Regularization of a Cauchy problem for the heat equation

Science & Technology Development, Vol 5, No.T20- 2017

Regularization of a Cauchy problem for the heat
equation
 Vo Van Au
University of Science, VNU-HCM
Can Tho University of Technology
 Nguyen Hoang Tuan
University of Education, Ho Chi Minh
(Received on 5th December, 2016, accepted on 28 th November, 2017)

ABSTRACT
In this paper, we study a Cauchy problem for the
heat equation with linear source in the form
ut ( x, t )  uxx ( x, t )  f ( x, t ), u(L, t )  (t ), ux (L, t )   (t ), ( x, t )  (0, L)  (0,2 ).
This problem is ill-posed in the sense of Hadamard. To
regularize the problem, the truncation method is
proposed to solve the problem in the presence of noisy

Cauchy
data
and
satisfying





satisfying
        and that
f

f  ( x, )  f ( x, )   . We give some error estimates
between the regularized solution and the exact solution
under some different a-priori conditions of exact
solution.

Key words: elliptic equation, ill-posed problem, cauchy problem, regularization method, truncation method
INTRODUCTION
In this paper, the temperature u( x, t ) for
( x, t )  [0, L]  [0, 2 ] is sought from known boundary
temperature u( L, t )   (t ) and heat flux ux ( L, t )   (t )
measurements satisfying the following problem:
0  x  L, 0  t  2 ,

ut ( x, t )  uxx ( x, t )  f ( x, t ),

u( L, t )   (t ),
u ( L, t )   (t ), , 
where
x

0  t  2 ,
0  t given
 2 ,

(1)

are
functions (usually in
L (0, 2 ) ) and f is a given linear heat source which
may depend on the independent variables ( x, t ) .

2

Note that we have no initial condition prescribed at
t 0 and moreover, the Cauchy data  and  are
perturbed so as to contain measurement errors in the
form of the input noisy Cauchy data   and   (also in
L2 (0, 2 ) ) satisfying

 

 

,

(2)
where
denotes the L2 (0, 2 ) -norm and  0
is a small positive number representing the level of
noise.

Trang 184

It is well-known that, at least in the linear case, the
problem (1) has at most one solution using classical
analytical sideways continuation for the parabolic heat
equation. The existence of solution also holds, in the
case f 0 . However, the problem is still ill-posed in
the sense that the solution, if it exists, does not depend
continuously on the data. Any small perturbation in the
observation data can cause large errors in the solution

u( x, t ) for x [0, L). Therefore, most classical
numerical methods often fail to give an acceptable
approximation of the solution. Thus regularization
techniques are required to stabilize the solution [3].
In recent years, the homogeneous sideways heat
equation, i.e., f 0 in the first equation in (1), has
been researched by many authors and various methods
have been proposed, e.g. the difference regularization
method [8], the boundary element Tikhonov
regularization method [5], the Fourier method [9], the
quasi-reversibility method [1, 6], the wavelet, wavelet-


TAẽP CH PHAT TRIEN KH&CN, TAP 20, SO T5- 2017
Galerkin and spectral regularization methods [2, 7], the
conjugate gradient method [4], to mention only a few.

regularization ones and show that the solution of our
regularized problem converges to the solution of the
original linear problem (if such solution exists), as the
regularization parameter tends to zero. In the nonhomogeneous problem, we have many choises of
stability terms for regularization. However, in the case
of non-homogeneous problem, the main solution u is
complicated and is defined by a linear integral equation
whose the right-hand side depends on the independent
variables ( x, t ). In this paper, we develop a truncation
method to solve in a stable manner this linear integral
equation.

To the best of our knowledge, the Cauchy problem

for the linear sideways heat equation has not yet been.
Therefore, in the present paper, we propose a new
method that is based on linear integral equation to
regularize problem (1) under two a priori conditions on
the exact solution.
As will be shown in next section, for the linear
sideways heat problem (1), its solution (exact solution)
can be represented as an integral equation which
contains some instability terms. In order to restore the
stability we replace these instable terms by some
THE MAIN RESULTS
Let

denote the inner product in L2 (0, 2 ), and

0 represent the noise level in (2). For

(t ), exp( int ) exp(int ), where (t ), exp( int )

have the Fourier series (t )
n

L2 (0, 2 ) -norm of

1
2

is

2

L2 (0, 2 ), we

2

(t ) exp( int )dt. The
0

2

2

(t ),exp( int ) .
(3)

n

The principal value of

in is

(1 i) n
in
(1 i) n

2
2

,

n

0,

,

n

0.
(4)

Suppose that the solution of problem (1) is represented as a Fourier series
un ( x) exp(int ), with un ( x)

u( x, t )

u( x, t ), exp( int )

n

1
2

2

u( x, t ) exp( int ) d t.
0

From (1), we have the following systems of second-order ordinary differential equations:

d 2 un
( x) inun ( x) f n ( x),
0 x L,

d x2


t (0, 2 ),
un ( L) n (t ), exp(int ) ,
d u
n ( L) n (t ), exp( int ) , t (0, 2 ),

dx

1
where f n ( x) f ( x, t ),exp(int )
2

(5)

2

f ( x, t ) exp(int ) d t

for all n .

0

Trang 185

Science & Technology Development, Vol 5, No.T20- 2017
For n  \{0}, multiplying the first equation in (5) by
obtain
un ( x)  cosh  ( L  x) in  un ( L) 

sinh  ( L  x) in 
in

sinh  (  x) in 
in
L

un ( L)  

and integrating both sides from x to L, we

sinh  (  x) in 
in

x

f n ( ) d  , n  \{0}.

In the case n  0, multiplying the first equation in (5) by   x and integrating both sides from x to L, we obtain

(6)

L

u0 ( x)  u0 ( L)  ( L  x)u0 ( L)   (  x) f 0 ( ) d  .

(7)

x

From (6) - (7) the exact form of u is given by
L


sinh  ( L  x) in 
sinh  (  x) in 
u( x, t )   cosh  ( L  x) in  n 
n  
f n ( ) d   exp(int )
in
in

n \{0} 
x


 (0 , 0 , f 0 )( x),

(8)

L

where (0 , 0 , f 0 )( x)  0  ( L  x) 0   (  x) f 0 ( ) d  . In a few sentences, we present a brief introduction
x

Fourier truncated method. From equation (8), it can be observed that cosh  ( L  x) in  ,

sinh  ( L  x) in 
in

sinh  (  x) in 

and

are unbounded, as n tends to infinity, so in order to guarantee the convergence of the solution u
in
given by (8), the coefficient (n , n ) must decay rapidly. But such a decay usually cannot occur for the measured
data (n , n ) . Hence, a natural way is to eliminate the high frequencies and consider the solution u for n  N ,
where N is a positive integer; this is the so-called Fourier truncated method, and N plays the role of a
regularization parameter satisfying lim N  . We define the following two operators:
 0

QN+ ( , , f )( x, t ) 

QN ( , , f )( x, t ) 


Q

n  N

1
2

( , , f )( x) exp(int )



 exp  ( L  x)

n  N

Q

n  N

1
2

+
N ,n




N , n




exp  ( L  x) in 
in

in   n 

exp  (  x) in 
in

x

exp  ( L  x) in 

To approximate u, we introduce the regularized solution

Trang 186

L

n  


f n ( ) d   exp(int ),


(9)

( , , f )( x) exp(int )

 exp  ( L  x)

n  N

in   n 

in

L

n  
x

exp  (  x) in 
in


f n ( ) d   exp(int ). (10)



TAẽP CH PHAT TRIEN KH&CN, TAP 20, SO T5- 2017
uN ( x, t )



sinh ( L x) in L sinh ( x) in
n
f n ( ) d exp(int )
cosh ( L x) in n
in
in

n N , n 0
x

QN ( , , f )( x, t ) (0 , 0 , f 0 )( x).

(11)

Our these results would be applied after any necessary minor modifications have been made.
Lemma 1. For n
\{0} and n M , we have the following inequalities:

cosh ( L

x) in

exp

M
(L
2

sinh (

x) in

exp

M
(
2

x) , (12)
x) .
(13)

Proof. For n \{0}, n M , one has
cosh ( L

x) in

exp ( L

x) in

1
exp ( L
2
1
exp
2

exp
2

1
exp
2

x) in

M
(L
2

(L

1
2

x)

x) in

(L

x) in
M
(L
2

exp

1
exp
2

n
(L
2

x)

1
exp
2

n
(L
2

x)

1
exp
2

n
(L
2

x)

1
exp
2

n
(L
2

x)

x) ,

and

sinh ( L

x) in

exp ( L

x) in

in

exp ( L

x) in

2 in
exp ( L

x) in

exp

2 n
1
exp

2

M
(L
2

(L

x) in

2 n
x)

1
2

exp

M
(L
2

x) ,

as required.
Lemma 2. For n N , we have

u ( x)
1
QN , n ( , , f )( x) un ( x) n .

2
in

(14)

Proof. Differentiating (6) with respect to x gives
L
cosh ( L x) in
cosh ( x) in
u ( x)
n
sinh ( L x) in n
n
f n ( ) d .
in
in
in
x

(15)

Adding (15) to (6), we infer that

Trang 187


Science & Technology Development, Vol 5, No.T20- 2017
un ( x) 

un ( x)

 exp  ( L  x) in  n 

exp  ( L  x) in 

in
from which complete the proof.

in

L

n  
x

exp  (  x) in 
in

f n ( ) d  ,

The following theorem comes from the regularization uN provides the error estimates in the L2 -norm when the
exact solution belongs to new spaces Gs , (s 0) . Here Gs is presented by

Gs (0, 2 )



L2 (0, 2 ) :

n exp  2 n

2s

2

 (t ),exp( int )

,

n

(16)

and this norm is given by



Gs ((0,2 )

n

2s

exp  2 n

 (t ),exp( int )

2
L2 (0,2 )

.

(17)

For a Hilbert space X, we denote

L (0, L; X )

 :[0, L]

X esssup  ( )
0  L

X

,
(18)

and



L (0, L; X )

esssup  ( ) X .
0  L

Theorem 1. Assume that problem (1) has a weak solution u  C [0, T ]; L2 (0, 2 )  . Choose N  0 such that


N  
lim N1  lim  exp  L

   0.
 0
 0
2 



(19)

(20)

(a). Suppose that the problem (1) has a solution u satisfying

u L 0, L;G0 (0,2 )  ux L 0, L;G0 (0,2 )  E1 ,
L
L

(21)

for some known constant E1  0. Then


N 
uNε  ( x, )  u( x, )  P 2  2 E12 exp   x
,
2 

where P  6 2 exp  L 2 N  



(22)



6 L exp L 2 N  1
.
2 N
2

(b). Suppose that the problem (1) has a solution u satisfying

u L 0, L;Gr (0,2 )  ux L 0, L;Gr (0,2 )  E2 ,
L
L

(23)

for r  0 and some known constant E2  0. Then


N 
uNε  ( x, )  u( x, )  P 2  2 N2 r E22 exp   x   .
2 

1
Corollary 1. Let us choose N 
ln 2   for   0 then
2
 

L  
2

Trang 188

(24)


TAẽP CH PHAT TRIEN KH&CN, TAP 20, SO T5- 2017
Estimate in (22) is calculated as follows
x

uN ( x, ) u( x, ) R 2 2 E12 L ,
where R 6

2
L



6L



2
L

(25)

1

2
1
2
ln .
L

2. Estimate in (24) is calculated as follows

2
1
uN ( x, ) u( x, ) R 2 2
ln
L

4 r

x

E22 L .
(26)

Proof of the Theorem 1. The proof is divided into two parts.

Part a. Estimate the error (22) between the regularization uN and the exact solution u with a priori (21).
We rewrite u as
u( x, t )

L

sinh ( L x) in
sinh ( x) in
n
f n ( ) d exp(int )
cosh ( L x) in n
in
in

n N , n 0
x

+

(0 , 0 , f 0 )( x) QN ( , , f )( x, t ) QN ( , , f )( x, t ).



(27)

From (11) and (27), thanks to Parsevals relation, we obtain
uN ( x, ) u( x, )

2

2



2

n N , n 0

uN , n ( x) un ( x) 4



n N

QN ,n ( , , f )( x) QN ,n ( , , f )( x)

: J1 ( x )





: J 2 ( x )
2



2 (n , n , f n )( x) (0 , 0 , f 0 )( x) 4
: J 3 ( x )



n N

2

QN+ ,n ( , , f )( x) .
: J 4 ( x )

(28)

We now apply Lemma 1 and using the Holders inequality, we have
2

2
2
sinh ( L x) in


J1 ( x) 6
cosh ( L x) in n n
n n
in
n N , n 0


6

L



sinh ( x) in
in

n N , n 0 x

6



n N , n 0

6

2



2





2

f n ( ) f n ( ) d

exp ( L x) 2 N 2 exp ( L x) 2 N 2

n
n

n
n


L

(
L

x
)


x exp ( x) 2 N
n N , n 0



f


n

2
( ) f n ( ) d ,


(29)

where we have used the elementary inequality (a b c)2 3(a2 b2 c2 ).
Similarly, the second equation J 2 ( x) writes

Trang 189


Science & Technology Development, Vol 5, No.T20- 2017
2

2
2
exp  ( L  x) in 




J 2 ( x)  12  exp ( L  x) in  n   n 
 n  n

in
n  N 
L

 

 12

exp  (  x) in 
in

n  N x

 12

 exp  ( L  x)

n  N



 12

 f n ( )  f n ( )  d 








2
2
2 N  n   n  exp ( L  x) 2 N   n  n 

L







2

 ( L  x) exp  (  x)

n  N

2

2 N

f


n

x


2
( )  f n ( ) d  .


(30)

Thanks to Holder’s inequality and using the basic inequality ea  a, a  0, we deduce that
L


2
2
2
J 3 ( x)  6  0  0  ( L  x) 2  0  0  ( L  x)  (  x) 2 f 0 ( )  f 0 ( ) d  .
x











2
2
 6 exp ( L  x) 2 N 0  0  exp ( L  x) 2 N   0  0 


 6 ( L  x)  exp (  x) 2 N
x


Using Lemma 2, easy calculations show that



L

J 4 ( x)  4





 exp   x

n  N





n  N




in   exp  x

in  un ( x)  exp  x



u ( x ) 
in  n










2

in 

2

 2 exp  x 2 N

(31)

2

exp   x in  exp  x in  un ( x )  exp  x in 

un ( x)

2

(32)

in


2
  exp  L 2 n  un ( x)   exp  L 2 n
n  N
 n  N
2
L  0, L ;GL0 (0,2 ) 

 ux

2

L  0, L ;GL0 (0,2 ) 

 u ( x )
n

in

2





.


L


2
2
2
 6 exp ( L  x) 2 N            ( L  x)  exp (  L) 2 N  f  ( , )  f ( , ) d 
x


2
2


0
 2 exp  x 2 N
u 
 ux L  0, L;G 0 (0,2 ) 
L
 L  0, L;GL (0,2 ) 

L
 2

 6 exp ( L  x) 2 N   L 2  exp (  L) 2 N d   2 exp  x 2 N  E12
x



 2

L 2
 6 exp ( L  x) 2 N  
1  exp ( x  L) 2 N    2 exp  x 2 N  E12 , (33)
2 N





2









which can be rewritten as







Trang 190


2
( )  f 0 ( ) d   .


n  N

 2 exp  x 2 N  u

Combining (28), (29), (30), (31) and (32) we infer
uN  ( x, )  u( x, )


0

u ( x ) 
1
un ( x)  n


2
in 

2

QN , n ( , , f )( x)  4

n  N

f























(33)


TAẽP CH PHAT TRIEN KH&CN, TAP 20, SO T5- 2017

1


2
6 L 2 exp L 2 N 1

N

2
2
uN ( x, ) u( x, ) 6 exp L 2 N
2 E1 exp x .


2

2 N







(34)

Part (b). Estimate the error (24) between the regularization uN and the exact solution u with a priori (23).
By an argument analogous to the previous one, the estimates of J1 ( x), J 2 ( x), J 3 ( x) in the proof of part (a) remains
valid. Also, replace J 4 ( x) by following estimate
J 4 ( x) 4




n N



n



n

r

n N



2 r

2

un ( x)

2

exp x in









2 N2 r exp x 2 N
2 r

n N

2

u ( x )
1
un ( x) n


2
in

u ( x )
r

r
exp x in n exp x in un ( x ) n exp x in n

in

n exp x in un ( x ) n exp x in

2

r

n N

2 N



2

QN , n ( , , f )( x) 4

r


2r
n exp L 2 n
n N

exp x 2 N u

2
L 0, L ;GLr (0,2 )

ux

2

2

u ( x)
n

2

n

2r

in
exp L 2 n

u ( x )
n

in

n N

2

L 0, L ;GLr (0,2 )

2





.


(35)

Combining (28), (29), (30), (31) and (35), we get
We obtain
uN ( x, ) u( x, )

2



6 exp ( L x) 2 N






2 N 2 r exp x 2 N

6 exp ( L x) 2 N
6 exp ( L x) 2 N



uN


( x, ) u( x, ) 6 2 exp L





2 N

u





L


2
f ( , ) f ( , ) d




( L x) exp ( L) 2 N
x

2
L 0, L ;GLr (0,2 )

ux

2

L 0, L ;GLr (0,2 )

L

L 2 exp ( L) 2 N d 2 N 2 r exp x 2 N E22
x


2
2

L
1 exp ( x L) 2 N 2 N 2 r exp x 2 N E22 .

2 N





2







6 L 2
exp L

exp x

2 N







2 N



2 N



2 N

2 r


E


2
2

1

2






exp x


N

2

(36)

(37)
This completes the proof of the theorem.
CONCLUSION
In this paper, the Cauchy problem for the heat
equation has been solved by employing the truncation

method for a resulting linear integral equation.
Convergence and stability estimates, as the
regularization parameter tends to zero, are proved.

Trang 191

Science & Technology Development, Vol 5, No.T20- 2017

Chỉnh hóa bài toán Cauchy cho phương trình
nhiệt
 Võ Văn Âu
Trường Đại học Khoa học Tự nhiên, ĐHQG-HCM
Trường Đại học Kỹ thuật Công nghệ Cần Thơ
 Nguyễn Hoàng Tuấn
Trường Đại học Sư phạm Thành phố Hồ Chí Minh

TÓM TẮT
Trong bài báo này, chúng tôi nghiên cứu bài toán
trường hợp dữ liệu Cauchy  ,  và hàm nguồn f bị
Cauchy cho phương trình nhiệt với hàm nguồn tuyến
nhiễu
bởi
và
thỏa
mãn
 ,  
f


tính
thỏa
phương
trình:
và
       

f  ( x, )  f ( x, )   .
ut ( x, t )  uxx ( x, t )  f ( x, t ), u(L, t )  (t ), ux (L, t )   (t ), ( x, t )  (0, L)  (0,2 ). Chúng tôi đưa ra các đánh giá sai số giữa nghiệm
Đây là bài toán không chỉnh theo nghĩa của
chỉnh hóa và nghiệm chính xác dưới một số tính trơn
Hadamard. Để chỉnh hóa bài toán này, phương pháp
khác nhau của nghiệm chính xác.
chặt cụt được đề xuất để giải quyết bài toán trong
Từ khóa: phương trình Eliptic, bài toán không chỉnh, bài toán Cauchy, phương pháp chỉnh hóa, phương
pháp chặt cụt

TÀI LIỆU THAM KHẢO
[1]. L. Elden, Approximations for a Cauchy problem for
the heat equation, Inverse Problems, 3, 263–273
(1987).
[2]. L. Elden, F. Berntsson, T. Reginska, Wavelet and
Fourier methods for solving the sideways heat
equation, SIAM J. Sci. Comput., 21, 2187–2205
(2000).
[3]. D.N. Hao, Methods for Inverse Heat Conduction
Problems, Peter Lang, Frankfurt am Main (1998).
[4]. D.N. Hao, P.X. Thanh, D. Lesnic, B.T. Johansson,
A boundary element method for a multidimensional inverse heat conduction problem, Int.
J. Computer Math. 89, 1540–1554 (2012).
[5]. D. Lesnic, L. Elliott, D.B. Ingham, Application of
the boundary element method to inverse heat
conduction problems, Int. J. Heat Mass Transfer,
39, 1503–1517 (1996).

Trang 192

[6]. J.C. Liu, T. Wei, A quasi-reversibility
regularization method for an inverse heat
conduction problem without initial data, Appl.
Math. Comput., 219, 10866–10881 (2013).
[7]. T. Reginska, L. Elden, Solving the sideways heat
equation by a wavelet-Galerkin method, Inverse
Problems, 13, 1093–1106 (1997).
[8]. X.T. Xiong, C.L. Fu, H.F. Li, Central difference
method of a non-standard inverse heat conduction
problem for determining surface heat flux from
interior observations, Appl. Math. Comput. 173,
1265–1287 (2006).
[9]. X.T. Xiong, C.L. Fu, H.F. Li, Fourier regularization
method of a sideways heat equation for determining
surface heat flux, J. Math. Anal. Appl. 317, 331–
348 (2006).