Regularization and error estimate for the nonlinear backward heat problem using a method of integral equation

Nonlinear Analysis 71 (2009) 4167–4176

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Regularization and error estimate for the nonlinear backward heat problem using a method of integral equationI Dang Duc Trong a , Nguyen Huy Tuan b,∗ a

HoChiMinh City National University, Department of Mathematics and Informatics, 227 Nguyen Van Cu, Q. 5, HoChiMinh City, Viet Nam

b

Department of Mathematics and Informatics, Ton Duc Thang University, 98 Ngo Tat To street, Binh Thanh district, Hochiminh City, Vietnam

article

abstract

info

Article history: Received 12 March 2008 Accepted 20 February 2009

We consider the inverse time problem for the nonlinear heat equation in the form ut − uxx = f (x, t , u(x, t )),

(x, t ) ∈ (0, π ) × (0, T ),

u(0, t ) = u(π , t ) = 0 t ∈ (0, T ).

MSC: 35K05 35K99 47J06 47H10

The nonlinear problem is severely ill-posed. We shall use the method of integral equation to regularize the problem and to get some error estimates. We show that the approximate problems are well-posed and that their solution u (x, t ) converges on [0, T ] if and only if the original problem has a unique solution. We obtain several other results, including some explicit convergence rates. Some numerical tests illustrate that the proposed method is feasible and effective. © 2009 Elsevier Ltd. All rights reserved.

Keywords: Backward heat problem Nonlinearly ill-posed problem Quasi-boundary value methods Quasi-reversibility methods

1. Introduction There are several important ill-posed problems for parabolic equations. A classical example is the backward heat equation. In other words, it may be possible to specify the temperature distribution at a particular time, say t = T , and from this data the question arises as to whether the temperature distribution at any earlier time t < T can be retrieved. This is usually referred to as the backward heat conduction problem, or the final value problem. In the present paper we will consider the problem of finding the temperature u(x, t ), (x, t ) ∈ (0, π ) × [0, T ] such that ut − uxx = f (x, t , u(x, t )), u(0, t ) = u(π , t ) = 0, u(x, T ) = ϕ(x),

(x, t ) ∈ (0, π ) × (0, T ),

t ∈ (0, T ),

x ∈ (0, π ),

(1) (2) (3)

where ϕ(x), f (x, t , z ) are given. The systematic study of the backward heat problem is of rather recent origin, although consideration has already been given to such problems for several hundred years. In general, no solution, which satisfies the heat equation and the final data, exists. Further, even if a solution did exist, it would not be continuously dependent on the final data. It makes it difficult to do numerical calculations. Hence, a regularization is in order. The linear case was studied extensively in the past four decades by many methods. In the pioneering work [1] in 1967, the authors presented, in a heuristic way, the quasi-reversibility method

I Supported by the Council for Natural Sciences of Viet Nam.

∗

Corresponding author. E-mail address: [email protected] (N. Huy Tuan).

0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.02.092

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D. Duc Trong, N. Huy Tuan / Nonlinear Analysis 71 (2009) 4167–4176

(QR method). They approximated the problem by adding a ‘‘corrector’’ into the main equation. In fact, they considered the problem ut + Au −  A∗ Au = 0,

t ∈ [0, T ],

u(T ) = ϕ. The stability magnitude of the method is of order ec  ut + Au +  Aut = 0,

−1

. In [2,3], the problem is approximated with

t ∈ [0, T ],

u(T ) = ϕ. The method is useful if we cannot construct clearly the operator A∗ . However, the stability order in the case is quite large as in the original quasi-reversibility methods. In [4], using the method, so-called, of stabilized quasi-reversibility, the author approximated the problem with ut + f (A)u = 0,

t ∈ [0, T ],

u(T ) = ϕ. He showed that, with appropriate conditions on the ‘‘corrector’’ f (A), the stability magnitude of the method is of order c  −1 . Sixteen years after the work by Lattes–Lions, in 1983, Showalter presented the quasi-boundary value method (QBV method). He considered the problem ut − Au(t ) = Bu(t ),

t ∈ [0, T ],

u(0) = ϕ, and approximated the problem with ut − Au(t ) = Bu(t ),

t ∈ [0, T ],

u(0) +  u(T ) = ϕ. According to him, this method gives a better stability estimate than the others of the discussed methods. Clark and Oppenheimer, in their paper [5], used the quasi-boundary method to regularize the backward problem with ut + Au(t ) = 0,

t ∈ [0, T ],

u(T ) +  u(0) = ϕ. The authors showed that the stability estimate of the method is of order  −1 . Very recently, in [6], the modified quasiboundary method was used to solve linear parabolic equation. Although we have many works on the linear homogeneous case of the backward problem, the literature on the linear nonhomogeneous case and the nonlinear case of the problem are quite scarce. In [7,8], Trong and Tuan used the QR method and the eigenvalue-expansion method to regularize a 1D linear nonhomogeneous backward problem. Recently, in [9], the authors used the QBV method to regularize the backward problem in the nonlinear form on the right-hand side. However, t

in [9], the authors showed that the error between the approximate problem and the exact solution having the form C  T , is not near zero, if  fixed and t tends to zero. Hence, the convergence of the approximate solution is very slow when t is in a neighborhood of zero. Moreover, the authors cannot estimate the regularization error in which t = 0. They can only find a time t depending on  and derive the error between the exact solution u(., 0) and the approximate solution u (., t ). This makes it difficult to compute exactly the solution in the case where there is no change in the measure of final data. Informally, problem (1)–(3) can be transformed into an integral equation having the form u(x, t ) =

∞ X

e−(t −T )p

2



ϕp −



e−(T −s)p fp (u)(s)ds sin px 2

(4)

t

p=1

where ϕ(x) =

T

Z

P∞

p=1

ϕp sin(px), f (u)(x, t ) =

P∞

p=1 fp

(u)(t ) sin(px) are the expansions of ϕ and f (u), respectively. Since

t < T , we know from (4) that, when p becomes large, exp{(T − t )p2 } increases rather quickly. Thus, the term e−(t −T )p is the unstability cause. Hence, to regularize the problem, we have to replace the term by a better term. Naturally, we shall replace 2

2

this term by ( p2 + e−Tp ) u (x, t ) =

∞ X

t −T T

. In this paper, we shall approximate the problem (1)–(3) by the following integral equation: 2

( p2 + e−Tp )

t −T T



ϕp −

T

Z



e(s−T )p fp (u )(s)ds sin(px), 2

0 ≤ t ≤ T.

(5)

t

p=1

where  is a positive parameter such that  < eT and fp (u)(t ) =

ϕp =

2

2

π

hf (x, t , u(x, t )), sin(px)i = 2

Z

π

π

Z

f (x, t , u(x, t )) sin(px)dx,

(6)

o

π

ϕ(x) sin(px)dx, π 0 and h., .i is the inner product in L2 (0, π ). π

hϕ(x), sin(px)i =

2

(7)

D. Duc Trong, N. Huy Tuan / Nonlinear Analysis 71 (2009) 4167–4176

4169

The paper is organized as follows. In Section 2, we shall show that (5) is well-posed. Then, in Section 3, we estimate the error between an exact solution u0 of Problem (1)–(3) and the approximation solution u of (5). In fact, we shall prove that 

ku (., t ) − u0 (., t )k ≤ M 

!1− Tt

T

t T

,

1 + ln( T )

(8)

where k.k is the norm in L2 (0, π ) and C depends on u0 and f . Note that (8) is an interesting improvement of some latter results, such as in [10,7,9,8]. In fact, in most of the previous t

results, the errors often have the form C  T . This is one of their disadvantages in which t is zero. Our article will also give the error estimates that have the form M 

t T



T 1+ln( T )

1− Tt

, which is better than some known estimates. Moreover,the

convergence of the approximate solution at t = 0 is also given. This is a generalization of the known results in [4–14]. The notations about the usefulness and advantage of this method can be founded in Remark 3. Finally, a numerical experiment will be given in Section 4, which shows the efficiency of our method. 2. The well-posedness of problem (5) In the section, we shall study the existence, the uniqueness and the stability of a (weak) solution of Problem (5). In fact, one has Theorem 1. Let ϕ(x) ∈ L2 (0, π ),  ∈ (0, eT ) and let f ∈ L∞ ([0, π] × [0, T ] × R) satisfy

|f (x, y, w) − f (x, y, v)| ≤ k|w − v| for a k > 0 independent of x, y, w, v . Then Problem (5) has a unique weak solution u (x, t ) which is in C ([0, T ]; L2 (0, π )) ∩ L2 (0, T ; H01 (0, π )) ∩ C 1 (0, T ; H01 (0, π )). Furthermore, the solution depends continuously on ϕ in C ([0, T ]; L2 (0, π )). Proof. First, we have the following two inequalities which will be useful in the next results. Denote A(, s, t , p) = 2 2 t −T 2 t −T 2 e(s−T )p ( p2 + e−Tp ) T , B(, s, t , p) = ( p2 + e−Tp ) T , h(p) = ( p2 + e−Tp )−1 and a() =

to show that h(p) ≤ h

ln( T )

! =

T

T

 1 + ln( T )


=

a()



T . 1+ln( T )

It is not difficult

.

for 0 <  < eT . Hence, for 0 ≤ t ≤ s ≤ T and 0 <  < eT , we get the following estimate 2 2 t −T 2 t −s 2 2 s−T A(, s, t , p) = e(s−T )p ( p2 + e−Tp ) T = ( p2 + e−Tp ) T e(s−T )p ( p2 + e−Tp ) T 2

t −s T

2

t −s T

= ( p2 + e−Tp ) ≤ ( p2 + e−Tp ) ≤

s−T T

! s−T t

T

t −s T

2

(1 +  p2 eTp )

=

1 + ln(  ) T

t −s T

(a())

s−t T

(9)

and in the case s = T , we get B(, s, t , p) = ( p + e 2

−Tp2

)

t −T T

≤

! T −T t

T

t −T T

1 + ln(  ) 1

=

t −T T

(a())

T −t T

.

(10)

For w ∈ C ([0, T ]; L2 (0, π )), we set G(w)(x, t ) =

∞ X

B(, s, t , p)ϕp sin(px) −

∞ Z X

p=1

p=1

T

A(, s, t , p)fp (w)(s)ds sin(px). t

Denote by |||.||| the sup norm in C ([0, T ]; L2 (0, π )). We shall prove by induction, for every w, v ∈ C ([0, T ]; L2 (0, π )), m ≥ 1 then

kG (w)(., t ) − G (v)(., t )k ≤ m

m

2

!2m

kT

 1 + ln(  ) T




(T − t )m C m |||w − v|||2 . m!

(11)

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D. Duc Trong, N. Huy Tuan / Nonlinear Analysis 71 (2009) 4167–4176

For m = 1, using (9)–(10) and the Holder inequality, we have

kG(w)(., t ) − G(v)(., t )k2 = ≤

≤

=

= ≤

∞ Z πX

T

A(, s, t , p) fp (w)(s) − fp (v)(s) ds




2 p=1

t

∞ Z πX

T

2 p=1

T

Z

(A(, s, t , p))2 ds

2


2

fp (w)(s) − fp (v)(s)

ds

t

t

2 Z T ∞ 
2 πX a() fp (w)(s) − fp (v)(s) ds (T − t ) 2 p=1  t  2 Z TX ∞
2 π a() fp (w)(s) − fp (v)(s) ds (T − t ) 2  t p=1 2  Z TZ π a() (T − t ) (f (x, s, w(x, s)) − f (x, s, v(x, s)))2 dxds  0 t 2  Z TZ π a() (T − t ) |w(x, s) − v(x, s)|2 dxds k2  t 0 !2 T

= Ck2

 1 + ln( T )

(T − t )|||w − v|||2 .




Thus (11) holds. Suppose that (11) holds for m = j. We prove that (11) holds for m = j + 1. Using (10)–(11) again, we have j +1

kG

(w)(., t ) − G

j +1

(v)(., t )k = 2

≤

∞ Z πX

T


A(, s, t , p) fp (G (w))(s) − fp (G (v))(s) ds j

2 p=1

t

π

2



a()

2

 ≤

a()

 2

 

≤ 

≤

T t

(T − t )

∞ X

2

|fp (Gj (w))(s) − fp (Gj (v))(s)|2 ds

p=1

T

Z

kf (., s, Gj (w)(., s)) − f (., s, Gj (v)(., s))k2 ds t

a()

2



≤

(T − t )

Z

j

(T − t )L2

T

Z

kGj (w)(., s) − Gj (v)(., s)k2 ds t

a()

2



(T − t )k2j+2 !2j+2

kT


 1 + ln(  ) T



a()



2j Z

T t

( T − s) j dsC j |||w − v|||2 j!

(T − t )j+1 j+1 C |||w − v|||2 . (j + 1)!

Therefore, by the induction principle, we obtain

|||G (w) − G (v)||| ≤ m

!m

kT

m


 1 + ln(  ) T

T m/2

√

m!

C m |||w − v|||,

for all w, v ∈ C ([0, T ]; L2 (0, π )). We consider G : C ([0, T ]; L2 (0, π )) → C ([0, T ]; L2 (0, π )). It is easy to see that lim

m→∞

!m

kT

 1 + ln(  ) T




T m/2 C m

√

m!

= 0.

So, there exists a positive integer number m0 such that Gm0 is a contraction. It follows that the equation Gm0 (w) = w has a unique solution u ∈ C ([0, T ]; L2 (0, π )). We claim that G(u ) = u . In fact, one has G(Gm0 (u )) = G(u ). Hence Gm0 (G(u )) = G(u ). By the uniqueness of the fixed point of Gm0 , one has G(u ) = u , i.e., the equation G(w) = w has a unique solution u ∈ C ([0, T ]; L2 (0, π )). Step 2. The solution of the problem (5) depends continuously on ϕ in L2 (0, π ). Let u and v be two solutions of (5) corresponding to the final values ϕ and ω.

D. Duc Trong, N. Huy Tuan / Nonlinear Analysis 71 (2009) 4167–4176

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From (5), one has in view of the inequality (a + b)2 ≤ 2(a2 + b2 )

Z ∞ πX B(, s, t , p)(ϕp − ωp ) − 2

ku(., t ) − v(., t )k2 =

p=1

≤π

∞ X

2

T

A(, s, t , p)(fp (u)(s) − fp (v)(s)ds)

t

(B(, s, t , p)|ϕp − ωp |)2 + π

p=1

∞ Z X

T

A(, s, t , p)|fp (u)(s) − fp (v)(s)|ds

2

.

t

p=1

Using (9) and (10), we get the following inequality 2t

2t

ku(., t ) − v(., t )k2 ≤  T −2 (a())2− T kϕ − ωk2 Z

T

2t

2t



+ 2k2 (T − t ) T (a())− T

−2s T

2s

(a()) T ku(., s) − v(., s)k2 ds.

(12)

t

It follows that



−2t T

2t T

(a()) ku(., t ) − v(., t )k ≤  2

−2

(a()) kϕ − ωk + 2k (T − t ) 2

2

2

T

Z



−2s T

2s

(a()) T ku(., s) − v(., s)k2 ds.

t

Using Gronwall’s inequality we have t

t

2 2 ku(., t ) − v(., t )k ≤  T −1 (a())1− T exp !1(−k Tt (T − t ) )kϕ − ωk

T

t

=  T −1

exp(k2 (T − t )2 )kϕ − ωk.

1 + ln( T )

This completes the proof of the theorem.

 T

Remark 1. In [15,12,13,7], the stability of magnitude is e  . In [10] (see p. 5) and in [9] (see p. 238), the authors give better t

stability estimates than the latter discussed methods. They show that the stability estimate is of order M  T −1 . t

However, in our paper, we give a better estimation of the stability order, which is C  T −1



T 1+ln( T )

1− Tt . It is easy to see

that the order of the error is less than the order given in [10,9]. This proves the advantages of our method. 3. Regularization and error estimates We first have a uniqueness result Theorem 2. Let f be as in Theorem 1. Then Problem (1)–(3) has at most one (weak) solution u ∈ C ([0, T ]; L2 (0, π )) ∩ L2 (0, T ; H01 (0, π )) ∩ C 1 ((0, T ); L2 (0, π )). Proof. See p. 239 in [9].



Despite the uniqueness, Problem (1)–(3) is still ill-posed. Hence, we have to resort to a regularization. We have the following result. Theorem 3. Let f ,  be as in Theorem 1. Suppose that Problem (1)–(3) has a weak solution u(x, t ) ∈ W = C ([0, T ]; L2 (0, π )) ∩ L2 (0, T ; H01 (0, π )) ∩ C 1 ((0, T ); L2 (0, π )) which satisfies ∞ X

2

p4 e2Tp u2p (t ) < ∞

∀t ∈ [0, T ].

p=1

Then 

ku(., t ) − u (., t )kL2 (0,π) ≤

√ Me

k2 T (T −t )



t T

T 1 + ln( T )

!1− Tt .

(13)

2

where M = π supt ∈[0,T ] n=1 p4 e2Tp u2p (t ). Proof. It is easy to show that ∀x, α > 0 we have

P∞

1 − (x + 1)−α ≤ xα. In fact, we consider g (x) to be of the form g (x) = 1 − (x + 1)−α − xα.

(14)

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D. Duc Trong, N. Huy Tuan / Nonlinear Analysis 71 (2009) 4167–4176

Taking the derivative of it, we get g 0 (x) = α(x + 1)−α−1 − α ≤ 0. It follow that g (x) is a decreasing function on (0, ∞) then g (x) ≤ g (0) = 0. Suppose the Problem (1)–(3) has an exact solution u, we get the following formula u(x, t ) =

∞ X

T

Z

(e−(t −T )p ϕp − 2

e−(t −s)p fp (u)(s)ds) sin px. 2

(15)

t

p=1

Hence up (x, t ) = (e−(t −T )p ϕp − 2

T

Z

e−(t −s)p fp (u)(s)ds) 2

(16)

t

where up (t ) = π2 hu(x, t ), sin pxi. 2 t −T Multiplying (16) by (1 +  p2 eTp ) T we have: 2

(1 +  p2 eTp )

t −T T

2

up (x, t ) = (1 +  p2 eTp ) 2

= ( p2 + e−Tp )

t −T T

t −T T

= B(, s, t , p)ϕp −

e−(t −T )p ϕp − 2

ϕp − Z T

2

(1 +  p2 eTp )

t −T T

e−(t −s)p fp (u)(s)ds 2

t

T

Z

T

Z

2

(1 +  p2 eTp )

t −T T

2 2 e(s−T )p e(T −t )p fp (u)(s)ds

t

A(, s, t , p)fp (u)(s)ds.

(17)

t

From (5) we have: up (t ) = B(, s, t , p)ϕp −

T

Z

A(, s, t , p)fp (u (s))ds.

(18)

t

From (15), (17), (18) and using triangle inequality we have: t −T

t −T

|up (t ) − up (t )| ≤ |up (t ) − (1 +  p2 eTp ) T up (x, t )| + (1 − (1 +  p2 eTp ) T )|up (t )|   Z T t 2 ≤ A(, s, t , p)|fp (u (s)) − fp (u(s))|ds +  1 − p2 eTp |up (t )|. 2

2

T

t

Using the inequality (a + b) ≤ 2(a + b ), two inequalities (9), (10) and Lipschitz property of f given in Theorem 1, we get the estimate 2

∞ πX

ku(., t ) − u (., t )k2 = ≤π

∞ Z X p=1

≤ π (T − t )

2 p=1

2

|up (t ) − up (t )|2

T

ε

A(, s, t , p)|(fp (u (s))) − fp (u(s))|ds

2

 2 ∞   X t 2 Tp2 +π  1− p e |up (t )|

t

T

p=1

∞ Z X

T

A2 (, s, t , p)|(fp (uε (s))) − fp (u(s))|2 ds + π  2

 1−

t

p=1

≤ π (T − t )

2

∞ Z X

T



2t −2s T

(a())

2s−2t T



|(fp (u (s))) − fp (u(s))| ds + π  2

t

p=1

2

t

2 X ∞

T

∞ X

2

p4 e2Tp |up (t )|2

p=1 4 2Tp2

p e

|up (t )|2

p=1 2− 2t T

≤ π (T − t ) 2t /T (a())

T

Z

 −2s/T (a())

2s −2 T

t

∞ X

|(fp (u (s))) − fp (u(s))|2 ds

p=1

 2 X ∞ t 2 + π 2 1 − p4 e2Tp |up (t )|2 T

≤ 2(T − t )

2t /T

p=1 2t

(a())2− T 2t

2t

+  T (a())2− T π

∞ X

T

Z

2s

 −2s/T (a()) T −2 kf (., s, u(., s)) − f (., s, u (., s))k2 ds

t 2

p4 e2Tp |up (t )|2 .

t

(0 <  < 1 ⇒  ≤  T )

p=1

≤

2t T

2− 2t T

(a())

π

∞ X p=1

4 2Tp2

p e

|up (t )| + 2Tk 2

2

!

T

Z

 t

−2s T

(a())

2s −2 T



ku(., s) − u (., s)k

2

ds.

D. Duc Trong, N. Huy Tuan / Nonlinear Analysis 71 (2009) 4167–4176

4173

Hence



T

! 2tT −2

T

−2t



ku(., t ) − u (., t )k ≤ M + 2k T

1 + ln( T )

2

2

T

Z



−2s/T

t

! 2sT −2

T 1 + ln( T )

ku(., s) − u (., s)k2 ds

2

where M = π supt ∈[0,T ] p=1 p4 e2Tp u2p (t ). Using Gronwall’s inequality, we get:

P∞



−2t /T

! 2tT −2

T

ku(., t ) − u (., t )k2 ≤ Me2k

2 T (T −t )

1 + ln( T )

.

It follows that 

ku(., t ) − u (., t )k ≤ Me



This completes the proof of Theorem 3.

!2− 2tT

T

2k2 T (T −t ) 2t /T

2

.

1 + ln( T )

(19)

 t

Remark 2. (1) Notice that the convergence estimate in [10,9] given by the form C  T , does not give any useful information on the continuous dependence of the solution at t = 0. Actually, when t → 0+ , the accuracy of the regularized solution becomes progressively lower. At t = 0, it merely implies that the error is bounded by C , i.e, the convergence of the regularization solution at t = 0 is not obtained theoretically. Moreover, comparing (19) with the result obtained in [10,9], we know that estimate (19) is sharp and the best known estimate. (2) If t = 0, we have the following error estimate 

ku(., 0) − u (., 0)k ≤

√ Me

T

k2 T 2

! .

1 + ln( T )

(20)

We also know that the result (20) is not given in [9]. Hence, (20) proves that our method is the best regularization method for the nonlinear backward problem. In the case of nonexact data, one has Theorem 4. Assume that the exact solution u of (1)–(3) corresponding to ϕ satisfies the conditions as in Theorem 4. Let ϕ be a measured data such that

kϕ − ϕk ≤ . Then there exists a function u satisfying 

ku (., t ) − u(., t )k ≤ (1 +

√

M ) exp



3L2 T (T − t )



2



t T

!1− Tt

T

,

1 + ln( T )

for every t ∈ [0, T ]

where M is defined in Theorem 3. Proof. Let v  be the solution of problem (5) corresponding to ϕ and let w  be the solution of problem (5) corresponding to ϕ . Using Theorems 2 and 3, we get:

ku (., t ) − u(., t )k ≤ kw (., t ) − v  (., t )k + kv  (., t ) − u(., t )k !1− Tt ≤

t T

≤ (1 + for every t ∈ [0, T ].

T

−1

1 + ln( T )

√

exp(k (T − t ) )kϕ − ϕk + 2

2

T

t

M ) exp(k2 T (T − t )) T

1 + ln( T )

!1− Tt

√ Me

k2 T (T −t )



t T

T

!1− Tt

1 + ln( T )

,



4. Numerical experiment Example 1 (The Linear Case). Let us consider the linear backward heat problem

−uxx + ut = u(x, t ) + g (x, t ), (x, t ) ∈ (0, π ) × (0, 1) u(0, t ) = u(π , t ) = 0, t ∈ [0, 1], u(x, 1) = ϕ(x), x ∈ [0, π]

(21) (22) (23)

4174

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where g (x, t ) = et sin x, and u(x, 1) = ϕ0 (x) ≡ e sin x. The exact solution of the equation is u(x, t ) = et sin x. Especially



u x,

999



1000

≡ u(x) = exp



999 1000



sin x ≈ 2.715564905 sin x.

Let ϕ (x) ≡ ϕ(x) = ( + 1)e sin x. We have

kϕ − ϕk2 =

sZ

π



r

2 e2

sin xdx =  e 2

0

π 2

.

Applying the method introduced in this paper, we find the regularized solution u x, form

999 1000




≡ u (x) having the following

u (x) = vm (x) = w1,m sin x + w6,m sin 6x where

v1 (x) = ( + 1)e sin x w1,1 = ( + 1)e, w6,1 = 0, and

 1   a=   5000  tm = 1 − am m = 1, 2, . . . , 5  t −tm  2 −tm i2 m+t1m   ) wi,m+1 == ( i + e

wi,m −

2

Z

π

tm

e tm+1

(s−tm )i2

π

Z



!

(vm (x) + g (x, s)) sin ixdx ds ,

i = 1, 6.

0

Put a = ku − uk the error between the regularization solution u and the exact solution u. Letting  = 1 = 10−3 ,  = 2 = 10−7 ,  = 3 = 10−11 , we have the first table

 1 = 10−3 2 = 10−4 3 = 10−11

u

a

2.690070830 sin(x) − 0.05711984842 sin(6x)

0.004426319084

2.715564655 sin(x) − 0.005493673754 sin(6x)

0.00003043045132

2.718118869 sin(x) − 0.005602850301 sin(6x)

0.002585355932

Example 2 (The Nonlinear Case). Let us consider the nonlinear backward heat problem

−uxx + ut = f (u) + g (x, t ), (x, t ) ∈ (0, π ) × (0, 1) u(0, t ) = u(π , t ) = 0, t ∈ [0, 1], u(x, 1) = ϕ(x), x ∈ [0, π] where

 4 u    e10 e41    u+ − e−1 e−1 f (u) = 10 41 e e    u+   e−1  e − 1 0

g (x, t ) = 2e sin x − e sin x, t

4t

and u(x, 1) = ϕ0 (x) ≡ e sin x.

4

u ∈ [−e10 , e10 ] u ∈ (e10 , e11 ] u ∈ (−e11 , −e10 ]

|u| > e11

,

(24) (25) (26)

D. Duc Trong, N. Huy Tuan / Nonlinear Analysis 71 (2009) 4167–4176

4175

The exact solution of the equation is u(x, t ) = et sin x. Especially



u x,

999 100



≡ u(x) = exp



999



1000

sin x ≈ 2.715564905 sin x.

Let ϕ (x) ≡ ϕ(x) = ( + 1)e sin x. We have

kϕ − ϕk2 =

sZ

π



2 e2

sin xdx =  e 2

0

r

π 2

.

Applying the method introduced in this paper, we find the regularized solution u x, form

999 1000




≡ u (x) having the following

u (x) = vm (x) = w1,m sin x + w6,m sin 6x where

v1 (x) = ( + 1)e sin x w1,1 = ( + 1)e, w6,1 = 0, and

 1   a = 5000   tm = 1 − am m = 1, 2, . . . , 5  t −tm  2 −tm i2 m+t1m   ) wi,m+1 == ( i + e

wi,m −

2

π

Z

tm

e

(s−tm )i2

tm+1

π

Z

 !
v (x) + g (x, s) sin ixdx ds , 4 m

i = 1, 6.

0

Put a = ku − uk the error between the regularization solution u and the exact solution u. Letting  = 1 = 10−3 ,  = 2 = 10−4 ,  = 3 = 10−11 , we have the second table

 1 = 10−5 2 = 10−7 3 = 10−11

u

a

2.715833791 sin(x) − 0.005461493459 sin(6x)

0.0002987139108

2.715552177 sin(x) − 0.005518178192 sin(6x)

0.00004317829056

2.718264487 sin(x) − 0.005466473792 sin(6x)

0.002729464336

We have in view of the third error table in [9] (see p. 214)

 1 = 10−5 2 = 10−7 3 = 10−11

u

a

2.646937077 sin x − 0.002178680692 sin 3x

0.05558566020

2.649052245 sin x − 0.004495263004 sin 3x

0.05316693437

2.430605996 sin x − 0.0001718460902 sin 3x

0.3266494251

Looking at above two tables with comparison between other methods, we can see that the error results of the third table are smaller than those in the second table. This shows that our approach has a nice regularizing effect and gives a better approximation in comparison with the previous method, for example in [9]. Acknowledgments The authors would like to thank the referees for their valuable criticisms which led to the improved version of the paper. References [1] R. Lattès, J.-L. Lions, Méthode de Quasi-réversibilité et Applications, Dunod, Paris, 1967. [2] S.M. Alekseeva, N.I. Yurchuk, The quasi-reversibility method for the problem of the control of an initial condition for the heat equation with an integral boundary condition, Differ. Equ. 34 (4) (1998) 493–500. [3] R.E. Showalter, Quasi-reversibility of first and second order parabolic evolution equations, in: Improperly posed boundary value problems, Conf., Univ. New Mexico, Albuquerque, N. M., 1974, in: Res. Notes in Math., vol. 1, Pitman, London, 1975, pp. 76–84. [4] K. Miller, Stabilized quasi-reversibility and other nearly-best-possible methods for non-well posed problems, in: Symposium on Non-Well Posed Problems and Logarithmic Convexity, in: Lecture Notes in Mathematics, vol. 316, Springer-Verlag, Berlin, 1973, pp. 161–176.

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[5] G.W. Clark, S.F. Oppenheimer, Quasireversibility methods for non-well posed problems, Electron. J. Differential Equations 1994 (8) (1994) 1–9. [6] M. Denche, K. Bessila, A modified quasi-boundary value method for ill-posed problems, J. Math. Anal.Appl. 301 (2005) 419–426. [7] D.D. Trong, N.H. Tuan, Regularization and error estimates for nonhomogeneous backward heat problems, Electron. J. Differential Equations 2006 (4) (2006) 1–10. [8] D.D. Trong, N.H. Tuan, A nonhomogeneous backward heat problem: Regularization and error estimates, Electron. J. Differential Equations 2008 (33) (2008) 1–14. [9] D.D. Trong, P.H. Quan, T.V. Khanh, N.H. Tuan, A nonlinear case of the 1-D backward heat problem: Regularization and error estimate, Z. Anal. Anwendungen 26 (2) (2007) 231–245. [10] P.H. Quan, D.D. Trong, A nonlinearly backward heat problem: Uniqueness, regularization and error estimate, Appl. Anal. 85 (6–7) (2006) 641–657. [11] M. Denche, K. Bessila, Quasi-boundary value method for non-well posed problem for a parabolic equation with integral boundary condition, Math. Probl. Eng. 7 (2) (2001) 129–145. [12] H. Gajewski, K. Zaccharias, Zur regularisierung einer klass nichtkorrekter probleme bei evolutiongleichungen, J. Math. Anal. Appl. 38 (1972) 784–789. [13] N.T. Long, A.P.Ngoc. Ding, Approximation of a parabolic non-linear evolution equation backwards in time, Inverse Problems 10 (1994) 905–914. [14] R.E. Showalter, Cauchy problem for hyper-parabolic partial differential equations, in: Trends in the Theory and Practice of Non-Linear Analysis, Elsevier, 1983. [15] R.E. Ewing, The approximation of certain parabolic equations backward in time by Sobolev equations, SIAM J. Math. Anal. 6 (2) (1975) 283–294.