) = {A | A ∈ H1 (Ω ), A × n = 0 on ∂ Ω } The semi-norm on Ht1 (Ω ) is defined by i1 h 2 kukH1 (Ω ) := k∇ · uk2L2 (Ω ) + k∇ × uk2L2(Ω ) , t

which is equivalent to the standard H1 (Ω )-norm kukH1 (Ω ) , see Girault & Raviart (1986). The weak formulation of the Maxwell-Schrödinger system (1.2)-(1.4) can be specified as follows: find (ψ , A, φ ) ∈ H01 (Ω ) × Ht1 (Ω ) × H01 (Ω ) such that ∀t ∈ (0, T ),  ∂ψ 1  , ϕ ) + ((i∇ + A) ψ , (i∇ + A) ϕ ) + (V0ψ , ϕ ) + (φ ψ , ϕ ) = 0, ∀ϕ ∈ H01 (Ω ), −i(   ∂ t 2   2   ( ∂ A , v) + (∇ × A, ∇ × v) + (∇ · A, ∇ · v) + ( i ψ ∗ ∇ψ − ψ ∇ψ ∗ , v) ∂ t2 2   ∀v ∈ Ht1 (Ω ), + (|ψ |2 A, v) = 0,      2  ( ∂ φ , η ) + (∇φ , ∇η ) = (|ψ |2 , η ), ∀η ∈ H01 (Ω ) ∂ t2

(2.1)

with the initial conditions ψ0 ∈ H01 (Ω ), A0 ∈ Ht1 (Ω ), φ0 ∈ H01 (Ω ), At (·, 0) ∈ L2 (Ω ) and φt (·, 0) ∈ L2 (Ω ). Let M be a positive integer and let ∆ t = T /M be the time step. For any k=1,2,· · · , M, we introduce the following notation:

∂ U k = (U k − U k−1)/∆ t, k

U = (U k + U k−1 )/2,

∂ 2U k = (∂ U k − ∂ U k−1)/∆ t,

e k = (U k + U k−2 )/2, U

(2.2)


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k k for any given sequence {U k }M 0 and denote u = u(·,t ) for any given functions u ∈ C(0, T ; X) with a Banach space X. Let Th = {e} be a regular partition of Ω into triangles in R2 or tetrahedrons in R3 without loss of generality, where the mesh size h = maxe∈Th {diam(e)}. For a given partition Th , let Vhr , Vrh and Vhr denote the corresponding r-th order finite element subspaces of H01 (Ω ) , Ht1 (Ω ) and H01 (Ω ), respectively. Let Rh , πh and Ih be the conventional point-wise interpolation operators on Vhr , Vrh and Vhr , respectively. For convenience, assume that the function A and φ is defined in the interval [−∆ t, T ] in terms of the time variable t. We can compute A(·, −∆ t) by

A(·, −∆ t) = A(·, 0) − ∆ t

∂A (·, 0) = A0 − ∆ tA1, ∂t

(2.3)

which leads to an approximation to A−1 with second order accuracy. φ −1 can be approximated in a similar way. An alternating Crank-Nicolson Galerkin finite element approximation to the Maxwell-Schrödinger system (2.1) is formulated as follows:

ψh0 = Rh ψ0 ,

A0h = πh A0 ,

φh0 = Ih φ0 ,

0 A−1 h = Ah − ∆ t π h A1 ,

φh−1 = φh0 − ∆ tIhψ1 ,

and find (ψhk , Akh , φhk ) ∈ Vhr × Vrh × Vhr such that for k = 1, 2, · · · , M,  1 k k k   (i∇ + Ah )ψ kh , (i∇ + Ah )ϕ + (V0 + φ h )ψ kh , ϕ = 0, ∀ϕ ∈ Vhr , −i(∂ ψhk , ϕ ) +    2   i  2 k k−1 ∗ k−1 k−1 k−1 ∗ e k , ∇ × v) (ψh ) ∇ψh − ψh ∇(ψh ) , v + (∇ × A (∂ Ah , v) + h 2   k k−1 2 e k r e    +(∇ · Ah , ∇ · v) + |ψh | Ah , v = 0, ∀v ∈ Vh ,    2 k (∂ φh , η ) + (∇φehk , ∇η ) = (|ψhk−1 |2 , η ), ∀η ∈ Vhr .

(2.4)

(2.5)

k k e k and φek are defined in (2.2), and (ψ k−1 )∗ denotes the complex conjugate of Note that Ah , φ h , ψ kh , A h h h ψhk−1 . For convenience, we define the following bilinear forms:

B(A; ψ , ϕ ) = ((i∇ + A)ψ , (i∇ + A)ϕ ),

D(A, v) = (∇ · A, ∇ · v) + (∇ × A, ∇ × v),

(2.6)

i f (ψ , ϕ ) = (ϕ ∗ ∇ψ − ψ ∇ϕ ∗ ). 2 Then the variational form of the Maxwell-Schrödinger system (2.1) and its discrete system (2.2) can be reformulated as follows:  1 ∂ψ 1   −i( ∂ t , ϕ ) + 2 B(A; ψ , ϕ ) + (V0 ψ , ϕ ) + (φ ψ , ϕ ) = 0, ∀ϕ ∈ H0 (Ω ),    2 ∂ A (2.7) ( 2 , v) + D(A, v) + ( f (ψ , ψ ), v) + (|ψ |2A, v) = 0, ∀v ∈ Ht1 (Ω ),  ∂t   2   ( ∂ φ , η ) + (∇φ , ∇η ) = (|ψ |2 , η ), ∀η ∈ H01 (Ω ) ∂ t2

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and for k = 1, 2, · · · , M,  1 k k   −i(∂ ψhk , ϕ ) + B(Ah ; ψ kh , ϕ ) + (V0 ψ kh , ϕ ) + (φ h ψ kh , ϕ ) = 0, ∀ϕ ∈ Vhr ,   2 e k , v = 0, ∀v ∈ Vr , e k , v) + f (ψ k−1 , ψ k−1 ), v + |ψ k−1 |2 A (∂ 2 Akh , v) + D(A h h h  h h h    2 k r k k−1 2 e (∂ φh , η ) + (∇φh , ∇η ) = (|ψh | , η ), ∀η ∈ Vh .

(2.8)

In this paper we assume that the Maxwell-Schrödinger equations (2.7) has one and only one weak solution (ψ , A, φ ) and the following regularity conditions are satisfied:

ψ , ψt , ψtt ∈ L∞ (0, T ; H r+1 (Ω )), ψtttt ∈ L2 (0, T ; L 2 (Ω ));

ψttt ∈ L∞ (0, T ; H 1 (Ω )),

A, At , Att ∈ L∞ (0, T ; Hr+1 (Ω )),

Attt ∈ L∞ (0, T ; H1 (Ω )), Atttt ∈ L2 (0, T ; L2 (Ω ));

(2.9)

φ , φt , φtt ∈ L∞ (0, T ; H r+1 (Ω )), φttt ∈ L∞ (0, T ; H 1 (Ω )), φtttt ∈ L2 (0, T ; L2 (Ω )). For the initial conditions (ψ0 , A0 , A1 , φ0 , φ1 ) , we assume that

ψ0 ∈ H r+1 (Ω ) ∩ H01 (Ω ); A0 , A1 ∈ Hr+1 (Ω ) ∩ Ht1 (Ω ); φ0 , φ1 ∈ H r+1 (Ω ) ∩ H01 (Ω ).

(2.10)

We now give the main convergence result in this paper as follows: T HEOREM 2.1 Suppose that the Maxwell-Schrödinger coupled system (2.7) has a unique solution (ψ , A, φ ) satisfying (2.9) and (2.10). Let (ψhk , Akh , φhk ) be the fully discrete numerical solution of (ψ , A, φ ) defined in (2.8). Then there exist two positive constants h0 > 0 and ∆ t0 > 0, such that when h < h0 , ∆ t < ∆ t0 , we have the following error estimates: max kψhk − ψ k k2L 2 (Ω ) + k∇(ψhk − ψ k )k2L2 (Ω ) + kAkh − Ak k2L2 (Ω ) 16k6M

+k∇ · (Akh − Ak )k2L2 (Ω ) + k∇ × (Akh − Ak )k2L2 (Ω ) +kφhk − φ k k2L2 (Ω ) + k∇(φhk − φ k )k2L2 (Ω ) 6 C h2r + (∆ t)4 , r > 1,

(2.11)

where ψ k = ψ (·,t k ), Ak = A(·,t k ), φ k = φ (·,t k ), and C is a constant independent of h, ∆ t. 3. The proof of Theorem 2.1 In this section, we will give the proof of Theorem 2.1. 3.1 Preliminaries For convenience, we list some imbedding inequalities and interpolation inequalities in Sobolev spaces (see, e.g., Ladyzhenskaya et al. (1968) and Girault & Raviart (1986)), and use them in the sequel. kukL p 6 CkukH 1 ,

kvkL p 6 CkvkH1 ,

1 6 p 6 6 (d = 2, 3),

kvkH1 6 C(k∇ × vkL2 + k∇ · vkL2 ),

v ∈ Ht1 (Ω ),

(3.1) (3.2)


kuk2L3 6 kukL2 kukL6 ,

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where kukL p = kukL p (Ω ) , kukH 1 = kukH 1 (Ω ) , kvkL2 = kvkL2 (Ω ) and kvkH1 = kvkH1(Ω ) . The following identities will be used frequently in this paper. M−1

M

∑ (ak − ak−1)bk = aM bM − a0b1 − ∑ ak (bk+1 − bk ), k=1 M

k=1 M

(3.4)

∑ (ak − ak−1)bk = aM bM − a0b0 − ∑ ak−1 (bk − bk−1). k=1

k=1

Let (Rh ψ , πh A, Ih φ ) denote the interpolation function of (ψ , A, φ ) in Vhr × Vrh ×Vhr . Set eψ = Rh ψ − ψ , eA = πh A− A, eφ = Ih φ − φ . By applying standard finite element theory and the regualrity conditions in (2.9), we have keψ kL 2 + hkeψ kH 1 6 Chr+1 ,

keA kL2 + hkeA kH1 6 Chr+1 ,

keφ kL2 + hkeφ kH 1 6 Chr+1 , kRh ψ kL ∞ + kπhAkL∞ + kIhφ kL∞ + k∇Rhψ kL3 6 C,

(3.5)

where C is a constant independent of h. The following lemmas will be useful in the proof of Theorem 2.1. L EMMA 3.1 For the solution of (2.8), we have 2

kψhk k2L 2 = kψh0 kL 2 ,

k = 1, 2 · · · , M.

(3.6)

L EMMA 3.2 For k = 1, 2 · · · , M, the following identities hold for the bilinear functional B(A; ψ , ϕ ) defined in (2.6): e ψ , ϕ ) = (A + A) e ψϕ ∗ , A − A e e + 2( f (ψ , ϕ ), A − A), B(A; ψ , ϕ ) − B(A; h i 1 k k k−1 k k−1 k k k−1 2 1 Re B A ; ψ , ∂ ψ =− (A + A )|ψ | , (∂ A + ∂ A ) (3.7) 2 2 1 1 k − f (ψ k−1 , ψ k−1 ), (∂ Ak + ∂ Ak−1 ) + ∂ B(A ; ψ k , ψ k ). 2 2 Lemma 3.1 can be proved by choosing ϕ = ψ kh in (2.8)1 and taking the imaginary part. A direct calculation gives (3.7) in Lemma 3.2. Let θψk = ψhk − Rh ψ k , θAk = Akh − πh Ak , θφk = φhk − Ih φ k . By using the error estimates of the interpolation operators (3.5), we only need to estimate θψk , θAk and θφk . We will prove the following estimate: (3.8) kθψk k2H 1 + k∂ θAk k2L2 + kθAk k2H1 + k∂ θφk k2L2 + kθφk k2H 1 6 C∗ h2r + (∆ t)4 , where k = 0, 1, · · · , M. By using the regularity assumption of the initial conditions (2.10) and the error estimates of the interpolation operators (3.5), we get kθψ0 k2H 1 + k∂ θA0 k2L2 + kθA0 k2H1 + k∂ θφ0 k2L2 + kθφ0k2H 1 6 C0 h2r + (∆ t)4 . (3.9)

We use the mathematical inductive method to show (3.8). By (3.9), if we require C∗ > C0 , then (3.8) holds for k = 0. We assume that (3.8) holds for 0 6 k 6 m − 1. In the rest of this section, we will find C∗ such that (3.8) holds for k 6 m, where C∗ is independent of k , h , ∆ t.

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Subtracting (2.7) from (2.8), we obtain the following equations for θAk , θφk and θψk : 2 k−1 ∂ A k 2 k k 2 e ∂ θA , v + D(θA , v) = − ∂ πh A , v 2 ∂t k k−1 2 k−1 ek, v +D(Ak−1 − πg | A − |ψhk−1|2 A h A , v) + |ψ h k−1 k−1 k−1 k−1 + f (ψ , ψ ) − f (ψh , ψh ), v , ∀v ∈ Vrh ,

∂ 2 φ k−1 2 k k 2 k e ∂ θφ , η + ∇θφ , ∇η = − ∂ Ih φ , η 2 ∂t k k−1 2 k−1 2 + ∇(φ k−1 − If φ η ψ | − | ψ | , η ), ∇ + | , h h

(3.10)

(3.11) ∀η ∈ Vhr ,

! 1 1 ∂ ψ k− 2 k k k Ah ; θ ψ , ϕ = 2i ∂ Rh ψ − , ϕ + 2V0 ψ k− 2 − ψ kh , ϕ ∂t k k− 12 k k− 12 k− 12 k− 21 − Rh ψ ), ϕ ) + 2 φ − φ h ψ kh , ϕ +B(A ; (ψ ψ 1 k + B(Ak− 2 ; Rh ψ k , ϕ ) − B(Ah ; Rh ψ k , ϕ ) , ∀ϕ ∈ Vhr .

−2i(∂ θψk , ϕ ) + B

(3.12)

The key steps of the proof of (3.8) are now briefly described. In order to find C∗ and to show that (3.8) holds for k = m, we first take v = 1 (θAk − θAk−2 ) in (3.10) and η = 1 (θφk − θφk−2 ) in (3.11) and 2∆ t 2∆ t k obtain the estimates of θAm and θφm . Then we choose ϕ = θ ψ in (3.12) and give the estimate of kθψm kL 2 . Finally, we take ϕ = ∂ θψk in (3.12) and make use of the above estimates of θAm and θφm to derive the energy-norm estimate for θψm . Using the above estimates, we can complete the proof of (3.8). 3.2 Estimates for (3.10) If we set

k ∂ 2 Ak−1 2 k = − ∂ πh A , v , I2k (v) = D(Ak−1 − πg h A , v), 2 ∂t ek, v , I3k (v) = |ψ k−1 |2 Ak−1 − |ψhk−1 |2 A h k k−1 k−1 k−1 I4 (v) = f (ψ , ψ ) − f (ψh , ψhk−1 ), v , v ∈ Vrh ,

I1k (v)

then we rewrite (3.10) as follows: ∂ 2 θAk , v + D(θeAk , v) = I1k (v) + I2k (v) + I3k (v) + I4k (v).

1 (θ k − θ k−2 ) = 1 (∂ θ k + ∂ θ k−1 ) = ∂ θ k in (3.14) and get We take v = 2∆ A A A A 2 t A 1 k , 1 (∂ θ k + ∂ θ k−1 )) ∂ 2 θAk , (∂ θAk + ∂ θAk−1 ) + D(θf A A A 2 i i 2 1 h 1 h k∂ θAk k2L2 − k∂ θAk−1k2L2 + D(θAk , θAk ) − D(θAk−2 , θAk−2 ) = 2∆ t 4∆ t k k k k = I1k (∂ θ A ) + I2k (∂ θ A ) + I3k (∂ θ A ) + I4k (∂ θ A ),

(3.13)

(3.14)

(3.15)


which leads to

1 1 1 k∂ θAm k2L2 + D(θAm , θAm ) + D(θAm−1 , θAm−1 ) 2 4 4 1 1 1 = k∂ θA0 k2L2 + D(θA0 , θA0 ) + D(θA−1 , θA−1 ) 2 4 4 i m h k k k k k k +∆ t ∑ I1 (∂ θ A ) + I2 (∂ θ A ) + I3k (∂ θ A ) + I4k (∂ θ A ) k=1

6 Ch2r + ∆ t

m

∑

k=1

Here we have used the fact that

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(3.16)

h i k k k k I1k (∂ θ A ) + I2k (∂ θ A ) + I3k (∂ θ A ) + I4k (∂ θ A ) .

k∂ θA0 k2L2 + D(θA0 , θA0 ) + D(θA−1, θA−1 ) 6 Ch2r . m

k

Now we estimate ∑ I kj (∂ θ A ), j = 1, 2, 3, 4. Under the regularity assumption of A in (2.9), we can k=1 prove m m 2 k−1 ∂ A k k k 2 k |I ( | ∂ θ )| 6 ∂ A , ∂ θ − ∑ 1 A ∑ ∂ t2 A | k=1 k=1 m k + ∑ | ∂ 2 Ak − π h ∂ 2 Ak , ∂ θ A | (3.17) k=1

6

m C 2r+2 k h + (∆ t)4 + C ∑ k∂ θ A k2L2 . ∆t k=1 k

k We rewrite the term ∆ t ∑m k=1 I2 (∂ θ A ) as follows: m m 1 k 1 k k k−1 k k−1 k−2 ∆ t ∑ I2 (∂ θ A ) = ∆ t ∑ D A − (A + A ), (∂ θA + ∂ θA ) 2 k=1 2 k=1 m 1 + ∆ t ∑ D (Ak + Ak−2 ) − πh(Ak + Ak−2), (∂ θAk + ∂ θAk−1 ) . 4 k=1

and

(3.18)

Applying (3.4), the regularity assumption and Young’s inequality, we get m 1 1 ∆ t ∑ D Ak−1 − (Ak + Ak−2 ), (∂ θAk + ∂ θAk−1 ) 2 k=1 2 m 2r 1 1 6 C h + (∆ t)4 + D (θAm , θAm ) + D θAm−1 , θAm−1 + C∆ t ∑ D θAk , θAk 64 64 k=0

(3.19)

m 1 ∆ t ∑ D (Ak + Ak−2 ) − πh(Ak + Ak−2 ), (∂ θAk + ∂ θAk−1 ) 4 k=1 m 1 1 6 C h2r + (∆ t)4 + D (θAm , θAm ) + D θAm−1 , θAm−1 + C∆ t ∑ D θAk , θAk . 64 64 k=0

(3.20)

We thus have

m

∆t

k

∑ I2k (∂ θ A ) 6 C k=1

+

2r h + (∆ t)4 + C∆ t

m

∑D k=0

1 1 D (θAm , θAm ) + D θAm−1 , θAm−1 . 32 32

θAk , θAk .

(3.21)

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C. P. MA ET AL.

We observe that k e k ), ∂ θ k + |ψ k−1 |2 (πh A ek − A e k ), ∂ θ k I3k (∂ θ A ) = |ψ k−1 |2 (Ak−1 − πhA A A h k k−1 2 k−1 2 k−1 2 k−1 2 k k e , ∂ θ + (|ψ | − |Rhψ | )θe , ∂ θ k + (|ψ | − |Rh ψ | )πh A A A A e k , ∂ θ k + (|Rh ψ k−1 |2 − |ψ k−1 |2 )θek , ∂ θ k + (|Rh ψ k−1 |2 − |ψhk−1 |2 )πh A A A A h def

=

6

(3.22)

k

∑ I3k, j (∂ θ A ).

j=1

The first four terms in (3.22) can be estimated by a standard argument, i.e. 4

k

∑ |I3k, j (∂ θ A )| 6 C

j=1

We notice that

n

o k (∆ t)4 + h2r+2 + D(θeAk , θeAk ) + k∂ θ A k2L2 .

|Rh ψ k−1 |2 − |ψhk−1 |2 = (Rh ψ k−1 − ψhk−1 )(Rh ψ k−1 )∗ + (Rh ψ k−1 − ψhk−1)∗ ψhk−1 = −θψk−1 (Rh ψ k−1 )∗ − (θψk−1 )∗ Rh ψ k−1 − |θψk−1 |2

(3.23)

(3.24)

and obtain k

k

k

|I3k,5 (∂ θ A )| 6 Ckθψk−1 kL 6 k∂ θ A kL2 + Ckθψk−1 k2L 6 k∂ θ A kL2 , k k k |I3k,6 (∂ θ A )| 6 Ckθψk−1 kL 6 kθeAk kL6 k∂ θ A kL2 + Ckθψk−1 k2L 6 kθeAk kL6 k∂ θ A kL2 .

(3.25)

By using the assumption of the induction, we have

1 kθψk−1 kL 6 6 Ckθψk−1 kH 1 6 CC∗2 (∆ t)2 + hr .

1 If we choose some sufficiently small h and ∆ t such that CC∗2 (∆ t)2 + hr 6 1, then we get n o k k k |I3k,5 (∂ θ A )| 6 Ckθψk−1 kL 6 k∂ θ A kL2 6 C k∇θψk−1 k2L2 + k∂ θ A k2L2 , o n (3.26) k k k |I3k,6 (∂ θ A )| 6 CkθeAk kL6 k∂ θ A kL2 6 C D(θeAk , θeAk ) + k∂ θ A k2L2 .

Combining (3.22)-(3.26) implies m

k

o m n C 2r k h + (∆ t)4 + C ∑ D(θeAk , θeAk ) + k∇θψk−1k2L2 + k∂ θ A k2L2 .

∑ |I3k (∂ θ A )| 6 ∆ t

k=1

m

(3.27)

k=1

k

To estimate ∑ I4k (v), we rewrite I4k (∂ θ A ) as follows: k=1

k k I4k (∂ θ A ) = f (ψ k−1 , ψ k−1 ) − f (Rh ψ k−1 , Rh ψ k−1 ), ∂ θ A k def k k + f (Rh ψ k−1 , Rh ψ k−1 ) − f (ψhk−1 , ψhk−1 ), ∂ θ A = I4k,1 (∂ θ A ) + I4k,2 (∂ θ A ).

(3.28)


11 of 27

Observing i i f (ψ , ψ ) − f (ϕ , ϕ ) = (ψ ∗ ∇ψ − ψ ∇ψ ∗ ) − (ϕ ∗ ∇ϕ − ϕ ∇ϕ ∗ ) 2 2 i ∗ i ∗ = − (ϕ ∇(ϕ − ψ ) − ϕ ∇(ϕ − ψ ) ) + ((ϕ − ψ )∇ψ ∗ − (ϕ − ψ )∗ ∇ψ ) 2 2

(3.29)

and applying (3.5), we get k k I4k,1 (∂ θ A ) = f (ψ k−1 , ψ k−1 ) − f (Rh ψ k−1 , Rh ψ k−1 ), ∂ θ A i k = − (Rh ψ k−1 )∗ ∇(Rh ψ k−1 − ψ k−1 ) − Rhψ k−1 ∇(Rh ψ k−1 − ψ k−1 )∗ , ∂ θ A 2 i k + (Rh ψ k−1 − ψ k−1 )∇(ψ k−1 )∗ − (Rh ψ k−1 − ψ k−1 )∗ ∇ψ k−1 , ∂ θ A 2n o k 6 C h2r + k∂ θ A k2L2 .

(3.30)

Using (3.5), we similarly prove k k I4k,2 (∂ θ A ) = f (Rh ψ k−1 , Rh ψ k−1 ) − f (ψhk−1 , ψhk−1 ), ∂ θ A i k = − (ψhk−1 )∗ ∇(ψhk−1 − Rhψ k−1 ) − ψhk−1 ∇(ψhk−1 − Rh ψ k−1 )∗ , ∂ θ A 2 i k + (ψhk−1 − Rhψ k−1 )∇(Rh ψ k−1 )∗ − (ψhk−1 − Rh ψ k−1 )∗ ∇Rh ψ k−1 , ∂ θ A 2 i k = − (θψk−1 )∗ ∇θψk−1 − θψk−1 ∇(θψk−1 )∗ , ∂ θ A 2 i k − (Rh ψ k−1 )∗ ∇θψk−1 − Rh ψ k−1 ∇(θψk−1 )∗ , ∂ θ A 2 i k + θψk−1 ∇(Rh ψ k−1 )∗ − (θψk−1 )∗ ∇Rh ψ k−1 , ∂ θ A 2 k k 6 − f (θψk−1 , θψk−1 ), ∂ θ A + CkRhψ k−1 kL ∞ k∇θψk−1 kL2 k∂ θ A kL2

(3.31)

k

+Ck∇Rh ψ k−1 kL3 kθψk−1 kL 6 k∂ θ A kL2 o n k k 6 − f (θψk−1 , θψk−1 ), ∂ θ A + C k∇θψk−1 k2L2 + k∂ θ A k2L2 .

Hence we have m

k

∑ I4k (∂ θ A ) 6 k=1

o m n m Ch2r k k − ∑ f (θψk−1 , θψk−1 ), ∂ θ A + C ∑ k∇θψk−1 k2L2 + k∂ θ A k2L2 . ∆t k=1 k=1

(3.32)

Substituting (3.17), (3.21), (3.27) and (3.32) into (3.16), we get 1 7 7 k∂ θAm k2L2 + D(θAm , θAm ) + D(θAm−1 , θAm−1 ) 2 32 32 m 2r 4 6 C h + (∆ t) + C∆ t ∑ D(θAk , θAk ) + k∂ θAk k2L2 m−1

+C∆ t

k=0 m

∑ k∇θψk k2L2 − ∆ t ∑

k=0

k=1

k

f (θψk−1 , θψk−1 ), ∂ θ A .

(3.33)

12 of 27

C. P. MA ET AL.

We estimate the last term on the right side of (3.33). It follows from the definition of bilinear functional f (ϕ , ψ ) in (2.6) that i m m k ∑ f (θψk−1 , θψk−1 ), ∂ θ A = 4 ∑ (θψk−1 )∗ ∇θψk−1 − θψk−1∇(θψk−1 )∗ , (∂ θAk + ∂ θAk−1) . (3.34) k=1 k=1 1

For ∆ t 6 h 2 , by the assumption of the induction and the inverse inequalities, we have 1

1

1

1

1

k∇θψk−1 kL3 6 Ch− 2 k∇θψk−1 kL2 6 Ch− 2 C∗2 {hr + h} 6 CC∗2 h 2 . 1

1

We choose a sufficiently small h > 0 such that CC∗2 h 2 6 1 and obtain k∇θψk−1 kL3 6 1. Consequently m 1 m k | ∑ f (θψk−1 , θψk−1 ), ∂ θ A | 6 ∑ kθψk−1 kL 6 k∇θψk−1 kL3 k∂ θAk + ∂ θAk−1 kL2 2 k=1 k=1 m m 1 6 ∑ kθψk−1 kL 6 k∂ θAk + ∂ θAk−1 kL2 6 C ∑ k∇θψk−1 kL2 k∂ θAk + ∂ θAk−1 kL2 2 k=1 k=1 6C

m−1

m

k=0

k=0

(3.35)

∑ k∇θψk k2L2 + C ∑ k∂ θAk k2L2 ,

where C is a constant independent of h, ∆ t. 1 As for h 2 6 ∆ t, by the assumption of the induction, we discover 1 1 k∇θψk−1 kL2 6 C∗2 (∆ t)2 + (∆ t)2r 6 2C∗2 (∆ t)2 . 1

Now choose a sufficiently small ∆ t > 0 to find 2C∗2 ∆ t 6 1, in which case we have k∇θψk−1 kL2 6 ∆ t.

It follows that m 1 m k | ∑ f (θψk−1 , θψk−1 ), ∂ θ A | 6 ∑ kθψk−1 kL 3 k∇θψk−1 kL2 k∂ θAk + ∂ θAk−1 kL6 2 k=1 k=1 ∆t m 1 m kθψk−1 kL 3 k∂ θAk + ∂ θAk−1 kL6 6 ∑ kθψk−1 kL 3 kθAk − θAk−2 kL6 6 ∑ 2 k=1 2 k=1 m

m

(3.36)

6 C ∑ k∇θψk−1 k2L2 + C ∑ (D(θAk , θAk ) + D(θAk−2 , θAk−2 )) k=1

k=1 2r

6 Ch + C

m−1

∑

k∇θψk k2L2

k=0

m

+ C ∑ D(θAk , θAk ). k=0

Combining (3.35), (3.36) and (3.33) implies 1 7 7 k∂ θAm k2L2 + D(θAm , θAm ) + D(θAm−1 , θAm−1 ) 6 C h2r + (∆ t)4 2 32 32 m m−1 +C∆ t ∑ D(θAk , θAk ) + k∂ θAk k2L2 + C∆ t ∑ k∇θψk k2L2 . k=0

k=0

(3.37)


We choose a sufficiently small ∆ t > 0 such that C∆ t 6

1 8

13 of 27

and find

3 3 3 k∂ θAm k2L2 + D(θAm , θAm ) + D(θAm−1 , θAm−1 ) 6 C h2r + (∆ t)4 8 32 32 m−1 m−1 k k +C∆ t ∑ D(θA , θA ) + k∂ θAk k2L2 + C∆ t ∑ k∇θψk k2L2 , k=0

(3.38)

k=0

which leads to 3 3 3 k∂ θAm k2L2 + D(θAm , θAm ) + D(θAm−1 , θAm−1 ) 8 32 32 6 C h2r + (∆ t)4 + CC∗ h2r + (∆ t)4

(3.39)

by the assumption of the induction. Applying the assumption of the induction again and (3.39), for k = 1, 2, · · · , m, we deduce k∂ Akh kL2 + kAkhkL6 6 k∂ θAk kL2 + kθAk kL6 + k∂ πhAk kL2 + kπhAk kL6 o n 1 6 C k∂ θAk kL2 + D 2 (θAk , θAk ) + k∂ πh Ak kL2 + kπhAk kL6 6 C,

(3.40)

1

when k∂ θAk kL2 + D 2 (θAk , θAk ) 6 1.

3.3 Estimates for (3.11) Setting ∂ 2 φ k−1 2 k = − ∂ Ih φ , η , 2 ∂t k J2k (η ) = ∇(φ k−1 − If h φ ), ∇η , J3k (η ) = |ψhk−1 |2 − |ψ k−1 |2 , η , η ∈ Vhr , J1k (η )

(3.41)

we rewrite (3.11) as follows:

∂ 2 θφk , η + ∇θeφk , ∇η = J1k (η ) + J2k (η ) + J3k (η ).

(3.42)

k We take η = 21 (∂ θφk + ∂ θφk−1 ) = ∂ θ φ in (3.42) and obtain

k k ∂ 2 θφk , ∂ θ φ + ∇θeφk , ∇∂ θ φ 1 1 k∂ θφk k2L2 − k∂ θφk−1 k2L2 + k∇θφk k2L2 − k∇θφk−2 k2L2 = 2∆ t 4∆ t k k k k k k = J1 (∂ θ φ ) + J2 (∂ θ φ ) + J3 (∂ θ φ ).

(3.43)

14 of 27

C. P. MA ET AL.

Multiply (3.43) by ∆ t and sum k = 1, 2, · · · , m to discover 1 1 1 k∂ θφm k2L2 + k∇θφm k2L2 + k∇θφm−1 k2L2 2 4 4 1 1 1 0 2 0 2 = k∂ θφ kL2 + k∇θφ kL2 + k∇θφ−1 k2L2 2 4 4 o m n k k k k k +∆ t ∑ J1 (∂ θ φ ) + J2 (∂ θ φ ) + J3k (∂ θ φ ) k=1

6 C h2r + (∆ t)4 + ∆ t

m

∑

k=1

(3.44)

n o k k k J1k (∂ θ φ ) + J2k (∂ θ φ ) + J3k (∂ θ φ ) .

Applying the regularity assumption of φ in (2.9), we deduce m 2 k−1 m m ∂ φ k k k 2 k k ∑ |J1 (∂ θ φ )| 6 ∑ | ∂ t 2 − ∂ φ , ∂ θ φ | + ∑ | ∂ 2 φ k − Ih∂ 2 φ k , ∂ θ φ | k=1 k=1 k=1 m C 2r+2 k 2 4 h + (∆ t) + C ∑ k∂ θ φ kL2 . 6 ∆t k=1 k

(3.45)

k

m k k ∑m k=1 J2 (∂ θ φ ) can be bounded by a argument similar to the estimate of ∑k=1 I2 (∂ θ A ) in (3.19) and (3.20). m 1 k ∆ t ∑ J2k (∂ θ φ ) 6 C h2r + (∆ t)4 + k∇θφm k2L2 8 k=1 (3.46) m 1 m−1 2 k 2 + k∇θφ kL2 + C∆ t ∑ k∇θφ kL2 . 8 k=0

Observing

k k k J3k (∂ θ φ ) = |Rh ψ k−1 |2 − |ψ k−1 |2 , ∂ θ φ + |ψhk−1 |2 − |Rh ψ k−1 |2 , ∂ θ φ ,

and using (3.24), we get m

∆t

k

∑ J3k (∂ θ φ ) 6 Ch2r + C∆ t k=1

n o k 2 k−1 2 k∇ θ k + k ∂ θ k 2 2 ∑ ψ φ L L m

k=1 m n k k−1 2 k∇θψk−1 k2L2 kθψ kL 4 k∂ θ φ kL2 6 Ch2r + C∆ t +∆ t k=1 k=1 m m n k +∆ t kθψk−1 kL 4 k∂ θ φ kL2 6 Ch2r + C∆ t k∇θψk−1 k2L2 k=1 k=1 m

∑

∑

∑

∑

k

+ k∂ θ φ k2L2 k

o

(3.47)

o

+ k∂ θ φ k2L2 .

Here we have used kθψk−1 kL 4 6 Ckθψk−1 kH 1 6 1. Substituting (3.45), (3.46) and (3.47) into (3.44) yields

1 1 1 k∂ θφm k2L2 + k∇θφm k2L2 + k∇θφm−1 k2L2 6 C h2r + (∆ t)4 2 8 8 m m−1 +C∆ t ∑ k∇θφk k2L2 + k∂ θ kφ k2L2 + C∆ t ∑ k∇θψk k2L2 .

(3.48)

k∂ φhk kL2 + kφhk kL6 6 C,

(3.49)

k=0

k=0

Similarly to (3.40), for k = 1, 2, · · · , m, we can prove where C is a constant independent of h, ∆ t and C∗ .


15 of 27

3.4 Estimates for (3.12) We rewrite (3.12) as follows: k k −2i(∂ θψk , ϕ ) + B Ah ; θ ψ , ϕ = where Qk1 (ϕ )

5

∑ Qkj (ϕ ),

(3.50)

j=1

! 1 ∂ ψ k− 2 ,ϕ , = 2i ∂ Rh ψ − ∂t k

1

1

Qk3 (ϕ ) = B(Ak− 2 ; (ψ k− 2 1

1 Qk2 (ϕ ) = 2V0 ψ k− 2 − ψ kh , ϕ , 1 1 k − Rhψ k ), ϕ ), Qk4 (ϕ ) = 2 φ k− 2 ψ k− 2 − φ h ψ kh , ϕ , k

Qk5 (ϕ ) = B(Ak− 2 ; Rh ψ k , ϕ ) − B(Ah ; Rh ψ k , ϕ ). k

Taking ϕ = θ ψ in (3.50), and observing the imaginary part of the above equation, we have h 1 k 2 k k k k kθψ kL 2 − kθψk−1 k2L 2 = −Im Qk1 (θ ψ ) + Qk2(θ ψ ) + Qk3(θ ψ ) + Qk4(θ ψ ) ∆t i k k k k k k +Qk5 (θ ψ ) 6 |Qk1 (θ ψ )| + |Qk2(θ ψ )| + |Qk3(θ ψ )| + |Qk4(θ ψ )| + |Qk5(θ ψ )|.

(3.51)

It is obvious that

! 1 ∂ ψ k− 2 k k = 2i ∂ ψ − , θ ψ + Rh ∂ ψ k − ∂ ψ k , θ ψ , ∂t 1 k k k k k k Q2 (θ ψ ) = −2V0 θ ψ , θ ψ + 2V0 ψ k − Rhψ k , θ ψ + 2V0 ψ k− 2 − ψ k , θ ψ .

k Qk1 (θ ψ )

k

Using the error estimates (3.5) for the interpolation operator Rh and the regularity of ψ in (2.9), we give the following estimate k k k (3.52) |Qk1 (θ ψ )| + |Qk2(θ ψ )| 6 C (∆ t)4 + h2r+2 + Ckθ ψ k2L 2 . Note that

and

B(A; ψ , ϕ ) = (∇ψ , ∇ϕ ) + |A|2 ψ , ϕ + i (ϕ ∗ ∇ψ − ψ ∇ϕ ∗ , A)

6 k∇ψ kL2 k∇ϕ kL2 + kAk2L6 kψ kL 6 kϕ kL 2 + kAkL6 kψ kL 3 k∇ϕ kL2 +k∇ψ kL2 kϕ kL 3 6 Ck∇ψ kL2 k∇ϕ kL2 , ∀A ∈ L6 (Ω ), ψ , ϕ ∈ H01 (Ω ), k

1

k

1

1

k

Qk3 (θ ψ ) = B(Ak− 2 ; (ψ k − Rhψ k ), θ ψ ) + B(Ak− 2 ; (ψ k− 2 − ψ k ), θ ψ ). Hence we obtain k

k k |Qk3 (θ ψ )| 6 C h2r + (∆ t)4 + Ck∇θ ψ k2L2 .

To estimate Qk4 (θ ψ ), we rewrite it as follows: 1 1 1 1 1 k k k Qk4 (θ ψ ) = (ψ k− 2 − Rh ψ k− 2 )φ k− 2 , θ ψ + Rh (ψ k− 2 − ψ k )φ k− 2 , θ ψ 1 1 1 k k + (Rh ψ k − ψ kh )φ k− 2 , θ ψ + ψ kh (φ k− 2 − Ih φ k− 2 ), θ ψ 1 k k k k k + ψ kh Ih (φ k− 2 − φ ), θ ψ + ψ kh (Ih φ − φ h ), θ ψ .

(3.53)

(3.54) (3.55)

(3.56)

16 of 27

C. P. MA ET AL.

It follows from Lemma 3.1, the regularity assumption (2.9), the properties of the interpolation operators and (3.56) that o n k k k (3.57) |Qk4 (θ ψ )| 6 C h2r + (∆ t)4 + C k∇θ ψ k2L2 + k∇θ φ k2L2 . We observe that

h i k k k k k Qk5 (θ ψ ) = B(Ah ; Rh ψ k , θ ψ ) − B(πhA ; Rh ψ k , θ ψ ) h i k k k k + B(πh A ; Rh ψ k , θ ψ ) − B(A ; Rh ψ k , θ ψ ) i h 1 k k k + B(A ; Rh ψ k , θ ψ ) − B(Ak− 2 ; Rh ψ k , θ ψ ) def

k

k

(3.58)

k

k,2 k,3 = Qk,1 5 (θ ψ ) + Q5 (θ ψ ) + Q5 (θ ψ ).

It follows from Lemma 3.2 that k k k k k k ∗ Qk,1 ( θ ) = R ψ θ ) (A + π A ), θ ( h h h A ψ ψ 5 k ∗ k ∗ k k k +i (θ ψ ) ∇Rh ψ − Rh ψ ∇(θ ψ ) , θ A , k k k k k k k ∗ Qk,2 5 (θ ψ ) = Rh ψ (θ ψ ) (πh A + A ), πh A − A k k k k +i (θ ψ )∗ ∇Rh ψ k − Rh ψ k ∇(θ ψ )∗ , πh A − A , 1 k k k k− 21 k k ∗ Qk,3 ), A − Ak− 2 5 (θ ψ ) = Rh ψ (θ ψ ) (A + A 1 k k k +i (θ ψ )∗ ∇Rh ψ k − Rh ψ k ∇(θ ψ )∗ , A − Ak− 2 .

(3.59)

n o k k k k k |Qk,1 5 (θ ψ )| 6 Ckθ A kL6 kθ ψ kL 2 + kθ ψ kL 3 + k∇θ ψ kL2 n o k k k 6 C D(θ A , θ A ) + k∇θ ψ k2L2 , n o o n k k k k k 2 r 2r + k + k∇ θ )| 6 Ch θ k θ k θ k θ k k , 6 C h + k∇ |Qk,2 ( 2 3 2 2 ψ ψ ψ ψ ψ L L L 5 L n o k k k k 2 |Qk,3 kθ ψ kL 2 + kθ ψ kL 3 + k∇θ ψ kL2 5 (θ ψ )| 6 C(∆ t) o n k 6 C (∆ t)4 + k∇θ ψ k2L2 ,

(3.60)

o n k k k k |Qk5 (θ ψ )| 6 C h2r + (∆ t)4 + C D(θ A , θ A ) + k∇θ ψ k2L2 .

(3.61)

Using (3.40), we prove

and thus

From (3.51), summing over k = 1, 2, · · · , m and combining (3.52), (3.55), (3.57) and (3.61), we have kθψm k2L 2 6 C h2r + (∆ t)4 + C∆ t 6 C h2r + (∆ t)4 + C∆ t

m

∑

n o k k k k D(θ A , θ A ) + k∇θ ψ k2L2 + k∇θ φ k2L2

n k=1 o k k k 2 k 2 D( θ , θ ) + k∇ θ k + k∇ θ k . 2 2 ∑ ψ L φ L A A m

k=0

(3.62)


17 of 27

To proceed further, we take ϕ = ∂ θψk = ∆1t (θψk − θψk−1), k = 1, 2, · · · , m in (3.10), to find k k −2i(∂ θψk , ∂ θψk ) + B Ah ; θ ψ , ∂ θψk =

5

∑ Qkj (∂ θψk ).

(3.63)

j=1

We take the real part of (3.63) and use (3.7) to get 5 1 k k−1 B(Ah ; θψk , θψk ) − B(Ah ; θψk−1 , θψk−1 ) = ∑ Re Qkj (∂ θψk ) 2∆ t j=1 1 k 1 k−1 k k−1 k−1 2 (Ah + Ah )|θψ | , (∂ Ah + ∂ Ah ) + 2 2 1 k k−1 k−1 k−1 + f (θψ , θψ ), (∂ Ah + ∂ Ah ) , 2

(3.64)

which leads to 5 m 1 1 m 0 B(Ah ; θψm , θψm ) = B(Ah ; θψ0 , θψ0 ) + ∆ t ∑ ∑ Re Qkj (∂ θψk ) 2 2 j=1 k=1 m 1 k k k−1 (Ah + Ah )|θψk−1 |2 , ∂ Ah +∆ t ∑ k=1 2 m k +∆ t ∑ f (θψk−1 , θψk−1 ), ∂ Ah .

(3.65)

k=1

It follows from (3.40) that m m 1 k k k k−1 k k−1 k−1 2 6 + A )| θ | , ∂ A (A ∑ kAh + Ah kL6 kθψk−1 k2L 6 k∂ Ah kL2 ∑ 2 h h ψ h k=1 k=1 m

m

k=1

k=0

(3.66)

6 C ∑ kθψk−1 k2L 6 6 C ∑ k∇θψk k2L2 . Combining (3.35) and (3.36) gives m

∑ k=1

6

k f (θψk−1 , θψk−1 ), ∂ Ah = m

∑

k=1

f (θψk−1 , θψk−1 ),

m

∑

m k k f (θψk−1 , θψk−1 ), ∂ θ A + ∑ f (θψk−1 , θψk−1 ), ∂ πh A

k=1 k=1 m k k k−1 k−1 ∂ θ A + C k∇θψ kL2 kθψ kL 6 k∂ πh A kL3 k=1

∑

(3.67)

o m n 6 Ch2r + C ∑ k∇θψk k2L2 + D(θAk , θAk ) + k∂ θAk k2L2 . k=0

Substituting (3.66) and (3.67) into (3.65), we have 5 m 1 m B(Ah ; θψm , θψm ) 6 Ch2r + ∆ t ∑ ∑ Re Qkj (∂ θψk ) 2 j=1 k=1 m k 2 +C∆ t ∑ k∇θψ kL2 + D(θAk , θAk ) + k∂ θAk k2L2 . k=0

(3.68)

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C. P. MA ET AL.

k k We now proceed to estimate ∑m k=1 Re Q j (∂ θψ ) , j = 1, · · · , 5. By virtue of (3.4), we get ! 1 m m ∂ ψ k− 2 k k k k k−1 , θψ − θψ ∆ t ∑ Q1 (∂ θψ ) = 2i ∑ ∂ Rh ψ − ∂t k=1 k=1 ! ! 1 m− 21 2 ∂ ψ ∂ ψ = 2i ∂ Rh ψ m − , θψm − 2i ∂ Rh ψ 1 − , θψ0 ∂t ∂t ! 1 1 m−1 ∂ ψ k+ 2 ∂ ψ k− 2 k k+1 k −2i ∑ ∂ Rh ψ − ∂ Rh ψ − + , θψ . ∂t ∂t k=1

(3.69)

From (3.5) and (3.69), we deduce m

∑ Qk1 (∂ θψk )| 6 C

|∆ t

k=1

We observe that

m−1 h2r+2 + (∆ t)4 + Ckθψm k2L 2 + C∆ t ∑ kθψk k2L 2 .

(3.70)

k=1

1 ∆ tQk2 (∂ θψk ) = 2V0 ψ k− 2 − ψ kh , θψk − θψk−1 1 1 k = 2V0 ψ k− 2 − Rh ψ k , θψk − θψk−1 − 2V0 (θψ + θψk−1 ), θψk − θψk−1 2 def k,1 = J2 + J2k,2 .

It is not difficult to prove m−1 m | ∑ J2k,1 | 6 C h2r+2 + (∆ t)4 + Ckθψm k2L 2 + C∆ t ∑ kθψk k2L 2 .

m

(3.71)

k=1

k=1

We estimate | ∑ Re[J2k,2 ]| by k=1

m

| ∑ Re[J2k,2 ]| = | − V0(kθψm k2L 2 − kθψ0 k2L 2 )| 6 Ckθψm k2L 2 + Ch2r+2.

(3.72)

k=1

Hence we have m

|∆ t

m

m

m

k=1

k=1

k=1

∑ Re Qk2 (∂ θψk ) | 6 | ∑ Re[J2k,1 ]| + | ∑ Re[J2k,2 ]| 6 | ∑ J2k,1| k=1 m

+| ∑ k=1

Re[J2k,2 ]|

m−1 6 C h2r+2 + (∆ t)4 + Ckθψm k2L 2 + C∆ t ∑ kθψk k2L 2 .

(3.73)

k=1

From the definition of the bilinear functional B(A; ψ , ϕ ) in (2.6), we rewrite ∆ tQk3 (∂ θψk ) as follows: 1 ∆ tQk3 (∂ θψk ) = ∇(ψ k− 2 − Rhψ k ), ∇(θψk − θψk−1 ) 1 1 + |Ak− 2 |2 (ψ k− 2 − Rhψ k ), θψk − θψk−1 (3.74) 1 1 +i ∇(ψ k− 2 − Rh ψ k )Ak− 2 , θψk − θψk−1 1 1 −i (ψ k− 2 − Rhψ k )Ak− 2 , ∇θψk − ∇θψk−1 .


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k k Analogous to the estimate of ∑m k=1 Q1 (∂ θψ ), we use (3.4), (3.5), the regularity assumption (2.9) and Young’s inequality to prove m

|∆ t

m

1

∑ Qk3 (∂ θψk )| 6 C{h2r + (∆ t)4} + Ckθψmk2L 2 + 16 k∇θψm k2L2 + C∆ t ∑ k∇θψk k2L2 .

k=1

(3.75)

k=0

Due to space limitations, we omit the proof of (3.75). We observe that 1

1

1

k

1

k

k

k k

φ k− 2 ψ k− 2 − φ h ψ kh = φ k− 2 ψ k− 2 − Ih φ Rh ψ k − Rh ψ k θ φ − φ h θ ψ , and have

1 1 k ∆ tQk4 (∂ θψk ) = φ k− 2 ψ k− 2 − Ih φ Rh ψ k , θψk − θψk−1 k k k − Rh ψ k θ φ , θψk − θψk−1 − φ h θ ψ , θψk − θψk−1 .

(3.76)

The first two terms can be estimated as follows.

m 1 1 k | ∑ φ k− 2 ψ k− 2 − Ihφ Rh ψ k , θψk − θψk−1 | 6 C{h2r + (∆ t)4 } k=1

+Ckθψm k2L 2 + C∆ t

m

|∑ k=1

m

∑ kθψk k2L 2 ,

k=0 1 m | 6 Ch2r + Ckθψm k2L 2 + k∇θ φ k2L2 8 m +C∆ t ∑ kθψk k2L 2 + k∇θφk k2L2 + k∂ θφk k2L2 .

k Rh ψ k θ φ ,

θψk

(3.77)

− θψk−1

k=0

By applying (3.49), the last term on the right hand side of (3.76) can be bounded as follows:

1 m k φ h , |θψk |2 − |θψk−1 |2 ∑ 2 k=1 k=1 1 m 1 0 1 m k k−1 = φ h , |θψm |2 − φ h , |θψ0 |2 − ∑ φ h − φ h , |θψk−1 |2 2 2 2 k=1 m 1 m k 6 kφ h kL6 kθψm kL 3 kθψm kL 2 + Ch2r + C∆ t ∑ k∂ φ h kL2 kθψk−1 k2L 4 2 k=1 m

∑ Re

k k

φ h θ ψ , θψk − θψk−1

=

6 Ckθψm kL 3 kθψm kL 2 + Ch2r + C∆ t

(3.78)

m

∑ kθψk−1k2L 4 k=1

6 Ch

2r

+ Ckθψm k2L 2

m 1 + k∇θψm k2L2 + C∆ t ∑ k∇θψk k2L2 . 16 k=0

Combining (3.76)-(3.78) gives m

∆t

1 1 m 6 C{h2r + (∆ t)4 } + Ckθψmk2L 2 + k∇θ φ k2L2 + k∇θψm k2L2 8 16 m +C∆ t ∑ k∇θψk k2L2 + k∇θφk k2L2 + k∂ θφk k2L2 .

∑ Re Qk4(∂ θψk ) k=1

k=0

(3.79)

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C. P. MA ET AL.

∆ tQk5 (∂ θψk ) can be decomposed as follows: h i 1 k ∆ tQk5 (∂ θψk ) = B(Ak− 2 ; Rh ψ k , θψk − θψk−1 ) − B(A ; Rh ψ k , θψk − θψk−1) h i k k + B(A ; Rh ψ k , θψk − θψk−1) − B(πhA ; Rh ψ k , θψk − θψk−1 ) i h k k + B(πh A ; Rh ψ k , θψk − θψk−1 ) − B(Ah ; Rh ψ k , θψk − θψk−1 )

(3.80)

def

k,2 k,3 = Rk,1 5 + R5 + R5 .

Following the lines of the proof of (3.79) and using (3.7), we prove m m 2r k,2 4 | ∑ Rk,1 + Ckθψm k2L 2 5 | + | ∑ R5 | 6 C h + (∆ t) k=1

k=1

(3.81)

m 1 + k∇θψm k2L2 + C∆ t ∑ k∇θψk k2L2 . 16 k=1

m

To estimate | ∑ Rk,3 5 |, we rewrite it as follows: k=1

m

m

k=1

k=1 m

∑ Rk,3 5 = ∑

k

k

k

k

Rh ψ k (πh A + Ah )(πh A − Ah ), θψk − θψk−1

k k − ∑ i Rh ψ k (πh A − Ah ), ∇θψk − ∇θψk−1 k=1 m

k

+ ∑ i ∇Rh ψ k (πh A k=1 def

k − Ah ),

θψk − θψk−1

= K1 + K2 + K3 .

(3.82)

Note that k k k k k k k−1 R ψ π A + A )( π A − A ), θ − θ ( h ∑ h h h h ψ ψ k=1 m m m 0 0 0 = − Rh ψ m (πh A + Ah )θ A , θψm + Rh ψ 0 (πh A + Ah )θ A , θψ0 m k k k k−1 k−1 k−1 + ∑ Rh ψ k (πh A + Ah )θ A − Rhψ k−1 (πh A + Ah )θ A , θψk−1 . m

K1 =

(3.83)

k=1

By applying Young’s inequality and (3.40), we can estimate the first two terms on the right side of (3.83) by m m m 0 0 0 | Rh ψ m (πh A + Ah )θ A , θψm | + | Rh ψ 0 (πh A + Ah )θ A , θψ0 | m

m

m

6 kRh ψ m kL 6 kπh A + Ah kL6 kθ A kL6 kθψm kL 2 + Ch2r m

1

m

m

6 Ckθ A kH1 kθψm kL 2 + Ch2r 6 CD 2 (θ A , θ A )kθψm kL 2 + Ch2r 6

1 m m D(θ A , θ A ) + Ckθψmk2L 2 + Ch2r . 32

(3.84)


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Observing

k

k

k

Rh ψ k (πh A + Ah )θ A − Rh ψ k−1 (πh A k

= ∆ t Rh ψ (πh A

k

k−1

k k−1 k θA − θA + Ah ) ,

∆t

k−1

θψk−1

+∆ t

Rh ψ k − Rhψ k−1 k k k−1 (πh A + Ah )θ A , θψk−1 ∆t

+∆ t

k−1 Rh ψ k−1 θ A

k

πh A − πh A ∆t

k−1

k−1

+ Ah )θ A , θψk−1 !

k

!

k−1

A − Ah + h ∆t

(3.85)

!

!

, θψk−1 ,

and using (3.40), we get k k k k−1 k−1 k−1 | Rh ψ k (πh A + Ah )θ A − Rh ψ k−1 (πh A + Ah )θ A , θψk−1 | k

k

6 ∆ tkRh ψ k kL 6 kπh A + Ah kL6 k

1 k k−1 (θ A − θ A )kL2 kθψk−1 kL 6 ∆t

Rhψ k − Rh ψ k−1 k k k−1 kL 2 kπh A + Ah kL6 kθ A kL6 kθψk−1 kL 6 ∆t k k−1 k k−1 πh A − πh A A − Ah k−1 + h kL2 kθψk−1 kL 6 +∆ tkRhψ k−1 kL 6 kθ A kL6 k ∆t ∆t k k−1 6 C∆ t k∂ θ A kL2 + kθ A kH1 kθψk−1 kH 1 k k−1 k−1 6 C∆ t k∂ θ A k2L2 + D(θ A , θ A ) + k∇θψk−1k2L2 . +∆ tk

(3.86)

From (3.83)-(3.86), we thus have

m k k k k |K1 | = | ∑ Rh ψ k (πh A + Ah )(πh A − Ah ), θψk − θψk−1 | k=1

1 m m D(θ A , θ A ) + Ckθψm k2L 2 + Ch2r 16 m +C∆ t ∑ k∂ θAk k2L2 + D(θAk , θAk ) + k∇θψk k2L2 6

(3.87)

k=0

From (3.82), using (3.4) and integrating by parts, we get k k Rh ψ k (πh A − Ah ), ∇θψk − ∇θψk−1 k=1 m 0 = − Rh ψ m θ A , ∇θψm + Rh ψ 0 θ A , ∇θψ0 m k k−1 + ∑ Rh ψ k θ A − Rh ψ k−1 θ A , ∇θψk−1 k=1 m m = ∇Rh ψ m θ A , θψm + Rh ψ m ∇ · θ A , θψm m 0 k k−1 + Rh ψ 0 θ A , ∇θψ0 + ∑ Rh ψ k θ A − Rh ψ k−1 θ A , ∇θψk−1 . m

K2 =

∑

k=1

(3.88)

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Using (3.5) and Young’s inequality, we can estimate the first three terms on the right side of (3.88): m m 0 | ∇Rh ψ m θ A , θψm | + | Rh ψ m ∇ · θ A , θψm | + | Rh ψ 0 θ A , ∇θψ0 | m

m

6 k∇Rh ψ m kL3 kθ A kL6 kθψm kL 2 + kIhψ m kL ∞ k∇ · θ A kL2 kθψm kL 2 + Ch2r

(3.89)

1 m m + Ch2r 6 D(θ A , θ A ) + Ckθψm k2L 2 + Ch2r . 32

m 6 Ckθ A kH1 kθψm kL 2

The last term on the right side of (3.88) can be estimated by m k k−1 | ∑ Rh ψ k θ A − Rh ψ k−1 θ A , ∇θψk−1 | k=1

m

6 ∆t

1

k

∑ k ∆ t (Rh ψ k − Rhψ k−1 )kL 3 kθ A kL6 k∇θψk−1 kL2

k=1 m

+∆ t

k

∑ kRhψ k−1 kL ∞ k∂ θ A kL2 k∇θψk−1kL2 k=1 m

k k kθ A kH1 + k∂ θ A kL2 k∇θψk−1 kL2

6 C∆ t

∑

6 C∆ t

k=1 m

∑

k=0

(3.90)

D(θAk , θAk ) + k∂ θAk k2L2 + k∇θψk k2L2 .

We thus get m k k |K2 | = | − i ∑ Rh ψ k (πh A − Ah ), ∇θψk − ∇θψk−1 | k=1

1 m m D(θ A , θ A ) + Ckθψmk2L 2 + Ch2r 16 o m n +C∆ t ∑ D(θAk , θAk ) + k∂ θAk k2L2 + k∇θψk k2L2 . 6

(3.91)

k=0

We similarly prove m k k |K3 | = |i ∑ ∇Rh ψ k (πh A − Ah ), θψk − θψk−1 | k=1

1 m m 6 D(θ A , θ A ) + Ckθψm k2L 2 + Ch2r 16 m +C∆ t ∑ D(θAk , θAk ) + k∂ θAk k2L2 + k∇θψk k2L2 .

(3.92)

k=0

Adding (3.87), (3.91) and (3.92) together, we find m

3 m m D(θ A , θ A ) + Ckθψm k2L 2 + Ch2r 16 k=1 o m n +C∆ t ∑ D(θAk , θAk ) + k∂ θAk k2L2 + k∇θψk k2L2 .

| ∑ Rk,3 5 |6

k=0

(3.93)


It follows from (3.80), (3.81) and (3.93) that m 3 1 m m |∆ t ∑ Qk5 (∂ θψk )| 6 C h2r + (∆ t)4 + D(θ A , θ A ) + k∇θψm k2L2 16 16 k=1 m +Ckθψm k2L 2 + C∆ t ∑ D(θAk , θAk ) + k∂ θAk k2L2 + k∇θψk k2L2

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(3.94)

k=0

Substituting (3.70), (3.73), (3.75), (3.79) and (3.94) into (3.68), we get

1 3 m B(Ah ; θψm , θψm ) 6 C{h2r + (∆ t)4 } + k∇θψm k2L2 2 16 3 1 m m m + k∇θ φ k2L2 + D(θ A , θ A ) + Ckθψm k2L 2 8 16 o m n +C∆ t ∑ k∇θψk k2L2 + k∇θφk k2L2 + k∂ θφk k2L2 + D(θAk , θAk ) + k∂ θAk k2L2 .

(3.95)

5 1 m m k∇θψm k2L2 + kAh θψm k2L2 6 − f (θψm , θψm ), Ah + C h2r + (∆ t)4 16 2 3 1 m m m + k∇θ φ k2L2 + D(θ A , θ A ) + Ckθψm k2L 2 8 16 o m n +C∆ t ∑ k∇θψk k2L2 + k∇θφk k2L2 + k∂ θφk k2L2 + D(θAk , θAk ) + k∂ θAk k2L2 .

(3.96)

k=0

Since we have

m m m B(Ah ; θψm , θψm ) = k∇θψm k2L2 + kAh θψm k2L2 + f (θψm , θψm ), Ah ,

k=0

It follows from (3.40) and the interpolation inequality (3.3) that i m m | f (θψm , θψm ), Ah | = | (θψm )∗ ∇θψm − θψm ∇(θψm )∗ , Ah | 2 m 6 kθψm kL 3 k∇θψm kL2 kAh kL6 6 Ckθψm kL 3 k∇θψm kL2

1 1 k∇θψm k2L2 6 Ckθψm kL 2 kθψm kL 6 + k∇θψm k2L2 32 32 1 6 Ckθψm kL 2 k∇θψm kL2 + k∇θψm k2L2 32 1 m 2 m 2 6 Ckθψ kL 2 + k∇θψ kL2 . 16 6 Ckθψm k2L 3 +

We thus obtain 1 3 1 m m m k∇θψm k2L2 6 C h2r + (∆ t)4 + k∇θ φ k2L2 + D(θ A , θ A ) + Ckθψm k2L 2 4 8 16 m +C∆ t ∑ k∇θψk k2L2 + k∇θφk k2L2 + k∂ θφk k2L2 + D(θAk , θAk ) + k∂ θAk k2L2 .

(3.97)

(3.98)

k=0

Multiplying (3.62) with (C + 1) and adding to (3.98), we get 1 3 1 m m m k∇θψm k2L2 + kθψm k2L 2 6 C h2r + (∆ t)4 + k∇θ φ k2L2 + D(θ A , θ A ) 4 8 16 m +C∆ t ∑ k∇θψk k2L2 + k∇θφk k2L2 + k∂ θφk k2L2 + D(θAk , θAk ) + k∂ θAk k2L2 . k=0

(3.99)

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Adding (3.39) and (3.48) to (3.99), we end up with 1 1 1 kθψm k2L 2 + k∇θψm k2L2 + k∂ θφm k2L2 + k∇θφm k2L2 4 2 16 2r 1 1 m 2 m m + k∂ θA kL2 + D(θA , θA ) 6 C h + (∆ t)4 2 8 m +C∆ t ∑ k∇θψk k2L2 + k∇θφk k2L2 + k∂ θφk k2L2 k=1 +D(θAk , θAk ) + k∂ θAk k2L2 .

(3.100)

Now by applying the discrete Gronwall’s inequality and choosing a sufficiently small ∆ t such that C∆ t 6 21 , we conclude n o max kθψm k2L 2 + k∇θψm k2L2 + k∂ θφm k2L2 + k∇θφm k2L2 + k∂ θAm k2L2 + D(θAm , θAm ) 06k6m (3.101) 2r TC h + (∆ t)4 6 C exp (2TC) h2r + (∆ t)4 . 6 C exp 1 − C∆ t If we take C∗ > C exp (2TC), then (3.8) holds for k = m. By virtue of the mathematical induction, we complete the proof of (3.8). Furthermore, using the triangle inequality and the equality M

θAM = ∆ t ∑ ∂ θAk + θA0 , we can complete the proof of Theorem 2.1. k=1

4. Numerical testes In this section, we present numerical experiments to illustrate the error estimates. We consider the following Maxwell–Schrödinger’s equations:  ∂ψ 1  + (i∇ + A)2 ψ + V0 ψ + φ ψ = f (x,t), (x,t) ∈ Ω × (0, T ), −i    ∂t 2    ∂ 2A i   + ∇ × (∇ × A) − ∇(∇ · A) + ψ ∗ ∇ψ − ψ ∇ψ ∗ 2 ∂t 2   +|ψ |2 A = g(x,t), (x,t) ∈ Ω × (0, T ),      2   ∂ φ − ∆ φ − |ψ |2 = l(x,t), (x,t) ∈ Ω × (0, T ). ∂ t2

(4.1)

where the initial-boundary conditions are given in (1.3)-(1.4). Let Ω = (0, 1)3 , T = 4 and V0 = 5. The exact solution (ψ , A, φ ) of (4.1) is defined by

ψ (x,t) = (1 + 0.5t)eiπ t sin(2π x1) sin(2π x2) sin(2π x3), A(x,t) = cos(π t) cos(π x1 ) sin(π x2 ) sin(π x3 ), sin(π x1 ) cos(π x2 ) sin(π x3 ), sin(π x1 ) sin(π x2 ) cos(π x3 ) , φ (x,t) = t + sin(π t) x1 x2 x3 (1.0 − x1)(1.0 − x2)(1.0 − x3).

The functions f (x,t), g(x,t) and l(x,t) at the right hand side of (4.1) are chosen correspondingly to the exact solution (ψ , A, φ ).


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We partition a whole domain Ω into quasi-uniform tetrahedrons with M + 1 nodes in each direction and 6M 3 elements in total. The system (4.1) is solved by the proposed Crank-Nicolson Galerkin finite element scheme (2.8) with linear elements and quadratic elements, respectively. To confirm the con1 vergence rate of the proposed method, we take ∆ t = h 2 for the linear element method and ∆ t = h for the quadratic element method respectively. The numerical results for the linear element method and the quadratic element method at time t = 1.0, 2.0, 3.0, 4.0 are displayed in Tables 1 and 2, respectively. It clearly shows that they are in good agreement with the error estimates presented in Theorem 2.1. Table 1. H 1 error of linear FEM with h =

t 1.0 2.0 3.0 4.0 t 1.0 2.0 3.0 4.0 t 1.0 2.0 3.0 4.0

1 M

1

and ∆ t = h 2 .

M=25 4.9855e-01 6.7234e-01 4.6375e-01 5.8205e-01

kAkh − A(·,tk )kH1 M=50 2.3894e-01 3.2057e-01 2.2119e-01 3.0487e-01

M=100 1.1887e-01 1.7574e-01 1.0373e-01 1.4287e-01

Order 1.03 0.97 1.08 1.02

M=25 3.0718e-01 4.3713e-01 3.0004e-01 2.2543e-01

kψhk − ψ (·,tk )kH 1 M=50 1.4419e-01 2.1289e-01 1.4202e-01 1.2316e-01

M=100 7.4104e-02 1.1430e-01 6.9620e-02 6.0316e-02

Order 1.02 0.97 1.05 0.95

M=25 1.9412e-01 1.2473e-01 1.1031e-01 8.7695e-02

kφhk − φ (·,tk )kH 1 M=50 8.6435e-02 7.0233e-02 6.1394e-02 4.1380e-02

M=100 4.3191e-02 3.3148e-02 2.9217e-02 2.1815e-02

Order 1.07 0.96 0.96 1.00

R EFERENCES A HMED , I. & L I , E. (2012) Simulation of plasmonics nanodevices with coupled Maxwell and schrödinger equations using the FDTD method. Advanced Electromagnetics, 1, 76–83. C HEN , Z. & H OFFMANN , K. H. (1995) Numerical studies of a non-stationary Ginzburg-Landau model for superconductivity. Adv. Math. Sci. Appl., 5, 363–389. C HEW, W. C. (2014) Vector Potential Electromagnetics with Generalized Gauge for Inhomogeneous Media: Formulation. Progress In Electromagnetics Research, 149, 69–84. G AO , H. D., L I , B. Y. & S UN , W. W. (2014) Optimal error estimates of linearized Crank-Nicolson Galerkin FEMs for the time-dependent Ginzburg-Landau equations in superconductivity. SIAM J. Numer. Anal., 52, 1183–1202. G AO , H. D. & S UN , W. W. (2015) An efficient fully linearized semi-implicit Galerkin-mixed FEM for the dynamical Ginzburg-Landau equations of superconductivity. J. Comput. Phys., 294, 329–345. G AO , H. D. & S UN , W. W. (2015) A new mixed formulation and efficient numerical solution of Ginzburg-Landau

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Table 2. H 1 error of quadratic FEM with h = ∆ t =

t 1.0 2.0 3.0 4.0 t 1.0 2.0 3.0 4.0 t 1.0 2.0 3.0 4.0

1 M.

M=25 6.5062e-02 5.5496e-02 4.1903e-02 3.4765e-02

kAkh − A(·,tk )kH1 M=50 1.6679e-02 1.5018e-02 1.3291e-02 8.4562e-03

M=100 4.1805e-03 3.7016e-03 2.4825e-03 2.0952e-03

Order 1.98 1.95 2.04 2.03

M=25 1.1930e-02 1.0237e-02 2.2342e-02 1.3203e-02

kψhk − ψ (·,tk )kH 1 M=50 3.2391e-03 2.9786e-03 4.8413e-03 3.1681e-03

M=100 8.0967e-04 5.9668e-04 1.3683e-03 7.8548e-04

Order 1.94 2.05 2.01 2.04

M=25 3.1070e-02 2.7189e-02 3.1162e-02 2.5343e-02

kφhk − φ (·,tk )kH 1 M=50 7.8114e-03 8.2275e-03 6.9836e-03 6.4257e-03

M=100 1.8648e-03 1.8301e-03 1.7682e-03 1.6146e-03

Order 2.03 1.95 2.07 1.99

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