## 2 Preliminaries Concepts  
In this section, we present some fundamental and appropriate definitions and preliminary concepts related to the fractional calculus and the Shehu transformation.  
**Definition 1**.: _Shehu transform_  
_Shehu transformation is new and similar to other integral transformation which is defined for functions of exponential order [40]. We take a function in the set \(A\) define by_  
\[A=\{u(\tau):\exists,\rho_{1},\rho_{2}>0,|u(\tau)|<Me^{\frac{|\tau|}{\rho_{1}}},\ \ if\ \ \tau\in[0,\infty), \tag{1}\]  
_The Shehu transform which is represented by \(S(.)\) for a function \(u(\tau)\) is defined as_  
\[S\{u(\tau)\}=V(s,\mu)=\int_{0}^{\infty}u(\tau)e^{\frac{-s\tau}{\tau}}u(\tau)d \tau,\ \ \tau>0,\ \ s>0. \tag{2}\]The Shehu transform of a function \(u(\tau)\) is \(V(s,\mu)\): then \(u(\tau)\) is called the inverse of \(V(s,\mu)\) which is expressed as  
\[S^{-1}\left\{V(s,\mu)\right\}=u(\tau),\;\;for\;\;\tau\geq 0,\;\;S^{-1}\text{is \;inverse \; Shehu\; transform}. \tag{3}\]  
**Definition 2**.: _Shehu transform for nth derivatives_  
_The Shehu transformation for nth derivatives is defined as [40]_  
\[S\left\{u^{(n)}(\tau)\right\}=\frac{s^{n}}{\mu^{n}}V(s,\mu)-\sum_{k=0}^{n-1} \left(\frac{s}{\mu}\right)^{n-k-1}u^{(k)}(0). \tag{4}\]  
**Definition 3**.: _Caputo operator of fractional partial derivative_  
_The fractional Caputo operator is represented as [41]_  
\[D_{\tau}^{\beta}f(\tau)=\begin{cases}\frac{\partial^{n}f(\tau)}{\partial\tau^ {n}},\quad\beta=n\in N,\\ \frac{1}{\Gamma(n-\beta)}\int_{0}^{\tau}(\tau-\phi)^{n-\beta-1}f^{(n)}(\phi) \phi\phi,\quad n-1<\beta\leq n,\quad n\in N.\end{cases} \tag{5}\]  
**Definition 4**.: _Shehu transform for fractional order derivatives_  
_The Shehu transformation for the fractional order derivatives is expressed as_  
\[S\left\{u^{(\beta)}(\tau)\right\}=\frac{s^{\beta}}{\mu^{\beta}}V(s,\mu)-\sum_ {k=0}^{n-1}\left(\frac{s}{\mu}\right)^{\beta-k-1}u^{(k)}(0),\quad 0<\beta \leq n, \tag{6}\]  
_In Table 1 show different special functions of Shehu transformation._  
**Theorem 1**.: _If the function \(u(\tau)\) is piecewise continues at every finite interval of \(0\leq\tau\leq\beta\) and of exponential order \(\alpha\) for \(\tau>\beta\), then there's the Shehu transform \(u(s,\mu)\)[40]._  
Proof.: For any natural number \(\beta\), we deduct algebraically:  
\[\int_{0}^{\infty}exp(-\frac{s\tau}{\mu})u(\tau)d\tau=\int_{0}^{\beta}exp(- \frac{s\tau}{\mu})u(\tau)d\tau+\int_{\beta}^{\infty}exp(-\frac{s\tau}{\mu})u( \tau)d\tau, \tag{7}\]  
since the function \(u(\tau)\) continues in a piecewise manner at every finite interval \(0\leq\tau\leq\beta\), there's the first integral on the right hand side. We suggest the following situation to validate this statement,  
\begin{table}
\begin{tabular}{c c} \hline \hline
**Functional Form** & **Shehu Transform Form** \\ \hline \(1\) & \(\frac{u}{s}\) \\ \(t\) & \(\frac{u^{2}}{s^{2}}\) \\ \(e^{\tau}\) & \(\frac{s^{4}}{s-\mu^{2}}\) \\ \(\sin(\tau)\) & \(\frac{u}{s^{4}+u^{2}}\) \\ \(\cos(\tau)\) & \(\frac{u\mu s}{s^{4}+u^{2}}\) \\ \(\frac{\tau^{n}}{n!}\) for \(n=0,1,2,\cdots\) & \((\frac{u}{s})^{n+1}\) \\ \(\frac{\tau^{n}}{\Gamma(n+1)}\) for \(n=0,1,2,\cdots\) & \((\frac{u}{s})^{n+1}\) \\ \hline \hline \end{tabular}
\end{table}
Table 1: The Shehu transform of some special functions.  
\[\begin{split}&|\int_{\alpha}^{\infty}exp(-\frac{s\tau}{\mu})u(\tau)d \tau|\leq\int_{\alpha}^{\infty}|exp(-\frac{s\tau}{\mu})u(\tau)|d\tau\\ &\leq\int_{\alpha}^{\infty}exp(-\frac{s\tau}{\mu})|u(\tau)|d\tau \\ &\leq\int_{\alpha}^{\infty}exp(-\frac{s\tau}{\mu})Nexp(\beta\tau)d \tau\\ &=N\int_{\alpha}^{\infty}exp(-\frac{(s-\beta u)\tau}{u})d\tau\\ &=-\frac{Nu}{(s-\beta u)}\lim_{\gamma\longrightarrow\infty}\left[ exp(-\frac{(s-\beta u)\tau}{u})d\tau\right]_{0}^{\gamma}\\ &=\frac{Nu}{(s-\beta u)}.\end{split} \tag{8}\]  
The proof is complete.