This paper discusses and implements a newly developed technique using the He method with Sawi Transform. The main aim is to solve some initial and boundary problems. This combination exhibits an accurate strategy to obtain a precise solution for linear and nonlinear problems. To validate the proposed Hybrid method, a 4- examples are discussed, these including:  Burger’s equation, telegraph equation, Kelin-Gordan equation, Duffing oscillator with cubic nonlinear term. The obtained results improve the exactness and the accuracy of the proposed combinations, and the proposed method is capable to solve a large number of linear and nonlinear initial and boundary value problems.

Keywords: Sawi Transform, Burger’s equation, telegraph equation, Kelin, Gordan equation, Duffing oscillator, He’s Polynomial

Burger’s equation was presented for the time by Bateman in 1915.¹ It is followed by Hradyesh kumar Mishra and Atulya K. Nagar and it is solved using He-laplace method in 2012,² then it followed by Mahgoub, MAM and Al Shikhit’s solved using  Mahgoub transform in 2017,³ Mohand, Mohamed Zebir solved it via Mohand transform in 2021,⁴ then it followed by Sarah Rabie, Bachir Nour Kharrat, Ghada Joujeh, Abd Alulkader Joukhadar, solved using He-Mohand method in 2023.⁵

In work⁶ Muhammad Nadeem and fengquanlil using He-laplace method to solve telegraph equation in2019, then it followed by Sarah Rabie, Bachir Nour Kharrat, Ghada Joujeh, Abd Alulkader Joukhadar, solved using He-Mohand method in 2023.⁵ In 2010⁷ MAJafari and Aminataei followed Homotopy Perturbation method (HPM) to solve Kelin-Gorden equation,then in 2012² Hradyesh kumar Mishra and Atulya K. Nagar and it is solved using He-laplace method.

Duffing oscillator it followed by Durmaz S.Demibag SA Kayamo and it is solved using Energy Balance method in 2010,2012,^8,9 then Khan and Mirzabeigy it is solved using Improved accuracy of He’ Balance method in 2014.¹⁰

This section provides review some of the basic concepts, which needed for this paper (Table 1 & 2):

1. Definition of Sawi transform

Sawi Transform of the function F(t); t>0 was proposed by Mahgoub,¹¹ is given as:

$s\left[f\left(t\right)\right]=R\left(v\right)=\frac{1}{v^2}\underset{0}{\overset{\infty }{{\displaystyle \int }}}f\left(t\right)e^{-\frac{t}{v}}dt ,t\ge 0 ,k_1\le v\le k_2$   (1)

Where: (S) is Sawi Transform operator.

2. Some properties of Sawi transform¹²

1. linearity property of Sawi Transform: If $s\left\{f\left(t\right)\right\}=R\left(v\right)$ and $s\left\{G\left(t\right)\right\}=I\left(v\right)$ then  $s\left\{af\left(t\right)+bG\left(t\right)\right\}=as\left\{f\left(t\right)\right\}+b\left\{G\left(t\right)\right\}=aR\left(v\right)+bI\left(v\right)$
2. $s\left\{f^{\prime }\left(t\right)\right\}=\frac{R\left(v\right)}{v}-\frac{f\left(0\right)}{v^2}$
3. $s\left\{f^{″}\left(t\right)\right\}=\frac{R\left(v\right)}{v^2}-\frac{f\left(0\right)}{v^3}-\frac{f^{\prime }\left(0\right)}{v^2}$
4. $s\left\{f^{\left(n\right)}\left(t\right)\right\}=\frac{R\left(v\right)}{v^n}-\frac{f\left(0\right)}{v^{n+1}}-\frac{f^{\prime }\left(0\right)}{v^2}-\dots -\frac{f^{n-1}\left(0\right)}{v^2}$
5. $s\left\{f\left(t\right)*G\left(t\right)\right\}=v^{2}s\left\{f\left(t\right)\right\}.s\left\{G\left(t\right)\right\}=v^{2}R\left(v\right).I\left(v\right)$

In order to explain the proposed method let’s consider the following nonlinear functional equation:

$L\left(u\left(x\right)\right)+N\left(u\left(x\right)\right)=g\left(x\right)$   (2)

Where:

L and N are linear and nonlinear operator respectively.

g(x): is analytical function.

taking the Sawi Transform of equation (2) and obtain:

$s\left\{L\left(u\left(x\right)\right)+N\left(u\left(x\right)\right)-g\left(x\right)\right\}=0$   (3)

Then multiplying the (3) equation with lag range multiplier, say $\lambda \left(v\right)$ , we get:

$\lambda \left(v\right)s\left\{L\left(u\left(x\right)\right)+N\left(u\left(x\right)\right)-g\left(x\right)\right\}=0$   (4)

Therefore, the recurrence relation becomes:

$u_{n+1}\left(x,v\right)=u_n\left(x,v\right)+\lambda \left(v\right)\left\{s\left\{L\left(u_n\left(x\right)\right)\right\}+s\left\{N\left({\tilde{u}}_n\left(x\right)\right)-g\left(x\right)\right\}\right\}$   (5)

Taking the variation of equation (5) results in:

$\delta u_{n+1}\left(x,v\right)=\delta u_n\left(x,v\right)+\lambda \left(v\right)\delta \left\{s\left\{L\left(u_n\left(x\right)\right)\right\}+s\left\{N\left({\tilde{u}}_n\left(x\right)\right)-g\left(x\right)\right\}\right\}$   (6)

To identify the value of Lagrange multiplier $\lambda \left(v\right)$ with the help of Sawi Transform, it is revealed that ${\tilde{u}}_n$  is a restricted variable, i,e, $\delta {\tilde{u}}_n=0$

taking the inverse of Sawi Transform of equation (5) this results in:

$u_{n+1}(X,t)=u_n(X,t)+s^{-1}\left\{\text{λ}\left(\text{v}\right)\right\{s\{L(u_n(X)\left\}\right\}+s\{N({\tilde{u}}_n(X)\}-g(X)\}\}$   (7)

The following section presents a descriptive examples of the proposed method.

Consider Burger’s equation:

$u_t=u_{xx}-u{u}_x$   (8)

With initial condition of:

$u\left(x,0\right)=1-\frac{2}{x}$   (9)

taking the sawi transform of equation (8):

$s\left\{u_t-u_{xx}+u{u}_x\right\}=0$   (10)

Multiplying the equation (10) with $\lambda \left(v\right)$ results in:

$\lambda \left(v\right)s\left\{u_t-u_{xx}+u{u}_x\right\}=0$

The recurrence relation takes the form:

$u_{n+1}\left(x,v\right)=u_n\left(x,v\right)+\lambda \left(v\right)s\left\{\frac{\partial u_n}{\partial t}-\frac{{\partial }^2{u}_n}{\partial x^2}+u_n\frac{\partial u_n}{\partial x}\right\}$   (11)

taking the variation of equation (11):

$\delta u_{n+1}\left(x,v\right)=\delta u_n\left(x,v\right)+\lambda \left(v\right)\delta \left\{\frac{1}{v}{u}_n\left(x,v\right)-\frac{1}{v^2}{\tilde{u}}_n^{\text{'}}\left(x,0\right)\right\}+\lambda \delta s\left\{-\frac{{\partial }^2{\tilde{u}}_n}{\partial x^2}+u_n\frac{\partial {\tilde{u}}_n}{\partial x}\right\}$

$\delta u_{n+1}\left(x,v\right)=\delta u_n\left(x,v\right)+\lambda \frac{1}{v}\delta u_n$

In turn gives the value of λ becomes as follows:

$\begin{array}{l}0=1+\frac{1}{v}\lambda \\ \lambda =-v\end{array}$

Which:${\tilde{u}}_n$ is a restricted variable $\delta {\tilde{u}}_n=0$ and $\frac{\delta u_{n+1}}{\delta u_n}=0$ using the value of $\lambda =-v$ , will result in:

$u_{n+1}\left(x,v\right)=u_n\left(x,v\right)-vs\left\{\frac{\partial u_n}{\partial t}-\frac{{\partial }^2{u}_n}{\partial x^2}+u_n\frac{\partial u_n}{\partial x}\right\}$   (12)

Taking the inverse Sawi Ttransform of equation (12):

$u_{n+1}\left(x,t\right)=u_n\left(x,t\right)-s^{-1}\left\{v s\left\{-\frac{{\partial }^2{u}_n}{\partial x^2}+u_n\frac{\partial u_n}{\partial x}\right\}\right\}$

Applying He’s polynomial formula, yields:

$u_0+p{u}_1+\mathrm{..}=u_n-p{s}^{-1}\left\{vs\left\{\left(-\frac{{\partial }^2{u}_0}{\partial x^2}+u_0\frac{\partial u_0}{\partial x}\right)+p\left(-\frac{{\partial }^2{u}_1}{\partial x^2}+u_1\frac{\partial u_0}{\partial x}+u_0\frac{\partial u_1}{\partial x}\right)+\dots .\right\}\right\}$

Equating highest power of p will result in:

$u_0=1-\frac{2}{x}$

$u_1=-s^{-1}\left\{vs\left\{\left(-\frac{{\partial }^2{u}_0}{\partial x^2}+u_0\frac{\partial u_0}{\partial x}\right)\right\}\right\}=-\frac{2}{x^2}t$

$u_2=-s^{-1}\left\{vs\left\{\left(-\frac{{\partial }^2{u}_1}{\partial x^2}+u_1\frac{\partial u_0}{\partial x}+u_0\frac{\partial u_1}{\partial x}\right)\right\}\right\}=-\frac{2}{x^3}{t}^2$

Hence the series solution can expressed as:

$u\left(x,t\right)=u_0+u_1+\dots =1-\frac{2}{x}-\frac{2}{x^3}{t}^2-\dots =1-\frac{2}{x-t}$

Consider the following Telegraph’s equation:

$u_{xx}=\frac{1}{3}{u}_{tt }+\frac{4}{3}{u}_t+u$   (13)

With initial conditions:

$u\left(x,0\right)=e^x+1 u_t\left(x,0\right)=-3$   (14)

and boundary conditions:

$u\left(0,t\right)=e^{-3t}+1 u_x\left(0,t\right)=1$   (15)

Taking the Sawi Transform of equation (13):

$s\left\{-u_{xx}+\frac{1}{3}{u}_{tt }+\frac{4}{3}{u}_t+u\right\}=0$   (16)

Multiplying the equation (16)with $\lambda \left(v\right)$ :

$\lambda \left(v\right)s\left\{-u_{xx}+\frac{1}{3}{u}_{tt }+\frac{4}{3}{u}_t+u\right\}=0$   (17)

The recurrence relation takes the form:

$u_{n+1}\left(x,v\right)=u_n\left(x,v\right)+\lambda \left(v\right)s\left[\frac{1}{3}\frac{{\partial }^2{u}_n}{\partial t^2}-\frac{{\partial }^2{u}_n}{\partial x^2}+u_n+\frac{4}{3}\frac{\partial u_n}{\partial t}\right]$   (18)

Taking the variation of equation (18):

$\delta u_{n+1}=\delta u_n+\lambda \left(v\right)\delta s\left[\frac{1}{3}\frac{{\partial }^2{u}_n}{\partial t^2}-\frac{{\partial }^2{u}_n}{\partial x^2}+u_n+\frac{4}{3}\frac{\partial u_n}{\partial t}\right]$

$\delta u_{n+1}=\delta u_n+\frac{\lambda \delta }{3}\left\{\left(\frac{1}{v^2} u_n\left(x,v\right)-\frac{1}{v^2}\tilde{u}{\text{'}}_n{\left(x,0\right)}^{`}-\frac{1}{v^3}{\tilde{u}}_n\left(x,0\right)\right)\right\}+\lambda \delta \text{s}\left\{-\frac{{\partial }^2{\tilde{u}}_n}{\partial x^2}+{\tilde{u}}_n+\frac{4}{3}\frac{\partial {\tilde{u}}_n}{\partial t}\right\}$

$\delta u_{n+1}=\delta u_n+\lambda \frac{1}{3v^2}\delta u_n$

In turn gives the value of becomes as follows:

$\begin{array}{l}0=1+\lambda \frac{1}{3v^2}\\ \lambda =-3v^2\end{array}$

Which:${\tilde{u}}_n$ is a restricted variable $\delta {\tilde{u}}_n=0$ and $\frac{\delta u_{n+1}}{\delta u_n}=0$ using the value of $\lambda \left(v\right)=-3v^2$ in equation (18),will result in:

$u_{n+1}\left(x,v\right)=u_n\left(x,v\right)-3v^{2}s\left[\frac{1}{3}\frac{{\partial }^2{u}_n}{\partial t^2}-\frac{{\partial }^2{u}_n}{\partial x^2}+u_n+\frac{4}{3}\frac{\partial u_n}{\partial t}\right]$   (19)

Taking the inverse Sawi Transform of equation (19):

$u_{n+1}\left(x,t\right)=u_n\left(x,t\right)-s^{-1}[3v^{2}s\left[-\frac{{\partial }^2{u}_n}{\partial x^2}+u_n+\frac{4}{3}\frac{\partial u_n}{\partial t}\right]$   (20)

Applying He’s polynomial formula, yields:

$\begin{array}{l}{u}_0+p{u}_1+p^2{u}_2+\dots \\ =u_n-p{\text{s}}^{-1}\{3v^2\text{s}\left\{\begin{array}{l}\left(-\frac{{\partial }^2{u}_0}{\partial x^2}+u_0+\frac{4}{3}\frac{\partial u_0}{\partial t}\right)\\ +p\left(-\frac{{\partial }^2{u}_1}{\partial x^2}+u_1+\frac{4}{3}\frac{\partial u_1}{\partial t}\right)\\ +p^2\left(-\frac{{\partial }^2{u}_2}{\partial x^2}+u_2+\frac{4}{3}\frac{\partial u_2}{\partial t}\right)+\dots \end{array}\right\}\end{array}$

Equating highest power of p will result in:

$\begin{array}{l}{p}^0:u_0=e^x+1-3t\\ p^1:u_1=-{\text{s}}^{-1}\left\{3v^2\text{s}\left\{\left(-\frac{{\partial }^2{u}_0}{\partial x^2}+u_0+\frac{4}{3}\frac{\partial u_0}{\partial t}\right)\right\}\right\}=\frac{9t^2}{2}+\frac{3t^3}{2}\\ p^2:u_2=-{\text{s}}^{-1}\left\{3v^2\text{s}\left\{\left(-\frac{{\partial }^2{u}_1}{\partial x^2}+u_1+\frac{4}{3}\frac{\partial u_1}{\partial t}\right)\right\}\right\}=-6t^3-\frac{21}{8}{t}^4-\frac{9}{40}{t}^5\end{array}$

Hence the series solution can expressed as:

$u\left(x,t\right)=u_0+u_1+u_2+\dots =e^x+1-3t+\frac{9t^2}{2}-\frac{9t^3}{2}+\frac{27}{8}{t}^4+\dots =e^x+e^{-3t}$

Consider the following Kelin-Gorden equation:

$\frac{{\partial }^{2}u}{\partial {\text{t}}^2}+u+\frac{{\partial }^{2}u}{\partial {\text{x}}^2}=0$   (21)

With initial conditions:

$\begin{array}{l}u\left(x,0\right)=e^{-x}+x\\ \frac{\partial u}{\partial t}\left(x,0\right)=0\end{array}$   (22)

Taking the Sawi Transform of equation (21):

$\text{S}\left\{\frac{{\partial }^{2}u}{\partial {\text{t}}^2}+u+\frac{{\partial }^{2}u}{\partial {\text{x}}^2} \right\}=0$   (23)

Multiplying the equation (23) with  $\lambda \left(v\right)$ :

$\lambda \left(v\right)\text{ s}\left\{\frac{{\partial }^{2}u}{\partial {\text{t}}^2}+u+\frac{{\partial }^{2}u}{\partial {\text{x}}^2} \right\}=0$   (24)

The recurrence relation takes the form:   

$u_{n+1}=u_n+\lambda \text{ s}\left\{\frac{{\partial }^2{u}_n}{\partial {\text{t}}^2}+u_n+\frac{{\partial }^2{u}_n}{\partial {\text{x}}^2} \right\}$   (25)

Taking the variation of equation (25):

$\begin{array}{l}\delta u_{n+1}=\delta u_n+\lambda \delta \left\{\left(\frac{1}{v^2}{u}_n\left(x,v\right)-\frac{1}{v^2}{u}_n\left(x,0\right)-\frac{1}{v^3}{u}_n{\left(x,0\right)}^{`}\right)\right\}+\lambda \delta \text{s}\left\{{\tilde{u}}_n+\frac{{\partial }^2{\tilde{u}}_n}{\partial {\text{x}}^2}\right\}\\ \delta u_{n+1}=\delta u_n+\lambda \frac{1}{v^2}\delta u_n\end{array}$

in turn gives the value of λ becomes as follows:

$\begin{array}{l}0=1+\lambda \frac{1}{v^2}\\ \lambda =-v^2\end{array}$

Which:${\tilde{u}}_n$ is a restricted variable $\delta {\tilde{u}}_n=0$ and $\frac{\delta u_{n+1}}{\delta u_n}=0$ using the value of $\lambda \left(v\right)=-v^2$ , will result in:

$u_{n+1}=u_n-v^2\text{ s}\left\{ \frac{{\partial }^2{u}_n}{\partial {\text{t}}^2}+u_n+\frac{{\partial }^2{u}_n}{\partial {\text{x}}^2}\right\}$   (26)

Taking the inverse of Sawi Transform of equation (26):

$u_{n+1}=u_n-{\text{s}}^{-1}\left\{v^2\text{s}\left\{ \frac{{\partial }^2{u}_n}{\partial {\text{t}}^2}+u_n+\frac{{\partial }^2{u}_n}{\partial {\text{x}}^2}\right\}\right\}$

Applying He’s polynomial formula, yields:

$u_0+p{u}_1+p^2{u}_2+\dots =u_n-p{\text{s}}^{-1}\{v^2\text{s}\left\{ \left( u_0+\frac{{\partial }^2{u}_0}{\partial {\text{x}}^2}\right)+p\left(u_1+\frac{{\partial }^2{u}_1}{\partial {\text{x}}^2}\right)+\mathrm{...}\right\}$

Equating highest power of p will result in:

$\begin{array}{l}{u}_0=e^{-x}+x\\ u_1=-x\frac{t^2}{2}\\ u_2=x\frac{1}{4!}{t}^4\end{array}$

Hence the series solution can expressed as:

$u\left(t\right)=u_0+u_1+u_2+u_3+\dots =e^{-x}+x-x\frac{t^2}{2}+x\frac{1}{4!}{t}^4=e^{-x}+xcos\left(t\right)$

Consider Duffing oscillator with cubic nonlinear term:

$u^{″}+u+\epsilon u^3=0$   (27)

With initial conditions:

$\begin{array}{l}u\left(0\right)=A u^{\prime }\left(0\right)=0\\ u^{″}+u+\epsilon u^3+{\omega }^{2}u-{\omega }^{2}u=0\\ u^{″}+{\omega }^{2}u+g\left(u\right)=0 \left(28\right) ;g\left(u\right)=u+\epsilon u^3-{\omega }^{2}u\end{array}$   (28)

taking the Sawi Transform of equation (27):

$s\left\{u^{″}+{\omega }^{2}u+g\left(u\right)\right\}=0$   (29)

Multiplying the equation (29) with $\lambda \left(v\right)$ result in:

$\lambda \left(v\right)s\left\{u^{″}+{\omega }^{2}u+g\left(u\right)\right\}=0$   (30)

The recurrence relation takes the form:

$u_{n+1}\left(v\right)=u_n\left(v\right)+\lambda S\left\{\frac{d^2{u}_n}{d{t}^2}+{\omega }^2{u}_n+g\left({\tilde{u}}_n\right)\right\}$   (31)

Taking the variation of equation (31):

$\begin{array}{l}\delta u_{n+1}\left(v\right)=\delta u_n\left(v\right)+\lambda \delta \left\{\frac{1}{v^2}{u}_n-\frac{1}{v^3}{\tilde{u}}_n\left(0\right)-\frac{1}{v^2}\widetilde{u_n}\text{'}\left(0\right)+{\omega }^2{u}_n\}+s\lambda \delta \{g\left({\tilde{u}}_n\right)\right\}\\ \delta u_{n+1}\left(v\right)=\delta u_n\left(v\right)+\lambda \left\{\frac{1}{v^2}+{\omega }^2\right\}\delta u_n\end{array}$

In turn gives the value of λ becomes as follows:

$\lambda =-\frac{v^2}{v^2{\omega }^2+1}$

Notice that ${\tilde{u}}_n$ is a restricted variable $\delta {\tilde{u}}_n=0$ and $\frac{\delta u_{n+1}}{\delta u_n}=0$ using the value of $\lambda =-\frac{v^2}{v^2{\omega }^2+1}$ in equation (31):

$u_{n+1}\left(v\right)=u_n\left(v\right)-\frac{v^2}{v^2{\omega }^2+1} s\left\{u^{″}+u+\epsilon u^3\right\}$   (32)

Taking the inverse Sawi Transform of equation (32):

$u_{n+1}\left(t\right)=u_n\left(t\right)-s^{-1}\left\{\frac{v^2}{v^2{\omega }^2+1} S\left\{u^{″}+u+\epsilon u^3\right\}\right\}$

Applying He’s polynomial formula, yields:

$u_0+p{u}_1+\mathrm{..}=u_n\left(t\right)-p\left\{s^{-1}\left\{\frac{v^2}{v^2{\omega }^2+1} s\left\{\begin{array}{l}\left(u_0^{\text{'}\text{'}}+u_0+\epsilon u_0^3\right)\\ +p\left(u_1^{\text{'}\text{'}}+u_1+3\epsilon u_0^2{u}_1\right)+\dots \end{array}\right\}\right\}\right\}$

Equating highest power of p will result in:

$\begin{array}{l}{u}_0=A \text{cos}\left(\omega t\right)\\ u_1=\frac{1}{8}\frac{1}{{\omega }^2}(\cos \left(\omega t\right)\epsilon A^3\left(-1+{\cos }^2\left(\omega t\right)\right)+\frac{A}{8\omega }\left(-4+4{\omega }^2-3\epsilon A^2\right)t\sin \left(\omega t\right)\end{array}$   (33)

No secular-term in (33) requires that:

$\begin{array}{l}\frac{A}{8\omega }\left(-4+4{\omega }^2-3\epsilon A^2\right)=0\\ -4+4{\omega }^2-3\epsilon A^2=0\\ \omega =\sqrt{1+\frac{3}{4}\epsilon A^2}\end{array}$

For most of the applications which have been studied in literature, the present study has provided more precise solutions with fewer iteration, compared to other methods. For future research work, it is recommended to combines He-Sawi method with other integral transform such as: Foks, Abood, sumdu and Elzaki.

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