Fractional partial differential equations and novel double integral transform

Tarig M. Elzaki; Shams A. Ahmed; Mounirah Areshi; Mourad Chamekh

doi:10.1016/j.jksus.2022.101832

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Original article

04 2022

:34;

101832

doi:

10.1016/j.jksus.2022.101832

Fractional partial differential equations and novel double integral transform

Tarig M. Elzaki^{^a}, Shams A. Ahmed^{^b,^c}, Mounirah Areshi^{^d}, Mourad Chamekh^{^a,^e,⁎}

a Mathematics Department, College of Science and Arts, AlKamel, University of Jeddah, Saudi Arabia

b Department of Mathematic, Faculty of Sciences and Arts, Jouf University, Tubarjal, Saudi Arabia

c Department of Mathematic, University of Gezira, Sudan

d Department of Mathematic, Faculty of Sciences, University of Tabuk, Tabuk, Saudi Arabia

e University of Tunis El Manar, National Engineering School at Tunis, LAMSIN, B.P. 37, 1002 Tunis-Belvédère, Tunisia

⁎Corresponding author.

Received: 2021-4-5, Accepted: 2022-1-10,

Disclaimer:
This article was originally published by Elsevier and was migrated to Scientific Scholar after the change of Publisher.

Peer review under responsibility of King Saud University.

Abstract

We propose in this study a combined expression mainly based on the double transformation of Laplace and Sumudu (DLST), by developing some results associated with this proposed transformation. We can apply this double transformation to certain functions to achieve interesting results which can be used to solve certain classes of fractional partial differential equations (FPDE). The numerical results show that this double transformation can lead to an exact solution of linear FPDEs. Laplace-Sumudu transform; Laplace transform; Sumudu transform; Fractional partial differential equations.

Keywords

Double Laplace-Sumudu transform

Sumudu transform

Fractional partial differential equations

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1 Introduction

The double integral transform is nevertheless a new and renewed study (see for example Alderremy et al., 2020; Alderremy and Elzaki, 2018; Elzaki, 2012; Waleed et al., 2021 ) where previous work considered definitions, simple theories of PDEs (Weatugala, 1993; Belgacem and Karaballi, 2002; Belgacem and Karaballi, 2006 ; Belgacem and Karaballi, 2006b). The combination of the integral transform with other computational techniques, such as Differential Transform Method, Homotopy Perturbation Method, Adomian Method and Variational Iteration Method has been the subject of much work (Ahmed et al., 2020; Elbadri et al., 2020; Mohamed and Elzaki, 2020; Alderremy et al., 2020, Chamekh et al., 2019; Chamekh et al., 2021 ) to solve differential equations. In this article, we will develop some results using the proposed double integral transformation. As this transform is still under study, we will compare the solutions with solutions obtained with standard methods. For non-linear FPDEs the work needs even more effort in the future to adapt this transformation to these types of equations.

Here we propose to study the FPDEs in the form:

Let $c, d, e$ real constants and $L$ is the linear differential operator. We consider resolving the following problem:

(1)

c \frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} + d \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}} + e L φ (z, t) = s (z, t), z, t \geq 0,

with

m - 1 < δ \leq m, n - 1 < ς \leq n, m, n \in N,

with

s (x, t)

is a source term,with the initial conditions ICs:

(2)

\frac{\partial^{j} φ (x, 0)}{\partial t^{j}} = f_{j} (x), j = 0, 1, . . ., m - 1,

and the boundary conditions BCs:

(3)

\frac{\partial^{k} φ (0, t)}{\partial z^{k}} = h_{k} (t), k = 0, 1, . . ., n - 1

In what follows, we will consider DLST to solve the problem (1–3).

2 Preliminaries

Definition: 2.1

The DLST of $φ (z, t)$ , denoted by: $L_{z} S_{t} [φ (z, t)] = \bar{φ} (ρ, σ)$ and defined as:

(4)

L_{x} S_{t} [φ (z, t)] = \bar{φ} (ρ, σ) = \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} φ (z, t) d z d t .

Clearly, the linearity of the (DLST) is shown in the relationship below:

(5)

L_{z} S_{t} [η φ (z, t) + γ ψ (z, t)] = \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} [η φ (z, t) + γ ψ (z, t)] d z d t = \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} η φ (z, t) d z d t + \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} γ ψ (z, t) d z d t = \frac{η}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} φ (z, t) d x d t + \frac{γ}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} ψ (z, t) d z d t = η L_{z} S_{t} [φ (z, t)] + γ L_{z} S_{t} [ψ (z, t)],

where

η

and

γ

are constants.

Definition: 2.2

The inverse (DLST) $L_{z}^{- 1} S_{t}^{- 1} [φ (ρ, σ)] = φ (z, t)$ is defined by:

(6)

L_{x}^{- 1} S_{t}^{- 1} [\bar{φ} (ρ, σ)] = φ (z, t) = (\frac{1}{2 π i}) \int_{γ - i \infty}^{γ + i \infty} e^{ρ z} d ρ (\frac{1}{2 π i}) \int_{η - i \infty}^{η + i \infty} \frac{1}{σ} e^{\frac{t}{σ}} \bar{φ} (ρ, σ) d σ .

Definition: 2.3The $δ o r ζ$ th order Caputo derivative ( $δ > 0$ , $ς > 0$ ) of $φ (z, t)$ is given by:

(7)

\begin{matrix} \frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} = \frac{1}{Γ (n - δ)} \int_{0}^{t} {(t - τ)}^{n - δ - 1} \frac{\partial^{n} φ (z, τ)}{\partial τ^{n}} d τ, p - 1 < δ ⩽ p, p \in N, \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}} \\ = \frac{1}{Γ (q - ς)} \int_{0}^{z} {(z - ζ)}^{q - ς - 1} \frac{\partial^{q} φ (ζ, t)}{\partial ζ^{q}} d ζ, q - 1 < ς ⩽ q, q \in N \end{matrix}

Using the results proven in Sakamoto et al., 2011, the DLST for the Caputo derivatives can give by:

(8)

L_{z} S_{t} [\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}}] = σ^{- δ} \bar{φ} (ρ, σ) - \sum_{j = 0}^{n - 1} σ^{- δ + j} L_{z} [\frac{\partial^{j} φ (z, 0)}{\partial t^{j}}], L_{z} S_{t} [\frac{\partial^{ς} φ (z, t)}{\partial z^{ς}}] = ρ^{ς} \bar{φ} (ρ, σ) - \sum_{k = 0}^{q - 1} ρ^{ς - 1 - k} S_{t} [\frac{\partial^{k} φ (0, t)}{\partial x^{k}}]

Definition: 2.4

The Mittag-Leffer functions $E_{δ, ς} (x)$ , $δ > 0$ and $ς > 0$ , is,

(9)

E_{δ, ς} (z) = \sum_{k = 0}^{\infty} \frac{z^{k}}{Γ (δ k + ς)}, z \in C, R e (δ) > 0, R e (ς) > 0 .

The single transform (ST) of $z^{ς - 1} E_{δ, ς} (λ z^{δ})$ and $t^{ς - 1} E_{δ, ς} (λ t^{δ})$ takes the form:

(10)

L_{x} [z^{ς - 1} E_{δ, ς} (λ z^{δ}),] = \frac{ρ^{δ - ς}}{ρ^{δ} - λ}, |λ| < |ρ^{δ}|, S_{t} [t^{ς - 1} E_{δ, ς} (λ t^{δ}),] = \frac{σ^{ς - 1}}{1 - λ σ^{δ}}, |λ| < |σ^{δ}| .

3 Fundamental properties of (DLST)

Sr. No	$φ (x, t)$	$L_{x} S_{t} [φ (x, t)] = \bar{φ} (ρ, σ)$
1	$1$	$\frac{1}{ρ}$
2	$z^{c} t^{d}$	$\frac{c! d!}{p^{c + 1}} σ^{d}$
3	$e^{cz + d t}$	$\frac{1}{(ρ - c) (1 - d σ)}$
4	$\sin (c z + d t)$	$\frac{c + d σ ρ}{(ρ^{2} + c^{2}) (1 + d^{2} σ^{2})}$
5	$\cos (c z + d t)$	$\frac{ρ - c d σ}{(ρ^{2} + c^{2}) (1 + d^{2} σ^{2})}$
6	$\sinh (c z + d t)$	$\frac{c + d σ ρ}{(ρ^{2} - c^{2}) (1 - d^{2} σ^{2})}$
7	$\cosh (c z + d t)$	$\frac{ρ + c d σ}{(ρ^{2} - c^{2}) (1 - d^{2} σ^{2})}$
8	$J_{0} (c \sqrt{zt})$ , $J_{0}$ is the zero order Bessel function	$\frac{4}{4 ρ + σ c^{2}}$
9	$φ (z - δ, t - ε) H (z - δ, t - ε)$	$e^{- ρ δ - \frac{ε}{σ}} \bar{ϕ} (ρ, σ)$
10	$(φ * * ψ) (z, t)$	$σ \bar{ϕ} (ρ, σ) \bar{ψ} (ρ, σ)$
11	$f (z) g (t)$	$L_{x} [f (x)] S_{t} [g (t)]$

3.1

3.1 Existence condition for the DLST

If $φ$ is an exponential order a and $b$ ; $z, t$ tend to infinity and if it exist a nonnegative real, $K : \forall z > Z, t > T;$

Then:

(11)

|φ (z, t)| = K e^{az + b t},

and we write:

(12)

φ (z, t) = O {(e}^{az + b t}) a s z, t t e n d t o i n f i n i t y

Or,

(13)

\lim_{z \to \infty, t \to \infty} e^{- ρ z - \frac{t}{σ}} |φ (z, t)| = K \lim_{z \to \infty, t \to \infty} e^{- (ρ - a) z - (\frac{1}{σ} - b) t} = 0, ρ > a, \frac{1}{σ} > b .

Then

φ

is an exponential order as

x, t

tend to infinity.

Theorem 3.1.1

If a function $φ$ on the interval $(0, Z)$ and $(0, T)$ of exponential order $c$ and $d$ , then the DLST of $φ$ well-defined for all $ρ$ and $\frac{1}{σ}$ supplied $Re [ρ] > a$ and $Re [\frac{1}{σ}] > b .$

Proof.

We find, from the Def. 2.1,

(14)

|\bar{φ} (ρ, σ)| = |\frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} φ (z, t) d z d t| \leq K \int_{0}^{\infty} e^{- (ρ - a) z} d z \int_{0}^{\infty} \frac{1}{σ} e^{- (\frac{1}{σ} - b) t} d t = \frac{K}{(ρ - a) (1 - b σ)}, Re [ρ] > a, Re [\frac{1}{σ}] > b .

So, from Eq. (14) we’ve got,

\lim_{z \to \infty, t \to \infty} |\bar{φ} (ρ, σ)| = 0, o r \lim_{z \to \infty, t \to \infty} \bar{φ} (ρ, σ) = 0 .

3.2

3.2 Basic derivatives properties of the DLST

If, $\bar{φ} (ρ, σ) = L_{z} S_{t} [ϕ (z, t)]$ , then:

(15)

L_{z} S_{t} [\frac{\partial φ (z, t)}{\partial z}] = ρ \bar{φ} (ρ, σ) - S [φ (0, t)] .

(16)

L_{z} S_{t} [\frac{\partial φ (z, t)}{\partial t}] = \frac{1}{σ} \bar{φ} (ρ, σ) - \frac{1}{σ} L (φ (z, 0))

(17)

L_{z} S_{t} [\frac{\partial^{2} φ (z, t)}{\partial z^{2}}] = ρ^{2} \bar{φ} (ρ, σ) - ρ S (φ (0, t)) - S (φ_{z} (0, t)) .

(18)

L_{z} S_{t} [\frac{\partial^{2} φ (z, t)}{\partial t^{2}}] = \frac{1}{σ^{2}} \bar{φ} (ρ, σ) - \frac{1}{σ^{2}} L (φ (z, 0)) - \frac{1}{σ} L (φ_{t} (z, 0)) .

(19)

L_{z} S_{t} [\frac{\partial^{2} φ (z, t)}{\partial z \partial t}] = \frac{ρ}{σ} \bar{φ} (ρ, σ) - \frac{ρ}{σ} L (φ (z, 0)) - S (φ_{t} (0, t)) .

Proof

For (16) $L_{z} S_{t} [\frac{\partial φ (z, t)}{\partial z}] = \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} \frac{\partial φ (z, t)}{\partial z} d z d t = \frac{1}{σ} \int_{0}^{\infty} e^{- \frac{t}{σ}} d t \{\int_{0}^{\infty} e^{- ρ z} \frac{\partial φ (z, t)}{\partial z} d z\}$

Let, $u = e^{- ρ z}, d v = \frac{\partial φ (z, t)}{\partial z} d z,$ thus: $L_{z} S_{t} [\frac{\partial ϕ (z, t)}{\partial z}] = \frac{1}{σ} \int_{0}^{\infty} e^{- \frac{t}{σ}} d t \{- φ (0, t) + ρ \int_{0}^{\infty} e^{- ρ z} φ (z, t) d x\} = ρ \bar{φ} (ρ, σ) - S (φ (0, t)) .$

Concerning (16) $L_{z} S_{t} [\frac{\partial φ (z, t)}{\partial t}] = \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} \frac{\partial φ (z, t)}{\partial t} d z d t = \frac{1}{σ} \int_{0}^{\infty} e^{- ρ z} d z \{\int_{0}^{\infty} e^{- \frac{t}{σ}} \frac{\partial ϕ (z, t)}{\partial t} d t\}$

Let, $u = e^{- \frac{t}{σ}}, d v = \frac{\partial φ}{\partial t} d t,$ then: $L_{z} S_{t} [\frac{\partial φ (z, t)}{\partial t}] = \frac{1}{σ} \int_{0}^{\infty} e^{- ρ x z} d z \{- φ (z, 0) + \frac{1}{σ} \int_{0}^{\infty} e^{- \frac{t}{σ}} φ (z, t) d t\} = \frac{1}{σ} \bar{φ} (ρ, σ) - \frac{1}{σ} L (φ (z, 0)) .$

It is easy to prove (17), (18), and (19).

Theorem 3.2.1

If $\bar{φ} (ρ, σ) = L_{z} S_{t} [φ (z, t)]$ , thus:

(20)

L_{z} S_{t} [φ (- δ + z, - ε + t) H (- δ + z, - ε + t)] = e^{- ρ δ - \frac{ε}{σ}} \bar{φ} (ρ, σ),

(21)

H (- δ + z, - ε + t) = \{\begin{matrix} 1, z > δ, t > ε \\ 0, o t h e r w i s e \end{matrix}\} .

Proof

Using Def. 2.1 $L_{z} S_{t} [φ (- δ + z, - ε + t) H (- δ + z, - ε + t)] = \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} φ (- δ + z, - ε + t) H (- δ + z, - ε + t) d z d t = \frac{1}{σ} \int_{δ}^{\infty} \int_{ε}^{\infty} e^{- ρ z - \frac{t}{σ}} φ (- δ + z, - ε + t) d x d t,$ Putting, $z - δ = q, t - ε = w = e^{- ρ δ - \frac{ε}{σ}} \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ q - \frac{w}{σ}} φ (q, w) d q d w = e^{- ρ δ - \frac{ε}{σ}} \bar{φ} (ρ, σ) .$

3.3

3.3 Convolution Theorem of DLST

3.4 Definition 3.3.1

The double convolution of $ϕ (x, t)$ and $ψ (x, t)$ is given,

(22)

(φ * * ψ) (z, t) = \int_{0}^{z} \int_{0}^{t} φ (- δ + z, - ε + t) ψ (δ, ε) d δ d ε .

Theorem 3.3.1 (Convolution Theorem)

If $L_{x} S_{t} [φ (z, t)] = \bar{φ} (ρ, σ)$ and $L_{z} S_{t} [ψ (z, t)] = \bar{ψ} (ρ, σ)$ then:

(23)

L_{z} S_{t} [(φ * * ψ) (z, t)] = σ \bar{φ} (ρ, σ) \bar{ψ} (ρ, σ) .

Proof

Using Def. 2.1 $L_{z} S_{t} [(φ * * ψ) (z, t)] = \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} (φ * * ψ) (z, t) d z d t = \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} \{\int_{0}^{z} \int_{0}^{t} φ (- δ + z, - ε + t) ψ (δ, ε) d δ d ε\} d z d t, = \frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} \{\int_{0}^{\infty} \int_{0}^{\infty} φ (- δ + z, - ε + t) H (- δ + z, - ε + t) ψ (δ, ε) d δ d ε\} d z d t = \int_{0}^{\infty} \int_{0}^{\infty} ψ (δ, ε) d δ d ε \{\frac{1}{σ} \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ z - \frac{t}{σ}} φ (- δ + z, - ε + t) H (- δ + z, - ε + t) d z d t\}, = \int_{0}^{\infty} \int_{0}^{\infty} ψ (δ, ε) d δ d ε \{e^{- ρ δ - \frac{ε}{σ}} \bar{φ} (ρ, σ)\} = \bar{φ} (ρ, σ) \int_{0}^{\infty} \int_{0}^{\infty} e^{- ρ δ - \frac{ε}{σ}} ψ (δ, ε) d δ d ε = σ \bar{ϕ} (ρ, σ) \bar{ψ} (ρ, σ)$

Using Theorem 3.2.1 to find

4 Principle of DLST method

In this part, we exercised DLST to find the solution to the Eq. (1).

First, apply DLST to Eq. (1), we get:

(24)

c [σ^{- δ} \bar{ϕ} (ρ, σ) - \sum_{j = 0}^{n - 1} σ^{- δ + j} L_{x} [\frac{\partial^{j} ϕ (x, 0)}{\partial t^{j}}]] + d [ρ^{ς} \bar{ϕ} (ρ, σ) - \sum_{k = 0}^{m - 1} ρ^{ς - 1 - k} S_{t} [\frac{\partial^{k} ϕ (0, t)}{\partial x^{k}}]] + e L_{x} S_{t} [L ϕ (x, t)] = \bar{g} (ρ, σ)

Furthermore, applying single (LT) to the ICs(2) and single (ST) to the BCs(3), we've got,

(25)

L_{x} [\frac{\partial^{j} ϕ (x, 0)}{\partial t^{j}}] = \bar{f_{j}} (ρ), j = 0, 1, . . ., n - 1, S_{t} [\frac{\partial^{k} ϕ (0, t)}{\partial x^{k}}] = \bar{h_{k}} (σ), k = 0, 1, \dots, m - 1 .

By replacing (25) in (24), we get

(26)

\bar{ϕ} (ρ, σ) = \frac{1}{(c σ^{- δ} + d ρ^{ς})} [\begin{matrix} c \sum_{j = 0}^{n - 1} σ^{- δ + j} \bar{f_{j}} (ρ) + d \sum_{k = 0}^{m - 1} ρ^{ς - 1 - k} \bar{h_{k}} (σ) - e L_{x} S_{t} [L ϕ (x, t)] \\ + \bar{g} (ρ, σ) \end{matrix}]

Taking $L_{x}^{- 1} S_{t}^{- 1} [\bar{ϕ} (ρ, σ)]$ of (26) to find the solution of (1);

(27)

ϕ (x, t) = L_{x}^{- 1} S_{t}^{- 1} [\frac{1}{(c σ^{- δ} + d ρ^{ς})} [\begin{matrix} c \sum_{j = 0}^{n - 1} σ^{- δ + j} \bar{f_{j}} (ρ) + d \sum_{k = 0}^{m - 1} ρ^{ς - 1 - k} \bar{h_{k}} (σ) \\ - e L_{x} S_{t} [L ϕ (x, t)] + \bar{g} (ρ, σ) \end{matrix}]]

5 Elucidative examples

Without forgetting the work of Podlubny 1999, who has proposed some analytical solutions for certain linear equations which can be calculated. We treat in this numerical part some physical applications using the DLST method. The objective is to illustrate the method and prove its applicability.

5.1

5.1 Fractional advection-diffusion equation

Putting, $c = 1, n = 1, s = 0, m = 2, d = - q_{1}, e = - q_{2}, L = \frac{\partial}{\partial x}$ , to find,

(28)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} = q_{1} \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}} + q_{2} \frac{\partial φ (z, t)}{\partial z}, 0 < δ \leq 1, 1 < ς \leq 2,

WithICs and BCs:

(29)

φ (z, 0) = f_{0} (z), φ (0, t) = h_{0} (t), φ_{z} (0, t) = h_{1} (t),

So,(27) yields the solution of (28)

(30)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(σ^{- δ} - q_{1} ρ^{ς} - q_{2} ρ)} [\begin{matrix} σ^{- δ} \bar{f_{0}} (ρ) - q_{1} ρ^{ς - 1} \bar{h_{0}} (σ) - q_{1} ρ^{ς - 2} \bar{h_{1}} (σ) \\ - q_{2} \bar{h_{0}} (σ) \end{matrix}]]

5.2

5.2 Fractional reaction-diffusion equation

Putting, $s = 0, m = 2,$ $n = 1, a = 1, b = - q,$ and $L φ (z, t) = φ (z, t)$ in Eq. (1), to get,

(31)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} + e φ (z, t) = q \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}}, 0 < δ \leq 1, 1 < ς \leq 2,

With, ICs and BCs:

(32)

φ (z, 0) = f_{0} (z), φ (0, t) = h_{0} (t), φ_{z} (0, t) = h_{1} (t),

So,(27) gives a solution of (31) as:

(33)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(σ^{- δ} + e - q ρ^{ς})} [σ^{- δ} \bar{f_{0}} (ρ) - q ρ^{ς - 1} \bar{h_{0}} (σ) - q ρ^{ς - 2} \bar{h_{1}} (σ)]] .

5.2.1

5.2.1 Fractional heat (diffusion) equation

If we choose, $e = 0$ in (31), we got,

(34)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} = q \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}}, 0 < δ \leq 1, 1 < ς \leq 2,

With, ICs and BCs:

(35)

φ (z, 0) = f_{0} (z), φ (0, t) = h_{0} (t), φ_{z} (0, t) = h_{1} (t),

So, (33) yields the solution of (34)

(36)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(σ^{- δ} - q ρ^{ς})} [σ^{- δ} \bar{f_{0}} (ρ) - q ρ^{ς - 1} \bar{h_{0}} (σ) - q ρ^{ς - 2} \bar{h_{1}} (σ)]] .

Example 1: Putting $q = 1$ and $ς = 2$ in (34), to yield,

(37)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} = \frac{\partial^{2} φ (z, t)}{\partial z^{2}}, 0 < δ \leq 1,

with;

(38)

φ (z, 0) = \sin z = f_{0} (z), φ (0, t) = 0 = h_{0} (t), φ_{z} (0, t) = E_{δ} (- t^{δ}) = h_{1} (t)

Replacing,

\bar{f_{0}} (ρ) = \frac{1}{ρ^{2} + 1}, \bar{h_{0}} (σ) = 0, \bar{h_{1}} (σ) = \frac{1}{1 + σ^{δ}},

in (36) and simplifying, we obtain,

(39)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(ρ^{2} + 1) (1 + σ^{δ})}] = \sin z E_{δ} (- t^{δ}) .

When,

δ = 1

, we get,

(40)

φ (z, t) = e^{- t} \sin z .

5.3

5.3 Fractional telegraph equation

Putting $m = 2, n = 2, d = - q, s = 0,$ and $L = c_{0} + c_{1} \frac{\partial}{\partial t}$ , we’ve got,

(41)

c \frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} + c_{1} \frac{\partial φ (z, t)}{\partial t} + c_{0} φ (z, t) = q \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}}, 1 < δ, ς \leq 2,

With, ICs and BCs:

(42)

φ (z, 0) = f_{0} (z), φ_{t} (z, 0) = f_{1} (z), φ (0, t) = h_{0} (t), φ_{z} (0, t) = h_{1} (t),

So, (27) gives the solution of (41) in the form,

(43)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(c_{0} + c_{1} σ^{- 1} + c σ^{- δ} - q ρ^{ς})} [\begin{matrix} c_{1} σ^{- 1} \bar{f_{0}} (ρ) + c σ^{- δ} \bar{f_{0}} (ρ) + c σ^{- δ + 1} \bar{f_{1}} (ρ) \\ - q ρ^{ς - 1} \bar{h_{0}} (σ) - q ρ^{ς - 2} \bar{h_{1}} (σ) \end{matrix}]] .

Example 2: If we choose, $c_{0} = c_{1} = c = q = 1$ and $δ = 2$ in (41), we get,

(44)

φ (z, t) + \frac{\partial φ (z, t)}{\partial t} + \frac{\partial^{2} φ (z, t)}{\partial t^{2}} = \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}}, 1 < ς \leq 2,

(45)

φ (0, t) = h_{0} (t) = e^{- t}, φ_{z} (0, t) = h_{1} (t) = e^{- t},

Replacing,

\bar{f_{0}} (ρ) = (\frac{1}{ρ} + \frac{1}{ρ^{2}}) \frac{ρ^{ς}}{ρ^{ς} - 1}, \bar{f_{1}} (ρ) = - (\frac{1}{ρ} + \frac{1}{ρ^{2}}) \frac{ρ^{ς}}{ρ^{ς} - 1}, \bar{h_{0}} (σ) = \bar{h_{1}} (σ) = \frac{1}{1 + σ},

in (43) and simplify it, to find:

(46)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(1 + σ)} (\frac{1}{ρ} + \frac{1}{ρ^{2}}) \frac{ρ^{ς}}{ρ^{ς} - 1}] = e^{- t} [E_{ς} (z^{ς}) + z E_{ς, 2} (z^{ς})],

When

ς = 2

, we get

(47)

φ (z, t) = e^{z - t} .

5.3.1

5.3.1 Fractional wave equation

If we choose, $c_{0} = c_{1} = 0, c = 1$ in (41), then, we get,

(48)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} = q \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}}, 1 < δ, ς \leq 2,

With ICs and BCs:

(49)

φ (z, 0) = f_{0} (z), φ_{t} (z, 0) = f_{1} (z), φ (0, t) = h_{0} (t), φ_{z} (0, t) = h_{1} (t),

Then, (43) gives a solution of (48) as;

(50)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(σ^{- δ} - q ρ^{ς})} [σ^{- δ} \bar{f_{0}} (ρ) + σ^{- δ + 1} \bar{f_{1}} (ρ) - q ρ^{ς - 1} \bar{h_{0}} (σ) - q ρ^{ς - 2} \bar{h_{1}} (σ)]] .

Fractional Klein–Gordon equation

Putting, $n = m = 2, c = 1, d = - 1,$ and $L φ = φ$ , to find,

(51)

\frac{\partial^{δ} φ}{\partial t^{δ}} - \frac{\partial^{ς} φ}{\partial z^{ς}} + e φ (z, t) = s (z, t), 1 < δ, ς \leq 2 .

With ICs and BCs:

(52)

φ (z, 0) = f_{0} (z), φ_{t} (z, 0) = f_{1} (z), φ (0, t) = h_{0} (t), φ_{z} (0, t) = h_{1} (t) .

So, (27) yields the solution of (51),

(53)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(e + σ^{- δ} - ρ^{ς})} [\begin{matrix} σ^{- δ} \bar{f_{0}} (ρ) + σ^{- δ + 1} \bar{f_{1}} (ρ) - ρ^{ς - 1} \bar{h_{0}} (σ) \\ - ρ^{ς - 2} \bar{h_{1}} (σ) + \bar{s} (ρ, σ) \end{matrix}]] .

Example 3: If we choose, $e = - 1, ς = 2,$ and $g (x, t) = 0$ in (51), we have got

(54)

\frac{\partial^{δ} φ (x, t)}{\partial t^{δ}} - \frac{\partial^{2} φ (x, t)}{\partial z^{2}} - φ (z, t) = 0, 1 < δ \leq 2,

with;

(55)

φ (z, 0) = \sin z + 1 = f_{0} (x), φ_{t} (z, 0) = 0 = f_{1} (z), φ (0, t) = E_{δ} (t^{δ}) = h_{0} (t), φ_{z} (0, t) = 1 = h_{1} (t) .

Substituting,

\bar{f_{0}} (ρ) = \frac{1}{ρ^{2} + 1} + \frac{1}{ρ}, \bar{f_{1}} (ρ) = 0, \bar{h_{0}} (σ) = \frac{1}{1 - σ^{δ}}, \bar{h_{1}} (σ) = 1,

in (53) and simplifying, to obtain,

(56)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{ρ^{2} + 1} + \frac{1}{ρ (1 - σ^{δ})}] = \sin z + E_{δ} (t^{δ}) .

5.4

5.4 Fractional Burger’s equation

Choosing, $= 2, n = 1, d = - 1, c = 1, e = 1, L = \frac{\partial}{\partial z}$ , we obtain,

(57)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} - \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}} + \frac{\partial φ (z, t)}{\partial z} = s (z, t), 0 < δ \leq 1, 1 < ς \leq 2,

and

(58)

φ (z, 0) = f_{0} (z), φ (0, t) = h_{0} (t), z (0, t) = h_{1} (t),

then we have the following solution,

(59)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(σ^{- δ} - ρ^{ς} + ρ)} [\begin{matrix} σ^{- δ} \bar{f_{0}} (ρ) + \bar{h_{0}} (σ) - ρ^{ς - 1} \bar{h_{0}} (σ) \\ - ρ^{ς - 2} \bar{h_{1}} (σ) + \bar{s} (ρ, σ) \end{matrix}]] .

Example 4: If we choose, $ς = 2,$ and $s (z, t) = 0$ in (57) to get,

(60)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} - \frac{\partial^{2} φ (z, t)}{\partial z^{2}} + \frac{\partial φ (z, t)}{\partial z} = 0, 0 < δ \leq 1,

with;

(61)

φ (z, 0) = e^{- z} = f_{0} (z), φ (0, t) = E_{δ} (2 t^{δ}) = h_{0} (t), φ_{z} (0, t) = - E_{δ} (2 t^{δ}) = h_{1} (t) .

Replacing, $\bar{f_{0}} (ρ) = \frac{1}{ρ + 1}, \bar{h_{0}} (σ) = \frac{1}{1 - 2 σ^{δ}}, \bar{h_{1}} (σ) = - \frac{1}{1 - 2 σ^{δ}},$ in (59) and simplifying, to obtain,

(62)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(ρ + 1) (1 - 2 σ^{δ})}] = e^{- z} E_{δ} (2 t^{δ}) .

5.5

5.5 Fractional Fokker-Planck equation

Putting, $m = 2, n = 1, d = - 1, c = 1, e = - 1, L = \frac{\partial}{\partial z},$ and $s (z, t) = 0$ , to get,

(63)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} = \frac{\partial^{ς} φ (z, t)}{\partial z^{ς}} + \frac{\partial φ (z, t)}{\partial z}, 0 < δ \leq 1, 1 < ς \leq 2,

with ICs and BCs:

(64)

φ (z, 0) = f_{0} (z), φ (0, t) = h_{0} (t), φ_{z} (0, t) = h_{1} (t),

Then:

(65)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(σ^{- δ} - ρ^{ς} - ρ)} [σ^{- δ} \bar{f_{0}} (ρ) - ρ^{ς - 1} \bar{h_{0}} (σ) - ρ^{ς - 2} \bar{h_{1}} (σ) - \bar{h_{0}} (σ)]] .

Example 5: If we choose, $ς = 2$ in (63) then, we find:

(66)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} = \frac{\partial^{2} φ (z, t)}{\partial z^{2}} + \frac{\partial φ (z, t)}{\partial z}, 0 < δ \leq 1,

with,

(67)

φ (z, 0) = z = f_{0} (z), φ (0, t) = \frac{t^{δ}}{Γ (1 + δ)} = h_{0} (t), φ_{z} (0, t) = 1 = h_{1} (t) .

Replacing, $\bar{f_{0}} (ρ) = \frac{1}{ρ^{2}}, \bar{h_{0}} (σ) = σ^{δ}, \bar{h_{1}} (σ) = 1,$ in (65) and simplify it, to find:

(68)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{ρ^{2}} + \frac{σ^{δ}}{ρ}] = z + \frac{t^{δ}}{Γ (1 + δ)} .

5.6

5.6 Fractional Korteweg-de Vries (KdV) equation

Putting, $m = 1, n = 1, c = 1,$ and $L = \frac{\partial^{3}}{\partial z^{3}}$ in (1), to get,

(69)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} + d \frac{\partial^{ς} φ (x, t)}{\partial z^{ς}} + e \frac{\partial^{3} φ (φ, t)}{\partial z^{3}} = s (z, t), 0 < δ, ς \leq 1,

with ICs and BCs:

(70)

φ (z, 0) = f_{0} (z), φ (0, t) = h_{0} (t), φ_{z} (0, t) = h_{1} (t), φ_{zz} (0, t) = h_{2} (t),

Then:

(71)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(σ^{- δ} + d ρ^{ς} + e ρ^{3})} [\begin{matrix} σ^{- δ} \bar{f_{0}} (ρ) + e ρ^{2} \bar{h_{0}} (σ) + e ρ \bar{h_{1}} (σ) \\ + e \bar{h_{2}} (σ) + d ρ^{ς - 1} \bar{h_{0}} (σ) + \bar{s} (ρ, σ) \end{matrix}]] .

Example 6: If we choose, $d = 2, e = 1, ς = 1,$ and $s (x, t) = 0$ in (69) we have:

(72)

\frac{\partial^{δ} φ (z, t)}{\partial t^{δ}} + 2 \frac{\partial φ (z, t)}{\partial z} + \frac{\partial^{3} φ (z, t)}{\partial z^{3}} = 0, 0 < δ \leq 1,

with,

(73)

φ (z, 0) = f_{0} (z) = \sin z, φ (0, t) = - t^{δ} E_{2 δ, δ + 1} (- t^{2 δ}) = h_{0} (t), φ_{z} (0, t) = E_{2 δ, 1} (- t^{2 δ}) = h_{1} (t), φ_{zz} (0, t) = t^{δ} E_{2 δ, δ + 1} (- t^{2 δ}) = h_{2} (t) .

Replacing,

\bar{f_{0}} (ρ) = \frac{1}{ρ^{2} + 1}, \bar{h_{0}} (σ) = - \frac{σ^{δ}}{1 + σ^{2 δ}}, \bar{h_{1}} (σ) = \frac{1}{1 + σ^{2 δ}}, \bar{h_{2}} (σ) = \frac{σ^{δ}}{1 + σ^{2 δ}},

in (71) and simplifying, to obtain,

(74)

φ (z, t) = L_{z}^{- 1} S_{t}^{- 1} [\frac{1}{(ρ^{2} + 1) (1 + σ^{2 δ})} - \frac{ρ σ^{δ}}{(ρ^{2} + 1) (1 + σ^{2 δ})}] = (\sin z) E_{2 δ, 1} (- t^{2 δ}) - (\cos z) t^{δ} E_{2 δ, δ + 1} (- t^{2 δ}) .

6 Discussions

We are going look in this section the numeric evaluation of obtained results of fractional equations that have proposed to be solve. Moreover, we will discuss the numerical behavior of the solution resulting from a fractional differential equation and compare it with that of the equation with integer derivative.

When δ = 1, the closed form solution of each Example 3, 4, 5 and 6 is simply calculated. We have opted to examine the numerical results for the values of δ = 0.95, 0.85 and 0.75. We have noticed that the solutions obtained for the different fractional values of δ are in coordination with the solution of closed form for α = 1, as shown in Figs. 1–4 . It suffices to clearly notice that when α approaches 1, the solution resulting from the fractional equation approaches this exact solution.

Solutions of Fractional Korteweg-de Vries (KdV).

7 Conclusions

This article deals with the approach of a new double transformation. The objective is based on the simplicity of the technique used to solve a large class of fractional equations of mathematical physics. The examples show that DLST can be a good alternative to treat many LFPDEs that appear in science. However, it should be noted that the solutions obtained are valid while the inverse of this transform exists.

Acknowledgement

This work was funded by the University of Jeddah , Saudi Arabia, under grant No. (UJ-20-101-DR ). The authors, therefore, acknowledge with thanks the University technical and financial support.

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

References

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Show Sections

[1] Ahmed S., Elbadri M., Mohamed Z.M., . A new efficient method for solving two-dimensional nonlinear system of Burger’s differential equation. J. Abstr. Appl. Anal. 2020:1-7.
[Google Scholar]

[2] Alderremy A.A., Chamekh M., Jeday F., . Semi-analytical solution for a system of competition with production a toxin in a chemostat. J. Math. Computer Sci.. 2020;20(02):155-160.
[Google Scholar]

[3] Belgacem F.B.M., Karaballi A.A., . Analytical investigations of the Sumudu transform and applications to integral production equations. Math. Probl. Eng. (3):103-118.
[Google Scholar]

[4] Belgacem F.B.M., Karaballi A.A., . Sumudu transform fundamental properties investigations and applications. J. Appl. Math. Stochast. Anal. 2006:1-23.
[Google Scholar]

[5] Chamekh M., Elzaki T.M., Brik N., . Semianalytical solution for some proportional delay differential equations. SN Appl. Sci.. 2019;1:148.
[Google Scholar]

[6] Chamekh M., Latrach M.A., Elzaki T.M., . Approximate analytical solutions for some obstacle problems. J. King Saud Univ. Sci.. 2021;33(2):101259.
[Google Scholar]

[7] Elbadri M., Ahmed S.A., Abdalla Y.T., Hdidi W., . A new solution of time-fractional coupled KdV equation by using natural decomposition method. J. Abstr. Appl. Anal. 2020:1-9.
[Google Scholar]

[8] Elzaki T.M., . Double Laplace variational iteration method for solution of nonlinear convolution partial differential equations. Arch. Sci.. 2012;65(12):588-593.
[Google Scholar]

[9] Alderremy A.A., Elzaki T.M., . On the new double integral transform for solving singular system of hyperbolic equations. J. Nonlinear Sci. Appl.. 2018;11(10):1207-1214.
[Google Scholar]

[10] Belgacem F.B.M., Karaballi A.A., . Sumudu transform fundamental properties investigations and applications. J. Appl. Math. Stochast. Anal.. 2006;2006:1-23.
[Google Scholar]

[11] Mohamed M.Z., Elzaki T.M., . Applications of new integral transform for linear and nonlinear fractional partial differential equations. J. King Saud Univ. Sci.. 2020;32(1):544-549.
[Google Scholar]

[12] Podlubny, I., 1999 Fractional Differential Equations, vol. 198 of Mathematics in Science and Engineering, Academic Press, San Diego.

[13] Weatugala G.K., . Sumudu transforms - a new integral transform to solve differential equations and control engineering problems. Math. Eng.. 1993;61:319-329.
[Google Scholar]

[14] Waleed M.O., Elzaki T.M., Siddig N.A.A., . Modified double conformable laplace transform and singular fractional pseudo-hyperbolic and pseudo-parabolic equations. J. King Saud Univ. Sci.. 2021;33:101378
[Google Scholar]

Fractional partial differential equations and novel double integral transform

Abstract

Keywords

Double Laplace-Sumudu transform

Sumudu transform

Fractional partial differential equations

1 Introduction

2 Preliminaries

3 Fundamental properties of (DLST)

3.1 Existence condition for the DLST

3.2 Basic derivatives properties of the DLST

3.3 Convolution Theorem of DLST

4 Principle of DLST method

5 Elucidative examples

5.1 Fractional advection-diffusion equation

5.2 Fractional reaction-diffusion equation

5.2.1 Fractional heat (diffusion) equation

5.3 Fractional telegraph equation

5.3.1 Fractional wave equation

5.4 Fractional Burger’s equation

5.5 Fractional Fokker-Planck equation

5.6 Fractional Korteweg-de Vries (KdV) equation

6 Discussions

7 Conclusions

Acknowledgement

Declaration of Competing Interest

References

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