Iterative finite element approximations of solutions to parabolic equations with nonlocal conditions

Nonlinear Analysis 50 (2002) 433– 454www.elsevier.com/locate/na

Iterative �nite element approximations ofsolutions to parabolic equations with

nonlocal conditions

Dennis Jackson∗

Department of Mathematical Sciences, Florida Institute of Technology, 150 West University Boulevard,Melbourne, FL 32901-6975 USA

Received 25 February 2000; accepted 4 December 2000

1. Introduction

The following nonlocal parabolic equation will be studied:

ut + Au=f(x; t) on Q;

u |�=0;

∫ T0

0h(x; �)u(x; �) d�+ g(T1; : : : ; TN ; u)= (x);

(1.1)

where 0¡T06T; � is a bounded domain in Rn with a smooth boundary �;Q=� ×(0; T ), and A is a second order elliptic operator,

Au=−n∑

i; j=1

@@xi

(aij(x)

@u@xj

)+

n∑i=1

ai(x)@u@xi

+ a0(x)u (1.2)

with aij(x); ai(x)∈C∞( 2�), and h(t; x)∈C∞( 2Q0) where Q0 = [0; T0]× �.

∗ Corresponding author. Tel.: +1-404-674-8091; fax: +1-404-674-7412.E-mail address: jackson@zach.�t.edu (D. Jackson).

0362-546X/02/$ - see front matter c© 2002 Elsevier Science Ltd. All rights reserved.PII: S0362 -546X(01)00741 -6

434 D. Jackson / Nonlinear Analysis 50 (2002) 433– 454

The function g: C0((0; T ];L2(�)) → L2(�) will include functions of the form

g(T1; : : : ; TN ; u)=N1∑i=1

hi(x)u(x; Ti) +N∑

i=N1+1

∫ Ti+Ki

Ti

hi(x; �)u(x; �) d�; (1.3)

with 0¡Ti6T for 16 i6N , Ti + Ki6T and 0¡Ti for N1 + 16 i6N; hi(x)∈C∞( 2�), and hi(x; t)∈C∞( 2� × [Ti; Ti + Ki]).An example of a problem of the form (1.1) is the problem of recovering the ini-

tial concentration of a chemical reactant from measurements of the pointwise amountof reaction over a period of time. Let u(x; t) represent the concentration of the re-actant. Assume it reacts with the region proportionally to the concentration at a rateh(x; t)u(x; t). Also, assume the concentration of the chemical satis�es a diDusion equa-tion, the amount of chemical removed when it reacts is at a rate k(x; t)u(x; t), and (x)is the amount of the reaction from time 0 when the chemical is suddenly mixed intothe region to time T0. The problem can be represented by the system

ut + Au=f(x; t)− k(x; t)u;

u |�=0;∫ T0

0h(x; �)u(x; �) d�= (x):

(1.4)

Nonlocal parabolic problems of the form

ut + Au=f(u);

u |�= g(x);

u(x; 0) + g(T1; : : : ; TN ; u)= (x);

(1.5)

where there is a condition connecting the initial value of u to values at later times,have been studied by several authors. Existence and uniqueness of solutions have beenshown by Byszewski [2], Chabrowski [3], Hess [5], Jackson [7], Kerelov [8] andVabishchewich [15]. In [6], the author obtains error estimates for the semidiscrete �niteelement approximation of equations of the form (1.5) where the parabolic equation islinear. Lin [10] uses an iterative �nite diDerence method to approximate solutions tononlocal problems of the form (1.5).In this paper, it will be shown that system (1.1) has a unique solution under suit-

able conditions on the data. We then approximate the system by a nonlocal system ofordinary diDerential equations and obtain error estimates for the semidiscrete approxi-mation of the solution. Finally, we use a fully discrete iterative procedure to approx-imate u(x; 0). The solution to system (1.1) can then be approximated using standardnumerical methods.

2. Existence and uniqueness of solutions

In this section, the Banach �xed point theorem is used to show the existence anduniqueness of solutions to Eqs. (1.1).

D. Jackson / Nonlinear Analysis 50 (2002) 433– 454 435

Let a(u; v) be de�ned by

a(u; v)=n∑

i; j=1

∫�

aij(x)@u@xi

@v@xj

dx +n∑

i=1

∫�

ai(x)@u@xi

v dx +∫�a0(x)uv dx: (2.1)

Assume for u∈H 10 (�),

a(u; u)¿ �‖u‖21; (2.2)

where Hs(�) and Hs0(�) are the usual Sobolev spaces with norms ‖ ‖s and ‖ ‖= ‖ ‖0,

see [1] or [11].If inequality (2.2) is not satis�ed, but

a(u; u) + 0‖u‖2¿ �‖u‖21;replace u by e 0tw in Eqs. (1.1) and A by 2A=A+ 0I .

Conditions (1.2) and (2.2) imply A with domain, D(A)=H 2(�)∩H 10 (�), generates

an analytic semigroup S(t)= e−At . The fractional powers of A are de�ned. If A satis-�es the inequality (2.2), f∈D(A%); %¿ 0 and 0¡%¡�, one can �nd a constant M'

depending continuously on � for each ' such that

‖A%S(t)f‖6Ma'e−at

t'‖f‖D(A%′ ); (2.3)

where '= % − %′; '¿ 0. See Pazy [12]. In the case A is positive self-adjoint withsmallest eigenvalue �, we can take Ma

' =(�'=(� − a))'e−'.Let the spaces W*;% be de�ned by

W*;a = {f∈C0((0; T ];L2(�)) | sup0¡t6T

t1+*‖f(t)‖¡∞} (2.4)

with Wa =W0; a and norms

‖f‖W*; a = sup0¡t6T

t1+*eat‖f(t)‖: (2.5)

Let a be a constant such that inequality (2.3) is satis�ed. We will assume g :Wa →L2(�) such that for u; v∈Wa,

‖g(T1; : : : ; TN ; u)− g(T1; : : : ; TN ; v)6 k*1 ‖u− v‖W*; a ; *=0: (2.6)

If g is of the form (1.3), then inequality (2.6) is satis�ed for *¿ 0 with constantk*1 given by

k*1 =

N1∑i=1

mi

T 1+*i

e−aTi +N∑

i=N1+1

∫ Ti+Ki

Ti

Hi(�)�1+* ea� d�; (2.7)

where mi =supx∈�|hi(x)| and Hi(t)= supx∈�|hi(x; t)|.Assume

h(x; t)= 1 + k(x; t) (2.8)


with k(x; 0)=0, and

‖e−AT0‖¡ 1 (i:e:; Ma0 e

−aT0 ¡ 1): (2.9)

If A is self-adjoint and condition (2.2) is satis�ed, inequality (2.9) holds for allT0; T0 ¿ 0. If condition (2.9) is not satis�ed, replace u by e 0tv for 0¿ 0 large enough.This, however, can increase the size of k*

1 for the new nonlocal function g.If u∈C0([0; T ];L2(�)) satis�es system (1.1) with f∈L1(0; T ;L2(�)), then

u(x; t)= S(t)u(0) +∫ t

0S(t − �)f(x; �) d�: (2.10)

The third condition in the system can be written as∫ T0

0S(�)u(0) d�+

∫ T0

0

∫ �

0h(x; �)S(�− �′)f(x; �′) d�′ d�

+∫ T0

0k(x; �)S(�)u(0) d�+ g(T1; : : : ; TN ; u)= (x): (2.11)

Inequalities (2.2) and (2.9) imply A−1 and (I − e−AT0 )−1 exist. Let

F(u)≡ (I − e−AT0 )−1[ (x)− g(T1; : : : ; TN ; u)−

∫ T0

0k(x; �)u(�) d�

]

−∫ T0

0

∫ �

0(I − e−AT0 )−1(e−A(�−�′)f(x; �′)) d�′ d�: (2.12)

Since A−1 exists,∫ T0

0S(�)u(0) d�=A−1(I − e−AT0 )u(0):

Thus, if the right-hand side of identity (2.12) is a member of D(A), Eqs. (2.10) and(2.11) are equivalent to

u(t)= S(t)AF(u) +∫ t

0S(t − �)f(x; �) d�: (2.13)

Recall, for 0¡.¡ 1; C.([0; T ];L2(�))= {f∈C0([0; T ];L2(�))| ‖f(t) − f(s)‖6L | t − s|.} for some L¿ 0, where L is independent of s and t.We will use the following lemmas:

Lemma 2.1. Let f∈C0([0; T ];D(A%)); %¿ 0, such that A%f∈C.([0; T ];L2(�)) forsome .; 0¡.¡ 1. Let h∈C∞( 2Q) and assume for each t; 06 t6T0; h(x; t)v∈D(A%+1) if v∈D(A%+1). Then∫ T0

0

∫ �

0h(x; �)e−A(�−�′)f(x; �′) d�′ d�∈D(A%+1):


Proof. Let f and h be as above. De�ne v(t) by

v(t)=∫ t

0e−A(t−�)f(x; �) d�: (2.14)

Since

A%v(t)=∫ t

0e−A(t−�)A%f(x; �) d�;

we have A%v∈C0([0; T ];D(A)). See Pazy [12, Theorem 4:3:5]. The lemma follows,since hv∈C0([0; T ];D(A%+1)).

Lemma 2.2. If w∈D(A%); %¿ 0; 2% �= integer+12 ; and g∈C∞( 2Q) such that g(x; �)v∈

D(Ak+1) for all �; 06 �6T0; and v∈D(Ak+1); where k is an integer such that− 1

4 ¡k − %¡ 34 ; then∫ T0

0g(x; �)S(�)w d�∈D(A%+1):

Proof. Assume the above hypotheses are satis�ed. Let %=m+ %′; where m is a non-negative integer, and 06 %′ ¡ 1: We have

Am+1(g(x; �)S(�)w) =∑

|*|62m

l*(x; �)D*(S(�)w)

+n∑

j=1

mj(x; �)@@xj

(AmS(�)w) + g(x; �)Am+1S(�)w; (2.15)

where l*(x; �)∈C∞( 2Q) and

mj(x; �)= (m+ 1)n∑

i=1

(aij(x) + aji(x))@g(x; �)@xi

:

Let 0¡2′ ¡2: Since S(�)w∈C0((0; T ];D(A.)) for all .¿ 0;∫ T0

2′ g(x; �)S(�)w d�; and∫ T0

2 g(x; �)S(�)w d�∈D(A%+1):Also, since

D(A%)=

{H 2%(�) if 06 %¡ 1

4 ;

H 2%(�) ∩ H 10 (�) if 1

4 ¡%6 1; % �= 34 ;

see Fujiwara [4],∫ 2

2′g(x; �)Am+1S(�)w d�=−

∫ 2

2′g(x; �)

dd�

(S(�)Amw) d�

= g(x; 2′)S(2′)Amw − g(x; 2)S(2)Amw

+∫ 2

2′g�(x; �)S(�)Amw d�


and S(�)w∈C0([0; T ];H 2%(�));∥∥∥∥A%+1∫ 2

2′g(x; �)S(�)w d�

∥∥∥∥6C∥∥∥∥Am+1

∫ 2

2′g(x; �)S(�)w d�

∥∥∥∥2%′

6∑

|*|62m

∫ 2

2′‖l*(x; .)D*(S(�)w)‖2%′ d�

+n∑

j=1

C∫ 2

2′‖AmS(�)w‖1+2%′ d�

+ ‖ g(x; 2′)S(2′)w − g(x; 2)S(2)w‖2%′

+∫ 2

2′‖g�(x; �)S(�)Amw‖2%′ d�

6C(2− 2′) +n∑

j=1

C∫ 2

2′‖A1=2S(�)Am+%′w‖ d�

6C(2− 2′) +n∑

j=1

C∫ 2

2′

1�1=2

‖S(�)w‖D(A%) d�

6C(2− 2′):

The lemma follows since A%+1 is a closed operator.

Lemma 2.3. If w∈D(A%); %¿ 0; 2% �= integer+12 ; g(x; t)∈C∞( 2Q); such that g(x; t)v∈

D(Ak+1) for all t; 06 t6T0; and v∈D(Ak+1); where − 14 ¡k − %¡ 3

4 : Then∫ T0

0�g(x; �)AS(�) d�∈D(A%+1):

Proof. The proof follows from Lemma 2.2, since

∫ T0

0�g(x; �)AS(�)w d�=− T0g(x; T0)S(T0)w −

∫ T0

0(�g(x; �))�S(�)w d�:

We can now prove our main existence and uniqueness theorem for the solution ofEqs. (1.1).Assume inequality (2.9) holds. Then (I − e−AT0 )−1 exists. Let

L1 = ‖(I − e−AT0 )−1‖: (2.16)

Theorem 2.4. Let (x)∈D(A%=2+1); %¿ 0; % �= integer + 12 , f(x; t)∈C0([0; T ];

D(A%=2)) such that A%=2f∈C.([0; T ];L2(�)) for some .; 0¡.¡ 1: Also assume


conditions (1:2); (2:2); (2:3); (2:6) and (2:9): Assume g(T1; : : : ; TN ; u)∈D(A%=2+1),u∈C0([0; T ]; D(A%=2+1)); and let k(t)=maxx∈�|k(x; t)|: Then if

M%1 L1k01 +Ma

1 L1

∫ T0

0�−1k(�)e−a� d�¡ 1; (2.17)

Eqs. (1.1) have a unique solution u; such that

u∈C0([0; T ];D(A%=2)) ∩ C1((0; T ];D(A%=2)) ∩ C0((0; T ];D(A%=2+1)): (2.18)

Proof. Assume the above hypotheses are satis�ed. For u∈W0; a; de�ne 3(u) by

3(u)=AS(t)F(u) +∫ t

0S(t − �)f(x; �) d�; (2.19)

where F(u) is de�ned by identity (2.12). Since F(u)∈L2(�); and ‖AS(t)F(u)‖6 (Ma

1 =t)‖F(u)‖; it follows that 3(u)∈W0; a:Let u; v∈W0; a: Then

‖teat(3(u)(t)− 3(v)(t))‖6Ma

1 L1‖g(T1; : : : ; TN ; u)− g(T1; : : : ; TN ; v‖

+Ma1 L1

∥∥∥∥∫ T0

0k(x; �)(u(�)− v(�)) d�

∥∥∥∥6Ma

1 L1k01‖u− v‖W0; a +Ma1 L1

∫ T0

0�−1 k(x; �)

ea�‖�ea�(u(�)− v(�))‖ d�

6Ma1 L1k01‖u− v‖W0; a +Ma

1 L1

∫ T0

0�−1k(�)e−a� d�‖u− v‖W0; a

6d‖u− v‖W0; a ;

where 0¡d¡ 1:Thus by the Banach �xed point theorem, there is a unique u∈W 1

0; a such thatu=3(u):Then by Eq. (2.19)

∫ T0

0k(x; �)u(�) d�=

∫ T0

0k(x; �)AS(�)F(u) d�+

∫ T0

0k(x; �)v(�) d�;

where v(t) is de�ned by Eq. (2.14). From the above and Lemmas 2.1 and 2.3, it fol-lows that if F(u)∈D(Ai); 06 i6 %=2+ 1 then

∫ T0

0 k(x; �)u(�) d�∈D(Ai+1): ThereforeF(u)∈D(A%=2+1); which implies conclusions (2.18).

If inequality (2.17) is not satis�ed, system (1.1) may not have a unique solution.


Example. Consider the system

ut − uxx =0; if 0¡x¡ 1;

u(0; t)= 0; u(1; t)= 0;∫ 1

0u(x; �) d�− bu(x; 2)= (x) (2.20)

for a= 62− 12 ; Ma

1 = 262e−1: Also L1 = (1−e−62)−1 and k01 = (b=2)e−262+1: Thus, Eqs.

(2.20) have a unique solution if (x)∈H 2(0; 1)∩H 10 (0; 1) and b62e−262

=(1−e−62)¡ 1:

But if b=(1 − e−62)=62e−262

; then u(x; t)=Ae−62t sin 6x satis�es Eqs. (2.20) with (x) ≡ 0 for all values of A and the system has no solution if (x)= sin 6x:

3. The semidiscrete approximation

In this section, we approximate system (1.1) by a nonlocal system of diDeren-tial equations. Under suitable conditions we show the system we use to approximatethe nonlocal problem (1.1) also has a unique solution. Finally, we obtain error esti-mates for the semidiscrete approximations, using semigroup methods similar to those ofLasiecka [9].Let {Vh} be a family of �nite dimensional vector spaces such that Vh ⊆ H 1(�) and

inf8∈Vh

{‖f − 8‖+ h‖f − 8‖1}6Chs‖f‖s (3.1)

for f∈Hs(�) ∩ H 10 (�); 16 s6 r; where C is independent of h.

Assume there are linear operators Ah from Vh into itself for h small enough suchthat for fh; gh ∈Vh and 2¿ 0

(Ahfh; fh)¿ �′‖fh‖21 (3.2)

with

�6 �′ + 2; (3.3)

(Ahfh; gh)6C‖fh‖1‖gh‖1 (3.4)

and

‖(PhA−1 − A−1h Ph)f‖6Ch%+2‖A%=2f‖; 06 %6 r − 2; (3.5)

where Ph is the L2 projection of L2(�) onto Vh and C is independent of h in eachcase.The above conditions are satis�ed with �′ in inequality (3.2) equal to � in inequality

(2.2) for the standard Galerkin method. They are also satis�ed if Nitsche’s method is


used with Vh ∈H 1(�); Vh|� ⊆ H 1(�) such that for 26 s6 r and f∈H 2(�)

inf8∈Vh

{‖f − 8‖+ h‖f − 8‖1 + h1=2‖f − 8‖L1(�) + hs=2‖f − 8‖H 1(�)}6Chs‖f‖s

and Ah :Vh → Vh is de�ned by

(Ahfh; gh)= a(fh; gh)−(@fh

@n; gh)L2(�)

+ *h−1(fh; gh)L2(�)

with conditions (3.2) and (3.3) satis�ed for * large enough, see [12] or [9].We approximate Eqs. (2.1) by the semidiscrete system on Vh:

u′h(t) + Ahuh =Phf(x; t); (3.6)

∫ T0

0Ph(h(x; �)uh(�)) d�+ Phg(T1; : : : ; TN ; uh)=Ph (x):

Inequality (3.2) implies that for Sh(t)= e−Aht

‖AihSh(t)fh‖6M

′a1e−at

t‖fh‖; i=0; 1; (3.7)

where M′a1 depends on a and �′ and not h with M

′a1 → Ma

1 in (2.3) as �′ → �:Let

Wh =W0; a ∩ C0((0; T ];Vh) (3.8)

with norm ‖fh‖Wh = ‖fh‖W0; a for fh ∈Wh:

Theorem 3.1. Let the conditions of Theorem 2:4 be satis7ed. Also assume conditions(3:2) with 2 small enough such that (2:17) is satis7ed if Ma

1 is replaced by M′a1 and

L1 replaced by L2; L2¿Lh1 = ‖(Ih − e−AhT0 )−1‖ for each h. Then for each h there is

a unique uh ∈C0([0; T ];Vh) which satis7es Eqs. (3.6).

Proof. De�ne 3h :Wh → Wh by

3h(uh)= Sh(t)AhFh(uh) +∫ t

0Sh(t − �)Phf(x; �) d�;

where

Fh(uh)≡ (Ih − e−AhT0 )−1[Ph (x)− Phg(T1; : : : ; TN ; uh)−

∫ T0

0Ph(k(x; �)uh(�) d�

]

−∫ T0

0

∫ �

0(Ih − e−AhT0 )−1e−Ah(�−�′)Phf(x; �′) d�′ d�: (3.9)

It follows as in the proof of Theorem 2.4, that 3h has a unique �xed point uh ∈Wh: Butsince Ah is bounded, it is easy to show uh ∈C0([0; T ];Vh) and uh satis�esEqs. (3.6).


Under assumptions (3.1)–(3.5), with 06 %6 s6 r; 06 %′6 r − 2; u0 ∈D(A%=2);and f∈L∞(0; T ;D(A%′=2)); we have

‖(S(t)− Sh(t)Ph)u0‖6 Chs

t(s−%)=2 ‖A%=2u0‖ (3.10)

and ∥∥∥∥∫ t

0(S(t − �)− Sh(t − �)Ph)f(x; �) d�

∥∥∥∥6Ch%′+2 ln1h‖f‖L∞(0;T ;D(A%′ =2)): (3.11)

We will need the following lemmas:

Lemma 3.2. If u0 ∈D(As=2); 16 s6 r;

‖[(I − S(T0))−1 − (Ih − Sh(T0))−1Ph]u0‖6Chs‖u0‖D(As=2): (3.12)

Proof. We have

‖[(I − S(T0))−1 − (Ih − Sh(T0))−1Ph]u0‖6 ‖(Ih − Sh(T0))−1Ph(Ih − Sh(T0))Ph − I + S(T0)(I − S(T0))−1u0‖+ ‖(I − Ph)(I − S(T0))−1u0‖

6 ‖(Ih − Sh(T0))−1‖ ‖(S(T0)− Sh(T0)Ph)(I − S(T0))−1u0‖+Chs‖(I − S(T0))−1u0‖D(As=2)

6Chs‖(I − S(T0))−1u0‖D(As=2)

6Chs‖u0‖D(As=2);

since (Ih − Sh(T0))−1 is bounded with a bound independent of h. The boundednessdepends on �′ in inequality (3.2).Using an argument similar to one in [9], we have the following lemma.

Lemma 3.3. If u0 ∈D(A%=2+1); 06 %6 s6 r and Ah satis7ed conditions (3:1)–(3:5);then for 0¡t6T0; *=max{1; (s− %)=2}

‖(AS(t)− AhSh(t)Ph)u0‖6 Chs

t*‖u0‖D(A1+%=2): (3.13)

Proof. First assume s6 % + 2. Then we can use error estimates (3.7) and (3.10) toshow

‖(AS(t)− AhSh(t)Ph)u0‖6 ‖(S(t)− Sh(t)Ph)Au0‖+ ‖Ah(t)Sh(t)(A−1

h Ph − PhA−1)Au0‖

6Chs

t(s−%)=2 ‖Au0‖D(A%=2) +M ′a

1

ths‖Au0‖D(A%=2)

6Chs

t*‖u0‖D(A%=2+1):


If %+ 26 s we can use the above and estimate (3.10) to show

‖(AS(t)− AhSh(t)Ph)u0‖

6∥∥∥(AS ( t

2

)− AhSh

( t2

)Ph

)S( t2

)u0∥∥∥

+∥∥∥AhSh

( t2

)Ph

(S( t2

)− Sh

( t2

)Ph

)u0∥∥∥

6Chs

t

∥∥∥∥A(s−%−2)=2S( t2

)u0‖D(A(%+2)=2) +

ct

hs

t(s−2−%)=2 ‖u0∥∥∥∥D(A%=2+1)

6Ch2

t(s−%)=2 ‖u0‖D(A%=2+1):

We can now prove our main approximation theorem.

Theorem 3.4. Assume the hypotheses of Theorems 2:4 and 3:1 are satis7ed with06 %6 s6 r; 06 s; assume conditions (3:1)–(3:5), and assume for *=max{0; (s−%− 2)=2}

L2M ′a1

(T*k*

1 +∫ T0

0

k(�)�1+* e

−a� dt)

¡ 1; (3.14)

where k(�)=maxx∈ 2�|k(x; �)|. Let u(t) satisfy system (1:1) and uh(t) system (3:6).Then

‖u(t)− uh(t)‖6C ;f;uhs(

1t1+* + 1 + d ln

1h

)

where d=1 if %6 s− 2; d=0 if %¿s− 2; and C ;f;u is a constant depending on u;but not h.

Proof. Since u(t) satis�es Eq. (2.13) and uh(t) satis�es

uh(t)= Sh(t)AhFh(uh) +∫ t

0Sh(t − �)Phf(x; �) d�; (3.15)

where Fh(uh) is de�ned by identity (3.9), it follows

‖u(t)− uh(t)‖6 ‖[(I − e−AT0 )−1AS(t)− (Ih − e−AhT0 )−1AhSh(t)Ph] ‖+ ‖[(I − e−AT0 )−1AS(t)g(T1; : : : ; TN ; u)− (Ih − e−AhT0 )−1

·AhSh(t)Phg(T1; : : : ; TN ; uh)]‖

+∥∥∥∥(I − e−AT0 )−1

∫ T0

0(AS(t)(k(x; �)e−A�u(0) d�


− (I − e−AhT0 )−1∫ T0

0AhSh(t)Ph(k(x; �)e−Ah�Phuh(0) d�

∥∥∥∥+∥∥∥∥∫ T0

0

∫ �

0(I − e−AT0 )−1S(t)A(h(x; �)e−A(�−�′)f(x; �′)

− (Ih − e−AhT0 )−1AhSh(t)Ph(h(x; �)e−Ah(�−�′)Phf(x; �′)) d�′ d�∥∥∥

+∥∥∥∥∫ t

0(S(t − �)− Sh(t − �)Ph)f(x; �) d�

∥∥∥∥= I1 + I2 + I3 + I4 + I5:

We can use Lemmas 3.2 and 3.3 to show

I16 ‖((I − e−AT0 )−1 − (Ih − e−AhT0 )−1Ph)S(t)A (x)‖+ ‖(Ih − e−AhT0 )−1Ph(AS(t)− AhSh(t)Ph) ‖

6Chs‖A(s−%)=2S(t)A1+%=2 ‖+ Chs

t1+* ‖ ‖D(A%=2+1)

6Chs

t1+* ‖ ‖D(A%=2+1): (3.16)

Similarly,

I26 ‖((I − e−AT0 )−1 − (Ih − e−AhT0 )−1Ph)S(t)Ag(T1; : : : ; TN ; u)‖+ ‖(Ih − e−AhT0 )−1Ph(AS(t)− AhSh(t)Ph)g(T1; : : : ; TN ; u)‖

6Chs 1t(s−%)=2 ‖g(T1; : : : ; TN ; u‖D(A1+%=2)

6Chs

t1+* ‖g(T1; : : : ; TN ; u)‖D(A%=2+1) + Lh1M ′a

1

te−atk*

1 ‖u− uh‖W*; a (3.17)

and also using estimate (3.11),

I3 + I46∥∥∥∥((I − e−AT0 )−1 − (Ih − e−AhT0 )−1Ph)S(t)A)

·(∫ T0

0k(x; �)e−A�u(0) d�+

∫ T0

0

∫ �

0h(x; �)e−A(�−�′)f(x; �′) d�′ d�

)∥∥∥∥+∥∥∥∥(Ih − e−AhT0 )−1Ph(AS(t)− AhSh(t)Ph)

·(∫ T0

0k(x; �)e−A�u(0) d�+

∫ T0

0

∫ �

0h(x; �)e−A(�−�′)f(x; �′) d�′ d�

)∥∥∥∥+∥∥∥∥(Ih − e−AhT0 )−1AhSh(t)Ph

∫ T0

0k(x; �)(u(x; �)− uh(x; �)) d�

∥∥∥∥


+∥∥∥∥(Ih − e−AhT0 )−1AhSh(t)Ph

∫ T0

0

∫ �

0(e−A(�−�′)

− e−Ah(�−�′)Ph)f(x; �′) d�′ d�∥∥∥

6Chs(

1t1+*

)∥∥∥∥∫ T0

0k(x; �) e−A�u(0) d�

+∫ T0

0

∫ �

0h(x; �)e−A(�−�′)f(x; �′) d�′ d�

∥∥∥∥D(A%=2+1)

+L2M ′a

1 e−at

t

∫ T0

0

k(�)�1+* e−a�‖u− uh‖W*; a d�

+Cth%+2 ln 1

h‖f‖L∞(0;T ;D(A%=2)): (3.18)

Finally, with the use of estimate (3.11),

I56Ch%+2 ln1h‖f‖L∞(0;T ;D(A%=2)): (3.19)

Let d=1 if %= s− 2 and d=0 if %¿s− 2. Then

‖u(t)− uh(t)‖6C ;f;uhs(

1t1+* + 1 + d ln

1h

)

+L2M′a1e−at

t

(K*1 +

∫ T0

0

k(�)�1+* e

−a� d�)‖u− uh‖W*; a ; (3.20)

where C ;f;u is the constant formed from the constants in inequalities (3.16)–(3.19).Thus by multiplying inequality (3.20) by t1+*eat and taking the supremum, it follows

from inequality (3.14)

‖u(t)− uh(t)‖W*; a 6C ;f;u;T hs(1 + d ln

lh

): (3.21)

The theorem follows from inequalities (3.20) and (3.21). Under additional assump-tions, error estimates can be obtained up to time t=0.

Let

P1h =A−1

h PhA: (3.22)

Then we have the following lemma.

Lemma 3.5. If Ah satis7es conditions (3.1)–(3.5); and 06 %6 s6 r; 16 s; then forf∈D(A%=2+1);

‖(AS(t)− AhSh(t)P1h)f‖6

Chs

t(s−%)=2 ‖f‖D(A%=2+1): (3.23)


Proof. The lemma follows from inequality (3.10) since

‖(AS(t)− AhSh(t)P1h)f‖ = ‖(S(t)A− Sh(t)PhA)f‖

6Chs

t(s−%)=2 ‖Af‖D(A%=2):

We will assume

h(x; t)= 1 + k(t)= h(t) (3.24)

and approximate system (1.1) by

u′h + Ahuh =Phf(x; t);∫ T0

0h(�)uh(�) d�+ Phg(T1; : : : ; TN ; uh)=P1

h (x): (3.25)

System (3.24) has a unique solution under the same conditions as Lemma 3:1.

Theorem 3.6. Assume the hypotheses of Theorem 3:4 are satis7ed, and assume uh(t)is the solution to system (3.15) where g(T1; : : : ; TN ; u) is de7ned by (1.3) with thehi; 16 i6N , independent of x. Assume inequality (3:14) is satis7ed with *=0 andk01 given by (2.7). Then with s= %;

‖u(t)− uh(t)‖6Chs(‖ ‖D(As=2+1) + ‖u(0)‖D(As=2)

+∥∥∥∥∫ T0

0

∫ �

0h(�)S(�− �′)f(x; �′) d�′ d�

∥∥∥∥D(As=2+1)

+‖f‖L∞(0;T ;D(As=2+1))): (3.26)

Proof. Let I ′i ; 16 i6 5; be analogous to the Ii in Theorem 3.4. We can use Lemma3.5 instead of Lemma 3.3 to show

I ′16Chs‖ ‖D(As=2+1) : (3.27)

Also,

I ′26 ‖((I − e−AT0 )−1 − (Ih − e−AhT0 )−1Ph)S(t)Ag(T1; : : : ; TN ; u)‖+ ‖(Ih − e−AT0 )−1Ph((S(t)− Sh(t)Ph)Ag(T1; : : : ; TN ; uh)‖

+L2

∥∥∥∥∥ Sh(t)Ph

N∑i=1

[hi(AS(Ti)− AhSh(Ti)Ph)u(0) + AhSh(Ti)Ph(u(0)− uh(0))

+∫ ti

0(AS(Ti − �)− AhSh(Ti − �)Ph)f(x; �) d�

∥∥∥∥∥


+L2‖Sh(t)Ph

(N∑

i=Ni+1

∫ Ti+Ki

Ti

hi(�)(Au(�)− Ahuh(�)) d�

)

6Chs‖g(T1; : : : ; TN ; u‖D(As=2+1) +N1∑i=1

Chs‖u(0)‖D(As=2)

+L2‖Sh(t)‖Ni∑i=1

Ma1 e

−aTi‖u(0)− uh(0)‖

+L2‖Sh(t)‖N∑

i=Ni+1

(∫ Ti+Ki

Ti

hi(�)Chs

�‖u(0)‖D(As=2) d�

+∫ Ti+Ki

Ti

|hi(�)| ‖AhSh(�)Ph(u(0)− uh(0))‖ d�

+∥∥∥∥∫ Ti+Ki

Ti

hi(�)∫ �

0(AS(�− �′)− AhSh(�− �′)Ph)f(x; �′) d�′ d�

∥∥∥∥6Chs(‖g(T1; : : : ; TN ; u)‖D(As=2+1) + ‖u(0)‖D(As=2) + ‖f‖L∞(0;T ;D(As=2+1))

+L2Ma1 ‖Sh(t)‖

(N1∑i=1

e−ati

ti+

N∑i=N1+1

∫ Ti+Ki

Ti

|hi(�)|e−a�

�d�

)

·‖u(0)− uh(0)‖:Likewise,

I ′4 + I ′56∥∥∥∥((I − e−AT0 )−1 − (Ih − e−AhT0 )−1Ph)S(t)

(∫ T0

0k(�)AS(�)u(0) d�

+A∫ T0

0

∫ �

0h(�)S(�− �′)f(x; �′) d�′ d�

∥∥∥∥+L2

∥∥∥∥S(t)− Sh(t)Ph)A∫ T0

0

∫ �

0h(�)S(�− �′)f(x; �′) d�′ d�

∥∥∥∥+L2‖Sh(t)‖

∥∥∥∥∫ T0

0

∫ �

0h(�)(AS(�− �′)− AhSh(�− �′)Ph)f(x; �′) d�′ d�

∥∥∥∥+L2

∥∥∥∥(S(t)− Sh(t)Ph)∫ T0

0k(�)AS(�)u(0) d�

∥∥∥∥+L2‖Sh(t)‖

∫ T0

0|k(�)| ‖(AS(�)− AhSh(�)Ph)u(0)‖ d�

+L2‖Sh(t)‖∫ T0

0|k(�)| ‖AhSh(�)Ph(u(0)− uh(0))‖ d�


6Chs(‖u(0)‖D(As=2) +∥∥∥∥∫ T0

0

∫ �

0h(�)S(�− �′)f(x; �) d�′ d�

∥∥∥∥D(As=2+1)

+ ‖f‖L∞(0;T0;D(As=2+1)) + L2‖Sh(t)‖∫ T0

0Ma

1|k(�)|

�e−a� d�‖u(0)− uh(0)‖:

Finally,

I5 =∥∥∥∥∫ t

0(S(t − �)− Sh(t − �)Ph)f(x; �) d�

∥∥∥∥6Chs‖f‖L∞(0;T ;D(As=2+1)):

Thus,

‖u(t)− uh(t)‖6Cf; ;uhs + Lh1M

a1 ‖Sh(t)‖

∫ T0

0

|k(�)|�

e−a� d�

+

(N1∑i=1

e−ati

ti+

N∑i=N1+1

∫ Ti+Ki

Ti

|hi(�)|�

e−a� d�

)‖u(0)− uh(0)‖:

(3.28)

Let t=0 in inequality (3.28). Therefore, since ‖Sh(0)‖=1; by inequality (3.13) with*=0, we have

‖u(0)− uh(0)‖6Cf; ;fhs: (3.29)

The theorem follows from inequalities (3.27) and (3:38).

4. An iterative approximation

Finally in this section, we will use contraction mappings to construct a fully discreteiterative scheme to approximate u(0) where u is the solution to system (1.1). Onceu(0) is approximated, u(t) can be approximated by standard numerical methods forinitial value problems.Assume the hypotheses of Theorem 3:5 and let Gh :Vh → Vh be de�ned by

Gh(vh) = e−AhT0vh +[PhA − Phg(T1; : : : ; TN ; uh;Ah;vh;Af)

−Ah

∫ T0

0k(�)uh;vh; 0 (�) d�

−∫ T0

0

∫ �

0h(�)e−(�−�′)AhPhAf(x; �′) d�′ d�

]; (4.1)

where uh;vh;f is the solution to the problem

u′h + Ahuh =Phf(x; t);

uh(0)= vh: (4.2)


Let

d=M ′0e

−aT0 ; (4.3)

where M ′0 is the constant in inequality (3.7). Then by an easy argument similar to the

proof of Theorem 2.4, we have:

Lemma 4.1. If

2L=d+ k01Ma1 +Ma

1

∫ T0

0|k(�)|e

−a�

�d�¡ 1; (4.4)

then there is a unique vh ∈Vn such that

vh =Gh(vh): (4.5)

Note that if vh is the �xed point in Eq. (4.5), uh;vh;P1hf

is the solution to system(3.25). Also inequality (4.4) implies inequality (3.14) with *=0.

Let {Uih;u0 ;f} be a fully discrete approximation of

ut + Au=f(x; t);

u |� =0;

u(0)= u0; (4.6)

such that if u is the solution to the boundary value problem (4.6), then

‖u(ti)− Uih;u0 ;f‖6 c(hst%−s=2

i + kpt%−pi )‖u0‖D(A%)

+chs‖f‖H1 + kp‖f‖H2 ; (4.7)

where

‖f‖H1 =p−1∑i=0

sup06t6T

‖f(i)(t)‖s−2i

and

‖f‖H2 =p−1∑i=0

sup06t6T

‖f(i)(t)‖2p−2i +∫ T

0‖f(p)(s)‖ ds

and p¿ 1; r¿ 2; and 16 s6 r.Also assume there is a constant c0 such that for 06 i and

‖AihU

nh;vn; 0‖6 c0

tin‖vh‖; (4.8)

where the ci0 are independent of h, see Thomee [13,14].Let tM =T; tn0 =T0; tMi =Ti and tM ′

1=Ti + Ki.

Let

Int(f)=M∑i=1

wif(ti) (4.9)


be a quadrature method on [0; T ] such that for f∈Cp([0; T ]; Vh)∥∥∥∥∫ T

0f(�) d�− Int(f)

∥∥∥∥6 ckp‖f‖Cp([0;T ];L2(�)): (4.10)

For vh ∈Vh, let

2G(vh) = (1 + k(T0))UM0h;vh;0

+ PhA

−M0∑i=0

wik ′(ti)Uih;vh;0 −

M0∑i=1

wih(ti)Phf(x; ti)

+M0∑i=1

wih(ti)Uih;Phf(x;0);Phft −

N1∑j=1

hiPhf(x; Ti)

−N1∑j=1

hiUMih;Ahvh−Phf(x;0);−ft

−N∑

j=N1+1

M ′j∑

i=Mj

wihj(ti)(Phf(x; ti)− Uih;Phfh(x;0)−Ahvh;ft ): (4.11)

Let k =max16i6M{ti − ti−1}.

Lemma 4.2. Assume k()∈Cp+1[0; T0]; hi()∈Cp+1[Ti; Ti + Ki] if N1 + 16 i6N ,sup06t6T0

|k ′(t)|6 k0; f∈Hi; i=1; 2 and the conditions of Theorem 3:5 with inte-ger p;p¿ s=2 instead of s=2 with inequality (4:4) are satis7ed. Also assume for ksmall enough there is an a; 0¡a¡ 1 such that

2L2 = (1 + k(T0))d+M0∑i=0

wik0c0 +N1∑j=1

|hj|Tj

c0 +N∑

j=Ni+1

Kjhjc0Tj6 a¡ 1; (4.12)

where d is the constant de7ned by (4:3) and hj =maxTj6t6Tj+Kj |hj(t)|; Mj+16 j6N .Then for k small enough, 2Gh is a contraction on Vh.

Proof. Let uh; vh ∈Vh. Then with wh = uh − vh and hi =maxT16t6Ti+Ki |hi(t)| for N1 +16 i6N; and integrating by parts for

∫ T0

0 k(�)uh; uh;0(�) d� we have

‖ 2Gh(un)− 2Gh(vn)‖6 (1 + k(T0))(‖UM0h;wh;0

− e−T0Ahwh‖+ ‖e−T0Ahwh‖)

+

∥∥∥∥∥M0∑i=0

wik ′(ti)Uih;wh;0

∥∥∥∥∥+∥∥∥∥∥

N1∑j=1

hiAhUMj

h;wh;0

∥∥∥∥∥+

N∑j=N1+1

∥∥∥∥∥M ′

i∑i=Mi

wih(ti)AhU ih;wh;0

∥∥∥∥∥


6Ck‖wh‖+ d‖wh‖+M0∑i=0

wik0c0‖wh‖

+N1∑j=1

|hj|Tj

c0‖wh‖+N∑

j=Ni+1

Kihjc0Tj

‖wh‖:

Note that condition (4.12) is analogous to inequality (4.4).Integrating by parts

Gh(vh) = (1 + k(T0))e−AhT0vh + PhA

−∫ T0

0h(�)(Phf(x; �)− e−�AhPhf(x; 0)−

∫ �

0e−(�−�′)AhPhf�(x; �′) d�′) d�

−∫ T0

0k ′(�)Uh;vh; 0 (�) d�

−Phg(T1; : : : ; TN ;Phf(x; t)− Uh;Phf(x;0)−Avh;ft ):

Theorem 4.3. Assume the conditions of Lemma 4:2; then

‖u(0)− uh‖6 c ;f(hs + ks=2); (4.13)

where u is the solution to system (1:1) and uh is the 7xed point of 2Gh.

Proof. Let vh be the �xed point for Gh. Then for hi =maxTi6t6Ti+Ki |hi(t)| for N1 +16 i6N ,

‖vh − uh‖6 ‖Gh(vh)− 2Gh(vh)‖+ ‖ 2Gh(vh)− 2G(uh)‖6 2L2‖vh − uh‖+ |1 + k(T0)| ‖e−AhT0vh − UM0

h1 ;vh;0‖

+∥∥∥∥∫ T0

0k ′(T0)(e−Ah�vh − e−A�u(0)) d�

∥∥∥∥+

∥∥∥∥∥∫ T0

0k ′(�)e−A�u(0) d�−

M0∑i=0

wik ′(ti)e−Ati u(0)

∥∥∥∥∥+

∥∥∥∥∥M0∑i=0

wik ′(ti)(e−Ati u(0)− Uih;vh;0 )

∥∥∥∥∥+

∥∥∥∥∥∫ T0

0h(�)Phf(x; �) d�−

M0∑i=1

wih(ti)Phf(x; ti)

∥∥∥∥∥+

∥∥∥∥∥h(�)uh;Phf(x;0);ft (�)−M0∑i=0

wih(ti)uh;Phf(x;0);ft (ti)

∥∥∥∥∥


+

∥∥∥∥∥M0∑i=0

(wih(ti)(uh;Phf(x;0);ft (ti)− Uih;Phf(x;0);ft )

∥∥∥∥∥+

N1∑j=1

|hj| ‖e−TiAh(Ahvh − UMih;Ahvh;0

)‖

+N1∑j=1

|hj| ‖uh;Phf(x;0);ft (tj)− UMj

h;Phf(x;0);ft‖

+N∑

j=N1+1

∥∥∥∥∥∥∫ Tj+Kj

Tj

hj(�)Ahuh;vh;0 (�) d�−Mj∑

i=Mj

wihj(ti)Ahuh;vh;0 (ti)

∥∥∥∥∥∥+

N∑j=N1+1

M ′j∑

i=Mj

wihj‖Ahuh;vh;0 (ti)− AhU ih;vh;0‖

+N∑

j=N1+1

∥∥∥∥∥∥∫ Tj+Kj

Tj

hj(�)uh;Phf(x;0);ft (�) d�−M ′

j∑i=Mj

wihj(ti)uh;Phf(x;0);ft (ti)

∥∥∥∥∥∥+

N∑j=N1+1

M ′j∑

i=Mj

wihj‖uh;Phf(x;0);ft (ti)− Uih;Phf(x;0);ft‖

+N∑

j=N1+1

Mj∑i=Mj

∥∥∥∥∥∥∫ Tj+Kj

Tj

hj(�)Phf(x; �) d�−Mi

j∑i=Mj

wihj(ti)Phf(x; ti)

∥∥∥∥∥∥ :

Since ‖u(0)− vh(0)‖6 hs‖u(0)‖D(As=2), we have

‖vh − uh‖6 11− 2L2

(c | 1 + k(T0)|ks=2 + c

∫ T0

0|k ′(�)| d� hs‖u(0)‖s

+ ckp‖k ′(t)e−Atu(0)‖Cp([0;T;L2(�))

+ cks(hs + ks=2)‖u(0)‖s + ckp‖h(t)Phf‖Cp([0;T0]; L2(�))

+ ckp‖h(t)uh;Phf(x;0);ft‖Cp([0;T0];L2(�))

+ c(hs‖ft‖H1 + ks=2‖ft‖H2 )

+N1∑j=1

c|hj|Tj

‖u(0)‖s +N1∑j=1

c|hj|(hs(‖ft‖H1 + ks=2‖ft‖H2 )

+N1∑j=1

c(hs + ks=2)‖f(x; 0)‖D(As=2)


Table 1Backward diDerence method

h k Number of e= ‖u(0)− vh‖iterations

0.1 0.01 2 0.036980.02 0.02 5 0.06559740.02 0.001 5 0.0035736

+N∑

j=N1+1

ckp‖uh;vh;0‖Cp([Tj;Tj+Kj]; L2(�))

+N∑

j=N1+1

(cks=2

Tj‖u(0)‖s + ckp‖uh;Phf(x;0);ft‖Cp([Tj;Tj+Kj]; L2(�))

+ c(hs‖ft‖H1 + ks=2‖ft‖H2 + c(hs + ks=2)‖f(x; 0)|D(As=2)

+ ckp‖hj(t)Phf‖Cp([Tj;Tj+Kj]; L2(�))

6 c(f; )(hs + ks=2):

The theorem follows using Theorem 3:5

‖u(0)− uh‖6 ‖u(0)− vh‖+ ‖vh − uh‖6 c(f; )(h2p + hs + ks=2)

6 c(f; )(hs + ks=2):

We now give some numerical results.

Example. The nonlocal problem

ut − uxx =− 26(1− t) cos(6 x) + (62(1− t)− 1) x sin(6 x);

u(0; t)= 0; u(1; t)= 0;∫ 1

0(1 + �)u(x; �) d�+ u(x; 2)=− 1

3x sin(6x)

has the solution u(x; t)= (1−t)x sin 6x. The backward diDerence method and the Crank–Nicolson–Galerkin method were used with trapezoidal integration in the iterative for-mula (see Tables 1 and 2)

vn+1h = 2G(vnh); v0h =

*0 :

We let e represent the L2 error of the diDerence of u(0) and the given number ofiterations of vh. The results are given in Table 1.


Table 2Cranck–Nicholson method

h k Number of e= ‖u(0)− vh‖iterations

0.1 0.05 Diverges0.1 1

30 7 0.0071700.02 1

150 diverges0.02 0.005 6 0.0002993550.01 0.002 6 0.000076229

The Crank–Nicholson methods satis�es inequalities (4.7) and (4.8) under the con-dition k6 %h2 for a �xed constant % independent of h and k [13,14]. Thus if k is toolarge compared to h, the method may fail as our results in Table 2 show.

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Iterative finite element approximations of solutions to parabolic equations with nonlocal conditions

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