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.



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: [email protected] (T. Le Minh).
0096-3003/$ - see front matter Ó 2012 Elsevier Inc. All rights reserved.
http://dx.doi.org/10.1016/j.amc.2012.11.069


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]

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


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;


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


aðsÞds

0

aðsÞdsþa

.

From (11), we get

Z

T

aðsÞds P

0

Z

T

aðsÞds 6

t

Z
Z


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

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

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


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


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


6070

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


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





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



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.



 !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


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


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:


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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