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Notes on a new approximate solution of 2-D heat equation backward

in time

Nguyen Huy Tuana,d,⇑, Dang Duc Trongb, Pham Hoang Quanc

a

Institute for Computational Science and Technology, Quarter 6, Linh Trung Ward, Thu Duc District, HoChiMinh City, Viet Nam

b

Department of Mathematics, University of Natural Science, Vietnam National University, 227 Nguyen Van Cu, Q.5, HoChiMinh City, Viet Nam

c

Department of Mathematics and Applications, Sai Gon University, 273 An Duong Vuong, Q.5, HoChiMinh City, Viet Nam

d

Department of Mathematics, Hochiminh University of Technology, 144/24 Dien Bien Phu, P.25, Binh Thanh Dist., HoChiMinh City, Viet Nam

a r t i c l e i n f o

Article history:

Received 16 September 2010

Received in revised form 29 April 2011

Accepted 8 May 2011

Available online 20 May 2011

Keywords:

Backward heat problem

Nonhomogeneous heat equation

Ill-posed problem

Quasi-boundary value method

Quasi-reversibility method

Error estimate

a b s t r a c t

In this paper, we consider a backward heat problem that appears in many applications This problem is ill-posed The solution of the problem as the solution exhibits unstable depen-dence on the given data functions Using a new regularization method, we regularize the problem and get some new error estimates Some numerical tests illustrate that the pro-posed method is feasible and effective This work is a generalization of many recent papers, including the earlier paper [A new regularized method for two dimensional nonhomogene-ous backward heat problem, Appl Math Comput 215(3) (2009) 873–880] and some other authors such as Chu-Li Fu et al.[1–3], Campbell et al.[4]

1 Introduction

Let T be a positive number We consider the problem of ﬁnding the temperature u(x, y, t), (x, y, t) 2X [0; T] such that

ut uxx uyy¼ f ðx; y; tÞ; ðx; y; tÞ 2X ð0; TÞ;

uð0; y; tÞ ¼ uðp;y; tÞ ¼ uðx; 0; tÞ ¼ uðx;p;tÞ ¼ 0; ðx; y; tÞ 2X ð0; TÞ;

uðx; y; TÞ ¼ gðx; yÞ; ðx; yÞ 2X;

8

>

whereX= (0,p) (0,p) and g(x, y), f(x, y, t) are given The problem is called the backward heat problem (BHP for short), or the ﬁnal-boundary value problem

It is known in general that the problem is ill-posed, i.e., a solution does not always exist, and in the case of existence, it does not depend continuously on the given datum In fact, from a small noise contaminated physical measurement, the corresponding solution may have a large error It makes the numerical computation difﬁcult Hence, a regularization is in order

The homogeneous backward heat problems, i.e the case f = 0, has been studied by many authors in recent years In a few words, we mention Ames and Payne [5], Lattes and Lions [6], Showalter [7], who approximated the BHP by quasi-reversibility method; Tautenhahn and Schroter [8]who established an optimal error estimate for a BHP; Seidman [9]

⇑Corresponding author at: Institute for Computational Science and Technology, Quarter 6, Linh Trung Ward, Thu Duc District, HoChiMinh City, Viet Nam E-mail address: tuanhuy_bs@yahoo.com (N.H Tuan).

Contents lists available atScienceDirect Applied Mathematical Modelling

j o u r n a l h o m e p a g e : w w w e l s e v i e r c o m / l o c a t e / a p m

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who established an optimal ﬁltering method; and Hao[10]who studied a modiﬁcation method We also refer to various other works of Chang et al.[11], Chu-Li Fu et al.[1Ờ3], Campbell et al[4], Lien et al.[12], Murniz et al.[13], Dokuchaev

et al.[14], and Engl et al.[15] Recently, the 1-D version of the problem in a infinite strip has been considered in[2,10] Phys-ically, this problem arises from the requirement of recovering the heat temperature at some earlier time using the knowl-edge about the final temperature The problem is also involved to the situation of a particle moving in a environment with constant diffusion coefficient (see[16]) when one asks to determine the particle position history from its current place The interest of backward heat equations also comes from financial mathematics, where the celebrated BlackScholes model[17]

for call option can be transformed into a backward parabolic equation whose form is related closely to backward heat equa-tions Although there are many papers on the homogeneous backward heat equation, the result on the inhomogeneous case

is very scarce while the inhomogeneous case is, of course, more general and nearer to practical application than the homo-geneous one Shortly, it allows the appearance of some heat source which is inevitable in nature

In the present paper, a modified method is used for solving 2-D backward heat problem, which will improve the stability results in some previous papers In many earlier works, we find that only logarithmic type estimates in L2-norm are avail-able; and estimates of Hỏlder type are very rate (seeRemark 3for more detail comparisons) In our method, corresponding to different levels of the smoothness of the exact solution, the convergence rates will be improved gradually Under some suit-able conditions on the exact solution u, we shall introduce the error estimate of orderp, (p > 0) This is a significant improve-ment in comparison with[1,10,18,8,12,3] Some comments on the usefulness of this method are given in Remarks The remainder of the paper is organized as follows Section2is devoted to a short presentation of a ill-posed problem(1)

In Section3, we shall construct the regularized solution and show that it works even with very weak condition on the exact solution Under some assumptions on the exact solution, some error estimates are derived Finally, a numerical experiment is given in Section4to illuminate the effect of our method

2 The ill-posed backward heat problem

Throughout this paper, we denote h,i, kk by the inner product and the norm in L2(X) Let us ﬁrst make clear what a weak solution of the Problem(1)is We call a function u 2 C ơ0; T; H1

đXỡ

\ C1đđ0; Tỡ; H2đXỡỡ to be a weak solution for Problem(1)if

d

dthuđ; ; tỡ; WiL 2 đXỡ huđ; ; tỡ;DWiL2 đXỡỬ hf đ; ; tỡ; WiL2 đXỡ; đ2ỡ

for all functions W 2 H2đXỡ \ H1đXỡ In fact, it is enough to choose W in the orthogonal basis {sin(nx)sin(my)}n,mP1and the formula(2)reduces to

unmđtỡ Ử eđTtỡđn 2 ợm 2 ỡgnm

Z T

t

eđstỡđn 2 ợm 2 ỡfnmđsỡds; 8m; n P 1; đ3ỡ

which may also be written formally as

uđx; y; tỡ Ử X1

n;mỬ1

eđtTỡđn 2 ợm 2 ỡgnm

Z T

t

eđtsỡđn 2 ợm 2 ỡfnmđsỡds

where

fnmđsỡ Ử 4

p2

Z p

0

Z p

0

f đx; y; sỡ sin nx sin mydx dy;

gnmỬ 4

p2

Z p

0

Z p

0

gđx; yỡ sin nx sin mydx dy;

unmđtỡ Ử 4

p2

Zp

0

Z p

0

uđx; y; tỡ sin nx sin mydx dy:

Note that if the exact solution u is smooth then the exact data (f,g) is smooth also However, the real data, which come from practical measure, is often discrete and non-smooth We shall therefore always assume that f 2 L2(0,T);L2(X)) and g 2 L2(X) and the error of the data is given on L2only Note that the expression(4)is the solution of problem(1)if it exists In the following theorem, we provide a condition of its existence

Theorem 1 Problem(1)has a unique solution u if and only if

X1

n;mỬ1

eTđn 2 ợm 2 ỡgnm

Z T

0

esđn 2 ợm 2 ỡfnmđsỡds

Proof Suppose the Problem (1) has a solution u 2 Cđơ0; T; H1

đXỡỡ \ C1đđ0; Tỡ; L2đXỡỡ, then u can be formulated in the frequency domain

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n;mỬ1

eđtTỡđn 2 ợm 2 ỡgnm

Z T

t

eđtsỡđn 2 ợm 2 ỡfnmđsỡds

This implies that

unmđ0ỡ Ử eTđn2ợm 2 ỡgnm

Z T

0

Then

kuđ; ; 0ỡk2Ử X1

n;mỬ1

ZT

0

<1:

If(5)holds, then deﬁnev(x,y) be as the function

vđx; yỡ Ử X1

n;mỬ1

Z T

0

sin nx sin my 2 L2đXỡ:

Consider the problem

ut uxx uyyỬ f đx; y; tỡ;

uđ0; y; tỡ Ử uđp;y; tỡ Ử uđx; 0; tỡ Ử uđx;p;tỡ Ử 0; t 2 đ0; Tỡ;

uđx; y; 0ỡ Ửvđx; yỡ; đx; yỡ 2 đ0;pỡ đ0;pỡ:

8

>

It is clear that(8)is the direct problem so it has a unique solution u (see[16]) We have

n;mỬ1

etđn 2 ợm 2 ỡ<vđx; yỡ; sin nx sin my > ợ

Z t

0

eđstỡđn 2 ợm 2 ỡfnmđsỡds

Let t = T in(9), we have

uđx; y; Tỡ Ử eTđn 2 ợm 2 ỡ eTđn 2 ợm 2 ỡgnm

Z T

0

ợ

Z T

0

eđsTỡđn 2 ợm 2 ỡfnmđsỡds

sin nx sin my

Ử X1

n;mỬ1

gnmsin nx sin my Ử gđx; yỡ:

Hence, u is the unique solution of(1) h

Theorem 2 Problem(1)has at most one solution in Cđơ0; T; H1đXỡỡ \ C1đđ0; Tỡ; L2đXỡỡ

Proof Let u(x, y, t), v(x, y, t) be two solutions of Problem (1) such that u;v2 Cđơ0; T; H1đXỡỡ \ C1đđ0; Tỡ; L2đXỡỡ Put w(x,y,t) = u(x,y,t) v(x,y,t) Then w satisﬁes the equation

wt wxx wyyỬ 0; đx; y; tỡ 2X đ0; Tỡ;XỬ đ0;pỡ đ0;pỡ;

wđ0; y; tỡ Ử wđp;y; tỡ Ử wđx; 0; tỡ Ử wđx;p;tỡ Ử 0; đx; y; tỡ 2X ơ0; T:

wđx; y; Tỡ Ử 0; x; y 2X:

Now, setting Gđtỡ ỬRp

0

Rp

0w2đx; y; tỡdx dyđ0 6 t 6 Tỡ By direct computation we get

G0đtỡ Ử 2

Z p

0

Z p

0

wđx; y; tỡwtđx; y; tỡdx dy Ử 2

Z p

0

Z p

0

wđx; y; tỡđwxxđx; y; tỡ ợ wyyđx; y; tỡỡdx dy:

Using the Green formula, we obtain

G0

đtỡ Ử 2

Z p

0

Z p

0

đw2

xđx; y; tỡ ợ w2

Taking the derivative of G0(t), one has

G00đtỡ Ử 4

Zp

0

Z p

0

wxđx; y; tỡwxtđx; y; tỡ ợ wyđx; y; tỡwytđx; y; tỡ

dx dy:

Using the technique of integration by parts, we get

G00

đtỡ Ử 4

Z pZ p

wxxđx; y; tỡwtđx; y; tỡ ợ wyyđx; y; tỡỡwtđx; y; tỡ

dx dy Ử 4

Z pZ p

đwxxđx; y; tỡ ợ wyyđx; y; tỡỡ2dx dy: đ11ỡ

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Now, from(10)and applying the Holder inequality, we have

Zp

0

Z p

0

w2

xđx; y; tỡ ợ w2

yđx; y; tỡ

dx dy Ử

Zp

0

Z p

0

wđx; y; tỡđwxxđx; y; tỡ ợ wyyđx; y; tỡỡdx dy

6

Z p

0

Z p

0

w2đx; y; tỡdx dy

0

Z p

0

đwxxđx; y; tỡ ợ wyyđx; y; tỡỡ2dxdy

: đ12ỡ

Thus(10)Ờ(12)imply

đG0đtỡỡ26GđtỡG00

đtỡ:

Hence Theorem 11 ([16], p 65) gives G(t) = 0 This implies that u(x, y, t) =v(x, y, t) The proof is completed In spite of the uniqueness, the problem(1)is still ill-posed and a regularization is necessary In next section, we shall establish the approx-imation problem h

3 Regularization and error estimate

In this section, we introduce a regularized problem as of integral equation and investigate the error estimate between the regularization solution and the exact one Let> 0, a > 0, b > 0 Let gbe a measured data satisfying kg gk 6 Starting from the ideas mentioned in the paper of Clark and Oppeinheimer[19], we consider the following approximate problem

u

t uxx uyyỬ X1

n;mỬ1

fnmđtỡ

1 ợ beaTđn 2 ợm 2 ỡsin nx sin my; đx; y; tỡ 2X đ0; Tỡ; đ13ỡ

uđ0; y; tỡ Ử uđp;y; tỡ Ử uđx; 0; tỡ Ử uđx;p;tỡ Ử 0 đx; y; tỡ 2X ơ0; T; đ14ỡ

uđx; y; Tỡ Ử X1

n;mỬ1

gnm

1 ợ beaTđn 2 ợm 2 ỡsin nx sin my; đx; yỡ 2X; đ15ỡ

where fnm(t), gnmare deﬁned by

fnmđtỡ Ử 4

p2

Z p

0

Z p

0

f đx; y; tỡ sin nx sin mydx dy;

gnmỬ 4

p2

Z p

0

Z p

0

gđx; yỡ sin nx sin mydx dy:

and b is a regularization parameter depending onand it is chosen latter The real number a P 1 is a constant.The case f = 0,

a = 1 is considered in[19] The major reason to choose a P 1 is explained inRemark 3

Theorem 3 Let f 2 L2((0,T);L2(X)) and g 2 L2(X) Then Problem(13)Ờ(15)has uniquely a weak solution u2 Cđơ0; T; L2đXỡ\

L2đ0; T; H1đXỡỡ \ C1đ0; T; H1đXỡỡ deﬁned as follows:

n;mỬ1

eđTtỡđn 2 ợm 2 ỡ

1 ợ beaTđn 2 ợm 2 ỡ gnm

Z T

t

The solution depends continuously on g in C([0,T];L2(X))

Proof The proof is divided into two steps In Step 1, we prove the existence and the uniqueness of a solution of problem

(13)Ờ(15) In Step 2, the stability of the solution is given

Step 1 The existence and the uniqueness of a solution of Problem(13)Ờ(15)

We divide this step into two parts

Part A If u2 Cđơ0; T; L2đXỡ \ L2đ0; T; H1đXỡỡ \ C1đ0; T; H1đXỡỡ satisﬁes(16)then uis solution of(13)Ờ(15)

We have

n;mỬ1

Z T

t

sin nx sin my:

We can verify directly that u2 Cđơ0; T; L2đXỡ \ C1đđ0; Tỡ; H1đXỡỡ \ L2đ0; T; H1đXỡỡỡ In fact, u2 C1đđ0; T; H1đXỡỡỡ Moreover, one has

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utđx; y; tỡ Ử X1

n;mỬ1

đn2ợ m2ỡeđTtỡđn 2 ợm 2 ỡ

Z T

t

đn2ợ m2ỡeđstTỡđn 2 ợm 2 ỡ

1 ợ beaTđn 2 ợm 2 ỡ fnmđsỡdsợ fnmđtỡ

1 ợ beaTđn 2 ợm 2 ỡ

sin nx sin my

Ử 4

p2

X1 n;mỬ1

đn2ợ m2ỡ < uđx; y; tỡ; sin nx sin my >

sin nx sin my ợ X1

n;mỬ1

fnmđtỡ

1 ợ beaTđn 2 ợm 2 ỡsin nx sin my

Ử uxxđx; y; tỡ ợ uyyđx; y; tỡ ợ X1

n;mỬ1

fnmđtỡ

1 ợ beaTđn 2 ợm 2 ỡsin nx sin my;

and

uđx; y; Tỡ Ử X1

n;mỬ1

gnm

1 ợ beaTđn 2 ợm 2 ỡsin nx sin my:

So uis the solution of(13)Ờ(15)

Part B Problem(13)Ờ(15)has at most one solution Cđơ0; T; H1đXỡỡ \ C1đđ0; Tỡ; L2đXỡỡ

Let u(x, y, t),v(x, y, t) be two solutions of Problem(4)Ờ(6)such that u;v2 Cđơ0; T; H1đXỡỡ \ C1đđ0; Tỡ; L2đXỡỡ Put w(x, y, t) = u(x,

-y, t) v(x, y, t) UsingTheorem 2, we get u(x, y, t) =v(x, y, t) The proof is completed Since Part A and Part B, we complete the proof of Step 1

Step 2 The solution of the problem(13)Ờ(15)depends continuously on g in L2(X)

First, to prove Step 2, we need the following lemma h

Lemma 1 For m,n,b > 0,0 6 c 6 d, we have

eđn 2 ợm 2 ỡc

Proof of Lemma 1 We have

eđn 2 ợm 2 ỡc

1 ợ beđn 2 ợm 2 ỡdỬ e

đn 2 ợm 2 ỡc

1 ợ beđn 2 ợm 2 ỡd

1 c6 eđn 2 ợm 2 ỡc

đ1 ợ beđn 2 ợm 2 ỡdỡc

6bc:

For 0 6 t 6 s 6 T, if letc = T t,d = aT in(17), then

If we let c = s t,d = aT in(17)then

eđstỡđn 2 ợm 2 ỡ

Let u andvbe two solutions of(13)Ờ(15)corresponding to the ﬁnal values g and h From the deﬁnitions of u andvwe have

n;mỬ1

Z T

t

sin nx sin my

and

vđx; y; tỡ Ử X1

n;mỬ1

1 ợ beaTđn 2 ợm 2 ỡ hnm

Z T

t

sin nx sin my;

where

gnmỬ 4

p2

Z p

0

Z p

0

gđx; yỡ sin nx sin my;

hnmỬ 4

p2

Z p

0

Z p

0

hđx; yỡ sin nx sin mydx dy:

Hence

uđx; y; tỡ vđx; y; tỡ Ử X1

1 ợ beaTđn 2 ợm 2 ỡđgnm hnmỡ sin nx sin my: đ20ỡ

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Using(18) and (20)and the inequality (a + b)262(a2+ b2), we obtain

kuð; ; tÞ vð; ; tÞk2¼p2

4

X1 n;m¼1

eðTtÞðn 2 þm 2 Þ

1 þ beaTðn 2 þm 2 Þðgnm hnmÞ

2

6p2

4 b

2t2T

aT X1 n;m¼1

ðjgnm hnmjÞ26b2t2TaT kg hk2:

Therefore

This completes the proof of theorem h

Remark 1 As shown in the Introduction, several regularization methods are established in the literature The stability of their solutions with respect to the variation of the ﬁnal value (at t = T) is controlled by inequalities with coefﬁcients of dif-ferent order of magnitude For instance, in[20], the order of magnitude is eT In[19], the authors give some better stability estimates than the latter discussed methods They show that the stability estimate is of order Mt

T 1 In[21], the authors improve the previous results by a better estimation of the stability order, that is A¼ C1 1þlnT T

ð Þ

ð From(21), if we set b = then the stability order is B¼ C2 1

a For a > 1, we have

lim

!0

B

A¼ C2

C1Tlim!01 1

a 1 þ ln T

¼ 0:

It is easy to see that the order Aof the error is less than the above order Bof stability estimate This proves the advantages of our method

Theorem 4 Assume that there exists a positive number P1such that

p2

4

X1

n;m¼1

e2tðn 2 þm 2 ÞjunmðtÞj2<P2; 0 6 t 6 T: ð22Þ

Let g2 L2(X) be measured data satisfying

kg gk 6:

Letv(,,t) be the solutions of Problem(13)–(15)corresponding to the ﬁnal data g Let b =a, then one has

kuð; ; tÞ vð; ; tÞk 6 ðP1þ 1Þt

Proof Suppose the problem(1)has an exact solution u Then u can be written as

uðx; y; tÞ ¼ X1

m;n¼1

eðtTÞðm 2 þn 2 Þ gnm

Z T

t

eðTsÞðn 2 þm 2 ÞfnmðsÞds

Hence

unmðtÞ ¼ eðtTÞðm 2 þn 2 Þ gnm

Z T

t

where unmðtÞ ¼4

p 2huðx; y; tÞ; sin nx sin myi It follows from(16)that

unmðtÞ ¼ e

ðTtÞðn 2 þm 2 Þ

1 þ beaTðn 2 þm 2 Þ gnm

Z T

t

eðsTÞðn 2 þm 2 ÞfnmðsÞds

Hence, we deduce from(25) and (26)that

u

nmðtÞ unmðtÞ ¼ eðTtÞðn 2 þm 2 Þ e

1 þ beaTðn 2 þm 2 Þ

!

gnm

Z T

t

eðsTÞðn 2 þm 2 ÞfnmðuÞðsÞds

¼ be

1 þ beaTðn 2 þm 2 ÞeaTðn 2 þm 2 Þ gnm

Z T

t

bþ eaTðn 2 þm 2 ÞunmðtÞ

¼ be

tðn 2 þm 2 Þ

bþ eaTðn 2 þm 2 Þetðn 2 þm 2 ÞunmðtÞ: ð27Þ

Using(18), we obtain

junmðtÞ unmðtÞj 6 baTtetðn 2 þm 2 ÞunmðtÞ: ð28Þ

Trang 7

This implies that

kuð; ; tÞ uð; ; tÞk2¼p2

4

X1 n;m¼1

junmðtÞ unmðtÞj26p2

4 b

2t

aT X1 n;m¼1

e2tðn 2 þm 2 ÞjunmðtÞj26P2

Since the condition kg gk 6and(21)give

kuð; ; tÞ vð; ; tÞk 6 btTaTkggk 6 btTaT: ð30Þ

Combining(29), (30)and using the triangle inequality, we have

kuð; ; tÞ vð; ; tÞk 6 kuð; ; tÞ uð; ; tÞk þ kuð:; tÞ vð; tÞk 6 P1baTt þ btTaT:

Notice that, from b =a, then

kuð; ; tÞ vð; ; tÞk 6 P1t

Tþt

T¼ ðP1þ 1Þt

T:

This completes the proof ofTheorem 4 h

Remark 2 (1) If f = 0, then from(32), we get

unmðtÞ ¼ eðtTÞðn 2 þm 2 Þgnm:

Therefore

unmð0Þ ¼ eTðn 2 þm 2 Þgnm¼ etðn 2 þm 2 ÞunmðtÞ:

This implies that

p2

4

X1

n;m¼1

e2tðn 2 þm 2 ÞjunmðtÞj2¼ kuð; ; 0Þk2:

The condition(22)is acceptable Moreover, the assumptions(22)can be replaced by assumptions on f and g Thus, since

unmðtÞ ¼ eðtTÞðm 2 þn 2 Þ gnm

Z T

t

ð31Þ

we get

etðm 2 þn 2 ÞunmðtÞ ¼ eTðm 2 þn 2 Þgnm

Z T

t

IfP1

n;m¼1e2Tðm 2 þn 2 Þg2

nm<1 andRT

0

P1 n;m¼1e2sðn 2 þm 2 Þf2

nmðsÞds < 1 then(22)is holds

2.In t = 0, the error(12)does not converges to zero when?0 This is of the same order as in[19] The error estimate here is not good at t = 0 because the condition of the exact solution u is so weak In practical applications we may expect that the exact solution is smoother In these cases the explicit error estimate is available in the next Theorem

Theorem 5 Assume that there exist positive numbers k,P2such that

p2

4

X1

n;m¼1

Letv(,,t) be deﬁned inTheorem 4 Let b ¼aT

Tþk, then one has

kuð; ; tÞ vð; ; tÞk 6 P2þTþkt

Proof Hence, we deduce from(25) and (26)that

u

nmðtÞ unmðtÞ ¼ eðTtÞðn 2 þm 2 Þ e

1 þ beaTðn 2 þm 2 Þ

!

gnm

Z T

t

¼ be

1 þ beaTðn 2 þm 2 ÞeaTðn 2 þm 2 Þ gnm

Z T

t

¼ be

ðaTkÞðn 2 þm 2 Þ

1 þ beaTðn 2 þm 2 Þekðn 2 þm 2 ÞunmðtÞ: ð35Þ

This implies that

junmðtÞ unmðtÞj 6 baTkekðn 2 þm 2 ÞunmðtÞ: ð36Þ

Trang 8

Using(36), we get the estimate

kuð; ; tÞ uð; ; tÞk2¼p2

4

X n; m ¼ 11junmðtÞ unmðtÞj2 6 p2

4b

2k X1 n;m¼1

e2kðn 2 þm 2 ÞjunmðtÞj2 6 P22b2k: ð37Þ

Since the condition kg gk 6and(21)give

kuð; ; tÞ vð; ; tÞk 6 btTaTkg gk 6 btTaT: ð38Þ

From(37) and (38), we have

kuð; ; tÞ vð; ; tÞk 6 kuð; ; tÞ uð; ; tÞk þ kuð; ; tÞ vð; ; tÞk 6 P2baTkþ btTaT:

Notice that from b ¼aT

Tþk, then

kuð; ; tÞ vð; ; tÞk 6 P2k

TþkþtþkTþk:

This completes the proof ofTheorem 5 h

Remark 3 (1) We emphasize once more that the error(34)(k > 0) is of Hölder type for all t 2 [0;T] It is easy to see that the convergence rate ofp, (0 < p) is more rapid than the that of lnð1

Þ

q

ðq > 0Þ when?0 Hence, the convergence rate in this paper is better than that of the regularizations proposed in some recent papers such as Clark and Oppenheimer[19], Denche and Bessila[22], ChuLiFu et al.[1–3], Tautenhahn[18,8]

2 For t = 0 in(34), we get

kuð; ; 0Þ vð; ; 0Þk 6 ðP2þ 1Þk

The rate of convergence at t = 0 isk

Tþk Since 0 < k 6 (a 1)T, the fastest convergence isa1

a Hence it can approachfor the large constant a This proves that(39)is sharp estimate

4 A numerical experiment

In this section, we establish some numerical tests for a example To compare some different methods, we give the numer-ical tests with same parameter regularization We consider the problem

ut uxx uyy¼ f ðx; y; tÞ ¼ 3etsin x sin y ð40Þ

and

The exact solution of the latter equation is

uðx; y; tÞ ¼ etsin x sin y:

Note that uðx; y; 1=2Þ ¼ ffiffiffi

e p sinðxÞ sinðyÞ 1:648721271 sinðxÞ sinðyÞ Let gpbe the measured ﬁnal data

gpðx; yÞ ¼ e sinðxÞ sinðyÞ þ1

psinðpxÞ sinðpyÞ:

So that the data error, at the ﬁnal time, is

FðpÞ ¼ kgp gk ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Z p

0

Z p

0

1

p2sin2ðpxÞ sin2ðpyÞdx dy

s

¼ p

2p:

The solution of(40) and (41)corresponding the ﬁnal value gpis

upðx; y; tÞ ¼ etsinðxÞ sin y þ1

pe

p 2 ð1tÞsinðpxÞ sinðpyÞ:

The error at t = 0 is

OðpÞ :¼ kupð; ; 0Þ uð; ; 0Þk ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Z p

0

Z p

0

e2p 2

p2 sin2ðpxÞ sin2ðpyÞ dx dy

s

¼e

p 2

p

2:

Then, we notice that

lim

p!1FðpÞ ¼ lim

p!1kgp gk ¼ lim

n!1

1 p

p

2¼ 0;

lim

p!1OðpÞ ¼ lim

p!1kupð; ; 0Þ uð; ; 0Þk ¼ lim

p!1

ep 2

p

2¼ 1:

Trang 9

From the two equalities above, we see that(40) and (41)is an ill-posed problem Let a ¼101 Approximating the problem as in

(13)–(15), the regularized solution is

vðx; y; tÞ ¼ X1

n;m¼1

etðn 2 þm 2 Þ

eðn2 þm2 Þ100 þ eðn 2 þm 2 ÞÞ

gnm

Z1

t

eðst1Þðn 2 þm 2 Þ

eðn2 þm2 Þ100 þ eðn 2 þm 2 Þ

fnmðsÞds

! sin nx sin my 0 6 t 6 T: ð42Þ

Hence, we have

vðx; y; tÞ ¼ e

12t

e1

50þ e2sin x sin y 3

Z 1

t

e3s2t2

e1

50þ e2ds

sin x sin y þ1

p

e2tp 2

ep2þ e2p 2 sinðpxÞ sinðpyÞ 0 6 t 6 1: ð43Þ

It follows that

vðx; y;1

2Þ ¼

1

e501þ e2 3

Z 1

1

e3s3

e501þ e2ds

! sin x sin y þ1

p

ep 2

ep2þ e2p 2 sin px sin py: ð44Þ

Let a= kv uk be the error between the regularized solutionvand the exact solution u

Let

¼1¼ 103

ffiffiffiffi

p

2

r

; ¼2¼ 104

ffiffiffiffi

p

2

r

;

¼3¼ 105 ffiffiffiffiffiffiffi

2p

p

; ¼4¼ 108

ffiffiffiffi

p

2

r

;

we haveTable 1

We note that the new method in this article give a better approximation than the previous method in[24] To prove this,

we continue to approximate this problem in 2-D case by the method given in[24] Infact, by reconsider the Table numerical

in[24](see p 879), we have theTable 2

By applying that method given in[24], we have the following regularized solution

vðx; y; tÞ ¼ X1

n;m¼1

etðn 2 þm 2 Þ

þ eðn 2 þm 2 ÞÞgnm

Z 1

t

etðn 2 þm 2 Þ

sþ esðn 2 þm 2 ÞfnmðsÞds

! sinðnxÞ sinðmyÞ:

Hence, we have

vðx; y; tÞ ¼ e

12t

þ e2sin x sin y 3

Z 1

t

es2t

sþ e2sds

sin x sin y þ1

p

e2tp 2

þ e2p 2 sinðpxÞ sinðpyÞ 0 6 t 6 1 ð45Þ

Table 1

The error of method in this paper.

1 ¼ 10 3 ffiffiffip

2

p 1.64870174813729sinxsiny + 2.706068625 10442978 sin(10 3 x)sin(10 3 y) 0.00003066665669

2 ¼ 10 4 ffiffiffip

2

p 1.64872107547220sinxsiny + 4.464249273 1044298034 sin(10 4 x)sin(10 4 y) 3.078760801 10 7

3 ¼ 10 5 ffiffiffip

2

3 ¼ 10 10 ffiffiffip

2

Table 3

The error of method in [23]

1 ¼ 103 ffiffiffip

2

p 1.735336020sin(x) sin(y) + 2.630490937 10434295

sin(1000x)sin(1000y) 0.1360541296

2 ¼ 104 ffiffiffip

2

p 1.674100247sin(x)sin(y) + 5.147719556 1043429449 sin(10 4

x)sin(10 4

3 ¼ 105 ffiffiffiffiffiffiffi

2p

p

1.658588375sinxsiny + 1.392549021 10 1085736205 sin(5.10 4

x)sin(5.10 4

4 ¼ 108 ffiffiffip

2

Table 2

The error of method in [24]

1 ¼ 10 3 ffiffiffip

2

p 1.618739596sin(x)sin(y) + 1.315245468 10434301 sin(10 3 x)sin(10 3 y) 0.04709510496

2 ¼ 10 4 ffiffiffip

2

p 1.645673218sinxsiny + 0.2573859778 1043429456 sin(10 4 x)sin(10 4 y) 0.004787870457

3 ¼ 10 5 ffiffiffip

2

p 1.648110757sinxsiny + 2.785098042 101085736215

sin(5.10 4

x)sin(5.10 4

3 ¼ 10 10 ffiffiffip

2

Trang 10

It follows that

vðx; y;1

2Þ ¼ ¼

1

þ e2sin x sin y 3

Z 1

1

es1

sþ e2sds

! sin x sin y þ1

p

ep 2

þ e2p 2sinðpxÞ sinðpyÞ 0 6 t 6 1 ð46Þ

Then, we have theTable 3

Fig 1 The exact solution.

Fig 2 The regularized solution with1 in Table 1

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