# Applications of Karry-Kalim-Adnan Transformation (KKAT) to Mechanics and Electrical Circuits

Authors: Karry Iqbal; Muhammad Kalim; Adnan Khan
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, we have used the new integral transformation known as Karry-Kalim-Adnan transformation (KKAT) to solve the ordinary linear differential equations as well as partial differential equations. We have used KKAT to solve the problems of engineering and sciences specially electrical and mechanical problems. Also, we have established the comparison between KKAT and existing transformations. We have determined the KKAT of error functions and complementary error function. The fundamental purpose of this research paper is to transform the given problem into the easier form to get its solution.


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## Body

## 1. Introduction

Integral transformation is a very important tool to solve the differential equations for engineers and scientists. Integral transforms are widely used to determine the solution of differential equations with the initial values. Integral transforms have played a great role in engineering, chemistry, biology, astronomy, social sciences, radio physics, and nuclear science. KKAT is closely related to Laplace transformation [1, 2], Fourier transformation [3], Sumudu transformation [4], Elzaki transformation [5], Aboodh transformation [6], Mahgoub transformation [7], etc. Mohand and Laplace transformations [8] were studied and used by Aggarwal and Chaudhary to solve the system of advanced differential equations. Sedeeg has used the Kamal transformation [9], to solve the linear ordinary differential equations having constant coefficients. Aggarwal et al. [10] determined the Elzaki transformation of error function.In general integral transformation, we take a set of functions offx in the region −∞<a≤x≤b<∞, and then, we select a field function κx,β. So the integral transformation [4] is defined by
(1)Tfx=Fβ=∫abfxκx,βdx,where the function κx,β is known as the kernel of the transformation T and β is a parameter. There are different kinds of transformations depending upon the kernel and the limits a and b.This field of study is very important for the researchers. In classical analysis, some special types of the integral transformations as mentioned above have been extensively studied and used in solving several problems of engineering and mathematical sciences. Gupta [11] has used Rohit transform to analyze the boundary value problems of science and engineering. All transformations correspond to different types of kernel κx,β. The limits of the integral are also different in each case.Historically, Pierre Simon de Laplace (1749-1827) [12] initiated the integral transformations. Joseph Fourier (1768-1830) used the modern mathematical theory on heat conduction, Fourier series, integral transforms, and inverse Fourier transformation.


### 1.1. Laplace Transformation

The Laplace transformation [1] for fx, x≥0, is given by
(2)Lfx=∫0∞e−βxfxdx=Aβ.The parameterβ belongs to real or complex plane.


### 1.2. Elzaki Transformation

The Elzaki transformation [1, 5] for the function fx, x≥0, is given by
(3)Efx=β∫0∞e−x/βfxdx=Cβ,β∈−β1β2,β1,β2>0.


### 1.3. Sumudu Transformation

The Sumudu transformation [4] for the function fx, x≥0, is given by
(4)Sfx=∫0∞e−xfβxdx=Dβ,0<β1≤β≤β2.


### 1.4. Rohit Transformation

The Rohit transformation [11] for the function fx, x≥0, is given by
(5)Rfx=β3∫0∞e−βxfxdx=Iβ.The parameterβ belongs to real or complex plane.The Kernels of well-known transformations are given in Table1.Table 1
Kernel of well-known transformations.
Sr. no.TransformationKernel1Laplace transformation [1]κx,β=e−βx2Sumudu transformation [4]κx,β=1/βe−1/βx3Elzaki transformation [1, 5]κx,β=βe−1/βx4Rohit transformation [11]κx,β=β3e−βx5Karry Kalim Adnan Transformationκx,β/α=1/αβe−β/αx


## 1.1. Laplace Transformation

The Laplace transformation [1] for fx, x≥0, is given by
(2)Lfx=∫0∞e−βxfxdx=Aβ.The parameterβ belongs to real or complex plane.


## 1.2. Elzaki Transformation

The Elzaki transformation [1, 5] for the function fx, x≥0, is given by
(3)Efx=β∫0∞e−x/βfxdx=Cβ,β∈−β1β2,β1,β2>0.


## 1.3. Sumudu Transformation

The Sumudu transformation [4] for the function fx, x≥0, is given by
(4)Sfx=∫0∞e−xfβxdx=Dβ,0<β1≤β≤β2.


## 1.4. Rohit Transformation

The Rohit transformation [11] for the function fx, x≥0, is given by
(5)Rfx=β3∫0∞e−βxfxdx=Iβ.The parameterβ belongs to real or complex plane.The Kernels of well-known transformations are given in Table1.Table 1
Kernel of well-known transformations.
Sr. no.TransformationKernel1Laplace transformation [1]κx,β=e−βx2Sumudu transformation [4]κx,β=1/βe−1/βx3Elzaki transformation [1, 5]κx,β=βe−1/βx4Rohit transformation [11]κx,β=β3e−βx5Karry Kalim Adnan Transformationκx,β/α=1/αβe−β/αx


## 2. Definition of KKAT

A transformation defined for function of exponential order from setS (see [4, 5])
(6)S=fx:∃P;a1,a2>0,fx<Pex/ai,x∈=1i×0,∞,where constant P is a finite number and a1,a2 may be finite or infinite.KKAT is represented by operatorK. and is defined by
(7)Kfx=1β∫0∞fαxe−βxdx,x≥0;α,β∈a1a2,where α and β are constants and α,β≠0.Kfx can also be written in the form
(8)Kfx=1αβ∫0∞fxe−β/αxdx=Fβα.Also(9)fx=K−1Fβα.


### 2.1. Some Useful Results of KKAT

Result 1.
If(10)fx=1,then
(11)K1=1αβ∫0∞e−β/αxdx.
By evaluating integral, we get(12)K1=1β2.Result 2.
If(13)fx=xneλx,then
(14)Kxneλx=1αβ∫0∞xne−β/α−λxdx.
We assume that(15)βα−λx=t⇒dx=dtβ/α−λ.
Thus, we get(16)Kxneλx=1αββ/α−λn+1∫0∞tne−tdt.
Hence,(17)Kxneλx=Γn+1αnββ−λαn+1,forn>−1.Result 3.
If(18)fx=sinλx,then
(19)Ksinλx=1αβ∫0∞e−β/αxsinλxdx.
By using(20)∫eaxsinbxdx=eaxa2+b2asinbx−bcosbx,1αβ∫0∞e−β/αxsinλxdx=1αβe−β/αx−β/α2+λ2−βαsinλx−λcosλx0∞.
Hence, we get(21)Ksinλx=λαββ2+λ2α2.Thus, the KKAT of some important functions is given in Table2 as follows.Table 2
KKAT of some important functions.
Sr. no.fxFβ/α111/β22xn,n∈Nn!αn/βn+23xneλx,n∈Nn!αn/ββ−λαn+14xn,n>−1Γn+1αn/βn+25xneλx,n>−1Γn+1αn/ββ−λαn+16eλx1/ββ−λα7sinλxλα/ββ2+λ2α28cosλx1/β2+λ2α29sinhxα/ββ2−α210coshλx1/β2−λ2α211eλxcoshxβ−λα/ββ−λα2−α212eλxsinhxα/ββ−λα2−α213sinλx+cosλxλα+β/ββ2+λ2α214f′xβ/αKfx−f0/αβ15f′′xβ2/α2Kfx−f′0/αβ−f0/α216eλxfxβ−λα/βFα,β−λα17xfx−αd/dβFα,β−α/βKfx18xf′x−αd/dββ/αKfx−f0/αβ−Kfx+f0/β2


## 2.1. Some Useful Results of KKAT

Result 1.
If(10)fx=1,then
(11)K1=1αβ∫0∞e−β/αxdx.
By evaluating integral, we get(12)K1=1β2.Result 2.
If(13)fx=xneλx,then
(14)Kxneλx=1αβ∫0∞xne−β/α−λxdx.
We assume that(15)βα−λx=t⇒dx=dtβ/α−λ.
Thus, we get(16)Kxneλx=1αββ/α−λn+1∫0∞tne−tdt.
Hence,(17)Kxneλx=Γn+1αnββ−λαn+1,forn>−1.Result 3.
If(18)fx=sinλx,then
(19)Ksinλx=1αβ∫0∞e−β/αxsinλxdx.
By using(20)∫eaxsinbxdx=eaxa2+b2asinbx−bcosbx,1αβ∫0∞e−β/αxsinλxdx=1αβe−β/αx−β/α2+λ2−βαsinλx−λcosλx0∞.
Hence, we get(21)Ksinλx=λαββ2+λ2α2.Thus, the KKAT of some important functions is given in Table2 as follows.Table 2
KKAT of some important functions.
Sr. no.fxFβ/α111/β22xn,n∈Nn!αn/βn+23xneλx,n∈Nn!αn/ββ−λαn+14xn,n>−1Γn+1αn/βn+25xneλx,n>−1Γn+1αn/ββ−λαn+16eλx1/ββ−λα7sinλxλα/ββ2+λ2α28cosλx1/β2+λ2α29sinhxα/ββ2−α210coshλx1/β2−λ2α211eλxcoshxβ−λα/ββ−λα2−α212eλxsinhxα/ββ−λα2−α213sinλx+cosλxλα+β/ββ2+λ2α214f′xβ/αKfx−f0/αβ15f′′xβ2/α2Kfx−f′0/αβ−f0/α216eλxfxβ−λα/βFα,β−λα17xfx−αd/dβFα,β−α/βKfx18xf′x−αd/dββ/αKfx−f0/αβ−Kfx+f0/β2


## 3. Linearity Property of KKAT

IfG1β/α,G2β/α are KKAT of g1x,g2x∈A, then KKAT of ag1x+bg2x is aG1β/α+bG2β/α, where a,b are any constant.Proof.
By definition of KKAT from equation (8), we get
(22)Kgx=Gβα=1αβ∫0∞e−β/αxgxdx,Kag1x+bg2x=1αβ∫0∞e−β/αxag1x+bg2xdx=a1αβ∫0∞e−β/αxg1xdx+b1αβ∫0∞e−β/αxg2xdx.
Hence, we obtain(23)Kag1x+bg2x=aG1βα+bG2βα.


## 4. Inverse of KKAT

IfFβ/α be the KKAT of fx, then fx is called the inverse KKAT of Fβ/α. The inverse KKAT is expressed in equation (9)
(24)K−1Fβα=fx.


### 4.1. KKAT Laplace Duality and KKAT Inversion Integral

IfFs and Fβ/α are Laplace transform and KKAT of function fx, respectively, then αβFβ/α=Fs by assuming β/α=s.Iffx be the inverse KKAT of Fβ/α and all the singularities of Fβ/α in the complex plane s=β/α=x+iy lie to the left of the line x=x°, then
(25)fx=12πilimx∘⟶∞∫x∘−iRx∘+iRβeβ/αxFβαdβ,fx=∑jRj,where Rj is residue of βeβ/αxFβ/α at the poles β=βj.


### 4.2. Verification of Inverse Function

To verify the inverse of KKAT, the following examples are given.Example 1.
IfK1=Fβ/α=1/β2, then K−11/β2=1, where fx=1 and Fβ/α is defined in equation (8).Proof.
If(26)Fβα=1β2,then the function βeβ/αxFβ/α=eβ/αx/β has simple pole at β=0.
Thus, we have(27)ft=K−11β2=R0,where
(28)R0=residueofβeβ/αxFβαatβ=0,also
(29)R0=limβ⟶0β2eβ/αxβ2,thus
(30)R0=1,therefore
(31)fx=K−11β2=1.Example 2.
If(32)Keλx=Fβα=1ββ−λα,then
(33)K−11ββ−λα=eλx.Proof.
If(34)Fβα=1ββ−λα,then the function βeβ/αxFβ/α=eβ/αx/β−λα has simple pole at β=λα,thus
(35)ft=K−1Fβα=R0,where
(36)R0=residueofβeβ/αxFβα,atβ=λα,also
(37)R0=limβ⟶λαβ−λαeβ/αxβ−λα,hence
(38)R0=eλx,therefore
(39)fx=eλx.Example 3.
IfKsinλx=Fβ/α=λα/ββ2+λ2α2, then K−1λα/ββ2+λ2α2=sinλx.Proof.
SinceFβ/α=λα/ββ2+λ2α2, then the function βeβ/αxFβ/α=λαeβ/αx/β2+λ2α2 has simple pole at β=±λαi; thus,
(40)ft=K−1Fβα=R0+R1,where
(41)R0=residueofβeβ/αxFβαatβ=λαi,also
(42)R0=limβ⟶λαiβ−λαiλαeβ/αxβ−λαiβ+λαi,thus
(43)R0=eiλx2i.
And(44)R1=residueofβeβ/αxFβαatβ=−λαi,thus
(45)R1=−e−iλx2i,hence
(46)ft=R0+R1=eiλx2i−e−iλx2i=sinλx,therefore
(47)ft=sinλx.The inverse of KKAT is given in Table3.Table 3
Inverse KKAT.
Sr. no.K−1Fβ/αfx1K−11/β212K−1n!αn/βn+2xn3K−11/ββ−λαeλx4K−1λα/ββ2+λ2α2sinλx5K−11/β2+λ2α2cosλx6K−1α/ββ2−α2sinhx7K−11/β2−α2coshx


## 4.1. KKAT Laplace Duality and KKAT Inversion Integral

IfFs and Fβ/α are Laplace transform and KKAT of function fx, respectively, then αβFβ/α=Fs by assuming β/α=s.Iffx be the inverse KKAT of Fβ/α and all the singularities of Fβ/α in the complex plane s=β/α=x+iy lie to the left of the line x=x°, then
(25)fx=12πilimx∘⟶∞∫x∘−iRx∘+iRβeβ/αxFβαdβ,fx=∑jRj,where Rj is residue of βeβ/αxFβ/α at the poles β=βj.


## 4.2. Verification of Inverse Function

To verify the inverse of KKAT, the following examples are given.Example 1.
IfK1=Fβ/α=1/β2, then K−11/β2=1, where fx=1 and Fβ/α is defined in equation (8).Proof.
If(26)Fβα=1β2,then the function βeβ/αxFβ/α=eβ/αx/β has simple pole at β=0.
Thus, we have(27)ft=K−11β2=R0,where
(28)R0=residueofβeβ/αxFβαatβ=0,also
(29)R0=limβ⟶0β2eβ/αxβ2,thus
(30)R0=1,therefore
(31)fx=K−11β2=1.Example 2.
If(32)Keλx=Fβα=1ββ−λα,then
(33)K−11ββ−λα=eλx.Proof.
If(34)Fβα=1ββ−λα,then the function βeβ/αxFβ/α=eβ/αx/β−λα has simple pole at β=λα,thus
(35)ft=K−1Fβα=R0,where
(36)R0=residueofβeβ/αxFβα,atβ=λα,also
(37)R0=limβ⟶λαβ−λαeβ/αxβ−λα,hence
(38)R0=eλx,therefore
(39)fx=eλx.Example 3.
IfKsinλx=Fβ/α=λα/ββ2+λ2α2, then K−1λα/ββ2+λ2α2=sinλx.Proof.
SinceFβ/α=λα/ββ2+λ2α2, then the function βeβ/αxFβ/α=λαeβ/αx/β2+λ2α2 has simple pole at β=±λαi; thus,
(40)ft=K−1Fβα=R0+R1,where
(41)R0=residueofβeβ/αxFβαatβ=λαi,also
(42)R0=limβ⟶λαiβ−λαiλαeβ/αxβ−λαiβ+λαi,thus
(43)R0=eiλx2i.
And(44)R1=residueofβeβ/αxFβαatβ=−λαi,thus
(45)R1=−e−iλx2i,hence
(46)ft=R0+R1=eiλx2i−e−iλx2i=sinλx,therefore
(47)ft=sinλx.The inverse of KKAT is given in Table3.Table 3
Inverse KKAT.
Sr. no.K−1Fβ/αfx1K−11/β212K−1n!αn/βn+2xn3K−11/ββ−λαeλx4K−1λα/ββ2+λ2α2sinλx5K−11/β2+λ2α2cosλx6K−1α/ββ2−α2sinhx7K−11/β2−α2coshx


## 5. Application of KKAT on the Integral Function

Ifgx=∫0xfzdz, then K∫0xfzdz=α/βFβ/α.Proof.
It is given that(48)gx=∫0xfzdz.
By using fundamental theorem of calculus,(49)g′x=fx.
We know that(50)Kg′x=βαKgx−g0αβ.
Butg0=0 and g′x=fx,thus
(51)Kfx=βαKgx,hence
(52)K∫0xfzdz=αβFβα.


## 6. Application of KKAT on Derivatives

LetKfx be KKAT of fx; then,
(53)Kf′x=βαKfx−f0αβ,Kf′′x=β/αKf′x−f′0/αβ,Kfnx=β/αKfn−1x−fn−10/αβ.


## 7. Applications of KKAT to Mechanics

### 7.1. Application No. 1

Let a bodyA of mass 1 gram moves on x-axis. It is attracted towards the origin O with a force equals to 4x. Also, assume that initially it is at rest when x=5; then, determine its position by considering
(i)
no any other forces acting on it(ii)
damping force or in other words resistance to the particle is equal to the 8 times the velocity at any instant [12]Solution: see Figure1.Figure 1From Figure1, for x>0, the net force towards left is given by −4x, while for x<0, the net force towards right is given by −4x. Thus, for both cases, the net force is equal to −4x.By Newton’s second law of motion,(54)mass×acceleration=netforce,d2xdt2=−4x,d2xdt2+4x=0.The initial conditions are(55)x0=5,x′0=0.Applying KKAT and using initial conditions on equation (54), we get
(56)Kx″+4Kx=0,β2α2+4Kx−5α2=0.Now, by using inverse KKAT on equation (56), we obtain
(57)Xt=5cos2t.


### 7.2. Application No. 2

The relationship between resistive force of air and the velocity of a freely falling body is given bydv/dt=g−λv, where vt is the velocity at any time t. Initially, the body is at rest [11].Solution: see Figure2.Figure 2The equation for the motion of body moving under constant gravitational acceleration (as shown in Figure2) is
(58)dvdt=g−λv.Applying KKAT on equation (58), we get
(59)Kdvdt=Kg−λv,βαKvt−v0αβ=gβ2−λKv,βαKvt=gβ2−λKv,βα+λKv=gβ2,Kv=gβ21λ−β/λβ+λα=gβ21λ−βλβ+λα,Kv=gλ1β2−1ββ+λα.Applying inverse KKAT both sides on equation (59), we get
(60)vt=gλ1−e−λt.


### 7.3. Application No. 3

#### 7.3.1. KKAT and Simple Harmonic Motion (SHM)

Let a body of massm executing SHM (as shown in Figure 3) and z be the displacement from the mean position at any time. The equation for this motion is given by d2z/dx2=+λ2z=0, where λ=k/m and k is constant of proportionality. For t=0, z0=0,z′0=1. We use KKAT to find the displacement at any time t [11].Figure 3Solution: see Figure3.The equation for SHM is given by(61)d2zdx2+λ2z=0.Applying KKAT and initial conditions on equation (61), we get
(62)Kd2zdx2+λ2z=0,Kd2zdx2+λ2Kz=0,β2α2Kz−z′0αβ−z0α2+λ2Kz=0,β2α2+λ2Kz−1αβ=0,Kz=1βαβ2+λ2α2.Applying inverse KKAT both sides on equation (62), we get
(63)z=1λK−1λαββ2+λα2,zt=1λsinλt.By using the value ofλ, we get
(64)zt=mksinkmt.


## 7.1. Application No. 1

Let a bodyA of mass 1 gram moves on x-axis. It is attracted towards the origin O with a force equals to 4x. Also, assume that initially it is at rest when x=5; then, determine its position by considering
(i)
no any other forces acting on it(ii)
damping force or in other words resistance to the particle is equal to the 8 times the velocity at any instant [12]Solution: see Figure1.Figure 1From Figure1, for x>0, the net force towards left is given by −4x, while for x<0, the net force towards right is given by −4x. Thus, for both cases, the net force is equal to −4x.By Newton’s second law of motion,(54)mass×acceleration=netforce,d2xdt2=−4x,d2xdt2+4x=0.The initial conditions are(55)x0=5,x′0=0.Applying KKAT and using initial conditions on equation (54), we get
(56)Kx″+4Kx=0,β2α2+4Kx−5α2=0.Now, by using inverse KKAT on equation (56), we obtain
(57)Xt=5cos2t.


## 7.2. Application No. 2

The relationship between resistive force of air and the velocity of a freely falling body is given bydv/dt=g−λv, where vt is the velocity at any time t. Initially, the body is at rest [11].Solution: see Figure2.Figure 2The equation for the motion of body moving under constant gravitational acceleration (as shown in Figure2) is
(58)dvdt=g−λv.Applying KKAT on equation (58), we get
(59)Kdvdt=Kg−λv,βαKvt−v0αβ=gβ2−λKv,βαKvt=gβ2−λKv,βα+λKv=gβ2,Kv=gβ21λ−β/λβ+λα=gβ21λ−βλβ+λα,Kv=gλ1β2−1ββ+λα.Applying inverse KKAT both sides on equation (59), we get
(60)vt=gλ1−e−λt.


## 7.3. Application No. 3

### 7.3.1. KKAT and Simple Harmonic Motion (SHM)

Let a body of massm executing SHM (as shown in Figure 3) and z be the displacement from the mean position at any time. The equation for this motion is given by d2z/dx2=+λ2z=0, where λ=k/m and k is constant of proportionality. For t=0, z0=0,z′0=1. We use KKAT to find the displacement at any time t [11].Figure 3Solution: see Figure3.The equation for SHM is given by(61)d2zdx2+λ2z=0.Applying KKAT and initial conditions on equation (61), we get
(62)Kd2zdx2+λ2z=0,Kd2zdx2+λ2Kz=0,β2α2Kz−z′0αβ−z0α2+λ2Kz=0,β2α2+λ2Kz−1αβ=0,Kz=1βαβ2+λ2α2.Applying inverse KKAT both sides on equation (62), we get
(63)z=1λK−1λαββ2+λα2,zt=1λsinλt.By using the value ofλ, we get
(64)zt=mksinkmt.


## 7.3.1. KKAT and Simple Harmonic Motion (SHM)

Let a body of massm executing SHM (as shown in Figure 3) and z be the displacement from the mean position at any time. The equation for this motion is given by d2z/dx2=+λ2z=0, where λ=k/m and k is constant of proportionality. For t=0, z0=0,z′0=1. We use KKAT to find the displacement at any time t [11].Figure 3Solution: see Figure3.The equation for SHM is given by(61)d2zdx2+λ2z=0.Applying KKAT and initial conditions on equation (61), we get
(62)Kd2zdx2+λ2z=0,Kd2zdx2+λ2Kz=0,β2α2Kz−z′0αβ−z0α2+λ2Kz=0,β2α2+λ2Kz−1αβ=0,Kz=1βαβ2+λ2α2.Applying inverse KKAT both sides on equation (62), we get
(63)z=1λK−1λαββ2+λα2,zt=1λsinλt.By using the value ofλ, we get
(64)zt=mksinkmt.


## 8. Applications of KKAT to Electrical Circuits

### 8.1. Application No. 4

A capacitor of capacitance0.04 Farad, inductor of inductance 1 Henry, and resistor of 8 ohms are connected in series with battery of emf E=150 V (as shown in Figure 4) at t=0 current and charge are zero. Determine charge and current at any time [12].Figure 4
RLC circuit.Solution: see Figure4.The voltageVRt across the resistor R when current It is flowing is given by
(65)VRt=RI.The voltageVLt across the inductor L when current It is flowing given by the relation
(66)VLt=LdIdt.The voltageVCt across the capacitor C is flowing given by
(67)VCt=QC.By Kirchhoff second law,(68)VRt+VLt+VCt=E,LdIdt+RI+QC=E,Ld2Qdt2+RdQdt+QC=E,d2Qdt2+8dQdt+Q.04=150,d2Qdt2+8dIdt+25Q=150.Using KKAT and initial conditions on equation (68), we get
(69)Kd2Qdt2+8KdQdt+25KQ=150K1,β2α2KQ−Q′0αβ−Q0α2+8βαKQ−Q0αβ+25KQ=150β2,β2α2KQ+8βαKQ+25KQ=150β2,β2α2+8βα+25KQ=150β2,β2+8αβ+25α2α2KQ=150β2,KQ=150β2.α2β2+8αβ+25α2,KQ=1β26−6β2+48αββ2+8αβ+25α2,KQ=1β26−6β2+24αββ2+8αβ+25α2−24αββ2+8αβ+25α2,KQ=1β26−6β2+24αββ+4α2+9α2−24αββ+4α2+9α2,KQ=6β2−6β2+24αββ2β+4α2+9α2−24αββ2β+4α2+9α2,KQ=6β2−6β+24αββ+4α2+9α2−24αββ+4α2+9α2,KQ=6β2−6β+24αββ+4α2+9α2−β+4α83αββ+4αβ+4α2+9α2.Applying inverse KKAT both sides on equation (69), we obtain
(70)Q=K−16β2−6β+24αββ+4α2+9α2−β+4α83αββ+4αβ+4α2+9α2=K−16β2−K−16β+24αββ+4α2+9α2−K−1β+4α83αββ+4αβ+4α2+9α2=K−16β2−K−16β+4αββ+4α2+9α2−K−1β+4α83αββ+4αβ+4α2+9α2,Qt=6−6e−4tcos3t−8e−4tsin3t.Now, by differentiating equation (70), we get
(71)I=dQtdt=ddt6−6e−4tcos3t−8e−4tsin3t,I=50e−4tsin3t.


## 8.1. Application No. 4

A capacitor of capacitance0.04 Farad, inductor of inductance 1 Henry, and resistor of 8 ohms are connected in series with battery of emf E=150 V (as shown in Figure 4) at t=0 current and charge are zero. Determine charge and current at any time [12].Figure 4
RLC circuit.Solution: see Figure4.The voltageVRt across the resistor R when current It is flowing is given by
(65)VRt=RI.The voltageVLt across the inductor L when current It is flowing given by the relation
(66)VLt=LdIdt.The voltageVCt across the capacitor C is flowing given by
(67)VCt=QC.By Kirchhoff second law,(68)VRt+VLt+VCt=E,LdIdt+RI+QC=E,Ld2Qdt2+RdQdt+QC=E,d2Qdt2+8dQdt+Q.04=150,d2Qdt2+8dIdt+25Q=150.Using KKAT and initial conditions on equation (68), we get
(69)Kd2Qdt2+8KdQdt+25KQ=150K1,β2α2KQ−Q′0αβ−Q0α2+8βαKQ−Q0αβ+25KQ=150β2,β2α2KQ+8βαKQ+25KQ=150β2,β2α2+8βα+25KQ=150β2,β2+8αβ+25α2α2KQ=150β2,KQ=150β2.α2β2+8αβ+25α2,KQ=1β26−6β2+48αββ2+8αβ+25α2,KQ=1β26−6β2+24αββ2+8αβ+25α2−24αββ2+8αβ+25α2,KQ=1β26−6β2+24αββ+4α2+9α2−24αββ+4α2+9α2,KQ=6β2−6β2+24αββ2β+4α2+9α2−24αββ2β+4α2+9α2,KQ=6β2−6β+24αββ+4α2+9α2−24αββ+4α2+9α2,KQ=6β2−6β+24αββ+4α2+9α2−β+4α83αββ+4αβ+4α2+9α2.Applying inverse KKAT both sides on equation (69), we obtain
(70)Q=K−16β2−6β+24αββ+4α2+9α2−β+4α83αββ+4αβ+4α2+9α2=K−16β2−K−16β+24αββ+4α2+9α2−K−1β+4α83αββ+4αβ+4α2+9α2=K−16β2−K−16β+4αββ+4α2+9α2−K−1β+4α83αββ+4αβ+4α2+9α2,Qt=6−6e−4tcos3t−8e−4tsin3t.Now, by differentiating equation (70), we get
(71)I=dQtdt=ddt6−6e−4tcos3t−8e−4tsin3t,I=50e−4tsin3t.


## 9. Connection between KKAT and Some Useful Transforms

In this paper, we have established the connection between KKAT and some useful integral transforms [1, 4, 5, 11] (Laplace, Elzaki, Sumudu, and Rohit).


### 9.1. Connection between KKAT and Laplace Transform

If KKAT of functionfx is Kfx=Fβ/α and Laplace transform of fx isLfx=Aβ defined in equation (2), then Fβ/α=1/αβAβ/α.Proof.
From equation (8)
(72)Kfx=1αβ∫0∞fxe−β/αxdx=Fβα=1αβ∫0∞fxe−β/αxdx,Fβα=1αβAβα.


### 9.2. Connection between KKAT and Elzaki Transform

If KKAT of functionfx is Kfx=Fβ/α and Efx=Cβ defined in equation (3), then
(73)Fβα=1α2Cαβ.Proof.
From equation (8)
(74)Kfx=1αβ∫0∞fxe−β/αxdx=Fβα=1α2αβ∫0∞fxe−x/α/βdx,Fβα=1α2Cαβ.


### 9.3. Connection between KKAT and Sumudu Transform

If KKAT of functionfx is Kfx=Fβ/α and Sfx=Dβ defined in equation (4), then
(75)Fβα=1β2Dαβ.


### 9.4. Connection between KKAT and Rohit Transform

If KKAT of functionfx is Kfx=Fβ/α from equation (6) and kfx=Iβ defined in equation (5), then
(76)Fβα=α2β4Iβ.


## 9.1. Connection between KKAT and Laplace Transform

If KKAT of functionfx is Kfx=Fβ/α and Laplace transform of fx isLfx=Aβ defined in equation (2), then Fβ/α=1/αβAβ/α.Proof.
From equation (8)
(72)Kfx=1αβ∫0∞fxe−β/αxdx=Fβα=1αβ∫0∞fxe−β/αxdx,Fβα=1αβAβα.


## 9.2. Connection between KKAT and Elzaki Transform

If KKAT of functionfx is Kfx=Fβ/α and Efx=Cβ defined in equation (3), then
(73)Fβα=1α2Cαβ.Proof.
From equation (8)
(74)Kfx=1αβ∫0∞fxe−β/αxdx=Fβα=1α2αβ∫0∞fxe−x/α/βdx,Fβα=1α2Cαβ.


## 9.3. Connection between KKAT and Sumudu Transform

If KKAT of functionfx is Kfx=Fβ/α and Sfx=Dβ defined in equation (4), then
(75)Fβα=1β2Dαβ.


## 9.4. Connection between KKAT and Rohit Transform

If KKAT of functionfx is Kfx=Fβ/α from equation (6) and kfx=Iβ defined in equation (5), then
(76)Fβα=α2β4Iβ.


## 10. Comparison with Laplace, Sumudu, Elzaki, and Rohit Transformations

We have presented KKAT and integral transforms (Laplace, Elzaki, Sumudu, and Rohit) of some functions in Table4.Table 4
Some important functions with KKAT, Laplace, Sumudu, Elzaki, and Rohit Transformations.
Sr. no.fxKKATLAPLACESUMUDUELZAKIROHITKfx=Fβ/αLfx=AβSfx=DβEfx=Cβkfx=Iβ111/β21/β1β2β22xn,n∈Nn!αn/βn+2n!/βn+1n!βnn!βn+2n!/βn−23eλx1/ββ−λα1/β−λ1/1−λββ2/1−λββ3/β−λ4sinλxλα/ββ2+λ2α2λ/β2+λ2λβ/1+λ2β2λβ3/1+λ2β2λβ3/β2+λ25cosλx1/β2+λ2α2β/β2+λ21/1+λ2β2β2/1+λ2β2β4/β2+λ26sinhxα/ββ2−α21/β2−1β/1−β2β3/1+β2β3/β2−17coshx1/β2−α2β/β2−11/1−β2β2/1+β2β4/β2−1


## 11. Error Function and Complementary Error Function

Error function and complementary error function [2, 10] are the special functions in mathematics, science, and engineering and it is defined as
(77)erft=2π∫0te−x2dx,(78)erfct=2π∫t∞e−x2dx.The sum of error function and complementary error function is equal to unity.(79)erft+erfct=1.


## 12. KKAT of Error Function

By using equation (77),
(80)erft=2π∫0te−x2dx=2π∫0t1−x21!+x42!−x63!+⋯dx=2πx−x31!.3+x52!.5−x73!.7+⋯t0=2πt1/2−t3/21!.3+t5/22!.5−t7/23!.7+⋯,Kerft=K2πt1/2−t3/21!.3+t5/22!.5−t7/23!.7+⋯=2πΓ3/2α1/2β5/2−Γ5/2α3/2β7/21!.3+Γ7/2α5/2β9/22!.5−Γ9/2α7/2β11/23!.7+⋯=2Γ3/2πα1/2β5/2−3α3/2β7/21!.3.2+5.3α5/2β9/22!.5.4−7.5.3α7/2β11/23!.7.8+⋯=2π.π2.α1/2β5/21−3α1!.3.2β+5.3α22!.5.4β2−7.5.3α33!.7.8β3+⋯,Kerft=α1/2β5/21+αβ−1/2,Kerft=1β2αα+β1/2.


## 13. KKAT of Complementary Error Function

By using equation (79), we get
(81)erft+erfct=1,erfct=1−erft.Applying KKAT both sides on equations (81), we get
(82)Kerfct=K1−erft.Using linear property of KKAT on equation (82), we get
(83)Kerfct=K1−Kerft=1β2−α1/2β5/21+αβ−1/2,Kerfct=1β2−1β2αα+β1/2,Kerfct=1β21−αα+β1/2.


## 14. Applications of KKAT to Partial Differential Equations

Suppose the partial differential equation contains unknown variablevy,x. Then, we will take KKAT of vy,x w.r.t. the variable x.
(84)Kvy,x=Vy,βα.For example,(85)Ke−λxsinπy=sinπyββ+λα,Ksinx+y=Ksinxcosy+cosxsiny,Ksinx+y=αcosyββ2+α2+sinyβ2+α2.Similarly,(86)K∂v∂y=1αβ∫0∞e−β/αx∂v∂ydx=1αβ.∂∂y∫0∞e−β/αxvy,xdx=∂∂y1αβ∫0∞e−β/αxvy,xdx,K∂v∂y=∂∂yVy,βα.And(87)K∂v∂x=βαKvy,x−vy,0αβ.


## 15. Example

Use KKAT to solve(88)∂2v∂y2=∂v∂x,0<y<a,0≤x<∞,(89)v0,x=1,v1,x=1,x>0,(90)vy,0=1+sinπy,<0y<1.Solution: the above example is one-dimensional heat problem. It describes the conduction of heat through a rod of unit length. The end points of rod are kept at zero temperature and initial temperature conditions are given.Taking KKAT both sides on equation (88), we get
(91)K∂2v∂y2=K∂v∂x.Using equation (84) and equation (86), we obtain
(92)∂2∂y2Vy,βα=βαVy,βα−vy,0αβ,∂2∂y2Vy,βα−βαVy,βα=−1+sinπyαβ.Equation (92) is a nonhomogenous linear second-order differential equation, and its solution is given by
(93)Vy,βα=Vc+Vp,where
(94)Vc=ceβ/αx+de−β/αx,Vp=−1D2−β/α1+sinπyαβ,Vp=1β2+sinπyπ2α+ββ.Thus, equation (93) becomes
(95)Vy,βα=ceβ/αx+de−β/αx+1β2+sinπyπ2α+ββ.Applying KKAT on conditions of equation (89), we get
(96)v0,x=1⇒V0,βα=1β2,(97)v1,x=1⇒V1,βα=1β2.Using equation (96) and equation (97) in equation (87), we get
(98)c=0,d=0.Thus, equation (95) becomes
(99)Vy,βα=1β2+sinπyπ2α+ββ.Applying inverse KKAT both sides on equation (99), we get
(100)K−1Vy,βα=K−11β2+K−1sinπyπ2α+ββ,vy,x=1+e−π2xsinπy.


## 16. Conclusion

In this paper, we have introduced the new integral transformation KKAT and reached at the conclusion that this transformation is easier to apply for difficult problems. Furthermore, it is compared with other transformations and found that KKAT is easier to apply and get the better results. This transformation is more effective and convenient to find the solution of a linear system of ordinary differential equations and partial differential equations. In future work, it will be useful for solving the problems where the other transformations fail and get the analytic problems in complex analysis where the Laplace transformation or other transformations fail to get exact results.



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*Source: 8722534-2022-07-04.xml*