DSpace at VNU: A modified integral equation method of the semilinear backward heat problem

Tuan, Regularization and error estimate for the non-linear backward heat problem using a method of integral equation, Nonnon-linear Anal.. Ingham, An iterative boundary element method fo

A modified integral equation method of the semilinear backward

heat problem

a

Department of Mathematics and Application, Saigon University, 273 An Duong Vuong, Ho Chi Minh City, Viet Nam

b

Department of Mathematics, University of Natural Science, Vietnam National University, 227 Nguyen Van Cu, Q5, Ho Chi Minh City, Viet Nam

a r t i c l e i n f o

Keywords:

Backward heat problem

Ill-posed problem

Nonlinear heat

Contraction principle

a b s t r a c t

The paper is concerned with the non-linear backward heat equation in the rectangle domain The problem is severely ill-posed We shall use a modified integral equation method to regularize the nonlinear problem The error estimates of Hölder type of the reg-ularized solutions are obtained Numerical results are presented to illustrate the accuracy and efficiency of the method This work is a generalization of many earlier papers, includ-ing the recent paper [D.D Trong, N.H Tuan, Regularization and error estimate for the non-linear backward heat problem using a method of integral equation, Nonnon-linear Anal 71 (9) (2009) 4167–4176]

Ó 2010 Elsevier Inc All rights reserved

1 Introduction

The backward heat conduction problem (BHCP) is that of finding the distribution of temperature from the final (time) data The problem is ill-posed in the sense of Hadamard In fact, for a given final data, we are not sure that a solution of the problem exists In case a solution exists, it may not depend continuously on the final data This is a typical example of the inverse and ill-posed problems For its applications we refer to various excellent literature, e.g Lattes and Lions[14]and Tikhonov and Arsenin[21] To find approximate solutions for this problem, many approaches have been investigated Lattes and Lions [14], Showalter [20], Ames and Hughes [1] and Miller [17] used quasi-reversibility method Schroter and Tautenhahn[29] established an optimal error estimate for a special BHCP A mollification method has been studied by Hao in[12] Kirkup and Wadsworth used an operator-splitting method in[16] This problem was also investigated by many other authors Dokuchaev[4], Engl et al.[5], Hassanov and Mueller[13], Lesnic et al.[15], and Yildiz et al.[30,31] Although there are many works on the homogeneous case and the linear inhomogeneous cases, the literature on the nonlinear case of the backward heat problem is quite scarce In 2005, Quan and Dung[18]offered a regularized solution

by semi-group method However, they gave an error estimate only in a very specific case in which the exact solution has

a finite Fourier series expansion In 2007, Trong et al.[23]used the quasi-boundary value method to treat the nonlinear case and attained an error estimate of ordert

T for each t > 0 This estimate is good at any fixed t > 0 but useless at t ¼ 0 Very recently, Trong and Tuan[26]improved this method to give an error estimate of ordert

Tðlnð1=ÞÞTt1for all t 2 ½0; T For the literature on nonhomogeneous and nonlinear backward heat, we refer the reader to the results in Fu et al.[8], Trong and his group[19,22–27] However, the error estimate in the mentioned papers is still of logarithmic order

In practice, we get the datauby measuring at discrete nodes Hence, instead ofu, we shall get an inexact datau

satisfying

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

kuuk 6;

where the constant>0 represents a bound on the measurement error, k  k denotes the L2-norm The major object of this paper is to provide a new regularization method to estimate the Hỏlder estimates on [0, 1] We prove that under some suit-able conditions, the approximate solutionvand the exact solution u satisfy the estimate

where C is a constant dependent on u and m is a constant independent of t, u

The rest of the paper is organized as follows In the next section, we shall state nonlinear BHCP Then, we review the reg-ularization method and give some estimates In Section3, we prove the main results Finally, in Section4, numerical exam-ples are tested to verify the efficacy of the our method

2 Mathematical problem and regularization

2.1 The inverse problem

Let u = u(x, t) be the distribution of temperature on the interval đ0;pỡ at the time t and let f đx; t; uỡ be the heat source which may not be linear in u In fact, we assume that f 2 L1

đơ0;p  ơ0; 1  Rỡ and that

jf đx; y; wỡ  f đx; y;vỡj 6 kjw vj;

where k > 0 independent of x, t, u From the theory of heat conduction, one has the equation

ut uxxỬ f đx; t; uđx; tỡỡ; đx; tỡ 2 đ0;pỡ  đ0; 1ỡ; đ2ỡ subject to the boundary condition

uđ0; tỡ Ử uđp;tỡ Ử 0; t 2 đ0; 1ỡ:

The inverse problem is to determine the distribution uđx; tỡ from the final data

If a solution exists, then it is the unique solution to the problems(2) and (3)([23], Theorem 3.1, p 239) For systems(2) and (3), there is no guarantee that the solution exists In the simplest case f = 0, the problems(2) and (3)has a unique solution if and only if

X1

nỬ1

e2n 2

u2<1;

whereunỬ2

p

Rp

0uđxỡ sinđnxỡdx (see[2]) If f Ử f đx; tỡ, (see[28, p.43, Lemma 1]) then the problems(2) and (3)has a unique solution if and only if

X1

nỬ1

en 2

un

Z 1 0

esn 2

fnđsỡds

<1;

where fnđsỡ Ử2

p

Rp

0f đx; sỡ sinđnxỡdx When f Ử f đx; t; uỡ, we do not know any general condition under which the problems(2) and (3)is solvable In[23], we presented a simple way to check the existence of solution to the systems(2) and (3)(see The-orem 3.2a, p 239) The main purpose of this paper is to find a stable computation method to approximate the exact solution when it exists Recently, we studied this problem in some previous work, for example[26] However, the error estimates in

[26]is of logarithmic order, which is not good enough (see[26,Theorem 3, p 4171]) This is a disadvantage point of that paper Here we improve the results in[26,28]by a new regularization method The main idea is to transform the problem into a new form

2.2 Regularization

As well known, problems(2) and (3)can be transformed to the following integral equation (see[6])

uđx; tỡ ỬX1

nỬ1

eđt1ỡn 2

un

Z1 t

eđ1sỡn 2

fnđuỡđsỡds

where

uđxỡ ỬX1

unsinđnxỡ;

f đuỡđx; tỡ ỬX1

nỬ1

fnđuỡđtỡ sinđnxỡ are the expansions ofuand f(u), respectively The term eđt1ỡn 2

is the unstability cause Hence, in order to regularize the problem, we have to replace this term by the better one Naturally, we shall replace this term byợ epn 2t1

p

where p is

a real number, p P 1 Thus, we shall approximate the problem(4)by the following problem

uđx; tỡ ỬX1

nỬ1

ợ epn 2

 t1

p

un

Z 1 t

eđs1ỡn 2

fnđuỡđsỡds

whereis a positive parameter and

fnđuỡđtỡ Ử2

phf đx; t; uđx; tỡỡ; sinđnxỡi Ử

2

p

Z p

o

f đx; t; uđx; tỡỡ sinđnxỡdx; đ6ỡ

unỬ2

phuđxỡ; sinđnxỡi Ử

2

p

Z p

0

and h; i is the inner product in L2

đ0;pỡ If p Ử 1 then the approximation problem 5 has been studied in[26]

We denote W Ử Cđơ0; 1; L2đ0;pỡỡ \ L2đ0; 1; H1đ0;pỡỡ \ C1đ0; 1; H1đ0;pỡỡ The main result of paper is

Theorem 1 Let;M > 0;u2 L2đ0;pỡ and letu2 L2đ0;pỡ be a measured data such that

kuuk 6:

Then the problem

wđx; tỡ ỬX1

nỬ1

ợ epn 2

 t1

p

u

n

Z 1 t

eđs1ỡn 2

fnđwỡđsỡds

sinđnxỡ; 0 6 t 6 1;

has a unique solution w2 W

Moreover, if problem(4)has a unique solution u 2 W satisfying

2psup

06t61

X1

nỬ1

e2đt1ợpỡn 2

with unđtỡ Ử2

p

Rp

0uđx; tỡ sin nxdx, then

kwđ; tỡ  uđ; tỡk 6 ffiffiffiffiffi

M

p

ợ ffiffiffi 2 p

Remark

1 Clark and Oppenheimer[2]considered the following assumptions on the exact solution

Under this very weak condition(10)they obtained an error estimate of ordert

T Here, we give a comparison between our results with the results in[2] Note that when p = 1 and f = 0, then(8)becomes(10)

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X1

pỬ1

eđ2t2ợ2pỡn 2

u2đtỡ

v

u

Ử

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

X1 nỬ1

e2tn 2

u2đtỡ

s

Ử kuđ; 0ỡk 6 M:

Moreover, the error estimate is then of ordert, which is the same as that in[2]

2 When p > 1 the condition(8)is very strict However, the error estimate(9)is then of orderp1=p This error estimate is much better than the logarithmic order estimates obtained in most of previously known results In work in progress, we are considering the possibility of getting similar estimates like that of(9)under less strict conditions than that of(8) When t Ử 0, we get

kuđ; 0ỡ  wđ; 0ỡk 6 ffiffiffiffiffi

M

p

ợ ffiffiffi 2 p

ek 2

p1p : The rate of convergence at t Ử 0 isp1p Hence, for p is large, the termp1p can approach This is a strong point of our method

3 The error(9)is of order O p1p

for all t 2 ơ0; 1 As we know, the convergence rate ofp1p;đp > 1ỡ is faster than that of the logarithmic order ln 1



 

 q

đq > 0ỡ when! 0 In most of known results, the error between the exact solution and the regularized solution is of the logarithmic order ln1



 m

, where m > 0 This type of order is also investigated in many recent papers, such as[2,3,7Ờ11,19,22Ờ27] Combining the above information, the reader can see that our method is effec-tive and useful

3 Proof of the main result

First we prove some useful results

Lemma 1 For 0 6 t 6 s 6 1 6 p and>0, we have

đaỡ eđs1ỡn 2

ợ epn 2

 t1

p

6ts

p;

đbỡ ợ epn 2t1

p

6t1p:

Proof of Lemma 1 Proof a We have

eđs1ỡn 2

ợ epn 2

 t1

p

Ửợ epn 2ts

p

eđs1ỡn 2

ợ epn 2

 s1

p

Ửợ epn 2ts

p

đ1 ợepn 2

6ợ epn 2ts

p

6ts

Proof b Let s Ử 1 inLemma 1a, we get theLemma 1b h

Lemma 2 For all x > 0; 0 <a<1 we have

1  đx ợ 1ỡ a

Proof of Lemma 2 Since1

a>1, and xa<đx ợ 1ỡa, we obtain

1 ợ x 6 1 ợ x1 ađx ợ 1ỡa6ơ1 ợ x1 ađx ợ 1ỡa1a:

This implies

1  đx ợ 1ỡaỬđx ợ 1ỡ

a

 1

đx ợ 1ỡa 6

1 ợ x1 ađx ợ 1ỡa 1

đx ợ 1ỡa Ử x

1 a: 

Lemma 3 Letu2 L2đ0;pỡ Then problem(5)has a unique weak solution uđx; tỡ 2 W

Proof For w 2 Cđơ0; 1; L2đ0;pỡỡ, define

Gđwỡđx; tỡ Ử /đx; tỡ X1

nỬ1

Z 1 t

eđs1ỡn 2

ợ epn 2

 t1

p

fnđwỡđsỡds

sinđnxỡ and

/đx; tỡ ỬX1

nỬ1

ợ epn 2

 t1

p

unsinđnxỡ:

We claim that, for every w;v2 Cđơ0; 1; L2đ0;pỡỡ; m P 1, we obtain

kGmđwỡđ; tỡ  Gmđvỡđ; tỡk26 k



 2m đ1  tỡm

where jjj  jjj is the sup norm in Cđơ0; 1; L2đ0;pỡỡ We shall prove the latter inequality by induction

For m Ử 1, usingLemmas 1 and 2, we have

kGđwỡđ; tỡ  Gđvỡđ; tỡk2Ửp

2

X1 nỬ1

Z1 t

eđs1ỡn 2

ợ epn 2

 t1

p

fnđwỡđsỡ  fnđvỡđsỡ

6p

2

X1 nỬ1

Z 1 t

eđs1ỡn 2

ợ epn 2

 t1

p

ds

Z 1 t

fnđwỡđsỡ  fnđvỡđsỡ

6p

2

X1 nỬ1

1

2đ1  tỡ

Z 1 t

fnđwỡđsỡ  fnđvỡđsỡ

đ ỡ2ds Ửp

2

1

2đ1  tỡ

Z 1 t

X1 nỬ1

fnđwỡđsỡ  fnđvỡđsỡ

Ử 1

2đ1  tỡ

Z 1 t

Z p

0

f đx; s; wđx; sỡỡ  f đx; s;vđx; sỡỡ

6k2

2đ1  tỡ

Z 1 t

Z p

0 jwđx; sỡ vđx; sỡj2dxds Ử Ck

2

2đ1  tỡjjjw vjjj2: đ15ỡ Thus(14)holds for m Ử 1

Suppose that(14)holds for m Ử j We prove that(14)holds for m Ử j ợ 1 UsingLemmas 1 and 2again, we have

kGjợ1đwỡđ; tỡ  Gjợ1đvỡđ; tỡk2Ửp

2

X1 nỬ1

Z 1 t

eđs1ỡn 2

ợ epn 2

 t1

p

fnđGjđwỡỡđsỡ  fnđGjđvỡỡđsỡ

ds

6p

2

1

2

X1 nỬ1

Z 1 t

jfnđGjđwỡỡđsỡ  fnđGjđvỡỡđsỡjds

6p

2

1

2đ1  tỡ

Z 1 t

X1 nỬ1

jfnđGjđwỡỡđsỡ  fnđGjđvỡỡđsỡj2ds

6 1

2đ1  tỡ

Z1 t

kf đ; s; Gjđwỡđ; sỡỡ  f đ; s; Gjđvỡđ; sỡỡk2ds

6 1

2đ1  tỡk2

Z 1 t

kGjđwỡđ; sỡ  Gjđvỡđ; sỡk2ds

6 1

2đ1  tỡk2 k



 2jZ 1 t

đ1  sỡj j! dsC j jjjw vjjj26 k



 2đjợ1ỡ

đ1  tỡjợ1

đj ợ 1ỡ! jjjw vjjj2: đ16ỡ Therefore, by the induction principle, we have

jjjGmđwỡ  Gmđvỡjjj 6 k



 m 1 ffiffiffiffiffiffi m!

p jjjw vjjj for all w;v2 Cđơ0; 1; L2đ0;pỡỡ

We consider G : Cđơ0; 1; L2đ0;pỡỡ ! Cđơ0; 1; L2đ0;pỡỡ Since

lim

m!1

k



 m

1 ffiffiffiffiffiffi m!

p Ử 0;

there exists a positive integer number m0such that Gm 0is a contraction It follows that the equation Gm 0đwỡ Ử w has a unique solution u2 Cđơ0; 1; L2đ0;pỡỡ

We claim that Gđuỡ Ử u In fact, one has GđGm0đuỡỡ Ử Gđuỡ Hence Gm0đGđuỡỡ Ử Gđuỡ By the uniqueness of the fixed point of Gm 0, one has Gđuỡ Ử u, i.e., the equation Gđwỡ Ử w has a unique solution u2 Cđơ0; 1; L2đ0;pỡỡ h

Lemma 4 The solution of the problem(5)depends continuously onuin L2đ0;pỡ

Let u andvbe two solutions of(5)corresponding to the final valuesuandx, respectively

From(5)one has in view of the inequality đa ợ bỡ262đa2ợ b2ỡ

kuđ; tỡ vđ; tỡk2Ửp

2

X1 nỬ1

ợ epn 2

 t1

p

đunxnỡ 

Z 1 t

eđs1ỡn 2

ợ epn 2

 t1

p

đfnđuỡđsỡ  fnđvỡđsỡdsỡ

2

6pX1 nỬ1

ợ epn 2

 t1

p

junxnj

ợpX1 nỬ1

Z 1 t

eđs1ỡn 2

ợ epn 2

 t1

p

jfnđuỡđsỡ  fnđvỡđsỡjdsỡ

: FromLemmas 1 and 2, we get

kuđ; tỡ vđ; tỡk2622t2p kuxk2ợ 2k2đ1  tỡ2tp

Z 1

 2s

pkuđ; sỡ vđ; sỡk2ds:

2t

pkuð; tÞ vð; tÞk2622

pkuxk2þ 2k2ð1  tÞ

Z 1 t

 2s

pkuð; sÞ vð; sÞk2ds:

Using Gronwall’s inequality we have

kuð; tÞ vð; tÞk 6 ffiffiffi

2

p

t1

pexpfk2ð1  tÞ2gkuxk:

Proof of Theorem 1 Letv be the solution of problem(5)corresponding touand let wbe the solution of problem(5)

corresponding tou

Using the triangle inequality, we get

kwð; tÞ  uð; tÞk 6 kwð; tÞ vð; tÞk þ kvð; tÞ  uð; tÞk: ð17Þ

We divide the proof into two steps

Step 1 Estimate kwð; tÞ vð; tÞk

UsingLemma 2, we get

kwð; tÞ  uð; tÞk 6 kwð; tÞ vð; tÞk 6 ffiffiffi

2

p

t1

pexpðk2ð1  tÞ2Þkuuk 6 ffiffiffi

2

p

ek 2 ð1tÞt1þpp : ð18Þ Step 2 Estimate kuð; tÞ vð; tÞk

Suppose the problems(2) and (3)has a unique solution u, we get the following formula

uðx; tÞ ¼X1

n¼1

eðt1Þn 2

un

Z1 t

eðtsÞn 2

fnðuÞðsÞds

Hence

unðtÞ ¼ eðt1Þn 2

un

Z 1 t

eðtsÞn 2

Multiplying(20)by ð1 þepn 2

Þt1p we have:

1 þepn 2

 t1

p

unðx; tÞ ¼ 1 þ epn 2t1

p

eðt1Þn 2

un

Z 1 t

1 þepn 2

 t1

p

eðtsÞn 2

fnðuÞðsÞds ¼þ en 2t1

un



Z 1 t

1 þepn 2

 t1

p

eðs1Þn 2

eð1tÞn 2

fnðuÞðsÞds ¼þ epn 2t1

p

un

Z1 t

eðs1Þn 2

þ epn 2

 t1

p

fnðuÞðsÞds: ð21Þ From(5)we have:

u

nðtÞ ¼þ epn 2t1

p

un

Z 1 t

eðs1Þn 2

þ epn 2

 t1

p

Because of(21) and (22), usingLemmas 1 and 2, we have:

junðtÞ  unðtÞj 6 junðtÞ  ð1 þepn 2

Þt1punðtÞj þ ð1  ð1 þepn 2

Þt1pÞjunðtÞj

6

Z 1 t

eðs1Þn 2

þ epn 2

 t1

p

jfnðuÞðsÞ  fnðuÞðsÞjds

þt1þpp eðt1þpÞn 2

junðtÞj Applying the inequality ða þ bÞ262ða2þ b2Þ andLemma 1, we obtain

kuð; tÞ  uð; tÞk2¼p

2

X1 n¼1

junðtÞ  unðtÞj2

6pX1 n¼1

Z 1 t

eðs1Þn 2

þ epn 2

 t1

p

jfnðuÞðsÞ  fnðuÞðsÞds

þ 22t2þ2pp pX1

n¼1

e2ðt1þpÞn 2

junðtÞj2

6pX1 n¼1

Z 1 t

e2ðs1Þn 2

þ epn 2

 2 t1

p

jfnðuÞðsÞ  fnðuÞðsÞj2ds þ 22t2þ2pp pX1

n¼1

e2ðt1þpÞn 2

junðtÞj2

6pX1 n¼1

Z 1 t

2 ts

pjfnðuÞðsÞ  fnðuÞðsÞj2ds þ 22t2þ2pp pX1

n¼1

e2ðt1þpÞn 2

junðtÞj2

622t p

Z 1 t

 2s

pkf ð; s; uð; sÞÞ  f ð; s; uð; sÞÞk2ds þ 2p 2t2þ2pp X1

e2ðt1þpÞn 2

junðtÞj2 ð23Þ

kuð; tÞ  uð; tÞk2622t

p

Z 1 t

 2s

pkf ð; s; uð; sÞÞ  f ð; s; uð; sÞÞk2ds þ 22t2þ2pp pX1

n¼1

e2ðt1þpÞn 2

junðtÞj2: This implies

 2t

pkuð; tÞ  uð; tÞk262þ2pp M þ 2k2

Z 1 t

 2s

pkuð; sÞ  uð; sÞk2ds:

Using Gronwall’s inequality, we get

 2t

pkuð; tÞ  uð; tÞk262þ2pp Me2k2ð1tÞ

:

It follows that

kuð; tÞ  uð; tÞk 6 ffiffiffiffiffi

M

p

ek 2

From(17), (18) and (24), we obtain

kwð; tÞ  uð; tÞk 6 kwð; tÞ vð; tÞ þ kvð; tÞ  uð; tÞk 6 ffiffiffi

2

p

t1

pexpðk2ð1  tÞ2Þkuuk þ ffiffiffiffiffi

M

p

ek2ð1tÞt1þpp

6 ffiffiffiffiffi M

p

þ ffiffiffi 2 p

ek2ð1tÞt1þpp ; for every t 2 ½0; 1 This completes the proof ofTheorem 1 h

4 Numerical experiment

In this section, we will describe a numerical implementation of problems(2) and (3)

Example 1 (The linear case) Let us consider the linear backward heat problem

uxxþ ut¼ uðx; tÞ þ gðx; tÞ; ðx; tÞ 2 ð0;pÞ  ð0; 1Þ;

uð0; tÞ ¼ uðp;tÞ ¼ 0; t 2 ½0; 1;

uðx; 1Þ ¼uðxÞ; x 2 ½0;p;

where

gðx; tÞ ¼ etsin x

and

uðx; 1Þ ¼u0ðxÞ  e sin x:

The exact solution of the equation is

uðx; tÞ ¼ etsin x:

Especially

u x;999

100

 uðxÞ ¼ exp 999

1000

sin x  2:715564905 sin x:

LetuðxÞ uðxÞ ¼ ðþ 1Þe sin x We have

kuuk2¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Z p

0

2e2sin2xdx

s

¼e

ffiffiffiffi

p

2

r : Applying the method introduced in this paper, we find the regularized solution uðx;999

1000Þ  uðxÞ having the following form

uðxÞ ¼vmðxÞ ¼ w1;msin x þ w6;msin 6x;

where

v1ðxÞ ¼ ðþ 1Þe sin x;

w1;1¼ ðþ 1Þe; w6;1¼ 0;

Let p ¼ 2 and

a ¼ 1

5000;

tm¼ 1  am m ¼ 1; 2; ; 5;

wi;mþ1¼ ðþ e2i 2

Þtmþ1tm2 wi;m2

p

Rt m

eðst m Þi 2 Rp

vmðxÞ þ gðx; sÞ

ð Þ sin ix dx

ds

; i ¼ 1; 6:

8

>

>

Let a be the error between the regularization solution u and the exact solution u, i.e., a¼ ku uk and

¼1¼ 103;¼2¼ 107;¼3¼ 1011, we lead to the first table

1¼ 103 2:718275536 sinðxÞ  0:005455669367 sinð6xÞ 0.002740395328

2¼ 104 2:715835736 sinðxÞ  0:005459510466 sinð6xÞ 0.0003006372545

3¼ 1011 2:715562882 sinðxÞ  0:005504563418 sinð6xÞ 0.00003232321842

Example 2 (The nonlinear case) Let us consider the nonlinear backward heat problem

uxxþ ut¼ f ðuÞ þ gðx; tÞ; ðx; tÞ 2 ð0;pÞ  ð0; 1Þ;

uð0; tÞ ¼ uðp;tÞ ¼ 0; t 2 ½0; 1;

uðx; 1Þ ¼uðxÞ; x 2 ½0;p;

where

f ðuÞ ¼

u2 u 2 ½e10;e10;

e 10

e1u þe 21 e1 u 2 ðe10;e11;

e 10

e1u þe 21

e1 u 2 ðe11;e10;

0 juj > e11;

8

>

<

>

:

gðx; tÞ ¼ 2etsin x  e2tsin2x;

and

uðx; 1Þ ¼u0ðxÞ  e sin x:

The exact solution of the equation is

uðx; tÞ ¼ etsin x:

In particular

u x;999

100

 uðxÞ ¼ exp 999

1000

sin x  2:715564905 sin x:

LetuðxÞ uðxÞ ¼ ðþ 1Þe sin x We have

kuuk2¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Z p

0

2e2sin2xdx

s

¼e

ffiffiffiffi

p

2

r : Applying the method introduced in this paper, we find the regularized solution u x;999

1000

 uðxÞ having the following form

uðxÞ ¼vmðxÞ ¼ w1;msin x þ w6;msin 6x;

where

v1ðxÞ ¼ ðþ 1Þe sin x;

w1;1¼ ðþ 1Þe; w6;1¼ 0

and

a ¼ 1

5000

tm¼ 1  am; m ¼ 1; 2; ; 5;

wi;mþ1¼ ðþ e2i2Þtmþ1tm2 wi;m2

p

Rt m

tmþ1eðst m Þi2 Rp

0 v2

mðxÞ þ gðx; sÞ

sin ix dx

ds

; i ¼ 1; 6:

8

>

>

Letting¼1¼ 103; ¼2¼ 104; ¼3¼ 1011, we have the table

1¼ 103 2:718264487 sinðxÞ  0:005466473792 sinð6xÞ 0.002729464336

2¼ 104 2:715833791 sinðxÞ  0:005461493459 sinð6xÞ 0.0002987139108

3¼ 1011 2:715552177 sinðxÞ  0:005518178192 sinð6xÞ 0.00004317829056

This project was supported by National Foundation for Science and Technology Development (NAFOSTED), Code: 101.01-2010.10 The authors thank the editor and the referees for their valuable comments leading to the improvement of our man-uscript The authors thank Truong Trung Tuyen in the Indiana University and Nguyen Minh Quan in the University of Buffalo for their most helpful comments on English grammar

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