Semi-discrete / Fully discrete Finite Element Methods for Heat-Wave ...

Semi-discrete / Fully discrete Finite Element Methods for Heat-Wave Interaction System Xu Zhang

(School of Mathematics, Sichuan University, Chengdu 610064, PR China Email:[email protected]) Guo-zhu Yu

(School of Mathematics, Sichuan University, Chengdu 610064, PR China Email:[email protected]) Xiao-ping Xie

(School of Mathematics, Sichuan University, Chengdu 610064, PR China Email:[email protected])

Abstract In this paper, we consider the finite element method for the coupled heat-wave system. Semidiscrete and fully discrete finite element formulations for the interaction are defined. Existence and uniqueness of finite element solutions are proved, and both semi-discrete and fully discrete error estimates in energy norm are obtained. Numerical experiments are done to verify the theoretical analysis. Key words: Heat-wave, Fluid-structure interaction, Semi-discrete, Fully discrete, Finite element, Error estimates

1.Introduction The analysis of multiphysics problems, and specifically the modeling of fluid-structure interactions (FSI), has attracted increasing attention during recent years. Possible applications include FSI modeling development in biomechanics [10], in numerical aeroelastic simulation [11,25], in designing the structural components of rocket engines [13], in large deformation [18,20], and also in nuclear fuel racks [17]. Besides, the FSI modeling, combined with the solid-rigid contact, is particularly important for the dynamics and impact of heart valves [14,15]. In recent decades, a significant research effort has been devoted to the development of numerical study of the FSI modeling. Q. Du, M. Gunzburger, L. Hou and J. Lee presented a semi-discrete finite element formulation of FSI modeling, verified a discrete inf-sup condition and established the existence of a finite element pressure in [4,5]. The Arbitrary Lagrangian-Eulerian (ALE) method, combining the classical Lagrangian and Eulerian viewpoints, is a suitable procedure for the analysis of FSI Modeling. Particular care must be taken to properly couple the fluid and the structure on the interface [6,8,12,16]. Team for Advanced Flow Simulation and Modeling developed the space-time FSI techniques that have been applied to a wide range of 3D computation of FSI problems [21]. L. Zhang and M. Gay presented a detailed derivation of immersed Finite Element Method for the solution of fluid-structure interaction problems in [22]. Moreover, K. Bathe and H. Zhang discussed some further development to solve fluid flows with structural interactions [21]. The FSI modeling describes the dynamics of fluids in contact with the elastic structures with natural transmission conditions at a common interface. These conditions couple the heat unknown with the 1

velocity of the wave solution. The wave-heat (HW) modeling is a linearized simplified version of more complex models which arise in fluid-structure interaction. The HW modeling is studied in recent years both analytically and computationally. X. Zhang and E. Zuazua analyzed long-time behavior of a coupled system consists of HW interaction in [23]. Applications include the production and absorption of sound occurring when unsteady flow interacts with solid bodies, biomedical flows in flexible pipes, piston problems, biomechanical systems and novel mechanical structures, and so forth. e.g. [2,9,19]. This paper is devoted to study of the conforming finite element method (FEM) for the HW modeling. Much effort has been done in recent years and has led to some significant results, little has concerned about the fully-discrete finite element formulation of the HW modeling. The main goal of this paper is to establish the discrete weak formation, the existence/uniqueness of the discrete solutions and the error estimation of both semi-discrete and fully discrete approximation. For simplification, we combine the fluid and the structure unknowns into a single velocity variety to further the discussion. We ultimately demonstrate that the displacement of structure and the velocity of fluid are of O (h + τ) accuracy. . The rest of paper is arranged as follows. In section 2, we introduce some notations of the HW model and recall some known properties. In section 3, we present the semi-discrete finite element approximation and derive error estimates. In section 4, we define the fully-discrete finite element approximation and consider the error estimates. In section 5, we present numerical experiments to test the accuracy of the conclusions in Section 4.

2.Notations and weak formulations 2.1 Notations We now describe the HW model and some notations: Assume that the wave and the heat occupy two neighboring open Lipschitz domains, Ω1 ⊂ Rd and ¯1SΩ ¯ 2 , i.e. Ω is the entire wave-heat region. Moreover, Ω2 ⊂ Rd where d = 2, 3 . Ω is the interior of Ω T we let γ = ∂Ω1 ∂Ω2 denote the interface between wave and heat, and let Γ1 = ∂Ω1 \γ, Γ2 = ∂Ω2 \γ, respectively, denote the parts of the heat and wave boundaries, excluding the interface γ. And ni (i = 1, 2) denotes the unit outward normal vector to Ωi (i = 1, 2).

Fig 1 Geometric description of the fluid-structure interaction model.

The model we discuss are the following coupled heat-wave system:

(2.1)

                              

vt − ∆v = f1 utt − ∆u = f2 v=0 u=0 ∂u ∂v = − ∂n ut = v, ∂n 1 2 v(0) = v0 u(0) = u0 , ut (0) = u1

2

in (0, T ) × Ω1 in (0, T ) × Ω2 on (0, T ) × Γ1 on (0, T ) × Γ2 on (0, T ) × γ in Ω1 in Ω2

This is a simplified and linearized model of fluid-structure interaction, which consists of a wave and a heat equation coupled through an interface with transmission conditions. In system (2.1) v can be viewed as the velocity of the fluid, while u and ut represent the displacement and velocity of the structure respectively. Moreover, f1 and f2 denotes the given body force and the given loading force and v0 is the given initial fluid velocity, while u0 and u1 are the given initial structure displacement and velocity. The fifth equation in (2.1) indicates that the velocity of fluid and structure are continuous across the interface γ. In this paper, the letter C, with or without subscript, denote generic constants that may not be the same at different occurrences. To avoid writing these constants repeatedly, by x . y (y & x) we mean that there exist a constant C such that x ≤ Cy. H m (K) and Hm (K) denote respectively the standard and the vector-valued Sobolev spaces with order m defined on the region K. Particularly, H01 (K) ≡ {u ∈ H 1 (K) : u|∂Ω = 0} and H01 (K) is similarly defined. Since the functions v, u also depend on time, we introduce the some Sobolev space involving time. Let H be any of the Hilbert spaces listed above, then L2 (0, T ; H), C(0, T ; H), H 1 (0, T ; H), with corresponding norm equipped, are defined as follows: R 1/2 o T def n L2 (0, T ; H) = u; kukL2 (0,T ;H) < ∞ and kukL2 (0,T ;H) = 0 kuk2H dt , o def n C(0, T ; H) = u; kukC(0,T ;H) < ∞ and k f kC(0,T ;H) = sup k f kH , t∈[0,T ] 1/2 R n o T def H 1 (0, T ; H) = u; kukH 1 (0,T ;H) < ∞ and k f kH 1 (0,T ;H) = 0 k f k2H + k∇ f k2H dt . To simplify the notations, we define several function spaces and the corresponding norms: def

Hi = [H01 ]|Ωi And we also define: ( v def (2.2) ξ = ut

in Ω1 in Ω2

def

k · kHi = k · k1,Ωi ( def

ξ0 =

v0 u1

i = 1, 2

in Ω1 in Ω2

In addition, we define the bilinear forms: Z def ai (u, v) = ∇u · ∇v dΩ

and

( def

and

f =

∀u, v ∈ Hi

Ωi

def

H = H1 ⊗ H2 . f1 f2

in Ω1 in Ω2

i = 1, 2

Obviously, the bilinear forms ai (·, ·) are continuous, i.e. (2.3)

|ai (u, v)| . kuk1,Ωi kvk1,Ωi

∀u, v ∈ Hi

i = 1, 2.

On the other hand, the bilinear forms ai (·, ·) are coercive by Korn’s inequalities [1], i.e. (2.4)

ai (u, u) & kuk21,Ωi

∀u ∈ Hi

i = 1, 2.

2.2 Weak formulation Now we are going to introduce the weak formulation for (2.1). It is followed from the regularity assumption:   f1 ∈ L2 (0, T ; L2 (Ω1 ))       f2 ∈ L2 (0, T ; L2 (Ω2 ))    v0 ∈ H1 (2.5)      u0 , u1 ∈ H2      v0 |γ = u1 |γ 3

Under the above assumption, a weak formulation for (2.1) can be derived by multiplying on both sides with any function η ∈ H01 (Ω), and integrating by parts. The weak problem of (2.1) is to seek a pair of (v, u) ∈ L2 (0, T ; X1 ) × L2 (0, T ; X2 ) such that: (2.6)

(vt , η)Ω1 + a1 (v, η) + (utt , η)Ω2 + a2 (u, η) = ( f1 , η)Ω1 + ( f2 , η)Ω2

∀η ∈ H01 (Ω).

with the initial and boundary conditions: (2.7)

v|t=0 = v0 , Z

u|t=0 = u0 ,

ut |t=0 = u1 ,

t

(2.8) 0

v(s)|γ ds = u(t)|γ − u0 |γ

a.e. t.

Noticing the definition (2.2), the HW system (2.1) is equivalent to the following form: Find a ξ ∈ L2 ((0, T ); L2 (Ω)) such that: Z 2

(2.9)

ξ|Ω1 ∈ L (0, T ; H1 ),

0

Z (2.10)

(ξt , η)Ω + a1 (ξ, η) + a2 Z

(2.11)

ξ(0) = ξ0 ,

t

t 0

ξ(s)|Ω2 ds ∈ L2 (0, T ; H2 ),

 ξ(s)ds, η = ( f , η)Ω − a2 (u0 , η),

t

Z




ξ(s)|Ω1 |γ ds =

0

t


ξ(s)|Ω2 |γ ds.

0

Theorem 2.1 Assume that f1 , f2 , u0 , u1 and v0 satisfy (2.5). Then there exists a unique pair of (v, u) ∈ L2 (0, T ; H1 ) × L2 (0, T ; H2 ) for (2.6)-(2.8).

Proof. Existence and uniqueness of the solution (v, u) ∈ L2 (0, T ; H1 ) × L2 (0, T ; H2 ) in (2.1) are provided in [23]. Therefore, such (v, u) is also a solution of the weak formulation (2.6)-(2.8). What remains to check is the uniqueness of the solution for (2.6)-(2.8). Assume further that there exists another pair of (v∗ , u∗ ) ∈ L2 (0, T ; H1 ) × L2 (0, T ; H2 ) that solves (2.6)-(2.8), then we have: (2.12)

(vt − v∗t , η)Ω1 + a1 (v − v∗ , η) + (utt − u∗tt , η)Ω2 + a2 (u − u∗ , η) = 0.

Take

( η=

We get

v − v∗ ut − u∗t

∀η ∈ H01 (Ω)

in Ω1 in Ω2

1d 1d 1d kut − u∗t k20,Ω2 + ku − u∗ k21,Ω2 + kv − v∗ k20,Ω1 + a1 (v − v∗ , v − v∗ ) = 0. 2 dt 2 dt 2 dt

Note that a1 (v − v∗ , v − v∗ ) = |v − v∗ |21,Ω1 , integration on both sides yields the following, Z kut −

u∗t k20,Ω2

+ ku −

Therefore, we have v = v∗ , and u = u∗ .

u∗ k21,Ω2

+ kv −

v∗ k20,Ω1

t

+ 0

|v − v∗ |21,Ω1 dt = 0.

2

3.Semi-discrete finite element approximation 3.1 Semi-discrete finite element discretization

4

In what follows we let Ω1 ⊂ R2 and Ω2 ⊂ R2 be both polyhedral domains with a triangulation T h , with mesh size h = max {hi }, and hi denotes the diameter of the triangle Ki . We also assume that the Ki ∈T h

triangulation do not cross the interface. For every 0 < h < 1, we assume that H h to be linear finite ¯ : ξ|K ∈ P1 (K), ∀K ∈ T h }. And similarly with section 2, we element space, i.e. H h = {ξ ∈ H01 (Ω) ∩ C(Ω) set def Hih = H h |Ωi , i = 1, 2 Assume that Hih (i = 1, 2). satisfy the standard approximation property [3], i.e. (3.1) (3.2)

inf ku − uh k0,Ωi . hr+1 kukr+1,Ωi

∀u ∈ Hr+1 (Ωi ) ∩ Hi ,

r = 0, 1.

inf ku − uh k1,Ωi . hr kukr+1,Ωi

∀u ∈ Hr+1 (Ωi ) ∩ Hi ,

r = 0, 1.

uh ∈Hih uh ∈Hih

The semi-discrete finite element problem of (2.6)-(2.8) is as follows: Seek a pair (vh , uh ) ∈ C 1 (0, T ; H1h ) × C 2 (0, T ; H2h ) such that: ∀η h ∈ H h , (3.3)

(∂t vh , η h )Ω1 + a1 (vh , η h ) + (∂tt uh , η h )Ω2 + a2 (uh , η h ) = ( f1 , η h )Ω1 + ( f2 , η h )Ω2 ,

(3.4)

vh |γ = ∂t uh |γ ,

(3.5)

vh |t=0 = v0,h

uh |t=0 = u0,h

∂t uh |t=0 = u1,h ,

where v0,h ∈ H1h , u0,h ∈ H2h and u1,h ∈ H2h are projection approximations of v0 , u0 and u1 , respectively, such that, (3.6)

∀ηh ∈ H2h ,

a2 (u0,h , ηh ) = a2 (u0 , ηh )

(3.7)

a1 (v0,h , ηh ) + (u1,h , ηh )Ω2 = a1 (v0 , ηh ) + (u1 , ηh )Ω2

∀ηh ∈ H h .

By the definition above, the discrete initial value v0,h u0,h and u1,h uniquely exist. The approximation of them will be discussed afterwards which is conducive to derivation the error estimate of the semi-discrete finite element approximation . Similarly with the last section, we define, ( vh in Ω1 def (3.8) ξh = ∂t uh in Ω2 Then it is equivalent to rewrite the semi-discrete formation (3.3)-(3.5) as follows: Z t  (3.9) (∂t ξh , ηh )Ω + a1 (ξh , ηh ) + a2 ξh (s) ds, ηh = ( f , ηh )Ω − a2 (u0,h , ηh ), 0

def

(3.10)

(

ξ0,h = ξh (0) =

v0,h u1,h

in Ω1 in Ω2

def

where (u, v)Ω = (u, v)Ω1 + (u, v)Ω2 . From (3.8) and (3.10), we have (3.11)

vh = ξh |Ω1

Z uh =

t 0

ξh (s)|Ω2 ds + u0,h .

 Let ϕ j Nj=1 be a basis of H h . (3.10) implies ξ0,h ∈ H h , so that we write: 5

(3.12)

ξ0,h =

N X

c jϕ j.

j=1

And the semi-discrete solution ξh for(3.7)-(3.8) can be written as follows: (3.13)

ξh =

N X

g j (t)ϕ j .

j=1

Therefore, (3.9)-(3.10) is equivalent to the following initial value problem for linear system of ordinary  differential equations for g j (t) Nj=1 :

(3.14)

 N Rt N N  P P P    a1 (ϕ j , ϕi ) g j (t) + a2 (ϕ j , ϕi ) 0 g j (s) ds (ϕ j , ϕi )Ω g0j (t) +    j=1 j=1 j=1      = ( f1 , ϕi )Ω1 + ( f2 , ϕi )Ω2 − a2 (u0,h , ϕi ) i = 1, . . . , N         g (0) = c i = 1, . . . , N i i 

Lemma 2.1 There exists a unique series of the solution g j (t) Proof. We set h j (t) =

Rt 0

g j (s) ds

( j = 1, . . . , N)

N

j=1

for (3.14).

then, (3.14) is equivalent to the following

ordinary differential equations:  N N  P P    (ϕ j , ϕi )Ω g0j (t) + a1 (ϕ j , ϕi ) h0j (t)    j=1 j=1     N N  P P    = − a2 (ϕ j , ϕi ) h j (t) − c j a1 (ϕ j , ϕi ) + ( f , ϕi )Ω − a2 (u0,h , ϕi )    j=1 j=1  (3.15)      h0i (t) = gi (t) − ci i = 1, . . . , N        gi (0) = ci i = 1, . . . , N        hi (0) = 0 i = 1, . . . , N

i = 1, . . . , N

Let g = (g1 , . . . , gN )T , h = (h1 , . . . , hN )T , A1 = {a1 (ϕ j , ϕi )}N×N A2 = {a2 (ϕ j , ϕi )}N×N and B = {(ϕ j , ϕi )Ω }N×N , then we can use matrix form to rewrite (3.15) as follows: ! ! ! !   B A1 g0 I 0 g   = + F    0 I h0 0 −A2 h (3.16)       (gT , hT ) = (c , . . . , c , 0, . . . , 0) 1 N where F denotes the corresponding term of the right side of (2.12), and I denotes the unit matrix. The matrices A1 , B are both positive definite as the function set {ϕ1 , . . . , ϕN } is linearly independent. There! B A1 fore, the matrix is also positive definite hence invertible. Thus, standard theories for constant 0 I coefficient system of linear ordinary differential equations ensure that the above system (3.14) has a unique solution {g1 , . . . , gN , h1 , . . . , hN )on [0, T ]. Eliminating {h1 , . . . , hN } in the solution, we get the  unique solution g j (t) Nj=1 for (3.14) 2 The above lemma immediately yields the existence and the uniqueness of the semi-discrete finite element solution ξh satisfying (3.7)-(3.8). By the equivalence between (3.1)-(3.3) and(3.7)-(3.8), we immediately obtain the corresponding results for (vh , uh ), i.e.

6

Theorem 3.2 Assume that f1 , f2 , u0 , u1 and v0 satisfy (2.5). Then, there exists a unique pair of (vh , uh ) ∈ C 1 (0, T ; H1h ) × C 2 (0, T ; H2h ) satisfying (3.3)-(3.5).

3.2 Error estimates We now derive the error estimates for semi-discrete finite element formulation. Recall the discrete initial value v0,h u0,h and u1,h defined in (3.7)-(3.8). Standard interpolation argument and the approximation properties [3] yield:

Lemma 3.3 Assume that v0 ∈ H1 , u1 ∈ H2 and u0 ∈ H2 . Then, there exists a unique triplet v0,h ∈ H1h , u0,h ∈ H2h and u1,h ∈ H2h satisfying (3.6)-(3.7).

Moreover, if v0 ∈ H1 ∩ Hr+1 (Ω1 ), u1 ∈ H2 ∩ Hr+1 (Ω2 ) and u0 ∈ H2 ∩ Hr+1 (Ω2 ) for some r ∈ [0, k] we have the following approximation: (3.17)

kv0,h − v0 k1,Ω1 + ku1,h − u1 k0,Ω2 + ku0,h − u0 k1,Ω2 . hr (kv0 kr+1,Ω1 + hku1 kr+1,Ω2 + ku0 kr+1,Ω2 )

Next, we turn to another useful tool that is also useful in deducing error estimates. Assume that H h = ¯ : ξ|K ∈ P1 (K), ∀K ∈ T h }. We define L2 − projection operator Ph : L2 (Ω) −→ H h , {ξ ∈ H01 (Ω) ∩ C(Ω) for any η ∈ L2 (Ω), Ph η is the solution of : (3.18)

∀wh ∈ H h .

(η − Ph η, wh )Ω = 0

Applying Green’s function together with Riesz-Thorin Theorem [24], we have the following lemma:

Lemma 3.4 Under the definition (3.18), the following estimate for the projector Ph holds: (3.19)

kη − Ph ηk1,Ω . hr kηkr+1,Ω

(3.20)

kη − Ph ηk0,Ω . hr+1 kηkr+1,Ω

∀η ∈ H with η|Ωi ∈ Hr+1 (Ωi ), r = 0, 1. ∀η ∈ H with η|Ωi ∈ Hr+1 (Ωi ), r = 0, 1.

Now we turn to prove the following error estimates for semi-discrete finite element approximation of HW system.

Theorem 3.5

Assume that f1 , f2 , u0 , u1 and v0 satisfy (2.5). Let (v, u) be the solution of (2.6)(2.8); and (vh , uh ) be the solution of (3.3)-(3.5). Assume further that v ∈ L2 (0, T ; Hr+1 (Ω1 )), ∂t v ∈ L2 (0, T ; Hr−1 (Ω1 )), ∂t u ∈ L2 (0, T ; Hr+1 (Ω2 )), ∂tt u ∈ L2 (0, T ; Hr−1 (Ω2 )), v0 ∈ Hr+1 (Ω1 ), u1 ∈ Hr+1 (Ω2 ) and u0 ∈ Hr+1 (Ω2 ), Then,

(3.21)

kv(t) − vh (t)k0,Ω1 + kv − vh kL2 (0,T ;H1 ) + k∂t u(t) − ∂t uh (t)k0,Ω2 + ku(t) − uh (t)k1,Ω2  . hr kv0 kr+1,Ω1 + ku1 kr+1,Ω2 + ku0 kr+1,Ω2 + kvkL2 (0,T ;Hr+1 (Ω1 ))  +kut kL2 (0,T ;Hr+1 (Ω2 )) + k∂t vkL2 (0,T ;Hr−1 (Ω1 )) + k∂tt ukL2 (0,T ;Hr−1 (Ω2 )) r = 1, 2

for all t ∈ [0, T ].

Proof. Let ξ and ξh be defined as (2.2) and (3.8), respectively. To simplify the notation, we set v˜ h (t) = {Ph ξ(t)}|Ω1 and w˜ h (t) = {Ph ξ(t)}|Ω2 , where Ph is the L2 − projection operator defined in (3.18). Subtracting continuous equations (2.9) from the semi-discrete equations (3.9), while applying the discrete initial value satisfying (3.6), we get: Z t  (3.22) (∂t ξ − ∂t ξh , ηh )Ω + a1 (ξ − ξh , ηh ) + a2 ξ(s) − ξh (s) ds, ηh = 0 ∀ηh ∈ H h . 0

This relation is equivalent to the following form: 7

(∂t v − ∂t vh , ηh )Ω1 + (∂tt u − ∂tt uh , ηh )Ω2 + a1 (v − vh , ηh ) + a2 (u − uh , ηh ) = 0 ∀ηh ∈ H h .

(3.23) Therefore,

(3.24) (∂t v − ∂t vh , v − vh )Ω1 + (∂tt u − ∂tt uh , ∂t u − ∂t uh )Ω2 + a1 (v − vh , v − vh ) + a2 (u − uh , ∂t u − ∂t uh ) = (∂t v − ∂t vh , v − v˜ h )Ω1 + (∂tt u − ∂tt uh , ∂t u − w˜ h )Ω2 + a1 (v − vh , v − v˜ h ) + a2 (u − uh , ∂t u − w˜ h ) +(∂t v − ∂t vh , v˜ h − vh )Ω1 + (∂tt u − ∂tt uh , w˜ h − ∂t uh )Ω2 + a1 (v − vh , v˜ h − vh ) + a2 (u − uh , ∂t w˜ h − ∂t uh ) = (∂t v − ∂t vh , v − v˜ h )Ω1 + (∂tt u − ∂tt uh , ∂t u − w˜ h )Ω2 + a1 (v − vh , v − v˜ h ) + a2 (u − uh , ∂t u − w˜ h ). The last equal sign holds due to the property of L2 − projection operator defined in (3.18). Moreover, considering the first two terms in the last equation of (3.24), the following equations hold: (∂t v − ∂t vh , v − v˜ h )Ω1 + (∂tt u − ∂tt uh , ∂t u − w˜ h )Ω2

(3.25)

= (∂t ξ(t) − ∂t ξh (t) , ξ(t) − Ph ξ(t))Ω = (∂t ξ(t) , ξ(t) − Ph ξ(t))Ω

(by (3.17))

= (∂t ξ(t) − ∂t Ph ξ(t), ξ(t) − Ph ξ(t))Ω

(by

(3.17))

1d (ξ(t) − Ph ξ(t), ξ(t) − Ph ξ(t))Ω 2 dt 1d 1d = kv − v˜ h k20,Ω1 + k∂t u − w˜ h k20,Ω2 . 2 dt 2 dt Similarly, the following equation holds: 1d 1d (3.26) (∂t v − ∂t vh , v − vh )Ω1 + (∂tt u − ∂tt uh , ∂t u − ∂t uh )Ω2 = kv − vh k20,Ω1 + k∂t u − ∂t uh k20,Ω2 . 2 dt 2 dt 1d (3.27) a2 (u − uh , ∂t u − ∂t uh ) = a2 (u − uh , u − uh ). 2 dt Combining (3.24)-(3.27), we get the following estimates: =

1d 1d 1d kv − vh k20,Ω1 + k∂t u − ∂t uh k20,Ω2 + a1 (v − vh , v − vh ) + a2 (u − uh , u − uh ) 2 dt 2 dt 2 dt

(3.28)

= (∂t v − ∂t vh , v − vh )Ω1 + (∂tt u − ∂tt uh , ∂t u − ∂t uh )Ω2 + a1 (v − vh , v − vh ) + a2 (u − uh , ∂t u − ∂t uh ) 1d 1d kv − v˜ h k20,Ω1 + k∂t u − w˜ h k20,Ω2 + a1 (v − vh , v − v˜ h ) + a2 (u − uh , ∂t u − w˜ h ) 2 dt 2 dt 1d . kv − v˜ h k20,Ω1 + kv(t) − vh (t)k21,Ω1 + kv(t) − v˜ h (t)k21,Ω1 2 dt 1d k∂t u − w˜ h k20,Ω2 + ku(t) − uh (t)k21,Ω2 + k∂t u(t) − w˜ h (t)k21,Ω2 . + 2 dt The coercivity of ai (·, ·), yields : =

d d d kv − vh k20,Ω1 + k∂t u − ∂t uh k20,Ω2 + kv − vh k21,Ω1 + ku − uh k21,Ω2 dt dt dt

(3.29)

d kv − v˜ h k20,Ω1 + kv(t) − vh (t)k21,Ω1 + kv(t) − v˜ h (t)k21,Ω1 dt d + k∂t u − w˜ h k20,Ω2 + ku(t) − uh (t)k21,Ω2 + k∂t u(t) − w˜ h (t)k21,Ω2 . dt Integration on both sides of (3.29) yields: .

(3.30)

kv(t) − vh (t)k20,Ω1 + k∂t u(t) − ∂t uh (t)k20,Ω2 kv − vh k2L2 (0,T ;H 1 (Ω1 )) + ku(t) − uh (t)k21,Ω2 8

. kv − v˜ h k20,Ω1 + k∂t u − w˜ h k20,Ω2 + kv(0) − v˜ h (0)k20,Ω1 + k∂t u(0) − w˜ h (0)k20,Ω2 +kv(0) − vh (0)k20,Ω1 + k∂t u(0) − ∂t uh (0)k20,Ω2 + ku(0) − uh (0)k21,Ω2 +kv − vh k2L2 (0,T ;H 1 (Ω1 )) + kv − v˜ h k2L2 (0,T ;H 1 (Ω1 )) + ku − uh k2L2 (0,T ;H 1 (Ω2 )) + k∂t u − w˜ h k2L2 (0,T ;H 1 (Ω2 )) Noticing the definition of (˜vh , w˜ h ), it is equivalent to put (3.29) as following: (3.31)

kv(t) − vh (t)k20,Ω1 + k∂t u(t) − ∂t uh (t)k20,Ω2 + kv − vh k2L2 (0,T ;H 1 (Ω1 )) + ku(t) − uh (t)k21,Ω2  . kv(0) − v0,h k20,Ω1 + k∂t u(0) − u1,h k20,Ω2 + ku(0) − u0,h k21,Ω2

 Z t +kξ0 − Ph ξ0 k20,Ω + kξ(t0 ) − Ph ξ(t0 )k20,Ω + kξ − Ph ξk2L2 (0,T ;H1 (Ω)) + ku(s) − uh (s)k21,Ω2 ds, 0

where t0 ∈ [0, T ] satisfying that: kξ(t0 ) −

Ph ξ(t0 )k20,Ω

= max kξ(t) −

Applying the estimate (3.19), we have: (3.32)

t∈[0,T ]

Ph ξ(t)k20,Ω .

kξ(t0 ) − Ph ξ(t0 )k20,Ω   . h2r kv(t0 )k2r,Ω1 + k∂t u(t0 )k2r,Ω2 )   . h2r kvk2L2 (0,T ;Hr+1 (Ω1 )) + k∂t vk2L2 (0,T ;Hr−1 (Ω1 )) + kuk2L2 (0,T ;Hr+1 (Ω2 )) + k∂t uk2L2 (0,T ;Hr−1 (Ω2 )) .

The last inequality follows from [7,pp. 288,Theorem 4]. Combining (3.17) and (3.32), dropping the first three terms on the left side of estimate (3.30), and applying the Gronwall’s inequality, we obtain that: (3.33)

ku(t) − uh (t)k21,Ω2  . h2r kv0 k2r,Ω1 + ku1 k2r,Ω2 + ku0 k2r,Ω2 +kvk2L2 (0,T ;Hr+1 (Ω1 )) + k∂t vk2L2 (0,T ;Hr−1 (Ω1 )) + kuk2L2 (0,T ;Hr+1 (Ω2 )) + k∂t uk2L2 (0,T ;Hr−1 (Ω2 ))

Hence, (3.21) follows from (3.31)-(3.33).



2

4.Fully-discrete finite element approximation 4.1 Fully-discrete finite element discretization We now consider a fully discrete scheme for (2.9)-(2.11). Suppose [0, T ] is partitioned into equal subintervals with time step τ = T/M, where M is a positive integer. We denote tn = nτ (0 ≤ n ≤ M) in the following discussion. The fully-discrete problem of (2.9)-(2.11) is as follows: Find ξhn ∈ H h (0 ≤ n ≤ M) such that: (4.1)


(∂¯t ξhn , ηh )Ω + a1 (ξhn , ηh ) + a2 Ln (ξh ), ξ h = ( f , ηh )Ω − a2 (u0,h , ηh ) (

(4.2)

ξh0

= ξ0,h =

v0,h u1,h

in Ω1 in Ω2

where, the operator ∂¯t denotes the backward difference operator, i.e. (4.3)

∂¯t ξn = (ξn − ξn−1 )/τ, 9

∀ηh ∈ Hh ,

and Ln denotes the composite trapezoidal operator, i.e. 1 1 Ln (ξh ) = τξh0 + τξh1 + . . . + τξhn−1 + τξhn . 2 2

(4.4)

Under the fully-discrete form (4.1)-(4.2), un and vn should be the following form: vnh = ξhn |Ω1 ,

(4.5)

unh = Ln (ξh |Ω2 ) + u0,h .

Theorem 4.1 Assume that f1 , f2 , u0 , u1 and v0 satisfy (2.5). Then, there exists a unique solution ξh ∈ H h (0 ≤ n ≤ M) satisfying (4.1)-(4.2).

Proof. Under the definitions of ∂¯t and Ln , the fully-discrete form (4.1)-(4.2) is equivalent to the following: (4.6) (4.7)

=

1 τ 1 τ

ξhn , ηh


Ω

ξhn−1 , ηh




+ a1 (ξhn , ηh ) + 2τ a2 (ξhn , ηh ) Ω


− 2τ a2 2ξhn−1 + . . . + 2ξh1 + ξh0 , ηh + ( f , ηh )Ω − a2 (u0,h , ηh )

∀ηh ∈ Hh .

ξh0 = ξ0,h .

N P  Recall that ϕ j Nj=1 are a basis of H h , therefore ξhn = ϕ j (x) g j (tn ). Let g = (g1 , . . . , gN )T , h = j=1

(h1 , . . . , hN )T , A1 = {a1 (ϕ j , ϕi )}N×N A2 = {a2 (ϕ j , ϕi )}N×N and B = {(ϕ j , ϕi )Ω }N×N , then we can use matrix form to rewrite (4.6)-(4.7) as follows: Find g = (g1 , . . . , gN )T such that: 1

τ  A2 g = F, τ 2 where F is the corresponding term of the right side of (4.6). Since matrices A1 , A2 , B are all positive definite, the coefficient matrix 1τ B + A1 + 2τ A2 is positive definite, therefore, invertible. Thus, there exists a unique g = (g1 , . . . , gN )T satisfying (4.8). 2 (4.8)

B + A1 +

Theorem 4.1 immediately yields the existence and the uniqueness of the pair (vn , un ) ∈ H1h × H2h (0 ≤ n ≤ M)

4.2 Error estimates In what follows, we turn to the error estimates of the fully-discrete form (4.1)-(4.2). This is the main result of this section.

Theorem 4.2 Assume that f1 , f2 , u0 , u1 and v0 satisfy (2.5). Let (v, u) be the solution of continuous

weak problem (2.6)-(2.8);and ξhn be the solution of fully-discrete problem (4.1)-(4.2). Assume further that v ∈ L2 (0, T ; Hr+1 (Ω1 )), ∂t v ∈ L2 (0, T ; Hr−1 (Ω1 )),∂t u ∈ L2 (0, T ; Hr+1 (Ω2 )), ∂tt u ∈ L2 (0, T ; Hr−1 (Ω2 )), v0 ∈ Hr+1 (Ω1 ), u1 ∈ Hr+1 (Ω2 ) and u0 ∈ Hr+1 (Ω2 ), Then, the following error estimates hold:

(4.10)

n P kun − unh k1,Ω2 + kvn − vnh k1,Ω1 i=0  . (hr + τ) kv0 kr+1,Ω1 + ku1 kr+1,Ω2 + ku0 kr+1,Ω2 + kvkL2 (0,T ;Hr+1 (Ω1 ))

+kut kL2 (0,T ;Hr+1 (Ω2 )) + k∂t vkL2 (0,T ;Hr−1 (Ω1 )) + k∂tt ukL2 (0,T ;Hr−1 (Ω2 )) (0 ≤ n ≤ M) where τ is the time step,

def

def

def

and ξn = ξ(tn ), vn = v(tn ), un = u(tn ) 10



Proof. Subtract (4.1)-(4.2) from (3.7)-(3.8), and let t = tn , then we have
0 = ∂t ξh (tn ) − ∂¯t ξhn , ηh Ω + a1 (ξh (tn ) − ξhn , ηh ) + a2 (4.11)

Z

tn 0

ξh (s)ds − Ln (ξh ) , ηh






= ∂t ξh (tn ) − ∂¯t ξh (tn ) , ηh Ω + ∂¯t ξh (tn ) − ∂¯t ξhn , ηh Ω + a1 (ξh (tn ) − ξhn , ηh ) Z tn

+a2 ξh (s)ds − Ltn (ξh ) , ηh + a2 Ltn (ξh ) − Ln (ξh ) , ηh , 0

where

def 1 2 τ ξh (t0 ) + 2ξh (t1 ) ξhn and set ηh = en in

Ltn (ξh ) =

Let en = ξh (tn ) −


+ · · · + 2ξh (tn−1 ) + ξh (tn ) . (4.11), we have

1 1 n n (e , e )Ω + a1 (en , en ) + τ a2 (en , en ) τ 2 Z  tn 
= − ∂t ξh (tn ) − ∂¯t ξh (tn ) , en Ω − a2 ξh (s)ds − Ltn (ξh ) , en 0


1 1 + (en−1 , en )Ω − τ a2 e0 + 2e1 + · · · + 2en−1 , en τ 2 Moreover, the following operator error estimates holds: Z tn ξh (s) ds − Ltn (ξh ) . τ2 k∂t ξh (tn ) − ∂¯t ξh (tn )k0,Ω . τ 0

Thus, the following inequity holds: 1 n 2 1 ke k0,Ω + ken k21,Ω1 + τ ken k21,Ω2 τ 2 n−1  1 1 1 nX j 2  1 n 2 1 1 ke k1,Ω2 + ke k0,Ω + τ ken k21,Ω2 + Cτ3 ≤ ken−1 k20,Ω + ken k20,Ω + τ ken k21,Ω2 + τ 4τ 2 2 2 j=0 4τ 8

(4.12)

Simplify (4.12), we have X 1 n 2 1 ke k0,Ω + ken k21,Ω1 + τ ken k21,Ω2 . ken−1 k20,Ω + T ke j k21,Ω2 + τ3 τ τ j=0 n−1

(4.13)

Note that ken−1 k20,Ω ≤ ken−1 k21,Ω1 + ken−1 k21,Ω2 , drop the first term of (4.13) and multiply with τ2 on both sides, then we have n−1 X (4.14) τ2 ken k21,Ω1 + τ3 ken k21,Ω2 . τ5 + (τ + T τ2 ) (ke j k21,Ω1 + ke j k21,Ω2 ). j=0

Applying the following version of Gronwall inequity: an + bn ≤ cn + λ

n−1 X

aj

⇒

an + bn ≤ cn exp(nλ),

j=0

we have Thus, (4.15)

τ2 ken k21,Ω1 + τ3 ken k21,Ω2 . τ5 exp (n(τ + T τ2 )) . τ5 . ken k21,Ω1 + τ ken k21,Ω2 . τ3

Dropping the first term on the left side of (4.15) and recalling the definition of en , then applying the triangle inequality, we obtain the following estimate,



Z tn

n

(4.16) kuh (tn ) − uh k1,Ω2 ≤ + kLtn (ξh ) − Ln (ξh )k1,Ω2 ξh (s)ds − Ltn (ξh )

1,Ω2

0

11



. τ2 + T max

|ξh (tn ) − ξhn |

1,Ω . τ 0≤n≤M

2

Combining with the estimate (3.23), and applying the triangle inequality, we then have, kun − unh k1,Ω2 ≤ kun − uh (tn )k1,Ω2 + kuh (tn ) − unh k1,Ω2 . hr + τ

(4.17)

Dropping the second term on the left side of (4.15) and summing over n, we have that, n X

(4.18)

kvh (ti ) − vih k1,Ω1 . τ2

i=0

Dropping the first, third and fourth term on the left side of the estimate (3.26), we have, n X

(4.19)

kvi − vh (ti )k21,Ω1 . Ln (kvn − vh (tn )k21,Ω1 )

i=0

Z . 0

T

kv(s) − vh (s)k21,Ω1 ds + o (τ2 )

. (h2r + τ2 ) Applying (4.18) and the triangle inequality, we have the following estimate, (4.20)

n X

kvi − vih k21,Ω1 ≤

i=0

n X

kvi − vh (ti )k21,Ω1 +

i=0

n X

kvh (ti ) − vih k1,Ω1 . (h2r + τ2 )

i=0

Combining (4.17) and (4.20), we obtain the estimate (4.10).

2

5.Numerical Experiments In the numerical experiments, we set [0, T ] = [0, 1], Ω1 = [0, 1] × [0, 1] and Ω2 = [1, 2] × [0, 1]. In addition, let v0 = x1 (x1 − 1)x2 (x − 2 − 1) u0 = u1 = (x1 − 1)(x1 − 2)x2 (x2 − 1) f1 = x1 (x1 − 1)x2 (x2 − 1)(1 + 2t) − 2[x1 (x1 − 1) + x2 (x2 − 1)](1 + t + t2 ) f2 = 2(x1 − 1)(x1 − 2)x2 (x2 − 1)(1 + 2t) − 2[(x1 − 1)(x1 − 2) + x2 (x2 − 1)](1 + t + t2 ) While, the exact solution are: v = x1 (x1 − 1)x2 (x2 − 1)(1 + t + t2 ) u = −(x1 − 1)(x1 − 2)x2 (x2 − 1)(1 + t + t2 ) In space direction, piecewise linear interpolation is used ; and in time direction, the backward difference operation is applied to approximate the temporal partial derivative, and the composite trapezoidal operation is used to approximate the integration. The tables listed below are based on the energy and L2 norms. We take t equal to 0.25,0.5,0.75 and 1, respectively. Table 1,2 show the relative error concerning the energy norm and Table 3,4 show the the relative error with regard to the L2 norm Numerical results indicate that the relative error with energy norm k · k1,Ωi are of O(h + τ) accuracy, and the relative error with the L2 norm k · k0,Ωi are of O(h2 + τ) accuracy as we induced in the above analysis. Hence, our discretionary schemes (3.7)-(3.8) and (4.1)-(4.2) are efficient and reliable.

12

t 0.25 0.50 0.75 1.00

Table 1: Error estimate of τ = h = 1/4 τ = h = 1/8 0.2471 0.1225 0.2470 0.1224 0.2468 0.1224 0.2468 0.1223

kvh − vk1,Ω1 /kvk1,Ω1 τ = h = 1/16 τ = h = 1/32 0.0611 0.0306 0.0611 0.0305 0.0611 0.0305 0.0610 0.0305

t 0.25 0.50 0.75 1.00

Table 2: Error estimate of τ = h = 1/4 τ = h = 1/8 0.2524 0.1253 0.2524 0.1253 0.2523 0.1253 0.2523 0.1253

kuh − uk1,Ω2 /kuk1,Ω2 τ = h = 1/16 τ = h = 1/32 0.0625 0.0313 0.0625 0.0313 0.0625 0.0313 0.0625 0.0313

t 0.25 0.50 0.75 1.00

Table 3: Error estimate of kvh − vk0,Ω1 /kvk0,Ω1 τ = 1/4, h = 1/2 τ = 1/16, h = 1/4 τ = 1/64, h = 1/8 0.2892 0.0688 0.0169 0.2866 0.0691 0.0170 0.2853 0.0690 0.0170 0.2845 0.0687 0.0169

t 0.25 0.50 0.75 1.00

Table 4: Error estimate of kuh − uk0,Ω2 /kuk0,Ω2 τ = 1/4, h = 1/2 τ = 1/16, h = 1/4 τ = 1/64, h = 1/8 0.2629 0.0629 0.0155 0.2695 0.0644 0.0158 0.2804 0.0686 0.0169 0.2869 0.0716 0.0179

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Int.