# Two Dimensional Kamal Transform of Space of Distribution of Boehmians DOI : http://dx.doi.org/10.17577/IJERTV9IS040696 Text Only Version

#### Two Dimensional Kamal Transform of Space of Distribution of Boehmians

*Arun J. Ghanwat

MESs, Shri Dnyaneshwar Mahavidyalaya Newasa, Dist. Ahmednagar

P. G. Department of Mathematics, New Arts, Commerce and Science College, Ahmednagar (Maharashtra).

Abstract:- In this paper we have tried to define two dimensional Kamal transform namely the double Kamal transform of bounded support with the help of Zemanian technique. The integrable Boehmian of double Kamal transform is developed using Mikusinkis theory with some of the properties. The double Kamal transform is extended to generalized function, some properties n distribution sense are obtained and tried to extend this class to integrable Boehmians.

Keywards: Convolution theorem, Double Kamal transform , Kamal Transform, Integrable Boehmian.

INTRODUCTION

In the field of engineering the ordinary and partial differential equations plays vital role. To solve the differential equation integral transforms are used suitably to convert into simpler form and obtain the solutions easily. The double Natural transform are studied in [2-4],  deals with applications of [2-4] where as  gives the applications of Kamal transform for single variable functions. The two dimensional Kamal transform of function of two variables can be used to solve partial differential equations.

Defintion 1: The Kamal transform for a single variable function of exponential order in the set

||

1

1

2

2

S = {(): , 1, 2 > 0, |()| < , (1)[0, )}; where the constant M is finite number, , may be finite

or infinite is defined by 

0

0

K [f(t)] = G(v) =

(), 0, 1 2.

1 + 1

The inverse Kamal transform is () = ( ) , 0.

2

Definition 2: The two dimensional Kamal transform namely the Double Kamal transform of a function (, ), , + of two

(+)

variables is defined by  as K2{(, ); (, )}= G(, )=0 0

(, ), provided the integral exists and (, )

can be expressed in the convergent series on a set A of functions continuous and of exponential order.

(|| || )

A = {: |(, )| <

+

, (1)x [0, ), = 1, 2; , ,

> 0 }; where the constant M should be finite

number, , may be finite or infinite and v is a variable of transform.

Let , be integrable functions of two variables then by using  the double convolution of , is given by (

)(, )= (, )( , ).

0 0

DOUBLE KAMAL TRANSFORM OF BOUNDED SUPPORT

Let J be compact subset of I = (0, ). denote the test function space of all complex valued infinitely smooth function

spaces of E (I x I) such that

+

, for k, m are non-zero positive integers. Then E (I x I) is a vector

| (, )| <

(,)

=

=

space under the usual operations. Also D (I x I) E (I x I). A sequence {} 1 converges in E (I x I) and for each non-negative

+

k, m, (

+

) converges to (

) uniformly on every compact subset I x I.

The space E (I x I) is complete and E(I x I) dual of E(I x I) consists of distribution of compact support. By the definition of

(+)

+

(+)

double Kamal transform the Kernel

positive real numbers u, v.

is smooth and satisfies | (

(,)

)| <

1. for all

Hence for every f E(I x I) we define the distributional double Kamal transform of compact support with the help of

(+)

Zemanian  as ((, ), (, ))= (, ), for (, ) IxI (2)

Theorem 1: The distributional double Kamal transform is linear.

Proof: For any constants c, d we have

(+)

(+)

(+)

((, ) + (, )) = (, ) + (, ),

= (, ),

+ (, ),

(+) (+)

= (, ),

+ (, ), = c ((, ))+d ((, )).

Theorem 2: Let h be distribution in E(I x I) and let be defined by

(, ) = { ( 1, 2 ), 1 , 2 0 , < 1 , < 2

(1

2)

Then (, ) =

+

[h(x, t)]

Proof: Here (, ) is a member of E(I x I), by translation property

(+)

(1+2)

1

2

(, ) =< ( 1, 2 ),

> = (, ),

= (, ), . . .

1 2

1 2

=

( + )

(, ),

( + )

=

( + )

[h(x, t)].

Theorem 3: Let (, ) E(I x I) and [h(x, t)] be the distributional double Kamal transform then

(+)

(+)

th th

[(x, t)] = (, ),

derivatives w. r. t. x & t.

and

[(x, t)] = (, ),

; where

and

are k

and n

Theorem 4: Let (, ) E(I x I) and [(x, t)] be the distributional double Kamal transform then

((, )) [(, )] = [(, ), (

1+

, )]

1+

Proof:

[(, )](, ) = (, ),

(+)

(

(, )) [(, )]=

(, ),

(+)

(+)

(1+)

(1+)

= (, ),

= (, ),

= (, ), .

= [(, ), ( ,

)].

1+ 1+

Theorem 5: Let (, ) E(I x I) and [(x, t)] be the distributional double Kamal transform of (, ) then

[(, )] = 1 [(, ) (1 . 1) (, )].

(+)

1 1 (+) 1

(+)

Proof: [(, )] = (, ),

= (, ), .

=

(, ),

= 1 [(, ) (1 . 1) (, )].

Theorem 6: Let (, ) E(I x I) and [(x, t)] be the distributional double Kamal transform of (, ) then (1) [(, )](, ) = 2 [(, )] + [(, )](, )

(2) [(, )](, ) = 2 [(, )] + [(, )](, )

Proof: Let (, ) E(I x I). Then by theorem 3 and equation (2) we have

[(, )] =

((, ),

(+)

=

=

)

(+)(1)

( + )

( + )( 2)

1

( + ) 1

( + )

(, ),

(

) = (, ),

= 2 (, ),

(, ),

= 1

2

2 [(, )] = [(, )](, ) [(, )](, )

[(, )](, ) = 2 [(, )] + [(, )](, )

Similarly we can prove the part (2).

[(, )](, ) 1 [(, )](, )

Theorem 7: Let (, ) E(I x I) and [(x, t)] be the distributional double Kamal transform of (, ) then {( ,

(

)

)

). ( , )} =

( , ) =

1, > , >

{

0, < , <

Proof: By using the definition of distributional double Kamal transform

{( , ). ( , )} = ( , ). ( , ,

(+)

For > , >

(+

+

)

{( , ). ( , )} = ( , ),

+

( + ) ( + )

=

( , ),

( + )

= {( , )}(, ).

CONVOLUTION OF DOUBLE KAMAL TRANSFORM

Let (, ), (, ) be distributions of compact support in E(I x I). We define the convolution of (, ), (, ) as (

)(, ), (, ) = (, ), (1, 2), ( + 1, + 2; (3) where E(I x I) if (1, 2), ( + 1, + 2 E (I x I).

Theorem 8: Let (, ) E(I x I) and E(I x I). If (, ) = (1, 2), ( + 1, + 2; where , =1,2,3,. Then is

+

infinitely differentiable and

+

(()) = ( , ),

(( + , + )

This result is proved in , Pp-130.

1 2 1 2

Theorem 9: (The Convolution theorem):- If [(, )] , [(, )] are distributional transforms of (, ) , (, )

respectively then [( )(, )] = [(, )]. [(, )]

(+)

+ +

1 2

1 2

( + )

Proof: [( )(, )] = ( )(, ),

= (, ), (1, 2),

1

2

1

2

= (, ), (1, 2),

(+)

= (, ), (1, 2),

1 2

1 2

( + )

.

=(, ),

. (1, 2),

= [(, )]. [(, )].

INTEGRABLE BOEHMIAN FOR DOUBLE KAMAL TRANSFORM

Let 1be the space of Lebesgue integrable functions on +; + is the set of positive real numbers then by , a sequence

=

=

{} 1 of continuous real valued functions over +x + is called a delta sequence if and only if

(1) (, ) = 1

0 0

2. | | < K, for some positive K and all .

0 0

(3) |(,)|<|(, )| 0 as for every > 0.

The space of all integrable Boehmians is denoted by 11, which is a convolution algebra with (1) [ ] = []

(2) [ ] + [] = ()+()

()

(3) [ ] [] = [ ]

(+)

Lemma: If [

] 11 then the sequence K[(, )] = 0

0

(, ) converges to each compact set x in

+x +.

Proof: We have {} 1 then its double Kamal transform () = converges uniformly to the function of (u, v). Hence for

=

each J x J, is positive on J x J and

K[ (, )] = K[ (, ) ] (4)

= on J x J i.e. K K as on J x J. (1 as ).

Now, we will define the double Kamal transform of an integrable Boehmians as

[ ] = lim

(5); where the limit is over compact subsets of x .

+ +

Thus, we get a continuous function of double Kamal transform of an integrable Boehmians.

Theorem 10: The double Kamal transform of an integrable Boehmians is well defined.

1 2

1 2

Proof: Let = [ ], = [

] 11 such that 1 = 2. Then

[ ] = []

=

, , . Appling double Kamal transform

( ) = (m n) = (n m)

From equation (5) and theorem (9)

lim

= lim

[ ] = [].

n

n

Theorem 11: The double Kamal transform of an integrable Boehmian is linear on compact sets.

Proof: Let , 11 be such that

= [ ], = [] then + = ()+()

()

( + ) = lim ( ) + lim ( )

= lim () + lim ()

= ()+ ().

Also, (c) = ] = lim

[

= c (); where c and = [ ].

Theorem 12: (Convolution theorem for integrable Boehmians)

Let [

] , [

] 11 then ([

] [

]) = ([

]) . ([

])

Proof:

([ ] []) = [()] = lim (

) = lim (( ). ( ))

()

= lim ( ). lim ( ) = ([ ]) . ([]) .

Theorem 13: If = [ ] 11 and () = 0 in then = 0 in 11 .

[

[

Proof: Let =

] and () = 0 then by the definition of , we have

lim hn = 0 on compact sets, again by definition of Kamal transform 0 a.e. in 11 and

Equivalently = [] is zero equivalence class in 1 1 .

is zero quotient of function.

Theorem 14: The generalized double Kamal transform is one-one mapping from 11 into space of continuous functions.

Proof: Let (1) = (2), 1, 2 11 . (1 2) = 0 in , theorem (13)

1 2 = 0 1 = 2.

Theorem 15: The generalized double Kamal transform : 11 is continuous with respect to – converence.

Proof: Let as in 11

Then the – convergence (( ) ) = [

] 0 , for some 11 , and 0 as .

{( ) } = [] = lim (

) = lim

0 as in .

As 0, () – () 0 as ()() as .

RESULTS

We have tried to obtain double Kamal transform of bounded support. The linearity property, Convolution is studied in distributional view and also Boehmian spaces are studied. The convergence and are defined in double Kamal transformation.

CONCLUSIONS

In this paper, the two dimensional Kamal transform namely double Kamal transform of bounded support is obtained. Also the extended double Kamal transform of distributional space along with some of its properties is studied. The convolution of double Kamal transform of Boehmians and extension of it to integrable Boehmians is derived.

REFERENCES

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