This problem is ill-posed in the sense that the solution if it exists does not depend continuously on the final data.. This is usually referred to as the backward heat conduction problem,
Trang 1On a backward heat problem with time-dependent coefficient:
Regularization and error estimates
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
b
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
a r t i c l e i n f o
Keywords:
Backward problem
Fourier transform
Ill-posed problems
Heat equation
Time-dependent coefficient
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
1 Introduction
There are several important ill-posed problems for parabolic equations A classical example is the backward heat equa-tion In other words, it may be possible to specify the temperature distribution at a particular time t < T from the temper-ature 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;p ½0; T such that
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 regulariza-tion is in order
In the special case of the problem(1)–(3)with aðtÞ ¼ 1, the problem becomes
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
0096-3003/$ - see front matter Ó 2012 Elsevier Inc All rights reserved.
⇑ Corresponding author.
E-mail address: lmtriet2005@yahoo.com.vn (T Le Minh).
Contents lists available atSciVerse ScienceDirect
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
Trang 2method 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
By applying the Fourier method, we can find the form of the solution of(7)–(9)
uðx; tÞ ¼X1
m¼1
exp m2Rt
0aðsÞds
bþ exp m2RT
0aðsÞds
The regularized solution(13)based on modifying the solution(10)of the problem(7)–(9)(noting that whena¼ 0, the solution(13)is the solution(10)) In this paper, we use the regularized solution(13)directly InTheorem 2.2, we can get the error estimate of Holder type for all t by using an appropriate parameteraP0 In fact, the error estimate for the case
0 < t < T is as follows
uð; tÞ vð; tÞ
k k 6 ð1 þ A1Þp2 tþpq2 Tþqaa:
In this case, we can choosea¼ 0 and require a soft condition of the exact solution u
A1¼ uð; 0Þk k < 1:
On the other hand, the error estimate for the case t ¼ 0 is as follows
uð; tÞ vð; tÞ
k k 6 ð1 þ A1Þq2 Tþqpaa:
In order to get the the error estimate of Holder type, we choosea>0 and require a strong condition of the exact solution u
A1¼ p
2
X1
m¼1
exp 2m 2a
umð0Þ
!1
<1:
It requires the exact solution u is smooth enough The remainder of the paper is divided into two sections In Section2, we establish the regularized solution and estimate the error between an exact solution u of problem(1)–(3)and the regularized solution uwith the Holder type Finally, a numerical experiment will be given in Section3
2 Regularization and main results
We denote that k k is the norm in L2ð0;pÞ Let h; i be the inner product in L2ð0;pÞ and gbe the measured data satisfying
kgðÞ gðÞk 6 Let aðtÞ : ½0; T ! R be the differentiable function for every t and satisfy
Suppose that Problems(1)–(3)have an exact solution u then u can be formulated as follows
uðx; tÞ ¼X1
m¼1
exp m2Rt
0aðsÞds
exp m2RT
0aðsÞds
Let b > 0 andaP0, we shall approximate the solution of the backward heat problem(1)–(3)by the regularized solution
as follows
vðx; tÞ ¼X1
m¼1
exp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
We note that b depends onesuch that lim
e !0bðeÞ ¼ 0 andais an arbitrarily nonnegative number
Next, we consider some lemmas which is useful to the proof of theorems
Trang 3Lemma 2.1 Let x 2 R;c>0, 0 6 a 6 b, and b–0 then
exa
1 þcexb6c a
Proof of Lemma 2.1 We have
exa
xa
ð1 þcexbÞað1 þcexbÞ1a
ð1 þcexbÞa
6c a
:
This completes the proof ofLemma 2.1 h
Lemma 2.2 Let aðtÞ satisfy(11)and 0 < b < 1 Then for m P 1, one has
exp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
for allaP0
Proof of Lemma 2.2 FromLemma 2.1, we obtain
exp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
b
cðtÞ
where cðtÞ ¼
RT
0 aðsÞdsRt
0 aðsÞds
RT
0 aðsÞdsþ a
From(11), we get
Z T
0
aðsÞds P
Z T 0
pds ¼ pT;
Z T
t
aðsÞds 6
Z T t
qds ¼ qðT tÞ:
Hence we get cðtÞ 6qðTtÞ
pTþ a Thus, since (15) gives
exp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
b
qðTtÞ pTþ a
¼ bqðtTÞpTþ a:
This completes the proof ofLemma 2.2 h
Lemma 2.3 Let aðtÞ satisfy(11)and 0 < b < 1 Then for m P 1, one has
bexp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
for allaP0
Proof of Lemma 2.3 From (15), we have
bexp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
b
cðtÞ
¼ b1cðtÞ:
From(11), we get
bexp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
This completes the proof of Lemma2.3 h
Trang 4Theorem 2.1 Let 0 < b < 1 andaP0 If v and w in Y are defined by(13)corresponding to the final values g and h in L2
ð0;pÞ, respectively then
vð; tÞ wð; tÞ
Proof of Theorem 2.1 Fromvand w are defined by(13)corresponding to the final values g and h in L2ð0;pÞ, we have
vðx; tÞ ¼X1
m¼1
exp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
and
wðx; tÞ ¼X1
m¼1
exp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
where
gm¼2
p
Z p
0
gðxÞ sinðmxÞdx; hm¼2
p
Z p
0
By usingLemma 2.2, we obtain
vð; tÞ wð; tÞ
2
X1 m¼1
exp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
2
2b
2qðtTÞ pTþ aX1 m¼1
gm hm
j j2¼ b2qðtTÞpTþ akg hk2: ð20Þ
Therefore, we get
vð; tÞ wð; tÞ
This completes the proof ofTheorem 2.1 h
Theorem 2.2 Let b ¼p;aP0 and g; g2 L2ð0;pÞ satisfy g gk k 6 If we suppose that u is the solution of problem(1)–(3)
such that
A1¼ p
2
X1
m¼1
exp 2m 2a
umð0Þ
!1
then one has for every t 2 ½0; T
uð; tÞ vð; tÞ
wherevðx; tÞ is defined by(13)corresponding to the noisy data gðxÞ
Proof of Theorem 2.2 Let ube defined by(13)with exact data g Using the triangle inequality, we get
uð; tÞ vð; tÞ
For the term uk ð; tÞ vð; tÞk, using(16), we estimate it as follows
uð; tÞ vð; tÞ
From(12), we get the mth Fourier sine coefficient of u
umðtÞ ¼
exp m2Rt
0aðsÞds
exp m2RT
0aðsÞds
Since(13), we get
Trang 5mðtÞ ¼
exp m2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
From(26) and (27)and usingLemma 2.3, we obtain
umðtÞ umðtÞ
2Rt
0aðsÞds
exp m2RT
0aðsÞds
2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
0
@
1
Agm
2 Rt
0aðsÞds þa
exp m2 RT
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
2 Rt
0aðsÞds þa
bþ exp m2 RT
0aðsÞds þa
umð0Þ
j j 6 bqTþptþaaexp m 2a
umð0Þ
This follows that
uð; tÞ uð; tÞ
2
X1 m¼1
umðtÞ umðtÞ
6b2ðptþqTþ aaÞp 2
X1 m¼1
exp 2m 2a
umð0Þ
j j2:
Hence, we obtain
uð; tÞ uð; tÞ
where A2
¼p
2
P1
m¼1exp 2m 2a
umð0Þ
j j2 From(24), (25),(29)and b ¼p, we have
uð; tÞ v
ð; tÞ
k k 6 A1bptþqTþaaþ bqðtTÞpTþ a6A1 p
ptþ a qTþ a
þ p
qðtTÞ pTþ a
6A1p2 tþpq2 TþqaaþptþqTþaa6ð1 þ A1Þq2 Tþqp2 tþpaa:
This completes the proof ofTheorem 2.2 h
3 Numerical experiment
Consider the linear homogeneous parabolic equation with time-dependent coefficient
utðx; tÞ ¼ aðtÞuxxðx; tÞ; ðx; tÞ 2 ½0;p ð0; 1;
uð0; tÞ ¼ uðp;tÞ ¼ 0; t 2 ½0; 1;
where
and
uðx; 1Þ ¼ gexðxÞ ¼sin x
Hence, we obtain
gex
Z p
0
sin x
e2
2ds
!1
¼
ffiffiffiffi p 2
r
e2
and
1 6 aðtÞ 6 3
for all t 2 ½0; 1
The exact solution of the equation is
Trang 6uðx; tÞ ¼X1
m¼1
exp m2Rt
0aðsÞds
exp m2R1
0aðsÞds
where gm¼2
p
Rp
0gexðxÞ sinðmxÞdx
Let t ¼ 0, from(32), we have
uðx; 0Þ ¼X1
m¼1
1
e2m 2gmsinðmxÞ:
Consider the measured data
geðxÞ ¼ 1 þ e
ffiffiffi
p
2
p
e2
!
then we have
gegex e
ffiffiffi
p
2
pe2
Z p
0
sin x
e2
2
dx
!1
¼e:
Leta¼ 0, from(13) and (33), we have the regularized solution for the case t ¼ 0
veðx; 0Þ ¼X1
m¼1
1
bþ e2m 2gmesinðmxÞ;
where gme¼2
p
Rp
0geðxÞ sinðmxÞdx anda¼ 0
Letebee1¼ 101;e2¼ 102;e3¼ 103;e4¼ 104;e5¼ 105and b ¼e1
, respectively Let
ReðtÞ ¼ vei ð;tÞuð;tÞ
uð; tÞ
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
e1¼ 101 6.411579e001 5.1157575999e001
e2¼ 102 5.914401e001 4.7190616771e001
e3¼ 103 4.215352e001 3.3634022181e001
e4¼ 104 2.549425e001 2.0341697917e001
e5¼ 105 1.372796e001 1.0953450889e001
We have the following table for the case t ¼ 0:1
e1¼ 101 5.743711e001 5.1155245814e001
e2¼ 102 5.298322e001 4.7188475240e001
e3¼ 103 3.776257e001 3.3632499109e001
e4¼ 104 2.283862e001 2.0340773067e001
e5¼ 105 1.229797e001 1.0952947987e001
where uk exð; 0Þk ¼ ð2ð1=2Þ pð1=2ÞÞ=2 ’ 1:2533 and uk exð; 0Þk ¼pð1=2Þ ð1=ð2 expð11=50ÞÞÞð1=2Þ ’ 1:1228
We have the following graph of the exact solution uexð; tÞ and of the regularized solutionve ið; tÞ; i ¼ 1; 2:
Trang 7We have the following graph of the regularized solutionve ið; tÞ; i ¼ 3; 4; 5:
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
Trang 8All 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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