DSpace at VNU: On a backward heat problem with time-dependent coefficient: Regularization and error estimates

Applied Mathematics and Computation 219 (2013) 6066–6073

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Applied Mathematics and Computation
journal homepage: www.elsevier.com/locate/amc

On a backward heat problem with time-dependent coefficient:
Regularization and error estimates
Nguyen Huy Tuan a, Pham Hoang Quan b, Dang Duc Trong c, Le Minh Triet b,⇑
a

Department of Applied Mathematics, Faculty of Science and Technology, Hoa Sen University, Quang Trung Software Park, Dist. 12, Ho Chi Minh City, Viet Nam
Department of Mathematics and Applications, SaiGon University, 273 An Duong Vuong st., Dist. 5, Ho Chi Minh City, Viet Nam
c
Department of Mathematics, University of Natural Science, Vietnam National University, 227 Nguyen Van Cu st., Dist. 5, Ho Chi Minh City, Viet Nam
b

a r t i c l e

i n f o

a b s t r a c t
In this paper, we consider a homogeneous backward heat conduction problem which
appears in some applied subjects. This problem is ill-posed in the sense that the solution
(if it exists) does not depend continuously on the final data. A new regularization method
is applied to formulate regularized solutions which are stably convergent to the exact ones
with Holder estimates. A numerical example shows that the computational effect of the
method is all satisfactory.
Ó 2012 Elsevier Inc. All rights reserved.

Keywords:
Backward problem
Fourier transform
Ill-posed problems
Heat equation
Time-dependent coefficient

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 t < T from the temperature data at the final time t ¼ T. This is usually referred to as the backward heat conduction problem, or the final value
problem. In the present paper, we consider the problem of finding the temperature uðx; tÞ; ðx; tÞ 2 ½0; pŠ  ½0; TŠ such that

ut ðx; tÞ ¼ aðtÞuxx ðx; tÞ;
uð0; tÞ ¼ uðp; tÞ ¼ 0;
uðx; TÞ ¼ gðxÞ;

ðx; tÞ 2 ½0; pŠ  ð0; TŠ;
t 2 ½0; TŠ;

x 2 ½0; pŠ;

ð1Þ
ð2Þ
ð3Þ

where aðtÞ; gðxÞ are given. The problem is called the backward heat problem, (BHP for short), the backward Cauchy problem
or the final value problem. In general, the solution of the problem does not exist. Further, even if the solution existed, it
would not be continuously dependent on the final data. It makes difficult to do numerical calculations. Hence, a regularization is in order.
In the special case of the problem (1)–(3) with aðtÞ ¼ 1, the problem becomes

ut ðx; tÞ ¼ uxx ðx; tÞ;

ðx; tÞ 2 ½0; pŠ  ð0; TŠ;

uð0; tÞ ¼ uðp; tÞ ¼ 0;
uðx; TÞ ¼ gðxÞ;

t 2 ½0; TŠ;

x 2 ½0; pŠ:

ð4Þ
ð5Þ
ð6Þ

The problem (4)–(6) has been considered by many authors using different methods. The mollification method has been
studied in [4]. An iterative algorithm with regularization techniques has been developed to approximate the BHP by
Jourhmane and Mera in [10]. Kirkup and Wadsworth have given an operator-splitting method in [9]. Quasi-reversibility
⇑ Corresponding author.
E-mail address: (T. Le Minh).
0096-3003/$ - see front matter Ó 2012 Elsevier Inc. All rights reserved.
/>

T. Nguyen Huy et al. / Applied Mathematics and Computation 219 (2013) 6066–6073

6067

method has been used by Lattes and Lions [1], Miller [2] and the other authors [3,13,11]. The boundary element method has
been also used by some authors (see [6,8]). All of them were devoted to computational aspects. However, few authors gave
their error estimates from the theoretical viewpoint for the BHP except Schroter and Tautenhahn [5], Yildiz and Ozdemir [7]
and Yildiz et al. [12]

Although we have many works on (4)–(6), however to the author’s knowledge, so far there are few results about
(1)–(3). The major object of this paper is to provide a regularization method to establish the Holder estimates. In fact,
we decided to regularize the exact problem by using the form of (13) directly. It can be called the quasi-solution method
(but it based on the quasi-boundary value method). By using quasi boundary value method, we have the regularized
problem as follows

ut ðx; tÞ ¼ aðtÞuxx ðx; tÞ;
uð0; tÞ ¼ uðp; tÞ ¼ 0;

ðx; tÞ 2 ½0; pŠ  ð0; TŠ;

ð7Þ

t 2 ½0; TŠ;

uðx; TÞ þ buðx; 0Þ ¼ gðxÞ;

ð8Þ

x 2 ½0; pŠ:

ð9Þ

By applying the Fourier method, we can find the form of the solution of (7)–(9)

n
o
Rt
exp Àm2 0 aðsÞds
n

o g m sinðmxÞ;
uðx; tÞ ¼
R
2 T aðsÞds
m¼1 b þ exp Àm
0
1
X

ðx; tÞ 2 ½0; pŠ  ½0; TŠ:

ð10Þ

The regularized solution (13) based on modifying the solution (10) of the problem (7)–(9) (noting that when a ¼ 0, the
solution (13) is the solution (10)). In this paper, we use the regularized solution (13) directly. In Theorem 2.2, we can get the
error estimate of Holder type for all t by using an appropriate parameter a P 0. In fact, the error estimate for the case
0 < t < T is as follows
p2 tþpa

kuðÁ; tÞ À v  ðÁ; tÞk 6 ð1 þ A1 Þq2 Tþqa :
In this case, we can choose a ¼ 0 and require a soft condition of the exact solution u

A1 ¼ kuðÁ; 0Þk < 1:
On the other hand, the error estimate for the case t ¼ 0 is as follows
pa

kuðÁ; tÞ À v  ðÁ; tÞk 6 ð1 þ A1 Þq2 Tþqa :
In order to get the the error estimate of Holder type, we choose a > 0 and require a strong condition of the exact solution u

A1 ¼

1
pX

2 m¼1

2
È
É
exp 2m2 a um ð0Þ

!12
< 1:

It requires the exact solution u is smooth enough. The remainder of the paper is divided into two sections. In Section 2, we
establish the regularized solution and estimate the error between an exact solution u of problem (1)–(3) and the regularized
solution u with the Holder type. Finally, a numerical experiment will be given in Section 3.
2. Regularization and main results
We denote that k Á k is the norm in L2 ð0; pÞ. Let hÁ; Ái be the inner product in L2 ð0; pÞ and g  be the measured data satisfying
kg  ðÁÞ À gðÁÞk 6 . Let aðtÞ : ½0; TŠ ! R be the differentiable function for every t and satisfy

0 < p 6 aðtÞ 6 q;

0 6 t 6 T:

ð11Þ

Suppose that Problems (1)–(3) have an exact solution u then u can be formulated as follows

n

o
Rt
1 exp Àm2
aðsÞds
X
0
n
o g m sinðmxÞ;
uðx; tÞ ¼
R
2 T aðsÞds
m¼1 exp Àm
0

ðx; tÞ 2 ½0; pŠ  ½0; TŠ:

ð12Þ

Let b > 0 and a P 0, we shall approximate the solution of the backward heat problem (1)–(3) by the regularized solution
as follows

n
R
o
t
exp Àm2 0 aðsÞds þ a
n
R
o g m sinðmxÞ;
v ðx; tÞ ¼

T
2
aðsÞds þ a
m¼1 b þ exp Àm
0
1
X

ðx; tÞ 2 ½0; pŠ  ½0; TŠ:

We note that b depends on e such that lim bðeÞ ¼ 0 and a is an arbitrarily nonnegative number.
e!0
Next, we consider some lemmas which is useful to the proof of theorems.

ð13Þ


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T. Nguyen Huy et al. / Applied Mathematics and Computation 219 (2013) 6066–6073

Lemma 2.1. Let x 2 R; c > 0, 0 6 a 6 b, and b–0 then

exa
a
6 cÀb :
1 þ cexb

ð14Þ

Proof of Lemma 2.1. We have

exa
exa
exa
Àab
¼
:
a
a 6
a 6 c
xb
1À
1 þ ce
ð1 þ cexb Þb ð1 þ cexb Þ b ð1 þ cexb Þb
This completes the proof of Lemma 2.1. h
Lemma 2.2. Let aðtÞ satisfy (11) and 0 < b < 1. Then for m P 1, one has

n
R
o
t
exp Àm2 0 aðsÞds þ a
qðtÀTÞ
n
R
o 6 b pTþa
T
2
b þ exp Àm 0 aðsÞds þ a

for all a P 0.
Proof of Lemma 2.2. From Lemma 2.1, we obtain

n
R
o
t
 cðtÞ
exp Àm2 0 aðsÞds þ a
1
n
R
o 6
;
T
b
2
b þ exp Àm 0 aðsÞds þ a
RT
where cðtÞ ¼

0

Rt

aðsÞdsÀ

RT
0

aðsÞds

0

aðsÞdsþa

.

From (11), we get

Z

T

aðsÞds P

0

Z

T

aðsÞds 6

t

Z
Z

T

pds ¼ pT;

0
T

qds ¼ qðT À tÞ:

t

Hence we get cðtÞ 6 qðTÀtÞ
. Thus, since (15) gives
pTþa

n
R
o
t
 qðTÀtÞ
exp Àm2 0 aðsÞds þ a
qðtÀTÞ
1 pTþa
n
R
o 6
¼ b pTþa :
T
b
2
b þ exp Àm 0 aðsÞds þ a

This completes the proof of Lemma 2.2. h
Lemma 2.3. Let aðtÞ satisfy (11) and 0 < b < 1. Then for m P 1, one has

n
R
o
t
b exp Àm2 0 aðsÞds þ a
ptþa
n
R
o 6 bqTþa
T
2
b þ exp Àm 0 aðsÞds þ a
for all a P 0.
Proof of Lemma 2.3. From (15), we have

n
R
o
t
 cðtÞ
b exp Àm2 0 aðsÞds þ a
1
n
R
o 6 b
¼ b1ÀcðtÞ :

T
b
2
b þ exp Àm 0 aðsÞds þ a
From (11), we get

n
R
o
t
b exp Àm2 0 aðsÞds þ a
ptþa
n
R
o 6 bqTþa :
T
2
b þ exp Àm 0 aðsÞds þ a
This completes the proof of Lemma 2.3. h

ð15Þ


T. Nguyen Huy et al. / Applied Mathematics and Computation 219 (2013) 6066–6073

6069

Theorem 2.1. Let 0 < b < 1 and a P 0. If v and w in Y are defined by (13) corresponding to the final values g and h in L2 ð0; pÞ,
respectively then
qðtÀTÞ

kv ðÁ; tÞ À wðÁ; tÞk 6 b pTþa kg À hk:
Proof of Theorem 2.1. From

ð16Þ

v and w are defined by (13) corresponding to the final values g and h in L2 ð0; pÞ, we have

n
R
o
t
exp Àm2 0 aðsÞds þ a
n
R
o g m sinðmxÞ;
v ðx; tÞ ¼
T
2
aðsÞds þ a
m¼1 b þ exp Àm
0

ðx; tÞ 2 ½0; pŠ  ½0; TŠ

ð17Þ

n
R
o

t
exp Àm2 0 aðsÞds þ a
n
R
o hm sinðmxÞ;
wðx; tÞ ¼
T
2
aðsÞds þ a
m¼1 b þ exp Àm
0

ðx; tÞ 2 ½0; pŠ  ½0; TŠ;

ð18Þ

1
X

and
1
X

where

gm ¼

2

Z p

p

0

gðxÞ sinðmxÞdx;

hm ¼

2

Z p

p

0

hðxÞ sinðmxÞdx:

ð19Þ

By using Lemma 2.2, we obtain

n



R

o

2

t

1  exp Àm2
1
aðsÞds þ a
X

2qðtÀTÞ
0
pX
p 2qðtÀTÞ
2
2


n


o
kv ðÁ; tÞ À wðÁ; tÞk ¼
À
h
Þ
jg m À hm j ¼ b pTþa kg À hk :
b pTþa
ðg
m  6

R
m

T
2 m¼1b þ exp Àm2
2

aðsÞds þ a
m¼1
0
2

ð20Þ

Therefore, we get
qðtÀTÞ

kv ðÁ; tÞ À wðÁ; tÞk 6 b pTþa kg À hk:

ð21Þ

This completes the proof of Theorem 2.1. h
p

Theorem 2.2. Let b ¼ q ; a P 0 and g; g  2 L2 ð0; pÞ satisfy kg À g  k 6 . If we suppose that u is the solution of problem (1)–(3)
such that

A1 ¼

1

pX

2 m¼1

È
É
exp 2m2 a jum ð0Þj2

!12
< 1;

ð22Þ

then one has for every t 2 ½0; TŠ
p2 tþpa

kuðÁ; tÞ À v  ðÁ; tÞk 6 ð1 þ A1 Þq2 Tþqa ;
where

v  ðx; tÞ

ð23Þ

is defined by (13) corresponding to the noisy data

g  ðxÞ.

Proof of Theorem 2.2. Let u be defined by (13) with exact data g. Using the triangle inequality, we get

kuðÁ; tÞ À v  ðÁ; tÞk 6 kuðÁ; tÞ À u ðÁ; tÞk þ ku ðÁ; tÞ À v  ðÁ; tÞk:

ð24Þ

For the term ku ðÁ; tÞ À v  ðÁ; tÞk, using (16), we estimate it as follows
qðtÀTÞ

qðtÀTÞ

ku ðÁ; tÞ À v  ðÁ; tÞk 6 b pTþa kg  ðÁÞ À gðÁÞk 6 b pTþa :

ð25Þ

From (12), we get the mth Fourier sine coefficient of u

n
o
Rt
exp Àm2 0 aðsÞds
n
o gm :
um ðtÞ ¼
RT
exp Àm2 0 aðsÞds
Since (13), we get

ð26Þ


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T. Nguyen Huy et al. / Applied Mathematics and Computation 219 (2013) 6066–6073

n
R
o
t
exp Àm2 0 aðsÞds þ a
n
R
o g m :
um ðtÞ ¼
T
b þ exp Àm2 0 aðsÞds þ a


ð27Þ

From (26) and (27) and using Lemma 2.3, we obtain

n
o
n
R
o 1 
0
R
t


2 t

exp Àm2 0 aðsÞds þ a


  exp Àm 0 aðsÞds


um ðtÞ À u ðtÞ ¼ @
A
n
o
n


o
g
À
RT
RT
m
m


 exp Àm2 0 aðsÞds
b þ exp Àm2 0 aðsÞds þ a
n
R
o
t
b exp Àm2 0 aðsÞds þ a
n

R
o
n
R
o jg m j
¼
T
T
b þ exp Àm2 0 aðsÞds þ a
exp Àm2 0 aðsÞds þ a
n
R
o
t
b exp Àm2 0 aðsÞds þ a
È
É
È
É
ptþa
n
R
o exp m2 a jum ð0Þj 6 bqTþa exp m2 a jum ð0Þj:
¼
T
2
b þ exp Àm 0 aðsÞds þ a

ð28Þ

This follows that
2

kuðÁ; tÞ À u ðÁ; tÞk ¼

1 
pX


2 m¼1

1
2
2ðptþaÞ p X
È
É
um ðtÞ À um ðtÞ 6 b qTþa
exp 2m2 a jum ð0Þj2 :
2 m¼1

Hence, we obtain
ptþa

kuðÁ; tÞ À u ðÁ; tÞk 6 A1 bqTþa :
where A21 ¼

p P1
2

m¼1

È

ð29Þ

É

exp 2m2 a jum ð0Þj2 .
p

From (24), (25), (29) and b ¼ q , we have
ptþa

qðtÀTÞ

kuðÁ; tÞ À v  ðÁ; tÞk 6 A1 bqTþa þ b pTþa  6 A1

ptþa
 p qTþ
a



q

þ

 p qðtÀTÞ
pTþa



q

p2 tþpa
q2 Tþqa

 6 A1 

ptþa

p2 tþpa

þ qTþa 6 ð1 þ A1 Þq2 Tþqa :

This completes the proof of Theorem 2.2. h

3. Numerical experiment
Consider the linear homogeneous parabolic equation with time-dependent coefficient

ðx; tÞ 2 ½0; pŠ  ð0; 1Š;

ut ðx; tÞ ¼ aðtÞuxx ðx; tÞ;
uð0; tÞ ¼ uðp; tÞ ¼ 0;

t 2 ½0; 1Š;

where

aðtÞ ¼ 2t þ 1

ð30Þ

and

uðx; 1Þ ¼ g ex ðxÞ ¼

sin x
:
e2

Hence, we obtain

kg ex k ¼

 !12 rffiffiffiffi
Z p
sin x2
p À2


 e2  ds ¼ 2 e
0

and

1 6 aðtÞ 6 3
for all t 2 ½0; 1Š.
The exact solution of the equation is

ð31Þ


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T. Nguyen Huy et al. / Applied Mathematics and Computation 219 (2013) 6066–6073

n
o
Rt
1 exp Àm2
aðsÞds
X
0
n
o g m sinðmxÞ;
uðx; tÞ ¼
R
2 1 aðsÞds
m¼1 exp Àm
0
Rp

where g m ¼ p2

ð32Þ

g ex ðxÞ sinðmxÞdx.

0

Let t ¼ 0, from (32), we have

uðx; 0Þ ¼

1
X

1
g sinðmxÞ:
À2m2 m
e
m¼1

Consider the measured data

e

!

1 þ pffiffipffi À2 g ex ðxÞ;
e
2

g e ðxÞ ¼

ð33Þ

then we have

 !12
Z p


sin x2
e




g eÀgex  ¼ ppffiffiffi À2
 e2  dx ¼ e:
e
0
2
Let a ¼ 0, from (13) and (33), we have the regularized solution for the case t ¼ 0

v e ðx; 0Þ ¼

1
X
m¼1

where g me ¼ p2
Let

Rp
0

1

g sinðmxÞ;
b þ eÀ2m2 me

g e ðxÞ sinðmxÞdx and a ¼ 0.
1

e be e1 ¼ 10À1 ; e2 ¼ 10À2 ; e3 ¼ 10À3 ; e4 ¼ 10À4 ; e5 ¼ 10À5 and b ¼ e3 , respectively. Let
Re ðtÞ ¼



v e ðÁ;tÞÀuðÁ;tÞ 
i

kuðÁ; tÞk

be the relative error between the exact solution and the regularized solution at the time t.
Then we shall make the comparison between the absolute error and the relative error in the case t ¼ 0 and t ¼ 0:1. We
have the following table for the case t ¼ 0.

v e ðÁ;0ÞÀu

e

i

À1

e1 ¼ 10
e2 ¼ 10À2

e3 ¼ 10À3
e4 ¼ 10À4
e5 ¼ 10À5

ex ðÁ;0Þ




Re ð0Þ

6.411579eÀ001

5.1157575999eÀ001

5.914401eÀ001

4.7190616771eÀ001

4.215352eÀ001

3.3634022181eÀ001

2.549425eÀ001

2.0341697917eÀ001

1.372796eÀ001

1.0953450889eÀ001

We have the following table for the case t ¼ 0:1.

e
e1 ¼ 10À1
e2 ¼ 10À2
e3 ¼ 10À3
e4 ¼ 10À4
e5 ¼ 10À5


v e ðÁ;0:1ÞÀu
i

ex ðÁ;0:1Þ




Re ð0:1Þ

5.743711eÀ001

5.1155245814eÀ001

5.298322eÀ001

4.7188475240eÀ001

3.776257eÀ001

3.3632499109eÀ001

2.283862eÀ001

2.0340773067eÀ001

1.229797eÀ001

1.0952947987eÀ001

where kuex ðÁ; 0Þk ¼ ð2ð1=2Þ Ã pð1=2ÞÞ=2 ’ 1:2533 and kuex ðÁ; 0Þk ¼ pð1=2Þ Ã ð1=ð2 à expð11=50ÞÞÞð1=2Þ ’ 1:1228.
We have the following graph of the exact solution uex ðÁ; tÞ and of the regularized solution v ei ðÁ; tÞ; i ¼ 1; 2:


6072

T. Nguyen Huy et al. / Applied Mathematics and Computation 219 (2013) 6066–6073

We have the following graph of the regularized solution

v e ðÁ; tÞ; i ¼ 3; 4; 5:
i

Now, the figure can represent visually the exact solution and the regularized solution at initally time t = 0. Now, the figure
can represent visually the exact solution and the regularized solution at the time t = 0.1.


T. Nguyen Huy et al. / Applied Mathematics and Computation 219 (2013) 6066–6073

6073

Acknowledgement
All authors were supported by the National Foundation for Science and Technology Development (NAFOSTED). We thank
the referees for constructive comments leading to the improved version of the paper.
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