solving systems of fractional partial differential...

[Aghili et. al., Vol.5 (Iss.12): December, 2017] ISSN- 2350-0530(O), ISSN- 2394-3629(P)

DOI: 10.5281/zenodo.1146189

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Science

SOLVING SYSTEMS OF FRACTIONAL PARTIAL DIFFERENTIAL

EQUATIONS VIA TWO DIMENSIONAL LAPLACE TRANSFORMS

A. Aghili *1

, M.R. Masomi 2

*1, 2 Department of Applied Mathematics, Faculty of Mathematical Sciences, University of

Guilan, P.O. Box, 1841, Rasht-Iran

Abstract

In this article, the authors used two dimensional Laplace transform to solve non - homogeneous

sub - ballistic fractional PDE and homogeneous systems of time fractional heat equations.

Constructive examples are also provided.

Mathematics Subject Classification 2010: 26A33; 34A08; 34K37; 35R11; 44A10.

Keywords: Fractional Partial Differential Equations; Fractional Diffusion Equations; Two

Dimensional Laplace Transform; Systems of Heat Equations.

Cite This Article: A. Aghili, and M.R. Masomi. (2017). “SOLVING SYSTEMS OF

FRACTIONAL PARTIAL DIFFERENTIAL EQUATIONS VIA TWO DIMENSIONAL

LAPLACE TRANSFORMS.” International Journal of Research - Granthaalayah, 5(12), 406-

420. https://doi.org/10.5281/zenodo.1146189.

1. Introduction and Preliminaries

The problem related to partial differential equation commonly can be solved by using a special

integral transform, thus many authors solved the boundary value problems by using single

Laplace transform. Furthermore, two dimensional Laplace transforms in the classical sense for

solving linear second order partial differential equations were used by Ditkin [11], Brychkov

[13]. The Laplace transform, it can be fairly said, stands first in importance among all integral

transforms for which there are many specific examples in which other transforms prove more

expedient. The Laplace transform is the most powerful in dealing with both initial boundary

value problems and transforms [1-9]. The two dimensional Laplace transforms is a powerful tool

in applied mathematics and engineering.

The fractional derivative is one of the most interdisciplinary fields of mathematics, with many

applications in physics and engineering and deals with extensions of derivatives and integrals to

non-integer orders. It represents a powerful tool in applied mathematics to study a myriad of

problems from different fields of science and engineering, with many breakthrough results found

in mathematical physics, finance, hydrology, biophysics, thermodynamics, control theory,

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DOI: 10.5281/zenodo.1146189


statistical mechanics, astrophysics, cosmology and bioengineering [13, 14, 15, 16]. Several

definitions have been proposed for a fractional derivative. We deal with Caputo fractional

derivatives only. In this section, we present the definition of this derivative.

Let us take f an arbitrary integrable function. By ( )a tI f t we denote the fractional integral of

f with order 0 on [ 0 , ]t defined as follows

1

1 ( )( ) .

( ) ( )

t

a ta

f xI f t dx

t x

The above integral is sometimes called the left-sided fractional integral.

For an arbitrary real number 0 ( 1 , )n n n N the Caputo fractional derivatives are

defined as

( )( )

1

1 ( )( ) ( ) .

( ) ( )

nt

C n n

a t a t na

f xD f t I f t dx

n t x

The Caputo fractional derivative is a regularization in the time origin for the Riemann-Liouville

fractional derivative by incorporating the relevant initial conditions. The major utility of the

Caputo fractional derivative is caused by the treatment of differential equations of the fractional

order for physical applications, where the initial conditions are usually expressed in terms of a

given function and its derivatives of integer (not fractional order), even if the governing equation

is of fractional order. If care is taken, the results obtained using the Caputo formulation can be

recast to the Riemann-Liouville version and vice versa.

Let ( )f t be a function of t specified for 0t . Then the Laplace transform of function ( )f t is

defined by

0

{ ( )} ( ) : ( ).s tL f t e f t dt F s

If { ( )} ( )L f t F s , then 1{ ( )}L F s is given by

1

( ) ( ) , ( 0)2

c it s

c if t e F s ds t

i

Where ( )F s is analytic in the region Re( )s c and ( ) 0f t for 0t .

This result is called complex inversion formula. It is also known as Bromwich's integral formula.

When 1n n , we get the following [14, 15]

11 ( )

0

0

{ ( )} ( ) (0).n

C k k

t

k

L D f t s F s s f



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Example 1.1. Let us solve the following fractional Volterra equation of convolution Type. The

Laplace transform provides a useful technique for the solution of such

Equations

2

0

0

(2 (t ) ( ) (2 ), 1, (0) 0.

tt

I D d J t

Solution. Upon taking the Laplace transform of the given equation, we obtain

1

( )( (s)) ,s se e

ss s

Solving the above equation, leads to

2

1(s)

se

s

.

By applying the inverse Laplace transform, we get the formal solution

1

2

1

1(t) (2 2 ).

2

tJ t

Lemma 1.1 Let us consider the following system of fractional singular integro - differential

equations of convolution-type with the Bessel kernel:

20 1 1 2

0

20 2 2 1

0

( ) ( ) ( ) (2 ( )) ( )

,

( ) ( ) ( ) (2 ( )) ( )

xc

x

xc

x

x tD g x f x J a x t g t dt

a

x tD g x f x J a x t g t dt

a

Where 1 2(0) (0) 0 , 1,0 1g g and 1 2( ), ( )f x f x are known functions.

Then the above system has the following formal solutions



DOI: 10.5281/zenodo.1146189


2

1 1

0

2

0

2

1

0

( 1)2

2 ( 1)

0

(2 1)( 1)

(2 1) 2 2(2 1)( 1)

0

( 1)2

2 (

( ) ( ) ( ) (2 2 ( )) ( )2

( 1) ( ) (2 (2 1)( )) ( )(2 1)

(0) ( ) ( )2

k

k

k

k

x kk

k

kxk

k

kk

k

x tg x J ka x t f t dt

ak

x tJ a k x t f t dt

k a

xf J

ak

2

0

1)

(2 1)( 1)

(2 1) 2 2(2 1)( 1)

(2 2 )

(0) ( 1) ( ) (2 (2 1) ).(2 1)

k

k

k

k

k

kax

xf J a k x

k a

2 1

0

2

2

0

1

0

(2 1)( 1)

(2 1) 2 2(2 1)( 1)

0

( 1)2

2 ( 1)

0

(2 1)(

(2 1) 2

( ) ( 1) ( ) (2 (2 1)( )) ( )(2 1)

( ) ( ) (2 2 ( )) ( )2

(0) ( 1) ( )(2 1)

k

k

k

k

k

kxk

k

x kk

k

k

k

x tg x J a k x t f t dt

k a

x tJ ka x t f t dt

ak

xf

k a

2

2

0

1)2

(2 1)( 1)

( 1)2

2 ( 1)

(2 (2 1) )

(0) ( ) ( ) (2 2 )2k

k

kk

k

J a k x

xf J kax

ak

Proof - In order to solve the above system, by introducing we can rewrite the above system of

partial fractional integro - differential equations in the following form

00

2( ) ( ) ( ) (2 ( )) ( )x

cx

x tD g x f x i J a x t g t d t

a

Where (0) 0, 1, 0,0 1g a .

By applying the Laplace transform of both sides of the above equation term - wise we obtain

0 0

1

11

( 1) ( 1) 1

exp( )( ) ( ) ( )

1 1( ) ( ) ( )

exp( ){1 }

exp( )

( ) ( ) ( )( )

exp( ) exp( )k k

as

as

as

k k

k kka kas s

s G s F s i G ss

G s F s F si

ss iss

F s i isF s

s s s



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Now, using the fact that

1 2 2 12

1

exp( ){ } ( ) (2 ),( ) ( 1) ,( ) ( 1)k k k k

as x

L J ax i i ias

Upon taking the inverse Laplace transform of the above term, yields

0

0

( 1)

2( 1)

0

( 1)

2( 1)

( ) ( ) ( ) (2 ( )) ( )

(0) ( ) ( ) (2 ).

k

k

kxk

k

k

kk

x tg x i J ka x t f t dt

ak

xf i J kax

ak

Finally, by taking the real and imaginary part of the above relation, we finally obtain the

solutions of the system in the following forms

2

1 1

0

2

0

2

1

0

( 1)2

2 ( 1)

0

(2 1)( 1)

(2 1) 2 2(2 1)( 1)

0

( 1)2

2 (

( ) ( ) ( ) (2 2 ( )) ( )2

( 1) ( ) (2 (2 1)( )) ( )(2 1)

(0) ( ) ( )2

k

k

k

k

x kk

k

kxk

k

kk

k

x tg x J ka x t f t dt

ak

x tJ a k x t f t dt

k a

xf J

ak

2

0

1)

(2 1)( 1)

(2 1) 2 2(2 1)( 1)

(2 2 )

(0) ( 1) ( ) (2 (2 1) ).(2 1)

k

k

k

k

k

kax

xf J a k x

k a

Similarly we get

2 1

0

2

2

0

1

0

(2 1)( 1)

(2 1) 2 2(2 1)( 1)

0

( 1)2

2 ( 1)

0

(2 1)(

(2 1) 2

( ) ( 1) ( ) (2 (2 1)( )) ( )(2 1)

( ) ( ) (2 2 ( )) ( )2

(0) ( 1) ( )(2 1)

k

k

k

k

k

kxk

k

x kk

k

k

k

x tg x J a k x t f t dt

k a

x tJ ka x t f t dt

ak

xf

k a

2

2

0

1)2

(2 1)( 1)

( 1)2

2 ( 1)

(2 (2 1) )

(0) ( ) ( ) (2 2 ).2k

k

kk

k

J a k x

xf J kax

ak



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Definition 1.1 Let ( , )f x y be a continuous function defined on the square[0, ) [0, ) ,

which is of exponential order, that is, for some , R

, 0

| ( , )|sup .

x yx y

f x y

e

Then the two dimensional Laplace transforms of ( , )f x y is defined as

20 0

{ ( , )} ( , ) : ( , ).p x q yL f x y e f x y dx dy F p q

If 2{ ( , )} ( , )L f x y F p q , then 1

2 { ( , )}L F p q is given by

2

1( , ) ( , ) .

(2 )

i c ix p y q

i c if x y e F p q dp dq

i

Definition 1.2 The one-dimensional convolution of ( )f x and ( )g x is as

0( * )( ) ( ) ( ) ,

x

f g x f x z g z dz

Also, the two-dimensional convolution of ( , )f x y and ( , )g x y is given by

0 0( ** )( , ) ( , ) ( , ) .

x y

f g x y f x y g d d

2. Evaluation of Integrals and Solution to Fractional P.D.E. By Means of the Two

Dimensional Laplace Transform

In this section, we evaluate the integrals that their evaluations are not an easy task. But, by means

of two dimensional Laplace transform we can evaluate these integrals.

Lemma 2.1 The following integral relations hold true

20

0

2 1 2) ( ) (2),

tan tana I bei d J

Where 3

40( ) Im( ( ) )

i

bei z J z e

and

300

2 2 cos ( 1)) ( ) .

( !)

n

n

t dtb I bei

t nt



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Proof. ( )a : By change of variable tant , we get

20

2 1 2( ) .

( 1) tI bei dt

t t

Using the fact that

( , )

22 2

2

1{ (2 )} ,

1p q x y

L t beit

p qt

Let us assume that

2 20

2 1( , ) ( 2 ) ,

( 1 )

x yI x y t bei dt

t t t

Then, we get

( , )

2 2 2 02 2

2 2

2 1 1 1{ ( , )} ,

1 11( 1) ( )

p q dtL I x y

p q p q qt t p

p q q

The inverse two dimensional Laplace transform of the above relation yields

0( , ) (2 ),I x y J x y

Which concludes that?

0(1,1) ( 2).I I J

Proof ( )b : Similarly, we obtain

30

( 1).

( !)

n

n

In

Theorem 2.1 (Sub Ballistic Fractional PDE) The following non - homogeneous partial

fractional differential equations [17].

0 0( , ) ( , ) 1 : , 0C C

x tD u x t D u x t x t

Where 0 , 1 with the boundary conditions (0, ) ( ,0 ) 0u t u x ,

We get the following formal solution



DOI: 10.5281/zenodo.1146189


1

0

( 1)( , ) .

(1 ) ( 2 )

n n n

n

x tu x t

n n

Proof: At first, we assume that 0 1 and 0 1 , then applying the two-dimensional

Laplace transform term wise to P.D.E, yields

1( , ) ( , ) ,p U p q q U p q

p q

with 2{ ( , )} ( , )L u x t U p q . Therefore

1

( 1) 11 0

1 1 ( 1)( , ) .

( )(1 )

n n

nn

qU p q

qp q p q pp q

p

The inverse two-dimensional Laplace transforms yields

( 1)

0

( 1)( , ) .

( 1) (1 )

n n n

n

x tu x t

n n

For the special case 0.5 , we have

2( , ) ( ),

xu x t x

t x t

And the figure is shown as follows.



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Note that when 0 1 and 1 , by using the two-dimensional Laplace transform, we may

calculate ( , )U p q as follows,

1

22 0

1 1 ( 1)( , ) ,

( )(1 )

n n

nn

pU p q

pp q p q qp q

q

So that

1

0

( 1)( , ) .

(1 ) ( 2 )

n n n

n

x tu x t

n n

For 1 and 0.5 , we have the following

In special case, let us take 1 , the Laplace transform with respect to t gives

1( , ) ( , ) ,xU x q qU x q

q

Therefore

2

1( , ) (1 ).q xU x q e

q

At this point, the inverses Laplace transform yields

:( , ) ( ) ( ) .

:

t x tu x t t t x H t x

x x t



DOI: 10.5281/zenodo.1146189


3. Solving System of Time Fractional Heat Equations

In this section, the authors considered certain homogeneous system of time fractional heat

equations which is a generalization to the problem of thermal effects on fluid flow and hydraulic

fracturing from well bores and cavities in low permeability formations [12]. In this work, only

the Laplace transformation is considered as it is easily understood and being popular among

engineers and scientists.

Problem3.1 Solving the homogeneous systems of time fractional heat equations [12]

2

2

0 2

C

t

uD u k

x

22 2

0 02

C C

t t

vD v k D u

x

Where 0 1 , 0 x , 0t with the boundary conditions,

(0, ) 1 , lim ( , ) , (0, ) 1 , lim ( , )xx x

u t u x t v t v x t

And the initial conditions are ( ,0) ( ,0) 0u x v x .

Solution: Application of the Laplace transform to the first equation and the initial conditions,

yields 2( , ) ( , ).x xs U x s k U x s

Now, the boundary conditions give

2

10 2

( )1 1

( , ) ,!

ns

xk

nn

x

kU x s es n

s



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That leads to

2

0

( )

( , ) .!

(1 )2

nn

n

x

tku x tnn

The special case 1 gives

( , ) ( ).2

xu x t erfc

k t

Similarly, applying the Laplace transform of the second equation and using the initial conditions,

we arrive at

2

1( , ) ( , ) .s

xk

x xV x s s V x s s e

Using the boundary conditions, solution of the above differential equation is as

22 2

1 12 2

( , ) .2

2

sx

kk x

V x s es

s k s

The Laplace transform inversion formula leads to the following formal solution

2 22 2 2 2 2

0

( )

( , ) .! 2 2

(1 ) (1 ) (1 )2 2 2 2 2

n n nn

n

xt t x tkv x t k

n n nn k

For the special case 1 , we obtain

2

22 2

4( , ) ( ) ( ) 2 .2 2 2

x

k tx t xv x t x erfc k e

k t k t

We have shown the solutions in the following figures for 1,0.5 and 1k



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4. Effect of a Uniform Overburden on the Passage of a Thermal Wave

In this section, the authors considered certain homogeneous time fractional heat equations which

are a generalization of the problem of the effect of a uniform overburden on the passage of a

thermal wave and the temperatures in the underlying rock studied by D.S. Parasnis [16].

Problem 4.1 Let us find the fractional heat flow in a two-layer earth [12]

2

10 1 1 2

( , )( , ) : 0 , 0C

t

u x tD u x t a x h t

x

2

20 2 2 2

( , )( , ) : , 0C

t

u x tD u x t a h x t

x

Where 0 1 and ia denotes the diffusivity of the layer, iu denotes the temperature and x

denotes the distance down from the surface. At the earth's surface, there is a diurnal cycle

1 0(0, ) sinu t T t , while at the interface between the two layers,

1 21 2 1 2

( , ) ( , )( , ) ( , ) ,

u h t u h tu h t u h t k k

x x

Where ik denotes the thermal conductivity. We also require that 2lim | ( , )|x

u x t

. The initial

conditions are 1 2( ,0) ( ,0) 0u x u x .



DOI: 10.5281/zenodo.1146189


Solution: Taking the Laplace transform of two equations, we get

2

2

( , ) s( , ) 0 ( 1,2).i

i

i

d U x sU x s i

d x a

Therefore, using the initial and boundary conditions, we get the following

101 2 2

1

1

exp( 2 )

( , ) exp( )

1 exp( 2 )

sh

aT w sU x s x

s w a sh

a

10

2 2

1

exp( )

1 exp( 2 )

sx

aT w

s w sh

a

And

2 1 20

2 2 2

1

exp ( )(1 )

( , ) ,

1 exp( 2 )

s s sh x

a a aT wU x s

s w sh

a

Where 1 2 2 1

1 2 2 1

k a k a

k a k a

. To invert 1U and 2U , we observe that

0 1

1

1( 1) exp( 2 ),

1 exp( 2 )

n n

n

sn h

ash

a

if1

exp( 2 ) 1s

ha

. Then we get the following relations

101 2 2

0 1

( , ) ( 1) exp ( 2 2 )n n

n

T w sU x s n h h x

s w a

0

2 20 1

( ) exp ( 2 )n

n

T w sn h x

s w a

And

0 12 2 2

0 2 1

(1 )( , ) ( ) exp ( 2 1) ( ) .n

n

T w a sU x s n h h x

s w a a



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We can show (by using convolution) that the inverse of

2 2( , ) ( ) exp( )

wF z s z s

s w

Is as

02

1( , ) sin( ( ) ) ( ,0; ) .

2

t zf z t t W d

t

Where (.,.)W stands for the Wright function. The final solutions are

1

1 0 0

0 01 1

2 2 2( , ) ( 1) ( , ) ( ) ( , )n n n

n n

n h h x n h xu x t T f t T f t

a a

And

1

2

2 0

0 1

(2 1) ( )

( , ) (1 ) ( ) ( , ).n

n

an h h x

au x t T f t

a

5. Conclusion

In this paper, analytic solution of the space - time fractional sub - ballistic and heat equations are

derived using the integral transform method. The authors implemented two dimensional Laplace

transform to solve non -homogeneous sub ballistic fractional PDE and homogeneous systems of

time fractional heat equations. Illustrative examples are given.

References

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[2] A. Aghili; M.R. Masomi. The Double Post-Widder Inversion Formula for Two Dimensional

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*Corresponding author.

E-mail address: armanaghili@ yahoo.com/ masomirasool@ yahoo.com


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