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 )

Trang 178

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

Trang 182

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

[1]. H. Cheng, C.L. Fu, G.H. Zheng, J. Gao, A

regularization for a Riesz-Feller space-fractional

backward diffusion problem, Inverse Probl. Sci.

Eng., 22, 860–872 (2014).

[2]. O.P. Agrawal, Solution for a fractional diffusionwave equation defined in a bounded domain,

Nonlinear Dynamics, 29, 145–155 (2002).

[3]. R. Metzler, J. Klafter, The random walk’s guide to

anomalous diffusion: a fractional dynamics

approach, Physical Reports, 339, 1–77 (2000).

[4]. WR. Schneider, W. Wyss, Fractional diffusion and

wave equations, Journal of Mathematical Physics,

30, 134– 144 (1989).

[5]. F. Mainardi, Y. Luchko, G. Pagnini, The

fundamental solution of the space-time fractional

diffusion equation, Fract. Cacl. Appl. Anal., 4, 153–

192 (2001).

[6]. Z.Q. Zhang, T. Wei, An optimal regularization

method for space-fractional backward diffusion

problem, Math. Comput. Simulation, 92, 14–27

(2013).

[7]. G.H. Zheng, T. Wei, Two regularization methods

for solving a Riesz-Feller space-fractional backward

diffusion problem, Inverse Problems, 26, 115017

(2010).

[8]. J. Zhao, S. Liu, T. Liu, An inverse problem for

space-fractional backward diffusion problem, Math.

Methods Appl. Sci., 37, 1147– 1158 (2014).

Trang 183

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 ]

Trang 172

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

Trang 174

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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(10)

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 )

Trang 178

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)

Trang 180

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

Trang 181

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

Trang 182

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

[1]. H. Cheng, C.L. Fu, G.H. Zheng, J. Gao, A

regularization for a Riesz-Feller space-fractional

backward diffusion problem, Inverse Probl. Sci.

Eng., 22, 860–872 (2014).

[2]. O.P. Agrawal, Solution for a fractional diffusionwave equation defined in a bounded domain,

Nonlinear Dynamics, 29, 145–155 (2002).

[3]. R. Metzler, J. Klafter, The random walk’s guide to

anomalous diffusion: a fractional dynamics

approach, Physical Reports, 339, 1–77 (2000).

[4]. WR. Schneider, W. Wyss, Fractional diffusion and

wave equations, Journal of Mathematical Physics,

30, 134– 144 (1989).

[5]. F. Mainardi, Y. Luchko, G. Pagnini, The

fundamental solution of the space-time fractional

diffusion equation, Fract. Cacl. Appl. Anal., 4, 153–

192 (2001).

[6]. Z.Q. Zhang, T. Wei, An optimal regularization

method for space-fractional backward diffusion

problem, Math. Comput. Simulation, 92, 14–27

(2013).

[7]. G.H. Zheng, T. Wei, Two regularization methods

for solving a Riesz-Feller space-fractional backward

diffusion problem, Inverse Problems, 26, 115017

(2010).

[8]. J. Zhao, S. Liu, T. Liu, An inverse problem for

space-fractional backward diffusion problem, Math.

Methods Appl. Sci., 37, 1147– 1158 (2014).

Trang 183

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