A non-local problem with discontinuous matching condition for loaded mixed type
equation involving the Caputo fractional derivative.
K.S.SADARANGANI1. O.KH. ABDULLAEV2
1. University of Las-Palmas de Gran Canaria. Las Palmas. Spain. [email protected]
2. National University of Uzbekistan. Tashkent. Uzbekistan. [email protected]
Abstract: Abstract: This research work devoted to investigations an existence and uniqueness
of solution of non-local boundary value problem with discontinuous matching condition for the
loaded equation. Considering parabolic-hyperbolic type equation is involve the Caputo
fractional derivative and loaded part joins in Riemann-Liouville integrals. The uniqueness of
solution is proved by the method of integral energy and the existence is proved by the method
of integral equations.
Key words and phrases: Loaded equation, parabolic-hyperbolic type, Caputo fractional
derivative, existence and uniqueness of solution, non-local condition, discontinuous matching
condition, integral energy, integral equations.
I.
Introduction and formulation of a problem.
It is well known that, fractional derivatives have been successfully applied to
problems in system biology [1], physics [2-5] and hydrology [6],[7]. Physical models
fractional differential operators have recently renewed attention from scientist which is
mainly due to applications as models for physical phenomena exhibiting anomalous
diffusion.
Note, than investigations of fractional analogs of main ODE and PDEs appear as a
result of the mathematic models for real-life processes [8], and have recently been
proved to be valuable tools in the modeling of many phenomena in various fields of
science and engineering [9],[10].
In the monographs of A.A.Kilbas, et al. [11], Miller and Ross [12], Podlubny [13],
S.G. Samko, et al. [14] we can see significant development of fractional differential
equations.
Very recently some basic theory for the initial boundary value problem (BVP)s of
fractional differential equations involving a Riemann-Liouville differential operator of
order 0    1 has been discussed by Lakshmikantham and Vatsala [15, 16]. In a series
of papers (see [17, 18]) the authors considered some classes of initial value problems for
functional differential equations involving Riemann-Liouville and Caputo fractional
derivatives of order 0    1 :
Well known that most fractional differential equations do not have exact analytic
solutions, so approximation and numerical techniques must be used. The numerical
solutions based on finite difference methods and several spectral algorithms for
fractional differential equations were reported in Refs. [19–25].
It should be noted that problems for a class of fractional differential system and
for the non-line differential equations with integral conditions was investigated in works
[26-30] and BVPs for the mixed type equations involving the Caputo and the RiemannLiouville fractional differential operators were investigated by many authors, see for
instance [31-33].
_____________
2010 Mathematics Subject Classification: 35M10.
BVPs with discounting matching conditions for the loaded equations with
fractional derivative have not been investigated yet.
This paper deals the existence and uniqueness of solution of the non-local
problem with discontinuous matching condition for loaded mixed type equation:
1


 1
u xx  С Doy u  p ( x, y )  (t  x) u (t ,0)dt , at y  0

x
(1)
0
1
u  u  q ( x, y ) (t  x  y ) 1 u (t ,0)dt , at y  0
yy

 xx
x y

involving the Caputo fractional derivative operator [35]:
y
1

( y  t )  f (t )dt ,
(2)
С Doy f 

(1   ) 0
where 0   ,  ,   1 .
Definition. Riemann-Liouville integral-differential operator fractional order  (   R ),
with starting from the point a is represented as follows [35]:
x
si gn( x  a )
f (t )

Dax f ( x) 
dt ,   0 ;
(3)
( ) a x  t  1
Dax f ( x)  f ( x) ,   0 ;
d k  k

k
Dax f ( x)  si gn ( x  a ) k Dax f ( x) , k  1    k , k  N .
dt
Definition. Caputo differential operator fractional order  (   0 ) is represented as
follows [35]:

k
 k ( k )
( x) , k  1    k , k  N .
С Dax f ( x )  si gn ( x  a ) Dax f
Let’s,  is domain, bounded with segments: A1 A2  {( x, y ) : x  1, 0  y  h} ,
B1 B2  {( x, y ) : x  0, 0  y  h} ,
characteristics : A1C : x  y  1 ;
B2 A2  {( x, y ) : y  h, 0  x  1} at the y  0 , and
B1C : x  y  0 of the equation (1) at y  0 , where
1 1
A1 1;0  , A2 1; h  , B1  0;0  , B2  0; h  , C  ;   .
2
Introduce designations:  ( x ) 
2
x 1
x 1 2
i
, i  1 .
2
2
 1

    ( y  0) ,     ( y  0) , I1   x :  x  1 , I 2   y : 0  y  h .
 2

In the domain of  the following problem is investigated.
Problem I. To find a solution u ( x, y ) of the equation (1) from the following class
of functions:


W  u ( x, y ) : u( x, y )  C ()  C 2 ( ) , u xx  C    , C Doy u  C    
satisfies boundary conditions:
u ( x, y )
A1 A2
  ( y) , 0  y  h
(4),
  ( y) , 0  y  h ,
u ( x, y )
B1B2
(5)
d
u  ( x )   a ( x )u y ( x,0)  b( x)u x ( x,0)  c ( x )u ( x,0)  d ( x ), x  I1 .
dx
and gluing condition:
lim y1 u y ( x, y )   ( x)u y ( x, 0), ( x,0)  A1B1
y 0
(6)
(7)
where  ( y ),  ( y ) , a ( x) , b( x ) , c( x ) , d ( x) and  ( x) ( ( x)  0) are given functions.
II.
Results and Discussion
The Uniqueness of solution of the Problem I.
In the sequel, we assume that q ( x, y )   q1 ( x  y ) q2 ( x  y ) . In fact, that equation (1) at
y  0 and on the characteristics coordinate   x  y and   x  y totally looks like:
u 
q1 ( )q2 ( ) 1
(t   ) 1 u (t ,0)dt .

4

(8)
Let’s enter designations: u ( x,0)   ( x), 0  x  1 ; u y ( x, 0)    ( x ) , 0  x  1 ;
lim y1 u y ( x, y )    ( x) , 0  x  1 .
y 0
Known, that solution of the Cauchy problem for equation (1) in the domain   can be
represented as follows:
x y
x y
x y
1
 ( x  y)   ( x  y) 1
1

u( x, y ) 
   (t )dt   q1 ( )d   q2 ( ) d  (t   ) 1 (t )dt . (9)
2
2 x y
4 x y


After using condition (6) and taking (3) into account from (9) we will get:
 2a( x)  1  ( x)  ( )q1 ( x)q2 ( x) Dx1  ( x)  1  2b( x)  ( x) 
2c( x) ( x)  2d ( x)
(10)
1
where q2 ( x )   q2 ( )d .
x
Considering designations and gluing condition (7) we have
  ( x )   ( x )  ( x ) .
Further from Eq. (1) at y  0 taking (2), (11) into account, and
(11)
lim D0y1 f ( y )  ( )lim y1 f ( y )
y 0
y 0
we get [28]:
 ( x)   ( x)( )  ( x)  (  ) p( x,0) Dx1  ( x)  0
(12)
Theorem 1. If satisfies conditions
 (0)q1 (0)q2 (0)
(13)
 0, p(0,0)  0, p( x,0)  0 ;
2a(0)  1
 1  2b( x)

 q1 ( x) q 2 ( x)

 ( х )c ( x )
(14)
0, 
 ( х)   0 ;
 ( х)   0 ,

2a( x )  1
 2a( x)  1

 2a ( x)  1

then, the solution u ( x, y ) of the Problem I is unique.
Proof. Known, that, if homogeneous problem has only trivial solution, then we
can state that original problem has unique solution. For this aim we assume that the
Problem I has two solutions, then denoting difference of these as u ( x, y ) we will get
appropriate homogenous problem.
Equation (12) we multiply to  ( x) and integrated from 0 to 1:
1
1
1
 ( x) ( x)dx  ( )   ( x) ( x)
0

0
( x)dx   (  )  ( x) p ( x,0) Dx1 ( x)dx  0 .
(15)
0
We will investigate the integral
1
1
I  ( )   ( x) ( x) ( x) dx   (  )  ( x) p ( x,0) Dx1  ( x)dx .

0
0
Taking (10) into account d ( x)  0 we get:
1
1
1  2b( x)   ( x )
 ( )( ) q1 ( x ) q 2 ( x )

I

(
x
)

(
x
)
D

(
x
)
dx


(

)
x1
0 2a( x)  1
0 2a( x)  1  ( x) ( x)dx 
2
1
1
 ( x )c ( x ) 2
 ( ) 
 ( x) dx  (  )  ( x) p ( x,0) Dx1 ( x)dx 
2a ( x)  1
0
0
( ) q1 ( x) q 2 ( x)
( ) 1  2b( x)
 1


(
x
)

(
x
)
dx
(
t

x
)

(
t
)
dt

 ( x ) d  2 ( x)  



2 0 2a ( x)  1
2 0 2a ( x)  1
x
1
1
1
1
1
1
 ( x )c ( x ) 2
( ) 
 ( x) dx   ( x) p ( x,0) dx  (t  x)  1 (t )dt . (16)
2a( x)  1
0
0
x
Considering  (1)  0,  (0)  0 (which deduced from the conditions (4), (5) in
homogeneous case) and on a base of the formula [34]:
x t



1
z
 1

    cos 0
2
cos  z  x  t   dz , 0    1
After some simplifications from (16) we will get:
( )q1 (0) q2 (0) (0)
I

4(2a(0)  1)(1   )sin
2
2
2
 1
 1
 
0 z  0  (t ) cos ztdt    0  (t ) sin ztdt   dz 




2
2
 1
 
q1 ( x)q2 ( x)   1
 

z dz   ( x)
    (t ) cos ztdt     (t ) sinztdt  dx 
 0

x
2
a
(
x
)

1

  x
0
 x
 
4(1   )sin
2
( )

1

1
1
( ) 2 
1  2b( x) 
 ( x )c ( x ) 2

 ( x)   ( x)
 ( x) dx 
 dx  2( ) 

2 0
2
a
(
x
)

1
2
a
(
x
)

1


0

p (0,0)

2(1   )sin
2

z
0

2
2
 1
 1
 
   (t ) cos ztdt      (t ) sin ztdt   dz 
 0
 0
 
2
2

1
 1
 1
 
1



0 z dz 0 x  p( x,0)  x  (t ) cos ztdt    x  (t )sinztdt  dx

2sin
(1   )


2
(17)
Thus, due to conditions (13),(14) from (17) we infer that  ( x )  0 . Hence, based
on the solution of the first boundary problem for Eq.(1) [32],[33] by using conditions

(4) and (5) we will get u( x, y)  0 in  .Further, from functional relations (10), taking
into account  ( x )  0 we deduce that   ( x)  0 . Consequently, based on the solution (9)

we obtain u( x, y)  0 in closed domain  .
The existence of solution of the Problem I.
Theorem 2. If satisfies conditions (13), (14) and
 ( y ),  ( y )  C  I 2   C1  I 2  ,
 


(18)
 
(19)
p( x,0)  C A1B1  C 2  A1 B1 
 
q ( x, y )  C    C 2   , a( x), b( x), c ( x), d ( x)  C1 I1  C 2  I1 
then the solution of the investigating problem is exist.
Proof. Taking (10) into account, from Eq.(12) we will obtain
(20)
 ( x )  A( x ) ( x )  f ( x )  B ( x ) ( x )
where
( )( ) ( x)q1 ( x)q2 ( x) 
2( ) ( x) d ( x)
(21)
f ( x) 
Dx1  ( x)  (  ) p( x,0) Dx1 ( x) 
2(2a( x)  1)
2a( x)  1
 ( ) ( x)(1  2b ( x ))
2( ) ( x)c( x )
, B( x) 
(22)
A( x ) 
2a ( x)  1
2a( x )  1
Solution of equation (20) together with conditions
 (0)   (0) ,  (1)   (0)
(23)
has a form
1
 (0)   (0) 
 ( x )  A1 ( x)    B(t ) (t )  f (t ) A1(t )dt 

A
(1)
1
x

1
x
A ( x)
A (t )
A (t )
 1
 B(t ) (t )  f (t )  1 dt    B(t ) (t )  f (t )  1 dt   (0) (24)

A1 (1) 0
A1 (t )
A1 (t )
0
where
x
t

A1 ( x)   exp   A( z )dz dt
(25)
0
0

Further, considering (21) and using (3) from (24) we will get:
1
1
1

( )  (t )q1 (t )q2 (t )
 1

 ( x )  A1 ( x)   A1(t ) B(t ) (t )dt 
A
(
t
)
dt
(
s

t
)

(
s
)
ds

1


2
2
a
(
t
)

1
x
x
t

1
1
1
A ( x) A (t )
 A1 ( x)  A1(t ) p(t ,0) dt  ( s  t )  1 ( s) ds  1  1 B (t ) (t )dt 
A1 (1) 0 A1(t )
x
t
1
1
( ) A1 ( x ) A1 (t ) (t ) q1 (t ) q 2 (t )

dt  ( s  t ) 1 ( s)ds 

2 A1 (1) 0 (2a (t )  1) A1(t )
t
1
1
x
A ( x) A1 (t )
A1 (t )
 1
 1
p
(
t
,0)
dt
(
s

t
)

(
s
)
ds

t
0 A1(t ) B(t ) (t )dt 
A1 (1) 0 A1(t )
x
1
( ) A1 (t ) (t ) q1 (t ) q 2 (t )

dt  ( s  t ) 1 ( s )ds 

2 0 (2a (t )  1) A1(t )
t
x
1
A (t )
  1 p (t ,0) dt  ( s  t )  1 ( s) ds  f1 ( x)
A(t )
0 1
t
(24)
where
x
1

A1 ( x )  2( ) d (t ) A1 (t ) (t )
d (t ) A1(t ) (t )
f1 ( x)   1 
dt  2 ( ) A1 ( x) 
dt 

A1 (1)  0 A1(t )(2a (t )  1)
2a (t )  1

x
1
A1 ( x) 2( ) d (t ) A1 (t ) (t )
A1 ( x )
(25)

dt

 (0)   (0)    (0)
A1 (1) x A1(t )(2a (t )  1)
A1 (1)
After some simplifications (24) we will rewrite on the form:
1
s
1

2 (t )
( )
 1  (t ) q1 (t ) q

 ( x )  A1 ( x)   A1(t ) B (t ) (t )dt 

(
s
)
ds
(
s

t
)
A
(
t
)
dt

1


2
2
a
(
t
)

1
x
x
x

1
s
1
A ( x ) A1 (t )
 A1 ( x )   ( s ) ds  ( s  t )  1 A1(t ) p (t ,0) dt  1
B (t ) (t )dt 


A
(1)
A
(
t
)
1
x
x
0 1
1
s
 2 (t )
( ) A1 ( x)
 1 A1 (t ) (t ) q1 (t ) q


(
s
)
ds
(
s

t
)
dt 
0
2 A1 (1) 0
(2a (t )  1) A1(t )
1
s
x
A ( x)
A (t )
A (t )
 1  ( s )ds  ( s  t )  1 1 p (t ,0) dt   1 B (t ) (t ) dt 
A1 (1) 0
A1(t )
A(t )
0
0 1
x
s
1
x

A1 (t ) (t )q1 (t )q2 (t )
A1 (t ) (t ) q1 (t ) q2 (t )
( ) 

dt


(
s
)
ds
dt
  ( s)ds 

1
1




2 0
A
(
t
)(2
a
(
t
)

1)(
s

t
)
A
(
t
)(2
a
(
t
)

1)(
s

t
)
0 1
x
0 1

x
s
1
x
A (t )( s  t )  1
A (t )( s  t )  1
  ( s) ds  1
p (t ,0)dt    ( s )ds  1
p (t ,0) dt  f1 ( x)


A
(
t
)
A
(
t
)
1
1
0
0
x
0
i.e. totally, we have integral equation:
1
 ( x)   K ( x, t ) (t )dt  f1 ( x) .
(26)
0
Here
 K ( x, s);
K ( x, t )   1
 K 2 ( x, s );
0  t  x,
x  t  1.

 A1 ( x)
  ( ) s
 2 (t )
A1 ( s)
 1 A1 (t ) (t ) q1 (t ) q
K 1 ( x, s )  
 1 
(
s

t
)
dt

B
(
s
)




A
(1)
2
(2
a
(
t
)

1)
A
(
t
)
A
(
s
)
 1

1
1
0

s
 A ( x) 
A (t )
 1
 1  ( s  t )  1 1 p (t ,0) dt
A1(t )
 A1 (1)
0
s


 2 (t )
( )
 1  (t ) q1 (t ) q


K 2 ( x, s )  A1 ( x)  A1 ( s ) B ( s) 
(
s

t
)
A
(
t
)
dt

1
2 x
2a (t )  1


(27)
(28)
s
 A1 ( x )  ( s  t )  1 A1(t ) p (t ,0) dt 
x
A1 ( x) A1 ( s)
B ( s) 
A1 (1) A1( s)
s
A (t ) (t )q1 (t ) q2 (t )
( ) A1 ( x)

( s  t ) 1 1
dt 

2 A1 (1) 0
(2a (t )  1) A1(t )
x
x

A1 ( x)  A1 (t )( s  t )  1
( ) A1 (t ) (t ) q1 (t ) q 2 (t )
 1

( s  t ) dt  1 
p (t ,0)dt (29)

2 0 A1(t )(2a(t )  1)
A1 (1)  0
A1(t )

Due to class (18), (19) of the given functions and after some evaluations, from (28),
(29) and (25), (27) we will conclude, that K ( x, t )  const , f1 ( x)  const .
Since kernel K ( x, t ) is continuous and function in right-side F ( x) is continuously
differentiable, solution of integral equation (26) we can write via resolvent-kernel:
1
 ( x )  f1 ( x )   ( x, t ) f1 (t ) dt ,
(30)
0
where ( x, t ) is the resolvent-kernel of K ( x, t ) .
Unknown functions   ( x) and   ( x ) we will found accordingly from (10) and
(11):
1
1
1
q1 ( x) q 2 ( x)
q1 ( x) q 2 ( x)
 1
 ( x) 
(t  x) dt  (t , s) f1 ( s )ds 
(t  x) 1 f1 (t )dt 


2 1  2a ( x )  x
2  2a ( x )  1 x
0

1
1  2b( x)
1  2b( x) ( x, t )
2c ( x )

f1( x) 
f
(
t
)
dt

f1 ( x) 
1
2a( x)  1
2a ( x)  1 0 x
2a( x)  1
1
2c ( x )
2d ( x )

( x, t ) f1 (t ) dt 

2a( x)  1 0
2a ( x )  1
and   ( x )   ( x)  ( x) .
Solution of the Problem I in the domain   we write as follows [33],[36]
y
y
1
u ( x, y )   G ( x, y,0, ) ( ) d   G ( x, y,1, ) ( ) d   G0 ( x   , y ) ( ) d  
0
0
y 1
0
1
   G ( x, y,0, ) p ( )d  d  (t   )  1 (t ) dt
(31)

0 0
y
Here
1
G0 ( x   , y ) 
  G ( x, y,  , )d ,

(1   ) 0
( y   ) 21   1, 2  x    2n  1, 2  x    2n
G ( x, y , , ) 
 e1, 2   ( y   ) 2   e1, 2   ( y   ) 2
2
n  






 
Is the Green’s function of the first boundary problem Eq. (1) in the domain   with the
Riemanne-Liouville fractional differential operator instead of the Caputo ones [35],
zn
e ( z)  
n 0 n ! (   n )

1,
1,
is the Wright type function [36].
III. Conclusion
If satisfy conditions (13), (14), (18) and (19), than the solution of the Problem I is
unique and exist, and this solution in the domains  and  will be found by the
formulate (9) and (31) respectively.
Competing interests
The authors declare that they have no competing interests.
Author's contributions
The two authors have participated into the obtained results. The collaboration of each
one cannot be separated in different parts of the paper. All of them have made
substantial contributions to the theoretical results. The two authors have been involved
in drafting the manuscript and revising it critically for important intellectual content.
All authors have given final approval of the version to be published.
Acknowledgements
The authos was grateful to the reviewers for useful suggestions which improve the
contents of this paper.
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