DSpace at VNU: Notes on a new approximate solution of 2-D heat equation backward in time

Applied Mathematical Modelling 35 (2011) 5673–5690

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Applied Mathematical Modelling
journal homepage: www.elsevier.com/locate/apm

Notes on a new approximate solution of 2-D heat equation backward
in time
Nguyen Huy Tuan a,d,⇑, Dang Duc Trong b, Pham Hoang Quan c
a

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

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 dependence 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 proposed 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 nonhomogeneous 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].
Ó 2011 Elsevier Inc. All rights reserved.

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

8
ðx; y; tÞ 2 X Â ð0; TÞ;
>
< ut À uxx À uyy ¼ f ðx; y; tÞ;
uð0; y; tÞ ¼ uðp; y; tÞ ¼ uðx; 0; tÞ ¼ uðx; p; tÞ ¼ 0; ðx; y; tÞ 2 X Â ð0; TÞ;
>
:
uðx; y; TÞ ¼ gðx; yÞ;
ðx; yÞ 2 X;

ð1Þ

where X = (0, p) Â (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 quasireversibility 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: (N.H. Tuan).
0307-904X/$ - see front matter Ó 2011 Elsevier Inc. All rights reserved.
doi:10.1016/j.apm.2011.05.010


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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

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. [1–3], 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]. Physically, this problem arises from the requirement of recovering the heat temperature at some earlier time using the knowledge 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 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 application than the homogeneous 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 available; and estimates of Hölder type are very rate (see Remark 3 for 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 suitable conditions on the exact solution u, we shall introduce the error estimate of order p, (p > 0). This is a significant improvement 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. Section 2 is devoted to a short presentation of a ill-posed problem (1).
In Section 3, 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 Section 4 to illuminate the effect of our method.
2. The ill-posed backward heat problem
Throughout this paper, we denote hÁ,Ái, kÁk 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 ½0; TŠ; H10 ðXÞ \ C 1 ðð0; TÞ; H2 ðXÞÞ to be a weak solution for
Problem (1) if

d
huðÁ; Á; tÞ; WiL2 ðXÞ À huðÁ; Á; tÞ; DWiL2 ðXÞ ¼ hf ðÁ; Á; tÞ; WiL2 ðXÞ ;
dt

ð2Þ

for all functions W 2 H2 ðXÞ \ H10 ðXÞ. In fact, it is enough to choose W in the orthogonal basis {sin(nx)sin(my)}n,mP1 and the
formula (2) reduces to

unm ðtÞ ¼ eðTÀtÞðn

2 þm2 Þ

g nm À

Z

T

eðsÀtÞðn

2 þm2 Þ

fnm ðsÞds;

8m; n P 1;

ð3Þ


fnm ðsÞds sin nx sin my;

ð4Þ

t

which may also be written formally as

Z
1 
X
2
2
eÀðtÀTÞðn þm Þ g nm À

uðx; y; tÞ ¼

fnm ðsÞ ¼

4

2 þm2 Þ

eÀðtÀsÞðn

t

n;m¼1

where

T

Z pZ p

f ðx; y; sÞ sin nx sin mydx dy;
Z p0 Z p0
4
g nm ¼ 2
gðx; yÞ sin nx sin mydx dy;
p 0 0
Z pZ p
4
unm ðtÞ ¼ 2
uðx; y; tÞ sin nx sin mydx dy:

p

2

p

0

0

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 L2 only. 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

Z
1 
X
2
2
eTðn þm Þ g nm À
n;m¼1

T

esðn

2 þm2 Þ

2
fnm ðsÞds < 1:

ð5Þ

0

Proof. Suppose the Problem (1) has a solution u 2 Cð½0; TŠ; H10 ðXÞÞ \ C 1 ðð0; TÞ; L2 ðXÞÞ, then u can be formulated in the
frequency domain


N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Z
1 
X
2
2
eÀðtÀTÞðn þm Þ g nm À

uðx; y; tÞ ¼

T

eÀðtÀsÞðn

2 þm2 Þ


fnm ðsÞds sin nx sin my:

5675

ð6Þ

t

n;m¼1

This implies that
2 þm2 Þ

unm ð0Þ ¼ eTðn

g nm À

Z

T

esðn

2 þm2 Þ

fnm ðsÞds:

ð7Þ

0

Then
1 
X

kuðÁ; Á; 0Þk2 ¼

eTðn

2 þm2 Þ

g nm À

n;m¼1

Z

T

2 þm2 Þ

esðn

2
fnm ðsÞds < 1:

0

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

Z
1 
X
2
2
eTðn þm Þ g nm À

v ðx; yÞ ¼

T

esðn

2 þm2 Þ


fnm ðsÞds sin nx sin my 2 L2 ðXÞ:

0

n;m¼1

Consider the problem

8
>
< ut À uxx À uyy ¼ f ðx; y; tÞ;
uð0; y; tÞ ¼ uðp; y; tÞ ¼ uðx; 0; tÞ ¼ uðx; p; tÞ ¼ 0; t 2 ð0; TÞ;
>
:
uðx; y; 0Þ ¼ v ðx; yÞ; ðx; yÞ 2 ð0; pÞ Â ð0; pÞ:

ð8Þ

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


Z t
1 
X
2
2
2
2
eÀtðn þm Þ < v ðx; yÞ; sin nx sin my > þ
eðsÀtÞðn þm Þ fnm ðsÞds sin nx sin my:

uðx; y; tÞ ¼

ð9Þ

0

n;m¼1

Let t = T in (9), we have



Z
2
2
2
2
uðx; y; TÞ ¼ eÀTðn þm Þ eTðn þm Þ g nm À

T

2 þm2 Þ

esðn

 Z
fnm ðsÞds þ

0

1
X

¼

T

eðsÀTÞðn

2 þm2 Þ


fnm ðsÞds sin nx sin my

0

g nm sin nx sin my ¼ gðx; yÞ:

n;m¼1

Hence, u is the unique solution of (1). h
Theorem 2. Problem (1) has at most one solution in Cð½0; TŠ; H10 ðXÞÞ \ C 1 ðð0; TÞ; L2 ðXÞÞ.
Proof. Let u(x, y, t), v(x, y, t) be two solutions of Problem (1) such that u; v 2 Cð½0; TŠ; H10 ðXÞÞ \ C 1 ðð0; TÞ; L2 ðXÞÞ. Put
w(x,y,t) = u(x,y,t) À v(x,y,t). Then w satisfies the equation

ðx; y; tÞ 2 X Â ð0; TÞ; X ¼ ð0; pÞ Â ð0; pÞ;

wt À wxx À wyy ¼ 0;

wð0; y; tÞ ¼ wðp; y; tÞ ¼ wðx; 0; tÞ ¼ wðx; p; tÞ ¼ 0;
wðx; y; TÞ ¼ 0;
Now, setting GðtÞ ¼

G0 ðtÞ ¼ 2

Rp Rp
0

0

Z pZ p
0

ðx; y; tÞ 2 X  ½0; TŠ:

x; y 2 X:
w2 ðx; y; tÞdx dyð0 6 t 6 TÞ. By direct computation we get

wðx; y; tÞwt ðx; y; tÞdx dy ¼ À2

0

Z pZ p
0

wðx; y; tÞðwxx ðx; y; tÞ þ wyy ðx; y; tÞÞdx dy:

0

Using the Green formula, we obtain

G0 ðtÞ ¼ À2

Z pZ p
0

0

ðw2x ðx; y; tÞ þ w2y ðx; y; tÞÞdx dy:

ð10Þ

Taking the derivative of G0 (t), one has

G00 ðtÞ ¼ À4

Z pZ p
À
0

Á
wx ðx; y; tÞwxt ðx; y; tÞ þ wy ðx; y; tÞwyt ðx; y; tÞ dx dy:

0

Using the technique of integration by parts, we get

G00 ðtÞ ¼ 4

Z pZ p
À
0

0

Á
wxx ðx; y; tÞwt ðx; y; tÞ þ wyy ðx; y; tÞÞwt ðx; y; tÞ dx dy ¼ 4

Z pZ p
0

0

ðwxx ðx; y; tÞ þ wyy ðx; y; tÞÞ2 dx dy:

ð11Þ


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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

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

Z p Z p
0

0

Z pZ p

w2x ðx; y; tÞ þ w2y ðx; y; tÞ dx dy ¼ À
wðx; y; tÞðwxx ðx; y; tÞ þ wyy ðx; y; tÞÞdx dy
0

0

Z p Z p
6
0

0

2

w ðx; y; tÞdx dy

12 Z p Z p
0

12
ðwxx ðx; y; tÞ þ wyy ðx; y; tÞÞ dxdy :
2

ð12Þ

0

Thus (10)–(12) imply

ðG0 ðtÞÞ2 6 GðtÞG00 ðtÞ:
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 approximation 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 g be a measured data satisfying kg À gk 6 . Starting
from the ideas mentioned in the paper of Clark and Oppeinheimer [19], we consider the following approximate problem
1
X

ut À uxx À uyy ¼

n;m¼1

fnm ðtÞ
sin nx sin my; ðx; y; tÞ 2 X Â ð0; TÞ;
1 þ beaTðn2 þm2 Þ

u ð0; y; tÞ ¼ u ðp; y; tÞ ¼ u ðx; 0; tÞ ¼ u ðx; p; tÞ ¼ 0 ðx; y; tÞ 2 X  ½0; TŠ;
1
X
g nm
sin nx sin my; ðx; yÞ 2 X;
u ðx; y; TÞ ¼
aTðn2 þm2 Þ
1
þ
be
n;m¼1

ð13Þ
ð14Þ
ð15Þ

where fnm(t), gnm are defined by

fnm ðtÞ ¼
g nm

Z pZ p

4

f ðx; y; tÞ sin nx sin mydx dy;
Z p0 Z p0
4
¼ 2
gðx; yÞ sin nx sin mydx dy:

p

p

2

0

0

and b is a regularization parameter depending on  and 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 in Remark 3.
Theorem 3. Let f 2 L2((0,T);L2 (X)) and g 2 L2 (X). Then Problem (13)–(15) has uniquely a weak solution u 2 Cð½0; TŠ; L2 ðXÞ\
L2 ð0; T; H10 ðXÞÞ \ C 1 ð0; T; H10 ðXÞÞ defined as follows:

u ðx; y; tÞ ¼

1
X
n;m¼1



Z T
2
2
eðTÀtÞðn þm Þ
ðsÀTÞðn2 þm2 Þ
g

À
e
f
ðsÞds
sin nx sin my:
nm
nm
2
2
1 þ beaTðn þm Þ
t

ð16Þ

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)–(15). In Step 2, the stability of the solution is given.
Step 1. The existence and the uniqueness of a solution of Problem (13)–(15).
We divide this step into two parts.
Part A If u 2 Cð½0; TŠ; L2 ðXÞ \ L2 ð0; T; H10 ðXÞÞ \ C 1 ð0; T; H10 ðXÞÞ satisfies (16) then u is solution of (13)–(15).
We have

u ðx; y; tÞ ¼

1
X
n;m¼1




Z T
2
2
eðTÀtÞðn þm Þ
ðsÀTÞðn2 þm2 Þ
g
À
e
f
ðsÞds
sin nx sin my:
nm
1 þ beaTðn2 þm2 Þ nm
t

We can verify directly that u 2 Cð½0; TŠ; L2 ðXÞ \ C 1 ðð0; TÞ; H10 ðXÞÞ \ L2 ð0; T; H10 ðXÞÞÞ. In fact, u 2 C 1 ðð0; TŠ; H10 ðXÞÞÞ. Moreover,
one has


N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690
1
X

ut ðx; y; tÞ ¼

2

n;m¼1

¼À

1
X

4

p

2

Àðn2 þ m2 ÞeðTÀtÞðn þm Þ
g nm À
1 þ beaTðn2 þm2 Þ

2

À

Z
t

T


2
2
ðn2 þ m2 ÞeðsÀtÀTÞðn þm Þ
fnm ðtÞ
sin nx sin my
fnm ðsÞds þ

2 þm2 Þ
2 þm2 Þ
aTðn
aTðn
1 þ be
1 þ be

1
X
Á
ðn2 þ m2 Þ < u ðx; y; tÞ; sin nx sin my > sin nx sin my þ

n;m¼1

n;m¼1

¼ uxx ðx; y; tÞ þ uyy ðx; y; tÞ þ

1
X
n;m¼1

5677

fnm ðtÞ
sin nx sin my
1 þ beaTðn2 þm2 Þ

fnm ðtÞ
sin nx sin my;

1 þ beaTðn2 þm2 Þ

and
1
X

u ðx; y; TÞ ¼

n;m¼1

g nm
sin nx sin my:
1 þ beaTðn2 þm2 Þ

So u is the solution of (13)–(15).
Part B. Problem (13)–(15) has at most one solution Cð½0; TŠ; H10 ðXÞÞ \ C 1 ðð0; TÞ; L2 ðXÞÞ.
Let u(x, y, t), v(x, y, t) be two solutions of Problem (4)–(6) such that u; v 2 Cð½0; TŠ; H10 ðXÞÞ \ C 1 ðð0; TÞ; L2 ðXÞÞ. Put w(x, y, t) = u(x, y, t) À v(x, y, t). Using Theorem 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)–(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
2

2

c
eðn þm Þc
6 bÀd :
1 þ beðn2 þm2 Þd

ð17Þ

Proof of Lemma 1. We have
2

2

2

2

2

2

eðn þm Þc
eðn þm Þc
eðn þm Þc
Àdc
¼À
:
c 6 b
Ác À
Á c6
2 þm2 Þd
2 þm2 Þd d
2 þm2 Þd 1Àd
ðn
d
ðn

ðn
1 þ beðn2 þm2 Þd
ð1 þ be
Þ
1 þ be
1 þ be
For 0 6 t 6 s 6 T, if letc = T À t,d = aT in (17), then
2

2

tÀT
eðTÀtÞðn þm Þ
6 b aT :
1 þ beaTðn2 þm2 Þ

ð18Þ

If we let c = s À t,d = aT in (17) then
2

2

tÀs
eðsÀtÞðn þm Þ
6 b aT :
1 þ beaTðn2 þm2 Þ

ð19Þ

Let u and v be two solutions of (13)–(15) corresponding to the final values g and h. From the definitions of u and v we have

uðx; y; tÞ ¼

1
X
n;m¼1



Z T
2
2
eðTÀtÞðn þm Þ
2
2
g nm À
eðsÀTÞðn þm Þ fnm ðsÞds sin nx sin my
2 þm2 Þ
aTðn
1 þ be
t

and

v ðx; y; tÞ ¼

1
X
n;m¼1



Z T
2
2
eðTÀtÞðn þm Þ
ðsÀTÞðn2 þm2 Þ
h
À
e
f
ðsÞds
sin nx sin my;
nm
nm
1 þ beaTðn2 þm2 Þ
t

where

g nm ¼
hnm ¼

4

p2
4

p2

Z pZ p
Z0 p Z0 p
0

gðx; yÞ sin nx sin my;
hðx; yÞ sin nx sin mydx dy:

0

Hence

uðx; y; tÞ À v ðx; y; tÞ ¼

1
X
n;m¼1

2

2

eðTÀtÞðn þm Þ
ðg À hnm Þ sin nx sin my:
1 þ beaTðn2 þm2 Þ nm

ð20Þ


5678

N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Using (18) and (20) and the inequality (a + b)2 6 2(a2 + b2), we obtain





1 þ be



2
2
2 p2 2tÀ2T X
1 
1
2tÀ2T
p2 X

 eðTÀtÞðn þm Þ
aT
kuðÁ; Á; tÞ À v ðÁ; Á; tÞk ¼
ðg
À
h
Þ
6
ðjg nm À hnm jÞ2 6 b aT kg À hk2 :

b


nm
nm
2
2
aTðn þm Þ

2

4

n;m¼1

4

n;m¼1

Therefore
tÀT

kuðÁ; Á; tÞ À v ðÁ; Á; tÞk 6 b aT kg À hk:

ð21Þ

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 difT
ferent order of magnitude. For instance, in [20], the order of magnitude is e . In [19], the authors give some better stability

t
estimates than the latter discussed methods. They show that the stability estimate is of order M T À1 . In [21], the authors
T
improve the previous results by a better estimation of the stability order, that is A ¼ C 1  1þln
.
From
(21), if we set b = 
ð
ðTÞ
1
then the stability order is B ¼ C 2 Àa . For a > 1, we have

lim
!0


 
B
C2
T
1
¼ 0:
¼
lim 1Àa 1 þ ln
A C 1 T !0


It is easy to see that the order A of the error is less than the above order B of stability estimate. This proves the advantages of
our method.
Theorem 4. Assume that there exists a positive number P1 such that

1
p2 X

4

2 þm2 Þ

e2tðn

junm ðtÞj2 < P21 ;

0 6 t 6 T:

ð22Þ

n;m¼1

Let g 2 L2(X) be measured data satisfying

kg  À gk 6 :
Let



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

kuðÁ; Á; tÞ À v  ðÁ; Á; tÞk 6 ðP1 þ 1ÞT :

ð23Þ

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

uðx; y; tÞ ¼

1
X

2 þn2 Þ

eÀðtÀTÞðm



g nm À

Z

T

2 þm2 Þ

eÀðTÀsÞðn


fnm ðsÞds sin nx sin my:

ð24Þ

t

m;n¼1

Hence

unm ðtÞ ¼ eÀðtÀTÞðm

2 þn2 Þ


Z
g nm À

T

eÀðTÀsÞðn

2 þm2 Þ


fnm ðsÞds ;

ð25Þ

t

where unm ðtÞ ¼ p42 huðx; y; tÞ; sin nx sin myi. It follows from (16) that

unm ðtÞ ¼



Z T
2
2
eðTÀtÞðn þm Þ
2
2
g nm À
eðsÀTÞðn þm Þ fnm ðsÞds :
2 þm2 Þ
aTðn
1 þ be
t

ð26Þ

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

!

Z T
2
2
eðTÀtÞðn þm Þ
ðsÀTÞðn2 þm2 Þ
g
e
À
À

e
f
ðuÞðsÞds
nm
nm
2
2
nm ðtÞ À unm ðtÞ ¼
1 þ beaTðn þm Þ
t


Z T
2
2
beðTÀtÞðn þm Þ aTðn2 þm2 Þ
b
ðsÀTÞðn2 þm2 Þ
e
g
À
e
f
ðsÞds
¼
u ðtÞ
¼
nm
nm
2

2
ÀaTðn2 þm2 Þ nm
b
þ
e
1 þ beaTðn þm Þ
t
ðTÀtÞðn2 þm2 Þ

u

2

¼

2

beÀtðn þm Þ
2
2
etðn þm Þ unm ðtÞ:
b þ eÀaTðn2 þm2 Þ

ð27Þ

Using (18), we obtain
t

2 þm2 Þ

junm ðtÞ À unm ðtÞj 6 baT etðn

unm ðtÞ:

ð28Þ


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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

This implies that

kuðÁ; Á; tÞ À u ðÁ; Á; tÞk2 ¼

1
p2 X

4

junm ðtÞ À unm ðtÞj2 6

n;m¼1

p2
4

2t

baT

1
X

2 þm2 Þ

e2tðn

2t

junm ðtÞj2 6 P 21 baT :

ð29Þ

n;m¼1

Since the condition kg À gk 6  and (21) give
tÀT

tÀT

ku ðÁ; Á; tÞ À v  ðÁ; Á; tÞk 6 b aT kg Àg k 6 b aT :

ð30Þ

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

tÀT

kuðÁ; Á; tÞ À v  ðÁ; Á; tÞk 6 kuðÁ; Á; tÞ À u ðÁ; Á; tÞk þ ku ð:; tÞ À v  ðÁ; tÞk 6 P1 baT þ b aT :
Notice that, from b = a, then
t

t

t

kuðÁ; Á; tÞ À v  ðÁ; Á; tÞk 6 P 1 T þ T ¼ ðP1 þ 1ÞT :
This completes the proof of Theorem 4. h
Remark 2. (1) If f = 0, then from (32), we get
2 þm2 Þ

unm ðtÞ ¼ eÀðtÀTÞðn

g nm :

Therefore
2 þm2 Þ

unm ð0Þ ¼ eTðn

2 þm2 Þ

g nm ¼ etðn

unm ðtÞ:

This implies that
1

p2 X

4

e2tðn

2 þm2 Þ

junm ðtÞj2 ¼ kuðÁ; Á; 0Þk2 :

n;m¼1

The condition (22) is acceptable. Moreover, the assumptions (22) can be replaced by assumptions on f and g. Thus, since
2 þn2 Þ

unm ðtÞ ¼ eÀðtÀTÞðm


g nm À

Z

T

2 þm2 Þ

eÀðTÀsÞðn


fnm ðsÞds

ð31Þ

t

we get
2 þn2 Þ

etðm

2 þn2 Þ

unm ðtÞ ¼ eTðm

g nm À

Z

T

esðn

2 þm2 Þ

fnm ðsÞds:

ð32Þ

t

R T P1

P1

2Tðm2 þn2 Þ 2
g nm
n;m¼1 e

2

2

2
< 1 and 0 n;m¼1 e2sðn þm Þ fnm
ðsÞds < 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.

If

Theorem 5. Assume that there exist positive numbers k,P2such that
1
p2 X

4

2 þm2 Þ

e2kðn

junm ðtÞj2 < P22 :

ð33Þ

n;m¼1

aT
Let v(Á,Á,t) be defined in Theorem 4. Let b ¼ Tþk , then one has


 k
t
kuðÁ; Á; tÞ À v  ðÁ; Á; tÞk 6 P2 þ Tþk Tþk :

ð34Þ

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

!

Z T
2
2
eðTÀtÞðn þm Þ
ðsÀTÞðn2 þm2 Þ
g
e
À
À

e
f
ðsÞds
nm
nm
2
2
nm ðtÞ À unm ðtÞ ¼
1 þ beaTðn þm Þ
t


Z T
2
2
2
2
ðTÀtÞðn þm Þ
be
beðaTÀkÞðn þm Þ kðn2 þm2 Þ
aTðn2 þm2 Þ
ðsÀTÞðn2 þm2 Þ
e
g
À
e
f
ðsÞds
¼
e

unm ðtÞ:
¼
nm
nm
1 þ beaTðn2 þm2 Þ
1 þ beÀaTðn2 þm2 Þ
t

u

ðTÀtÞðn2 þm2 Þ

ð35Þ

This implies that
k

2 þm2 Þ

junm ðtÞ À unm ðtÞj 6 baT ekðn

unm ðtÞ:

ð36Þ


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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Using (36), we get the estimate

kuðÁ; Á; tÞ À u ðÁ; Á; tÞk2 ¼

p2 X
4

n; m ¼ 11 junm ðtÞ À unm ðtÞj2 6

p2
4

2k

baT

1
X

2 þm2 Þ

e2kðn

2k

junm ðtÞj2 6 P22 baT :

ð37Þ

n;m¼1

Since the condition kg À gk 6  and (21) give
tÀT

tÀT

ku ðÁ; Á; tÞ À v  ðÁ; Á; tÞk 6 b aT kg  À gk 6 b aT :

ð38Þ

From (37) and (38), we have
k

tÀT

kuðÁ; Á; tÞ À v  ðÁ; Á; tÞk 6 kuðÁ; Á; tÞ À u ðÁ; Á; tÞk þ ku ðÁ; Á; tÞ À v  ðÁ; Á; tÞk 6 P 2 baT þ b aT :
aT

Notice that from b ¼ Tþk , then
tþk

k

kuðÁ; Á; tÞ À v  ðÁ; Á; tÞk 6 P2 Tþk þ Tþk :
This completes the proof of Theorem 5. h
Remark 3. (1) We emphasize once more that the error (34) À(k > 0)
ÁÀqis of Hölder type for all t 2 [0;T]. It is easy to see that the
convergence rate of p, (0 < p) is more rapid than the that of lnð1Þ 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
k

kuðÁ; Á; 0Þ À v  ðÁ; Á; 0Þk 6 ðP2 þ 1ÞTþk :

ð39Þ

k
Tþk

The rate of convergence at t = 0 is  . Since 0 < k 6 (a À 1)T, the fastest convergence is
large constant a. This proves that (39) is sharp estimate.



aÀ1
a

. Hence it can approach

 for the

4. A numerical experiment
In this section, we establish some numerical tests for a example. To compare some different methods, we give the numerical tests with same parameter regularization . We consider the problem

ut À uxx À uyy ¼ f ðx; y; tÞ ¼ 3et sin x sin y

ð40Þ

uðx; y; 1Þ ¼ gðx; yÞ ¼ e sin x sin y:

ð41Þ

and

The exact solution of the latter equation is

uðx; y; tÞ ¼ et sin x sin y:
Note that uðx; y; 1=2Þ ¼

pffiffiffi
e sinðxÞ sinðyÞ % 1:648721271 sinðxÞ sinðyÞ. Let gpbe the measured final data

g p ðx; yÞ ¼ e sinðxÞ sinðyÞ þ

1
sinðpxÞ sinðpyÞ:
p

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

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Z pZ p
1
p
2
2
FðpÞ ¼ kg p À gk ¼
sin ðpxÞ sin ðpyÞdx dy ¼
:

2
2p
0
0 p
The solution of (40) and (41) corresponding the final value gpis

1 2
up ðx; y; tÞ ¼ et sinðxÞ sin y þ ep ð1ÀtÞ sinðpxÞ sinðpyÞ:
p
The error at t = 0 is

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Z p Z p 2p2
2
e
ep p
2
2
OðpÞ :¼ ku ðÁ; Á; 0Þ À uðÁ; Á; 0Þk ¼
sin
ðpxÞ
sin
ðpyÞ
dx
dy
¼
:
p2
p 2
0

0
p

Then, we notice that

lim FðpÞ ¼ lim kg p À gk ¼ lim

p!1

p!1

n!1

1p
¼ 0;
p2
2

lim OðpÞ ¼ lim kup ðÁ; Á; 0Þ À uðÁ; Á; 0Þk ¼ lim

p!1

p!1

p!1

ep p
¼ 1:
p 2

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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

From the two equalities above, we see that (40) and (41) is an ill-posed problem. Let a ¼ 101
. Approximating the problem as in
100
(13)–(15), the regularized solution is

v  ðx; y; tÞ ¼

1
X

eÀtðn
ðn2 þm2 Þ
100

e

n;m¼1

Hence, we have

v  ðx; y; tÞ ¼

e

þ eÀðn2 þm2 Þ Þ

g nm À

þ eÀ2

1

2

¼

e

1
50

þ eÀ2

À3

Z
1
2

1

e

ðn2 þm2 Þ
100

e

þ eÀðn2 þm2 Þ

fnm ðsÞds sin nx sin my 0 6 t 6 T:

ð42Þ


2
1
eÀ2tp
ds sin x sin y þ
sinðpxÞ sinðpyÞ 0 6 t 6 1:
2
p
p e50 þ eÀ2p2
e þ eÀ2
!

e3sÀ3
1
50

eðsÀtÀ1Þðn

1
50

t

1

!

2 þm2 Þ

1

e3sÀ2tÀ2

It follows that

1
v  ðx; y; Þ

Z
t

Z
sin x sin y À 3

e1À2t
1
50

2 þm2 Þ

þ eÀ2

ds sin x sin y þ

ð43Þ

2

1
eÀp
sin px sin py:
p2
p e50 þ eÀ2p2

ð44Þ

Let a = kv À uk be the error between the regularized solution v and the exact solution u.
Let

 ¼ 1 ¼ 10À3
 ¼ 3

rffiffiffiffi

p

 ¼ 2 ¼ 10À4

;

2

pffiffiffiffiffiffiffi
À5
¼ 10
2p;

rffiffiffiffi

 ¼ 4 ¼ 10À8

p

;
2
rffiffiffiffi

p
2

;

we have Table 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 the Table 2.
By applying that method given in [24], we have the following regularized solution

v  ðx; y; tÞ

¼

1
X
n;m¼1

2

2

eÀtðn þm Þ
g À
 þ eÀðn2 þm2 Þ Þ nm

Z

1

t

!
2
2
eÀtðn þm Þ
f ðsÞds sinðnxÞ sinðmyÞ:
s þ eÀsðn2 þm2 Þ nm

Hence, we have

v  ðx; y; tÞ ¼

e1À2t

sin x sin y À 3
 þ eÀ2

Z
t

1


esÀ2t
1
ds
sin x sin y þ
p
s þ eÀ2s

2

eÀ2tp
sinðpxÞ sinðpyÞ 0 6 t 6 1
 þ eÀ2p2

ð45Þ

Table 1
The error of method in this paper.


1
2

3
3

pffiffiffi
¼ 10À3 p2
pffiffiffi
¼ 10À4 p2
ffiffiffi
À5 pp
¼ 10
2
ffi
À10 pffiffi
p
¼ 10
2

v

a

1.64870174813729sinxsiny + 2.706068625 Â 10À442978sin(103x)sin(103y)

0.00003066665669

1.64872107547220sinxsiny + 4.464249273 Â 10À44298034sin(104x)sin(104y)

3.078760801 Â 10À7

1.64872126874784sinxsiny

3.141592652 Â 10À9

1.64872127070013sinxsiny

1.5707963267948 Â 10À19

Table 2
The error of method in [24].


1
2
3
3

pffiffiffi
¼ 10À3 p2
ffi
À4 pffiffi
p
¼ 10
2
ffiffiffi
À5 pp
¼ 10
2
pffiffiffi
¼ 10À10 p2

v

a

1.618739596sin(x)sin(y) + 1.315245468 Â 10À434301sin(103x)sin(103y)

0.04709510496

1.645673218sinxsiny + 0.2573859778 Â 10À43429456sin(104x)sin(104y)

0.004787870457

1.648110757sinxsiny + 2.785098042 Â 10À1085736215sin(5.104x)sin(5.104y)

0.0009589931490

1.648720965sinxsiny

4.806636760 Â 10À7

Table 3
The error of method in [23].


1
2
3
4

pffiffiffi

¼ 10À3 p2
ffiffiffi
À4 pp
¼ 10
2
pffiffiffiffiffiffi
ffi
À5
¼ 10
2p
ffip
À8 pffiffi
¼ 10
2

v

a

1.735336020sin(x) sin(y) + 2.630490937 Â 10À434295sin(1000x)sin(1000y)

0.1360541296

1.674100247sin(x)sin(y) + 5.147719556 Â 10À43429449sin(104x)sin(104y)

0.03986520228

1.658588375sinxsiny + 1.392549021 Â 10À1085736205sin(5.104x)sin(5.104y)

0.01549921072

1.648834493sin(x)sin(y)

0.0001778487018


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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Fig. 1. The exact solution.

Fig. 2. The regularized solution with

1 in Table 1.

It follows that

v  ðx; y;

1
1
Þ¼¼
sin x sin y À 3
2
 þ eÀ2

Then, we have the Table 3.

Z

1
2

1

!
esÀ1
1
ds
sin x sin y þ
p
s þ eÀ2s

2

eÀp
sinðpxÞ sinðpyÞ 0 6 t 6 1
 þ eÀ2p2

ð46Þ


N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Fig. 3. The regularized solution with

2 in Table 1.

Fig. 4. The regularized solution with

3 in Table 1.

5683

We have the graphics of the exact solution and the regularized solution in Tables 1–3. These are displayed on a rectangular domain (0,p) Â (0,p) (See Figs. 1–15).
Now we consider the random measured data case.
Let grandom(x) = esin xsin y + e⁄randN⁄sin(randN⁄x)⁄sin(randN ⁄y) be the random measured data, where randN = ceil(NÁ⁄rand). The function ceil(NÁ⁄rand) returns the random value between 0 and N.


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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Fig. 5. The regularized solution with

4 in Table 1.

Fig. 6. The regularized solution with

1 in Table 2.

With the random measured data grandom, we get the regularized solution

v eran ðx; y; tÞ ¼

e1À2t
1
50

ee þ eÀ2

sin x sin y À 3

 sinðrandNyÞ:

Z
t

1

e3sÀ2tÀ2
1
50

ee þ eÀ2


ds sin x sin y þ randN

2

eeÀ2tðrandNÞ
ðrandN Þ2
50

ee

þ eÀ2ðrandNÞ

2

sinðrandNxÞ


N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Fig. 7. The regularized solution with

2 in Table 2.

Fig. 8. The regularized solution with

3 in Table 2.

It follows that
e

v ran



!

Z 1
2
1
1
e3sÀ2
eeÀðrandNÞ
¼ 1

x; y;
sin x sin y À 3
ds sin x sin y þ randN ðrandNÞ2
sinðrandNxÞ
1
1
2
ee50 þ eÀ2
ee50 þ eÀ2
2
ee 50 þ eÀ2ðrandNÞ2
 sinðrandNyÞ:

We have the following error table

5685


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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Fig. 9. The regularized solution with

4 in Table 2.

Fig. 10. The regularized solution with

1 in Table 3.

Looking at three above tables with comparison between three methods, we can see the error results of the Table 1 are
smaller than many times in the Tables 2 and 3. This shows that our approach has a nice regularizing effect and give a better
approximation with comparison to the previous method established in [23,24]. (See Table 4).


N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Fig. 11. The regularized solution with

2 in Table 3.

Fig. 12. The regularized solution with

3 in Table 3.

5687

5. Conclusion
We have considered a regularization problem for a nonhomogeneous backward heat equations in bounded domain,
namely Problem (1). In many earlier works on the backward problem, while one may obtain an Hölder type error estimate


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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Fig. 13. The regularized solution with

Fig. 14. The random regularized solution

v eran

4 in Table 3.

at t ¼ 12 and e = 10À5.

at any fixed t > 0, an explicit error estimate at t = 0 is still difficult and was given in logarithm type only. The present paper
proposes a regularized solution with error estimate of Hölder type for all t 2 [0, T]. Moreover, our regularization is enough for
a numerical setting and the numerical results seems satisfactory. In the future, we will consider the regularized problem for a
nonlinear backward heat equation.


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N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

Fig. 15. The random regularized solution

v eran

at t ¼ 12 and e = 10À5.

Table 4
The error of method in this paper with random measured data.


¼
¼
¼
¼

¼

pffiffiffi
10À3 p2
ffi
À4 pffiffi
p
10
2
ffiffiffi
À5 pp
10
2
pffiffiffi
10À8 p2
pffiffiffi
10À10 p2

randN (ramdom measured data)

a

1270

5.4673 Â 10À5

9134

5.6474 Â 10À7

9134

5.8564 Â 10À9

976

4.8573 Â 10À15

8148

5.6826 Â 10À19

Acknowledgments
The authors thank the editor and the reviewers for their valuable comments leading to the improvement of our manuscript. The authors are also extremely grateful for the support given by Vietnamese National Foundation for Science and
Technology Development (NAFOSTED). The authors thank Tra Quoc Khanh for his most helpful comments on numerical
results.

References
[1] C.-L. Fu, Z. Qian, R. Shi, A modified method for a backward heat conduction problem, Appl. Math. Comput. 185 (2007) 564–573.
[2] C.-L. Fu, X.-T. Xiong, Z. Qian, Fourier regularization for a backward heat equation, J. Math. Anal. Appl. 331 (1) (2007) 472–480.
[3] X.-T. Xiong, C.L. Fu, Z. Qian, X. Gao, Error estimates of a difference approximation method for a backward heat conduction problem, Int. J. Math. Math.
Sci. (2006) 1–9.
[4] B.M. Campbell H, R. Hughes, E. McNabb, Regularization of the backward heat equation via heatlets, Electron. J. Diff. Eqns. 130 (2008) 1–8.
[5] K.A. Ames, L.E. Payne, Asymptoic for two regularizations of the Cauchy problem for the backward heat equation, Math. Models Methods Appl. Sci.
(1998) 187–202.
[6] R. Lattes, J.-L. Lions, Méthode de Quasi-réversibilité et Applications, Dunod, Paris, 1967.
[7] R.E. Showalter, The final value problem for evolution equations, J. Math. Anal. Appl. 47 (1974) 563–572.
[8] U. Tautenhahn, T. Schroter, On optimal regularization methods for the backward heat equation, Z. Anal. Anwend. 15 (1996) 475–493.
[9] T.I. Seidman, Optimal filtering for the backward heat equation, SIAM J. Numer. Anal. 33 (1996) 162–170.
[10] D.N. Hao, N.V. Duc, Stability results for the heat equation backward in time, J. Math. Anal. Appl. 353 (2) (2009) 627–641.

[11] C.W. Chang, C.S. Liu, J.R. Chang, A quasi-boundary semi-analytical approach for two-dimensional backward heat conduction problems, CMC: Comput.
Mater. Continua 15 (2010) 45–66.
[12] D.D. Trong, T.N. Lien, Regularization of a discrete backward problem using coefficients of truncated Lagrange polynomials, Electron. J. Diff. Eqns. 51
(2007) 1–14.


5690

N.H. Tuan et al. / Applied Mathematical Modelling 35 (2011) 5673–5690

[13] B.W. Muniz, A comparison of some inverse methods for estimating the initial condition of the heat equation, J. Comput. Appl. Math. 103 (1999) 145–
163.
[14] N. Dokuchaev, Regularity for some backward heat equations, J. Phys. A: Math. Theor. 43 (2010).
[15] H.W. Engl, M. Hanke, A. Neubauer, Regularization of Inverse Problems, Mathematics and Its Applications (2000).
[16] L.C. Evans, Partial Differential Equations, Amer. Math. Soc., Providence, RI, 1998.
[17] F. Black, M. Scholes, The pricing of options and corporate liabilities, J. Polit. Econ. 81 (3) (1973).
[18] U. Tautenhahn, Optimality for ill-posed problems under general source conditions, Numer. Funct. Anal. Optim. 19 (1998) 377–398.
[19] G.W. Clark, S.F. Oppenheimer, Quasireversibility methods for non-well posed problems, Electron. J. Diff. Eqns. 1994 (8) (1994) 1–9.
[20] D.D. Trong, N.H. Tuan, Regularization and error estimates for nonhomogeneous backward heat problems, Electron. J. Diff. Eqns. (2006) 1–10.
[21] D.D. Trong, N.H. Tuan, Regularization and error estimate for the nonlinear backward heat problem using a method of integral equation, Nonlinear Anal.
71 (9) (2009) 4167–4176.
[22] M. Denche, K. Bessila, A modified quasi-boundary value method for ill-posed problems, J. Math. Anal. Appl. 301 (2005) 419–426.
[23] N.H. Tuan, D.D. Trong, A new regularized method for two dimensional nonhomogeneous backward heat problem, Appl. Math. Comput. 215 (3) (2009)
873–880.
[24] 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. Anwend.
26 (2) (2007) 231–245.