DSpace at VNU: The truncation method for a two-dimensional nonhomogeneous backward heat problem

Using the trunca-tion method for Fourier series we propose a simple regularized solutrunca-tion which not only works on a very weak condition on the exact data but also attains, due to t

The truncation method for a two-dimensional nonhomogeneous

backward heat problem

a

Department of Mathematical Sciences, University of Copenhagen, Denmark

b

Faculty of Mathematics and Computer Sciences, University of Science, Vietnam National University, HoChiMinh City, Viet Nam

c

Faculty of Mathematics, SaiGon University, HoChiMinh City, Viet Nam

a r t i c l e i n f o

Keywords:

Backward heat problem

Ill-posed problem

Nonhomogeneous heat

Truncation method

Error estimate

a b s t r a c t

We consider the backward heat problem

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

uðx; y; TÞ ¼ gðx; yÞ; ðx; yÞ 2X;

with the homogeneous Dirichlet condition on the rectangleX= (0,p)  (0,p), where the data f and g are given approximately The problem is severely ill-posed Using the trunca-tion method for Fourier series we propose a simple regularized solutrunca-tion which not only works on a very weak condition on the exact data but also attains, due to the smoothness

of the exact solution, explicit error estimates which include the approximation ðlnð1ÞÞ3=2pffiffiffi

in H2(X) Some numerical examples are given to illuminate the effect of our method

Ó 2010 Elsevier Inc All rights reserved

1 Introduction

LetX= (0, 1)  (0, 1) be a heat conduction rectangle Given the heat source f(x, y, t) on (x, y, t) 2X [0, T] and the final tem-perature u(x, y, T) at some time T > 0, we consider the problem of recovering the temtem-perature distribution u(x, y, t) from the backward heat problem

ut uxx uyy¼ f ðx; y; tÞ; ðx; y; tÞ 2X ð0; TÞ; ð1Þ uð0; y; tÞ ¼ uðp;y; tÞ ¼ uðx; 0; tÞ ¼ uðx;p;tÞ ¼ 0; t 2 ð0; TÞ; ð2Þ

Since f and g come from measurement, they are in general non-smooth and only approximate values This is a typical exam-ple of the inverse and ill-posed problem since although this problem has at most one solution (seeTheorem 1in Section2

the solution does not always exist, and in the case of existence, it does not depend continuously on the given data The insta-bility makes the numerically calculus difficult and hence a regularization is in order

The homogeneous backward heat problems, i.e the case f = 0, was extensively considered by many authors using many approach, e.g the original quasi-reversibility method of Lattès and Lions[10], the quasi-boundary value problem method

[15], the quasi-solution method of Tikhonov and Arsenin[16], the logarithmic convexity method[1]and the C-regularized semi-groups technique[7] Physically, this problem arises from the requirement of recovering the heat temperature at some

0096-3003/$ - see front matter Ó 2010 Elsevier Inc All rights reserved.

* Corresponding author.

E-mail address: tuanhuy_bs@yahoo.com (N.H Tuan).

Contents lists available atScienceDirect Applied Mathematics and Computation

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 m c

earlier time using the knowledge about the final temperature The problem is also involved to the situation of a particle mov-ing in a environment with constant diffusion coefficient (see[6]) 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 Black–Scholes model[2]for call option can be transformed into a backward parabolic equation whose form is related closely

to backward heat equations 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 appli-cation than the homogeneous one Shortly, it allows the appearance of some heat source which is inevitable in nature Let us mention here some approaches and their technical difficulties of many earlier works In the method of quasi-reversibility, the main ideas is of replacing the unbounded operator A (in our case is D) by a perturbed one A In the ori-ginal method in 1967, Lattès and Lions[10]proposed A(A) = A eA*A, i.e adding a ‘‘corrector” into the original operator, to obtain a well-posed problem The essential difficulty of the quasi-reversibility method is due to the appearance of the sec-ond-order operator A*A which produces serious difficulties on the numerical implementation In addition, the stability mag-nitude of the approximating problem, i.e the error introduced by a small change in the final value, is of order eTwhich is very

estimate better than the one of the quasi-reversibility method discussed above The main ideas of this method is of adding an appropriate ‘‘corrector” into the final data (instead of the main equation) Using this method, Clark and Oppenheimer[3], and very recently Denche and Bessila[4], regularized the backward heat problem by replacing the final condition by

uðTÞ þuð0Þ ¼ g

and

uðTÞ u0ð0Þ ¼ g;

respectively This method, in general, gives the stability estimate of order1

Although there are many papers on the homogeneous case of the backward problem, we only find a few result on the

a one-dimensional inhomogeneous linear problem by the quasi-reversibility method As we mention before, the stability magnitude of the method is of order eT In their work the error between the approximate problem and the exact solution is

ðT  tÞ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

8

t4kuð:; 0Þk2þ t2 @4f ðx; tÞ

@x4













2

L 2 ð0;T;L 2 ð0;pÞÞ

v

u

which is very large when t becomes small In 2007, Trong et al.[19]used an improved version of quasi-boundary value

meth-od to regularize the one-dimensional version of(1)–(3)for a nonlinear heat source f = f(x, t, u) Their error estimate ist/Tfor

t > 0 and (ln(1/))1/4for t = 0

One of the essential requirements of the previous works on inhomogeneous problem, e.g.[17,19], is

X1

k¼1

e2Tk 2

g2

where gkis the coefficient of the Fourier series of the final datum u(.,T) = g, i.e

gk¼2

p

Z p

0

gðxÞ sinðkxÞdx:

While such a condition is reasonable in homogeneous problems, it is not necessarily true in the inhomogeneous case For example, consider the problem

ut uxx¼ f ðx; tÞ  etx; ðx; tÞ 2 ð0;pÞ  ð0; TÞ;

uð0; tÞ ¼ uðp;tÞ ¼ 0; t 2 ð0; TÞ:

Corresponding to the final value u(x,1) = g(x)  ex, the equation has a (unique) solution u(x,t) = etx However, by direct com-putation we find that gk¼ 2eð1Þ kþ1

X1

k¼1

e2Tk 2

g2

k¼ 4e2X1

k¼1

e2Tk 2

k2 >1:

In the present paper, we do not need condition(4) In fact, we shall give a simple and convenient way to construct the reg-ularized method which works with very weak assumption on the exact solution

Let us give a simple analysis for the ill-posedness of the problem(1)–(3) This problem may be rewritten formally as

uðx; y; tÞ ¼ X1

m;n¼1

eðTtÞðm 2 þn 2 Þ gmn

Z T

t

eðsTÞðm 2 þn 2 ÞfmnðsÞds

where gmnand fnm(t) are the coefficient of the Fourier-sin expansion of g and f(., , t), i.e.

gmn:¼ 4

p2

Z

X

gðx; yÞ sinðmxÞ sinðnyÞdxdy;

fmnðtÞ :¼ 4

p2

Z

X

f ðx; y; tÞ sinðmxÞ sinðnyÞdxdy:

If t < T then eðTtÞðm 2 þn 2 Þincreases very fast when m2+ n2becomes large Thus the term eðtTÞðm 2 þn 2 Þis the source of instability

It is a natural think to recover the stability of problem(5)is to filter all high frequencies In the present paper, we simply

do that by using the truncated regularization method, namely taking the sum(5)only for m2+ n26Mwith an appropriate regularization parameter M The truncated regularization method is a very simple and effective method for solving some ill-posed problems and it has been successfully applied to some inverse heat conduction problems[5,8,13] However, in many earlier works, we find that only logarithmic type estimates in L2-norm are available; and estimates of Hölder type are very rate (seeRemarks 5 and 6for more detail comparisons) In our method, corresponding to different levels of the smoothness

of the exact solution, the convergence rates will be improved gradually In particular, if we impose a condition similar to(4)

then the error estimate in H2(X) is ðlnðÞÞ3=2pffiffiffi

, which is better than any Hölder estimate of orderq

with q 2 (0, 1/2) We mention that our regularized solution in all case is unique, and all error estimates are valid for all t 2 [0, T]

The remainder of the paper is organized as follows In Section2we shall construct the regularized and show that it works even with very weak condition on the exact solution In Section3, many error estimates are derived, in both of the usual cases such as the exact solution u in H1

ðXÞ or H2(X), and the special cases when the exact solution is very smooth Some numerical experiments are given in Section4to illuminate the effect of our method

2 Regularized solution

Let us first make clear what a weak solution of the problem(1)–(3)is As follows we shall write u(t) = u(., , t) for short We call a function u 2 C([0, T]; L2(X)) \ C1((0, T); L2(X)) to be a weak solution for the problem(1)–(3)if

d

dthuðtÞ; WiL 2 ðXÞ huðtÞ;DWiL2 ðXÞ¼ huðtÞ;DWiL2 ðXÞ; ð6Þ

for all function Wðx; yÞ 2 H2ðXÞ \ H1ðXÞ In fact, it is enough to choose W in the orthogonal basis {sin(mx)sin(ny)}m,nP1and the formula(6)reduces to

umnðtÞ ¼ eðTtÞðm 2 þn 2 Þgmn

Z T

t

eðstÞðm 2 þn 2 ÞfmnðsÞds; 8m; n P 1; ð7Þ

which may also be written formally as(5)

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 L1(0, T; L1(X)) and

g 2 L1(X), and the error of the data is given on L1only Note that(7)still makes sense with such data, and this formula gives immediately the uniqueness

Theorem 1 (Uniqueness) For each f 2 L1(0, T; L1(X)) and g 2 L1(X), the problem (1)–(3)has at most one (weak) solution

u 2 C([0, T]; L2(X)) \ C1((0, T); L2(X))

In spite of the uniqueness, the problem is still ill-posed and a regularization is necessary For each> 0, introduce the truncation mapping P:L1ðXÞ ! C1ðXÞ \ H1ðXÞ

Pwðx; yÞ ¼ X

m;nP1;m 2 þn 2 6 M

wmnsinðmxÞ sinðnyÞ; with M¼lnð1Þ

In fact, Pis a finite-dimensional orthogonal projection on L2(X), but it works on L1(X) as well We shall approximate the original problem by the following well-posed problem

Theorem 2 (Well-posed problem) For each f 2 L1(0, T; L1(X)) and g 2 L1(X), let w 2 L1(0, T; L2(X)) defined by

wmnðtÞ ¼ eðTtÞðm 2 þn 2 ÞðPgÞmn

Z T

t

eðstÞðm 2 þn 2 ÞðPf ÞmnðsÞds; 8m; n P 1: ð9Þ

Then w = Pw and it depends continuously on (f, g), i.e if wiis the solution with respect to (fi, gi), i = 1, 2, then

kw1ðtÞ  w2ðtÞkL2 ðXÞ6

2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnðe1Þ p

p ffiffiffiffiffiffi 2T

p etT 2T kg1 g2kL1 ðXÞþ kf1 f2kL1 ð0;T;L 1 ðXÞÞ

:

Proof Note that w(t) is well-defined because wmn(t) = 0 if m2+ n2> M This fact also implies that w = Pw Now for two

kw1ðtÞ  w2ðtÞk2L2 ðXÞ¼p2

4 X

m;nP1

jw1;mnðtÞ  w2;mnðtÞj2

¼p2

4 X

m;nP1;m 2 þn 2 6 M

eðTtÞðm 2 þn 2 Þðg1 g2Þmn

Z T

t

eðstÞðm 2 þn 2 Þðf1 f2ÞmnðsÞds



2

6 4

p2

X

m;nP1;m 2 þn 2 6 M

eðTtÞM ekg1 g2kL 1

ðXÞþ

Z T

t

eðTtÞM ekf1ðsÞ  f2ðsÞkL 1

ðXÞds



2

6 4

p2Mee2ðTtÞM e kg1 g2kL1

ðXÞþ kf1 f2kL1

ð0;T;L1ðXÞÞ

:

Here we have used jvmnj 6 jvjL1 ð X Þand the fact that

#fðm; nÞ 2 Z2jm; n P 1; m2þ n26Meg 6 Me:

Reviewing the value of Me, we have the desired estimate h

Remark 1 A significant convenience of our method is that it is very easy to compute and represent explicitly the solution w defined by(9) Moreover, this solution is very smooth because wðtÞ ¼ PwðtÞ 2 C1

ðXÞ \ H1ðXÞ for all t 2 [0, T]

Remark 2 The stability magnitude of our well-posed problem is of order ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

lnðe1Þ p

etT 2T It is much better, especially when

t = 0, than the stability magnitudes given by quasi-reversibility method and quasi-boundary value method, for example,

tT

T in[3,19]and (ln(1))1in[4,18]

Our regularized solution is the solution produced directly by the well-posed problem in the previous section from the given data which works even on a very weak assumption on the exact solution

Theorem 3 (Regularized solution) Assume that the problem(1)–(3)has at most one (weak) solution u 2 C([0, T]; L2(X)) \

C1((0, T); L2(X)) corresponding to f 2 L1(0, T; L1(X)) and g 2 L1(X) Let fand gbe measured data satisfying

kf f kL 1 ð0;T;L 1 ðXÞÞ6;kg gkL 1 ðXÞ6:

Define the regularized solution u2 L1(0,T; L2(X)) from fand gas in(9) Then for each t 2 ½0; T; uðtÞ 2 C1ðXÞ \ H1ðXÞ and lime?0u(t) = u(t) in L2(X)

Proof We shall use the notations Pand Mdefined in(8) Note that uðtÞ ¼ PuðtÞ 2 C1ðXÞ \ H1ðXÞ as inRemark 1 More-over using the stability inTheorem 2we find that

kueðtÞ  uðtÞkL2 ðXÞ6kPueðtÞ  PeuðtÞkL2 ðXÞþ kPeuðtÞ  uðtÞkL2 ðXÞ6

4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnðe1Þ p

p ffiffiffiffiffiffi 2T

p eTþt2T þp

2 X

m;nP1;m 2 þn 2 >M e

jumnðtÞj2

0

@

1 A

1=2

ð10Þ

and it must converge to 0 as?0 To obtain the convergence of the second term in the right-hand side of(10), we note that

p2

4

X

m;nP1

jumnðtÞj2¼ kuðtÞk2L2 ðXÞ<1

In the above theorem, we did not give an error estimate because the condition of the exact solution u is so weak (we even did not require uðtÞ 2 H1ðXÞ) However in practical application we may expect that the exact solution is smoother In these cases many explicit errors estimates are available in the next section An essential point here is that the regularized solution

is the same in any case This is a substantial pleasure for practical application because even if someones do not know how good the exact solution is they are always ensured that the regularized solution works as well as possible without any fur-ther adjustment

3 Error estimates

From the usual viewpoint from variational method, it is natural to assume that uðtÞ 2 H1ðXÞ for all t 2 [0, T] Moreover, if f

is smooth and u is a classical solution for the heat Eq.(1)then uðtÞ 2 H2

ðXÞ \ H1ðXÞ for all t 2 [0, T] For these two cases we have the following explicit error estimates

Theorem 4 (Error estimate for usual cases) Let u, u as inTheorem 4and let t 2 [0, T]

(i) Assume that uðtÞ 2 H1

ðXÞ Then lim

 !0uðtÞ ¼ uðtÞ in H1ðXÞ and

kueðtÞ  uðtÞkL2 ðXÞ6

4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnðe1Þ p

p ffiffiffiffiffiffi 2T

p eTþt 2T þ

ffiffiffiffiffiffi 2T p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnðe1Þ

p kruðtÞkL2 ðXÞ:

(ii) Assume that uðtÞ 2 H2ðXÞ \ H1ðXÞ Then lim?0u(t) = u(t) in H2(X) and

kueðtÞ  uðtÞkL2

ðXÞ6

4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnðe1Þ p

p ffiffiffiffiffiffi 2T

p eTþt2T þ 2T

lnðe1ÞkuðtÞkH 2

ðXÞ;

kueðtÞ  uðtÞkH1 ðXÞ6

2 lnðe1Þ

pT eTþt 2Tþ

ffiffiffiffiffiffi 2T p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnðe1Þ

p kuðtÞk2H2 ðXÞ:

Here we use the norm

kwk2H1¼ krwk2

L 2¼ kwxk2L2þ kwyk2L2; kwk2H2¼ kwk2L2þ kwk2H1þ kwxxk2L2þ kwxyk2L2þ kwyxk2L2þ kwyyk2L2:

Proof

(i) By using the integral by part and the Parseval equality, it is straightforward to check that if uðtÞ 2 H1ðXÞ then

p2

4

X

m;nP1

ðm2þ n2ÞjumnðtÞj2¼ kruðtÞk2L2

X

m;nP1;m 2 þn 2 >M e

jumnðtÞj26 1

Me

X

m;nP1

ðm2þ n2ÞjumnðtÞj2¼ 4

p2MekruðtÞk2L2 ðXÞ:

Substituting the latter inequality into the estimate(10)in the proof ofTheorem 3, we obtain the error estimate in L2

To prove the convergence in H1we use the identity(11)and the stability ofTheorem 2again

krueðtÞ ruðtÞk2

L 2 ðXÞ¼p2

4 X

m;nP1

ðm2þ n2Þjue;mnðtÞ  umnðtÞj2

¼p2

4 X

m;nP1;m 2 þn 2 6 M e

ðm2þ n2Þjue;mnðtÞ  umnðtÞj2þp2

4 X

m;nP1;m 2 þn 2 >M e

ðm2þ n2ÞjumnðtÞj2

6p2

4 X

m;nP1;m 2 þn 2 6M e

MejðPeueÞmnðtÞ  ðPeuÞmnðtÞj2þp2

4 X

m;nP1;m 2 þn 2 >M e

ðm2þ n2ÞjumnðtÞj2

6MekPeueðtÞ  PeuðtÞk2L2

ðXÞþp2

4 X

m;nP1;m 2 þn 2 >M e

ðm2þ n2ÞjumnðtÞj2

64ðlnðe1ÞÞ2

p2T2 eTþtT þp2

4 X

m;nP1;m 2 þn 2 >M e

ðm2þ n2ÞjumnðtÞj2: ð12Þ

The second term in the right-hand side in(12)converges to 0 as?0 because the convergence in(11) Thus the conver-gence in H1has been proved

(ii) We now assume that uðtÞ 2 H2ðXÞ \ H1ðXÞ We have an identity similar to(11)

p2

4

X

m;nP1

ðm2þ n2Þ2jumnðtÞj2¼ kuxxðtÞk2L2 ðXÞþ kuxyðtÞk2L2 ðXÞþ kuyxðtÞk2L2 ðXÞþ kuyyðtÞk2L2 ðXÞ: ð13Þ

The error estimate in L2(X) follows(10)and the following inequality

X

m;nP1;m 2 þn 2 >M e

jumnðtÞj26 1

M2e

X

m;nP1

ðm2þ n2Þ2jumnðtÞj26 4

p2M2ekuðtÞk

2

H 2

ðXÞ:

Similarly, from(12)and the estimate

X

m;nP1;m 2 þn 2 >Me

ðm2þ n2ÞjumnðtÞj26 1

Me

X

m;nP1

ðm2þ n2Þ2jumnðtÞj26 4

p2MekuðtÞk2H 2

ðXÞ;

we find that

krueđtỡ ruđtỡk2L2

đXỡ6

p2đlnđe1ỡỡ2 4T2 eTợtT ợ 1

Mekuđtỡk2H2

đXỡ:

Using the inequality a ợ b 6 ffiffiffi

a p

ợ ffiffiffi b p

we obtain the error estimate in H1 Finally we prove the convergence in H2(X) Similarly to(12)we have

kđue uỡxxđtỡk2L2ợ kđue uỡxyđtỡk2L2ợ kđue uỡyxđtỡk2L2ợ kđue uỡyyđtỡk2L2

Ửp2

4

X

m;nP1

đm2ợ n2ỡ2jue;mnđtỡ  umnđtỡj2

6p2

4

X

m;nP1;m 2 ợn 2 6 M e

M2

ejue;mnđtỡ  umnđtỡj2ợp2

4 X

m;nP1;m 2 ợn 2 >M e

đm2ợ n2ỡ2jumnđtỡj2

6M2

ekPeueđtỡ  Peuđtỡk2L2 đXỡợp2

4 X

m;nP1;m 2 ợn 2 >M e

đm2ợ n2ỡ2jumnđtỡj2

62đlnđe1ỡỡ3

p2T3 eTợtT ợp2

4 X

m;nP1;m 2 ợn 2 >M e

as?0 due to the convergence in(13) h

conse-quence, we shall immediately obtain a uniform convergence whenever the corresponding uniform condition is imposed For example, if the exact solution u is in Cđơ0; T; H1

đXỡỡ or Cđơ0; T; H2đXỡ \ H1đXỡỡ then we have the estimates kue ukCđơ0;T;L2 đ X

and kue ukCđơ0;T;H1 đ X , respectively, in the same form of estimates inTheorem 4

Remark 4 The error estimates inTheorem 4work well no matter t > 0 or t = 0 In many earlier works, we find that the error

kuđ0ỡ  uđ0ỡkL2 đ X ỡis often not given (e.g.[17]) and an explicit error estimate in H1đXỡ is not available (e.g.[3,7,4,17,19,18])

InTheorem 4(ii), an error estimate in H2(X) is not given because we do not have enough information on the exact solu-tion (we just know uđtỡ 2 H2đXỡ \ H1đXỡ) However, when u is smoother then an explicit estimate in H2(X) may be derived In the last theorem, we shall give the error estimates in some special cases when the exact solution is very good We see from the proof ofTheorem 4that the facts uđtỡ 2 H1

đXỡ and uđtỡ 2 H2đXỡ \ H1đXỡ are equivalent to

X

m;nP1

đm2ợ n2ỡkjumnđtỡj2<1

with k = 1,2, respectively We shall see that from the latter condition with k > 2 we may improve the estimate, and in par-ticular give an error estimate in H2(X) We next consider a stronger condition similar to (in fact, weaker than)

sup

t2ơ0;T

X

m;nP1

which is a two-dimensional version of the condition(4)in[14] Such a condition seems essential to solve the nonlinear prob-lem Although it is quite strict for the linear case, as we discussed in the first section, if the above condition(15)holds then

we have a very good convergence rate which is of order đlnđ1ỡỡ3=2pffiffiffi

Theorem 5 Error estimate for special casesLet u, uas inTheorem 3and let t 2 [0, T]

(i) Assume that

Ekđtỡ Ử X1

n;mỬ1

đn2ợ m2ỡku2

for some constant k > 2 Then

kueđtỡ  uđtỡkL2 đXỡ6

4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnđe1ỡ p

p ffiffiffiffiffiffi 2T

p eTợt2T ợp ffiffiffiffiffiffiffiffiffiffiffi

Ekđtỡ p 2

2T lnđe1ỡ

 k

2

;

kueđtỡ  uđtỡkH1 đXỡ6

2 lnđe1ỡ

pT eTợt2T ợp ffiffiffiffiffiffiffiffiffiffiffi

Ekđtỡ p 2

2T lnđe1ỡ

 k1

2

;

kueđtỡ  uđtỡkH2 đXỡ6

2đlnđe1ỡỡ3=2

pT ffiffiffi T

p eTợt 2T ợ3p ffiffiffiffiffiffiffiffiffiffiffi

Ekđtỡ p 2

2T lnđe1ỡ

 k2

2

:

Here we assume6e2Tfor the estimate in H2(X)

(ii) Assume that

FrðtÞ ¼ X

m;nP1

e2rðm 2 þn 2 ÞjumnðtÞj2<1

for some constant r > 0 Then

kueðtÞ  uðtÞkL 2

ðXÞ6

4 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lnðe1Þ p

p ffiffiffiffiffiffi 2T

p eTþt 2T þp ffiffiffiffiffiffiffiffiffiffi

FrðtÞ p

2 er 2T;

kueðtÞ  uðtÞkH1 ðXÞ6

2 lnðe1Þ

pT eTþt2Tþp ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

FrðtÞ lnðe1Þ p

2 ffiffiffiffiffiffi 2T

p e2Tr;

kueðtÞ  uðtÞkH2 ðXÞ6

6ðlnðe1ÞÞ3=2

pT ffiffiffi T

p eTþt 2Tþ3p ffiffiffiffiffiffiffiffiffiffi

FrðtÞ p lnðe1Þ 4T er

2T:

Here we assume6e2Tfor the estimate in H1

ðXÞ, and6e4Tfor the estimate in H2(X)

Proof

(i) We use the same way of the proof ofTheorem 4 We shall prove the error estimates in H2(X) (the other ones are sim-ilar and easier) From

X

m;nP1;m 2 þn 2 >M e

ðm2þ n2Þ2jumnðtÞj26 1

Mk2e

X

m;nP1

ðm2þ n2ÞkjumnðtÞj26EkðtÞ

Mk2e

kðue uÞxxðtÞk2L2þ kðue uÞxyðtÞk2L2þ kðue uÞyxðtÞk2L2þ kðue uÞyyðtÞk2L262ðlnðe1ÞÞ3

p2T3 eTþtT þ p2Ek

4Mk2e

6 4

pM

3=2

e eTþt2Tþp ffiffiffiffiffiffiffiffiffiffiffi

EkðtÞ p

 k2 2

e

!2

:

Using

kwkH 26kwkL 2þ kwkH 1þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

kwxxk2L2þ kwxyk2L2þ kwyxk2L2þ kwyyk2L2

q

ð17Þ

and MP1 we conclude the desired estimate in H2(X)

(ii) From(10)and

X

m;nP1;m 2 þn 2 >M e

jumnðtÞj26e2rM e X

m;nP1

e2rðm 2 þn 2 ÞjumnðtÞj26FrðtÞer

T

we get the error estimate in L2(X)

Note that the function n ´ en/n is increasing when n P 1 Thus

ðm2þ n2Þ 6 Mee2rðm 2 þn 2 M e Þ when m2þ n2>MeP1:

It implies that

X

m;nP1;m 2 þn 2 >Me

ðm2þ n2ÞjumnðtÞj26Me

X

m;nP1

e2rðm 2 þn 2 M e ÞjumnðtÞj26MeFrðtÞer

T:

The error estimate in H1ðXÞ follows the above estimate and(12)

Similarly, because the function n ´ en

/n2is increasing when n P 2, we find that

ðm2þ n2Þ26M2

ee2rðm 2 þn 2 M e Þ if m2

þ n2>MeP2:

It follows that

X

m;nP1;m 2 þn 2 >Me

ðm2þ n2ÞjumnðtÞj26M2

e

X

m;nP1

e2rðm 2 þn 2 M e ÞjumnðtÞj26M2

eFrðtÞer

T:

kðue uÞxxðtÞk2L2þ kðue uÞxyðtÞk2L2þ kðue uÞyxðtÞk2L2þ kðue uÞyyðtÞk2L262ðlnðe1ÞÞ3

p2T3 eTþtT þ M2eFrðtÞer

T

6 4

pM

3=2

e eTþt2T þp ffiffiffiffiffiffiffiffiffiffi

FrðtÞ p

2 Mee2Tr

!2

:

Using(17)again and MP1 we conclude the error estimate in H2(X) h

Remark 5 If(15)holds, i.e.P

m;nP1e2Tðm 2 þn 2 ÞjumnðtÞj2<1, then applyingTheorem 5in the case r = T we get

sup

t2½0;T

kueðtÞ  uðtÞkL2 ðXÞ6C ffiffiffi

e

p

; sup

t2½0;T

kueðtÞ  uðtÞkH1 ðXÞ6C lnðe1Þ ffiffiffi

e

p

; sup

t2½0;T

kueðtÞ  uðtÞkH2 ðXÞ6Cðlnðe1ÞÞ3=2 ffiffiffi

e

p :

Notice that in[11], under a similar condition, Liu gave the error estimate (see Theorem 3.3, p 466)

kga g0kL2 ðXÞ6C 1 1t0T

Let t0= 0, we get

kga g0kL2

ðXÞ6C ffiffiffiffiffi

1

p

Thus at t = 0, our method gave the same order error in L2-norm as the method of Liu[11] However, the strong point of our paper is that the error estimates in H1ðXÞ or H2(X) established, and also of Hölder type (in fact, they are better than any esti-mate of orderqwith q 2 (0, 1/2)) They are not given in[11]

Remark 6 The truncated regularization method is a very simple and effective method for solving some ill-posed problems and it has been successfully applied to some inverse heat conduction problems[5,8,13] Recently, in[14]many applications for a model of the Helmholtz equation are introduced and a Fourier method was applied for solving a Cauchy problem for the Helmholtz equation In[9], Fu and his group used the truncated method to solve the backward heat in the unbounded region and established the logarithmic order of the form

kuð:; tÞ  ud;n maxk 6 E1tT lnE

d

 ðtTÞs 2T

E 1

TlnEþ ln ln E s

!s 2

0

@

1

And in[20], the authors gave the following estimates

kwb;abð:; :; tÞ  uð:; :; tÞk 6 bt=T ln1

b

  a ðTtÞ 2T

expðk2ðT  tÞ2Þ þ Q ðb; t; uÞ ln1

b

 a

2!

And in[21], Trong and Tuan only established the logarithmic form as follows

kuð:; :; tÞ  uð:; :; tÞk 6 C

1 þ ln T



ð20Þ

Note that the errors(18)–(20)are the same order asTheorem 5(i) However,the logarithmic type estimate is, in general, much worse than any Hölder type estimate, i.e.qfor some q > 0 InTheorem 5(ii) we also establish this type of estimates, which are not given in[9,20,21] It worth mentioning that our regularized solution is unique, in all cases This proves that our method is effective

Remark 7 Sometime, it is also important to consider the 2-D backward heat for a general two-dimensional domain, e.g

[12] In this case to apply the truncation method, we need to consider the spectral problem of operator Din this domain (with homogeneous Dirichlet boundary condition) However, this question is not always solvable explicitly and this is a dis-advantage point of our method

4 Numerical experiments

In this section we give some numerical experiments for our method For simplicity, we shall recover the initial temper-ature at t = 0 from the final data at T = 1

Example 1 Consider the problem

utDu ¼ f ðx; y; tÞ  3etsinðxÞ sinðyÞ;

with the final condition

uðx; y; 1Þ ¼ gðx; yÞ  e sinðxÞ sinðyÞ:

Problem(1)–(3)with exact data (f, g) has the exact solution

uðx; y; tÞ ¼ etsinðxÞ sinðyÞ:

For any n = 1, 2, , let us take the measured data

fn¼ f ; gnðx; yÞ ¼ gðx; yÞ þ n sinðnxÞ sinðnyÞ:

Then Problem(1)–(3)with measured data (f, g) has corresponding solution

eunðx; y; tÞ ¼ uðx; y; tÞ þ1

ne

n 2 ð1tÞsinðnxÞ sinðnyÞ:

We see that

kgnðx; yÞ  gðx; yÞkL1 ðXÞ¼4

n! 0;

k~unð:; :; 0Þ  uð:; :; 0ÞkL2

ðXÞ¼e

n 2

n ! þ1

It means that if n is large then a small error of data might cause a large error of solutions Therefore, the problem is really unstable and hence a regularization is necessary Using the regularization ofTheorem 2correspodinge= 4/n, we see that when n > 4e4then the regularized solution at t = 0 is

ueðx; yÞ ¼ sinðxÞ sinðyÞ;

which coincides the exact solution u(., , 0) In this example, our method works very well because the exact solution’s form is

of a truncated Fourier series

Example 2 Consider the problem

utDu ¼ f ðx; y; tÞ  sinðxÞðty3pty2 6ty þ 6y þ 2tp 2pÞ

with the final condition

uðx; y; 1Þ ¼ gðx; yÞ  0:

The exact solution of the latter equation is

uðx; y; tÞ ¼ ð1  tÞ sinðxÞy2ðp yÞ:

For any n = 1, 2, , take the measured data

fn¼ f ; gnðx; yÞ ¼ 1

4nsinðnxÞ sinðnyÞ:

Then the disturbed solution is

~

unðx; y; tÞ ¼ uðx; y; tÞ þ 1

4ne

n 2 ð1tÞsinðnxÞ sinðnyÞ:

We see that

kgnðx; yÞ  gðx; yÞkL1 ðXÞ¼1

n! 0;

k~unð:; :; 0Þ  uð:; :; 0ÞkL2

ðXÞ¼e

n 2

4n! þ1:

Table 1

Errors between disturbed solution, regularized solution and exact solution.

10 5

10 9

4 sinðxÞ sinðyÞ  3 sinðxÞ sinð2yÞ þ 4

10 15

4 sinðxÞ sinðyÞ  3 sinðxÞ sinð2yÞ þ 4

27 sinðxÞ sinð3yÞ  3

16 sinðxÞ sinð4yÞ exp(10 30

Thus the problem in this case is also unstable We now compute the regularized solutions by using the regularization method introduced in the previous sections withe= 1/n The effect of our regularization is represented viaTable 1, where we denote

by u:¼ u(., , 0) the regularized value at t = 0,eu:¼eunð:; :; 0Þ the disturbed value (with= 1/n), and u0:¼ u(., , 0) the exact value We can see that while the errors between the disturbed solution and the exact solution is extremely large, the error between the regularized solution and the exact solution is acceptable, even in H1-norm

Acknowledgments

Most part of the work was done when the first author was a student of University of Science, Vietnam National University

at HoChiMinh City The second and the third authors are supported by the Council for Natural Sciences of Vietnam We thank the referees for constructive comments leading to the improved version of the paper

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