## 4 Applications and Discussion  
**Example 1**. : _Consider the (2+time fractional-order) dimensional hyperbolic wave model:_  
\[u_{\tau}^{\beta}(\Im,\Re,\tau)=\frac{1}{12}\Im^{2}u_{\Im\Im}(\Im,\Re,\tau)+ \frac{1}{12}\Re^{2}u_{\Re\Re}(\Im,\Re,\tau),\quad\beta\in(1,2) \tag{19}\]  
_with initial conditions_  
\[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]  
_If \(\beta=2\), then the exact solution of Equation (19) is_  
\[u(\Im,\Re,\tau)=\Im^{4}\sinh(t)+\Re^{4}\cosh(t), \tag{21}\]  
_Taking the Shehu transform of Equation (19) we obtain_  
\[\frac{s^{\beta}}{\mu^{\beta}}\left\{V(s,\mu)-\frac{\mu}{s}u(0)-\frac{\mu^{2}}{ s^{2}}u^{{}^{\prime}}(0)\right\}=S\left\{\frac{1}{12}\Im^{2}u_{\Im\Im}(\Im,\Re, \tau)+\frac{1}{12}\Re^{2}u_{\Re\Re}(\Im,\Re,\tau)\right\}. \tag{22}\]  
_Simplifying Equation (22), we get_  
\[V(s,\mu)=\frac{\mu^{\beta}}{s^{\beta}}S\left\{\frac{1}{12}\Im^{2}u_{\Im\Im}( \Im,\Re,\tau)+\frac{1}{12}\Re^{2}u_{\Re\Re}(\Im,\Re,\tau)\right\}+\frac{\mu}{s }u(0)+\frac{\mu^{2}}{s^{2}}u^{{}^{\prime}}(0). \tag{23}\]  
_Applying inverse Shehu transform, we get_  
\[u(\Im,\Re,\tau)=u(0)+u^{{}^{\prime}}(0)\tau+S^{-1}\left\{\frac{\mu^{\beta}}{s^ {\beta}}S\left\{\frac{1}{12}\Im^{2}u_{\Im\Im}(\Im,\Re,\tau)+\frac{1}{12}\Re^{ 2}u_{\Re\Re}(\Im,\Re,\tau)\right\}\right\}. \tag{24}\]  
_Thus we get the following recursive scheme_  
\[u_{0}(\Im,\Re,\tau)=u(0)+u^{{}^{\prime}}(0)\tau=\Im^{4}+\Re^{4}\tau, \tag{25}\]  
\[u_{m+1}(\Im,\Re,\tau)=S^{-1}\left\{\frac{\mu^{\beta}}{s^{\beta}}S\left\{\frac{ 1}{12}\Im^{2}u_{m\Im\Im}(\Im,\Re,\tau)+\frac{1}{12}\Re^{2}u_{m\Re\Re}(\Im,\Re,\tau)\right\}\right\}. \tag{26}\]  
_Using Equation (26), for \(m=0,1,2,3,\cdots\) we get the following values_\[u_{1}(\Im,\Re,\tau)=\Im^{4}\frac{\tau^{\beta}}{\beta!}+\Re^{4} \frac{\tau^{\beta+1}}{(\beta+1)!}, \tag{27}\] \[u_{2}(\Im,\Re,\tau)=\Im^{4}\frac{\tau^{2\beta}}{(2\beta)!}+\Re^{4} \frac{\tau^{2\beta+1}}{(2\beta+1)!},\] \[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]0 \[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]1 \[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]2  
_Now using the values of \(u_{0},u_{1},u_{2},u_{3},\cdots\), we get Shehu transformation solution for example 1_  
\[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]3 \[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]4  
_After simplification, we get_  
\[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]5 \[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]6  
_In particular, when \(\beta\to 2\), the analytical solution of Shehu transform become as_  
\[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]7  
_which provide the close form solution as_  
\[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]8  
_Figures 1 and 2 represent the exact and analytical solutions of Example 1. The solutions-graphs have confirmed the closed contact between the exact solution and the analytical solution obtained by the proposed method. In Figure 3, the solution of Example 1 are calculated at different fractional-order \(\beta\) of the derivative. It is investigated that the solutions at different fractional-orders \(\beta\) are convergent to an integer-order solution of Example 1. Figure 4 represent the solution verses time graph for Example 1. It is observed that as the time fractional-order varies toward time integer-order, the time fractional-order solutions also approaches to the solution of an integer-order problem of Example 1. All the above solution analysis of Example 1 indicate that SDM is an efficient and effective method to solve fractional-order partial differential equations that are frequently arising in science and engineering._\[u(\Im,\Re,0)=\Im^{4},\qquad u_{\tau}(\Im,\Re,0)=\Re^{4}. \tag{20}\]9 Represents the exact solution of Example 1 at \(\beta=2\). \[u(\Im,\Re,\tau)=\Im^{4}\sinh(t)+\Re^{4}\cosh(t), \tag{21}\]0 Represents the solution at different fractional order of Example 1.