Regularization for a Riesz-Feller space fractional backward diffusion problem with a time-dependent coefficient
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Science & Technology Development, Vol 5, No.T20- 2017
Regularization for a Riesz-Feller space
fractional backward diffusion problem with a
time-dependent coefficient
Dinh Nguyen Duy Hai
University of Science, VNU-HCM
Ho Chi Minh City University of Transport
(Received on 5th December 2016, accepted on 28 th November 2017)
ABSTRACT
In the present paper, we consider a backward
0,2 . This problem is ill-posed, i.e., the solution
problem for a space-fractional diffusion equation
(if it exists) does not depend continuously on the data.
(SFDE) with a time-dependent coefficient. Such the
Therefore, we propose one new regularization solution
problem is obtained from the classical diffusion
to solve it. Then, the convergence estimate is obtained
equation by replacing the second-order spatial
under a priori bound assumptions for exact solution.
derivative with the Riesz-Feller derivative of order
Key words: space-fractional backward diffusion problem, Ill-posed problem, Regularization, error
estimate, time-dependent coefficient
INTRODUCTION
The fractional differential equations appear more
and more frequently in physical, chemical, biology and
engineering applications. Nowadays, fractional
diffusion equation plays important roles in modeling
anomalous diffusion and subdiffusion systems [2],
description of fractional random walk, unification of
diffusion [3], and wave propagation phenomenon [4]. It
is well known that the SFDE is obtained from the
classical diffusion equation in which the second-order
space derivative is replaced with a space-fractional
partial derivative.
Let : [0, T ]
is a continuous function on
satisfying (t )0 . In this paper, we consider a
[0, T ]
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backward problem for the following nonlinear SFDE
with a time-dependent coefficient
ut ( x, t ) (t ) x D F ( x, t , u ( x, t )), ( x, t ) (0, T ),
u ( x, t ) x 0, t (0, T ),
u ( x, T) G ( x ), x ,
(1)
where the fractional spatial derivative
D
x
is the
Riesz-Feller
fractional
derivative
of
order
and
skewness
(0 2)
(| | min{ , 2 }, 1) defined in [5], as
follows:
TAẽP CH PHAT TRIEN KH&CN, TAP 20, SO T5- 2017
(1 ) ( )
x D f ( x)
sin
2
2
d f ( x)
, 2.
x D0 f ( x )
dx 2
0
f ( x s ) f ( x)
( )
ds sin
1
s
2
f ( x s ) f ( x)
ds , 0 2,
s1
0
Here, we wish to determine the temperature u ( x, t ) from temperature measurements G ( x ). Since the
measurements usually contain an error, we now could assume that the measured data function G ( x) satisfies
G G
, where the constant 0 represents the noise level. Moreover, assume there hold the following a
L ( )
priori bound
2
u (, 0)
2
L ( )
E, E .
(2)
We assume that F satisfies the Lipschitz condition
F ( x, t , z1 ) F ( x, t , z2 ) L2 ( ) K F z1 z2 L2 ( )
(3)
for some constant K F independent of x, t , z1 , z2 with
1
K F 0, .
(4)
T
In case of the source function F 0 and (t ) 1,
The remainder of this paper is organized as
Problem (1) has been proposed by some authors. Zheng
follows. In Section 2, we propose the regularizing
and Wei [7] used two methods, the spectral
scheme for Problem (1). Then, in Section 3, we show
regularization and modified equation methods, to solve
that the regularizing scheme of Problem (1) is wellthis problem. In [6], they developed an optimal
posed. Finally, the convergence estimate is given in
modified method to solve this problem by an a priori
Section 4.
and an a posteriori strategy. In 2014, Zhao et al [8]
REGULARIZATION FOR PROBLEM (1)
applied a simplified Tikhonov regularization method to
Let G ( ) denote the Fourier transform of the
deal with this problem. After then, a new regularization
integrable function G ( x ), which defined by
method of iteration type for solving this problem has
1
been introduced by Cheng et al [1]. Although we have
G ( ) :
exp( ix )G ( x) dx, i 1.
many works on the linear homogeneous case of the
2
backward problem, the nonlinear case of the problem is
In terms of the Fourier transform, we have the
following
properties for the Riesz-Feller spacequite scarce. For the nonlinear problem, the solution u
fractional
derivative
[5]
is complicated and defined by an integral equation such
D (G )( ) ( )G ( ),
that the right hand side depends on u. This leads to
x
studying nonlinear problem is very difficult, so in this
where
paper we develop a new appropriate technique.
( ) | |
cos isign( ) sin .
2
2
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(5)
Science & Technology Development, Vol 5, No.T20- 2017
t
We define the function k (t ) by k (t )
1
ds.
(s)
By taking a Fourier transform to Problem (1), we transform Problem (1) into the following differential equation
0
ut ( , t ) (t ) ( )u ( , t ) F ( , t , u ( , t)),
u ( , T) G ( ).
(6)
The solution to equation (6) is given by
T
uˆ ( , t ) exp( ( )( k (T ) k (t )))[Gˆ ( ) exp( ( )( k ( s) k (T ))) Fˆ ( , s, u ( , s)) ds].
(7)
t
From (7), applying the inverse Fourier transform, we get
T
1
ˆ ( ) exp( ( )( k ( s ) k (T ))) Fˆ ( , s, u (, s )) ds] exp(ix ) d .
u ( x, t )
exp
(
(
)(
k
(
T
)
k
(
t
))
)
[
G
2
t
(8)
From which when becomes large, the terms
therefore recovering the scalar (temperature, pollution)
u ( x, t ) from the measured data G ( x ) is severely illexp ( )( k (T ) k (t )) increases rather quickly:
small errors in high-frequency components can blow up
posed. In this note, we regularize Problem (1) by the
and completely destroy the solution for 0 t T ,
problem
U ( , t )
exp
( )( k (T ) k (t ))
T
G ( )
exp
( )( k ( s ) k (t ))
k (T )
k (T )
t
1 exp | | cos
1 exp | | cos
2
2
k (T )
exp | | cos
2
exp ( k ( s ) k (t )) ( ) Fˆ ( , s, U ( , s )) ds,
k (T )
1 exp | | cos
2
where is regularization parameter.
t
0
THE WELL POSEDNESS OF PROBLEM (9)
First, we consider the following Lemma which is used in the proof of the main results.
Lemma 1. Let t , s [0, T ].
1) If s t , then we have
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Fˆ ( , s, U ( , s )) ds
(9)
TAẽP CH PHAT TRIEN KH&CN, TAP 20, SO T5- 2017
(
exp ( )( k ( s ) k (t ))
a)
(
1 exp | | cos
(
k (T )
2
(
k ( t ) k (T )
)
k (T )
.
( )k (T ))
1 exp | | cos
2
2) If s t , then we have
exp( ( )( k ( s ) k (t ) k (T )))
1 exp(| | cos( )k (T ))
c)
.
( )k (T ))
exp ( )( k (T ) k (t ))
b)
k (t )k ( s )
)
k (t )k ( s )
k (T )
.
2
Proof. First, we prove (a). In fact, we have
exp(| | cos(
exp( ( )( k ( s ) k (t )))
1 exp(| | cos(
2
exp( | | cos(
exp(| | cos(
[ exp( | |
cos(
2
2
) k (T ))
)( k ( s ) k (t ) k (T )))
2
k ( s )k (t )
) k (T ))]
[ exp( | |
k (T )
cos(
2
k (T ) k ( s ) k ( t )
) k (T ))]
k (T )
k (t )k ( s )
1
)( k ( s ) k (t ) k (T )))
2
) k (T ))
exp( | | cos( ) k (T ))
2
k ( s )k (t )
k (T )
.
k (T )
As an immediate consequence of (a), making the change s T , we have (b).
Next, we prove (c). In fact from (b), we obtain
exp( ( )( k (T ) ( k (t ) k ( s))))
1 exp(| | cos(
2
k ( t ) k ( s ) k (T )
k (T )
) k (T ))
it follows that
exp( ( )( k ( s ) k (t ) k (T )))
k (t )k ( s )
(
1 exp | | cos(
This completes the proof.
2
) k (T )
k (T )
.
)
We are now in a position to prove the following theorem.
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Science & Technology Development, Vol 5, No.T20- 2017
Theorem 1. Suppose m 0,
1
KFT
2
2
1 . Let G L2 ( ) and F satisfies (3) then Problem (9) is well-posed.
Proof. We divide it into two steps.
Step1. The existence and the uniqueness of a solution of Problem (9).
Let us define the norm on C ([0; T ]; L2 ( )) as follows
k (t )
‖ h‖ 0 sup
‖ h(t )‖
k (T )
0 t T
, for all h C ([0; T ]; L ( )).
2
2
L (
)
It is easily be seen that ‖ ‖ 0 is a norm of C ([0; T ]; L2 ( )) .
For v C ([0; T ]; L2 ( )) , we consider the following function
A(v )( x, t )
1
2
2
t
0
2
exp(| | cos(
1
t
2
1 exp(| | cos(
exp( ( )( k ( s ) k (t )))
T
1
B ( x, t )
1 exp(| | cos(
)k (T ))
2
)k (T ))
2
Fˆ ( , s, v ) exp(i x ) dsd
)k (T ))
exp(( k ( s ) k (t )) ( )) Fˆ ( , s, v ) exp(i x ) dsd ,
where
B ( x, t )
(
)
exp ( )( k (T ) k (t ))
(
G ( ) exp(i x ) d .
( )k (T ))
1 exp | | cos
2
We claim that, for every v1 , v2 C ([0; T ]; L ( ))
2
‖ A(v1 ) A(v2 )‖ 0 K F T‖ v1 v2 ‖ 0 .
First, by Lemma 1 and (3), we have two following estimates for all t [0, T ]
T
exp( ( )( k ( s ) k (t )))
ˆ
ˆ
J1
F ( , s , v1 ) F ( , s, v2 ) ds d
t 1 exp(| | cos( )k (T ))
2
2
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TAẽP CH PHAT TRIEN KH&CN, TAP 20, SO T5- 2017
2
exp( ( )( k ( s ) k (t )))
T
(T t )
1 exp(| | cos(
t
2
F ( , s , v1 ) F ( , s , v2 ) dsd
)k (T ))
2
2 ( k ( t ) k ( s ))
T
(T t )
2k (t )
2
F ( , s , v1 ) F ( , s , v2 ) dsd
k (T )
k (T )
2
t
k (T )
k (T )
v1 (, s ) v2 (, s )
2
2
L (
)
ds
t
2 k ( s )
2k (t )
2 k ( s )
T
K F (T t )
K F (T t ) sup
2
2
k (T )
0 s T
2k (t )
v1 (, s ) v2 (, s )
)
2
2
L (
k (T )
K F (T t ) v1 v2
2
2
2
0
(11)
and
2
t exp(| | cos( 2 )k (T ))
J2
exp(( k ( s ) k (t )) ( )) F ( , s , v1 ) F ( , s, v2 ) ds d
0 1 exp(| | cos(
)k (T ))
2
t
t
0
(
exp | | cos
2
( )k (T ))
(
2
(
1 exp | |
exp ( k ( s ) k ( t )) ( )
cos
2
F ( , s , v1 ) F ( , s , v2 ) dsd
)
( )k (T ))
2
t
t
2 ( k ( t ) k ( s ))
2k (t )
2
F ( , s , v1 ) F ( , s , v2 ) dsd
k (T )
k (T )
2
0
k (T )
k (T )
v1 (, s ) v2 (, s )
2
2
L (
)
ds
(12)
0
2 k ( s )
2k (t )
2 k ( s )
t
KFt
K F t sup
2
2
k (T )
0 s T
2k (t )
v1 (., s ) v2 (., s )
2
2
L (
)
k (T )
K F t v1 v2
2
2
For 0 t T , using the inequality ( a b) (1 m) a 1
2
2
1
2
.
0
b for all real numbers a and
2
m
b and m 0, we
obtain
2k (t )
A(v1 )(, t ) A(v2 )(, t )
By choosing m
T t
t
2
2
L ( )
(1 m)
k (T )
2k (t )
1
2
2
2
K F t v1 v2 0 1 k (T ) K F (T t ) v1 v2 0 .
m
2 2
2
, we have
2 k ( t )
k (T )
A(v1 )(, t ) A(v2 )(, t )
2
2
L (
K F T v1 v2 0 , for all t (0, T ).
2
)
2
2
(13)
On the other hand, letting t 0 in (11), we have
A(v1 )(, 0) A(v2 )(, 0)
2
K F T v1 v2 0 .
(14)
K F T v1 v2 0 .
(15)
2
2
L (
)
2
2
By letting t T in (12), we have
A(v1 )(, T ) A(v2 )(, T )
2
2
2
L (
2
)
2
2
Combining (13), (14) and (15), we obtain
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Science & Technology Development, Vol 5, No.T20- 2017
2 k ( t )
k (T )
‖ A(v1 )(, t ) A(v2 )(, t )‖
2
K F T ‖ v1 v2 ‖ 0 , for all t [0, T ]
2
2
L (
)
2
2
which leads to (10). Since K F T 1, A is a contraction. It follows that the equation A(v) v has a unique solution
U C ([0; T ]; L ( )) .
2
Step 2. The solution of Problem (9) continuously depends on the data.
Let V , W be two solutions of Problem (9) corresponding to the final values GV and GW . By straightforward
computation, we write
exp( ( )( k (T ) k (t )))
W ( , t ) V ( , t )
1 exp(| | cos(
(
t
(
0
( ) GV ( ))
) [Fˆ ( , s, V
cos( )k (T ))
( , s )) Fˆ ( , s, W ( , s )) ds
]
2
)k (T ))
W
1 exp | |
exp(| | cos(
t
2
(G
exp ( )( k ( s ) k (t ))
T
) k (T ))
2
1 exp(| | cos(
exp(( k ( s ) k (t )) ( )) [ Fˆ ( , s , W ( , s )) Fˆ ( , s , V ( , s ))]ds
.
) k (T ))
2
Now applying Lemma 1, we get
k ( t ) k (T )
W ( , t ) V ( , t )
k (t )k ( s )
T
GW ( ) GV ( )
k (T )
k (T )
Fˆ ( , s, V ( , s )) Fˆ ( , s, W ( , s )) ds
t
k (t )k ( s )
t
Fˆ ( , s , V ( , s )) Fˆ ( , s, W ( , s )) ds
k (T )
0
k ( t ) k (T )
k (T )
T
GW ( ) GV ( )
k (t )k ( s )
k (T )
Fˆ ( , s , V ( , s )) Fˆ ( , s , W ( , s ))
ds.
0
Since m 0,
1
2
KFT
2
1
T 1 m
1 , we have that 0 K F
. From the inequality
( a b) 1
2
for all real number a, b and m 0, we get
W (, t ) V (, t )
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2
2
L ( )
W (, t ) V (, t )
2
2
L ( )
1 2
2
a (1 m)b
m
(16)
TAẽP CH PHAT TRIEN KH&CN, TAP 20, SO T5- 2017
1
1
m
1
1
m
T k ( tk)(Tk)( s )
(1 m)
F ( , s , V ( , s )) F ( , s, W ( , s )) ds d
)
0
2 k ( t )2 k (T )
GW GV
k (T )
2
2
2
L (
2 k ( t )2 k (T )
GW GV
k (T )
2 k ( t ) 2 k ( s )
T
2
(1 m) K F T
2
2
L (
)
2
W (, s ) V (, s )
k (T )
2
L (
)
ds.
0
This leads to
2 k ( t )
2
W (, t ) V (, t )
k (T )
2
L (
)
1 2
1 GW GV
m
2 k ( s )
T
2
(1 m) K F T
2
2
L (
)
W (, s ) V (, s )
k (T )
2
2
L (
)
ds. (17)
0
2 k ( t )
Set Z (t )
k (T )
W (, t ) V (, t )
2
2
L (
)
, t [0, T ]. Since W , V C ([0, T ]; L ( )), we see that the function Z is
2
continuous on [0, T ] and attains over there its maximum M at some t0 [0, T ]. Let M max Z (t ). From (17), we
t[0,T ]
obtain
M 1
1 2
GW GV
m
2
2
L ( )
(1 m) K F T M ,
2
2
or equivalently
1 2
2
2
1 (1 m) K F T M 1 m GW GV
2
2
.
L ( )
This implies that for all t [0, T ]
2 k ( t )
k (T )
W (, t ) V (, t )
2
2
L (
)
1 1 2 G G
W
V
m
M
2
2
2
2
L (
)
.
1 (1 m) K F T
Thus, we obtain
1
W (, t ) V (, t )
2
L ( )
1
k ( t ) k (T )
m
2
2
1 (1 m) K F T
k (T )
GW GV
2
L ( )
, t [0, T ].
(18)
This completes the proof of Step 2 and also the proof of the theorem.
CONVERGENCE ESTIMATE
Now we are ready to state the main result
Theorem 2. Let m 0,
u (, 0)
2
L (
)
1
2
KFT
2
2
1 . Suppose that Problem (1) has a unique solution u C ([0, T ]; L ( )) satisfying
E with E and the regularization parameter
E
then we have the estimate
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Science & Technology Development, Vol 5, No.T20- 2017
1
U (, t ) u (, t )
2
L ( )
1
k (t )
m
2
k (T ) E
2
2
1 (1 m) K F T
1
k (t )
k (T )
.
Proof. Assuming that u is a solution of Problem (9) corresponding to the final values G , we shall estimate
u (, t ) u (, t )
2
L (
)
. First we have
uˆ ( , t ) exp( ( )( k (T ) k (t ))) Gˆ ( ) exp( ( )( k ( s) k (T ))) Fˆ ( , s, u ( , s)) ds
T
1 exp(| | cos(
2
ˆ
ˆ
G ( ) exp( ( )( k ( s ) k (T ))) F ( , s, u ( , s )) ds
t
)k (T ))
T
exp( ( )( k (T ) k (t ))) exp(| | cos(
t
exp( ( )( k (T ) k (t )))
1 exp(| | cos(
2
2
(19)
)k (T ))
ˆ
ˆ
G ( ) exp( ( )( k ( s ) k (T ))) F ( , s, u ( , s )) ds .
t
)k (T ))
T
On the other hand, we get
uˆ ( , T ) Gˆ ( ) exp( k (T ) ( )) uˆ ( , 0) exp(k ( s ) ( )) Fˆ ( , s, u (, s)) ds .
T
0
This implies that
T
Gˆ ( ) exp( ( )( k ( s ) k (T ))) Fˆ ( , s, u ( , s )) ds
t
(20)
t
exp( k (T ) ( ))uˆ ( , 0) exp(( k ( s ) k (T )) ( )) Fˆ ( , s, u ( , s )) ds.
0
Combining (19) and (20), we obtain
uˆ ( , t )
T
ˆ
G
(
)
exp( ( )( k ( s ) k (T ))) Fˆ ( , s, u ( , s )) ds
t
1 exp(| | cos( )k (T ))
(
)
exp ( )( k (T ) k (t ))
2
exp( k (t ) ( )) exp(| | cos(
(
(
)k (T ))
uˆ ( , 0)
2
2
2
( )k (T ))
1 exp | | cos
exp(| | cos(
)k (T ))
( )k (T ))
1 exp | | cos
t
exp((k (s ) k (t ))
( ) Fˆ ( , s , u ( , s ))ds .
)
0
2
(21)
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TAẽP CH PHAT TRIEN KH&CN, TAP 20, SO T5- 2017
It follows from (9) and (21) that u ( , t ) u ( , t ) B1 B2 B3 ,
where
(
t
) [F ( , s, u
cos( )k (T ))
exp ( )( k ( s ) k (t ))
T
B1
(
1 exp | |
exp( k (t ) ( )) exp(| | cos(
1 exp(| | cos(
exp(| | cos(
B3
( , s )) F ( , s, u ( , s )) ds,
]
2
B2
2
1 exp(| | cos(
2
)k (T ))
)k (T ))
2
)k (T ))
2
u ( , 0),
)k (T ))
t
exp((k ( s) k (t ))
( ))[ F ( , s , u ( , s )) F ( , s, u ( , s ))]ds.
0
This leads to
u ( , t ) u ( , t ) | B1 | | B2 | | B3 |
k (t )k ( s )
T
k (t )
t
k (t ) k ( s )
F ( , s , u ( , s )) F ( , s , u ( , s )) ds k (T ) | u (, 0) |
k (T )
t
F ( , s , u ( , s ) ) F ( , s , u ( , s ) ) ds
k (T )
0
k (t )
k (t )k ( s )
T
k ( T ) | u (, 0) |
F ( , s, u ( , s )) F ( , s , u ( , s )) ds.
k (T )
0
(22)
Using this and (16), we conclude that
u (, t ) u (, t )
2
2
L ( )
u (, t ) u (, t )
2
2
L ( )
2k (t )
T k ( tk)(Tk)( s )
1 k (T )
2
1
u (, 0) L ( ) (1 m)
F ( , s, u ( , s )) F ( , s, u ( , s )) ds d
m
0
2
2
1
1
m
2k (t )
k (T )
2k (t ) T
u (, 0)
2
2
(1 m) K F T
2
L (
)
k (T )
2 k ( s )
k (T )
u (, s ) u (, s )
2
2
L (
)
ds,
0
and thus
2 k ( t )
k (T )
u (, t ) u (, t )
2
2
L (
)
1
1
u (, 0)
m
T
2
2
L (
2
)
(1 m) K F T
2 k ( s )
k (T )
u (, s ) u (, s )
2
2
L (
)
ds.
0
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Science & Technology Development, Vol 5, No.T20- 2017
Since u, u C ([0, T ]; L2 ( )), the function u (, t ) u (, t )
2
L (
is continuous on 0, T . Therefore, there exists a
)
2 k ( t )
positive N max t[ 0,T ]
u (, t ) u (, t )
k (T )
N 1
2
2
L ( )
. This implies that
1
2
‖ u (, 0)‖ L (
m
2
(1 m) K F T N ,
2
)
2
that is,
2 k ( t )
1 1 ‖ u (, 0)‖ 2
L ( )
m
N
.
2
2
2
u (, t ) u (, t )
k (T )
2
2
L (
)
1 (1 m) K F T
Hence, we obtain the error estimate
1
u (, t ) u (, t )
2
L ( )
E
1
k (t )
m
k (T ) .
2
2
1 (1 m) K F T
On the other hand, using estimate (18), we get
1
u (, t ) U (, t )
2
L ( )
1
m
2
2
1 (1 m) K F T
1
k ( t ) k (T )
k (T )
G G
2
L ( )
1
m
2
2
1 (1 m) K F T
k ( t ) k (T )
k (T )
.
From the triangle inequality and these estimates, we obtain
U (, t ) u (, t )
1
2
L (
)
1
m
2
2
1 (1 m) K F T
With
E
U (, t ) u (, t )
)
1
k ( t ) k (T )
k (T )
2
L (
u (, t ) u (, t )
2
L (
)
1
k (t )
m
E
k (T ) .
2
2
1 (1 m) K F T
, then we have the estimate
1
U (, t ) u (, t )
2
L (
)
2
1
k (t )
m
k (T ) E
2
2
1 (1 m) K F T
1
k (t )
k (T )
.
This completes the proof.
Remark 1. If (t ) 1 and F ( x, t , u ) 0 then Problem (1) becomes a homogeneous problem. The error estimate in
t
Theorem 2 is of order T . It is similar to the homogeneous case in [1, 6, 8].
CONCLUSION
In this paper, we use the new regularization
solution to slove a Riesz-Feller space-fractional
backward diffusion problem with a time-dependent
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coefficient. The convergence result has been obtained
under a priori bound assumptions for the exact solution
and the suitable choices of the regularization parameter.
TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T5- 2017
Acknowledgements: The author desires to thank the
handling editor and anonymous referees for their most
helpful comments on this paper.
Chỉnh hóa cho bài tốn khuếch tán ngược cấp
phân số khơng gian Riesz-Feller với hệ số phụ
thuộc thời gian
Đinh Nguyễn Duy Hải
Trường Đại học Khoa học Tự nhiên, ĐHQG-HCM
Trường Đại học Giao thơng Vận tải Tp Hồ Chí Minh
TĨM TẮT
Trong bài báo này, chúng tơi xét một bài tốn
chỉnh, nghĩa là nghiệm (nếu tồn tại) khơng phụ thuộc
ngược cho phương trình khuếch tán cấp phân số khơng
liên tục vào dữ liệu. Vì vậy, chúng tơi đưa ra một
gian với hệ số phụ thuộc thời gian. Bài tốn này có
nghiệm chỉnh hóa mới để giải bài tốn này. Sau đó,
được từ phương trình khuếch tán cổ điển bằng cách
ước lượng hội tụ thu được dưới một giả định bị chặn
thay đạo hàm bậc hai biến khơng gian bằng đạo hàm
tiên nghiệm cho nghiệm chính xác.
Riesz-Feller với 0,2 . Đây là bài tốn khơng
Từ khóa: bài tốn khuếch tán ngược cấp phân số khơng gian, bài tốn khơng chỉnh, chỉnh hóa, ước lượng
lỗi, hệ số phụ thuộc thời gian
TÀI LIỆU THAM KHẢO
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