Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 372906, 21 pages
http://dx.doi.org/10.1155/2013/372906
Research Article
Fully Discrete Finite Element Approximation for the Stabilized
Gauge-Uzawa Method to Solve the Boussinesq Equations
Jae-Hong Pyo
Department of Mathematics, Kangwon National University, Chuncheon 200-701, Republic of Korea
Correspondence should be addressed to Jae-Hong Pyo; jhpyo@kangwon.ac.kr
Received 9 February 2013; Accepted 11 March 2013
Academic Editor: Jinyun Yuan
Copyright © 2013 Jae-Hong Pyo. This is an open access article distributed under the Creative Commons Attribution License, which
permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The stabilized Gauge-Uzawa method (SGUM), which is a 2nd-order projection type algorithm used to solve Navier-Stokes
equations, has been newly constructed in the work of Pyo, 2013. In this paper, we apply the SGUM to the evolution Boussinesq
equations, which model the thermal driven motion of incompressible fluids. We prove that SGUM is unconditionally stable, and
we perform error estimations on the fully discrete finite element space via variational approach for the velocity, pressure, and
temperature, the three physical unknowns. We conclude with numerical tests to check accuracy and physically relevant numerical
simulations, the Bénard convection problem and the thermal driven cavity flow.
1. Introduction
The stabilized Gauge-Uzawa method (SGUM) is a 2nd-order
projection type method to solve the evolution Navier-Stokes
equations. In this paper, we extend SGUM to the evolution
Boussinesq equations: given a bounded polygon Ω in R𝑑 with
𝑑 = 2 or 3,
u𝑡 + (u ⋅ ∇) u + ∇𝑝 − 𝜇Δu + 𝜅𝜇2 g𝜃 = f,
∇ ⋅ u = 0,
in Ω,
𝜃𝑡 + u ⋅ ∇𝜃 − 𝜆𝜇Δ𝜃 = 𝑏,
in Ω,
(1)
in Ω,
with initial conditions u(x, 0) = u0 , 𝜃(x, 0) = 𝜃0 in Ω,
vanishing Dirichlet boundary conditions u = 0, 𝜃 = 0
on 𝜕Ω, and pressure mean-value ∫Ω 𝑝 = 0. The forcing
functions f and 𝑏 are given, and g is the vector of gravitational
acceleration. The nondimensional numbers 𝜇 = Re−1 and
𝜆 = Pr−1 are reciprocal of the Reynolds and Prandtl
numbers, respectively, whereas 𝜅 is the Grashof number.
The Boussinesq system (1) describes fluid motion due to
density differences which are in turn induced by temperature
gradients: hot and thus less dense fluid tends to rise against
gravity and cooler fluid falls in its place. The simplest
governing equations are thus the Navier-Stokes equations for
motion of an incompressible fluid, with forcing 𝜅𝜇2 g𝜃 due to
buoyancy and the heat equation for diffusion and transport
of heat. Density differences are thus ignored altogether except
for buoyancy.
The projection type methods are representative solvers for
the incompressible flows, and the Gauge-Uzawa method is
a typical projection method. The Gauge-Uzawa method was
constructed in [1] to solve Navier-Stokes equations and
extended to more complicated problems, the Boussinesq
equations in [2] and the nonconstant density Navier-Stokes
equations in [3]. However, most of studies for the GaugeUzawa method have been limited only for the first-order
accuracy backward Euler time marching algorithm. The
second-order Gauge-Uzawa method using BDF2 scheme was
introduced in [4] and proved superiority for accuracy on the
normal mode space, but we could not get any theoretical
proof via energy estimate even stability and we suffer from
weak stability performance on the numerical test. Recently,
we construct SGUM in [5] which is unconditionally stable
for semidiscrete level to solve the Navier-Stokes equations.
The goal of this paper is to extend SGUM to the Boussinesq
equations (1), which model the motion of an incompressible
viscous fluid due to thermal effects [6, 7]. We will estimate
errors and stability on the fully discrete finite element
space. The main difficulties in the fully discrete estimation
arise from losing the cancellation law due to the failing of
2
Journal of Applied Mathematics
the divergence free condition of the discrete velocity function.
The strategy of projection type methods computes first an
artificial velocity and then decomposes it to divergence free
velocity and curl free functions. However, the divergence
free condition cannot be preserved in discrete finite element
space, and so the cancellation law (12) can not be satisfied any
more. In order to solve this difficulty, we impose the discontinuous velocity on across interelement boundaries to make
fulfill discrete divergence free velocity (12) automatically. We
will discuss this issue at Remark 2 below. This discontinuity
makes it difficult to treat nonlinear term and to apply the
integration by parts, because the discontinuous solution is not
included in 𝐻1 (Ω). So we need to hire technical skills in proof
of this paper.
One more remarkable discovery is in the second numerical test at the last section which is the Bénard convection
problem with the same setting in [2]. In this performance,
we newly find out that the number of circulations depends
on the time step size 𝜏. We obtain similar simulation within
[2] for a relatively larger 𝜏, but the behavior is changed for
the small 𝜏. So we conclude that the numerical result of the
Bénard convection problem in [2] is not an eventual solution
and a more smaller time marching step is required to get the
desired simulation.
We will denote 𝜏 as the time marching size. Also we will
use 𝛿 as the difference of 2 consecutive functions for example,
for any sequence function 𝑧𝑛+1 ,
𝑛+1
𝛿𝑧
𝑛+1
=𝑧
𝑛
−𝑧 ,
𝛿𝛿𝑧𝑛+1 = 𝛿 (𝛿𝑧𝑛+1 ) = 𝑧𝑛+1 − 2𝑧𝑛 + 𝑧𝑛−1 , . . . .
(2)
(3)
∀wℎ ∈ Vℎ , ∀𝜙ℎ ∈ Pℎ .
1
⟨(uℎ ⋅ ∇) kℎ , wℎ ⟩
2
−
(5)
1
⟨(uℎ ⋅ ∇) wℎ , kℎ ⟩ ,
2
we now introduce the fully discrete SGUM to solve the
evolution Boussinesq equations (1).
Algorithm 1 (the fully discrete stabilized Gauge-Uzawa
method). Compute 𝜃ℎ1 , u1ℎ , and 𝑝ℎ1 via any method satisfying
Assumption 4 below, and set 𝜓ℎ1 = (−2𝜏/3)𝑝ℎ1 and 𝑞ℎ1 = 0.
Repeat for 1 ≤ 𝑛 ≤ 𝑁 = [𝑇/𝜏 − 1].
∗
𝑛
𝑛−1
Step 1. Set u∗ℎ = 2u𝑛ℎ − u𝑛−1
ℎ and 𝜃ℎ = 2𝜃ℎ − 𝜃ℎ , and then find
𝑛+1
̂ ℎ ∈ Vℎ as the solution of
u
1
𝑛
𝑛−1
⟨3̂
u𝑛+1
ℎ − 4uℎ + uℎ , wℎ ⟩
2𝜏
̂ 𝑛+1
+ ⟨∇𝑝ℎ𝑛 , wℎ ⟩ + N (u∗ℎ , u
ℎ , wℎ )
+
𝜇 ⟨∇̂
u𝑛+1
ℎ , ∇wℎ ⟩
= ⟨f (𝑡𝑛+1 ) , wℎ ⟩ ,
2
+ 𝜅𝜇
(4)
(6)
⟨g𝜃ℎ∗ , wℎ ⟩
∀wℎ ∈ Vℎ .
Step 2. Find 𝜓ℎ𝑛+1 ∈ Pℎ as the solution of
∀𝜙ℎ ∈ Pℎ .
(7)
𝑛+1
Step 3. Update u𝑛+1
∈ Pℎ according to
ℎ and 𝑞ℎ
𝑛+1
̂ 𝑛+1
=u
− 𝜓ℎ𝑛 ) ,
u𝑛+1
ℎ
ℎ + ∇ (𝜓ℎ
̂ 𝑛+1
⟨𝑞ℎ𝑛+1 , 𝜙ℎ ⟩ = ⟨𝑞ℎ𝑛 , 𝜙ℎ ⟩ − ⟨∇ ⋅ u
ℎ , 𝜙ℎ ⟩ ,
(8)
∀𝜙ℎ ∈ Pℎ . (9)
Step 4. Update pressure 𝑝ℎ𝑛+1 by
3𝜓ℎ𝑛+1
+ 𝜇𝑞ℎ𝑛+1 .
2𝜏
(10)
Step 5. Find 𝜃ℎ𝑛+1 ∈ Tℎ as the solution of
1
⟨3𝜃ℎ𝑛+1 − 4𝜃ℎ𝑛 + 𝜃ℎ𝑛−1 , 𝜙ℎ ⟩
2𝜏
𝑛+1
+ N (u𝑛+1
ℎ , 𝜃ℎ , 𝜙ℎ )
where P(𝐾), Q(𝐾), and R(𝐾) are spaces of polynomials with
degree bounded uniformly with respect to 𝐾 ∈ T [9, 10]. We
stress that the space Pℎ is composed of continuous functions
to ensure the crucial equality
⟨∇ ⋅ wℎ , 𝜙ℎ ⟩ = − ⟨wℎ , ∇𝜙ℎ ⟩ ,
:=
𝑝ℎ𝑛+1 = −
󵄨
Wℎ := {wℎ ∈ L (Ω) : wℎ 󵄨󵄨󵄨𝐾 ∈ P (𝐾) ∀𝐾 ∈ T} ,
2
󵄨
Tℎ := {𝜃ℎ ∈ 𝐻01 (Ω) ∩ 𝐶0 (Ω) : 𝜃ℎ 󵄨󵄨󵄨𝐾 ∈ Q (𝐾) ∀𝐾 ∈ T} ,
󵄨
Pℎ := {𝑞ℎ ∈ 𝐿20 (Ω) ∩ 𝐶0 (Ω) : 𝑞ℎ 󵄨󵄨󵄨𝐾 ∈ R (𝐾) ∀𝐾 ∈ T} ,
N (uℎ , kℎ , wℎ )
̂ 𝑛+1
⟨∇𝜓ℎ𝑛+1 , ∇𝜙ℎ ⟩ = ⟨∇𝜓ℎ𝑛 , ∇𝜙ℎ ⟩ + ⟨∇ ⋅ u
ℎ , 𝜙ℎ ⟩ ,
In order to introduce the finite element discretization, we
need further notations. Let 𝐻𝑠 (Ω) be the Sobolev space with
𝑑
𝑠 derivatives in 𝐿2 (Ω), set L2 (Ω) = (𝐿2 (Ω)) and H𝑠 (Ω) =
𝑑
(𝐻𝑠 (Ω)) , where 𝑑 = 2 or 3, and denote by 𝐿20 (Ω) the
subspace of 𝐿2 (Ω) of functions with vanishing mean value.
We indicate with ‖ ⋅ ‖𝑠 the norm in 𝐻𝑠 (Ω) and with ⟨⋅ , ⋅⟩
the inner product in 𝐿2 (Ω). Let T = {𝐾} be a shaperegular quasiuniform partition of Ω of mesh size ℎ into closed
elements 𝐾 [8–10]. The vector and scalar finite element spaces
are
Vℎ := Wℎ ∩ H10 (Ω) ,
Using the following discrete counterpart of the form
N(u, k, w) := ⟨(u ⋅ ∇)k, w⟩
(11)
+ 𝜆𝜇 ⟨∇𝜃ℎ𝑛+1 , ∇𝜙ℎ ⟩
= ⟨𝑏 (𝑡𝑛+1 ) , 𝜙ℎ ⟩ ,
∀𝜙ℎ ∈ Tℎ .
𝑛+1
is a
Remark 2 (discontinuity of u𝑛+1
ℎ ). We note that uℎ
discontinuous function across inter-element boundaries and
Journal of Applied Mathematics
3
that, in light of (7) and (8), u𝑛+1
ℎ is discrete divergence free in
the sense that
𝑛+1
u𝑛+1
⟨u𝑛+1
ℎ , ∇𝜙ℎ ⟩ = ⟨̂
ℎ + 𝛿∇𝜓ℎ , ∇𝜙ℎ ⟩ = 0,
∀𝜙ℎ ∈ Pℎ .
(12)
We now summarize the results of this paper along with
organization. We introduce appropriate Assumptions 1–5 in
Section 2 and introduce well-known lemmas. In Section 3, we
prove the stability result.
Theorem 3 (stability). The SGUM is unconditionally stable in
the sense that, for all 𝜏 > 0, the following a priori bound holds:
󵄩󵄩 𝑁+1 󵄩󵄩2 󵄩󵄩 𝑁+1 󵄩󵄩2 󵄩󵄩 𝑁+1
𝑁󵄩
󵄩󵄩2
󵄩 󵄩
󵄩󵄩uℎ 󵄩󵄩 + 󵄩󵄩u
󵄩0 󵄩 ̂ ℎ 󵄩󵄩0 + 󵄩󵄩2uℎ − uℎ 󵄩󵄩0
󵄩
󵄩2 󵄩
󵄩2
󵄩
+ 󵄩󵄩󵄩󵄩2𝜃ℎ𝑁+1 − 𝜃ℎ𝑁󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩𝜃ℎ𝑁+1 󵄩󵄩󵄩󵄩0
(13)
𝑁
󵄩
󵄩 𝑛+1 󵄩󵄩2
󵄩
uℎ 󵄩󵄩󵄩0 + 𝜆 󵄩󵄩󵄩󵄩∇𝜃ℎ𝑛+1 󵄩󵄩󵄩󵄩)
+ 𝜇𝜏 ∑ (󵄩󵄩󵄩󵄩∇̂
𝑛=1
𝑛=1
We then will carry out the following optimal error estimates through several lemmas in Section 4.
Theorem 4 (error estimates). Suppose the exact solution of (1)
is smooth enough and 𝜏 ≤ 𝐶ℎ. If Assumptions 1, 3, 4, and 5
mentioned later hold, then the errors of Algorithm 1 will be
bound by
(16)
𝑁
󵄩󵄩2
󵄩
󵄩2
󵄩
󵄩 + ℎ2 󵄩󵄩󵄩𝜃(𝑡𝑛+1 ) − 𝜃ℎ𝑛+1 󵄩󵄩󵄩 )
̂ 𝑛+1
𝜏 ∑ (󵄩󵄩󵄩󵄩u(𝑡𝑛+1 ) − u
ℎ 󵄩
󵄩1
󵄩
󵄩1
𝑛=1
(14)
≤ 𝐶 (𝜏2 + ℎ2 ) ,
(17)
󵄩 󵄩
‖𝜔‖2 ≤ 𝐶󵄩󵄩󵄩𝜉󵄩󵄩󵄩0 .
(18)
We impose the following properties for relations between
the spaces Vℎ and Pℎ .
Assumption 2 (discrete inf-sup). There exists a constant 𝛽 > 0
such that
⟨∇ ⋅ w , 𝑠 ⟩
inf sup 󵄩󵄩 󵄩󵄩 󵄩󵄩ℎ 󵄩󵄩ℎ ≥ 𝛽.
𝑠ℎ ∈Pℎ w ∈V 󵄩wℎ 󵄩 󵄩𝑠ℎ 󵄩
ℎ
ℎ 󵄩
󵄩1 󵄩 󵄩0
(19)
In order to launch Algorithm 1, we need to set (u1ℎ , 𝑝ℎ1 , 𝜃ℎ1 )
via any first-order methods which hold the following conditions.
Assumption 4 (the setting of the first step values). Let
(u(𝑡1 ), 𝑝(𝑡1 ), 𝜃(𝑡1 )) be the exact solution of (1) at 𝑡 = 𝑡1 . The
first step value (u1ℎ , 𝑝ℎ1 , 𝜃ℎ1 ) satisfies
𝑁
󵄩
󵄩2
𝜏 ∑ 󵄩󵄩󵄩󵄩𝜃(𝑡𝑛+1 ) − 𝜃ℎ𝑛+1 󵄩󵄩󵄩󵄩0 ≤ 𝐶 (𝜏4 + ℎ4 ) .
𝑛=1
Moreover, if Assumption 2 also hold, then one has
𝑁
󵄩
󵄩2
𝜏 ∑ 󵄩󵄩󵄩󵄩𝑝(𝑡𝑛+1 ) − 𝑝ℎ𝑛+1 󵄩󵄩󵄩󵄩0 ≤ 𝐶 (𝜏2 + ℎ2 ) .
in Ω,
Assumption 3 (shape regularity and quasiuniformity [8–10]).
There exists a constant 𝐶 > 0 such that the ratio between the
diameter ℎ𝐾 of an element 𝐾 ∈ T and the diameter of the
largest ball contained in 𝐾 is bounded uniformly by 𝐶, and
ℎ𝐾 is comparable with the meshsize ℎ for all 𝐾 ∈ T.
󵄩󵄩2
󵄩)
̂ 𝑛+1
u
ℎ 󵄩
󵄩0
≤ 𝐶 (𝜏4 + ℎ4 ) ,
𝑛=1
in Ω,
We remark that the validity of Assumption 1 is known if 𝜕Ω
is of class C2 [11, 12], or if 𝜕Ω is a two-dimensional convex
polygon [13] and is generally believed for convex polyhedral
[12].
𝑁
󵄩
󵄩2
󵄩
󵄩2
+ 𝐶𝜏 ∑ (󵄩󵄩󵄩󵄩f(𝑡𝑛+1 )󵄩󵄩󵄩󵄩−1 + 󵄩󵄩󵄩󵄩𝑏(𝑡𝑛+1 )󵄩󵄩󵄩󵄩−1 ) .
𝑛=1
∇ ⋅ k = 0,
Assumption 1 (regularity of (k, 𝑟) and 𝜔). The unique solutions {v, 𝑟} of (16) and 𝜔 of (17) satisfy
‖k‖2 + ‖𝑟‖1 ≤ 𝐶‖z‖0 ,
󵄩 󵄩2 󵄩
󵄩2 󵄩
󵄩2
≤ 󵄩󵄩󵄩󵄩u1ℎ 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩2u1ℎ − u0ℎ 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩2𝜃ℎ1 − 𝜃ℎ0 󵄩󵄩󵄩󵄩0
󵄩2
󵄩 󵄩2
󵄩
󵄩 󵄩2
+ 󵄩󵄩󵄩󵄩𝜃ℎ1 󵄩󵄩󵄩󵄩0 + 3󵄩󵄩󵄩󵄩∇𝜓ℎ1 󵄩󵄩󵄩󵄩0 + 2𝜇𝜏󵄩󵄩󵄩󵄩𝑞ℎ1 󵄩󵄩󵄩󵄩0
󵄩
+ 󵄩󵄩󵄩󵄩u(𝑡𝑛+1 ) −
−Δk + ∇𝑟 = z,
with boundary condition 𝜔 = 0.
We now state a basic assumption about Ω.
󵄩2
󵄩2
󵄩
󵄩
+ 3󵄩󵄩󵄩󵄩∇𝜓ℎ𝑁+1 󵄩󵄩󵄩󵄩0 + 2𝜇𝜏󵄩󵄩󵄩󵄩𝑞ℎ𝑁+1 󵄩󵄩󵄩󵄩0
󵄩
𝜏 ∑ (󵄩󵄩󵄩󵄩u(𝑡𝑛+1 ) −
In this section, we introduce 5 assumptions and well known
lemmas to use in proof of main theorems. We resort to a duality argument for
−Δ𝜔 = 𝜉,
𝑛=1
󵄩󵄩2
󵄩
u𝑛+1
ℎ 󵄩
󵄩0
2. Preliminaries
which is the stationary Stokes system with vanishing boundary condition k = 0, as well as Poisson’s equation
𝑁
󵄩󵄩2 󵄩󵄩
󵄩2 󵄩
󵄩2
󵄩
󵄩 + 󵄩󵄩𝛿𝛿𝜃ℎ𝑛+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩 )
+ ∑ (󵄩󵄩󵄩󵄩𝛿𝛿u𝑛+1
ℎ 󵄩
󵄩0 󵄩
󵄩0 󵄩
󵄩0
𝑁
We note that the condition 𝜏 ≤ 𝐶ℎ in Theorem 4
can be omitted for the linearized Boussinesq equations (see
Remark 16). Finally, we perform numerical tests in Section 5
to check accuracy and physically relevant numerical simulations, the Bénard convection problem and the thermal driven
cavity flow.
(15)
󵄩 󵄩
󵄩
󵄩󵄩
󵄩󵄩u (𝑡1 ) − u1ℎ 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝜃 (𝑡1 ) − 𝜃ℎ1 󵄩󵄩󵄩 ≤ 𝐶 (𝜏2 + ℎ2 ) ,
󵄩0 󵄩
󵄩0
󵄩
󵄩 󵄩󵄩 1
󵄩
󵄩󵄩 1
󵄩󵄩 1
1󵄩
1󵄩
󵄩
󵄩󵄩u(𝑡 ) − uℎ 󵄩󵄩 + 󵄩󵄩𝑝(𝑡 ) − 𝑝ℎ 󵄩󵄩 + 󵄩󵄩𝜃(𝑡 ) − 𝜃ℎ1 󵄩󵄩󵄩 ≤ 𝐶 (𝜏 + ℎ) .
󵄩1 󵄩
󵄩1 󵄩
󵄩1
󵄩
(20)
4
Journal of Applied Mathematics
Assumption 5 (approximability [8–10]). For each (w, 𝜂, 𝑠) ∈
H2 (Ω) × 𝐻2 (Ω) × 𝐻1 (Ω), there exist approximations
(wℎ , 𝜂ℎ , 𝑠ℎ ) ∈ Vℎ × Tℎ × Pℎ such that
󵄩󵄩
󵄩
󵄩
󵄩
2
󵄩󵄩w − wℎ 󵄩󵄩󵄩0 + ℎ󵄩󵄩󵄩w − wℎ 󵄩󵄩󵄩1 ≤ 𝐶ℎ ‖w‖2 ,
󵄩
󵄩
󵄩󵄩
󵄩
2󵄩 󵄩
󵄩󵄩𝜂 − 𝜂ℎ 󵄩󵄩󵄩0 + ℎ󵄩󵄩󵄩𝜂 − 𝜂ℎ 󵄩󵄩󵄩1 ≤ 𝐶ℎ 󵄩󵄩󵄩𝜂󵄩󵄩󵄩2 ,
󵄩󵄩
󵄩
󵄩󵄩𝑠 − 𝑠ℎ 󵄩󵄩󵄩0 ≤ 𝐶ℎ‖𝑠‖1 .
Lemma 9 (error estimates for Poisson’s equation [8, 10]). Let
𝜔 ∈ 𝐻01 (Ω) be the solution of Poisson’s equation (17), and, 𝜔ℎ ∈
Tℎ be its finite element approximation
⟨∇𝜔ℎ , ∇𝜙ℎ ⟩ = ⟨𝜉, 𝜙ℎ ⟩ ,
(21)
󵄩󵄩
󵄩
󵄩
󵄩
2
󵄩󵄩𝜔 − 𝜔ℎ 󵄩󵄩󵄩0 + ℎ󵄩󵄩󵄩𝜔 − 𝜔ℎ 󵄩󵄩󵄩1 ≤ 𝐶ℎ ‖𝜔‖2
󵄩 󵄩
≤ 𝐶ℎ2 󵄩󵄩󵄩𝜉󵄩󵄩󵄩0 ,
Lemma 5 (inverse inequality). If 𝐼ℎ denotes the Clément
interpolant, then
(22)
We finally state without proof several properties of the
nonlinear form N. In view of (5), we have the following
properties of N for all uℎ , kℎ , wℎ ∈ Vℎ :
N (uℎ , kℎ , wℎ ) = −N (uℎ , wℎ , kℎ ) ,
(23)
N (uℎ , kℎ , kℎ ) = 0,
⟨∇ ⋅ kℎ , 𝑠ℎ ⟩ = 0,
∀wℎ ∈ Vℎ ,
∀𝑠ℎ ∈ Pℎ .
= ⟨(u ⋅ ∇) kℎ , wℎ ⟩
(31)
= − ⟨(u ⋅ ∇) wℎ , kℎ ⟩ .
(24)
Applying Sobolev imbedding lemma yields the following useful results.
Then we can find the well-known lemmas in [8–10].
Lemma 7 (error estimates for mixed FEM). Let (v, 𝑟) ∈
H10 (Ω) × 𝐿20 (Ω) be the solutions of (16), and, (vℎ , 𝑟ℎ ) =
Sℎ (v, 𝑟) ∈ Vℎ × Pℎ be the Stokes projection defined by (24),
respectively. If Assumptions 1, 2, 3, and 5 hold, then
󵄩󵄩
󵄩
󵄩
󵄩
󵄩
󵄩
2
󵄩󵄩k − kℎ 󵄩󵄩󵄩0 + ℎ󵄩󵄩󵄩k − kℎ 󵄩󵄩󵄩1 + ℎ󵄩󵄩󵄩𝑟 − 𝑟ℎ 󵄩󵄩󵄩0 ≤ 𝐶ℎ (‖k‖2 + ‖𝑟‖1 ) ,
(25)
󵄩󵄩󵄨󵄨
󵄨󵄩 󵄩
󵄩
󵄩
󵄩
󵄩󵄩󵄨󵄨k − kℎ 󵄨󵄨󵄨󵄩󵄩󵄩 := 󵄩󵄩󵄩k − kℎ 󵄩󵄩󵄩L∞ (Ω) + 󵄩󵄩󵄩∇(k − kℎ )󵄩󵄩󵄩L3 (Ω) ≤ 𝐶‖z‖0 .
(26)
Proof. Inequality (25) is standard [8–10]. To establish (26), we
just deal with the L∞ -norm, since the other can be treated
similarly. If 𝐼ℎ denotes the Clément interpolant, then
‖k − 𝐼ℎ k‖L∞ (Ω) ≤ 𝐶‖k‖2 and
󵄩
󵄩󵄩
−𝑑/2 󵄩
󵄩󵄩𝐼ℎ k − kℎ 󵄩󵄩󵄩 2 ≤ 𝐶‖k‖2
󵄩󵄩𝐼ℎ k − kℎ 󵄩󵄩󵄩L∞ (Ω) ≤ 𝐶ℎ
󵄩
󵄩L (Ω)
(30)
∇ ⋅ u = 0 󳨐⇒ N (u, kℎ , wℎ )
Let now (kℎ , 𝑟ℎ ) ∈ Vℎ × Pℎ indicate the finite element
solution of (16); namely,
⟨∇kℎ , ∇wℎ ⟩ + ⟨∇𝑟ℎ , wℎ ⟩ = ⟨z, wℎ ⟩ ,
(29)
󵄩󵄩󵄩󵄨󵄨󵄨𝜔 − 𝜔ℎ 󵄨󵄨󵄨󵄩󵄩󵄩 ≤ 𝐶󵄩󵄩󵄩𝜉󵄩󵄩󵄩 .
󵄩 󵄩0
󵄩󵄨
󵄨󵄩
Lemma 6 (div-grad relation). If w ∈ H10 (Ω), then
‖∇ ⋅ w‖0 ≤ ‖∇w‖0 .
(28)
If Assumptions 1, 3, and 5 hold, then there exists a positive constant 𝐶 satisfying
The following elementary but crucial relations are derived
in [14].
󵄩󵄩 󵄩󵄩
−𝑑/6 󵄩
󵄩󵄩𝐼ℎ w󵄩󵄩󵄩 ,
󵄩󵄩𝐼ℎ w󵄩󵄩L3 (Ω) ≤ 𝐶ℎ
󵄩 󵄩0
󵄩
󵄩󵄩
2−𝑑/6
‖w‖2 .
󵄩󵄩w − 𝐼ℎ w󵄩󵄩󵄩L3 (Ω) ≤ 𝐶ℎ
∀𝜙ℎ ∈ Tℎ .
(27)
as a consequence of an inverse estimate. This completes the
proof.
Remark 8 (𝐻1 stability of 𝑟ℎ ). The bound ‖∇𝑟ℎ ‖0 ≤ 𝐶(‖k‖2 +
‖𝑟‖1 ) is a simple by-product of (16). To see this, we add and
subtract 𝐼ℎ 𝑞, use the stability of 𝐼ℎ in 𝐻1 , and observe that (25)
implies ‖ ∇(𝑟ℎ − 𝐼ℎ 𝑟)‖0 ≤ 𝐶ℎ−1 ‖ 𝑟ℎ − 𝐼ℎ 𝑟‖0 ≤ 𝐶.
Lemma 10 (bounds on nonlinear convection [1, 12]). Let
u, k ∈ H2 (Ω) with ∇ ⋅ u = 0, and let uh , kh , wh ∈ Vℎ . Then
󵄩 󵄩󵄩 󵄩
‖u‖1 󵄩󵄩󵄩kℎ 󵄩󵄩󵄩1 󵄩󵄩󵄩wℎ 󵄩󵄩󵄩1
{
{
󵄩 󵄩󵄩 󵄩
N (u, kℎ , wℎ ) ≤ 𝐶 { ‖u‖2 󵄩󵄩󵄩kℎ 󵄩󵄩󵄩1 󵄩󵄩󵄩wℎ 󵄩󵄩󵄩0
{
󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩
{ ‖u‖2 󵄩󵄩kℎ 󵄩󵄩0 󵄩󵄩wℎ 󵄩󵄩1 ,
󵄩 󵄩
󵄩 󵄩
N (uℎ , k, wℎ ) ≤ 𝐶󵄩󵄩󵄩uℎ 󵄩󵄩󵄩0 ‖k‖2 󵄩󵄩󵄩wℎ 󵄩󵄩󵄩1 .
(32)
(33)
In addition,
󵄩󵄩 󵄩󵄩 󵄩󵄩󵄨󵄨 󵄨󵄨󵄩󵄩 󵄩󵄩 󵄩󵄩
󵄩uℎ 󵄩 󵄩󵄨kℎ 󵄨󵄩 󵄩wℎ 󵄩
N (uℎ , kℎ , wℎ ) ≤ 𝐶 { 󵄩󵄩 󵄩 󵄩󵄩 󵄩0 󵄩󵄨󵄩󵄩 󵄨󵄩󵄩󵄩 󵄩󵄩󵄩 󵄩󵄩󵄩1
󵄩󵄩uℎ 󵄩󵄩L3 (Ω) 󵄩󵄩kℎ 󵄩󵄩1 󵄩󵄩wℎ 󵄩󵄩1 .
(34)
Remark 11 (Treatment of convection term). As we mention in
Remark 2, u𝑛+1
ℎ is discontinuous across inter-element bound∉ H1 (Ω). Thus, we can’t directly apply (32)
aries and so u𝑛+1
ℎ
anymore to treat the convection term. To solve this difficulty,
we apply (34) together with (26) and inverse inequality
Lemma 5.
We will use the following algebraic identities frequently
to treat time derivative terms.
Journal of Applied Mathematics
5
Lemma 12 (inner product of time derivative terms). For any
𝑁
sequence {𝑧𝑛 }𝑛=0 , one has
Before we estimate 𝐴 2 , we evaluate an inequality via choosing
𝜙ℎ = 𝛿𝑞ℎ𝑛+1 in (9) to get
2 ⟨3𝑧𝑛+1 − 4𝑧𝑛 + 𝑧𝑛−1 , 𝑧𝑛+1 ⟩
󵄩2
󵄩2 󵄩
󵄩2
󵄩
󵄩
= 𝛿󵄩󵄩󵄩󵄩𝑧𝑛+1 󵄩󵄩󵄩󵄩0 + 𝛿󵄩󵄩󵄩󵄩2𝑧𝑛+1 − 𝑧𝑛 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩𝛿𝛿𝑧𝑛+1 󵄩󵄩󵄩󵄩0 ,
(35)
󵄩󵄩 𝑛+1 󵄩󵄩2
𝑛+1
󵄩󵄩𝛿𝑞ℎ 󵄩󵄩 = − ⟨∇ ⋅ u
̂ 𝑛+1
ℎ , 𝛿𝑞ℎ ⟩
󵄩
󵄩0
󵄩󵄩 󵄩󵄩 𝑛+1 󵄩󵄩
󵄩
󵄩 󵄩󵄩𝛿𝑞ℎ 󵄩󵄩 .
̂ 𝑛+1
≤ 󵄩󵄩󵄩󵄩∇ ⋅ u
ℎ 󵄩
󵄩0 󵄩
󵄩0
(36)
̂ 𝑛+1
Lemma 6 derives ‖𝛿𝑞ℎ𝑛+1 ‖0 ≤ ‖∇ ⋅ u
u𝑛+1
ℎ ‖0 ≤ ‖∇̂
ℎ ‖0 , and so
(9) and (37) lead us to
2 ⟨𝑧𝑛+1 − 𝑧𝑛 , 𝑧𝑛+1 ⟩
𝑛+1 󵄩
󵄩2
󵄩
= 󵄩󵄩󵄩󵄩𝑧
𝑛󵄩
󵄩2
󵄩
󵄩󵄩 − 󵄩󵄩󵄩󵄩𝑧𝑛 󵄩󵄩󵄩󵄩20 + 󵄩󵄩󵄩𝑧𝑛+1 − 𝑧 󵄩󵄩 ,
󵄩0
󵄩0
󵄩
2 ⟨𝑧𝑛+1 − 𝑧𝑛 , 𝑧𝑛 ⟩
󵄩2 󵄩 󵄩2 󵄩
󵄩2
󵄩
= 󵄩󵄩󵄩󵄩𝑧𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩𝑧𝑛 󵄩󵄩󵄩0 − 󵄩󵄩󵄩󵄩𝑧𝑛+1 − 𝑧𝑛 󵄩󵄩󵄩󵄩0 .
(37)
We show that the SGUM is unconditionally stable via a standard energy method in this section. We start to prove stability
with rewriting the momentum equation (6) by using (8) and
(10) as follows:
1
𝑛
𝑛−1
⟨3u𝑛+1
ℎ − 4uℎ + uℎ , wℎ ⟩
2𝜏
+
− ⟨∇ (
2
2
𝐴 2 = −4𝜇𝜏 ⟨𝑞ℎ𝑛 , 𝛿𝑞ℎ𝑛+1 ⟩
󵄩 2 󵄩 󵄩2 󵄩
󵄩2
󵄩
= −2𝜇𝜏 (󵄩󵄩󵄩󵄩𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩𝑞ℎ𝑛 󵄩󵄩󵄩0 − 󵄩󵄩󵄩󵄩𝛿𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
(43)
󵄩
󵄩 2 󵄩 󵄩2
󵄩 𝑛+1 󵄩󵄩2
uℎ 󵄩󵄩󵄩0 .
≤ −2𝜇𝜏 (󵄩󵄩󵄩󵄩𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩𝑞ℎ𝑛 󵄩󵄩󵄩0 ) + 2𝜇𝜏󵄩󵄩󵄩󵄩∇̂
3. Proof of Stability
̂ 𝑛+1
N (u∗ℎ , u
ℎ , wℎ )
2
(42)
Clearly, we have
𝜏󵄩
󵄩2
󵄩 𝑛+1 󵄩󵄩2
uℎ 󵄩󵄩󵄩0 .
𝐴 3 ≤ 𝐶 󵄩󵄩󵄩󵄩f(𝑡𝑛+1 )󵄩󵄩󵄩󵄩−1 + 𝜏𝜇󵄩󵄩󵄩󵄩∇̂
𝜇
2
2
+ 𝜅𝜇
⟨g𝜃ℎ∗ , wℎ ⟩
3 𝑛+1
𝜓 − 𝜇𝑞ℎ𝑛 ) , wℎ ⟩
2𝜏 ℎ
(38)
󵄩2
󵄩
𝐴 4 ≤ 𝐶𝜅2 𝜇4 𝜏󵄩󵄩󵄩󵄩2𝜃ℎ𝑛 − 𝜃ℎ𝑛−1 󵄩󵄩󵄩󵄩0
󵄩󵄩2 󵄩󵄩
󵄩2
󵄩
󵄩 + 󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩 .
+ 𝐶𝜏󵄩󵄩󵄩󵄩u𝑛+1
ℎ 󵄩
󵄩0 󵄩
󵄩0
We now choose wℎ = 4𝜏̂
u𝑛+1
ℎ and apply (35) to obtain
󵄩2
󵄩2
󵄩
󵄩
𝛿󵄩󵄩󵄩󵄩𝜃ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝛿󵄩󵄩󵄩󵄩2𝜃ℎ𝑛+1 − 𝜃ℎ𝑛 󵄩󵄩󵄩󵄩0
󵄩2
󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿𝜃ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 4𝜏𝜆𝜇 󵄩󵄩󵄩󵄩∇𝜃ℎ𝑛+1 󵄩󵄩󵄩󵄩
(39)
4
𝑖=1
= 4𝜏 ⟨𝑏 (𝑡𝑛+1 ) , 𝜃ℎ𝑛 ⟩
where
̂ 𝑛+1
𝐴 1 := 6 ⟨∇𝜓ℎ𝑛+1 , u
ℎ ⟩,
̂ 𝑛+1
𝐴 3 := 4𝜏 ⟨f (𝑡𝑛+1 ) , u
ℎ ⟩,
≤𝐶
(40)
̂ 𝑛+1
𝐴 4 := −4𝜏𝜅𝜇2 ⟨g (2𝜃ℎ𝑛 − 𝜃ℎ𝑛−1 ) , u
ℎ ⟩.
(46)
𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩󵄩𝑏(𝑡 )󵄩󵄩 + 𝜆𝜇𝜏󵄩󵄩󵄩󵄩∇𝜃ℎ𝑛 󵄩󵄩󵄩󵄩20 .
󵄩−1
𝜆𝜇 󵄩
Inserting 𝐴 1 –𝐴 4 back into (39), adding with (46), and then
summing over 𝑛 from 1 to 𝑁 lead to (13) by help of discrete
2
𝑛+1 2
Gronwall inequality and the equality ‖̂
u𝑛+1
ℎ ‖0 = ‖uℎ ‖0 +
2
‖∇𝛿𝜓ℎ𝑛+1 ‖0 . So we finish the proof of Theorem 3.
We give thanks to (30) for eliminating the convection term.
𝑛+1
̂ 𝑛+1
In light of u
= u𝑛+1
ℎ
ℎ − ∇𝛿𝜓ℎ , (12) and (36) yield
𝐴 1 = −6 ⟨∇𝜓ℎ𝑛+1 , ∇𝛿𝜓ℎ𝑛+1 ⟩
󵄩2 󵄩
󵄩2
󵄩
󵄩2 󵄩
= −3 (󵄩󵄩󵄩󵄩∇𝜓ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩∇𝜓ℎ𝑛 󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩󵄩0 ) .
(45)
In conjunction with N(u∗ℎ , 𝜃ℎ𝑛+1 , 𝜃ℎ𝑛+1 ) = 0 which comes from
(30), if we choose 𝜙ℎ = 4𝜏𝜃ℎ𝑛+1 in (11) and use (35), then we
obtain
= ∑𝐴 𝑖 ,
̂ 𝑛+1
𝐴 2 := 4𝜏𝜇 ⟨𝑞ℎ𝑛 , ∇ ⋅ u
ℎ ⟩,
2
𝑛+1
𝑛+1
We now attack 𝐴 4 with ‖̂
u𝑛+1
ℎ ‖0 = ‖uℎ ‖0 + ‖∇𝛿𝜓ℎ ‖0 which
comes from (8) and (12). Then we arrive at
𝑛+1
) , wℎ ⟩ .
+ 𝜇 ⟨∇̂
u𝑛+1
ℎ , ∇wℎ ⟩ = ⟨f (𝑡
󵄩󵄩2
󵄩
󵄩
𝑛󵄩
󵄩󵄩2
󵄩 + 𝛿󵄩󵄩󵄩2u𝑛+1
𝛿󵄩󵄩󵄩󵄩u𝑛+1
ℎ 󵄩
󵄩0
󵄩 ℎ − uℎ 󵄩󵄩0
󵄩󵄩2
󵄩
󵄩 𝑛+1 󵄩󵄩2
󵄩 + 4𝜏𝜇󵄩󵄩󵄩∇̂
󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿u𝑛+1
ℎ 󵄩
󵄩0
󵄩 uℎ 󵄩󵄩0
2
(44)
(41)
4. Error Estimates
We prove Theorem 4 which is error estimates for SGUM of
Algorithm 1. This proof is carried out through several
lemmas. We start to prove this theorem with defining
𝑛+1
𝑛+1
), 𝑝(𝑡𝑛+1 )) ∈ Vℎ × Pℎ to be
(U𝑛+1
ℎ , 𝑃ℎ ) := Sℎ (u(𝑡
6
Journal of Applied Mathematics
the Stokes projection of the true solution at time 𝑡𝑛+1 . It means
𝑛+1
that (U𝑛+1
ℎ , 𝑃ℎ ) ∈ Vℎ × Pℎ is the solution of
⟨∇ (u (𝑡𝑛+1 ) − U𝑛+1
ℎ ) , ∇wℎ ⟩
𝜀ℎ𝑛+1 := 𝑃ℎ𝑛+1 +
+ ⟨∇ (𝑝 (𝑡𝑛+1 ) − 𝑃ℎ𝑛+1 ) , wℎ ⟩ = 0,
∀wℎ ∈ Vℎ ,
⟨∇ ⋅ U𝑛+1
ℎ , 𝜙ℎ ⟩ = 0,
We also define
Θ𝑛+1
ℎ
(47)
𝜙ℎ ∈ Pℎ .
And we denote notations G
𝑛+1
𝑛+1
:= u(𝑡
∀𝜙ℎ ∈ Tℎ .
)−
𝜂𝑛+1 := 𝜃 (𝑡𝑛+1 ) − Θ𝑛+1
ℎ .
(48)
𝑡𝑛+1
≤ 𝐶𝜏ℎ ∫
𝑡𝑛
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
̂ 𝑛+1 = E
̂ 𝑛+1 = G𝑛+1 = 0,
E
ℎ
(56)
on 𝜕Ω.
(49)
𝑛+1
⟨E𝑛+1
, ∇𝜙ℎ ⟩
ℎ , ∇𝜙ℎ ⟩ = ⟨E
= ⟨G𝑛+1 , ∇𝜙ℎ ⟩
(50)
= 0,
(57)
∀𝜙ℎ ∈ Pℎ .
Whence we deduce crucial orthogonality properties:
󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩
󵄩2
󵄩
󵄩2
󵄩󵄩𝛿G 󵄩󵄩 + ℎ2 󵄩󵄩󵄩𝛿G𝑛+1 󵄩󵄩󵄩 + ℎ2 󵄩󵄩󵄩𝛿𝑔𝑛+1 󵄩󵄩󵄩
󵄩
󵄩0
󵄩
󵄩1
󵄩
󵄩0
4
We readily obtain the following properties:
Moreover, from (12),
From Lemma 7, we can deduce
󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩
󵄩2
󵄩
󵄩2
󵄩󵄩G 󵄩󵄩 + ℎ2 󵄩󵄩󵄩G𝑛+1 󵄩󵄩󵄩 + ℎ2 󵄩󵄩󵄩𝑔𝑛+1 󵄩󵄩󵄩
󵄩
󵄩0
󵄩
󵄩1
󵄩
󵄩0
2
󵄩
󵄩
󵄩
󵄩2
≤ 𝐶ℎ4 (󵄩󵄩󵄩󵄩u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 + 󵄩󵄩󵄩󵄩𝑝(𝑡𝑛+1 )󵄩󵄩󵄩󵄩1 ) ,
(55)
̂ 𝑛+1 = E𝑛+1 + ∇𝛿𝜓𝑛+1 ,
E
ℎ
ℎ
ℎ
U𝑛+1
ℎ ,
𝑔𝑛+1 := 𝑝 (𝑡𝑛+1 ) − 𝑃ℎ𝑛+1 ,
3𝜓ℎ𝑛+1
.
2𝜏
̂ 𝑛+1 − E
̂ 𝑛+1 ,
=E
G𝑛+1 = E𝑛+1 − E𝑛+1
ℎ
ℎ
∈ Tℎ as the solution of
𝑛+1
⟨∇Θ𝑛+1
) , ∇𝜙ℎ ⟩ ,
ℎ , ∇𝜙ℎ ⟩ = ⟨∇𝜃 (𝑡
Additionally, we denote
(51)
󵄩2
󵄩󵄩 ̂ 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩
󵄩󵄩Eℎ 󵄩󵄩 = 󵄩󵄩Eℎ 󵄩󵄩 + 󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩 .
󵄩0 󵄩
󵄩0 󵄩
󵄩0
󵄩
(58)
In conjunction with the definition of |‖ ⋅ ‖| in (26), we can
derive
󵄩󵄩󵄨󵄨 𝑛+1 󵄨󵄨󵄩󵄩
󵄩󵄩󵄨󵄨G 󵄨󵄨󵄩󵄩 ≤ 𝐶.
(52)
󵄨󵄩
󵄩󵄨
We also point out that, owing to Lemma 6, 𝑞ℎ𝑛+1 ∈ Pℎ defined
in (9) satisfies
We now carry out error evaluation by comparing (65) with
(6)–(10) and then by comparing (82) and (11). We derive
strong estimates of order 1, and this result is instrumental in
proving weak estimates of order 2 for the errors
(59)
𝑛+1
E𝑛+1
:= U𝑛+1
ℎ
ℎ − uℎ ,
̂ 𝑛+1 := U𝑛+1 − u
̂ 𝑛+1
E
ℎ
ℎ
ℎ ,
𝑒ℎ𝑛+1 := 𝑃ℎ𝑛+1 − 𝑝ℎ𝑛+1 ,
(53)
Then, in conjunction with (50), we can readily get the same
accuracy for the errors
E𝑛+1 := u (𝑡𝑛+1 ) − u𝑛+1
ℎ ,
𝑒𝑛+1 := 𝑝 (𝑡𝑛+1 ) − 𝑝ℎ𝑛+1 ,
𝜗𝑛+1 := 𝜃 (𝑡𝑛+1 ) − 𝜃ℎ𝑛+1 .
= 𝜗𝑛+1 − 𝜗ℎ𝑛+1 ,
In conjunction with 𝜂𝑛+1 = 𝜃(𝑡𝑛+1 ) − Θ𝑛+1
ℎ
Assumption 5 leads to
󵄩
󵄩
󵄩
󵄩2
󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩󵄩𝜂 󵄩󵄩 + ℎ2 󵄩󵄩󵄩𝜂𝑛+1 󵄩󵄩󵄩 ≤ 𝐶ℎ4 󵄩󵄩󵄩𝜃(𝑡𝑛+1 )󵄩󵄩󵄩 ,
󵄩0
󵄩
󵄩1
󵄩
󵄩2
󵄩
𝑛+1
𝑡
󵄩
󵄩2
󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩󵄩𝛿𝜂 󵄩󵄩 + ℎ2 󵄩󵄩󵄩𝛿𝜂𝑛+1 󵄩󵄩󵄩 ≤ 𝐶𝜏ℎ4 ∫
󵄩0
󵄩
󵄩1
󵄩
𝑡𝑛
𝑛+1
𝜗ℎ𝑛+1 := Θ𝑛+1
ℎ − 𝜃ℎ .
̂ 𝑛+1 := u (𝑡𝑛+1 ) − u
̂ 𝑛+1
E
ℎ ,
󵄩󵄩 𝑛+1
󵄩
󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩󵄩𝑞ℎ − 𝑞ℎ𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩∇E
󵄩󵄩 .
󵄩
󵄩0 󵄩
󵄩0
(54)
󵄩󵄩 󵄩󵄩
󵄩󵄩𝜃𝑡 󵄩󵄩2 𝑑𝑡.
(60)
We now estimate the first-order accuracy for velocity and
temperature in Lemma 13 and then the 2nd-order accuracy
for time derivative of velocity and temperature in Lemma 15.
The result of Lemma 13 is instrumental to treat the convection
term in proof of Lemma 15. We will use Lemmas 13 and 15 to
prove optimal error accuracy in Lemma 17. Finally, we will
prove pressure error estimate in Lemma 18.
Lemma 13 (reduced rate of convergence for velocity and temperature). Suppose the exact solution of (1) is smooth enough.
Journal of Applied Mathematics
7
If Assumptions 3–5 hold, then the velocity and temperature
error functions satisfy
󵄩󵄩2
󵄩󵄩 ̂ 𝑁+1 󵄩󵄩2 󵄩󵄩 𝑁+1 󵄩󵄩2 󵄩󵄩 𝑁+1
󵄩
󵄩󵄩Eℎ 󵄩󵄩 + 󵄩󵄩Eℎ 󵄩󵄩 + 󵄩󵄩2Eℎ − E𝑁
ℎ󵄩
󵄩0 󵄩
󵄩0 󵄩
󵄩0
󵄩
We replace the pressure 𝑝ℎ𝑛 term in (6) with (10) and then subtract it from (65) to obtain
1
̂ 𝑛+1 − 4E𝑛 + E𝑛−1 , wℎ ⟩
⟨3E
2𝜏
󵄩2 4𝜏2 󵄩󵄩 𝑁+1 󵄩󵄩2
󵄩
󵄩󵄩∇𝜀 󵄩󵄩
+ 2𝜇𝜏󵄩󵄩󵄩󵄩𝑞ℎ𝑁+1 󵄩󵄩󵄩󵄩0 +
3 󵄩 ℎ 󵄩0
󵄩2 󵄩
󵄩2
󵄩
+ 󵄩󵄩󵄩󵄩𝜗ℎ𝑁+1 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩2𝜗ℎ𝑁+1 − 𝜗ℎ𝑁󵄩󵄩󵄩󵄩0
̂ 𝑛+1 , ∇wℎ ⟩
+ 𝜇𝜏 ⟨∇E
ℎ
= − ⟨∇ (𝛿𝑃ℎ𝑛+1 + 𝜀ℎ𝑛 ) , wℎ ⟩ + 𝜇 ⟨∇𝑞ℎ𝑛 , wℎ ⟩
𝑁
󵄩2
󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩
󵄩 + 𝜆󵄩󵄩󵄩∇𝜗ℎ𝑛+1 󵄩󵄩󵄩 )
+ 𝜇𝜏 ∑ (󵄩󵄩󵄩󵄩∇E
ℎ 󵄩
󵄩0
󵄩0
󵄩
̂ 𝑛+1
+ N (2u𝑛ℎ − u𝑛−1
ℎ ,u
ℎ , wℎ )
(61)
− N (u (𝑡𝑛+1 ) , u (𝑡𝑛+1 ) , wℎ )
𝑛=1
𝑁
󵄩󵄩2 󵄩󵄩
󵄩2 󵄩
󵄩2
󵄩
󵄩 + 󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛿𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩 )
+ ∑ (󵄩󵄩󵄩󵄩𝛿𝛿E𝑛+1
ℎ 󵄩
󵄩0 󵄩
󵄩0 󵄩
󵄩0
− 𝜅𝜇2 ⟨g (𝜃 (𝑡𝑛+1 ) − 2𝜃ℎ𝑛 + 𝜃ℎ𝑛−1 ) , wℎ ⟩
𝑛=1
+ ⟨R𝑛+1 , wℎ ⟩ .
≤ 𝐶 (𝜏2 + ℎ2 ) .
Proof. We resort to the Taylor theorem to write (1) as follows:
3u (𝑡𝑛+1 ) − 4u (𝑡𝑛 ) + u (𝑡𝑛−1 )
̂ 𝑛+1 = 4𝜏(E
̂ 𝑛+1 − G𝑛+1 ) = 4𝜏(E𝑛+1 +
Choosing wℎ = 4𝜏E
ℎ
ℎ
𝑛+1
∇𝛿𝜓ℎ ) in (66) and using (58) and (35) yield
󵄩󵄩2
󵄩
󵄩
𝑛󵄩
󵄩󵄩2
󵄩 + 𝛿󵄩󵄩󵄩2E𝑛+1
𝛿󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
󵄩 ℎ − Eℎ 󵄩󵄩0
󵄩󵄩2
󵄩2
󵄩
󵄩
󵄩 + 6󵄩󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿E𝑛+1
ℎ 󵄩
󵄩0
󵄩0
󵄩
󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩
+ 4𝜇𝜏󵄩󵄩󵄩󵄩∇E
ℎ 󵄩
󵄩0
2𝜏
+ (u (𝑡𝑛+1 ) ⋅ ∇) u (𝑡𝑛+1 ) + ∇𝑝 (𝑡𝑛+1 )
𝑛+1
− 𝜇Δu (𝑡
2
𝑛+1
) + 𝜅𝜇 g𝜃 (𝑡
(62)
)
= R𝑛+1 + f (𝑡𝑛+1 ) ,
7
𝑛=1
2𝜏
where
+ u (𝑡𝑛+1 ) ⋅ ∇𝜃 (𝑡𝑛+1 ) − 𝜆𝜇Δ𝜃 (𝑡𝑛+1 )
(63)
̂ 𝑛+1 ⟩ ,
𝐴 2 := −2 ⟨3G𝑛+1 − 4G𝑛 + G𝑛−1 , E
ℎ
where
𝑛+1
2
̂ 𝑛+1 ⟩ ,
𝐴 3 := −4𝜏 ⟨∇𝛿𝑃ℎ𝑛+1 , E
ℎ
𝑡
2
1
− ( ) ∫ u𝑡𝑡𝑡 (𝑠) (𝑡𝑛−1 − 𝑠) 𝑑𝑠,
4𝜏 𝑡𝑛−1
̂ 𝑛+1 ⟩ ,
𝐴 4 := −4𝜏 ⟨∇𝜀ℎ𝑛 , E
ℎ
u𝑡𝑡𝑡 (𝑠) (𝑠 − 𝑡𝑛 ) 𝑑𝑠
𝑛+1
1
𝑄𝑛+1 := ( ) ∫
𝜏 𝑡𝑛
(64)
2
𝜃𝑡𝑡𝑡 (𝑠) (𝑠 − 𝑡𝑛 ) 𝑑𝑠
are the truncation errors. In conjunction with the definition
𝑛+1
of the Stokes projection {U𝑛+1
ℎ , 𝑃ℎ }, we readily get a weak
formulation for (62), ∀wℎ ∈ Vℎ :
1
⟨3u (𝑡𝑛+1 ) − 4u (𝑡𝑛 ) + u (𝑡𝑛−1 ) , wℎ ⟩
2𝜏
𝑛+1
+ 𝜅𝜇 ⟨g𝜃 (𝑡
= ⟨R𝑛+1 , wℎ ⟩ + ⟨f (𝑡𝑛+1 ) , wℎ ⟩ .
We now estimate all the terms from 𝐴 1 to 𝐴 7 , respectively. To
tackle 𝐴 1 , we first add and subtract 2u(𝑡𝑛 ) − u(𝑡𝑛−1 ) to obtain
̂ 𝑛+1 )
𝐴 1 = − 4𝜏N (𝛿𝛿u (𝑡𝑛+1 ) , u (𝑡𝑛+1 ) , E
ℎ
̂ 𝑛+1 , E
̂ 𝑛+1 )
− 4𝜏N (2u𝑛ℎ − u𝑛−1
ℎ ,E
ℎ
+ N (u (𝑡𝑛+1 ) , u (𝑡𝑛+1 ) , wℎ ) + ⟨∇𝑃ℎ𝑛+1 , wℎ ⟩
2
̂ 𝑛+1 ⟩ ,
𝐴 5 := 4𝜇𝜏 ⟨∇𝑞ℎ𝑛 , E
ℎ
̂ 𝑛+1 ⟩ .
𝐴 7 := −4𝜏𝜅𝜇2 ⟨g (𝜃 (𝑡𝑛+1 ) − 2𝜃ℎ𝑛 + 𝜃ℎ𝑛−1 ) , E
ℎ
𝑛+1
𝜇 ⟨∇U𝑛+1
ℎ , ∇wℎ ⟩
(68)
̂ 𝑛+1 ⟩ ,
𝐴 6 := 4𝜏 ⟨R𝑛+1 , E
ℎ
𝑡
2
1
− ( ) ∫ 𝜃𝑡𝑡𝑡 (𝑠) (𝑡𝑛−1 − 𝑠) 𝑑𝑠
4𝜏 𝑡𝑛−1
+
̂ 𝑛+1 )
𝐴 1 := − 4𝜏N (u (𝑡𝑛+1 ) , u (𝑡𝑛+1 ) , E
ℎ
̂ 𝑛+1
̂ 𝑛+1
+ 4𝜏N (2u𝑛ℎ − u𝑛−1
ℎ ,u
ℎ , Eℎ ) ,
= 𝑄𝑛+1 + 𝑏 (𝑡𝑛+1 ) ,
𝑡𝑛+1
(67)
= ∑𝐴 𝑖,
3𝜃 (𝑡𝑛+1 ) − 4𝜃 (𝑡𝑛 ) + 𝜃 (𝑡𝑛−1 )
𝑡
1
R𝑛+1 := ( ) ∫
𝜏 𝑡𝑛
(66)
) , wℎ ⟩
(69)
̂ 𝑛+1 ) .
− 4𝜏N (2E𝑛 − E𝑛−1 , u (𝑡𝑛+1 ) , E
ℎ
(65)
̂ 𝑛+1 ̂ 𝑛+1
Because of N(2u𝑛ℎ − u𝑛−1
ℎ , Eℎ , Eℎ ) = 0, which comes from
(30), the second term of 𝐴 1 can be replaced by
̂ 𝑛+1 ) .
4𝜏N (2E𝑛 − E𝑛−1 − 2u (𝑡𝑛 ) + u (𝑡𝑛−1 ) , G𝑛+1 , E
ℎ
(70)
8
Journal of Applied Mathematics
If we now apply inequality (𝑎 + 𝑏)2 ≤ 4𝑎2 + (4/3)𝑏2 , then we
can get
If we apply Lemma 10, then we can readily obtain
󵄩
󵄩󵄩
󵄩
𝐴 1 ≤ 𝐶𝜏 (󵄩󵄩󵄩󵄩𝛿𝛿u (𝑡𝑛+1 )󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩u (𝑡𝑛+1 )󵄩󵄩󵄩󵄩2
󵄩󵄩
󵄩
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩
+󵄩󵄩󵄩󵄩2E𝑛 − E𝑛−1 󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩u (𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 ) 󵄩󵄩󵄩󵄩E
ℎ 󵄩
󵄩1
󵄩 󵄩󵄩󵄨󵄨 𝑛+1 󵄨󵄨󵄩󵄩
󵄩 𝑛
𝑛−1 󵄩
󵄩
+ 𝐶𝜏 (󵄩󵄩󵄩2E − E 󵄩󵄩󵄩0 󵄩󵄩󵄩󵄨󵄨󵄨G 󵄨󵄨󵄨󵄩󵄩󵄩
󵄩
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩󵄩
󵄩.
+󵄩󵄩󵄩󵄩2u(𝑡𝑛 ) − u(𝑡𝑛−1 )󵄩󵄩󵄩󵄩2 󵄩󵄩󵄩󵄩G𝑛+1 󵄩󵄩󵄩󵄩0 ) 󵄩󵄩󵄩󵄩E
ℎ 󵄩
󵄩1
(71)
󵄩󵄩2
󵄩
󵄩
𝐴 1 ≤ 𝐶𝜏 (󵄩󵄩󵄩󵄩2E𝑛ℎ − E𝑛−1
ℎ 󵄩
󵄩0
×∫
𝑡𝑛+1
𝑡𝑛+1
4
+ 𝐶ℎ ∫
𝑡𝑛−1
1 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
󵄩
󵄩0
4
󵄩2 󵄩
󵄩2
󵄩
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
+ 𝐶𝜏 ∫
×∫
𝑡𝑛+1
𝑡𝑛
= − 4𝜇𝜏 ⟨𝑞ℎ𝑛 , 𝛿𝑞ℎ𝑛+1 ⟩
(73)
+ 𝐶𝜏
󵄩2 󵄩 󵄩2 󵄩
󵄩2
󵄩
(󵄩󵄩󵄩󵄩∇𝜀ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩∇𝜀ℎ𝑛 󵄩󵄩󵄩0 − 󵄩󵄩󵄩󵄩∇𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
3󵄩
󵄩
𝑛 󵄩2
󵄩󵄩∇𝜀ℎ 󵄩󵄩󵄩0
󵄩2
󵄩
+ 𝐶𝜏󵄩󵄩󵄩󵄩∇𝛿𝑃ℎ𝑛+1 󵄩󵄩󵄩󵄩0 .
(78)
󵄩
󵄩2 󵄩 󵄩2
≤ − 2𝜇𝜏 (󵄩󵄩󵄩󵄩𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩𝑞ℎ𝑛 󵄩󵄩󵄩0 )
󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩.
+ 2𝜇𝜏󵄩󵄩󵄩󵄩∇E
ℎ 󵄩
󵄩0
Also we readily get
(74)
󵄩󵄩2
󵄩
󵄩 +
𝐴 6 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
4
+ 𝐶𝜏 ∫
𝑡𝑛+1
𝑡𝑛−1
1 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
󵄩0
4󵄩
󵄩󵄩
󵄩2
󵄩󵄩u𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0 𝑑𝑡.
(79)
The Hölder inequality and (60) yield
̂ 𝑛+1 ⟩
𝐴 7 = − 4𝜅𝜇2 𝜏 ⟨g (𝛿𝛿𝜃 (𝑡𝑛+1 ) − 2𝜗𝑛 + 𝜗𝑛−1 ) , E
ℎ
𝐴 4 = − 4𝜏 ⟨∇𝜀ℎ𝑛 , ∇𝛿𝜓ℎ𝑛+1 ⟩
4𝜏2
≤ −
3
󵄩2 󵄩 󵄩2 󵄩
󵄩2
󵄩
= − 2𝜇𝜏 (󵄩󵄩󵄩󵄩𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩𝑞ℎ𝑛 󵄩󵄩󵄩0 − 󵄩󵄩󵄩󵄩𝛿𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
󵄩2 󵄩
󵄩2
󵄩
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
8𝜏2
⟨∇𝜀ℎ𝑛 , ∇ (𝛿𝜀ℎ𝑛+1 − 𝛿𝑃ℎ𝑛+1 )⟩
3
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
̂ 𝑛+1
𝐴 5 = 4𝜇𝜏 ⟨𝑞ℎ𝑛 , ∇ ⋅ u
ℎ ⟩
In conjunction with the definition 𝜀ℎ𝑛+1 = 𝑃ℎ𝑛+1 + 3𝜓ℎ𝑛+1 /2𝜏,
𝐴 4 can be evaluated by
= −
(77)
In light of ∇ ⋅ u(𝑡𝑛+1 ) = 0 and (37), (9) and (59) yield
𝐴 3 = − 4𝜏 ⟨∇𝛿𝑃ℎ𝑛+1 , ∇𝛿𝜓ℎ𝑛+1 ⟩
1 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩 + 𝐶𝜏3
󵄩0
4󵄩
𝑡𝑛+1
𝑡𝑛
In order to estimate 𝐴 3 and 𝐴 4 , we note that the cancellation
̂ 𝑛+1 = E𝑛+1 + ∇𝛿𝜓𝑛+1
law (12) gives ⟨∇𝜀ℎ𝑛 , E𝑛+1
ℎ ⟩ = 0. Then Eℎ
ℎ
ℎ
yields
≤
4𝜏2 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛 󵄩󵄩2
(󵄩󵄩∇𝜀 󵄩󵄩 − 󵄩∇𝜀 󵄩 )
3 󵄩 ℎ 󵄩0 󵄩 ℎ 󵄩0
󵄩2
󵄩
󵄩 󵄩2
+ 𝐶𝜏3 󵄩󵄩󵄩∇𝜀ℎ𝑛 󵄩󵄩󵄩0 + 4󵄩󵄩󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩󵄩0
2
In light of (51) and (58), 𝐴 2 becomes
󵄩󵄩2
󵄩
󵄩 +
𝐴 2 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
So we arrive at
(72)
󵄩󵄩
󵄩2
󵄩󵄩u𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0 𝑑𝑡.
𝑡𝑛−1
(76)
𝐴4 ≤ −
󵄩2 󵄩
󵄩2
󵄩
+ 󵄩󵄩󵄩󵄩2G𝑛 − G𝑛−1 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩G𝑛+1 󵄩󵄩󵄩󵄩0 )
󵄩 ̂ 𝑛+1 󵄩󵄩2 𝐶𝜏
󵄩 +
+ 𝜇𝜏󵄩󵄩󵄩󵄩∇E
ℎ 󵄩
󵄩0
𝜇
󵄩󵄩2
4𝜏2 󵄩󵄩󵄩
3
󵄩󵄩∇𝛿𝑃ℎ𝑛+1 + ∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
󵄩󵄩0
3 󵄩󵄩
2𝜏
󵄩
󵄩2
󵄩2
󵄩
≤ 𝐶𝜏2 󵄩󵄩󵄩󵄩∇𝛿𝑃ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 4󵄩󵄩󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩󵄩0 .
=
Since we have ‖u(𝑡𝑛+1 )‖2 + ‖|G𝑛+1 |‖ ≤ 𝑀 according to (52),
we arrive at
4
4𝜏2 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩
󵄩
󵄩0
3
󵄩2 󵄩
󵄩2
󵄩
≤ 𝐶𝜅2 𝜇4 𝜏 (󵄩󵄩󵄩󵄩2𝜗ℎ𝑛 − 𝜗ℎ𝑛−1 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩2𝜂𝑛 − 𝜂𝑛−1 󵄩󵄩󵄩󵄩0 )
(75)
󵄩󵄩2
󵄩
󵄩 +
+ 𝜏󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
4
𝑡𝑛 +1
+ 𝐶𝜏 ∫
𝑡𝑛−1
1 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
󵄩0
4󵄩
󵄩2
󵄩󵄩
󵄩󵄩𝜃𝑡𝑡 (𝑡)󵄩󵄩󵄩0 𝑑𝑡.
(80)
Journal of Applied Mathematics
9
Inserting the above estimates into (67) gives
where
󵄩
󵄩󵄩2
󵄩
𝑛󵄩
󵄩󵄩2
󵄩 + 𝛿󵄩󵄩󵄩2E𝑛+1
𝛿󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
󵄩 ℎ − Eℎ 󵄩󵄩0
󵄩󵄩2 󵄩󵄩
󵄩2
󵄩
󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩 + 󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩 + 𝜇𝜏󵄩󵄩󵄩∇E
󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿E𝑛+1
ℎ 󵄩
󵄩0 󵄩
󵄩0
󵄩 ℎ 󵄩󵄩0
+
≤
𝐴 8 := − 4𝜏N (u (𝑡𝑛+1 ) , 𝜃 (𝑡𝑛+1 ) , 𝜗ℎ𝑛+1 )
𝑛+1 𝑛+1
+ 4𝜏N (u𝑛+1
ℎ , 𝜃ℎ , 𝜗ℎ ) ,
4𝜏2 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩2
󵄩
󵄩󵄩∇𝜀 󵄩󵄩 + 2𝜇𝜏󵄩󵄩󵄩𝑞ℎ𝑛+1 󵄩󵄩󵄩
󵄩0
󵄩
3 󵄩 ℎ 󵄩0
𝐴 9 := − 2 ⟨3𝜂𝑛+1 − 4𝜂𝑛 + 𝜂𝑛−1 , 𝜗ℎ𝑛+1 ⟩
4𝜏2 󵄩󵄩 𝑛 󵄩󵄩2
󵄩 𝑛 󵄩2
3󵄩
𝑛 󵄩2
󵄩∇𝜀 󵄩 + 𝐶𝜏 󵄩󵄩󵄩∇𝜀ℎ 󵄩󵄩󵄩0 + 2𝜇𝜏󵄩󵄩󵄩𝑞ℎ 󵄩󵄩󵄩0
3 󵄩 ℎ 󵄩0
󵄩󵄩2 󵄩󵄩 𝑛
󵄩󵄩2
󵄩
󵄩 + 󵄩󵄩2Eℎ − E𝑛−1
󵄩
+ 𝐶𝜏 (󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
ℎ 󵄩
󵄩0 󵄩
󵄩0
󵄩2 󵄩
󵄩2
󵄩
+󵄩󵄩󵄩󵄩2G𝑛 − G𝑛−1 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩G𝑛+1 󵄩󵄩󵄩󵄩0 )
+ 4𝜏 ⟨𝑄𝑛+1 , 𝜗ℎ𝑛+1 ⟩ .
(81)
󵄩2
󵄩
+ 𝐶𝜅2 𝜇4 𝜏󵄩󵄩󵄩󵄩2𝜂𝑛 − 𝜂𝑛−1 󵄩󵄩󵄩󵄩0
𝑛
+ 𝐶𝜏4 ∫
𝑡 +1
𝑡𝑛−1
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝜃𝑡𝑡 (𝑡)󵄩󵄩󵄩0 ) 𝑑𝑡
4
+ 𝐶 (𝜏 + ℎ ) ∫
𝑡𝑛+1
+ 4𝜏N (E𝑛+1 − u (𝑡𝑛+1 ) , 𝜂𝑛+1 , 𝜗ℎ𝑛+1 )
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
𝑡𝑛−1
On the other hand, the definition (48) of Θ𝑛+1
∈ Tℎ leads to a
ℎ
weak formula of (63) as, for all 𝜙ℎ ∈ Tℎ ,
1
⟨3𝜃 (𝑡𝑛+1 ) − 4𝜃 (𝑡𝑛 ) + 𝜃 (𝑡𝑛−1 ) , 𝜙ℎ ⟩
2𝜏
𝑛+1
+ N (u (𝑡
𝑛+1
) , 𝜃 (𝑡
) , 𝜙ℎ )
𝑛+1
𝑛+1
+ 𝑏 (𝑡
󵄩
󵄩󵄩
󵄩
󵄩󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩E𝑛+1 󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩𝜃(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩1
󵄩
󵄩
󵄩
󵄩
+ 𝐶𝜏 (󵄩󵄩󵄩󵄩E𝑛+1 󵄩󵄩󵄩󵄩L3 (Ω) 󵄩󵄩󵄩󵄩𝜂𝑛+1 󵄩󵄩󵄩󵄩1
󵄩 󵄩
󵄩
󵄩
󵄩󵄩
+󵄩󵄩󵄩󵄩u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 󵄩󵄩󵄩󵄩𝜂𝑛+1 󵄩󵄩󵄩󵄩0 ) 󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩1
≤
(82)
We now subtract (11) from (82) to derive
𝑛+1
𝑡
󵄩2
󵄩
󵄩
󵄩2
≤ 𝐶𝜏󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶ℎ4 ∫ 󵄩󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩2 𝑑𝑡
𝑡𝑛−1
1
⟨3𝜗𝑛+1 − 4𝜗𝑛 + 𝜗𝑛−1 , 𝜙ℎ ⟩
2𝜏
+ N (u (𝑡𝑛+1 ) , 𝜃 (𝑡𝑛+1 ) , 𝜙ℎ )
+ 𝐶𝜏4 ∫
(87)
󵄩󵄩
󵄩2
󵄩󵄩𝜃𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0 𝑑𝑡.
Inserting the above estimates into (84) gives
= ⟨𝑄𝑛+1 , 𝜙ℎ ⟩ .
󵄩2
󵄩2
󵄩
󵄩
𝛿󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝛿󵄩󵄩󵄩󵄩2𝜗ℎ𝑛+1 − 𝜗ℎ𝑛 󵄩󵄩󵄩󵄩0
󵄩2
󵄩2
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 3𝜆𝜇𝜏󵄩󵄩󵄩󵄩∇𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
Choosing 𝜙ℎ = 4𝜏𝜗ℎ𝑛+1 = 4𝜏(𝜗𝑛+1 − 𝜂𝑛+1 ) yields
= 𝐴 8 + 𝐴 9,
𝑡𝑛+1
𝑡𝑛−1
(83)
+ 𝜆𝜇 ⟨∇𝜗ℎ𝑛+1 , ∇𝜙ℎ ⟩
󵄩2
󵄩2
󵄩
󵄩
𝛿󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝛿󵄩󵄩󵄩󵄩2𝜗ℎ𝑛+1 − 𝜗ℎ𝑛 󵄩󵄩󵄩󵄩0
󵄩2
󵄩2
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 4𝜆𝜇𝜏󵄩󵄩󵄩󵄩∇𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
𝑡𝑛+1
We use ‖𝛿𝜂𝑛+1 ‖0 ≤ 𝐶𝜏ℎ4 ∫𝑡𝑛 ‖𝜃𝑡 (𝑡)‖22 𝑑𝑡 which is (60) to
attack 𝐴 9 :
󵄩󵄩
󵄩
󵄩
𝐴 9 ≤ 𝐶󵄩󵄩󵄩󵄩3𝜂𝑛+1 − 4𝜂𝑛 + 𝜂𝑛−1 󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩󵄩
󵄩
󵄩
+ 𝐶𝜏󵄩󵄩󵄩󵄩𝑄𝑛+1 󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
) , 𝜙ℎ ⟩ .
𝑛+1
− N (u𝑛+1
ℎ , 𝜃ℎ , 𝜙ℎ )
(86)
𝐶𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩2
󵄩
(󵄩󵄩󵄩E 󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝜂 󵄩󵄩󵄩0 ) + 𝜆𝜇𝜏󵄩󵄩󵄩󵄩∇𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 .
𝜆𝜇
2
+ 𝜆𝜇 ⟨∇Θ𝑛+1
ℎ , ∇𝜙ℎ ⟩
= ⟨𝑄
To estimate 𝐴 8 , we note N(E𝑛+1 − u(𝑡𝑛+1 ), 𝜗ℎ𝑛+1 , 𝜗ℎ𝑛+1 ) = 0
which comes from (30), then ‖E𝑛+1 ‖L3 (Ω) ≤ 𝐶ℎ−1/2 ‖E𝑛+1 ‖0
and ‖𝜂𝑛+1 ‖1 ≤ 𝐶ℎ‖𝜃(𝑡𝑛+1 )‖2 yield
𝐴 8 = − 4𝜏N (E𝑛+1 , 𝜃 (𝑡𝑛+1 ) , 𝜗ℎ𝑛+1 )
󵄩2
󵄩
+ 𝐶𝜅2 𝜇4 𝜏󵄩󵄩󵄩󵄩2𝜗ℎ𝑛 − 𝜗ℎ𝑛−1 󵄩󵄩󵄩󵄩0
2
(85)
(84)
󵄩2 𝐶𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 +
(󵄩󵄩E 󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝜂 󵄩󵄩󵄩0 )
𝜆𝜇 󵄩
4
4
𝑡𝑛+1
+ 𝐶 (ℎ + 𝜏 ) ∫
𝑡𝑛−1
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩𝜃𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩2 ) 𝑑𝑡.
(88)
10
Journal of Applied Mathematics
Adding 2 inequalities (81) and (88) and summing over 𝑛 from
1 to 𝑁 imply
8𝜏2
3
4𝜏2
≤ −
3
𝐴4 = −
󵄩󵄩 𝑁+1 󵄩󵄩2 󵄩󵄩 𝑁+1
󵄩󵄩2
󵄩󵄩Eℎ 󵄩󵄩 + 󵄩󵄩2Eℎ − E𝑁
󵄩
ℎ󵄩
󵄩
󵄩0 󵄩
󵄩0
󵄩2
󵄩󵄩 𝑁+1 󵄩󵄩2 󵄩󵄩 𝑁+1
+ 󵄩󵄩󵄩𝜗ℎ 󵄩󵄩󵄩0 + 󵄩󵄩󵄩2𝜗ℎ − 𝜗ℎ𝑁󵄩󵄩󵄩󵄩0
4𝜏2 󵄩󵄩 2 󵄩󵄩2
󵄩 󵄩2
󵄩󵄩∇𝜀 󵄩󵄩 + 𝐶𝜏2 󵄩󵄩󵄩∇𝜀ℎ1 󵄩󵄩󵄩
󵄩 󵄩0
3 󵄩 ℎ 󵄩0
2
󵄩󵄩
2󵄩
3
+ 4󵄩󵄩󵄩∇𝛿𝜓ℎ 󵄩󵄩󵄩󵄩0 + 𝐶𝜏
𝑛=1
󵄩2
󵄩
+󵄩󵄩󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
𝑡2
󵄩
󵄩2 󵄩
󵄩2
× ∫ (󵄩󵄩󵄩𝑝𝑡𝑡 (𝑡)󵄩󵄩󵄩1 + 󵄩󵄩󵄩u𝑡𝑡 (𝑡)󵄩󵄩󵄩2 ) 𝑑𝑡.
𝑡0
𝑁
󵄩2
󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩
󵄩 + 3𝜆󵄩󵄩󵄩∇𝜗ℎ𝑛+1 󵄩󵄩󵄩 )
+ 𝜇𝜏 ∑ (󵄩󵄩󵄩󵄩∇E
ℎ 󵄩
󵄩0
󵄩0
󵄩
In light of Assumption 4, we arrive at
𝑛=1
󵄩2 󵄩 󵄩2
󵄩 󵄩2 󵄩
≤ 󵄩󵄩󵄩󵄩E1ℎ 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩2E1ℎ − E0ℎ 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩𝜗ℎ1 󵄩󵄩󵄩󵄩0
󵄩2 4𝜏2 󵄩󵄩 1 󵄩󵄩2
󵄩
󵄩 󵄩2
󵄩󵄩∇𝜀ℎ 󵄩󵄩 + 2𝜇𝜏󵄩󵄩󵄩𝑞ℎ1 󵄩󵄩󵄩
+ 󵄩󵄩󵄩󵄩2𝜗ℎ1 − 𝜗ℎ0 󵄩󵄩󵄩󵄩0 +
󵄩
󵄩
󵄩 󵄩0
0
3
(89)
𝑁
󵄩2 󵄩
󵄩2
󵄩
+ 𝜏 ∑ (󵄩󵄩󵄩󵄩E𝑛+1 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩𝜂𝑛+1 󵄩󵄩󵄩󵄩0 )
𝑛=1
+ 𝐶𝜏 ∑
𝑛=1
󵄩
+ 󵄩󵄩󵄩󵄩2𝜗ℎ𝑛 −
+
×∫
𝑡𝑛+1
𝑡𝑛−1
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝜃𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0
󵄩
󵄩2
+󵄩󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩2 ) 𝑑𝑡
+ 𝐶 (𝜏2 + ℎ4 ) ∫
𝑡𝑛−1
+
󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2
(91)
4𝜏2 󵄩󵄩 2 󵄩󵄩2
󵄩2
󵄩
󵄩󵄩∇𝜀 󵄩󵄩 + 𝜇𝜆𝜏󵄩󵄩󵄩∇𝜗ℎ2 󵄩󵄩󵄩
󵄩0
󵄩
3 󵄩 ℎ 󵄩0
≤ 𝐶 (𝜏4 + ℎ4 ) .
We now start to estimate errors for time derivative of
velocity.
+ 𝐶 (ℎ4 + 𝜏4 )
𝑡𝑛+1
󵄩2
󵄩󵄩 ̂ 2 󵄩󵄩2 󵄩󵄩 2 󵄩󵄩2 󵄩󵄩 2
󵄩󵄩Eℎ 󵄩󵄩 + 󵄩󵄩Eℎ 󵄩󵄩 + 󵄩󵄩2Eℎ − E1ℎ 󵄩󵄩󵄩
󵄩0
󵄩 󵄩0 󵄩 󵄩0 󵄩
2
1 󵄩󵄩
󵄩2
󵄩
2󵄩
2
+ 󵄩󵄩󵄩𝛿𝛿Eℎ 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩∇𝛿𝜓ℎ 󵄩󵄩󵄩󵄩0
2
󵄩2 󵄩 󵄩2
󵄩
+ 󵄩󵄩󵄩󵄩2𝜗ℎ2 − 𝜗ℎ2 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩𝜗ℎ2 󵄩󵄩󵄩󵄩0
󵄩 ̂ 2 󵄩󵄩2
󵄩 󵄩2
󵄩 + 2𝜇𝜏󵄩󵄩󵄩𝑞ℎ2 󵄩󵄩󵄩
+ 𝜇𝜏󵄩󵄩󵄩󵄩∇E
ℎ󵄩
󵄩0
󵄩 󵄩0
󵄩2
𝜗ℎ𝑛−1 󵄩󵄩󵄩󵄩0
󵄩2
󵄩
+󵄩󵄩󵄩󵄩2𝜂𝑛 − 𝜂𝑛−1 󵄩󵄩󵄩󵄩0 )
󵄩 󵄩2
𝐶𝜏3 󵄩󵄩󵄩∇𝜀ℎ𝑛 󵄩󵄩󵄩0
(90)
≤ −
𝑁
󵄩󵄩2 󵄩󵄩
󵄩2
󵄩
󵄩 + 󵄩󵄩𝛿𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩
+ ∑ (󵄩󵄩󵄩󵄩𝛿𝛿E𝑁+1
ℎ 󵄩
󵄩0 󵄩
󵄩0
󵄩2
󵄩
(󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
⟨∇𝜀ℎ1 , ∇ (𝛿𝜀ℎ2 − 𝛿𝑃ℎ2 )⟩
󵄩2
󵄩 󵄩2 󵄩 󵄩2 󵄩
(󵄩󵄩󵄩󵄩∇𝜀ℎ2 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩󵄩∇𝜀ℎ1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩󵄩∇𝛿𝜀ℎ2 󵄩󵄩󵄩󵄩0 )
󵄩 󵄩2
󵄩
󵄩2
+ 𝐶𝜏2 󵄩󵄩󵄩󵄩∇𝜀ℎ1 󵄩󵄩󵄩󵄩0 + 𝐶𝜏2 󵄩󵄩󵄩󵄩∇𝛿𝑃ℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩2 4𝜏2 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩
󵄩󵄩∇𝜀 󵄩󵄩
+ 2𝜇𝜏󵄩󵄩󵄩󵄩𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 +
3 󵄩 ℎ 󵄩0
𝑁
initial errors for the case 𝑛 = 1, and so we have to recompute
again (77). We start to rewrite 𝐴 4 as
󵄩
󵄩2
+󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
Because of 𝑞1 = 0, we have 𝜀ℎ1 = 𝑃ℎ1 + (3/2𝜏)𝜓ℎ1 = 𝑒ℎ1 , and
2
thus, Assumption 4 yields ‖∇𝜀ℎ1 ‖0 ≤ 𝐶 and the first 4 terms
in the right-hand side can be bound by Assumption 4 and
properties E0 = 0 and 𝑞1 = 0 which are directly deduced from
the conditions in Algorithm 1. Moreover, the remaining terms
can be treated by the discrete Gronwall lemma. Finally, in
conjunction with (58), we conclude the desired result and
complete this proof.
Remark 14 (optimal estimation). In order to get optimal
accuracy, we must get rid of the terms of 𝐴 3 and 𝐴 4 by
applying duality argument in Lemma 17. To do this, we
first evaluate the errors for time derivative of velocity and
temperature in Lemma 15. Thus, we need to evaluate optimal
Lemma 15 (error estimate for time derivative of velocity
and temperature). Suppose the exact solution of (1) is smooth
enough and 𝜏 ≤ 𝐶ℎ. If Assumptions 3–5 hold, then the time
derivative velocity and temperature error functions satisfy
󵄩󵄩 𝑁+1 󵄩󵄩2 󵄩󵄩 ̂ 𝑁+1 󵄩󵄩2
󵄩󵄩𝛿Eℎ 󵄩󵄩 + 󵄩󵄩𝛿Eℎ 󵄩󵄩
󵄩
󵄩0 󵄩
󵄩0
󵄩󵄩2
󵄩󵄩
𝑁+1
󵄩
+ 󵄩󵄩󵄩2𝛿Eℎ − 𝛿E𝑁
ℎ󵄩
󵄩0
󵄩󵄩 𝑁+1 󵄩󵄩2 󵄩󵄩 𝑁+1 󵄩󵄩2
+ 2𝜇𝜏󵄩󵄩󵄩𝛿𝑞ℎ 󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝛿𝜗ℎ 󵄩󵄩󵄩0
󵄩2
󵄩
+ 󵄩󵄩󵄩󵄩2𝛿𝜗ℎ𝑁+1 − 𝛿𝜗ℎ𝑁󵄩󵄩󵄩󵄩0
𝑁
󵄩󵄩2 󵄩󵄩
󵄩2
󵄩
󵄩 + 󵄩󵄩∇𝛿𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
+ ∑ (󵄩󵄩󵄩󵄩𝛿𝛿𝛿E𝑛+1
ℎ 󵄩
󵄩0 󵄩
󵄩0
𝑛=2
+
󵄩2
󵄩
+󵄩󵄩󵄩󵄩𝛿𝛿𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
4𝜏2 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜀ℎ𝑁+1 󵄩󵄩󵄩
󵄩
󵄩0
3
𝑁
󵄩2
󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩
󵄩 + 𝜆󵄩󵄩󵄩∇𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩 )
+ 𝜇𝜏 ∑ (󵄩󵄩󵄩󵄩∇𝛿E
ℎ 󵄩
󵄩0
󵄩0
󵄩
𝑛=2
2
2
≤ 𝐶(𝜏 + ℎ2 ) .
(92)
Journal of Applied Mathematics
11
Remark 16 (the condition 𝜏 ≤ 𝐶ℎ). The assumption 𝜏 ≤
𝐶ℎ requires to control convection terms which are used
at only (100) and (117), so we can omit this condition for
the linearized Boussinesq equations. However, this condition
cannot be removed for nonlinear equation case, because (99)
must be bounded by ℎ.
In estimating convection terms, we will use Lemma 10
frequently without notice. We recall ‖u(𝑡)‖2 ≤ 𝐶 to obtain
Proof. We subtract two consecutive formulas of (66) and
̂ 𝑛+1 to obtain
impose wℎ = 4𝜏𝛿E
ℎ
(96)
𝐴 1,1 + 𝐴 1,2
󵄩
󵄩󵄩
󵄩
≤ 𝐶𝜏 (󵄩󵄩󵄩󵄩𝛿𝛿u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2
󵄩
󵄩󵄩
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩󵄩
+󵄩󵄩󵄩𝛿𝛿u(𝑡𝑛 )󵄩󵄩󵄩0 󵄩󵄩󵄩u(𝑡𝑛 )󵄩󵄩󵄩2 ) 󵄩󵄩󵄩󵄩𝛿E
ℎ 󵄩
󵄩1
𝜇𝜏 󵄩󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩󵄩∇𝛿Eℎ 󵄩󵄩
≤
󵄩0
6󵄩
𝑛+1
󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩
𝑛󵄩
󵄩󵄩2
󵄩󵄩𝛿Eℎ 󵄩󵄩 + 󵄩󵄩2𝛿E𝑛+1
ℎ − 𝛿Eℎ 󵄩
󵄩
󵄩0 󵄩
󵄩0
󵄩󵄩2
󵄩2
󵄩
󵄩
󵄩 + 6󵄩󵄩󵄩∇𝛿𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿𝛿E𝑛+1
ℎ 󵄩
󵄩0
󵄩0
󵄩
󵄩󵄩2
󵄩2 󵄩
󵄩
󵄩
− 󵄩󵄩󵄩𝛿E𝑛ℎ 󵄩󵄩󵄩0 − 󵄩󵄩󵄩󵄩2𝛿E𝑛ℎ − 𝛿E𝑛−1
ℎ 󵄩
󵄩0
󵄩󵄩 ̂ 𝑛+1 󵄩󵄩2
+ 4𝜇𝜏󵄩󵄩󵄩∇𝛿Eℎ 󵄩󵄩󵄩0
+
𝐶𝜏4 𝑡 󵄩󵄩
󵄩2
∫ 󵄩u (𝑡)󵄩󵄩 𝑑𝑡.
𝜇 𝑡𝑛−2 󵄩 𝑡𝑡 󵄩0
The result in Lemma 13, ‖2E𝑛 − E𝑛−1 ‖0 ≤ 𝐶(𝜏 + ℎ), is essential
to treat the next 2 convection terms. Invoking (33), we have
(93)
7
= ∑𝐴 𝑖,
𝑛=1
where
̂ 𝑛+1 )
𝐴 1 := 4𝜏N (u (𝑡𝑛 ) , u (𝑡𝑛 ) , 𝛿E
ℎ
𝑛−2 𝑛 ̂ 𝑛+1
̂ ℎ , 𝛿Eℎ )
− 4𝜏N (2u𝑛−1
ℎ − uℎ , u
𝐴 1,3 + 𝐴 1,4
󵄩
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩2E𝑛 − E𝑛−1 󵄩󵄩󵄩󵄩0
󵄩
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩
× 󵄩󵄩󵄩󵄩𝛿u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 󵄩󵄩󵄩󵄩𝛿E
ℎ 󵄩
󵄩1
󵄩
󵄩󵄩
+ 𝐶𝜏󵄩󵄩󵄩2𝛿E𝑛 − 𝛿E𝑛−1 󵄩󵄩󵄩󵄩0
󵄩
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩󵄩
× 󵄩󵄩󵄩u(𝑡𝑛 )󵄩󵄩󵄩2 󵄩󵄩󵄩󵄩𝛿E
ℎ 󵄩
󵄩1
𝜇𝜏 󵄩󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩󵄩∇𝛿Eℎ 󵄩󵄩
≤
󵄩0
6󵄩
̂ 𝑛+1 )
− 4𝜏N (u (𝑡𝑛+1 ) , u (𝑡𝑛+1 ) , 𝛿E
ℎ
+
̂ 𝑛+1
̂ 𝑛+1
+ 4𝜏N (2u𝑛ℎ − u𝑛−1
ℎ ,u
ℎ , 𝛿Eℎ ) ,
+
̂ 𝑛+1 ⟩ ,
𝐴 2 := −2 ⟨3𝛿G𝑛+1 − 4𝛿G𝑛 + 𝛿G𝑛−1 , 𝛿E
ℎ
̂ 𝑛+1 ⟩ ,
𝐴 3 := −4𝜏 ⟨∇𝛿𝛿𝑃ℎ𝑛+1 , 𝛿E
ℎ
̂ 𝑛+1 , 𝛿E
̂ 𝑛+1 )
G𝑛+1 + E
ℎ
ℎ
̂ 𝑛+1 ⟩ .
× ⟨g𝛿 (𝜃 (𝑡𝑛+1 ) − 2𝜃𝑛 + 𝜃𝑛−1 ) , 𝛿E
ℎ
We now estimate each term from 𝐴 1 to 𝐴 7 separately. The
convection term 𝐴 1 can be rewritten as follows:
̂ 𝑛+1 )
𝐴 1 = 4𝜏N (𝛿𝛿u (𝑡𝑛 ) , u (𝑡𝑛 ) , 𝛿E
− 4𝜏N (𝛿𝛿u (𝑡
) , u (𝑡
󵄩󵄩
󵄩2
󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 𝑑𝑡.
= −4𝜏N (2𝛿u𝑛ℎ − 𝛿u𝑛−1
ℎ ,
𝐴 7 := 4𝜅𝜇2 𝜏
𝑛+1
𝑛
𝐴 1,5 + 𝐴 1,6
̂ 𝑛+1 ⟩ ,
, 𝛿E
ℎ
𝑛+1
𝜇
̂ 𝑛+1 ̂ 𝑛+1
We note N(2u𝑛ℎ − u𝑛−1
ℎ , 𝛿Eℎ , 𝛿Eℎ ) = 0 which comes from
(30). Then we obtain
̂ 𝑛+1 ⟩ ,
𝐴 5 := 4𝜇𝜏 ⟨∇𝛿𝑞ℎ𝑛 , 𝛿E
ℎ
𝐴 6 := 4𝜏 ⟨𝛿R
𝐶𝜏2 (𝜏2 + ℎ2 )
𝑡𝑛−1
̂ 𝑛+1 ⟩ ,
𝐴 4 := −4𝜏 ⟨∇𝛿𝜀ℎ𝑛 , 𝛿E
ℎ
𝑛+1
𝐶𝜏 󵄩󵄩
󵄩2
󵄩󵄩2𝛿E𝑛 − 𝛿E𝑛−1 󵄩󵄩󵄩
󵄩
󵄩0
𝜇
×∫
(94)
(97)
𝑛−2
− 4𝜏N (2u𝑛−1
ℎ − uℎ ,
̂ 𝑛+1 )
𝛿G𝑛+1 , 𝛿E
ℎ
= 4𝜏N (2𝛿E𝑛 − 𝛿E𝑛−1 − 2𝛿u (𝑡𝑛 )
̂ 𝑛+1
) , 𝛿E
)
̂ 𝑛+1 , 𝛿E
̂ 𝑛+1 )
+ 𝛿u (𝑡𝑛−1 ) , G𝑛+1 + E
ℎ
ℎ
̂ 𝑛+1 )
+ 4𝜏N (2E𝑛−1 − E𝑛−2 , u (𝑡𝑛 ) , 𝛿E
̂ 𝑛+1 )
− 4𝜏N (2E𝑛 − E𝑛−1 , u (𝑡𝑛+1 ) , 𝛿E
(95)
+ 4𝜏N (2E𝑛−1 − E𝑛−2
− 2u (𝑡𝑛−1 ) + u (𝑡𝑛−2 ) ,
𝑛−2 ̂ 𝑛 ̂ 𝑛+1
)
+ 4𝜏N (2u𝑛−1
ℎ − uℎ , E , 𝛿E
̂ 𝑛+1 , 𝛿E
̂ 𝑛+1 ) .
− 4𝜏N (2u𝑛ℎ − u𝑛−1
ℎ ,E
̂ 𝑛+1 )
𝛿G𝑛+1 , 𝛿E
ℎ
:= 𝐵1 + 𝐵2 .
(98)
12
Journal of Applied Mathematics
To attack 𝐵1 , we first note Lemma 5 which is, for any wℎ ∈ Vℎ ,
‖wℎ ‖L3 (Ω) ≤ 𝐶ℎ−𝑑/6 ‖wℎ ‖0 . If we apply
󵄩
󵄩󵄩 ̂ 𝑛+1
󵄩󵄩E + G𝑛+1 󵄩󵄩󵄩
󵄩0
󵄩
󵄩
󵄩 ̂ 𝑛+1
+ √𝜏 + ℎ 󵄩󵄩󵄩󵄩E
+ G𝑛+1 󵄩󵄩󵄩󵄩1
Integral by parts leads to
̂ 𝑛+1 ⟩
𝐴 3 = 4𝜏 ⟨𝛿𝛿𝑃ℎ𝑛+1 , ∇ ⋅ 𝛿E
ℎ
≤
(99)
≤ 𝐶 (𝜏 + ℎ)
×∫
which is the result of Lemma 13, then we can conclude, in light
of (34),
󵄩
󵄩
𝐵1 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩2𝛿E𝑛 − 𝛿E𝑛−1 󵄩󵄩󵄩󵄩L3 (Ω)
󵄩 󵄩󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩 ̂ 𝑛+1
𝑛+1 󵄩
󵄩󵄩 󵄩󵄩∇𝛿Eℎ 󵄩󵄩
× 󵄩󵄩󵄩󵄩E
ℎ +G
󵄩1 󵄩
󵄩0
󵄩󵄩
󵄩
+ 𝐶𝜏󵄩󵄩󵄩2𝛿u(𝑡𝑛 ) − 𝛿u(𝑡𝑛−1 )󵄩󵄩󵄩󵄩2
󵄩 ̂ 𝑛+1
𝑛+1 󵄩
𝑛+1 󵄩
󵄩󵄩 󵄩󵄩󵄩∇𝛿E
󵄩
× 󵄩󵄩󵄩󵄩E
󵄩󵄩0 󵄩󵄩 ̂ ℎ 󵄩󵄩󵄩0
ℎ +G
(100)
𝜇𝜏 󵄩󵄩 ̂ 𝑛+1 󵄩󵄩2 𝐶𝜏
𝜏
󵄩󵄩∇𝛿Eℎ 󵄩󵄩 +
≤
(1 + )
󵄩0
6󵄩
𝜇
ℎ
󵄩2
󵄩
× 󵄩󵄩󵄩󵄩2𝛿E𝑛 − 𝛿E𝑛−1 󵄩󵄩󵄩󵄩0
𝑡
𝐶𝜏2
󵄩
󵄩2
(𝜏 + ℎ)2 ∫ 󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 𝑑𝑡.
𝜇
𝑡𝑛−2
𝜇𝜏 󵄩󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩󵄩∇𝛿Eℎ 󵄩󵄩 + 𝐶𝜏2 ℎ2
󵄩0
6󵄩
×∫
𝑡
𝑛+1
𝑡𝑛−1
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
𝐴 2 = − 2 ⟨3𝛿G
− 4𝛿G
+ 𝐶ℎ4 ∫
𝑡𝑛−2
=−
≤−
8𝜏2
⟨∇𝛿𝜀ℎ𝑛 , ∇ (𝛿𝛿𝜀ℎ𝑛+1 − 𝛿𝛿𝑃ℎ𝑛+1 )⟩
3
4𝜏2 󵄩󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩∇𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩∇𝛿𝜀ℎ𝑛 󵄩󵄩󵄩0
3 󵄩
󵄩2
󵄩
−󵄩󵄩󵄩󵄩∇𝛿𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
(104)
󵄩
󵄩2
+󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
󵄩󵄩2
4𝜏2 󵄩󵄩󵄩
3
󵄩󵄩∇𝛿𝛿𝑃ℎ𝑛+1 + ∇𝛿𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
󵄩󵄩0
󵄩
3 󵄩
2𝜏
󵄩
󵄩2
≤ 𝐶𝜏2 󵄩󵄩󵄩󵄩∇𝛿𝛿𝑃ℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩2
󵄩
+ 4󵄩󵄩󵄩󵄩∇𝛿𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩󵄩0 .
=
(105)
So we arrive at
4𝜏2 󵄩󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩∇𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩∇𝛿𝜀ℎ𝑛 󵄩󵄩󵄩0 )
3
󵄩2
󵄩
󵄩
󵄩2
+ 4󵄩󵄩󵄩󵄩∇𝛿𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶𝜏3 󵄩󵄩󵄩∇𝛿𝜀ℎ𝑛 󵄩󵄩󵄩0
𝐴4 ≤ −
𝑡𝑛+1
(106)
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡𝑡 (𝑡)󵄩󵄩󵄩2 +󵄩󵄩󵄩𝑝𝑡𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
𝐴 5 = − 4𝜇𝜏 ⟨𝛿𝑞ℎ𝑛 , 𝛿𝛿𝑞ℎ𝑛+1 ⟩
󵄩2 󵄩 󵄩2
󵄩
= − 2𝜇𝜏 (󵄩󵄩󵄩󵄩𝛿𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩𝛿𝑞ℎ𝑛 󵄩󵄩󵄩0
(102)
󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2
4𝜏2 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩
󵄩0
3 󵄩
Invoking (9), (37) and (60) lead to
𝑛
𝜇𝜏 󵄩󵄩
󵄩󵄩2
󵄩󵄩∇𝛿E𝑛+1
󵄩
ℎ 󵄩
󵄩0
6󵄩
𝑡𝑛+1
𝑛+1
𝐴 4 = −4𝜏 ⟨∇𝛿𝜀ℎ𝑛 , 𝛿E𝑛+1
ℎ + ∇𝛿𝛿𝜓ℎ ⟩
𝑡𝑛−1
̂ 𝑛+1 ⟩
+𝛿G𝑛−1 , 𝛿E
ℎ
≤
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
̂ 𝑛+1 = 𝛿E𝑛+1 +
In order to tackle 𝐴 4 , marking use of 𝛿E
ℎ
ℎ
𝑛+1
∇𝛿𝛿𝜓ℎ , we readily get
+ 𝐶𝜏4 ∫
In light of Hölder inequality, (51) yields
𝑛+1
𝑡𝑛−1
If we now apply inequality (𝑎 + 𝑏)2 ≤ 4𝑎2 + (4/3)𝑏2 , then we
can have
≤
We now estimate 𝐵2 using ‖2E𝑛−1 − E𝑛−2 ‖L3 (Ω)
−𝑑/6
𝑛−1
𝑛−2
𝐶ℎ
‖2E − E ‖0 ≤ 𝑀:
󵄩
󵄩
𝐵2 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩2E𝑛−1 − E𝑛−2 󵄩󵄩󵄩󵄩L3 (Ω)
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩
󵄩
× 󵄩󵄩󵄩󵄩𝛿G𝑛+1 󵄩󵄩󵄩󵄩1 󵄩󵄩󵄩󵄩∇𝛿E
ℎ 󵄩
󵄩0
󵄩󵄩
󵄩
𝑛−1
+ 𝐶𝜏󵄩󵄩󵄩2u(𝑡 ) − u(𝑡𝑛−2 )󵄩󵄩󵄩󵄩1
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩
󵄩
× 󵄩󵄩󵄩󵄩𝛿G𝑛+1 󵄩󵄩󵄩󵄩1 󵄩󵄩󵄩󵄩∇𝛿E
ℎ 󵄩
󵄩0
(101)
󵄩󵄩 𝑛+1 󵄩󵄩2 𝜇𝜏 󵄩󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩
󵄩
󵄩
󵄩
≤ 𝐶𝜏󵄩󵄩𝛿G 󵄩󵄩1 +
󵄩∇𝛿Eℎ 󵄩󵄩0
6󵄩
≤
𝑡𝑛+1
(103)
󵄩2
󵄩
󵄩
󵄩2
+ 𝐶𝜏3 󵄩󵄩󵄩∇𝛿𝜀ℎ𝑛 󵄩󵄩󵄩0 + 𝐶𝜏󵄩󵄩󵄩󵄩∇𝛿𝛿𝑃ℎ𝑛+1 󵄩󵄩󵄩󵄩0 .
𝑛
+
𝜇𝜏 󵄩󵄩 ̂ 𝑛+1 󵄩󵄩2 𝐶𝜏4
󵄩󵄩∇𝛿Eℎ 󵄩󵄩 +
󵄩0
6󵄩
𝜇
󵄩2
󵄩
−󵄩󵄩󵄩󵄩𝛿𝛿𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
󵄩
󵄩2 󵄩 󵄩2
≤ − 2𝜇𝜏 (󵄩󵄩󵄩󵄩𝛿𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 − 󵄩󵄩󵄩𝛿𝑞ℎ𝑛 󵄩󵄩󵄩0 )
󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩.
+ 2𝜇𝜏󵄩󵄩󵄩󵄩∇𝛿E
ℎ 󵄩
󵄩0
(107)
Journal of Applied Mathematics
13
Finally, the 𝐴 6 term becomes
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩
󵄩
𝐴 6 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩𝛿R𝑛+1 󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩𝛿E
ℎ 󵄩
󵄩0
󵄩󵄩 ̂ 𝑛+1 󵄩󵄩2
≤ 𝐶𝜏󵄩󵄩󵄩𝛿Eℎ 󵄩󵄩󵄩0
+ 𝐶𝜏4 ∫
𝑡𝑛+1
𝑡𝑛−1
To evaluate errors of 𝛿𝜗ℎ𝑛+1 , we subtract two consecutive
formulas (83) and choose by 𝜙ℎ = 4𝜏𝛿𝜗ℎ𝑛+1 = 4𝜏(𝛿𝜗𝑛+1 −
𝛿𝜂𝑛+1 ) as follows:
󵄩
󵄩2
󵄩2
󵄩
𝛿󵄩󵄩󵄩󵄩𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝛿󵄩󵄩󵄩󵄩2𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩2
󵄩2
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 4𝜆𝜇𝜏󵄩󵄩󵄩󵄩∇𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
(108)
󵄩2
󵄩󵄩
󵄩󵄩u𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0 𝑑𝑡.
= 𝐴 8 + 𝐴 9,
For the new term 𝐴 7 , we argue as in Lemma 13 to arrive at
𝐴 7 ≤ 𝐶𝜅2 𝜇4 𝜏
where
𝐴 8 := 4𝜏N (u (𝑡𝑛 ) , 𝜃 (𝑡𝑛 ) , 𝛿𝜗ℎ𝑛+1 )
󵄩2
󵄩
× (󵄩󵄩󵄩󵄩2𝛿𝜗ℎ𝑛 − 𝛿𝜗ℎ𝑛−1 󵄩󵄩󵄩󵄩0
󵄩2
󵄩
+󵄩󵄩󵄩󵄩2𝛿𝜂𝑛 − 𝛿𝜂𝑛−1 󵄩󵄩󵄩󵄩0 )
󵄩󵄩2
󵄩
󵄩 +
+ 𝜏󵄩󵄩󵄩󵄩𝛿E𝑛+1
ℎ 󵄩
󵄩0
+ 𝐶𝜏4 ∫
𝑡𝑛 +1
𝑡𝑛−2
− 4𝜏N (u𝑛ℎ , 𝜃ℎ𝑛 , 𝛿𝜗ℎ𝑛+1 )
(109)
1 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
󵄩
󵄩0
2
+
4𝜏
3
− 4𝜏N (u (𝑡𝑛+1 ) , 𝜃 (𝑡𝑛+1 ) , 𝛿𝜗ℎ𝑛+1 )
𝑛+1
𝑛+1
+ 4𝜏N (u𝑛+1
ℎ , 𝜃ℎ , 𝛿𝜗ℎ ) ,
+𝛿𝜂𝑛−1 , 𝛿𝜗ℎ𝑛+1 ⟩
+ 4𝜏 ⟨𝛿𝑄𝑛+1 , 𝛿𝜗ℎ𝑛+1 ⟩ .
We treat the 𝐴 8 term first by rewriting it as follows:
𝐴 8 = − 4𝜏N (E𝑛+1 , 𝜃 (𝑡𝑛+1 ) , 𝛿𝜗ℎ𝑛+1 )
𝑛+1
, 𝛿𝜗ℎ𝑛+1 )
− 4𝜏N (u𝑛+1
ℎ ,𝜗
󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩
󵄩
󵄩0
+ 4𝜏N (E𝑛 , 𝜃 (𝑡𝑛 ) , 𝛿𝜗ℎ𝑛+1 )
𝐶𝜏 󵄩󵄩
󵄩󵄩2
󵄩
≤
(󵄩󵄩2𝛿E𝑛ℎ − 𝛿E𝑛−1
ℎ 󵄩
󵄩0
𝜇 󵄩
4
:= ∑𝐴 8,𝑖 .
𝑖=1
2
4𝜏 󵄩󵄩
𝑛 󵄩2
2 4
󵄩∇𝛿𝜀ℎ 󵄩󵄩󵄩0 + 𝐶𝜅 𝜇 𝜏
3 󵄩
󵄩2
󵄩
× (󵄩󵄩󵄩󵄩2𝛿𝜗ℎ𝑛 − 𝛿𝜗ℎ𝑛−1 󵄩󵄩󵄩󵄩0
(110)
󵄩2
󵄩
+ 󵄩󵄩󵄩󵄩2𝛿𝜂𝑛 − 𝛿𝜂𝑛−1 󵄩󵄩󵄩󵄩0 )
𝑡𝑛+1
𝑡𝑛−1
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0 + 󵄩󵄩󵄩u𝑡𝑡 (𝑡)󵄩󵄩󵄩2
󵄩
󵄩2 󵄩
󵄩2
+ 󵄩󵄩󵄩𝑝𝑡𝑡 (𝑡)󵄩󵄩󵄩1 +󵄩󵄩󵄩𝜃𝑡𝑡 (𝑡)󵄩󵄩󵄩0 ) 𝑑𝑡
+ 𝐶𝜏
×∫
3󵄩
󵄩
𝑛 󵄩2
󵄩󵄩∇𝛿𝜀ℎ 󵄩󵄩󵄩0
𝑡𝑛+1
𝑡𝑛−2
2
2 2
+ 𝐶(𝜏 + ℎ )
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡𝑡 (𝑡)󵄩󵄩󵄩0 + 󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2
󵄩
󵄩2
+󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
In estimating convection terms, we will use Lemma 10
frequently without notice. We recall ‖𝜃(𝑡)‖2 ≤ 𝐶 and ‖E𝑛 ‖0 ≤
𝐶(𝜏 + ℎ) which is the result in Lemma 13 to obtain
𝐴 8,1 + 𝐴 8,3
= −4𝜏N (𝛿E𝑛+1 , 𝜃 (𝑡𝑛+1 ) , 𝛿𝜗ℎ𝑛+1 )
󵄩
󵄩󵄩2
󵄩 + 2𝜇𝜏󵄩󵄩󵄩󵄩𝛿𝑞ℎ𝑛 󵄩󵄩󵄩󵄩20 + 𝐶𝜏4
+ 𝜏󵄩󵄩󵄩󵄩𝛿E𝑛+1
ℎ 󵄩
󵄩0
×∫
(113)
4𝜏N (u𝑛ℎ , 𝜗𝑛 , 𝛿𝜗ℎ𝑛+1 )
+
󵄩2
󵄩
+ 󵄩󵄩󵄩󵄩2𝛿G𝑛 − 𝛿G𝑛−1 󵄩󵄩󵄩󵄩0 )
+
(112)
𝐴 9 := − 2 ⟨3𝛿𝜂𝑛+1 − 4𝛿𝜂𝑛
󵄩󵄩
󵄩2
󵄩󵄩𝜃𝑡𝑡 (𝑡)󵄩󵄩󵄩0 𝑑𝑡.
We now insert the above estimates into (93) to obtain
2
󵄩󵄩2
󵄩
󵄩
𝑛󵄩
󵄩 + 𝛿󵄩󵄩󵄩2𝛿E𝑛+1
󵄩󵄩
𝛿󵄩󵄩󵄩󵄩𝛿E𝑛+1
ℎ 󵄩
ℎ − 𝛿Eℎ 󵄩
󵄩0
󵄩0
󵄩
󵄩󵄩2
󵄩
󵄩 ̂ 𝑛+1 󵄩󵄩2
󵄩 + 𝜇𝜏󵄩󵄩󵄩∇𝛿E
󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿𝛿E𝑛+1
ℎ 󵄩
ℎ 󵄩
󵄩0
󵄩0
󵄩
2
2
󵄩
󵄩
󵄩󵄩
󵄩
+ 󵄩󵄩󵄩∇𝛿𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 2𝜇𝜏󵄩󵄩󵄩󵄩𝛿𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0
2
(111)
− 4𝜏N (E𝑛 , 𝛿𝜃 (𝑡𝑛+1 ) , 𝛿𝜗ℎ𝑛+1 )
󵄩󵄩
󵄩
󵄩
≤ 𝐶𝜏 (󵄩󵄩󵄩󵄩𝛿E𝑛+1 󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩𝜃(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2
󵄩
󵄩 󵄩
󵄩 󵄩󵄩
+󵄩󵄩󵄩E𝑛 󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩𝛿𝜃(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 ) 󵄩󵄩󵄩󵄩𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩1
󵄩2
󵄩
≤ 𝜆𝜇𝜏󵄩󵄩󵄩󵄩∇𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
+
𝐶𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2 𝐶𝜏2 (𝜏 + ℎ)2
󵄩󵄩𝛿E 󵄩󵄩 +
󵄩0
𝜆𝜇 󵄩
𝜆𝜇
𝑡𝑛+1
×∫
𝑡𝑛
󵄩󵄩
󵄩2
󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩2 𝑑𝑡.
(114)
14
Journal of Applied Mathematics
𝑛+1
𝑛+1
In conjunction with N(u𝑛+1
ℎ , 𝛿𝜗ℎ , 𝛿𝜗ℎ ) = 0 which comes
from (30), we rewrite 𝐴 8,2 + 𝐴 8,4 as
𝐴 8,2 + 𝐴 8,4
𝑛+1
= 4𝜏N (E𝑛+1
ℎ − Uℎ ,
𝛿𝜂𝑛+1 , 𝛿𝜗ℎ𝑛+1 )
𝑛+1
+ 4𝜏N (𝛿E𝑛+1
ℎ − 𝛿Uℎ ,
(115)
Inserting the above estimates into (111) yields
󵄩2
󵄩2
󵄩
󵄩
𝛿󵄩󵄩󵄩󵄩𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝛿󵄩󵄩󵄩󵄩2𝛿𝜗ℎ𝑛+1 − 𝛿𝜗ℎ𝑛 󵄩󵄩󵄩󵄩0
󵄩2
󵄩2
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝜆𝜇𝜏󵄩󵄩󵄩󵄩∇𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩2
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
+
𝐶𝜏 󵄩󵄩 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛 󵄩󵄩2
(󵄩󵄩𝛿E 󵄩󵄩 + 󵄩𝛿E 󵄩
𝜆𝜇 󵄩 ℎ 󵄩0 󵄩 ℎ 󵄩0
2
󵄩
󵄩
+󵄩󵄩󵄩󵄩𝛿G𝑛+1 󵄩󵄩󵄩󵄩0 ) + 𝐶(𝜏2 + ℎ2 )
𝜗𝑛 , 𝛿𝜗ℎ𝑛+1 )
:= 𝐵3 + 𝐵4 .
×∫
𝑛+1
Because we can readily get ‖E𝑛+1
ℎ − Uℎ ‖L3 (Ω) ≤ 𝑀 via
Lemmas 5 and 13, we conclude, in conjunction with (60),
×∫
𝑡𝑛+1
𝑡𝑛
𝑡𝑛−2
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1
󵄩
󵄩 2 󵄩 󵄩2
+󵄩󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝜃𝑡𝑡𝑡 󵄩󵄩󵄩0 ) 𝑑𝑡.
We now estimate optimal accuracy for velocity and
temperature.
(116)
󵄩󵄩
󵄩2
󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩2 𝑑𝑡.
Before we attack 𝐵4 , we note that the assumption 𝜏 ≤ 𝐶ℎ is
required to apply ‖𝛿E𝑛ℎ ‖L3 (Ω) ≤ 𝐶𝜏−1/2 ‖𝛿E𝑛ℎ ‖0 . We now imply
2
𝑡𝑛+1
Adding 2 equations (110) and (119) and then summing up 𝑛
from 2 to 𝑁 lead up to (92) and complete the proof.
󵄩
𝑛+1 󵄩
󵄩󵄩
𝐵3 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩E𝑛+1
ℎ − Uℎ 󵄩
󵄩L3 (Ω)
󵄩󵄩
󵄩
󵄩
× 󵄩󵄩󵄩󵄩𝛿𝜂𝑛+1 󵄩󵄩󵄩󵄩1 󵄩󵄩󵄩󵄩𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩1
󵄩2 𝐶𝜏2 ℎ2
󵄩
≤ 𝜆𝜇𝜏󵄩󵄩󵄩󵄩∇𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 +
𝜆𝜇
2
‖𝜗𝑛 ‖0 + 𝜏‖𝜗𝑛 ‖1 ≤ 𝐶(𝜏2 + ℎ2 ) and (51) to get
𝐵4 = 4𝜏N (𝛿G𝑛+1 − 𝛿u (𝑡𝑛+1 )
𝑛
𝑛+1
+𝛿E𝑛+1
ℎ , 𝜗 , 𝛿𝜗ℎ )
Lemma 17 (full rate of convergence for velocity and temperature). One denotes that (v𝑛+1 , 𝑟𝑛+1 ) and (vℎ𝑛+1 , 𝑟ℎ𝑛+1 ) are the
solutions of (16) and (24) with z = E𝑛+1
ℎ , respectively. And let
𝜔𝑛+1 and 𝜔ℎ𝑛+1 be solutions of (17) and (28) with 𝜉 = 𝜗ℎ𝑛+1 .
Let the exact solution of (1) be smooth enough and 𝜏 ≤ 𝐶ℎ. If
Assumptions 1 and 3–5 hold, then one has
󵄩󵄩 𝑁+1 󵄩󵄩2
󵄩󵄩∇kℎ 󵄩󵄩
󵄩0
󵄩
󵄩2
󵄩
󵄩2 󵄩
+ 󵄩󵄩󵄩󵄩∇ (2kℎ𝑁+1 − kℎ𝑁)󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩∇𝜔ℎ𝑁+1 󵄩󵄩󵄩󵄩0
󵄩
󵄩2
+ 󵄩󵄩󵄩󵄩∇ (2𝜔ℎ𝑁+1 − 𝜔ℎ𝑁)󵄩󵄩󵄩󵄩0
𝑁
󵄩2
󵄩
+ ∑ (󵄩󵄩󵄩󵄩∇𝛿𝛿kℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩
󵄩 󵄩
󵄩󵄩
󵄩3 )
≤ 𝐶𝜏 ((󵄩󵄩󵄩󵄩𝛿G𝑛+1 󵄩󵄩󵄩󵄩1 + 󵄩󵄩󵄩󵄩𝛿E𝑛+1
ℎ 󵄩
󵄩L (Ω)
𝑛=1
󵄩2
󵄩
+󵄩󵄩󵄩󵄩∇𝛿𝛿𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
󵄩
󵄩󵄩 󵄩 󵄩
󵄩 󵄩 󵄩
× 󵄩󵄩󵄩𝜗𝑛 󵄩󵄩󵄩1 + 󵄩󵄩󵄩𝛿u(𝑡𝑛 )󵄩󵄩󵄩2 󵄩󵄩󵄩𝜗𝑛 󵄩󵄩󵄩0 ) 󵄩󵄩󵄩󵄩𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩1
≤
𝐶𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩2
󵄩
󵄩󵄩𝛿E 󵄩󵄩 + 𝜆𝜇𝜏󵄩󵄩󵄩∇𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩
󵄩0
󵄩
𝜆𝜇 󵄩 ℎ 󵄩0
+
(117)
𝑁
𝑛=1
󵄩2
󵄩
+𝜆󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
𝜆𝜇
×∫
𝑡𝑛−1
≤ 𝐶 (𝜏4 + ℎ4 ) .
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
Proof. We choose wℎ = 4𝜏kℎ𝑛+1 ∈ Vℎ in (66) to obtain
󵄩2
󵄩
󵄩2
󵄩
𝛿󵄩󵄩󵄩󵄩∇vℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝛿󵄩󵄩󵄩󵄩∇ (2vℎ𝑛+1 − vℎ𝑛 )󵄩󵄩󵄩󵄩0
󵄩2
󵄩󵄩2
󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩∇𝛿𝛿vℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 4𝜇𝜏󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
In light of Hölder inequality, 𝐴 9 becomes
󵄩2
󵄩
𝐴 9 ≤ C𝜏󵄩󵄩󵄩󵄩𝛿𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶 (𝜏4 + ℎ4 )
×∫
𝑡𝑛+1
𝑡𝑛−2
󵄩
󵄩 2 󵄩 󵄩2
(󵄩󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝜃𝑡𝑡𝑡 󵄩󵄩󵄩0 ) 𝑑𝑡.
(120)
󵄩󵄩2
󵄩
󵄩
+ 2𝜇𝜏 ∑ (󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
𝐶𝜏2 (𝜏2 + ℎ2 )
𝑡𝑛+1
(119)
(118)
5
= ∑𝐴 𝑖 ,
𝑖=1
(121)
Journal of Applied Mathematics
15
In light of 𝜀ℎ𝑛+1 = 𝑃ℎ𝑛+1 + 3𝜓ℎ𝑛+1 /2𝜏, we can obtain
where
𝑛+1
̂ 𝑛+1
𝐴 1 := 4𝜏N (2u𝑛ℎ − u𝑛−1
ℎ ,u
ℎ , kℎ )
𝑛+1
− 4𝜏N (u (𝑡
𝐴 2 := ⟨3G
𝑛+1
𝑛+1
) , u (𝑡
󵄩2
󵄩󵄩
󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩
󵄩0
󵄩
) , kℎ𝑛+1 ) ,
− 4G + G𝑛−1 , kℎ𝑛+1 ⟩ ,
4𝜏2 󵄩󵄩
󵄩2
󵄩󵄩∇ (𝛿𝜀ℎ𝑛+1 − 𝛿𝑃ℎ𝑛+1 )󵄩󵄩󵄩
󵄩
󵄩0
9
󵄩
󵄩2
≤ 𝐶𝜏2 󵄩󵄩󵄩󵄩∇𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩󵄩0
𝑛
̂ 𝑛+1 , ∇𝑟𝑛+1 ⟩ ,
𝐴 3 := 4𝜇𝜏 ⟨E
ℎ
ℎ
=
(122)
𝐴 4 := 4𝜏 ⟨R𝑛+1 , kℎ𝑛+1 ⟩ ,
𝐴 5 := 4𝜅𝜇2 𝜏 ⟨g (𝜃 (𝑡𝑛+1 ) − 2𝜃ℎ𝑛 + 𝜃ℎ𝑛−1 ) , kℎ𝑛+1 ⟩ .
+ 𝐶𝜏3 ∫
We now estimate all the terms from 𝐴 1 to 𝐴 5 , respectively.
The convection term 𝐴 1 can be rewritten as follows:
𝑛+1
𝐴 1 = − 4𝜏N (𝛿𝛿u (𝑡
𝑛
𝑛+1
) , u (𝑡
𝑛−1
− 4𝜏N (2u (𝑡 ) − u (𝑡
) , kℎ𝑛+1 )
̂ 𝑛+1
),E
, kℎ𝑛+1 )
̂ 𝑛+1 , k𝑛+1 )
+ 4𝜏N (2E𝑛 − E𝑛−1 , E
ℎ
𝑡𝑛
(123)
𝜇𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2
(󵄩󵄩E 󵄩󵄩 + 󵄩󵄩G 󵄩󵄩󵄩0 )
2 󵄩 ℎ 󵄩0 󵄩
󵄩
󵄩2
+ 𝐶𝜇𝜏3 󵄩󵄩󵄩󵄩∇𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶𝜇𝜏4
− 4𝜏N (2E𝑛 − E𝑛−1 , u (𝑡𝑛+1 ) , kℎ𝑛+1 ) ,
×∫
𝑡𝑛+1
𝑡𝑛−1
×∫
+
(125)
≤
𝐶𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩󵄩∇k 󵄩󵄩 .
+
𝜇 󵄩 ℎ 󵄩0
+
𝐶𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩󵄩∇k 󵄩󵄩 .
𝜇 󵄩 ℎ 󵄩0
(128)
𝐶𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩󵄩∇k 󵄩󵄩 .
𝜇 󵄩 ℎ 󵄩0
󵄩
󵄩
𝐴 1,3 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩2E𝑛 − E𝑛−1 󵄩󵄩󵄩󵄩0
󵄩 ̂ 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛+1 󵄩󵄩
󵄩󵄩 󵄩󵄩k 󵄩󵄩
× 󵄩󵄩󵄩󵄩E
󵄩1 󵄩
󵄩2
󵄩
󵄩󵄩 𝑛
+ 𝐶𝜏󵄩󵄩󵄩2E − E𝑛−1 󵄩󵄩󵄩󵄩L3 (Ω)
󵄩
󵄩 ̂ 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛+1
󵄩󵄩 󵄩󵄩k − kℎ𝑛+1 󵄩󵄩󵄩
× 󵄩󵄩󵄩󵄩E
󵄩1 󵄩
󵄩1
󵄩󵄩 ̂ 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛+1 󵄩󵄩
≤ 𝐶𝜏 (𝜏 + ℎ) 󵄩󵄩󵄩E 󵄩󵄩󵄩1 󵄩󵄩󵄩k 󵄩󵄩󵄩2
󵄩 ̂ 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛+1 󵄩󵄩
󵄩󵄩 󵄩󵄩k 󵄩󵄩
+ 𝐶𝜏ℎ󵄩󵄩󵄩󵄩E
󵄩1 󵄩
󵄩2
󵄩
󵄩󵄩
󵄩
× 󵄩󵄩󵄩󵄩u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 󵄩󵄩󵄩󵄩kℎ𝑛+1 󵄩󵄩󵄩󵄩1
Because ∇ ⋅ (2u(𝑡𝑛 ) − u(𝑡𝑛−1 )) = 0 and 2u(𝑡𝑛 ) − u(𝑡𝑛−1 ) = 0 on
boundary, we can use (32) and so we get
󵄩
󵄩
𝐴 1,2 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩2u(𝑡𝑛 ) − u(𝑡𝑛−1 )󵄩󵄩󵄩󵄩2
󵄩 ̂ 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛+1 󵄩󵄩
󵄩󵄩 󵄩󵄩kℎ 󵄩󵄩
× 󵄩󵄩󵄩󵄩E
󵄩0 󵄩
󵄩1
𝜇𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2
≤
(󵄩󵄩E 󵄩󵄩 + 󵄩󵄩G 󵄩󵄩󵄩0
(126)
2 󵄩 ℎ 󵄩0 󵄩
󵄩2
󵄩
+󵄩󵄩󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩󵄩0 )
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡
If we apply ‖2E𝑛 − E𝑛−1 ‖L3 (Ω) ≤ 𝐶ℎ−𝑑/6 ‖2E𝑛 − E𝑛−1 ‖0 ≤
𝐶ℎ−𝑑/6 (𝜏 + ℎ) which can be derived from Lemmas 5 and 13,
then we can get, by the help of Lemma 10 and Assumption 1,
󵄩2
󵄩󵄩
󵄩󵄩u𝑡𝑡 (𝑡)󵄩󵄩󵄩0 𝑑𝑡,
𝜇𝜏 󵄩󵄩 𝑛 󵄩󵄩2 󵄩󵄩 𝑛 󵄩󵄩2
(󵄩E 󵄩 + 󵄩G 󵄩
2 󵄩 ℎ 󵄩0 󵄩 󵄩0
󵄩2 󵄩
󵄩2
󵄩
+󵄩󵄩󵄩𝛿E𝑛ℎ 󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝛿G𝑛 󵄩󵄩󵄩0 )
𝑡𝑛+1
𝑡𝑛
󵄩
󵄩 󵄩 󵄩
𝐴 1,4 ≤ 𝐶𝜏 (󵄩󵄩󵄩E𝑛 󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝛿E𝑛 󵄩󵄩󵄩0 )
≤
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡,
and so we can conclude
𝐴 1,2 ≤
and we denote by 𝐴 1,𝑖 , for 𝑖 = 1, 2, . . . , 4, the four terms in
the right-hand side. To estimate convection terms, we will use
frequently Lemma 10 without notice. Using ‖u(𝑡𝑛+1 )‖2 ≤ 𝑀,
we can readily get
󵄩
󵄩
𝐴 1,1 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩𝛿𝛿u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩0
󵄩
󵄩
󵄩󵄩
× 󵄩󵄩󵄩󵄩u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 󵄩󵄩󵄩󵄩kℎ𝑛+1 󵄩󵄩󵄩󵄩1
(124)
󵄩2
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩∇kℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶𝜇𝜏4
𝑡𝑛+1
(127)
(129)
𝐶
𝜏(𝜏 + ℎ)2
𝜇
󵄩 ̂ 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩 + 󵄩󵄩∇G 󵄩󵄩 )
× (󵄩󵄩󵄩󵄩∇E
ℎ 󵄩
󵄩0 󵄩
󵄩0
+
𝜇𝜏 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩󵄩E 󵄩󵄩 .
2 󵄩 ℎ 󵄩0
In conjunction with (51), we can have
𝐶 󵄩󵄩 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛 󵄩󵄩2
󵄩2
󵄩
(󵄩󵄩󵄩𝛿G 󵄩󵄩󵄩0 + 󵄩󵄩𝛿G 󵄩󵄩0 ) + 𝐶𝜏󵄩󵄩󵄩󵄩kℎ𝑛+1 󵄩󵄩󵄩󵄩0
𝜏
󵄩2
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩kℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶ℎ4
𝐴2 ≤
×∫
𝑡𝑛+1
𝑡𝑛−1
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡.
(130)
16
Journal of Applied Mathematics
󵄩2
󵄩
+ 𝐶𝜅2 𝜇4 𝜏󵄩󵄩󵄩󵄩2𝜗𝑛 − 𝜗𝑛−1 󵄩󵄩󵄩󵄩0
̂ 𝑛+1 = E𝑛+1 + ∇𝛿𝜓𝑛+1 and Assumption 1 give
The definition E
ℎ
ℎ
ℎ
us
𝐶
𝜏(𝜏 + ℎ)2
𝜇
󵄩 ̂ 𝑛+1 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2
󵄩 + 󵄩󵄩∇G 󵄩󵄩 )
× (󵄩󵄩󵄩󵄩∇E
ℎ 󵄩
󵄩0 󵄩
󵄩0
+
𝐴 3 = 4𝜇𝜏 ⟨∇𝛿𝜓ℎ𝑛+1 , ∇𝑟ℎ𝑛+1 ⟩
󵄩
󵄩󵄩2 𝐶𝜏 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜓ℎ𝑛+1 󵄩󵄩󵄩 .
󵄩 +
≤ 𝜇𝜏󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
󵄩0
𝜇 󵄩
(131)
+
+ 𝐶𝜇 (𝜏4 + ℎ4 )
If we apply (127) again, then we arrive at
󵄩󵄩2
󵄩
󵄩
𝐴 3 ≤ 𝜇𝜏󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
𝐶𝜏3 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩
+
󵄩
󵄩0
𝜇
4
𝑡
𝑛+1
𝐶𝜏
+
∫
𝜇 𝑡𝑛
𝑡𝑛−1
(132)
×∫
󵄩󵄩
󵄩2
󵄩󵄩∇𝑝𝑡 (𝑡)󵄩󵄩󵄩0 𝑑𝑡.
𝐴 4 = 4𝜏 ⟨R𝑛+1 , kℎ𝑛+1 ⟩
󵄩2
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩∇kℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶𝜏4
×∫
𝑡𝑛−1
𝑡𝑛+1
(133)
On the other hand, we choose 𝜙ℎ = 4𝜏𝜔ℎ𝑛+1 in (83) to obtain
󵄩
󵄩2
𝛿󵄩󵄩󵄩󵄩∇𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩
󵄩2
+ 𝛿󵄩󵄩󵄩󵄩∇ (2𝜔ℎ𝑛+1 − 𝜔ℎ𝑛 )󵄩󵄩󵄩󵄩0
󵄩2
󵄩
(136)
+ 󵄩󵄩󵄩󵄩∇𝛿𝛿𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩2
󵄩
+ 4𝜆𝜇𝜏󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
= 𝐴 6 + 𝐴 7,
≤
where
𝑛+1
𝑛+1
𝐴 6 := 4𝜏N (u𝑛+1
ℎ , 𝜃ℎ , 𝜔ℎ )
𝐴 5 = 4𝜅𝜇2 𝜏
𝑛
𝑛−1
× ⟨g (𝛿𝛿𝜃 (𝑡 ) + 2𝜗 −𝜗
󵄩2
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩∇kℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶𝜅2 𝜇4 𝜏
󵄩
󵄩2
× 󵄩󵄩󵄩󵄩2𝜗𝑛 − 𝜗𝑛−1 󵄩󵄩󵄩󵄩0 + 𝐶𝜅2 𝜇4 𝜏4
×∫
𝑡𝑛−1
󵄩
󵄩2 󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡𝑡 (𝑡)󵄩󵄩󵄩0 + 󵄩󵄩󵄩u𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩0 ) 𝑑𝑡.
(135)
𝐶𝜏3 ∫𝑡𝑛−1 ‖𝜃𝑡 (𝑡)‖20 𝑑𝑡; whence
𝑡𝑛+1
𝑡𝑛+1
𝑡𝑛−1
󵄩󵄩
󵄩2
󵄩󵄩u𝑡𝑡𝑡 (𝑠)󵄩󵄩󵄩0 𝑑𝑡.
For the new term 𝐴 5 , we observe ‖𝛿𝛿𝜃(𝑡𝑛+1 )‖0
𝑛+1
𝑡𝑛+1
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩u𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝑝𝑡 (𝑡)󵄩󵄩󵄩1 ) 𝑑𝑡
󵄩2
󵄩
+ 𝐶𝜏󵄩󵄩󵄩󵄩∇vℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶𝜏4
×∫
On the other hand, the truncation error term becomes
𝑡𝑛+1
𝐶𝜏3 󵄩󵄩
󵄩2
󵄩󵄩∇𝛿𝜀ℎ𝑛+1 󵄩󵄩󵄩
󵄩
󵄩0
𝜇
− 4𝜏N (u (𝑡𝑛+1 ) , 𝜃 (𝑡𝑛+1 ) , 𝜔ℎ𝑛+1 ) ,
) , kℎ𝑛+1 ⟩
𝐴 7 := − 2 ⟨3𝜂𝑛+1 − 4𝜂𝑛 + 𝜂𝑛−1 , 𝜔ℎ𝑛+1 ⟩
Invoking k0 = 0 and inserting the above estimates from 𝐴 1
and 𝐴 5 into (121) lead us to
󵄩2
󵄩
𝛿󵄩󵄩󵄩󵄩∇vℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩
󵄩2
+ 𝛿󵄩󵄩󵄩󵄩∇ (2vℎ𝑛+1 − vℎ𝑛 )󵄩󵄩󵄩󵄩0
󵄩2
󵄩
+ 󵄩󵄩󵄩󵄩∇𝛿𝛿vℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩󵄩2
󵄩
󵄩
+ 2𝜇𝜏󵄩󵄩󵄩󵄩E𝑛+1
ℎ 󵄩
󵄩0
𝜇𝜏 󵄩󵄩 𝑛 󵄩󵄩2 󵄩󵄩 𝑛+1 󵄩󵄩2
≤
(󵄩E 󵄩 + 󵄩󵄩G 󵄩󵄩󵄩0
2 󵄩 ℎ 󵄩0 󵄩
󵄩2 󵄩
󵄩2
󵄩 󵄩2 󵄩
+󵄩󵄩󵄩G𝑛 󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝛿E𝑛ℎ 󵄩󵄩󵄩0 + 󵄩󵄩󵄩𝛿G𝑛 󵄩󵄩󵄩0 )
+ 4𝜏 ⟨𝑄𝑛+1 , 𝜔ℎ𝑛+1 ⟩ .
(134)
󵄩󵄩
󵄩2
󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩0 𝑑𝑡.
(137)
In order to estimate 𝐴 6 , we note first ‖E𝑛+1 ‖L3 (Ω)
−1/2
𝑛+1
2
‖𝜗𝑛+1 ‖1
2
≤
2
≤ 𝐶((𝜏 + ℎ )/𝜏) which come
𝐶ℎ ‖E ‖0 and
from Lemmas 5 and 13, respectively. Then Lemma 10 and the
assumption 𝜏 ≤ 𝐶ℎ yield
𝐴 6 = − 4𝜏N (E𝑛+1 , 𝜃 (𝑡𝑛+1 ) , 𝜔ℎ𝑛+1 )
+ 4𝜏N (E𝑛+1 − u (𝑡𝑛+1 ) , 𝜗𝑛+1 , 𝜔ℎ𝑛+1 )
󵄩󵄩
󵄩
󵄩󵄩
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩E𝑛+1 󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩𝜃(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 󵄩󵄩󵄩󵄩𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩1
󵄩
󵄩
󵄩󵄩
󵄩
󵄩
+ 𝐶𝜏󵄩󵄩󵄩󵄩E𝑛+1 󵄩󵄩󵄩󵄩L3 (Ω) 󵄩󵄩󵄩󵄩𝜗𝑛+1 󵄩󵄩󵄩󵄩1 󵄩󵄩󵄩󵄩𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩1
󵄩󵄩
󵄩
󵄩
󵄩󵄩
+ 𝐶𝜏󵄩󵄩󵄩󵄩u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 󵄩󵄩󵄩󵄩𝜗𝑛+1 󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩󵄩𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩1
󵄩2
󵄩2
󵄩
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩∇𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝜇𝜏󵄩󵄩󵄩󵄩E𝑛+1 󵄩󵄩󵄩󵄩0
󵄩2 󵄩
󵄩2
󵄩
+ 𝜆𝜇𝜏 (󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩𝜂𝑛+1 󵄩󵄩󵄩󵄩0 ) .
(138)
Journal of Applied Mathematics
𝑛+1
2
Since ‖𝛿𝜂𝑛+1 ‖0 ≤ 𝐶ℎ4 𝜏 ∫𝑛
17
‖𝜃𝑡 (𝑡)‖22 𝑑𝑡, we can readily get
󵄩2
󵄩
𝐴 7 ≤ 𝐶𝜏󵄩󵄩󵄩󵄩∇𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶 (𝜏4 + ℎ4 )
×∫
𝑛+1
𝑛−1
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝜃𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩2 ) 𝑑𝑡.
Inserting the above estimates into (136) yields
󵄩
󵄩2
𝛿󵄩󵄩󵄩󵄩∇𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩
󵄩2
+ 𝛿󵄩󵄩󵄩󵄩∇ (2𝜔ℎ𝑛+1 − 𝜔ℎ𝑛 )󵄩󵄩󵄩󵄩0
󵄩2
󵄩2
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩∇𝛿𝛿𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 3𝜆𝜇𝜏󵄩󵄩󵄩󵄩𝜗ℎ𝑛+1 󵄩󵄩󵄩󵄩0
󵄩2
󵄩2
󵄩
󵄩
≤ 𝐶𝜏󵄩󵄩󵄩󵄩∇𝜔ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 𝜆𝜇𝜏󵄩󵄩󵄩󵄩𝜂𝑛+1 󵄩󵄩󵄩󵄩0
󵄩2
󵄩
+ 𝜇𝜏󵄩󵄩󵄩󵄩E𝑛+1 󵄩󵄩󵄩󵄩0 + 𝐶 (𝜏4 + ℎ4 )
×∫
𝑛+1
𝑛−1
(139)
𝐴1 ≤
≤
𝐶 󵄩󵄩 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛 󵄩󵄩 󵄩󵄩 󵄩󵄩
(󵄩󵄩𝛿E 󵄩󵄩󵄩0 + 󵄩󵄩𝛿E 󵄩󵄩0 ) 󵄩󵄩wℎ 󵄩󵄩0
𝜏 󵄩
𝐶 󵄩󵄩 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛 󵄩󵄩 󵄩󵄩
󵄩
(󵄩󵄩𝛿E 󵄩󵄩󵄩0 + 󵄩󵄩𝛿E 󵄩󵄩0 ) 󵄩󵄩∇wℎ 󵄩󵄩󵄩0 ,
𝜏 󵄩
󵄩 ̂ 𝑛+1 󵄩󵄩 󵄩󵄩
󵄩 󵄩󵄩∇wℎ 󵄩󵄩󵄩󵄩0 .
𝐴 2 ≤ 𝐶󵄩󵄩󵄩󵄩∇E
ℎ 󵄩
󵄩0
(140)
In conjunction with the discrete Gronwall inequality,
adding (135) and (140) and summing over 𝑛 from 1 to 𝑁 lead
to (120).
We now estimate the pressure error in 𝐿2 (0, 𝑇; 𝐿2 (Ω)).
This hinges on the error estimates for the time derivative of
velocity and temperature of Lemma 15.
Lemma 18 (pressure error estimate). Let the exact solution of
(1) be smooth enough and 𝜏 ≤ 𝐶ℎ. If Assumptions 1–5 hold,
then one has
𝑁
(141)
󵄩
󵄩
𝐴 3 ≤ 𝐶󵄩󵄩󵄩󵄩𝛿𝛿u (𝑡𝑛+1 )󵄩󵄩󵄩󵄩0
󵄩
󵄩󵄩 󵄩
×󵄩󵄩󵄩󵄩u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩2 󵄩󵄩󵄩wℎ 󵄩󵄩󵄩1
󵄩
󵄩󵄩
󵄩
≤ 𝐶󵄩󵄩󵄩󵄩𝛿𝛿u (𝑡𝑛+1 )󵄩󵄩󵄩󵄩1 󵄩󵄩󵄩∇wℎ 󵄩󵄩󵄩1 ,
󵄩
󵄩
𝐴 4 ≤ 𝐶󵄩󵄩󵄩󵄩2u (𝑡𝑛+1 ) − u (𝑡𝑛 )󵄩󵄩󵄩󵄩2
󵄩 ̂ 𝑛+1 󵄩󵄩 󵄩󵄩 󵄩󵄩
󵄩󵄩 󵄩󵄩wℎ 󵄩󵄩1
×󵄩󵄩󵄩󵄩E
󵄩0
󵄩󵄩 ̂ 𝑛+1 󵄩󵄩 󵄩󵄩
󵄩
≤ 𝐶󵄩󵄩󵄩E 󵄩󵄩󵄩0 󵄩󵄩∇wℎ 󵄩󵄩󵄩0 .
Proof. We first recall again inf-sup condition in Assumption 2. Consequently, it suffices to estimate ⟨𝑒𝑛+1 , ∇ ⋅ w⟩ in
terms of ‖∇w‖0 . In conjunction with (10), we can rewrite (66)
as
⟨𝑒ℎ𝑛+1 , ∇ ⋅ wℎ ⟩
1
⟨3E𝑛+1 − 4E𝑛 + E𝑛−1 , wℎ ⟩
2𝜏
+
󵄩
󵄩
𝐴 5 ≤ 𝐶󵄩󵄩󵄩󵄩2E𝑛 − E𝑛−1 󵄩󵄩󵄩󵄩L3 (Ω)
󵄩 𝑛+1 󵄩󵄩 󵄩󵄩 󵄩󵄩
̂ ℎ 󵄩󵄩󵄩1 󵄩󵄩wℎ 󵄩󵄩1
× 󵄩󵄩󵄩󵄩u
≤
𝑛−1
+ N (2E − E
̂ 𝑛+1
,u
ℎ , wℎ )
− 𝜇 ⟨∇𝛿𝑞ℎ𝑛+1 , wℎ ⟩ − ⟨R𝑛+1 , wℎ ⟩
+ 𝜅𝜇2 ⟨g (𝜃 (𝑡𝑛+1 ) − 2𝜃ℎ𝑛 + 𝜃ℎ𝑛−1 ) , wℎ ⟩
8
:= ∑𝐴 𝑖 .
𝑖=1
(145)
Integrate by parts and Hölder inequality yield
󵄩󵄩
󵄩
󵄩
𝐴 6 ≤ 𝐶󵄩󵄩󵄩󵄩𝛿𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 󵄩󵄩󵄩∇wℎ 󵄩󵄩󵄩0 .
󵄩
󵄩
𝐴 7 ≤ 󵄩󵄩󵄩󵄩R𝑛+1 󵄩󵄩󵄩󵄩−1 ‖∇w‖0 .
(146)
(147)
The new term 𝐴 8 can be bound by the Hölder inequality as
+ N (𝛿𝛿u (𝑡𝑛+1 ) , u (𝑡𝑛+1 ) , wℎ )
𝑛
𝐶 󵄩󵄩 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛 󵄩󵄩
(󵄩󵄩E 󵄩󵄩󵄩0 + 󵄩󵄩E 󵄩󵄩0 )
√ℎ 󵄩
󵄩
󵄩
× 󵄩󵄩󵄩∇wℎ 󵄩󵄩󵄩0 .
On the other hand, we have
̂ 𝑛+1 , ∇wℎ ⟩
𝜇 ⟨∇E
ℎ
̂ 𝑛+1 , wℎ )
+ N (2u (𝑡𝑛+1 ) − u (𝑡𝑛 ) , E
(144)
̂ 𝑛+1 − u(𝑡𝑛+1 )‖1 ≤ 𝐶 from Lemma 13,
In light of ‖̂
u𝑛+1
ℎ ‖1 = ‖E
we can have
𝑛=1
=
(143)
Terms 𝐴 3 and 𝐴 4 can be dealt with thanks to the aid of
Lemma 10 and ‖u(𝑡𝑛+1 )‖2 ≤ 𝑀 as follows:
󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩󵄩𝜃𝑡 (𝑡)󵄩󵄩󵄩2 + 󵄩󵄩󵄩𝜃𝑡𝑡𝑡 (𝑡)󵄩󵄩󵄩2 ) 𝑑𝑡.
󵄩
󵄩2
𝜏 ∑ 󵄩󵄩󵄩󵄩𝑒ℎ𝑛+1 󵄩󵄩󵄩󵄩0 ≤ 𝐶 (𝜏2 + ℎ2 ) .
We now proceed to estimate each term from 𝐴 1 to 𝐴 7
separately. We readily obtain
𝐴 8 = 𝜅𝜇2
(142)
× ⟨g (𝛿𝛿𝜃 (𝑡𝑛+1 )
−2𝜗𝑛 + 𝜗𝑛−1 ) , w𝑛+1
ℎ ⟩
󵄩
󵄩
≤ 𝐶𝜅𝜇2 (󵄩󵄩󵄩󵄩𝛿𝛿𝜃(𝑡𝑛+1 )󵄩󵄩󵄩󵄩0
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩2𝜗ℎ𝑛 − 𝜗ℎ𝑛−1 󵄩󵄩󵄩󵄩0
󵄩
󵄩
+󵄩󵄩󵄩󵄩2𝜂𝑛 − 𝜂𝑛−1 󵄩󵄩󵄩󵄩0 ) ‖∇w‖0 .
(148)
18
Journal of Applied Mathematics
Table 1: Error decay for Algorithm 1 with 𝜏 = ℎ and 𝜇 = 1.
𝜏=ℎ
‖𝐸‖0
‖𝐸‖𝐿∞
‖𝐸‖1
‖𝑒‖0
‖𝑒‖𝐿∞
‖𝜗‖0
‖𝜗‖𝐿∞
‖𝜗‖1
1/16
0.00198001
Order
0.00837795
Order
0.0279984
Order
0.0341734
Order
0.197857
Order
0.000215482
Order
0.000197687
Order
0.0167834
Order
1/32
0.000651734
1.603153
0.00282504
1.568326
0.00950852
1.558052
0.0123334
1.470303
0.107171
0.884544
5.3851𝑒 − 05
2.000522
4.98479𝑒 − 05
1.987613
0.00839145
1.000043
1/64
0.000188113
1.792684
0.000825997
1.774063
0.00280233
1.762594
0.00376493
1.711876
0.0453294
1.241396
1.34612𝑒 − 05
2.000166
1.25012𝑒 − 05
1.995466
0.0041957
1.000009
1/128
5.0822𝑒 − 05
1.888075
0.000225008
1.876160
0.00077052
1.862723
0.00105695
1.832716
0.0170497
1.410701
3.36511𝑒 − 06
2.000081
3.1314𝑒 − 06
1.997187
0.00209784
1.000007
1/256
1.32686𝑒 − 05
1.937437
5.90462𝑒 − 05
1.930060
0.000205078
1.909660
0.000283961
1.896142
0.00610541
1.481586
8.4091𝑒 − 07
2.000630
7.84502𝑒 − 07
1.996959
0.00104892
1.000000
5
𝜃=0
𝜕 𝜃 = 0
1
u0 = 0
𝜃0 = 0
𝜕 𝜃 = 0
𝜃=1
Figure 1: Example 20: initial and boundary values of Bénard convection problem.
Inserting the estimates from 𝐴 1 to 𝐴 7 back into (142) and
employing the discrete inf-sup condition in Assumption 2, we
obtain
󵄩
󵄩
𝐶󵄩󵄩󵄩󵄩𝑒𝑛+1 󵄩󵄩󵄩󵄩0
≤
1 󵄩󵄩 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛 󵄩󵄩
(󵄩󵄩𝛿E 󵄩󵄩󵄩0 + 󵄩󵄩𝛿E 󵄩󵄩0 )
𝜏 󵄩
+
1 󵄩󵄩 𝑛+1 󵄩󵄩 󵄩󵄩 𝑛 󵄩󵄩
(󵄩󵄩E 󵄩󵄩󵄩0 + 󵄩󵄩E 󵄩󵄩0 )
√ℎ 󵄩
󵄩
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝑞ℎ𝑛+1 󵄩󵄩󵄩󵄩0 + 󵄩󵄩󵄩󵄩∇E
ℎ 󵄩
󵄩0
(149)
󵄩
󵄩 󵄩 ̂ 𝑛+1 󵄩󵄩
󵄩󵄩
+ 󵄩󵄩󵄩󵄩𝛿𝛿u(𝑡𝑛+1 )󵄩󵄩󵄩󵄩1 + 󵄩󵄩󵄩󵄩E
󵄩0
󵄩
󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝑅𝑛+1 󵄩󵄩󵄩󵄩−1 + 󵄩󵄩󵄩󵄩𝛿𝛿𝜃(𝑡𝑛+1 )󵄩󵄩󵄩󵄩0
5. Numerical Experiments
We finally document 3 computational performances of
SGUM. The first is to check accuracy, and then the next 2
examples are physically relevant numerical simulations, the
Bénard convection problem and the thermal driven cavity
flow. We perform the last 2 examples under the same set
within [2], but we conclude with different numerical simulation for the second test, the Bénard convection problem, from
that of [2]. We impose Taylor-Hood (P2 − P1) in all 3
experiments.
Example 19 (mesh analysis). In this first experiment, we
choose square domain [0, 1] × [0, 1] and impose forcing term
the exact solution to become
𝑢 = 𝜋 (𝑡2 − 𝑡 + 1) 𝑥2 (1 − 𝑥)2 sin (2𝜋𝑦) ,
V = −2 (𝑡2 − 𝑡 + 1) 𝑥 (2𝑥 − 1) (𝑥 − 1) sin2 (𝜋𝑦) ,
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩2𝜗ℎ𝑛 − 𝜗ℎ𝑛−1 󵄩󵄩󵄩󵄩0
𝑝 = − (𝑡2 − 𝑡 − 1) cos (𝜋𝑥) (𝑦2 + 1) ,
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩2𝜂𝑛 − 𝜂𝑛−1 󵄩󵄩󵄩󵄩0 .
𝜃 = cos (𝑡) sin (𝜋𝑥) 𝑦 (1 − 𝑦) .
If we now square it, multiply it by 𝜏, and sum over 𝑛 from 1
to 𝑁, then Lemmas 13, 15, and 17 derive (141).
(150)
Table 1 is error decay with 𝜇 = 1 and 𝜏 = ℎ. We conclude
that the numerical accuracy of SGUM is optimal and consists
with the result of Theorem 4.
Journal of Applied Mathematics
19
Velocity
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
3
3.5
4
4.5
5
3
3.5
4
4.5
5
Temperature
1
0.5
0
0
0.5
1
1.5
2
2.5
Pressure
1
0.5
0
0.5
0
1
1.5
2
2.5
Figure 2: Example 20: streamlines of velocity and isolines of temperature and pressure, at time 𝑡 = 1.0. The nondimensional parameters are
𝜅 = 104 , 𝜆 = 1, 𝜇 = 1, and the discretization parameters are 𝜏 = 10−2 , ℎ = 2−4 .
Velocity
1
0.5
0
0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Temperature
1
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
3
3.5
4
4.5
5
3
3.5
4
4.5
5
Temperature
0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Pressure
1
0
0
0.5
1
1.5
2
2.5
Pressure
1
0.5
0
0
1
0.5
0
Velocity
1
0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0
0
0.5
1
1.5
2
2.5
Figure 3: Example 20: streamlines of velocity and isolines of
temperature and pressure, at time 𝑡 = 1.0. The nondimensional
parameters are 𝜅 = 104 , 𝜆 = 1, and 𝜇 = 1, and the discretization
parameters are 𝜏 = 10−3 , ℎ = 2−4 .
Figure 4: Example 20: streamlines of velocity and isolines of
temperature and pressure, at time 𝑡 = 1.0. The nondimensional
parameters are 𝜅 = 104 , 𝜆 = 1, and 𝜇 = 1, and the discretization
parameters are 𝜏 = 10−4 , ℎ = 2−6 .
Example 20 (Bénard convection). In order to explore the
applicability of the SGUM, we consider the Bénard convection on the domain Ω = [0, 5] × [0, 1] with forcing f = 0 and
𝑏 = 0. Figure 1 displays the initial and boundary conditions
for velocity u and temperature 𝜃, as already studied in [2].
Figures 2–4 are simulations at 𝑡 = 1 with the nondimensional
parameters 𝜅 = 104 , 𝜆 = 1, and 𝜇 = 1. Figure 2 is the result
for the case 𝜏 = 10−2 , ℎ = 2−4 which is the same condition
in [2], and so it displays similar behavior within [2] including
6 circulations in the velocity stream line. However, Figures 3
and 4, the higher resolution simulations with 𝜏 = 10−3 , ℎ =
2−4 , and 𝜏 = 10−4 , ℎ = 2−6 , display 8 circulations in
𝜕 𝜃 = 0
𝜃=
1
2
u0 = 0
𝜃0 = 0
𝜃=−
1
2
𝜕 𝜃 = 0
Figure 5: Example 21: initial and boundary values for thermal
driven cavity flow.
20
Journal of Applied Mathematics
𝑡 = 0.003
0
0.5
1
𝑡 = 0.01
0
0.5
1
0.5
𝑡 = 0.04
0
0.5
0.5
(a)
0.6
0.6
0.6
0.4
0.4
0.4
0.2
0.2
0.2
0
1
1
0
0
0.5
1
𝑡 = 0.01
1
0
0.8
0.8
0.6
0.6
0.6
0.4
0.4
0.4
0.2
0.2
0.2
0
0
0.5
1
𝑢max = 89.532748
0
0
0.5
1
𝑡 = 0.02
1
0
0.8
0.8
0.6
0.6
0.6
0.4
0.4
0.4
0.2
0.2
0.2
0
0.5
1
𝑢max = 69.691775
0
0
0.5
1
𝑡 = 0.04
1
0
0.8
0.8
0.6
0.6
0.6
0.4
0.4
0.4
0.2
0.2
0.2
0
0
0
0.5
1
𝑢max = 70.995722
0
0.5
1
𝑡 = 0.1
1
0
0.8
0.8
0.6
0.6
0.6
0.4
0.4
0.4
0.2
0.2
0.2
0
0.5
1
(b)
0
0
0.5
(c)
1
0
1
0.5
1
𝑡 = 0.02
0
0.5
1
𝑡 = 0.04
0
0.5
1
𝑡 = 0.1
1
0.8
0
0
1
0.8
0.5
𝑡 = 0.01
1
0.8
0
0
1
0.8
1
1
0.5
𝑢max = 75.041598
1
𝑡 = 0.1
0
0.8
0
𝑡 = 0.003
1
0.8
1
1
𝑡 = 0.003
1
0.8
1
𝑡 = 0.02
0
𝑢max = 27.065242
1
0
0.5
1
(d)
Figure 6: Example 21: time sequence 𝑡 = 0.003, 0.01, 0.02, 0.04, and 0.1 for the driven cavity. The first two columns are the streamlines and
vector fields for velocity, and the third and fourth ones are the contour lines for pressure and temperature, respectively. The nondimensional
parameters are 𝜅 = 105 , 𝜆 = 1, and 𝜇 = 1, and the discretization parameters are 𝜏 = 10−4 , ℎ = 2−5 . Note that 𝑢max stands for ‖u‖∞ .
the stream line. So we conclude that the high resolution result
is correct simulation and thus Figure 2 and the result in [2] are
not eventual simulation.
Example 21 (thermal driven cavity flow). We consider the
thermal driven cavity flow in an enclosed square Ω = [0, 1]2 ,
as already studied in several papers [2, 6, 7]. The experiment
is carried out with the same setting as in the work of
Gresho et al. [6], which is shown in Figure 5. Figure 6 displays
the evolution from rest (𝑡 = 0) to steady state (𝑡 = 0.2).
References
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method. I. The Navier-Stokes equations,” SIAM Journal on
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Journal of Applied Mathematics
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[10] V. Girault and P.-A. Raviart, Finite Element Methods for NavierStokes Equations, Springer, 1986.
[11] P. Constantin and C. Foias, Navier-Stokes Equations, The University of Chicago Press, Chicago, Ill, USA, 1988.
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