Lobachevskii Journal of Mathematics
http://ljm.ksu.ru
Vol. 16, 2004, 79 – 89
© A. Rashid
Abdur Rashid
THE PSEUDOSPECTRAL METHOD FOR
THERMOTROPIC PRIMITIVE EQUATION AND ITS
ERROR ESTIMATION
(submitted by A. Lapin)
ABSTRACT. In this paper, a pseudospectral method is proposed for solving
the periodic problem of thermotropic primitive equation. The strict error
estimation is proved.
________________
2000 Mathematical Subject Classification. 35Q35, 65M70,65N30..
Key words and phrases. Thermotropic primitive equation, pseudospectral
scheme, error estimation.
This work is supported by Gomal University, D.I.Khan, Pakistan..
|
1. Introduction
Thermotropic primitive equation is governed by the following differential
equations[1]:
∂U
∂t + U ∂U
∂x + V ∂U
∂y + ∂φ
∂x − νΔU − FV = 0,
∂V
∂t + U ∂V
∂x + V ∂V
∂y + ∂φ
∂y − νΔV + FU = 0,
∂φ
∂t + U ∂φ
∂x + V ∂φ
∂y + φ ∂U
∂x + ∂V
∂x = 0,
| (1.1) |
where U,V are the components of the speed in x, y directions respectively, g
is the acceleration of gravity, H is the height of the geopotential surface,
φ = gH, F is coriolis
parameter and ν
is the coefficient of friction.
There has been a rapid development in the spectral methods for the last
two decades. They have become important tools for numerical solutions of
partial differential equations, and have been widely applied to numerical
simulations in various fields [2-5]. Although the pseudospectral methods are
easier to implement for nonlinear partial differential equations, they are not
stable as the spectral ones due to ’aliasing’. Therefore some author proposed
the filtering technique [10-11] to remedy the deficiency of instability. Some
papers have also been devoted to theoretical study and numerical solutions of
(1.1) [6-9].
The aim of this paper is to consider the periodic initial boundary-value
problem for thermotropic primitive equation. A pseudospectral scheme with
restraint operator in combination with first order time differencing technique
is considered for thermotropic primitive equation. The stability and rate of
convergence for the approximate problem are proved.
2. The Pseudospectral Scheme
Let Ω = (x,y)∣− π < x,y < π and all functions
have the period 2π
for the variable x and y. The norm of the space
Lq (Ω) is denoted
by ⋅Lq(Ω).
In particular, the scaler product and the norm of
L2 (Ω) are denoted
by ⋅,⋅ and
⋅L2(Ω) respectively.
Let m1,m2 and N be
integers and m = m1 2 + m2 2.
Define
V N = Span ei(m1x+m2y)∣∣m∣≤ N,N > 0.
Let PN
be the orthogonal projection operator, i.e.
PNη,ψ = η,ψ,∀ψ ∈ V N.
For the pseudospectral approximation, we put the nodes
xj1,yj2 = 2πj1
2N + 1, 2πj2
2N + 1 , − N ≤ j1,j2 ≤ N,
and let Pc ˜ be the interpolation
operator, i.e. for η(x,y) ∈ C(Ω)
Pc ˜η xj1,yj2 = η xj1,yj2 , − N ≤ j1,j2 ≤ N.
Define Pc = PNPc ˜. To
weaken the nonlinear instability of computation, we follow the work of [11] to adopt the
filtering operator Rγ
with γ > 1,
i.e. if
η(x,y) = ∑
∣m∣≤Nηm1,m2ei(m1x+m2y),
then
Rγη(x,y) = ∑
∣m∣≤N 1 −∣m∣
N γ η
m1,m2ei(m1x+m2y).
Let τ
be the mesh spacing of the variable t and define
Sτ = t = kτ∣k = 0, 1, 2,⋅⋅⋅.
ηt(t) = η(t + τ) − η(t)
τ .
To approximate the nonlinear terms, we define
dα η,u,v = αd(1) η,u,v + (1 − α)d(2) η,u,v,0 ≤ α ≤ 1,
d(1) η,u,v = P
c u∂η
∂x + v∂η
∂y,
d(2) η,u,v = ∂
∂xPc(uη) + ∂
∂yPc(vη).
Let uN,vN,ϕN be the approximations
to U, V and φ respectively,
where for all (x,y) ∈ Ω
and t ∈ Sτ,
ηN(x,y,t) = ∑
∣m∣,∣n∣≤Nηm,nN(t)ei(mx+ny),η = u,v,ϕ.
The pseudospectral scheme for solving (1.1) is
utN + R
γd1∕2 Rγ(uN + δτu
tN),uN,vN + ∂
∂xϕN − νΔ(uN + στu
tN)
− F(vN + δτv
tN) = 0,
vtN + R
γd1∕2 Rγ(vN + δτv
tN),uN,vN + ∂
∂yϕN − νΔ(vN + στv
tN)
+ F(uN + δτu
tN) = 0,
ϕtN + R
γd0 Rγ(ϕN + δτϕ
tN),uN,vN
+ A(ϕN + δτϕ
tN,uN + δτu
tN,vN + δτv
tN) = 0,
| (2.1) |
where 0 ≤ δ ≤ 1,0 ≤ σ ≤ 1
and A(η,ξ,η∗) = P
c η ∂ξ
∂x + ∂η∗
∂x .
3. Some Lemmas
Lemma 1. [1].
For all η(x,y,t)
2 η(t),ηt(t) = η(t) 2
t − τ ηt(t) 2.
Lemma 2. [5].
For all η(x,y,t) ∈ V N,
then
∂η
∂x2 ≤ N2 η(t) 2, ∂η
∂y2 ≤ N2 η(t) 2.
Lemma 3. [10].
For all η(x,y,t) ∈ V N
and ξ(x,y,t) ∈ V N,
then
η(t)ξ(t)2 ≤ (2N + 1)2 η(t) 2 ξ(t) 2.
Lemma 4. [15].
For all η(x,y,t) ∈ Hβ(Ω)
and ξ(x,y,t) ∈ V N,
then
PNη(t) − η(t) HS(Ω) ≤ C1NS−β η(t)
Hβ(Ω),0 ≤ s ≤ β,
PCη(t) − η(t) HS(Ω) ≤ C2NS−β η(t)
Hβ(Ω),0 ≤ s ≤ β,β > 1,
Rγξ(t) − ξ(t) HS(Ω) ≤ C3NS−β ξ(t)
Hβ(Ω),0 ≤ s ≤ β,γ > β − s,
RγPNη(t) − η(t) HS(Ω) ≤ C4NS−β η(t)
Hβ(Ω),0 ≤ s ≤ β,γ > β − s,
where C1 − C4
are positive constants.
Lemma 5. [9].
Assume that the following conditions are fulfilled:
(i) ξ(t)
and η(t)
are non-negative functions defined on Sτ;
(ii) ρ,a,M1,M2,
and M3
are nonnegative constants;
(iii) A(x) is a function such that, if x ≤ M3,
then A(x) ≤ 0;
(iv) ξ(t) ≤ ρ + τ∑
t′=0t−τ M
1ξ(t′) + M
2Naξ2(t′) + A(ξ(t′))η(t′) ;
(v) ρe(M1+M2)T ≤ min M
3, 1
Na,
ξ(0) ≤ ρ,
t ≤ T.
Then
ξ(t) ≤ ρe(M1+M2)t.
In particular, if M2 = 0
and A(ξ(t′)) = 0,
then for all ρ
and n.
ξ(t) ≤ ρeM1t.
4. Error Estimation
For simplicity, we take δ = 0,
let UN = P
NU,V N = P
NV,
and φN = P
Nφ,
then (1.1) leads to
UtN + R
γd1∕2 RγUN,UN,V N + ∂
∂xφN − νΔ(UN + στU
tN) − FV N = 0,
V tN + R
γd1∕2 RγV N,UN,V N + ∂
∂yφN − νΔ(V N + στV
tN) + FUN = 0,
φtN + R
γd0 RγφN,UN,V N + A(φN,UN,V N) = 0,
| (4.1) |
where
G1N = U
tN −∂UN
∂t ,
G2N = R
γd1∕2 RγUN,UN,V N − P
N U ∂U
∂x + V ∂U
∂y ,
G3N = −νστΔU
tN,
G4N = V
tN −∂V N
∂t ,
G5N = R
γd1∕2 RγV N,UN,V N − P
N U ∂V
∂x + V ∂V
∂y ,
G6N = −νστΔV
tN,
G7N = φ
tN −∂φN
∂t ,
G8N = R
γd0 RγφN,UN,V N − P
N U ∂φ
∂x + V ∂φ
∂y,
G9N = A φN,UN,V N − P
N φ∂U
∂x + φ∂V
∂x .
Put
u˜ = u − UN,v˜ = v − V N,ϕ˜ = ϕ − φN.
Then from (1.1) and (2.1), we obtain
u˜tN + ξ
1N + ξ
2N + ∂
∂xϕN − νΔ u˜N + στu˜
tN − Fv˜N = ∑
l=13G
lN,
v˜tN + ξ
3N + ξ
4N + ∂
∂yϕN − νΔ v˜N + στv˜
tN + Fu˜N = ∑
l=46G
lN,
ϕ˜tN + ∑
k=59ξ
kN = ∑
l=79G
lN,
| (4.2) |
where
ξ1N = R
γd1∕2 Rγu˜N,UN,vN + R
γd1∕2 RγUN,u˜N,v˜N ,
ξ2N = R
γd1∕2 Rγu˜N,u˜N,v˜N ,
ξ3N = R
γd1∕2 Rγv˜N,UN,V N + R
γd1∕2 RγV N,u˜N,v˜N ,
ξ4N = R
γd1∕2 Rγv˜N,u˜N,v˜N ,
ξ5N = R
γd0 Rγϕ˜N,UN,V N + R
γd0 RγφN,u˜N,v˜N ,
ξ6N = R
γd0 Rγϕ˜N,u˜N,v˜N ,
ξ7N = −A ϕ˜N,u˜N,v˜N ,
ξ8N = −A ϕ˜N,UN,V N ,
ξ9N = −A φN,u˜N,v˜N .
We shall use the following notations
E = U,V,φ,EN = UN,V N,φN ,E˜N = u˜N,v˜N,ϕ˜N ,
E ˜ N (t)2 = u˜N(t) 2 + v˜N(t) 2 + ϕ˜N(t) 2,
E ˜ tN(t)2 = u˜
tN(t) 2 + v˜
tN(t) 2 + ϕ˜
tN(t) 2,
E ˜ N (t)
12 = u˜N(t)
12 + v˜N(t)
12 + ϕ˜N(t)
12.
Let Hβ(Ω)
be the Sobolev space equipped with the norm
⋅Hβ(Ω). In
particular L2(Ω) = H0(Ω),
we define E1 = U,V,φ
E1Hβ(Ω)2 = U(t)
Hβ(Ω)2 + V (t)
Hβ(Ω)2 + φ(t)
Hβ(Ω)2,
∣E1∣β = max 0≤t≤τ E1(t) Hβ(Ω).
Now we suppose
ρ(t) = ∥E˜N(0)∥2)+ντ(σ+q
2)∣E˜N(0)∣
12+τ∑
t′=0t−τ∥G
lN(t′)∥2,l = 1,..., 9
E˜(t) = ∥E˜N(t)∥2 + ντ(σ + q
2)∣E˜N(0)∣
12
+ τ∑
t′=0t−τr
1τ ∥E˜tN(t′)∥2 + ν(2 − 7ɛ)∣E˜N(t′)∣
12 .
Theorem 1. Suppose the following conditions are fulfilled
(i) δ = 0,τN2 < ∞,
(ii) σ > 1∕2orτN2 < 2
v(1−2σ),
(iii) for suitably small positive constant
M1 and all
t ≤ T , such
that ρ(T) ≤M1
N2 .
Then there exists a positive constant
M2 such that
for all t ∈ Sτ,
t ≤ T, we
have
E˜(t) ≤ ρ(t)eM2t.
Theorem 2. Assume that the conditions (i), (ii) of Theorem 1 are satisfied. In
addition E ∈ C2(0,T; m
00(Ω)),E
1 ∈ C(0,T; H52+r(Ω) ⋂
m0B+1(Ω)),
r > 0,β ≥ 1,
then
E˜(t) ≤ M3(τ2 + N−2β)eM4t.
Mℓ being positive constants
depending only on ∥∣E1∥∣5
2+r
and ν.
Now we define
∥η(t)∥mrβ = max 0≤s≤q 1
4π2 ∑
a+b=0β ∬
Ω ∂a+bη(x,y,t)
∂xa∂yb 2dxdy1∕2,
∥η∥Cq(0,T;mrβ(Ω)) = max 0≤s≤q max 0≤t≤T ∂sη(t)
∂ts mrβ(Ω),
Cq 0,T; m
rβ(Ω) = η(x,y,t)∣∥η∥
Cq(0,T;mrβ(Ω)) < ∞.
5. The Proof of Theorem 1
Let c
be a positive constant which may be different in different cases,
q
denote an undetermined positive constant and
ɛ > 0. Taking the scaler
product (4.2) with 2u˜N + qτu˜
tN,
we have
(∥u˜N(t)∥2)
t + τ(q − 1 − ɛ)∥u˜tN(t)∥2
+ (2u˜N(t) + qτu˜
tN(t),ξ˜
1N(t) + ξ˜
2N(t)
+ ∂φ˜N
∂x (t) + Fv˜N(t)) + 2ν∣u˜N(t)∣
12
+ ντ(σ + q
2)(∣u˜(t)∣12)
t + ντ2(σq − σ −q
2)∣u˜tN(t)∣
12
≤ c∥u˜N(t)∥2 + c(1 + q2τ
4ɛ ) ∑
l=13∥G
lN(t)∥2.
| (5.1) |
Similarly from the second and third formulas of (4.2), we have
(∥v˜N(t)∥2)
t + τ(q − 1 − ɛ)∥v˜tN(t)∥2
+ (2v˜N(t) + qτv˜
tN(t),ξ˜
3N(t) + ξ˜
4N(t)
+ ∂φ˜N
∂y (t) + Fu˜N(t)) + 2ν∣v˜N(t)∣
12
+ ντ(σ + q
2)(∣v˜(t)∣12)
t + ντ2(σq − σ −q
2)∣v˜tN(t)∣
12
≤ c∥v˜N(t)∥2 + c(1 + q2τ
4ɛ ) ∑
l=46∥G
lN(t)∥2
| (5.2) |
(∥ϕ˜N(t)∥2)
t + τ(q − 1 − ɛ)∥ϕ˜tN(t)∥2
+ (2ϕ˜N(t) + qτϕ˜
tN(t),∑
l=59ξ˜
lN(t))
≤ c∥ϕ˜N(t)∥2 + c(1 + q2τ
4ɛ ) ∑
l=79∥G
lN(t)∥2.
| (5.3) |
Putting (5.1)-(5.3) together, we get
(∥E˜N(t)∥2)
t + τ(q − 1 − ɛ)∥E˜tN(t)∥2
+ 2ν∣E˜N(t)∣
12 + ντ(σ + q
2)(∣E˜(t)∣12)
t
+ ντ2(σq − σ −q
2)∣E˜tN(t)∣
12 + ∑
l=16M
lN(t)
≤ c∥E˜N(t)∥2 + c(1 + q2τ
4ɛ ) ∑
l=19∥G
lN(t)∥2,
| (5.4) |
where
M1N(t) = (2u˜N(t) + qτu˜
tN(t),ξ
1N(t)) + (2v˜N(t) + qτv˜
tN(t),ξ
3N(t))
+ (2ϕ˜N(t) + qτϕ˜
tN(t),ξ
5N(t)) − qτ(u˜
tN(t),Fv˜N(t))
+ qτ(v˜tN(t),Fu˜N(t)),
M2N(t) = (2u˜N(t) + qτu˜
tN(t),ξ
2N(t)) + (2v˜N(t) + qτv˜
tN(t),ξ
4N(t))
+ (2ϕ˜N(t) + qτϕ˜
tN(t),ξ
6N(t)),
M3N(t) = (2u˜N(t) + qτu˜
tN(t), ∂
∂xϕN(t)) + (2v˜N(t) + qτv˜
tN(t), ∂
∂yϕ˜N(t)),
M4N(t) = (2ϕ˜N(t) + qτϕ˜
tN(t),ξ
7N(t)),
M5N(t) = (2ϕ˜N(t) + qτϕ˜
tN(t),ξ
8N(t)),
M4N(t) = (2ϕ˜N(t) + qτϕ˜
tN(t),ξ
9N(t)).
We now estimate ∣MlN(t)∣.
Because of the Schwarz inequality and embedding theorem, we have
∣M1N(t)∣ ≤ ɛν∣E˜N(t)∣
12 + ɛτ∥E˜
tN(t)∥2 + c
ɛν(1 + τq2N2)∣∥E
1N∥∣
5
2+r∥E˜N(t)∥2,
∣M2N(t)∣ ≤ ɛν∣E˜N(t)∣
12 + ɛτ∥E˜
tN(t)∥2 + cN2
ɛν (∥E˜N(t)∥2
+ ∣E˜N(t)∣
12 + ∣∥E
1N∥∣
5
2+r)∥E˜N(t)∥2,
∣M3N(t)∣ ≤ ɛτ∣E˜
tN(t)∣
12 + ɛν∣E˜N(t)∣
12 + c
ɛν(1 + τq2N2)∥E˜N(t)∥2,
∣M4N(t)∣ ≤ ɛτ∥E˜
tN(t)∥
12 + ɛν∣E˜N(t)∣
12 + c
ɛν(1 + τq2N2)∥E˜N(t)∥2,
∣M5N(t)∣ ≤ ɛτ∥E˜
tN(t)∥
12 + ɛν∣E˜N(t)∣
12 + cN2
ɛν (1 + τq2N2)∥E˜N(t)∥4,
∣M6N(t)∣ ≤ ɛτ∥E˜
tN(t)∥
12 + ɛν∣E˜N(t)∣
12 + c
ɛν∣∥E1∥∣5
2+r∥E˜N(t)∥2.
By substituting the above estimates into (5.4), we get
(∥E˜N(t)∥2)
t + τ(q − 1 − 7ɛ)∥E˜N(t)∥2 + ν(2 − 6ɛ)∣E˜N(t)∣
12
+ ντ(σ + q
2)(∣E˜N(t)∣
12)
t + ντ2(σq − σ −q
2)∣E˜N(t)∣
12
≤ HN(t) + A
1∥E˜N(t)∥2 + B
1∥E˜N(t)∥4 + B
2∣E˜N(t)∣
12,
| (5.5) |
where
A1 = c 1 + c
ɛν + c
ɛν(1 + τq2N2) ∣∥E
1∥∣52+r,
B1 = cN2
ɛν (1 + τq2N2) + cN2
ɛν ,
B2 = cN2
ɛν ∥E˜N(t)∥2,
HN(t) = c 1 + q2τ
4ɛ ∑
l=19∥G
lN∥2.
Now let ɛ be
suitably small, r1 > 0,
and
q1 = max 1 + r1 + 7ɛ, 2σ
2σ − 1 ,
q2 = r1 + 1 + 7ɛ + ντN2
2 ,
q3 = 2r1 + 2 + 14ɛ + 2σντN2 2 − ντN2(1 − 2σ) −1.
If σ > 1∕2, we
put q = q1,
and it follows from (5.5) that
(∥E˜N(t)∥2)
t + r1τ∥E˜N(t)∥2 + ν(2 − 6ɛ)∣E˜N(t)∣
12
+ ντ(σ + q
2)(∣E˜N(t)∣
12)
t
≤ HN(t) + A
1∥E˜N(t)∥2 + B
1∥E˜N(t)∥4 + B
2∣E˜N(t)∣
12.
| (5.6) |
If σ = 1∕2, we
put q = q2,
and so
τ(q − 1 − 7ɛ)∥E˜tN∥2 + ντ2(σq − σ −q
2)∣E˜tN(t)∣
12 ≥ r
1τ∣E˜tN(t)∣2.
Therefor (5.6) is still holds.
If σ < 1∕2,
τN2 < 2
ν(1−2σ), we
put q = q3,
and thus (5.12) holds.
By summing up (5.6) for t ∈ Sτ
, we get
(∥E˜N(t)∥2)
t + ντ(σ + q
2)(∣E˜N(t)∣
12)
t + τ∑
t′=0t−τr
1τ∥E˜tN(t′)∥2
+ ν(2 − 7ɛ)∣E˜N(t′)∣
12 ≤ ρ(t)
+ τ∑
t′=0t−τA
1E˜1N(t′) + B
1E˜12(t′)
+ B2∣E˜2N(t′)∣
12,
| (5.7) |
where
ρ(t) = ∥E˜N(0)∥2) + ντ(σ + q
2)∣E˜N(0)∣
12 + τ∑
t′=0t−τHN(t′)
from which and Lemma 5, the proof is completed.
6. The Proof of Theorem 2
We first have
∥GlN(t) ≤ cτ∥E∥
C2(0,T;m00(Ω)),l = 1, 4, 7.
From Lemma 4 and the embedding theorem, we get
∥GlN(t)∥≤ c∥∣E
1∥∣52+rN−β∥E
1∥m0β+1(Ω),l = 2, 5, 8.
It is easy to show that
∥GlN(t)∥≤ cτ ∂E
∂t (t) m20(Ω) + ∂E
∂t (t) m20(Ω) ,l = 3, 6.
We have also
∥G9N(t)∥≤ cN−β∥∣E
1∥∣52+r∥E(t)∥m0β+1(Ω),∥E˜N(0)∥≤ cN−β∥E(0)∥
m0β(Ω).
Therefore if the conditions of Theorem 1 are fulfilled, then
ρ(t) ≤ c τ2 + N−2β .
By combining the above estimations with Theorem 1, we complete the proof
of Theorem 2.
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DEPARTMENT OF MATHEMATICS, GOMAL UNIVERSITY, D.I.KHAN, PAKISTAN.
E-mail address: rashid_himat@yahoo.com
Received February 8, 2004; revised version May 27, 2004