defect correction method for time-dependent viscoelastic fluid flow

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Defect correction method for time-dependent viscoelastic fluid flowYunzhang Zhang a , Yanren Hou a & Baoying Mu aa School of Science , Xian Jiaotong University , Xian, 710049,ChinaPublished online: 11 Mar 2011.

To cite this article: Yunzhang Zhang , Yanren Hou & Baoying Mu (2011) Defect correction methodfor time-dependent viscoelastic fluid flow, International Journal of Computer Mathematics, 88:7,1546-1563, DOI: 10.1080/00207160.2010.521549

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International Journal of Computer MathematicsVol. 88, No. 7, May 2011, 1546–1563

Defect correction method for time-dependent viscoelasticfluid flow

Yunzhang Zhang*, Yanren Hou and Baoying Mu

School of Science, Xian Jiaotong University, Xian 710049, China

(Received 5 December 2009; revised version received 12 June 2010; accepted 27 August 2010)

A defect correction method for solving the time-dependent viscoelastic fluid flow, aiming at highWeissenberg numbers, is presented. In the defect step, the constitutive equation is computed with theartificially reduced Weissenberg parameter for stability, and the residual is considered in the correctionstep. We show the convergence of the method and derive an error estimate. Numerical experiments supportthe theoretical results and demonstrate the effectiveness of the method.

Keywords: viscoelastic fluid flow; finite element; time dependent; defect correction method; discontinu-ous Galerkin; error estimate; Weissenberg number

2000 AMS Subject Classification: 65N30

1. Introduction

Time-dependent calculations of viscoelastic flows are important to the understanding of manyproblems in non-Newtonian fluid mechanics, particularly those related to flow instabilities.

In this paper, we consider a defect correction method for time-dependent viscoelastic fluidflow obeying an Oldroyd-B type constitutive law. Defect-correction methods have been used veryeffectively in the numerical approximation of convection dominated flow equations, e.g. Navier–Stokes equations, convection-diffusion problems. For such problems, in order to avoid spuriousoscillations in the approximation, the defect step acts to regularize the differential equation. Thisregularization generates a lower order (in accuracy) approximation to the equations. The correctionsteps then recapture the accuracy lost (see [2,7,12,13,17] and the references therein).

As the constitutive equation of viscoelastic fluid flows has hyperbolic, nonlinear character, thenumerical simulation is a difficult problem. In addition, when the Weissenberg number increases,boundary layers for the stress develop. This will add to the difficulty of computing accurate numer-ical approximations. Therefore, there are many papers in developing stable numerical algorithmsfor high Weissenberg number flows (see [18,20] and the references therein) over the years.

*Corresponding author. Email: [email protected]

ISSN 0020-7160 print/ISSN 1029-0265 online© 2011 Taylor & FrancisDOI: 10.1080/00207160.2010.521549http://www.informaworld.com

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International Journal of Computer Mathematics 1547

A defect-correction method for steady viscoelastic and Oseen-viscoelastic fluid flow modelwas studied in [8,9,14]. The basic idea of the defect correction method for viscoelastic fluid flowis as follows. In the defect step, the Weissenberg number is reduced artificially by using a mesh-dependent parameter to obtain better convergence of an iteration scheme. Then, in a correctionstep, the initial approximation is improved by using the residual correction. This paper will extendthe defect-correction method in [8,9,14] to the time-dependent viscoelastic fluid flow.

The rest of the paper will be organized as follows. In Section 2, we present the transient vis-coelastic fluid flow model and its variational formulation. In Section 3, we give some mathematicalnotations and two lemmas used in the analysis. Defect correction algorithm is given in Section 4.In Section 5, we propose the error estimate for the defect correction method. Numerical resultsare presented in Section 6. Finally, some conclusions are drawn.

2. Model problem and the variational formulation

Let � ⊂ R2 be an open, simply-connected, bounded polygonal domain with C3 boundary �; let

T > 0 be a given final time. We consider the following Oldroyd’s problem:

λσt + σ + λ(u · ∇)σ + λga(σ, ∇u) − 2αD(u) = 0, in � × (0, T ),

Reut − ∇ · σ − 2(1 − α)∇ · D(u) + ∇p = f, in � × (0, T ),

div u = 0, in � × (0, T ),

u = 0, on � × (0, T ),

u(0, x) = u0(x), σ (0, x) = σ0(x), on � × {0}, (1)

where λ is the Weissenberg number of the viscoelastic fluid, Re is the Reynolds number and α isa number such that 0 < α < 1, which may be considered as the fraction of viscoelastic viscosity.The unknowns are u the fluid velocity vector, p the pressure and σ , which is the viscoelastic partof the total stress tensor σtot = σ + 2(1 − α)D(u) − pI. In (1), D(u) = 1/2(∇u + ∇uT) is therate of the strain tensor, and ga(σ, ∇u) is defined by

ga(σ, ∇u) = 1 − a

2(σ∇u + (∇u)Tσ) − 1 + a

2((∇u)σ + σ(∇u)T), ∀a ∈ [−1, 1]. (2)

The gradient of u is defined such that (∇u)i,j = ∂ui/∂xj . Proofs of the existence and uniquenessof solutions to Equation (1) can be found in [10].

We use the Sobolev spaces Wmp (D) with norms ‖ · ‖m,p,D if p < ∞, ‖ · ‖m,∞,D if p = ∞.

We denote the Sobolev space Wm2 by Hm with the norm ‖ · ‖m. If D = �, D is omitted, i.e.

(·, ·) = (·, ·)� and ‖ · ‖ = ‖ · ‖m.We define the function spaces for the velocity u, the pressure p and the stress σ , respectively:

X := H 10 (�)2 := {v ∈ H 1(�)2 : v = 0 on �},

Q := L20(�) =

{q ∈ L2(�);

∫�

q dx = 0

},

S := {τ = (τij ); τij = τji; τij ∈ L2(�); i, j = 1, 2},

V :={

v ∈ X;∫

�

q(∇ · v) dx = 0, ∀q ∈ Q

}.

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1548 Y. Zhang et al.

The corresponding weak formulation of problem (1) is given by: Find (σ, u, p) ∈ (S, X, Q)

such that for all (τ, v, q) ∈ (S, X, Q)

λ(σt , τ ) + (σ, τ ) + λ((u · ∇)σ, τ ) + λ(ga(σ, ∇u), τ ) − 2α(D(u), τ ) = 0,

Re(ut , v) + (σ, D(v)) + 2(1 − α)(D(u), D(v)) + (p, ∇ · v) = (f, v),

(q, ∇ · u) = 0. (3)

By virtue of the divergence free space V , the weak formulation (3) can be written as: Find(σ, u) ∈ (S, V ) such that for all (τ, v) ∈ (S, V )

λ(σt , τ ) + (σ, τ ) + λ((u · ∇)σ, τ ) + λ(ga(σ, ∇u), τ ) − 2α(D(u), τ ) = 0,

Re(ut , v) + (σ, D(v)) + 2(1 − α)(D(u), D(v)) = (f, v). (4)

3. Mathematical notations

Suppose T h is a uniformly regular triangulation of � such that � = {∪K : K ∈ T h} and assumethat there exist positive constants c1, c2 such that c1h � hK � c2ρK , where hK is the diameterof K , ρK is the diameter of the greatest ball included in K , and h = maxK∈T hhK . Throughoutthe paper, the constants c1, c2, C1, C2, . . . denote different constants which are independent ofmeshsize h and timestep k.

We use the classical Taylor–Hood FE for the approximation in space of (u, p) : P2-continuousin velocity, P1-continuous in pressure; and we consider P1-discontinuous approximation of thestresses. The corresponding FE spaces are

Xh = {v ∈ X ∩ C0(�)2; v|K ∈ P2(K)2, ∀K ∈ T h},Sh = {τ ∈ S; τ|K ∈ P1(K)2×2; ∀K ∈ T h},Qh = {q ∈ Q ∩ C0(�); q|K ∈ P1(K); ∀K ∈ T h},V h = {v ∈ Xh; (q, ∇ · v) = 0, ∀q ∈ Qh},

where Pm(K) denotes the space of polynomials of degree � m on K ∈ T h. It is well known thatthe Taylor–Hood pair (Xh, Qh) satisfies the inf-sup (or LBB) condition.

From Adams and Fournier [1] and Brenner and Scott [5], we have the following results.

Lemma 3.1 Let Ih denote the interpolation of v on Xh. Then for all v ∈ Wmp (�) ∩ Cr(�) and

0 � s � min{m, r + 1},‖v − Ihv‖s,∞ � Chm−s−d/p‖v‖Wm

p. (5)

Lemma 3.2 Let T h, 0 < h < 1, denote a quasi-uniform family of subdivisions of a polyhedraldomain � ⊂ Rd . Let (K, P, N) be a reference finite element such that P ⊂ Wl

p(K) ∩ Wmq (K),

where 1 � p � ∞, 1 � q � ∞, and 0 � m � l. For K ∈ T h, let (K, PK, NK) be the affineequivalent element, and Vh = {v : v is measurable and v|K ∈ PK, ∀K ∈ T h}. Then there existsC = C(l, p, q) such that for all v ∈ Vh⎡

⎣∑K∈Th

‖v‖p

Wlp(K)

⎤⎦

1/p

� Chm−l+min{0,(d/p)−(d/q)}⎡⎣∑

K∈Th

‖v‖q

Wmq (K)

⎤⎦

1/q

. (6)

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For discontinuous stress, we need to use an upwinding technique introduced in Lesaint andRaviart [16]: for this, we define ∂K−(u) = {x ∈ ∂K; u(x) · n(x) < 0}, where ∂K is the boundaryof K ∈ T h and n is the outward unit normal to ∂K , and

�h = {∪∂K : K ∈ T h} \ �, τ±(u)(x) = limε→0±

τ(x + εu(x)).

Also, for all functions in∏

K∈Th[H 1(K)]4, we define

(σ, τ )h =∑K∈Th

(σ, τ )K,

〈σ±, τ±〉h,u =∑K∈Th

∫∂K−(u)

(σ±(u), τ±(u))|n · u| ds,

σ± �2h,u = 〈σ±, σ±〉h,u, |τ‖0,�h =

⎛⎝∑

K∈Th

|τ |20,∂K

⎞⎠

1/2

.

The term ((u · ∇)σ, τ ) is approximated by means of an operator B on Xh × Sh × Sh, defined by

Bh(u, σ, τ ) = ((u · ∇)σ, τ )h + ( 12

)(∇ · uσ, τ) + 〈σ+ − σ−, τ+〉h,u

= −((u · ∇)τ, σ )h − ( 12

)(∇ · uτ, σ ) + 〈σ−, τ− − τ+〉h,u. (7)

Thus, we have

Bh(u, σ, σ ) = (1/2) σ+ − σ− �2h,u . (8)

The Backward Euler time discretization and (P2, P1, P1dc) finite-element approximation ofEquation (3) in space lead to the following system of equations for (uh,n+1, σ h,n+1, ph,n+1) ∈(Xh, Sh, Qh), ∀(v, τ, q) ∈ (Xh, Sh, Qh) at t = tn+1, n ≥ 0, with k := �t = ti+1 − ti

λ

(σh,n+1 − σh,n

k, τ

)+ (σ h,n+1, τ ) + λBh(uh,n, σ h,n+1, τ ) − 2α(D(uh,n+1), τ )

+λ(ga(σh,n, ∇uh,n), τ ) = 0,

Re

(uh,n+1 − uh,n

k, v)

+ (σ h,n+1, D(v)) + 2(1 − α)(D(uh,n+1), D(v))

−(ph,n+1, ∇ · v) = (f(tn+1), v),

(q, ∇ · uh,n+1) = 0. (9)

Existence of a solution to the problem (9) has been documented by Baranger and Wardi [4] underthe assumption that the continuous problem (4) yields a bounded solution

u ∈ C1([0, T ], H 3)2) ∩ C2([0, T ], (L2)2),

σ ∈ C1([0, T ], (H 2)2×2) ∩ C2([0, T ], (L2)2×2),

p ∈ L2([0, T ], H 2 ∩ L20) ∩ C0([0, T ], H 2).

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The error estimates

{k

N∑n=0

[‖uh,n − u(tn)‖20 + ‖D(uh,n − u(tn))‖2

0]}1/2

� C(k + h3/2), and

{k

N∑i=0

{‖σh,n − σ(tn)‖20}1/2

}� C(k + h3/2);

(N = T

k

), (10)

with the constant C independent of h and k, are also proved in Baranger and Wardi [4].

4. Defect correction method

Our defect correction method is described as follows.

Algorithm 1 Defect-correction methodStep 1. Solve the defected problem: Find (uh,n+1

1 , σh,n+11 , p

h,n+11 ) ∈ (Xh, Sh, Qh) such that for

all (τ, v, q) ∈ (Sh, Xh, Qh)

λ

(σ

h,n+11 − σ

h,n1

k, τ

)+ (σ

h,n+11 , τ ) + (λ − Eh2)Bh(uh,n

1 , σh,n+11 , τ )

− 2α(D(uh,n+11 ), τ ) + λ(ga(σ

h,n1 , ∇uh,n

1 ), τ ) = 0,

Re

(uh,n+1

1 − uh,n1

k, v

)+ (σ

h,n+11 , D(v)) + 2(1 − α)(D(uh,n+1

1 ), D(v))

− (ph,n+1i+1 , ∇ · v) = (f(tn+1), v),

(q, ∇ · uh,n+11 ) = 0, (11)

where E is chosen such that λ − Eh2 > 0.Step 2. For i = 1, 2, . . . , solve the correction problem: find (uh,n+1

i+1 , σh,n+1i+1 , p

h,n+1i+1 ) ∈

(Xh, Sh, Qh) such that for all (τ, v, q) ∈ (Sh, Xh, Qh)

λ

(σ

h,n+1i+1 − σ

h,ni

k, τ

)+ (σ

h,n+1i+1 , τ ) + (λ − Eh2)Bh(uh,n+1

i , σh,n+1i+1 , τ ) − 2α(D(uh,n+1

i+1 ), τ )

= −λ(ga(σh,n+1i , ∇uh,n+1

i ), τ ) − Eh2Bh(uh,n+1i , σ

h,n+1i , τ ),

Re

(uh,n+1

i+1 − uh,ni+1

k, v

)+ (σ

h,n+1i+1 , D(v)) + 2(1 − α)(D(uh,n+1

i+1 ), D(v))

− (ph,n+1i+1 , ∇ · v) = (f(tn+1), v),

(q, ∇ · uh,n+1i+1 ) = 0. (12)

The initial value approximations are taken to be uh,01 = uh,0

2 = u0: the elliptic projection of u0

onto V h, in the sense of a(u, v) = (D(u), D(v)), and σh,01 = σ

h,02 = σ0: the orthogonal projection

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of σ0 onto Sh (this is possible since V h and Sh are included in (L2(�))2 and (L2(�))4, resp.).For simplification of notations, we define

A((σ, u), (τ, v)) = 4α(1 − α)(D(u), D(v)) + 2α(σ, D(v)) − 2α(D(u), τ ) + (σ, τ ) (13)

and

Ak((σ, u), (τ, v)) = A((σ, u), (τ, v)) +(

2αRe

k

)(u, v) +

(λ

k

)(σ, τ ). (14)

Then, Step 1 can be written as: ∀(τ, v) ∈ (Sh, V h)

Ak((σh,n+11 , uh,n+1

1 ), (τ, v)) + (λ − Eh2)Bh(uh,n1 , σ

h,n+11 , τ )

= 2α(f(tn+1), v) − λ(ga(σh,n1 , ∇uh,n

1 ), τ ) +(

2αRe

k

)(uh,n

1 , v) +(

λ

k

)(σ

h,n1 , τ ). (15)

We find that

A((σ, u), (σ, u)) = ‖σ‖20 + 4α(1 − α)‖D(u)‖2

0; (16)

thus A, and consequently Ak , are coercive on Sh × V h(0 < α < 1).Therefore, we obtain existence and uniqueness of the solution of Step 1.In the same way, we can get the existence of a unique solution of Step 2.To ensure computability of the algorithm, we first prove stability of Step 1.In Equation (11), let τ = σ

h,n+11 , v = uh,n+1

1 , and we obtain

4α(1 − α)‖D(uh,n+11 )‖2

0 + ‖σh,n+11 ‖2

0 +(

2αRe

2k

)(‖uh,n+1

1 ‖20 − ‖uh,n

1 ‖20)

+ λ

2k(‖σh,n+1

1 ‖20 − ‖σh,n

1 ‖20) + 1

2(λ − Eh2) (σ

h,n+11 )+ − (σ

h,n+11 )− �2

h,uh,n1

� 2α(f(tn+1), uh,n+11 ) − λ(ga(σ

h,n1 , ∇uh,n

1 ), σh,n+11 ). (17)

As f ∈ L2(R+, H 1), u0 ∈ H 2 ∩ V , σ0 ∈ H 2 and the initial value approximations uh,01 = u0,

σh,01 = σ0, by induction hypotheses, we can obtain

k

n+1∑i=1

(‖D(uh,i1 )‖2

0 + ‖σh,i1 ‖2

0) + ‖uh,n+11 ‖2

0 + ‖σh,n+11 ‖2

0 � ‖u0‖20 + ‖σ0‖2

0 + G20 = G2, (18)

where the constant G = G(α, λ, λ − Eh2, T , f, �, u0, σ0) is independent of k and h.

5. Error estimates

The main results of this section are presented in the following theorem.

Theorem 5.1 Let f ∈ L2(R+, H 1), u0 ∈ H 2 ∩ V , σ0 ∈ H 2 , uh1, uh

2, σh1 , σ h

2 satisfy Equa-tions (11) and (12). Then if kh2 � C1 and h � h0, there exists a constant C =

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C(α, �, T , u, p, σ, f, λ − Eh2, λ) independent of h and k, such that

max0�n�N

‖σh,n1 − σ(tn)‖0 + max

0�n�N‖uh,n

1 − u(tn)‖0

+[k

N∑n=0

{‖D(uh,n1 − u(tn))‖2

0 + ‖σh,n1 − σ(tn)‖2

0}]1/2

� C(k + h3/2), (19)

max0�n�N

‖σh,n2 − σ(tn)‖0 + max

0�n�N‖uh,n

2 − u(tn)‖0

+[k

N∑n=0

{‖D(uh,n2 − u(tn))‖2

0 + ‖σh,n2 − σ(tn)‖2

0}]1/2

� C(k + h3/2). (20)

Proof The proof of Theorem 5.1 is established in two steps,

(I): By Principle of Mathematical Induction, we prove the error estimate (19).(II): Prove the error estimate given in Equation (20).(I). We first prove the error estimate (19) by Principle of Mathematical Induction.

For the error estimation, we introduce the bound

M = max{‖u‖C1(0,T ;(H 3)2), ‖σ‖C1(0,T ;(H 2)2×2), ‖p‖C0(0,T ;H 2), ‖u‖C2(0,T ;(L2)2), ‖σ‖C2(0,T ;(L2)2×2)}.

For 0 � n < N , u(tn), σ (tn), p(tn) are, resp. in H 3(�), H 2(�), H 2(�) and so, there exists(u(tn), p(tn)) ∈ V h × Qh [3,5] such that

‖(u − u)(tn)‖0 + h‖(u − u)(tn)‖1,2 � C2h2‖u(tn)‖3,2,

‖(p − p)(tn)‖0 � C3h2‖p(tn)‖2,2,

‖(σ − σ )(tn)‖ � C4h2‖σ(tn)‖2,2. (21)

We can define u(·) by the elliptic projection of u(·) on V h, such that a((u − u)(·), vh) = 0, ∀vh ∈V h, where a(u, v) = (D(u), D(v)); then Equation (21) and the following properties are satisfiedfor ∀u ∈ C1([0, T ], H 3):

dudt

=(

dudt

)∼,

∥∥∥∥(

dudt

)(s) −

(dudt

)∼(s)

∥∥∥∥0

� C5h3

∥∥∥∥(

dudt

)(s)

∥∥∥∥3,2

. (22)

Same properties are satisfied for ∀σ ∈ C1([0, T ], (H 2)4):

dσ

dt=(

dσ

dt

)∼,

∥∥∥∥(

dσ

dt

)(s) −

(dσ

dt

)∼(s)

∥∥∥∥0

� C6h2

∥∥∥∥(

dσ

dt

)(s)

∥∥∥∥2,2

. (23)

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As (σ, u, p) being the solution of Equation (1), ∀(τ, v) ∈ Sh × V h, we have

2αRe(ut (tn+1), v) + λ(σt (tn+1), τ ) + A((σ(tn+1), u(tn+1)), (τ, v)) + λBh(u(tn+1), σ (tn+1), τ )

= 2α(f(tn+1), v) + 2α(p(tn+1), ∇ · v) − λ(ga(σ (tn+1), ∇u(tn+1)), τ ). (24)

We denote eh,ij = uh,i

j − u(ti), εh,ij = σ

h,ij − σ (ti), ∀i, 0 � i � N, j = 1, 2. From Step 1,

∀(τ, v) ∈ Sh × V h, we have

2αRe

((e

h,n+11 − e

h,n1 )

k, v

)+ λ

((ε

h,n+11 − ε

h,n1 )

k, τ

)+ A((ε

h,n+11 , e

h,n+11 ), (τ, v))

+ (λ − Eh2)Bh(uh,n1 , ε

h,n+11 , τ )

= 2α(f (tn+1), v) − λ(ga(σh,n1 , ∇uh,n

1 ), τ ) − 2αRe

((u(tn+1) − u(tn))

k, v)

− λ

((σ (tn+1) − σ (tn))

k, τ

)+ A((−σ (tn+1), −u(tn+1)), (τ, v))

− (λ − Eh2)Bh(uh,n1 , σ (tn+1), τ ). (25)

We combine Equation (24) with Equation (25) to get

2αRe

((e

h,n+11 − e

h,n1 )

k, v

)+ λ

((ε

h,n+11 − ε

h,n1 )

k, τ

)+ A((ε

h,n+11 , e

h,n+11 ), (τ, v))

+ (λ − Eh2)Bh(uh,n1 , ε

h,n+11 , τ )

= 2αRe

(ut (tn+1) − (u(tn+1) − u(tn))

k, v)

+ λ

(σt (tn+1) − (σ (tn+1) − σ (tn))

k, τ

)

+ A(((σ − σ )(tn+1), (u − u)(tn+1)), (τ, v))

+ λ[(ga(σ (tn+1), ∇u(tn+1)), τ ) − (ga(σh,n1 , ∇uh,n

1 ), τ )] − 2α(p(tn+1), ∇ · v)

+ (λ − Eh2)[Bh(u(tn+1), σ (tn+1), τ ) − Bh(uh,n1 , σ (tn+1), τ )]

+ Eh2Bh(u(tn+1), σ (tn+1), τ ). (26)

Taking v = eh,n+11 and τ = ε

h,n+11 into Equation (26) and using Equations (16) and (8), we get

(αRe

k

)[‖eh,n+1

1 ‖20 − ‖eh,n

1 ‖20 + ‖eh,n+1

1 − eh,n1 ‖2

0] + λ

2k[‖εh,n+1

1 ‖20 − ‖εh,n

1 ‖20

+ ‖εh,n+11 − ε

h,n1 ‖2

0] + ‖εh,n+11 ‖2

0 + 4α(1 − α)‖D(eh,n+11 )‖2

0

+ 1

2(λ − Eh2) (ε

h,n+11 )+ − (ε

h,n+11 )− �2

h,uh,n1

� (26)1,n + · · · + (26)7,n with v = eh,n+11 , τ = ε

h,n+11 . (27)

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For controlling each term on the right-hand side of Equation (27), for C0 > 0 and (σ, u) denotethe solution of Equation (1), we define the ball Bm,h,k , for 0 � m � N , by

Bm,h,k =⎧⎨⎩(τ, v)i=0,...,m ∈ (S × X)m+1, max

0�i�m‖τi − σ(ti)‖0

+[k

m∑i=0

{‖τi − σ(ti)‖20 + ‖D(vi − u(ti))‖2

0}]1/2

� C0(k + h3/2)

⎫⎬⎭ .

By Equation (21), we have

‖σh,01 − σ(0)‖0 + k1/2{‖σh,0

1 − σ(0)‖0 + ‖D(uh,01 − u(0))‖0}

= ‖(σ − σ)(0)‖0 + k1/2{‖(σ − σ)(0)‖0 + ‖D(u − u)(0)‖0}� C7M(k + h3/2)h1/2(1 + k1/2).

It can be controlled by C0(k + h3/2), if M and h are small enough. Thus (σh,01 , uh,0

1 ) = (σ0, u0) ∈B0,h,k.

Now, our target is to prove that we can choose M0, C1, h0 such that for M � M0, h � h0, if(σ

h,n1 , uh,n

1 )0�n�m−1 ∈ Bm−1,h,k for a C0 = C0(M0, C1, h0), then (σh,n1 , uh,n

1 )0�n�m ∈ Bm,h,k forthe same C0; thus, for all m such that mk � T . So we can suppose that (σ

h,n1 , uh,n

1 )0�n�m−1 ∈Bm−1,h,k .

Multiplying Equation (27) by k and summing from n = 0 to m − 1, remarking that eh,01 = 0

and εh,01 = 0, then we have

αRe

[‖eh,m

1 ‖20 +

m−1∑n=0

‖eh,n+11 − e

h,n1 ‖2

0

]+(

λ

2

)[‖εh,m

1 ‖20 +

m−1∑n=0

‖εh,n+11 − ε

h,n1 ‖2

0

]

+ k

m−1∑n=0

‖εh,n+11 ‖2

0 + 4α(1 − α)k

m−1∑n=0

‖D(eh,n+11 )‖2

0

+ 1

2(λ − Eh2)k

m−1∑n=0

(εh,n+11 )+ − (ε

h,n+11 )− �2

h,uh,n1

� 2αRekm−1∑n=0

(ut (tn+1) − (u(tn+1) − u(tn))

k, eh,n+11

)

+ λk

m−1∑n=0

(σt (tn+1) − (σ (tn+1) − σ (tn))

k, εh,n+11

)

+ k

m−1∑n=0

A(((σ − σ )(tn+1), (u − u)(tn+1)), (εh,n+11 , e

h,n+11 ))

+ (λ − Eh2)k

m−1∑n=0

[Bh(u(tn+1), σ (tn+1), εh,n+11 ) − Bh(uh,n

1 , σ (tn+1), εh,n+11 )]

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+ λk

m−1∑n=0

[(ga(σ (tn+1), ∇u(tn+1)), εh,n+11 ) − (ga(σ

h,n1 , ∇uh,n

1 ), εh,n+11 )]

− 2αk

m−1∑n=0

(p(tn+1), ∇ · eh,n+11 )

+ Eh2k

m−1∑n=0

Bh(u(tn+1), σ (tn+1), εh,n+11 )

= (28)1 + · · · + (28)7. (28)

We now estimate each term on the right-hand side of Equation (28).We have that(for details see [4])

‖(28)1‖0 � 2αReM(k + C8h3)

[m−1∑n=0

k

]1/2 [m−1∑n=0

k‖eh,n+11 ‖2

0

]1/2

, (29)

‖(28)2‖0 � λM(k + C9h2)

[m−1∑n=0

k

]1/2 [m−1∑n=0

k‖eh,n+11 ‖2

0

]1/2

, (30)

‖(28)3‖0 � C10h2M

[m−1∑n=0

k

]1/2 [m−1∑n=0

k(‖εh,n+11 ‖2

0 + ‖D(eh,n+11 )‖2

0)

]1/2

, (31)

‖(28)4‖0 � C11(λ − Eh2)(h3/2 + k)M

[m−1∑n=0

k

]1/2

×[

m−1∑n=0

k{‖εh,n+11 ‖2

0+ [εh,n+11 ] �2

h,uh,n1

}]1/2

, (32)

‖(28)5‖0 � C12(λkM2 + λk2M2)

[m−1∑n=0

k

]1/2 [m−1∑n=0

k‖εh,n+11 ‖2

0

]1/2

+ C13λMC0(h3/2 + k)

[m−1∑n=0

k‖εh,n+11 ‖2

0

]1/2

+ C14λC20 (h3/2 + k)h1/2‖εh,m

1 ‖0 + C14λC30(h

3/2 + k)2h1/2, (33)

‖(28)6‖0 � 2αC15h2M

[m−1∑n=0

k

]1/2 [m−1∑n=0

k(‖D(eh,n+11 )‖2

0

]1/2

, (34)

‖(28)7‖0 � C16Eh2M2

[m−1∑n=0

k

]1/2 [m−1∑n=0

k‖εh,n+11 ‖2

0

]1/2

. (35)

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Combining Equations (28)–(35), we obtain the following estimate

αRe

[‖eh,m

1 ‖20 +

m−1∑n=0

‖eh,n+11 − e

h,n1 ‖2

0

]+ (λ/2)

[‖εh,m

1 ‖20 +

m−1∑n=0

‖εh,n+11 − ε

h,n1 ‖2

0

]

+ k

m−1∑n=0

‖εh,n+11 ‖2

0 + 4α(1 − α)k

m−1∑n=0

‖D(eh,n+11 )‖2

0

+ 1

2(λ − Eh2)k

m−1∑n=0

[εh,n+11 ] �2

h,uh,n1

� C17[2αReMk + 2αReMh2 + λMk + λMh2 + Mh2

+ (λ − Eh2)M(k + h3/2) + λM2(k + k2) + λMC0(k + h3/2)

+ Eh2M2][

m−1∑n=0

k

]1/2

×[

m−1∑n=0

k{‖D(eh,n+11 )‖2

0 + ‖εh,n+11 ‖2

0

+ [εh,n+11 ] �2

h,uh,n1

}]1/2

+ C14λC20 (h3/2 + k)h1/2‖εh,m

1 ‖0 + C14λC30(h

3/2 + k)2h1/2. (36)

Thus, we have

‖eh,m1 ‖2

0 + ‖εh,m1 ‖2

0 + k

m−1∑n=0

{‖εh,n+11 ‖2

0 + ‖D(eh,n+11 )‖2

0+ [εh,n+11 ] �2

h,uh,n1

}

� C18[2αReMk + 2αReMh2 + λMk + λMh2 + Mh2

+ (λ − Eh2)M(k + h3/2) + λM2(k + k2) + λMC0(k + h3/2)

+ Eh2M2][

m−1∑n=0

k{‖D(eh,n+11 )‖2

0 + ‖εh,n+11 ‖2

0+ [εh,n+11 ] �2

h,uh,n1

}]1/2

+ C19λC20 (h3/2 + k)h1/2‖εh,m

1 ‖0 + C19λC30(h

3/2 + k)2h1/2, (37)

where C18 = C17T1/2/C, C19 = C14/C, with C = min{1, (1/2)(λ − Eh3/2), 4α(1 − α), αRe}.

By triangle inequality, Cauchy–Schwarz inequality and recurrence assumption, we get

max0�n�m

‖εh,m1 ‖2

0 + max0�n�m

‖eh,m1 ‖2

0 + k

m−1∑n=0

{‖D(uh,n+11 − u(tn+1))‖2

0

+ ‖σh,n+11 − σ(tn+1)‖2

0+ [εh,n+11 ] �2

h,uh,n1

}

�(

8

3

)(k + h3/2)2{C2

18[2αReM + 2αReMh1/2 + λM + λMh1/2 + Mh1/2

+ (λ − Eh2)M + λM2 + λMC0 + λMC0h + λM2k + Eh1/2M2

+ λC0h1/2]2 + 2C2

19λ2C4

0h + 2C19λC30h

1/2 + 2M2(C22 + C2

4 )h4T }. (38)

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Now, for a given C0, we can choose M , h small enough such that

(8/3){C218[2αReM + 2αReMh1/2 + λM + λMh1/2 + Mh1/2

+ (λ − Eh2)M + λM2 + λMC0 + λMC0h + λM2k + Eh1/2M2

+ λC0h1/2]2 + 2C2

19λ2C4

0h + 2C19λC30h

1/2 + 2M2(C22 + C2

4 )h4T }� C2

0 . (39)

In this case, we can choose M0, h0, small enough to ensure that, when M � M0 and h � h0 aresatisfied, inequality (39) holds. Thus

max0�n�m

‖εh,n1 ‖0 + max

0�n�m‖eh,n

1 ‖0 +[k

m−1∑n=0

{‖D(uh,n+11 − u(tn+1))‖2

0

+‖σh,n+11 − σ(tn+1)‖2

0+ [εh,n+11 ] �2

h,uh,n1

}]1/2

� C0(k + h3/2). (40)

Thus, if M � M0 and h � h0 are satisfied, then (σh,n1 , uh,n

1 )0�n�m ∈ Bm,h,k and consequently∀m : 0 � m � N ,(σ h,n

1 , uh,n1 )0�n�m ∈ Bm,h,k . So, for all 0 � m � N , the inequality (40) is right.

We complete the proof of Equation (19).(II). We will show that the inequality (20) is true.Now, we combine Step 2 with Equation (24) and introduce the error in the correction step

approximation eh,i2 , ε

h,i2 , 0 � i � N . This gives

2αRe

((e

h,n+12 − e

h,n2 )

k, v

)+ λ

((ε

h,n+12 − ε

h,n2 )

k, τ

)

+ A((εh,n+12 , e

h,n+12 ), (τ, v)) + (λ − Eh2)Bh(uh,n+1

1 , εh,n+12 , τ )

= 2αRe

(ut (tn+1) − u(tn+1) − u(tn)

k, v

)+ λ

(σt (tn+1) − (σ (tn+1) − σ (tn))

k, τ

)

+ A(((σ − σ )(tn+1), (u − u)(tn+1)), (τ, v))

+ λ[(ga(σ (tn+1), ∇u(tn+1)), τ ) − (ga(σh,n2 , ∇uh,n

2 ), τ )] − 2α(p(tn+1), ∇ · v)

− Eh2Bh(uh,n1 , σ

h,n1 , τ ) − (λ − Eh2)Bh(uh,n+1

1 , σ (tn+1), τ )

+ λBh(u(tn+1), σ (tn+1), τ ). (41)

Comparing Equation (41) with Equation (26), we note that they are different only on the followingterms.

− Eh2Bh(uh,n+11 , σ

h,n+11 , τ ) − (λ − Eh2)Bh(uh,n+1

1 , σ (tn+1), τ )

+ (λ − Eh2)Bh(uh,n1 , σ (tn+1), τ )

= Eh2Bh(u − uh,n1 , σ (tn+1), τ ) − Eh2Bh(u, σ (tn+1), τ )

+ λBh(uh,n1 − uh,n+1

1 , σ (tn+1), τ ) + Eh2Bh(uh,n+11 , σ (tn+1) − σ

h,n+11 , τ ). (42)

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So we will only deal with these terms.

Eh2Bh(u − uh,n1 , σ (tn+1), τ )

� C20Eh2[‖u − uh,n1 ‖0,4‖∇σ‖0,4‖τ‖0 + ‖∇(u − uh,n

1 )‖0‖σ‖∞‖τ‖0]� C21Eh2[‖∇(u − uh,n

1 )‖0‖∇σ‖2‖τ‖0] � C21Eh2M[‖∇(u − uh,n1 )‖0‖τ‖0]. (43)

Similarly, we have

−Eh2Bh(u, σ (tn+1), τ ) � C22Eh2M2‖τ‖0, (44)

λBh(uh,n1 − uh,n+1

1 , σ (tn+1), τ ) = λBh(uh,n1 − u + u − uh,n+1

1 , σ (tn+1), τ )

� C23λM(‖∇(uh,n1 − u)‖0 + ‖∇(u − uh,n+1

1 )‖0)‖τ‖0. (45)

To bound the last Bh term of Equation (42), we use the second equation of (7).

Bh(uh,n+11 , σ (tn+1) − σ

h,n+11 , τ ) = −((uh,n+1

1 · ∇)(σ (tn+1) − σh,n+11 ), τ )h

− (1/2)((∇ · uh,n+11 )τ, σ (tn+1) − σ

h,n+11 )

+ 〈(σ (tn+1) − σh,n+11 )−, τ− − τ+〉h,u

h,n+11

. (46)

−((uh,n+11 · ∇)(σ (tn+1) − σ

h,n+11 ), τ )h � C24‖uh,n+1

1 ‖0,4‖∇(σ (tn+1) − σh,n+11 )‖0,4,h‖τ‖0

� C25h−5/2‖uh,n+1

1 ‖0‖σ (tn+1) − σh,n+11 ‖0‖τ‖0

� C25h−5/2G‖σ (tn+1) − σ

h,n+11 ‖0‖τ‖0. (47)

−1

2((∇ · uh,n+1

1 )τ, σ (tn+1) − σh,n+11 ) � C26h

−2G‖σ (tn+1) − σh,n+11 ‖0‖τ‖0. (48)

〈(σ (tn+1) − σh,n+11 )−, τ− − τ+〉h,u

h,n+11

� (σ (tn+1) − σh,n+11 )− �h,u

h,n+11

τ− − τ+ �h,uh,n+11

� C27h−1/2‖uh,n+1

1 ‖1/20 h−1/2‖σ (tn+1) − σ

h,n+11 ‖0 τ− − τ+ �h,u

h,n+11

� C28Gh−1‖σ (tn+1) − σh,n+11 ‖0 τ− − τ+ �h,u

h,n+11

. (49)

Therefore, combining Equations (42)–(49), we have

k

m−1∑n=0

[−Eh2Bh(uh,n+11 , σ

h,n+11 , τ ) − (λ − Eh2)Bh(uh,n+1

1 , σ (tn+1), τ )

+ (λ − Eh2)Bh(uh,n1 , σ (tn+1), τ )]

� k

m−1∑n=0

[C21Eh2M‖∇(u − uh,n1 )‖0‖τ‖0 + C22Eh2M2‖τ‖0

+ C23λM(‖∇(uh,n1 − u)‖0 + ‖∇(u − uh,n+1

1 )‖0)‖τ‖0

+ Eh2C25h−5/2G‖σ (tn+1) − σ

h,n+11 ‖0‖τ‖0

+ Eh2C26h−2G‖σ (tn+1) − σ

h,n+11 ‖0‖τ‖0

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+ Eh2C28Gh−1‖σ (tn+1) − σh,n+11 ‖0 τ− − τ+ �h,u

h,n+11

]

� C21Eh2M

(m−1∑n=0

k‖∇(u − uh,n1 )

)‖2

0)1/2

(m−1∑n=0

k‖τ‖20

)1/2

+ C22Eh2M2

(m−1∑n=0

k

)1/2 (m−1∑n=0

k‖τ‖20

)1/2

+ C23λM(‖∇(uh,n1 − u)‖0 + ‖∇(u − uh,n+1

1 )‖0)‖τ‖0

+ C23λM

(m−1∑n=0

k‖∇(u − uh,n+11 )‖2

0

)1/2 (m−1∑n=0

k‖τ‖20

)1/2

+ Eh−1/2C25G

(m−1∑n=0

k‖σ (tn+1) − σh,n+11 ‖2

0

)1/2 (m−1∑n=0

k‖τ‖20

)1/2

+ EC26G

(m−1∑n=0

k‖σ (tn+1) − σh,n+11 ‖2

0

)1/2 (m−1∑n=0

k‖τ‖20

)1/2

+ EhC28G

(m−1∑n=0

k‖σ (tn+1) − σh,n+11 ‖2

0

)1/2 (m−1∑n=0

k τ− − τ+ �2h,u

h,n+11

)1/2

. (50)

To conclude, repeat the proof of the first statement of Theorem 5.1, replacing uh,n1 ,σh,n

1 , eh,n1

and εh,n1 by uh,n

2 ,σh,n2 , e

h,n2 and ε

h,n2 , respectively, using Korn inequality, Equation (50) and

the bound [k∑m−1n=0 {‖D(uh,n+1

1 − u(tn+1))‖20 + ‖σh,n+1

1 − σ(tn+1)‖20]1/2 � C0(k + h3/2) from

Equation (40). Hence, we get

max0�n�m

‖εh,n2 ‖0 + max

0�n�m‖eh,n

2 ‖0 +[k

m−1∑n=0

{‖D(uh,n+12 − u(tn+1))‖2

0

+‖σh,n+12 − σ(tn+1)‖2

0+ [εh,n+12 ] �2

h,uh,n+11

}]1/2

� C0(k + h3/2). (51)

�

6. Numerical results

In this section, we present two series numerical examples to investigate the accuracy and effec-tiveness of the defect correction method. The method has been implemented using [11] thefinite-element software package FreeFem++ in 2-d. Linear systems are solved using the UMF-PACK solver. We use the stopping criterion defined by ‖σh

i − σhi−1‖ � 10−8 for the iterative

nonlinear solver in both the defect step and the correction step of the defect-correction algorithm.We also set the maximum number of iteration equal to 15. The first example is an analyticalsolution example to test the convergence results of the defect-correction method. The secondexample simulates viscoelastic flow through a 4:1 planar contraction, a prototypical problem forviscoelastic fluid flow.

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6.1 Analytical solution

Let � = [0, 1] × [0, 1]. Same as in [6,9,15], the right-hand side functions in Equation (1) wereappropriately given so that the true solution was

u =(

10x2(x − 1)2y(y − 1)(2y − 1) cos(t)−10x(x − 1)(2x − 1)y2(y − 1)2 cos(t)

),

p = 10(2x − 1)(2y − 1)cos(t),

σ = 2αD(u).

We denote ‖|v‖|∞,m = max0�i�T/k ‖vi‖m, ‖|v‖|0,m = [∑T/k

i=0 k‖vi‖2m](1/2) and λ1 = λ −

Eh2. We will compute ‖|ε1‖|∞,0 = ‖|σ − σh,n+11 ‖|∞,0, ‖|e2‖|0,1 = ‖|u − uh,n+1

2 ‖|0,1, etc.,respectively. Also, we let r be the experimental global rate of convergence givenby r = log(Er/Er ′)/ log(h/h′), where h and h′ denote two consecutive mesh sizes withcorresponding global errors Er and Er ′.

From λ = 5 to λ = 1000, the calculated convergence rates in Tables 1–4 confirm what ispredicted by Theorem 5.1 for (P 2, P 1, P 1dc) discretization in space. However, for the vis-coelastic fluid flow with sufficiently high Weissenberg numbers, the computed rates agree withthose predicted by Theorem 5.1.

6.2 4-to-1 planar contraction flow

The second example is a 4-to-1 contraction channel flow problem. This has been a long-standing benchmark problem for viscoelastic flow [6,19]. It is assumed that the channel

Table 1. Defect step approximation, Re = 1, a = 0, α = 0.5, k = 0.005, T = 0.045, λ = 5.0, λ1 = 4.0.

1/h ‖ε1‖∞,0 r ‖e1‖∞,0 r ‖ε1‖0,0 r ‖e1‖0,1 r

4 0.0669507 – 0.00283248 – 0.0141469 – 0.0117198 –8 0.0192394 1.799 0.000881781 1.684 0.00406679 1.799 0.00303882 1.947

16 0.004976 1.951 0.000233245 1.919 0.00105363 1.949 0.000756365 2.00632 0.00130813 1.927 0.0000588779 1.986 0.000270682 1.961 0.000188769 2.002

Table 2. Correction step approximation, Re = 1, a = 0, α = 0.5, k = 0.01, T = 0.1, λ = 8.0, λ1 = 4.0.

1/h ‖ε2‖∞,0 r ‖e2‖∞,0 r ‖ε2‖0,0 r ‖e2‖0,1 r

4 0.0669361 – 0.00359691 – 0.0210484 – 0.018313 –8 0.0192374 1.799 0.00116739 1.623 0.00605504 1.798 0.00480083 1.932

16 0.00497739 1.950 0.000315294 1.889 0.00157139 1.946 0.00120342 1.99632 0.00128328 1.956 0.0000845356 1.899 0.00040163 1.968 0.000303224 1.989

Table 3. Defect step approximation, Re = 1, a = 0, α = 0.5, k = 0.005, T = 0.045, λ = 1000, λ1 = 500.

1/h ‖ε1‖∞,0 r ‖e1‖∞,0 r ‖ε1‖0,0 r ‖e1‖0,1 r

4 0.0670171 – 0.00286338 – 0.014216 – 0.0117456 –8 0.0192853 1.797 0.000889592 1.687 0.00408753 1.798 0.0030438 1.948

16 0.00507442 1.926 0.000235374 1.918 0.0010645 1.941 0.000757442 2.00732 0.00157094 1.692 0.0000603321 1.964 0.000294561 1.854 0.000190473 1.992

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Table 4. Correction step approximation, Re = 1, a = 0, α = 0.5, k = 0.005, T = 0.045, λ = 1000, λ1 = 500.

1/h ‖ε2‖∞,0 r ‖e2‖∞,0 r ‖ε2‖0,0 r ‖e2‖0,1 r

4 0.0670142 – 0.00286365 – 0.0142157 – 0.011746 –8 0.0192643 1.799 0.000889713 1.686 0.00408579 1.799 0.0030441 1.948

16 0.00498912 1.949 0.000235428 1.918 0.00105738 1.950 0.000757344 2.00732 0.00126513 1.980 0.0000592782 1.990 0.000267401 1.983 0.000188787 2.004

lengths are sufficiently long for fully developed Poiseuille flow at both the inflow and out-flow boundaries. The computations were performed on a uniformly refined version of the meshshown in Figure 1 with �xmin = 0.0625 and �ymin = 0.015625. We denote �in = {(x, y) : x =0, 0 � y � 1} and �out = {(x, y) : x = 8, 0 � y � 0.25}. On this domain the velocity bound-ary conditions are u1 = 1/32(1 − y2), u2 = 0, on �in and u1 = 2((1/16) − y2), u2 = 0, on�out. On �in, specified boundary conditions for σ are given as follows: σ11 = (−αλ(a + 1)

(−y/16)2/((a2 − 1)λ2(−y/16)2 − 1), σ12 = σ21 = (−α(−y/16))/((a2 − 1)λ2(−y/16)2 − 1),σ22 = (−αλ(a − 1)(−y/16)2)/((a2 − 1)λ2(−y/16)2 − 1). Symmetry conditions are imposedon the bottom of the computational domain. The parameters Re, α, λ, λ1, and a are set to 1,8/9, 0.7, 0.5 and 1, respectively.

We perform the following study: starting from rest, we measure the time that approximationsolution reaches a steady state. The criterion to stop this process is the following:

max

{‖un+1

h − unh‖L2(�)

‖un+1h ‖L2(�)

,‖σn+1

h − σnh ‖L2(�)

‖σn+1h ‖L2(�)

}� 10−5,

where n + 1, n denote tn+1, tn, respectively.In Figure 2 we plot the evolution of the kinetic energy ‖un+1

h ‖20/2 and ‖σn+1

h ‖20/2 using k = 0.05

until it reaches its steady state, where we observe the convergence towards a steady state and alsothe absence of oscillations along the process.

1

0.5

00 2 4 6 8

Figure 1. Plot of 4:1 contraction domain geometry and sample contraction mesh.

0 1 2 3 4 5 6 7 85.05

5.1

5.15

5.2

5.25

5.3

5.35

5.4

5.45

5 .5 × 10−3

t

n+1

0.5*

||mh

|| 02

n+1

0.5*

||sh

|| 02

0 1 2 3 4 5 6 7 80

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

t

Figure 2. Evolution of ‖un+1h ‖2

0/2 (left) and ‖σn+1h ‖2

0/2 (right) in time with k = 0.05.

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0 0.05 0.1 0.15 0.2 0.250

0.02

0.04

0.06

0.08

0.1

0.12

0.14

y

steady problemtime dependent problem

steady problemtime dependent problem

0 0.05 0.1 0.15 0.2 0.25−5

−4

−3

−2

−1

0

1× 10−3

y

Figure 3. Horizontal (left) and vertical (right) velocity near reentrant corner with λ = 0.7, λ1 = 0.5. The marks ‘−’indicate results for steady problem and ‘+’ indicate results for time-dependent problem at t = 7.8 using k = 0.05.

Figure 4. Streamlines and magnitude of velocity contours for u with λ = 0.7, λ1 = 0.5. (Left): steady problem, (right):time-dependent problem at t = 7.8 using k = 0.05.

Figure 3 presents the horizontal and vertical velocity near re-entrant corner along the verticalline x = 4.0625 for λ = 0.7, λ1 = 0.5. From Figure 3, we see that the horizontal velocity is almostcontinuous, while the vertical velocity has high gradients near y = 0.23. It is easy to discover thatthe time-dependent problem converges towards the steady problem. Figure 4 shows streamlinesof the fluid with λ = 0.7, λ1 = 0.5. (Left): steady problem, (right): time-dependent problem att = 7.8 using k = 0.05. We see that they are just the same.

7. Conclusions

In this paper, we have extended the defect correction method to the time-dependent viscoelasticfluid flow.We establish a priori error estimates for the method, and provide numerical computationsto support the theoretical results and demonstrate the effectiveness of the method.

Further developments should extend the method to other non-Newtonian flow. Also, moreappropriate choice of parameter λ1 is currently under investigation.

Acknowledgements

This work was supported by the National Natural Science Foundation of China(10871156) and Jiaoda Founda-tion(2009xjtujc30). The authors would like to thank Prof J.S. Howell for meaty discussion of program code.

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The authors also thank the anonymous reviewer and editor for their valuable comments and suggestions that helpedto improve the results of the paper.

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defect correction method for time-dependent viscoelastic fluid flow

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