Applications of new integral transform for linear and nonlinear fractional partial differential equations

A function $h(\tau ),\tau >0,$ is told to be in the space $C_{\eta },\eta \in \mathbb{R},$ if the yonder be a real number $\sigma >\eta$ such that $h(\tau )={\tau }^{\sigma }{h}_1(\tau ),$ where $h_1(\tau )\in C\left[0,\infty \right),$ clearly $C_{\eta }\subset C_{v}ifv⩽\eta .$

let’s Riemann-Liouville fractional integral operator of order $\upsilon ⩾0,$ then $h(\tau )\in C_{\eta },\eta ⩾-1,$ is known as, accordingly (Miller and Ross, 1993; Podlubny, 1999; Mahdy et al., 2015);

Caputo fractional derivative of the left sided, $h,$ $h\in C_{-1}^n,n\in \mathrm{N}\cup \left\{0\right\},$ is known as, accordingly

we hold properties of the operator (Miller and Ross, 1993; Podlubny, 1999; Mahdy et al., 2015; Elzaki, 2011; Mohamed and Elzaki, 2014; Adomian, 1994; Bulut and Evans, 2002; El-Tawil et al., 2004; Arife and Yildirim, 2011; He, 1997; He, 1997; He, 1998; He, 1998; He, 1999; Wazwaz, 2007);

the new integral transform (NIT) is known as, accordingly (Elzaki, 2011; Mohamed and Elzaki, 2014); $$A=\left\{\left|h\left(\tau \right)\right|<N{e}^{\frac{\tau }{k_j}},if\tau \in {\left(-1\right)}^i\times \left[0,\infty \right)\right\}$$

Before the next formulation,

If $r-1<\sigma ⩽r,r\in \mathrm{N},$ then the new integral transform (NIT) of the fractional derivative $D_{\ast }^{\sigma }h(\xi ,\tau )$ is,

where $H(\xi ,\rho )$ be the (NIT) $h(\xi ,\tau )$ (Mohamed and Elzaki, 2014).

Consider the following one-dimensional linear inhomogeneous fractional wave equation (Adomian, 1994).

Subject to the initial condition:

Applying the (NIT) to both parties Eq. (4.1), to get:

Using given initial condition Eq. (4.3), become

For the nonlinear, function $Y\psi (\xi ,\tau ),$ the first Adomian polynomials (Adomian, 1994), to obtain:

Finally, by using the two parties of the Eq. (4.5), we get the repeated algorithm as the following

Applying the inverse (NIT) Eqs. (4.6) and (4.7), to obtain: $${\psi }_0(\xi ,\tau )={\mathrm{E}}^{-1}\left[{\rho }^{\sigma }\mathrm{E}\left[\frac{{\tau }^{1-\sigma }}{\mathrm{\Gamma }(2-\sigma )}sin\xi +\tau cos\xi \right]\right],$$

Consequently,

By excluding the noise terms and holding the part that contains the non-noise terms, we get the accurate solution of the Eq. (4.1). Starting by ${\psi }_0$ , and repeating the process twice, we obtain the accurate solution $\psi (\xi ,\tau )=\tau sin\xi$ .

Consider the following one-dimensional linear inhomogeneous fractional Burgers equation (Odibat and Momani, 2009);

Subject to the initial condition:

Applying the (NIT) to both parties Eq. (4.12), to get:

Using given initial condition Eq. (5.14) become,

The nonlinear function $Y\psi (\xi ,\tau ),$ the first Adomian polynomials (Adomian, 1994), to obtain;

Finally, by using the two parties of the Eq. (4.16), we get the repeated algorithm as the following:

Applying the inverse (NIT) Eqs. (4.17) and (4.18), to obtain: $${\psi }_0(\xi ,\tau )={\mathrm{E}}^{-1}\left[{\rho }^2{\xi }^2+{\rho }^{\sigma }\mathrm{E}\left[\frac{t^{2-\sigma }}{\mathrm{\Gamma }(3-\sigma )}+2\xi -2\right]\right],$$ $${\psi }_{r+1}(\xi ,\tau )=-{\mathrm{E}}^{-1}\left[{\rho }^{\sigma }\mathrm{E}\left[{\sum }_{r=0}^{\infty }{\psi }_{r\xi }(\xi ,\rho )-{\sum }_{r=0}^{\infty }{\psi }_{r\xi \xi }(\xi ,\rho )\right]\right].$$

Consequently,

Canceled noise terms of ${\psi }_0$ satisfy Eq. (4.12), we find that the exact solution is given by $\psi (\xi ,\tau )={\xi }^2+{\tau }^2.$

Consider the following nonlinear Time- Fraction KdV equation (Wazwaz, 2007)

Subject to the initial condition:

Applying the (NIT) to both parties Eq. (4.23), we will get:

Using the given initial condition, Eq. (4.25) become,

The first few components of $A_r$ polynomials are given by: $$\begin{array}{c}{A}_0={\psi }_{0\xi }^2,\hfill \\ A_1={(2{\psi }_0{\psi }_1)}_{\xi },\hfill \\ A_2={(2{\psi }_0{\psi }_2+{\psi }_1^2)}_{\xi },\hfill \end{array}$$

Finally, by using the two parties of the Eq. (4.27), we get the repeated algorithm as the following:

Applying the inverse (NIT) for Eqs. (4.28) and (4.29), to obtain: $${\psi }_0(\xi ,\tau )=6\xi ,$$ $${\psi }_{r+1}(\xi ,\tau )=-{\mathrm{E}}^{-1}\left[{\rho }^{\sigma }\mathrm{E}\left[\sum _{r=0}^{\infty }{A}_r+{\psi }_{r\xi \xi \xi }(\xi ,\tau )\right]\right].$$

Consequently,

The other components of the EDM competence are determined in an identical format, then approximate solution of (4.23) in sequence, $$\psi (\xi ,\tau )=6\xi \left[1+\frac{36}{\sigma !}{\tau }^{\sigma }+2\frac{{(36)}^2}{2\sigma !}{\tau }^{2\sigma }+{(36)}^3\left[\frac{4}{2\sigma !}+\frac{1}{{(\sigma !)}^2}\right]\frac{2\sigma !}{3\sigma !}{\tau }^{3\sigma }+\cdots \right]$$

And when $\sigma =1,$ we obtain the exact solution of the nonlinear KdV Equation (Wazwaz, 2007) (see Table 1 and Fig. 1).

Consider the following nonlinear Time- Fraction differential equation (Wazwaz, 2007);

Subject to initial condition:

Applying the (NIT) to both parties Eq. (4.31), we will get:

Using given initial condition Eq. (4.33) become,

The Nonlinear function $Y\psi (\xi ,\tau ),$ the first Adomian polynomials (Adomian, 1994), to will get

The first few components of $A_n(x,t)$ polynomials are given by:

Finally, by using the two parties of the Eq. (4.35), we get the repeated algorithm as the following

Applying the inverse (NIT) Eqs. (4.37) and (4.38), to obtain:

Consequently,

When $\sigma =2,$ Eq. (4.42) becomes:

On the side, we catch sight of that the development of the function $\psi (\xi ,\tau )=tan(\xi \tau ),$ according to the Taylor series in the vicinity of $\tau =0,$ is will get:

Therefore, we conclude that: $\psi (\xi ,\tau )=tan(\xi \tau ),$ that is the accurate solution of Eq. (4.31) in the status $\sigma =2$ (see Table 2 and Fig. 2).