302-311 SOURCE AND START IN DECREASAL EQUATIONS ' CORRECT BY IDENTIFYING THE INNER


302			AMERICAN	Journal	of	Public	Diplomacy	and	International	Studies																														www.grnjournal.us		 
  

  

AMERICAN Journal of Public Diplomacy and   
International Studies  

             Volume 01, Issue 09, 2023 ISSN (E): 2993-2157  

  
  
  

SOURCE AND START IN DECREASAL EQUATIONS ' 
CORRECT BY IDENTIFYING THE INNER FUNCTION 

PROBLEMS 
Latipova Shahnoza Salim daughter 

Asia International University 
General technician Department of Sciences teacher 

slatipova543@gmail.com  
 

Abstract: this article Caputo in the sense of fraction in order private derivative 
differential equation for mixed issue we learn This in the article  right issue solve and In 
Caputo's sense show that the solution of the partial differential equation of fractional order 
exists and is unique , and it is intended to obtain results related to the correct problem of 
determining the source function caught. 

Keywords: Caputo  derivative differential equation , exact problem, inverse problem, 
Cauchy issue , a fixed number. 

 

Let's say so. We are following 

                    (1.1) 
of a fractional equation in the Caputo sense 

, (1.2) 
the initial condition and the following  

, (1.3) 

, (1.4) 

a solution that satisfies the boundary conditions to find the issue let's see , here , – 

the given functions, – a constant number , – a fixed number,  through  In Caputo's 
sense - an ordered fraction is defined as an ordered derivative. 

 ( 1.1 ) - (1.4) is called a correct problem . 

3.1.1 - definition. If  function the following ,

  to the property have is , all conditions of (1.1) - (1.4). if 

satisfied , then this  to the function (1.1) - (1.4) of the problem the solution is called 
By finding a solution to this exact problem in the master 's thesis 
, the inverse problem of finding the source function is also studied.  

0 1r< <
2( , ) ( , ) = ( ), 0 , 0 < ;t xxD u x t a u x t f x x l t Tr - < < <

( , 0) = ( ), 0u x x x lj+ £ £

(0, ) 0, 0u t t T= £ £
( , ) 0, 0u l t t T= £ £

( )xj ( )f x
a T tD

r

r

( )( , ) [0, ] [0, ]u x t C l TÎ ´ ( , )tD u x tr

( )( , ) (0, ) (0,xxu x t C l TÎ ´
( , )u x t



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Let's assume that in problem (1.1) - (1.4) in addition to the function, 
the function is also unknown . To solve this problem we need an additional condition . We 
get the following condition as an additional condition : 

. (1.5) 

In this problem (1.1) - (1.5).  and to the problem of finding functions by finding 
the right side of the equation is called an inverse problem . 

the right problem 
The solution of the correct problem for the partial differential equation of order K is 

shown, that is, the solution of the correct problem (1.1) - (1.4) exists and is proved to be 
unique. 

To solve the problem (1.1) - (1.4), we prove the following theorem. 
Theorem 1.1. ,  functions continuous , fragmented - continuous to the 

derivative have and ,  conditions satisfactory be functions 
. _ Then the solution of problem (1.1) - (1.4) will be unique and it will look like this : 

 

. (1.6) 

Proof . Theorem to prove for private derivative equations in solving wide spread out of 
methods one o ' variables separation , that is _ From the Fourier method _ we use (1.1) – 
(1.4) problem solution 

 

for a view , here is a function 

               (1.7) 

                         (1.8) 

                        (1.9) 

                        (1.10) 

of the problem, and the function 

     (1.11) 

                       (1.12) 

(1.13) 

(1.14) 
the solution to the problem. 

To solve the problem (1.1) - (1.4), it is enough to solve the above two auxiliary 
problems. 

          As we have seen above, in this part, we will solve the problem (1.1) - (1.4) separately 
for two cases, homogeneous and non-homogeneous . 

We use the Fourier method to solve the problem (1.7) – (1.10). The solution 

( , )u x t ( )f x

( , ) = ( ), 0 <u x x Tt y t <
( , )u x t ( )f x

( )xj ( )f x
(0) ( ) 0lj j= = (0) ( ) 0f f l= =

2

,1
1

( , ) n
n

nau x t E t
l

r
r

pj
¥

=

é æ öæ ö= - +ê ç ÷ç ÷ç ÷è øê è øë
å

2

, 1 sinn
na nxf t E t
l l

r r
r r

p p
+

ùæ öæ ö+ - úç ÷ç ÷ç ÷è ø úè øû

( , ) ( , ) ( , )u x t v x t w x t= +
( , )v x t

2( , ) ( , ) = 0,t xxD v x t a v x tr - 0 , 0 < ;x l t T< < <
( , 0) = ( ), 0 ,v x x x lj+ £ £
(0, ) 0, 0 ,v t t T= £ £
( , ) 0, 0 .v l t t T= £ £

( , )w x t
2( , ) ( , ) = ( ),t xxD w x t a w x t f xr - 0 , 0 < ;x l t T< < <
( , 0) = 0, 0 ,w x x l+ £ £
(0, ) 0, 0 ,w t t T= £ £
( , ) 0, 0 .w l t t T= £ £



304			AMERICAN	Journal	of	Public	Diplomacy	and	International	Studies																														www.grnjournal.us		 
  

  

(1.15) 

we look for in the form , where - is a function of only a variable, and - is a 
function of only a variable. 

the solution in the form (1.15) into the equation (1.7) and form the following equation: 

 

Then this of equality two side  to being to send as a result , 

                                                    (1.16) 

we form the equation. For the function of the form (1.16) to be a solution of the equation (1.7) , 
from (1.16) consists of to be need , that is  manly of variables all in values must 
be appropriate . (1.13) is the left side of Eq only  to variable , right part only depends on the 
variable x . If we change something value choose  variable if we change (1.16) of 
equality right part and vice versa   of the variable something value choose variable if we 
change (1.16), the left part of the equation will be unchanged. This equality holds only if 
both sides of the equality are equal to a constant number. Therefore, the following equality 
holds: 

                   ( 1.17 ) 

here - is constant, and since there is no requirement for its sign, we take it with a minus sign 
for the convenience of further calculations. 

(1.17) from the equation   and   from scratch to identify different functions 

                   (1.18) 

                                           (1.19) 
we come to ordinary differential equations.  

(1.9) and ( 1.10 ), we have the following equality. 

 

 
we have equalities. Moreover, for the function since it is a nonzero function  

                                                                           (1.20) 

It follows that additional conditions are satisfied. Thus, to define a function, we formulate 

a simple eigenvalue problem: we need to find such values of the parameter as the result 

                                                         (1.21) 

the problem have a nontrivial solution. the characteristic value for such values of the 
parameter, and the corresponding nontrivial solution is called the characteristic function of 
the given problem . This given eigenvalue and eigenfunction problem is also called the 
Sturm–Liouville problem.  of the parameter We will consider the negative , zero and 
positive cases separately: 

( , ) ( ) ( ) 0v x t T t X x= × ¹
( )X x x ( )T t

t

2( ) ( ) ( ) ( )T t X x a T t X x¢ ¢¢=
( ) ( ) 0T t X x× ¹

2
( ) ( )

( )( )
T t X x

X xa T t
¢ ¢¢

=

0< < ;t T
t

x t
t x

2
( ) ( )

( )( )
T t X x

X xa T t
l

¢ ¢¢
= = -

l

( )X x ( )T t
2( ) ( ) 0T t a T tl¢ + =

( ) ( ) 0X x X xl¢¢ + =

(0, ) ( ) (0) 0v t T t X= × =
( , ) ( ) ( ) 0v l t T t X l= × =

( )T t ( )X x
(0) ( ) 0X X l= =

( )X x
l

( ) ( ) 0
(0) ( ) 0
X x X x
X X l

l¢¢ + =ì
í = =î

l

l



305			AMERICAN	Journal	of	Public	Diplomacy	and	International	Studies																														www.grnjournal.us		 
  

  

1) If  let it be Then the problem of eigenvalue and eigenfunction given by (1. 

21 ) will not have a non-zero solution. We will show it below. The solution  Let's 
search in the form , then we will have the following equalities 

                                 

 
and given of the matter the solution   in appearance will be Now 

using the boundary conditions 

 

we get the system and solve it: 

                                      

from   and   that we find So , if  if 

(3.1.18) problem  to the solution have it is 

2) If  let it be Then the given eigenvalue and eigenfunction problem (1.21) will 
not have a non-zero solution. If indeed _ _  to Eq have we will be , from this 

 to the solution we arrive at , if we use the boundary conditions 

Þ       and   from . 

So , if  if , (3.1.21 ) is the problem   to the solution have it is 

3) If  let it be Then the given eigenvalue and eigenfunction problem (1.21) will 

have a non-zero solution. If indeed if so , the solution  in appearance let's 
look for , then we will have the following equalities 

 
and  that account if we get , it is given of the matter the solution 

 in the form will be According to the boundary conditions 

 

0l <
( ) kxX x e=

2

2

0

0

kx kxk e e

k

k

l

l

l

+ =

+ =

= ± -

1 2( ) x xX x c e c el l- - -= +

1 2

1 2

(0) 0

( ) 0l l

X c c

X l c e c el l- - -

= + =ìï
í

= + =ïî

2 1

1 1 0l l

c c

c e c el l- - -

= -ìï
í

- =ïî

0   va     0l ll e el l- - -¹ - ¹ 1 0c = 2 0c = 0l <
( ) 0X x =
0l =

0l = ( ) 0X x¢¢ =

1 2( )X x c x c= +

1 2

1 2

(0) 0 0
( ) 0
X c c
X l c l c

= × + =ì
í = × + =î

2 0c = 0l ¹ 1 0c =

0l = ( ) 0X x =
0l >

0l > ( ) kxX x e=

2

2

0

0

kx kxk e e

k

k i

l

l

l

+ =

+ =

= ±

cos sini xe x i xl l l= ±

( ) cos sinX x A x B xl l= +

( ) ( )
( ) ( )

(0) cos 0 sin 0 0

( ) cos sin 0

X A B A

X l A l B l

l l

l l

ì = × + × = =ï
í

= + =ïî



306			AMERICAN	Journal	of	Public	Diplomacy	and	International	Studies																														www.grnjournal.us		 
  

  

will be From this  will be , to be for  to be need _ In 

that case  or , this on the ground because n= 1,2,3,…  and  

positive numbers . So , the given problem is non-trivial to the solution  

special only in values have and it will be as follows 

                                              (1.22) 

The solution is if we choose an arbitrary constant coefficient as one 

                                               ( 1.23) 

will appear. 

Now and , characteristic to value suitable special function  s for 

the following expressions we find : 

                                (1.24) 

If we transfer the expression (1.24) to the problem (1.7), the following equality is formed: 

. 

From this, 

 

we form the equation. So we come to the following issue: 

                              (1.25) 

(3.1.25) The solution of the Cauchy problem is, by virtue of (2.2.14), the following (see 
[Kilbas]): 

                          . (1.26) 

Since the sum of particular solutions is a solution 

 

function is also a solution. So (1.7) – (1.10) is a formal solution of the problem 

( )( ) sin 0X l B l l= = ( ) 0X x ¹ 0B ¹

( )sin 0l l =
l
npl = l l

2

n
n
l
pl l æ ö= = ç ÷
è ø

( ) sinn n
nxX x B
l

p
=

( ) sinn
nxX x
l

p
=

2

n
n
l
pl æ ö= ç ÷
è ø

( )nT t

1
( , ) ( ) sinn

n

nxv x t T t
l

p¥

=
= ×å

2
2

1 1
( ) sin ( ) sin = 0t n n

n n

nx n nxD T t a T t
l l l

r p p p¥ ¥

= =

æ ö× + ×ç ÷
è ø

å å

2
2

1
( ) ( ) sin = 0t n n

n

n nxD T t a T t
l l

r p p¥

=

é ùæ ö+ ×ê úç ÷
è øê úë û

å

2
2( ) ( ) 0,

( 0) ,

t n n

n n

nD T t a T t
l

T

r p

j

ì æ ö+ =ï ç ÷í è ø
ï + =î

2

,1( )n n
naT t E t
l

r
r

pj
æ öæ ö= -ç ÷ç ÷ç ÷è øè ø

1
( , ) ( ) ( )n n

n
v x t T t X t

¥

=
= ×å



307			AMERICAN	Journal	of	Public	Diplomacy	and	International	Studies																														www.grnjournal.us		 
  

  

                          ( 1.27) 

will appear. 
Now we show that this series is flat convergent. For this, the partial sum of the series 

(1.27). 

 

we define as 
If the Mittag–Leffler function 

 

from the estimate, (1.27) results in smooth convergence of the series. 
 of flat approachability  

 

at 

 

price and  of the function properties soon come comes out 

In addition, it follows that the equality (1.11) 

, So, it follows from the above considerations that the 
function (1.27) is a solution to the problem (1.11) - (1.14). 

In addition, it follows that the equality (1.7) 

, So, it follows from the above considerations that the 
function (1.27) is a solution of the problem (1.7) - (1.10). 

Now let's look at the non-homogeneous case. (1.11) – (1.14) and Fourier to solve the 
problem method we use , that is  function  

 

apparently _ _ we are looking for Him (1.11) to Eq take go let 's put and  to simplify the 
following equality harvest we do : 

. 

initial condition account if we get   condition harvest we do So the following 

2

,1
1

( , ) sinn
n

na nxv x t E t
l l

r
r

p pj
¥

=

æ öæ ö= -ç ÷ç ÷ç ÷è øè ø
å

2

,1
1

( , ) sin
j

j n
n

na nxV x t E t
l l

r
r

p pj
=

æ öæ ö= -ç ÷ç ÷ç ÷è øè ø
å

,
1| ( ) |
1

E z
zr µ - £

+

( , )xxv x t
2 22

,12
1

( , ) sinj n
n

n na nxV x t E t
x l l l

r
r

p p pj
¥

=

æ ö¶ æ ö æ ö= -ç ÷ç ÷ ç ÷ç ÷¶ è ø è øè ø
å

2 2

,1 1n naE t
l l

r
r

p pæ öæ ö æ ö- £ç ÷ç ÷ ç ÷ç ÷è ø è øè ø
( )xj

2( , ) ( , ) = 0t xxD v x t a v x tr -

( , ) ((0, ) (0, ))tD v x t C l Tr Î ´ > 0t

2

2( , ) = ( , )t j jD V x t V x t
x

r ¶
¶

( , ) ((0, ) (0, ))tD v x t C l Tr Î ´ > 0t

( , )w x t

1
( , ) ( ) sinn

n

nxw x t T t
l

p¥

=
= ×å

2
2( ) ( ) = ( )t n n n

nD T t a T t f x
l

r pæ ö+ ç ÷
è ø

( 0) 0nT + =



308			AMERICAN	Journal	of	Public	Diplomacy	and	International	Studies																														www.grnjournal.us		 
  

  

(1.28) 

let's get to the point. Solving it, we get the following formal solution (see [Kielbas]) 

. (1.29) 

If  function  Given that does not depend on , then 

 

we form the equation. If 

 

taking into account the situation, then we obtain the following solution for the problem (3.1.28): 

. 

Thus, we have the following formal solution to problem (1.11)–(1.14): 

. (1.30) 

Now we show that this series is flat convergent. For this, the partial sum of the series (1.30). 

 

we define as If the Mittag–Leffler function 

 

from the estimate, (1.30) results in smooth convergence of the series.  of flat 
approachability  

 

at 

 

price and  of the function properties soon come comes out 

2
2( ) ( ) = ( ),

( 0) 0,

t n n n

n

nD T t a T t f x
l

T

r pì æ ö+ï ç ÷í è ø
ï + =î

( )1
,

0

( ) (t )
t

n n nT t E f dr r
r rh l h h h-= - -ò

( )f x t

( ) ( )1 1
, ,

0 0

(t )
t t

n n n nE f d f E dr r r r
r r r rh l h h h h l h h- -- - = -ò ò

( ) ( )1
, , 1

0

t

n nE d t E tr r r r
r r r rh l h h l-

+- = -ò

2

, 1( )n n
naT t f t E t
l

r r
r r

p
+

æ öæ ö= -ç ÷ç ÷ç ÷è øè ø

2

, 1
1

( , ) sinn
n

na nxw x t f t E t
l l

r r
r r

p p¥

+
=

æ öæ ö= - ×ç ÷ç ÷ç ÷è øè ø
å

2

, 1
1

( , ) sin
j

j n
n

na nxW x t f t E t
l l

r r
r r

p p
+

=

æ öæ ö= - ×ç ÷ç ÷ç ÷è øè ø
å

,
1| ( ) |
1

E z
zr µ - £

+
( , )xxw x t

2 22

, 12
1

( , ) sin
j

j n
n

n na nxW x t f t E t
x l l l

r r
r r

p p p
+

=

æ ö¶ æ ö æ ö= - ×ç ÷ç ÷ ç ÷ç ÷¶ è ø è øè ø
å

2 2

, 1 1n naE t
l l

r
r r

p p
+

æ öæ ö æ ö- £ç ÷ç ÷ ç ÷ç ÷è ø è øè ø
( )f x



309			AMERICAN	Journal	of	Public	Diplomacy	and	International	Studies																														www.grnjournal.us		 
  

  

Moreover, equality (1.11)  it 

follows that from So, it follows from the above 
considerations that the function (1.30) is a solution to the problem (1.11) - (1.14). 

Summarizing these solutions, we get the following solution for problem (1.7) - (1.10) 
: 

     

(1.31) 

Thus , by the formula (1.6) . determined  function (1.7) - (1.10) will be the 
solution of the problem . 

 
 

 
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2

2
=1

( , ) = ( , ) ( )sin ,
j

t j j k
k

nxD W x t W x t f t
lx

r p¶
+

¶
å > 0t

( , ) ((0, ) (0, ))tD w x t C l Tr Î ´

2

,1
1

( , ) ( , ) ( , ) sinn
n

na nxu x t v x t w x t E t
l l

r
r

p pj
¥

=

æ öæ ö= + = - +ç ÷ç ÷ç ÷è øè ø
å

2

, 1
1

sin .n
n

na nxf t E t
l l

r r
r r

p p¥

+
=

æ öæ ö+ - ×ç ÷ç ÷ç ÷è øè ø
å

( , )u x t



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