Contents lists available atScienceDirect
Journal of Computational and Applied
Mathematics
journal homepage:www.elsevier.com/locate/cam
Split least-squares finite element methods for linear and nonlinear
parabolic problems
IHongxing Rui
a,∗, Sang Dong Kim
b, Seokchan Kim
caSchool of Mathematics, Shandong University, Jinan, Shandong, 250100, China
bDepartment of Mathematics, Kyungpook National University, Daegu 702-701, South Korea
cDepartment of Applied Mathematics, Changwon National University, Changwon 641-773, South Korea
a r t i c l e i n f o
Article history:
Received 9 June 2007
Received in revised form 5 March 2008
MSC: 65M12 65M15 65M60 Keywords: Split Least squares Finite element Error estimates Parabolic problem a b s t r a c t
In this paper, we propose some least-squares finite element procedures for linear and nonlinear parabolic equations based on first-order systems. By selecting the least-squares functional properly each proposed procedure can be split into two independent symmetric positive definite sub-procedures, one of which is for the primary unknown variableuand the other is for the expanded flux unknown variableσ. Optimal order error estimates are developed. Finally we give some numerical examples which are in good agreement with the theoretical analysis.
© 2008 Elsevier B.V. All rights reserved.
1. Introduction
The purpose of this paper is to consider the least-squares finite element procedures for linear and nonlinear parabolic problems written as first-order systems. It is well known that, compared to mixed element methods, the least-squares finite element method has two typical advantages as follows: it is not subject to the Ladyzhenkaya–Babuska–Brezzi [13, 1,4] consistency condition, so the choice of approximation spaces becomes flexible, and it results in a symmetric positive definite system.
Least-squares finite element methods for elliptic problems, based on first-order systems, were introduced by [12] where a least-squares residual minimization is introduced for the mixed system in primary unknown variableuand expanded unknown flux σ. Then an elegant theory for least-squares finite element approximation for general elliptic boundary value problems was established, see, for example, [12,10,11,16,5,6] and the references therein. Concerning the parabolic problems, [14] and [15] introduced the least-squares finite element procedure with semi-discretization in time and fully discrete scheme. They also established the a posterior error estimates and constructed adaptive algorithms.
In this paper we consider the least-squares finite element procedures for linear and nonlinear parabolic problems. Like [14,15] we define the least-squares functionals using weight-factors. By selecting different weight-factors we get different
IThe work was supported by the Korea Research Foundation under contract number KRF-2005-070-C00017, the National Natural Science Foundation of China Grant No. 10771124 and the Research Fund for Doctoral Program of High Education by the State Education Ministry of China No. 20060422006.
∗Corresponding author.
E-mail addresses:[email protected](H. Rui),[email protected](S.D. Kim),[email protected](S. Kim). 0377-0427/$ – see front matter©2008 Elsevier B.V. All rights reserved.
procedures. We show that all the procedures presented in this paper can be divided into two independent sub-procedures, one of which is for the primary unknown variableuand the other is for the expanded fluxσ. The key point used to explain the split of the procedure isLemma 2.1which was obtained by integration by parts. Similar results have been found and used by [8] to prove the coercivity of least-squares bilinear formats and by [2,3] to establish connections between least-squares and mixed methods. The last two papers also show that not only is the pressure the same as in the Galerkin method, but also the flux is the same as in the mixed method under some conditions on the finite element spaces.
In this paper three procedures were presented for linear parabolic problems. In the first procedure the sub-procedure for the primary unknownuis the same as the standard Galerkin finite element procedure. In the second procedure one of the sub-procedures is for the expanded fluxσonly. The third one is a procedure with second-order approximation in time increment. We give one procedure to deal with the nonlinear problem. For these schemes we give the optimal order error estimates. Finally we give some numerical examples.
The remainder of this paper is organized as follows. In Section2we introduce the split least-squares schemes for linear problems. In Section3, we establish the optimal order error estimates. In Section4, we give a least-squares finite element procedure for nonlinear problems. Finally in Section5we give some numerical examples.
Throughout this paper, the notations of standard Sobolev spaces L2
(
Ω)
, Hk(
Ω)
and associated normsk · k = k · kL2(Ω), k · kk= k · kHk(Ω)are adopted as those in [7]. For simplicity we usek · kL∞(Hm+1)andk · kL2(Hm+1)to representk · kL∞(J;(Hm+1(Ω))d)
andk · kL2(J;(Hm+1(Ω))d)respectively forJ =
(
0,
T)
andd ≤ 3. A constantC(with or without subscript) stands for a genericpositive constant independent of the mesh parameterhu,hσand
1
t, it may be different at different occurrence.2. Least-squares procedure for linear problems
In this section we present three least-squares finite element procedures for linear problems. For simplicity we just consider the homogeneous boundary condition. The same idea can be used to deal with problems with non-homogeneous boundary condition.
Consider the following parabolic problem on a bounded domainΩ⊂Rd
,
d=2,
3:
φ
ut−div(A∇
u)
=f,
inΩ×J,
u=0,
onΓD×J,
A∇u·n=0 onΓN×J,
(2.1) subject to the initial conditionu
(
x,
0)
=u0(
x)
onΩ×J,
(2.2)where
∂
Ω = ΓD∩ΓN, n is the outward unit normal vector,J =(
0,
T]is the time interval andφ
is a continuous functionsatisfying
φ
1 ≤φ
≤φ
2with two positive constantsφ
1andφ
2. We further assume thatA =(
aij(
x))
di,j=1is a bounded, symmetric and positive definite matrix inΩ, i.e., there exist positive constantsα
andβ
such that,α
kξk2≤(
Aξ,ξ)≤β
kξk2,
∀ξ ∈ Rd.
(2.3)In some applications, the problem(2.1)appears as a first-order system for bothuandσ = −A∇u
,
σ =(σ
1, . . . ,
σd)
, asfollows:
φ
ut+ divσ −f =0,
inΩ×J,
σ + A∇u=0,
inΩ×J,
u=0,
onΓD×J,
σ ·n=0 onΓN×J.
(2.4)For example, in the compressible miscible displacement problem [9],urepresents the pressure andσrepresents the Darcy velocity or flux. In this case the approximations to bothuandσare necessary. We consider the least-squares mixed element approximations for(2.4).
First we consider the first-order approximation in time increment. Let
1
t be a time increment. With tn = n1
t,un=u
(
tn,
·)
, putδ
tun:= un−un−11
t,
ρ
n 1:=φ(δ
tun−unt).
(2.5) It is clear thatρ
n 1=O Z tn tn−1 kuttkdt ! =O 4t Z tn tn−1 kuttk2dt !12 .
(2.6)Define two function spaces V= {v∈H1
(
Ω)
:v=0 onΓD}
,
(2.7)From(2.4)we know that forn≥1,
(
un,
σn)
∈V×W satisfy that (φ
−1 2(φ
un+1
tdivσn−Fn 1)
=0,
inΩ×J,
A−12(σ
n+A∇un)
=0,
inΩ×J,
(2.9) whereFn 1=φ
un−1+1
tfn+1
tρ
n1.For
(
v,
τ)
∈V×W, define the first kind of least-squares functionalJn1
(
v,
τ)
as follows.Jn1
(
v,
τ)
= kφ
−12(φ
v+1
tdivτ −F1n)
k2+1
tkA−12(τ +
A∇v
)
k2.
(2.10)The least-squares minimization problem corresponding to(2.9)is: find
(
un,
σn)
∈V×σn∈W such thatJn1
(
un,
σn)
= inf (v,τ)∈V×WJn
1
(
v,
τ). (2.11)Define the bilinear forma
(
u,
σ;v,
τ)
corresponding to the least-squares functionalJn1as
a
(
u,
σ;v,
τ)=(φ
−1(φ
u+1
tdivσ), φ
v+1
tdivτ)
+1
t(
A−1(σ + A∇
u),
τ + A∇v).
(2.12) The weak statement of the minimization problem(2.11)becomes: find(
un,
σn)
∈V×W such thata
(
un,
σn;v,
τ)
=(φ
−1F1n, φ
v+1
tdivτ),
∀(
v,
τ)
∈V×W.
(2.13) Noticing the definition ofFn1,(2.13)becomes
a
(
un,
σn;v,
τ)
=(
un−1+1
tφ
−1(
fn+ρ
n1), φ
v+1
tdivτ),
∀(
v,
τ)
∈V×W.
(2.14) Now we consider the second weak formulation different from(2.13). From(2.4)we have that forn≥1,(
un,
σn)
∈V×Wsatisfy that (
φ
−12(φ
un+1
tdivσn−Fn 1)
=0,
inΩ×J,
A−12(σ
n+A∇un−Gn)
=0,
in Ω×J,
(2.15)whereGn=σn−1+A∇un−1. For
(
v,
τ)
∈V×W, define the second kind of least-squares functionalJn2
(
v,
τ)
as follows.Jn2
(
v,
τ)
= kφ
−12(φ
v+1
tdivτ −Fn)
k2+1
tkA−12(τ + A∇
v−Gn
)
k2.
(2.16)The least-squares minimization problem corresponding to(2.15)is: find
(
un,
σn)
∈V×W such thatJn2
(
un,
σn)
= inf (v,τ)∈V×WJn
2
(
v,
τ). (2.17)Similarly to(2.14), the weak statement of(2.17)is: find
(
un,
σn)
∈V×W such thata
(
un,
σn;v,
τ)
=(
un−1+1
tφ
−1(
fn+ρ
n1), φ
v+1
tdivτ),
+1
t(A
−1σn−1+ ∇un−1,
τ + A∇v)
∀
(
v,
τ)
∈V×W.
(2.18)In order to approximate the formulations(2.14)and(2.18), we need to construct the finite element spaces. LetThuandThσ
be two families of regular finite element partitions of the domainΩ, which are either identical or not. Lethuandhσdenote
the largest of the diameters of the element inThuandThσ respectively. Based onThuandThσ, respectively, we construct the
finite element spaces Vh⊂Vand Wh⊂W with the following approximation properties:
inf vh∈Vh {kv−vhk +huk∇
(
v−vh)
k} ≤Chmu+1kvkm+1,
(2.19) inf τh∈Wh kτ − τhk ≤Chk+1 σ kτkk+1,
(2.20) inf τh∈Wh kdiv(τ − τ
h)
k ≤Chk1σkτkk1+1,
(2.21)forv∈V∩Hm+1
(
Ω)
andτ ∈W∩(
Hk1+1(
Ω))
d. It is clear that when assumption(2.20)holds we can deducek1=k, and when
Whis selected as any of the Raviart–Thomas mixed element space [17] we can choosek1=k+1. In this paper we always supposek1=k+1 when Whis any of the Raviart–Thomas mixed element space [17] andk1=kotherwise.
We select the initial approximationu0
h∈Vh,σ0h∈Whsuch that ( ku0−u0hkj≤Chm +1−j u ku0km+1
,
j=0,
1,
kσ0−σ0hk ≤Chσk+1kσ0kk+1,
(2.22) whereσ0=A∇u0. The first least-squares finite element procedure based on(2.14)reads as follows.Scheme (I). Forn≥1 find
(
unh
,
σnh)
∈Vh×Whsuch that a(
un h,
σ n h;vh,
τh)
=(
unh−1+1
tφ
−1fn, φ
v h+1
tdivτh),
∀(
vh,
τh)
∈Vh×Wh.
(2.23)Based on(2.18)the second least-squares finite element procedure reads as follows.
Scheme (II). Forn≥1 find
(
un h,
σ n h)
∈Vh×Whsuch that a(
unh,
σnh;vh,
τh)
=(
un −1 h +1
tφ
−1fn, φ
v h+1
tdivτh)
+1
t(
A−1σn−1 h + ∇u n−1 h,
τh+A∇vh),
∀(
vh,
τh)
∈Vh×Wh.
(2.24)Now let us mention about the bilinear forma
(
·,
·; ·,
·)
in the following lemma, which leads to decoupled systems.Lemma 2.1. For anyu
,
v∈V andσ,
τ ∈Wwe have that,a
(
u,σ;v,
τ)
=(φ
u,
v)
+1
t(A∇
u,
∇v)
+1
t(A
−1σ,
τ)
+1
t2(φ
−1divσ,
divτ).
(2.25)Proof. A direct calculation shows that
a
(
u,σ;v,
τ)
=(φ
u,
v)
+1
t(
A∇u,
∇v)
+1
t(
A−1σ,
τ)
+1
t2(φ
−1divσ,
divτ)
+1
t((
u,
divτ)
+(
v,
divσ)
+(
∇u,
τ)
+(
∇v,
σ)),
Integrating by parts shows that
(
u,
divτ)
+(
v,
divσ)
+(
∇u,
τ)
+(
∇v,
σ)
=0,
(2.26)which completes the proof.
UsingLemma 2.1, we have the decoupling equivalent form of each scheme (I) or (II) alternatively by puttingτh=0 and
vh=0 in(2.23)or(2.24).
Equivalent Form of Scheme (I). With the initial guess
(
u0h,
σ0h)
∈Vh×Wh, forn≥1 find(
unh,
σhn)
∈Vh×Whsuch that forallvh∈Vhandτh∈Wh
(φ
un h,
vh)
+1
t(
A∇unh,
∇vh)
=(φ
unh−1+1
tf n,
v h),
(2.27)(A
−1σn h,
τh)
+1
t(φ
−1divσn h,
divτh)
=(
un −1 h +1
tφ
−1fn,
divτ h).
(2.28)Equivalent Form of Scheme (II). With the initial guess
(
u0h
,
σ0h)
∈Vh×Wh, forn≥1 find(
unh,
σ nh
)
∈Vh×Whsuch that forallvh∈Vhandτh∈Wh
(φ
unh,
vh)
+1
t(
A∇uhn,
∇vh)
=(φ
unh−1+1
tf n,
v h)
+1
t(σ
nh−1+A∇u n−1 h,
∇vh)
(2.29)(
A−1σn h,
τh)
+1
t(φ
−1divσnh,
divτh)
=(
A−1σn −1 h,
τh)
+1
t(φ
−1fn,
divτh).
(2.30)Note that each Scheme (I) or (II) is split into two independent symmetric positive definite systems. Sub-procedure(2.27) is the same as the standard Galerkin finite element procedure for parabolic problems. Sub-procedure(2.30)is a procedure for the unknown fluxσn
hwith first-order approximation in time increment.
It clear that both problems(2.23)and(2.24)have a unique solution. Now we consider the second-order approximation in time increment. Let
ρ
n 2:=φ
δ
tun−u n−12 t +1 2div(σ
n+σn−1)
− divσn−12,
(2.31) which can be estimated asρ
n 2=O 1
t 3 2 Z tn tn−1(
|uttt|2+ |divσtt|2)
dt !12 .
(2.32)From(2.4)we know that forn≥1,
(
un,
σn)
∈V×W satisfy that
φ
−12φ
un+1
t 2 divσ n−Fn 2 =0,
inΩ×J,
A−12(σ
n+A∇un−Gn)
=0,
inΩ×J,
(2.33)whereGnis the same as in(2.15),
F2n=
φ
un−1+1
tfn−12 −1
t 2 divσn−1+
1
tρ
n2
.
(2.34)For
(
v,
τ)
∈V×W, define the least-squares functionalJn3
(
v,
τ)
as follows. Jn3(
v,
τ)
=φ
−12φ
v+1
t 2 divτ −F n 2 2 +1
t 2 kA −12(τ + A∇
v−Gn)
k2.
(2.35)The least-squares minimization problem corresponding to(2.33)is: find
(
un,
σn)
∈V×W such thatJn3
(
un,
σn)
= infv∈V,τ∈WJ
n
3
(
v,
τ).
(2.36)Define the bilinear formb
(
·,
·; ·,
·)
asb
(
u,
σ;v,
τ)
= u+1
t 2φ
−1divσ, φ
v+1
t 2 divτ +1
t 2(
A −1σ + ∇u,
τ + A∇v).
(2.37)Noticing the definition ofFn
2in(2.34), the weak statement of the minimization problem(2.36)is: find
(
un,
σn)
∈V×W such that b(
un,
σn;v,
τ)
= un−1+1
tφ
−1 fn−12 −1 2divσ n−1+ρ
n 2, φ
v+1
t 2 divτ,
+1
t 2(
A −1σn−1+ ∇un−1,
τ + A∇v)
∀(
v,
τ)
∈V×W.
(2.38) Then the corresponding least-squares finite element procedure reads as follows.Scheme (III). With the initial guess
(
u0h,
σ0h)
∈Vh×Wh, forn≥1 find(
unh,
σhn)
∈Vh×Whsuch thatb
(
un h,
σ n h;vh,
τh)
= un−1 h +1
tφ
−1fn−12 − 1 2divσ n−1 h, φ
vh+1
t 2 divτh +1
t 2(
A −1σn−1 h + ∇u n−1 h,
τh+A∇vh),
∀(
vh,
τh)
∈Vh×Wh.
(2.39)Similarly toLemma 2.1we know that the following lemma holds.
Lemma 2.2. For anyu
,
v∈V andσ,τ ∈Wwe have that,b
(
u,σ;v,
τ)
=(φ
u,
v)
+1
t 2(
A∇u,
∇v)
+1
t 2(
A −1σ,
τ)
+1
t 2 2(φ
−1divσ,
divτ).
(2.40)UsingLemma 2.2we have a decoupling equivalent form of Scheme (III).
Equivalent Form of Scheme (III). With the initial guess
(
u0h,
σ0h)
∈Vh×Wh, forn≥1 find(
unh,
σhn)
∈Vh×Whsuch that(φ
unh,
vh)
+1
t 2(
A∇u n h,
∇vh)
=φ
unh−1+1
tfn−12 −1
t 2 divσ n−1 h,
vh +1
t 2(σ
n−1 h +A∇u n−1 h,
∇vh),
∀vh∈Vh.
(2.41)(
A−1σn h,
τh)
+1
t 2(φ
−1divσn h,
divτh)
=(
A−1σnh−1,
τh)
+1
tφ
−1fn−12 −1 2divσ n−1 h,
divτh,
∀τh∈Wh.
(2.42)Then this scheme also can be split into two independent sub-procedures. Sub-procedure(2.42)is a procedure for the unknown fluxσn
hwith second-order approximation in time increment.
Remark 2.3. Results similar toLemma 2.1orLemma 2.2have been found and used by [8] to prove the coercivity of least-squares bilinear formats and by [2,3] to establish connections between least-squares and mixed methods.
3. Error estimates
In this section we give the error estimates for the schemes described in Section2. We first discuss the error estimate for Scheme (I) in the followingTheorem 3.1.
Theorem 3.1. Suppose
(
unh
,
σnh)
∈ Vh × Wh is the solution of Scheme(
I)
. Under the assumption ku0h − u0k =O
(
hum+1−jku0kHm+1−j),
j=0,
1, there exists a positive constantCindependent ofhu,hσand1
tsuch thatkunh−unks≤Chum+1−s
(
kukL∞(Hm+1)+ kutkL2(Hm+1))
+C1
tkuttkL2(L2),
s=0,
1,
(3.1) kσnh−σk +1
t12kdiv(σ
n h−σ n)
k ≤ C(
hk+1 σ kσnkk+1+1
t 1 2hk1 σkσnkk1+1+1
tkuttkL2(L2))
+Cmin{hmu, 1
t−12hm+1 u }(
kukL∞(Hm+1)+ kutkL2(Hm+1)).
(3.2)Proof. Since Scheme (I) is equivalent to(2.27)and(2.28), from the error estimates of the finite element method for parabolic problems (see [18] and [19] for example), we know that(3.1)holds.
We next considerσi h−σ
i, 1≤i≤n≤ T
1t. Subtracting(2.14)from(2.23)and settingvh=0, usingLemma 2.1, we have
(
A−1(
σi h−σ i),
τ h)
+1
t(φ
−1div(
σih−σ i),
divτ h)
=(
ui −1 h −u i−1,
divτ h)
−1
t(φ
−1ρ
i1,
divτh)
∀τh∈Wh = −(
∇(
uih−1−ui−1),
τh)
−1
t(φ
−1ρ
i 1,
divτh).
(3.3) LetσiI∈Whbe an interpolant ofσisuch that
( kσiI−σik ≤Chσk+1kσikk+1
,
kdiv(σ
i I−σ i)
k ≤Chk1 σkσikk1+1.
(3.4) Denote byξ
i σ=σih−σ i I.
(3.5)Settingτh=
ξ
iσ =σih−σiIin(3.3), and using the-inequality,(2.6), we havekA−12
ξ
i σk2+1
tkφ
− 1 2divξ
i σk2 =(
A−1(σ
i−σi I)
;ξ
i σ)
+1
t(φ
−1div(σ
i−σiI)
;divξ
i σ)
−(
∇(
uih−1−u i−1), ξ
i σ)
+1
t(φ
−1ρ
i1,
divξ
iσ)
≤ 1 2 kA−12ξ
i σk2+1
tkφ
− 1 2divξ
i σk2 +C h kσi−σiIk2+1
tkdiv(σ
i−σi I)
k 2+ k∇(
ui−1 h −u i−1)
k2+1
tkρ
i 1k2 i ≤ 1 2 kA−12ξ
i σk2+1
tkφ
− 1 2divξ
i σk2 +C1
t2kuttk2L2(L2) +Ch2(k+1) σ kσik2k+1+1
thσ2k1kσik2k1+1+ k∇(
u i−1 h −u i−1)
k2.
(3.6) By using −(
∇(
uhi−1−ui−1), ξ
iσ)
=(
uih−1−ui−1,
divξ
i σ)
≤ C1
t−12kui−1 h −u i−1k1
t12kφ
12divξ
i σk,
we have the following estimate instead of(3.6),kA−12
ξ
i σk2+1
tkφ
− 1 2divξ
i σk2≤ 1 2 kA−12ξ
i σk2+1
tkφ
− 1 2divξ
i σk2 +C1
t2kuttk2L2(L2) +C h2σ(k+1)kσikk2+1+1
th2σk1kσik2 k1+1+1
t−1kuih−1−u i−1k2.
(3.7)Then, using(3.1)and(3.7), and the positive definiteness ofA, we have that
k
ξ
i σk +1
t 1 2kdivξ
i σk ≤C(
hkσ+1kσikk+1+1
t 1 2hk1 σkσikk1+1+1
tkuttkL2(L2))
+Cmin{hmu, 1
t−12hm+1 u }(
kukL∞(Hm+1)+ kutkL2(Hm+1)).
(3.8)Combining(3.8)with(3.4)completes the proof.
For the error estimates for Scheme (II), for anyi≤n≤1T
twe define the auxiliary projectionu˜ i
h∈Vhsatisfying
(
A∇(
u˜ih−ui),
∇vh)
=0,
∀vh∈Vh.
(3.9)From this definition we have that
(
A∇δ
t(
u˜ih−ui
),
∇vh
)
=0,
∀vh∈Vh.
(3.10)From [7] it is easy to see that that
k ˜uih−uikj≤Chum+1−jkuikm+1
,
j=0,
1,
kδ
t(
u˜ih−u i)
k j≤Chmu+1−j 11
t Z ti ti−1 kutk2m+1dt !12,
j=0,
1.
(3.11) Theorem 3.2. Suppose(
unh
,
σnh)
∈Vh×Whis the solution of Scheme(
II)
. The initial guess satisfieskσ0h−σ0k =O(
huk1kσ0kHk1)
.Whenhu,hσand
1
tare sufficiently small, there exists a positive constantCindependent ofhu,hσand1
tsuch that kσnh−σnk + n X i=11
tkdiv(σ
i h−σ i)
k2 !12 ≤C(
hk1σkσk L∞(Hk1 +1)+h k+1 σ kσtkL2(Hk+1)+1
tkuttkL2(L2)).
(3.12)Further, ifu0h = ˜u0hholds there, we have that
kunh−unk ≤Chum+1
(
kukL∞(Hm+1)+ kutkL2(Hm+1))
+Chk1σkσ kL∞(Hk1 +1)+Chk+1
σ kσtkL2(Hk+1)+C
1
tkuttkL2(L2).
(3.13)Proof. Subtracting(2.18)from(2.24)we have that
a
(
uih−ui,
σih−σi;vh,
τh)
=(
uih−1−u i−1, φ
v h+1
tdivτh)
+1
t(
A−1(σ
ih−1−σ i−1)
+ ∇(
ui−1 h −u i−1),
τ h +A∇vh),
−1
t(φ
−1ρ
i 1, φ
vh+1
tdivτh),
∀(
vh,
τh)
∈Vh×Wh.
(3.14)Settingvh=0, usingLemma 2.1and the divergence theorem, we have forτh∈Whthat
(
A−1(σ
i h−σ i),
τ h)
+1
t(φ
−1div(σ
ih−σ i),
divτ h)
=(
A−1(σ
ih−1−σ i−1),
τ h)
−1
t(φ
−1ρ
i1,
divτh),
(3.15)which can be written as
(
A−1ξ
i σ,
τh)
+1
t(φ
−1divξ
iσ,
divτh)
=(
A−1ξ
iσ−1,
τh)
+1
tA−1δ
t(
σi−σiI),
τh +1
tφ
−1div(
σi−σi I),
divτh −1
t(φ
−1ρ
i1,
divτh)
≤1 2kA −12ξ
i−1 σ k2+ 1 2kA −12τ hk2+1
t 2 kA −12δ
t(σ
i−σiI)
k 2+1
t 2 kτhk 2+1
tkφ
−12div(σ
i−σi I)
k 2 +1
t 4 kφ
−12 divτhk2+1
tkφ
− 1 2ρ
i 1k2+1
t 4 kφ
−12 divτhk2,
(3.16)where the notation
δ
tis defined in(2.5).Note that
φ
is bounded below and above, 0< φ
1≤φ
≤φ
2. Then puttingτh=ξ
iσin(3.16)and using the-inequality wehave kA−12
ξ
i σk2+1
tkφ
− 1 2divξ
i σk2 ≤ 1+1
t 2 kA −12ξ
i σk2+1
t 2 kφ
−12 divξ
i σk2+ 1 2kA −12ξ
i−1 σ k2 +C1
thkδ
t(
σ − σI)
ik2+ kdiv(
σ − σ I)
ik2+ kρ
i1k2 i.
(3.17) Since kδ
t(σ − σ
I)
ik = 11
t Z ti ti−1(σ − σ
I)
tdt ≤Chkσ+1 11
t Z ti ti−1 kσtk2 k+1dt !12,
applying(3.4)and(2.6)to(3.17)we havekA−12
ξ
i σk2+1
tkφ
− 1 2divξ
i σk2 ≤ kA− 1 2ξ
i−1 σ k2+1
tkA− 1 2ξ
i σk2+C1
t2 Z ti ti−1 kuttk2dt +C " h2(σk+1) Zti ti−1 kσtk2 k+1dt+1
t hσ2k1kσk2L∞(Hk1 +1) #.
(3.18)Carrying out summation fori=1
,
2, . . . ,
nwe have that kA−12ξ
n σk2+ n X i=11
tkφ
−12divξ
n σk2 ≤ n X i=11
tkA−12ξ
i σk2+ kA− 1 2(σ
0 h−σ 0 I)
k 2+C1
t2Z T 0 kuttk2dt +C h2σ(k+1) Z T 0 kσtk2k+1dt+hσ2k1kσ k2L∞(Hk1 +1).
(3.19) Noticingσ0 h−σ0I =(σ
0h−σ0
)
+(σ
0−σ0I)
, using Gronwall’s inequality shows thatk
ξ
nσk2+ n X i=11
tkdivξ
i σk2≤C h h2σk1kσk2 L∞(Hk1 +1)+h 2(k+1) σ kσtk2L2(Hk+1)+1
t2kuttk2L2(L2) i.
(3.20)Combining with(3.4)completes the proof of(3.12). Now we consider the estimate ofuih−ui, fori≤n≤ T
1t. Lettingτh =0 in(3.14), usingLemma 2.1and the divergence
theorem lead to
(φ(
uih−ui),
vh)
+1
t(
A∇(
uih−u i),
∇v h)
=(φ(
uii−1−u i−1),
v h)
−1
t(ρ
i1,
vh)
+1
t(σ
ih−1−σ i−1,
∇v h)
+1
t(
A∇(
ui−1 h −u i−1),
∇v h),
∀vh∈Vh.
(3.21)With the use of the definition ofeu i h, we have
(φ(
uih− ˜uih),
vh)
+1
t(
A∇(
uih− ˜u i h),
∇vh)
=(φ(
uih−1− ˜uhi−1),
vh)
+1
t(φδ
t(
ui− ˜uih),
vh)
−1
t(
div(σ
i −1 h −σ i−1),
v h)
−1
t(ρ
i1,
vh)
+1
t(
A∇(
uih−1− ˜uhi−1),
∇vh),
∀vh∈Vh.
(3.22) Letξ
i u=u i h− ˜u i h,
η
i u=u i− ˜ui h.
(3.23)With the choicevh=
ξ
ui =uih− ˜uihin(3.22), it follows thatk
φ
12ξ
i uk 2+1
tkA1 2∇ξ
iuk2 ≤ 1+1
t 2 kφ
1 2ξ
i uk 2+1
t 2 kA 1 2∇ξ
iuk2+1 2kφ
1 2ξ
i−1 u k 2+1
t 2 kA 1 2∇ξ
iu−1k2 +C1
thkδ
t(
ui− ˜ui h)
k 2+ kdiv(σ
i−1 h −σ i−1)
k2+ kρ
i 1k2 i,
(3.24)which can be reduced to k
φ
12ξ
i uk 2+1
tkA1 2∇ξ
iuk2 ≤ kφ
12ξ
i−1 u k 2+1
tkφ
1 2ξ
i uk 2+1
tkA1 2∇ξ
iu−1k2 +C1
thkδ
t(
ui− ˜uih)
k 2+ kdiv(σ
i−1 h −σ i−1)
k2+ kρ
i 1k2 i.
(3.25)Summing(3.25)fromi=1 tonand noticing(3.12)we have
k
φ
12ξ
n uk 2+1
tkA12∇ξ
n uk 2 ≤Xn i=11
tkφ
12ξ
i uk 2+ kφ
12ξ
0 uk 2+1
tkA12∇ξ
0 uk 2+C(
h2(m+1) u kutk2L2(Hm+1) +1
t2kuttk2L2(L2))
+C n X i=11
tkdiv(σ
i−1 h −σ i−1)
k2 ≤ n X i=11
tkφ
12ξ
n uk2+C h h2(um+1)kutk2L2(Hm+1)+1
t2kuttk2L2(L) i +Chh2σk1kσk2 L∞(Hk1 +1)+h 2(k+1) σ kσtk2L2(Hk+1) i.
(3.26)Therefore we can apply Gronwall’s inequality to(3.26). Hence it follows that k
ξ
n uk +1
t 1 2k∇ξ
n uk ≤C h hm+1 u kutkL2(Hm+1)+1
tkuttkL2(L2) i +Chhk1 σkσ kL∞(Hk1 +1)+hσk+1kσtkL2(Hk+1) i.
Finally, combining(3.27)with(3.11)completes the proof.Remark 3.3. Instead ofu0h = ˜u0hif we supposeku0h−u0k
j=O
(
hmu+1−j)
, from the proof we know that replacing(3.13)we havea estimate
kunh−unk ≤C
(
hmu+1+1
t12hmu +hk1σ +
1
t).
Now we give the error estimate for Scheme (III). For this purpose, defineσ˜ih∈Whsuch that
(
σ˜ih−σi,
τh)
+(φ
−1div(
σ˜hi −σi),
divτh)
=0,
∀τh∈Wh.
(3.27)It is clear thatσ˜ihexist uniquely. By splittingσ˜ih−σiasσ˜i
h−σi=
(
σ˜ ih−σiI
)
+(σ
iI−σi)
and using(3.27), we have(
σ˜ih−σiI,
τh)
+(φ
−1div(
σ˜ih−σ iI
),
divτh)
=(σ
i−σiI,
τh)
+(φ
−1div(σ
i−σiI),
divτh)
≤ 1 2kσ i−σi Ik 2+1 2kτhk 2+ 1 2k
φ
−12 div(σ
i−σi I)
k + 1 2kφ
−12 divτhk2,
∀τh∈Wh.
(3.28) Letτh=σih−σ iI. We have the following error estimate,
k ˜σi
h−σ
ik + kdiv
(
σ˜i−σi)
k ≤Chk1σkσikk1+1
.
(3.29)From(3.28)we also get that
(δ
t(
σ˜ih−σi
I
),
τh)
+(φ
−1divδ
t(
σ˜ih−σ iI
),
divτh)
=(δ
t(σ
i−σiI),
τh)
+(φ
−1divδ
t(σ
i−σiI),
divτh),
∀τh∈Wh
.
(3.30)Letτh=
δ
t(σ
ih−σiI)
. We have the following error estimate similarly,k
δ
t(
σ˜i−σi)
k + kdivδ
t(
σ˜i−σi)
k ≤Chk1σ 11
t Z ti ti−1 kσtkk1+1dt !12.
(3.31)Theorem 3.4. Suppose
(
unh
,
σnh)
∈Vh×Whis the solution of Scheme (III). Under the assumptionkσ0h−σ0k =O(
hk+1 σ kσ0kHk+1
)
, then kσn h−σk + " n X i=11
tkdiv(
σi h−σ i+σi−1 h −σ i−1)
k2 #12 ≤Chhk1σkσk L∞(Hk1 +1)+h k+1 σ kσtkL∞(Hk+1) i +1
t2hku(3)t kL2(L2)+ kσttkL2(H1) i.
(3.32) Moreover, ifkdiv(σ
0h−σ0
)
k =O(
hk1σkdivσ0kHk1)
andu0h= ˜u0hwe havekunh−unk ≤Chmu+1hkukL∞(Hm+1)+ kutkL2(Hm+1) i +
1
t2hkut(4)kL2(L2)+ kσ( 3) t kL2(H1)+ ku( 3) t kL∞(L2)+ kσttkL∞(H1) i +hk1σkσ k L∞(Hk1 +1)+h k+1 σ kσtkL2(Hk1 +1)+ kσ kL∞(Hk+1).
(3.33)HereCdenotes a positive constantCindependent ofhu,hσand
1
t. Proof. First note that subtracting(2.38)from(2.39)leads tob
(
uih−ui,
σhi −σi;vh,
τh)
= uih−1−ui−1−1
t 2φ
−1(
div(σ
i−1 h −σ i−1)), φ
v h+1
t 2 divτh +1
t 2 A−1(σ
i−1 h −σ i−1)
+ ∇(
ui−1 h −u i−1),
τ h+A∇vh,
−1
tφ
−1ρ
i 2, φ
vh+1
t 2 divτh,
∀(
vh,
τh)
∈Vh×Wh.
(3.34)UsingLemma 2.2and(3.34)with a chosenvh=0, it follows that
1
t 2(
A −1(σ
i h−σ i),
τ h)
+1
t 2 2(φ
−1div(σ
i h−σ i),
divτ h)
=1
t 2(
u i−1 h −u i−1,
divτ h)
−1
t 2 2(φ
−1div(
σi−1 h −σ i−1),
divτ h)
+1
t 2(
A −1(σ
i−1 h −σ i−1),
τ h)
+1
t 2(
∇(
u i−1 h −u i−1),
τ h),
−(1
t)
2 2(φ
−1ρ
i 2,
divτh),
∀(
vh,
τh)
∈Vh×Wh.
(3.35)Let us split intoσi
h−σi=
ξ
σi −η
iσwhereξ
iσ =σih− ˜σ i
h∈Wh
,
η
iσ =σi− ˜σih.
(3.36)By the definition ofσ˜ihin(3.27), we have
(φ
−1div(η
i σ+η
iσ−1),
divτh)
= −(η
iσ+η
iσ−1,
τh).
It is clear that(
uih−1−ui−1,
divτh)
+(
∇(
uih−1−u i−1),
τ h)
=0.
Hence(3.35)reduces to: for allτh∈Wh
(A
−1(ξ
i σ−ξ
σi−1),
τh)
+1
t 2(φ
−1div(ξ
i σ+ξ
iσ−1),
divτh)
=1
t(
A−1δ
tη
iσ,
τh)
−1
t 2(η
i σ+η
iσ−1,
τh)
−1
t(φ
−1ρ
i2,
divτh).
(3.37)Lettingτh=
ξ
iσ+ξ
σi−1∈Whin(3.37)and using the Cauchy inequality, we havekA−12
ξ
i σk2− kA− 1 2ξ
i−1 σ k2+1
t 2 kφ
−12 div(ξ
i σ+ξ
iσ−1)
k2≤1
tkξ
iσ+ξ
iσ−1k2+1
t 4 kφ
−12 div(ξ
i σ+ξ
iσ−1)
k2 +1
t 1 2kA −1δ
tη
iσk2+ 1 8kη
i σ+η
iσ−1k2+ kφ
− 1 2ρ
i 2k2.
(3.38)Summing(3.38)fori=1
,
2, . . . ,
nwe can deduce that kA−12ξ
n σk2+ 1 2 n X i=11
tkφ
−12div(ξ
i σ+ξ
σi−1)
k2 ≤C n X i=11
tkξ
iσk2+Ckξ
0σk2+C n X i=11
t[kδ
tη
iσk2+ kη
iσk2+ kρ
i2k2] +C1
tkη
0σk2.
(3.39) Using Gronwall’s Lemma we can get that,k
ξ
nσk2+ n X i=11
tkdiv(ξ
i σ+ξ
iσ−1)
k2≤Chσ2(k+1)(
kσk2L∞(Hk+1)+ kσtk2L2(Hk+1))
+C1
t4(
ku( 3) t k2L2(L2)+ kdivσttk2L2(L2)).
(3.40)Combining with(3.4)completes the proof of(3.32).
Choosingτh= 11t
(ξ
σi −ξ
σi−1)
=δ
tξ
iσin(3.37)we have that1
tkA−12δ
tξ
iσk2+ 1 2kφ
−12 divξ
i σk2− 1 2kφ
−12 divξ
i−1 σ k2 =1
t(A
−1δ
tη
iσ, δ
tξ
iσ)
−1
t 2(η
i σ+η
iσ−1, δ
tξ
iσ)
−1
t(φ
−1ρ
i 2,
divδ
tξ
iσ),
≤C1
tkA−12δ
tξ
iσk(
kδ
tη
iσk + kη
iσ+η
i −1 σ k)
−(φ
−1ρ
i2,
divξ
i σ)
+(φ
−1ρ
i2−1,
divξ
i−1 σ)
−1
t(φ
−1δ
tρ
i2,
divξ
i−1 σ),
(3.41) where we have used the equivalence1
t(φ
−1ρ
i2,
divδ
tξ
iσ)
=(φ
−1ρ
i2,
divξ
iσ)
−(φ
−1ρ
i2−1,
divξ
iσ−1)
−1
t(φ
−1δ
tρ
i2,
divξ
iσ−1).
For convenience we introduce a notationρ
02and
δ
tρ
12 = ρ12−ρ02
1t . Making summation overi = 1
,
2, . . . ,
nand using theCauchy inequality result in
n X i=1
1
tkA−12δ
tξ
i σk2+ 1 2kφ
−1 2divξ
n σk2≤C n X i=11
tkA−12δ
tξ
i σk2(
kδ
tη
iσk + kη
iσ+η
i −1 σ k)
+ 1 2kφ
−1 2divξ
0 σk2 −(φ
−1ρ
n2,
divξ
n σ)
+(φ
−1ρ
02,
divξ
0σ)
+ n X i=21
t(φ
−1δ
tρ
i2,
divξ
i−1 σ)
+1
t(φ
−1δ
tρ
12,
divξ
0σ).
(3.42) Since(φ
−1ρ
02
,
divξ
0σ)
+1
t(φ
−1δ
tρ
12,
divξ
0σ)
=(φ
−1ρ
12,
divξ
0σ),
using the-inequality we have thatn X i=1
1
tkA−12δ
tξ
iσk2+ 1 2kφ
−12 divξ
n σk2≤ 1 2 n X i=11
tkA−12δ
tξ
iσk2+ 1 4kφ
−12 divξ
n σk2+Ckdivξ
0σk2 +C n X i=11
t(
kδ
tη
i σk2+ kη
iσk2)
+C(
kη
0σk2+ kρ
n2k2+ kρ
12k2)
+C n X i=21
tkδ
tρ
i 2k2+C n X i=11
tkφ
−12 divξ
i σk2.
(3.43) Moving the first two terms of the right-hand side to the left side, then Gronwall’s inequality results inkdiv
ξ
n σk2+ n X i=11
tkδ
tξ
iσk2 ≤ Ch2σ(k+1)(
kσ k2 L∞(Hk+1)+ kσtk2L2(Hk+1))
+Ch2σk1kdivσ0k2Hk1 +C1
t4(
kut(4)k2L2(L2)+ kσ (3) t k2L2(L2)+ ku (3) t k2L∞(L2)+ kσttk2L∞(L2)).
(3.44)Now we consider the estimate ofun h−u n. Choosingτ h=0 in(3.34)we have that,
(φ(
uih−ui),
vh)
+1
t 2(
A∇(
u i h−u i),
∇v h)
=(φ(
uih−1−u i−1),
v)
−1
t(ρ
i 2,
vh)
−1
t(
div(σ
ih−1−σ i−1),
v h)
+1
t 2(
A∇(
u i−1 h −u i−1),
∇v h),
∀vh∈Vh.
(3.45) Denote byξ
n u=u n h− ˜u n h,
η
n u=u n− ˜un h.
(3.46)From(3.45)we have that
(φξ
i u,
vh)
+1
t 2(
A∇ξ
i u,
∇vh)
=(φξ
i −1 u,
vh)
+1
t(φδ
tη
iu,
vh)
−1
t(
div(σ
ih−1−σ i−1),
v h)
−1
t(ρ
i2,
vh)
+1
t 2(A∇ξ
i−1 u,
∇vh),
∀vh∈Vh.
(3.47)Settingvh=
ξ
iuand using the Cauchy inequality we have thatk
φ
12ξ
i uk 2+1
t 2 kA 1 2∇ξ
i uk 2 ≤ 1 2kφ
1 2ξ
i uk 2+1 2kφ
1 2ξ
i−1 u k 2+1
t 6 kφ
1 2ξ
i uk 2+C1
tkδ
tη
iuk 2 +1
t 6 kφ
1 2ξ
i uk 2+C1
tkdiv(σ
i−1 h −σ i−1)
k2 +1
t 6 kφ
1 2ξ
i uk 2+C1
tkρ
i 2k2+1
t 4 kA 1 2∇ξ
i uk 2+1
t 4 kA 1 2∇ξ
i−1 u k 2 = 1+1
t 2 kφ
1 2ξ
i uk 2+1
t 4 kA 1 2∇ξ
iuk2+1 2kφ
1 2ξ
i−1 u k 2+1
t 4 kA 1 2∇ξ
iu−1k2 +C1
t[kδ
tη
uik2+ kdiv(σ
i−1 h −σ i−1)
k2+ kρ
i 2k2],
(3.48) then kφ
12ξ
i uk 2+1
t 2 kA 1 2∇ξ
iuk2 ≤ kφ
12ξ
i−1 u k 2+1
tkφ
1 2ξ
i uk 2+1
t 2 kA 1 2∇ξ
iu−1k2 +C1
t[kδ
tη
i uk 2+ kdiv(
σi−1 h −σ i−1)
k2+ kρ
i 2k2].
(3.49)Summing(3.49)overi=1
,
2, . . . ,
n, we have that kφ
12ξ
n uk 2+1
t 2 kA 1 2∇ξ
nuk2≤ n X i=11
tkφ
12ξ
n uk 2+ kφ
1 2ξ
0 uk 2+1
t 2 kA 1 2∇ξ
u0k2 +Ch2(m+1) u kutk2L2(Hm+1)+1
t2(
ku(3)t k2L2(L2)+ kdivσttk 2 L2(L2))
+C n X i=11
tkdiv(σ
i−1 h −σ i−1)
k2.
(3.50) Since div(σ
i−1 h −σ i−1)
= div(ξ
i−1 σ)
+ div(
σ˜i −1 h −σ i−1),
noticing(3.44)and(3.31), by Gronwall’s inequality shows thatkunh− ˜unhk +
1
t12k∇ξ
nuk ≤Chmu+1kutkL2(Hm+1)+C1
t2(
ku(t4)kL2(L2)+ kσ(t3)kL2(H1))
+C1
t2(
kut(3)kL∞(L2)+ kσttkL∞(H1))
+Chk1σkσkL∞(Hk1 +1)+Chk+1
σ
(
kσtkL2(Hk+1)+ kσkL∞(Hk+1)).
Combining with(3.11)completes the proof. 4. Least-squares procedure for nonlinear problems
In this section we give a least-squares finite element procedure for nonlinear parabolic problems. We consider the following problem on a bounded domainΩ⊂Rd:
φ(
u)
ut−div(
A(
u)
∇u)
=f(
u),
inΩ×J,
u=0,
onΓD×J,
A(
u)
∇u·n=0 onΓN×J,
(4.1)subject to the initial condition
u
(
x,
0)
=u0(
x)
onΩ×J.
(4.2)The coefficient
φ(
u)
is a strictly positive function and the coefficient matrixA(
u)
=(
aij(
u))
di,j=1is a bounded, symmetric and positive definite matrix, i.e., there exist two positive constantsφ
1andφ
2and two positive constantsα
andβ
such that, foru∈R1