For Review Onlyrdo.psu.ac.th/sjstweb/Ar-Press/61-Mar/24.pdf · For Review Only DIFFERENTIAL...

For Review Only

Differential Transformation Method for Vibration of

Membranes

Journal: Songklanakarin Journal of Science and Technology

Manuscript ID SJST-2017-0432.R1

Manuscript Type: Original Article

Date Submitted by the Author: 14-Feb-2018

Complete List of Authors: Mansilp, Kamonpad; King Mongkut's Institute of Technology Ladkrabang, mathematics Kasemsuwan, Jaipong; King Mongkut's Institute of Technology Ladkrabang, mathematics

Keyword: Vibration of Membrane, Differential Transformation Method, Analytical Solution, Approximate Solution.

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Songklanakarin Journal of Science and Technology SJST-2017-0432.R1 Mansilp

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DIFFERENTIAL TRANSFORMATION METHOD FOR VIBRATION OF

MEMBRANES

Original Article

Kamonpad Mansilp1*, and Jaipong Kasemsuwan

2

1,2Department of Mathematics, Faculty of Science,

King Mongkut's Institute of Technology Ladkrabang,

Bangkok 10520, Thailand.

* Corresponding author, Email address: [email protected]

Abstract

The purpose of this research is to apply the three-dimensional differential transform

method to estimate solutions to the equation of motion for vibration in a membrane with

certain boundary condition problems. The analytical solutions without an external force and

damping term under the initial and boundary conditions are presented. We found that the

comparison results between the analytical solution and the estimated solution are in good

agreement in case of no damping and the external force. Furthermore, the differential

transform method can be used to find approximated solutions taking into account both an

external force and damping term. This cannot be achieved via an analytical solution.

2010 Mathematics Subject Classification: 35G20, 35L05, 35L10 , 35L70.

Keywords: Vibration of Membrane, Differential Transformation Method, Analytical

Solution, Approximate Solution.

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1. Introduction

Many researchers have attempted to understand several phenomena occurring in

nature by applying knowledge from different field such as mechanical engineering, electrical

engineering, industrial engineering, energy and medicine. Most of these problems have been

studied by employing some form of mathematical modeling using both ordinary differential

equation (ODE) and partial differential equation (PDE). These problems require an accurate

solution, either as an analytical solution or approximate solution. The differential transform

method (DTM) is among the most effective mathematical methods for finding solutions to

these differential equations (Hatami, Ganji, & Sheikholeslami, 2017).

The differential transform method (DTM) is based on the high-order Taylor series

expansion. This method is a powerful tool for solving linear and non-linear ordinary

differential equations (Ayaz, 2004; Arikoglu, & Ozkol, 2006; Catal, 2008) and for solving

two-and three-dimensionals partial differential equations in both linear and non-linear

problems. DTM can be used to solve the differential equation subject to initial and boundary

conditions having both linear and non-linear terms within an acceptable error range.

The two-dimensional differential transform method (2D-DTM) is used to find the

solutions of both linear PDEs (Chen & Ho, 1999; Othman, & Mahdy, 2010; Ayaz, 2003;

Yang, Liu, & Bai, 2006) and nonlinear PDEs (Bildik, Konuralp, Bek, & Kucukarslan, 2006;

Kangalgil & Ayaz, 2009; Biazar & Eslami, 2010; Biazar, Eslami, & Islam, 2012).

Additionally, the three-dimensional differential transform method (3D-DTM) is

applied to find the solutions of linear and non-linear PDEs (Saravanan & Magesh, 2013;

Bagheri & Manafianheris, 2012). It is noted that the differential transform method can be

used to solve multidimensional PDEs such as the Westervelt equation (Jafari, Sadeghi, &

Biswas, 2012), heat-like and wave-like equations (Tabaei, Celuk, & Tabaei, 2012), and fuzzy

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partial differential equations (Mirzaee & Yari, 2015) as well as the linear and nonlinear

system of PDEs (Ayaz, 2004; Zedan & AliAlghamdi, 2012).

Many researchers are concerned with the use of non-linear PDEs by applying several

transform methods as follows. The Fitzhuah Nangumo (FN) equation is a mathematical

model for solving scientific and engineering problems by using q-HATM and the fractional

reduced differential transform method (FRDTM) which is based on DTM (Kumar, Singh, &

Baleanu, 2017).

The numerical solutions to non-linear fractional dynamical model of interpersonal and

romantic relationships are presented by applying q-homotopy analysis via the Sumudu

transform method (q-HASTM) (Singh, Kumar, Qurashi, & Baleanu, 2017).

Jeffery-Hamel flow in non-parallel walls, which relates to non-linear PDE and occurs

in fluid dynamics and other scientific applications, is solved by using an efficient hybrid

computational technique, the homotopy analysis transform method (HATM) (Singh, Rashidi,

Sushila, & Kumar, 2017).

The homotopy perturbation Sumudu transform method (HPSTM) and homotopy

analysis Sumuda transform method (HASTM) are more convenient than the homotopy

perturbation method (HPM) and the homotopy analysis method (HAM) since they produce a

comparative analytical study for a system of time factional non-linear differential equations

(Choi, Kumar, Singh, & Swroop, 2016).

Further examples of the application of PDEs by Laplace transform method to various

problem include Case I-application of drum head vibration solving by separation of variables

method, and Case II and III-application of signal transmission and application of chemical

communication in insects. All three cases give simulations on the solutions implemented by

MATLAB software (Ojwando, 2016).

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A modified He-Laplace method (MHLM) is applied to solve space and time nonlinear

fractional differential-difference equations (NFDDEs) (Prakash, Kothandapan, & Bharathi,

2016).

In our previous work, we studied the suspended vibrating string equation using 2D –

DTM. It was found that DTM can be applied to various problems of the suspended string

equation.

In this proposed work, we study further the vibration equation in three dimensions of

the motion of a membrane using the 3D-DTM. The oscillation of a membrane-like plate,

which is determined by the tension and insignificant resistance to the bending. The

differential transform method was applied to find solutions of the motion equation of a

membrane with an external force and damping term. We compare the results with an

analytical solution. We show in detail the derivation of the transformed formula of DTM

which is of the nth power form (to be shown in theorem 3.6). The obtained formula helps

simplify the use of DTM in solving non-linear PDEs. It is noted that the formula requires

high computation to find a complete solutions mainly due to the fact that the formula is in the

recursive form.

The equation of motion of a membrane

The equation of motion for the forced transverse vibration of a membrane, after (Rao,

2011) is as follows:

2 2 2

2 2 2( , , ) ( , ) ,

w w wP f x y t x y

x y tρ

∂ ∂ ∂+ + = ∂ ∂ ∂

(1)

where ( , , )f x y t is the pressure acting in the z direction(external force), P is the intensity of

tension at a point equal to production of the tensile stress and the thickness of the membrane,

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and ( , )x yρ is the mass per unit area. We assume that ( , , ) 0f x y t = , 1,P = and ( , ) 1.x yρ =

Then Eq.(1) leads to:

2 2 2

2 2 2.

w w w

x y t

∂ ∂ ∂+ =

∂ ∂ ∂

(2)

The initial conditions of (2) after (Rao, 2011) are:

( , ,0) sin sin , 0 , 0 ,

( , ,0) 0, 0 , 0 ,

x yw x y x a y b

a b

wx y x a y b

t

π π= ≤ ≤ ≤ ≤

∂= ≤ ≤ ≤ ≤

∂

we assumed that 1a = and 1b = .Therefore, the boundary conditions of the equation of motion

of a membrane are definded thus:

( ,0, ) 0, 0 1,

(0, , ) 0, 0 1,

( ,1, ) 0, 0 1,

(1, , ) 0, 0 1, .

w x t x

w y t y

w x t x

w y t y t

= ≤ ≤

= ≤ ≤

= ≤ ≤

= ≤ ≤ ∈�

(3)

An analytical solution

The following is the derivation of the analytical solution of the problem (2).

Considering the initial conditions ( , ,0) sin sinw x y x yπ π= and ( , ,0) 0,w

x yt

∂=

∂ the boundary

conditions are shown in (3).The general solution can be derived by the separation of variables

technique:

( , , ) ( ) ( ) ( ),w x y t X x Y y T t= (4)

Subject to the eigenvalue, 2 2,ω α− and

2β , can be written (2)

2( ) ( ) ( ),

( ) ( ) ( )

X x Y y T t

X x Y y T tω

′′ ′′ ′′+ = = −

2( ) ( ),

( ) ( )

X x Y y

X x Y yω

′′ ′′+ = −

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and

2 2( ) ( ),

( ) ( )

X x Y y

X x Y yω α

′′ ′′− = + =

2( ),

( )

X x

X xα

′′− =

then 2( ) ( ) 0,X x X xα′′ + =

2 2( ),

( )

Y y

Y yα ω

′′= −

assumed 2 2 2 ,α ω β− + = 2 2( ) ( ) ( ) 0,Y y Y yα ω′′ + − + =

then 2( ) ( ) 0,Y y Y yβ′′ + =

and

2( ),

( )

T t

T tω

′′=

then 2( ) ( ) 0,T t T tω′′ + =

2( ) ( ) 0,X x X xα′′ + = (5)

2( ) ( ) 0Y y Y yβ′′ + = , (6)

2( ) ( ) 0.T t T tω′′ + = (7)

We obtain a solution of (5) as 1 2( ) cos sin ,X x C x C xα α= + where 1C and 2C are arbitrary

constants. Subject to the boundary conditions (0) 0X = , we have 1 0.C = Then

2( ) sinX x C xα= full step the boundary condition (1) 0.X = Let 2 0C ≠ give ;m m Iα π= ∈ ,

we have that

( ) sinm mX x C m xπ= . (8)

We obtain a solution of (6) as 3 4( ) cos sinY y C x C yβ β= + .According to boundary

conditions (0) 0Y = , we have 3 0.C = Then the boundary conditions ( ) 0Y y = .

Letting 4 0C ≠ gives ;n n Iβ π= ∈ , then we see that:

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( ) sin .n nY y C n yπ= (9)

From

2 2 2

2 2

,

.m n

ω β α

ω π

= +

= +

We obtain a solution of (7) as

( ) cos sinmn mn mnT t A t B tω ω= + .

Then

2 2 2 2( ) cos sinmn mn mnT t A m n t B m n tπ π= + + + . (10)

According to (8), (9) and (10) can be written

( )( )

2 2

2 2

( , , ) ( ) ( ) ( ),

sin sin cos

sin sin sin ,

mn

mn m n

mn m n

W x y t X x Y y T t

A C C m x n y m n t

B C C m x n y m n t

π π π

π π π

=

= +

+ +

(11)

where mn mn m nF A C C= , mn mn m nH B C C= and for all ,m n I∈

By using the superposition principle (11) become

( )( )

1 1

2 2

1 1

2 2

( , , ) ( , , ),

sin sin cos ,

sin sin sin .

mn

m n

mn

m n

mn

w x y t w x y t

F m x n y m n t

H m x n y m n t

π π π

π π π

∞ ∞

= =

∞ ∞

= =

=

= +

+ +

∑∑

∑∑ (12)

According to the initial condition ( , ,0) sin sinw x y x yπ π= and ( , ,0) 0.w

x yt

∂=

∂ By using

Fourier series, we have

1 ; 1 1,

0; ,

2 .

mn

mn

F m and n

H m n I

ω π

= = =

= ∈

=

Substituting 11 1F = , 0mnH = and 2ω π=

in (12), we obtain the analytical solution:

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( , , ) sin sin cos 2 .w x y t x y tπ π π= (13)

In the case of the problem (2) with damping term are represented by

2 2 2

2 2 2.

w w w w

t x y t

∂ ∂ ∂ ∂= + −

∂ ∂ ∂ ∂

By similar calculation, we obtain the solution of , ( )m nT t as

2 2

2, , ,

1 4 1 4( ) ( cos sin )

2 2

t

m n m n m nT t e A t B tω ω− − −

= + . (14)

Considering (14), we can see that substituting 2ω π=

to (14), it make 21 4ω− be the

complex number. Therefore ( )T t are not solvable.

2. Three-Dimensional Differential transform method

The basic definitions and fundamental operations of differential transform are defined

below.

Definition 2.1 The three-dimensional differential transform of function ( , , )w x y t is defined

as:

(0,0,0)

1 ( , , )( , , )

! ! !

k h m

k h m

w x y tW k h m

k h m x y t

+ +∂=

∂ ∂ ∂0, 0 0.k h and m≥ ≥ ≥

Definition 2.2 The inverse three-dimensional differential transform of sequence

{ }, , 0

( , , )k h m

W k h m∞

=is defined as:

0 0 0

( , , ) ( , , ) .k h m

k h m

w x y t W k h m x y t∞ ∞ ∞

= = =

= ∑∑∑

(14)

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3. The fundamental operations of three-dimensional differential transform method .

From Table 1 theorem 3.1-3.5 shown in (Yang, Liu, & Bai, 2006).

The following is the derivation of theorem 3.6:

If 1 2 1( , , ) ( , , ) ( , , )... ( , , ) ( , , )n nv x y t w x y t w x y t w x y t w x y t−= then,

1 1 1 3 13 32 2 2

1 2 1 2 1 2 2 1 2 1 2 10 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 2 2 1 2 1 2 1

1 1 2 1 2 1 2 1 1

( , , ) ...

( , , ) ( , , )...

( , , ) ( , ,

n n n

n n n n n n

mk h m k h mk hk h m

k k h h m m k k h h m m

n n n n n n n n n n

V k h m

W k h m W k k h h m m

W k k h h m m W k k h h

− − −

− − − − − −= = = = = = = = = = = =

− − − − − − − − −

=

− − −

− − − − −

∑ ∑ ∑ ∑ ∑ ∑ ∑∑∑∑ ∑ ∑

1 ).nm m −−

For : , : ; , : , : ; ,n nV I W I n I v w n I x+ + + +→ → ∈ → → ∈ ∈� � � � � � �

and , , 0,1,2,3,...k h m =

By definition of the three-dimensional differential transform:

(0,0,0)

1 ( , , )( , , ) ,

! ! !

k h m

k h m

v x y tV k h m

k h m x y t

+ + ∂= ∂ ∂ ∂

we obtain 1 2 1( , , ) ( , , ) ( , , )... ( , , ) ( , , )n nv x y t w x y t w x y t w x y t w x y t−=

Then by using definition (13), we have:

1 1 1

1 1 1

1 1

1 2 1

(0,0,0)

0 0 0 1 1 1 1 1 1

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

! ! !

1 ! ! !

! ! ! ( )!( )! ( )!( )! ( )!( )!n n n

n n n

n n

k h m

n nk h m

k h m

k h m n n n n n n

k h m

k h

V k h m w x y t w x y t w x y t w x y tk h m x y t

k h m

k h m k k k h h h m m m

x y t

− − −

− − −

− −

+ +

−

= = = − − − − − −

+ +

∂= ∂ ∂ ∂

=− − −

∂

∂ ∂ ∂

∑ ∑ ∑

1

1 1 1

1 1 1

1 2

1

(0,0,0)

( , , ) ( , , )...

( , , ) ( , , )

n

n n n

n n n

m

k k h h m m

n nk k h h m m

w x y t w x y t

w x y t w x y tx y t

−

− − −

− − −

− + − + −

− − − −

∂∂ ∂ ∂

1 1 1

1 1 1

1 1 1

1 1 1

1 1 1

0 0 0

1 2 1

1 1 1 (0,0,0)

1 1 1

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

( )!( )!( )!

1

( )!( )!( )!

n n n

n n n

n n n

n n n

n n n

k h m

k h m

k h m

nk h m

n n n

k k h h m m

nk k h h m m

n n n

w x y t w x y t w x y tk h m x y t

wk k h h m m x y t

− − −

− − −

− − −

− − −

− − −

= = =

+ +

−− − −

− + − + −

− − −− − −

=

∂ ∂ ∂ ∂

∂− − − ∂ ∂ ∂

∑ ∑ ∑

(0,0,0)

( , , )x y t

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1 1 1

1 2 1 2 1 2

2 2 2

2 2 2

0 0 0 0 0 0

1 2 2

2 2 2 (0,0,0)

1 2 1 2 1 2

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

( )!( )!( )!

1

( )!( )!( )!

n n n

n n n n n n

n n n

n n n

k h mk h m

k k h h m m

k h m

nk h m

n n n

n n n n n n


k k h h m m

− − −

− − − − − −

− − −

− − −

= = = = = =

+ +

−− − −

− − − − − −

=

∂ ∂ ∂ ∂

− − −

∂×

∑ ∑ ∑ ∑ ∑ ∑

1 2 1 2 1 2

1 2 1 2 1 2

(0,0 ,0)

1 1 1

1 1 1

1

1 1 1 (0,0,0)

( , , )

1( , , )

( )!( )!( )!

n n n n n n

n n n n n n

n n n

n n n

k k h h m m

nk k h h m m

k k h h m m

nk k h h m m

n n n

w x y tx y t

w x y tk k h h m m x y t

− − − − − −

− − − − − −

− − −

− − −

− + − + −

−− − −

− + − + −

− − −− − −

∂ ∂ ∂

∂ − − − ∂ ∂ ∂

1 2 1 2 1 2

1 2 3 1 2 3 1 2 3

2 2 2

2 2 2

0 0 0 0 0 0 0 0 0

1 2 3

3 3 3 (0,0,0)

2

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

( )!( )!( )!

1

(

n n n n n n

n n n n n n n n n

n n n

n n n

k k h h m mk h m

k k k h h h m m m

k h m

nk h m

n n n

n n


k k

− − − − − −

− − − − − − − − −

− − −

− − −

= = = = = = = = =

+ +

−− − −

−

=

∂ ∂ ∂ ∂

−

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

2 3 2 3 2 3

2 3 2 3 2 3

1 2 1 2 1 2

1 2

3 2 3 2 3

2

(0,0,0)

1 2 1 2 1 2

)!( )!( )!

( , , )

1

( )!( )!( )!

n n n n n n

n n n n n n

n n n n n n

n n

n n n n

k k h h m m

nk k h h m m

n n n n n n

k k h h m m

k k

h h m m

w x y tx y t

k k h h m m

x y

− − − − − −

− − − − − −

− − − − − −

− −

− − − − −

− + − + −

−− − −

− − − − − −

− + − + −

−

− −

∂× ∂ ∂ ∂

− − −

∂×∂ ∂ 1 2 1 2

1 1 1

1 1 1

1

(0,0,0)

1 1 1

1 1 1

1 1 1 (0,0,0)

( , , )

1( , , ) ,

(1 )!(1 )!(1 )!

n n n n

n n n

n n n

nh h m m

k h m

nk h m

n n n

w x y tt

w x y tk h m x y t

− − − −

− − −

− − −

−− −

− + − + −

− − −− − −

∂

∂ − − − ∂ ∂ ∂

1 2 1 2 1 2

1 2 3 1 2 3 1 2 3

3 13 34 4 4

3 2 3 2 3 2

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0

1 1 1 1 2 2 1 2 1 2 1

1 1 2 1

( , , ) ...

( , , ) ( , , )...

( ,

n n n n n n

n n n n n n n n n

k k h h m mk h m

k k k h h h m m m

k h mk h m

k k h h m m

n n n n

V k h m


W k k h

− − − − − −

− − − − − − − − −= = = = = = = = =

= = = = = =

− − − −

=

− − −

−

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

∑∑∑∑ ∑ ∑

2 1 2 1 1 1, ) ( , , ).n n n n n n nh m m W k k h h m m− − − − − −− − − − −

since 1 2 1( , , ) ( , , ) ( , , )... ( , , ) ( , , )n nv x y t w x y t w x y t w x y t w x y t−= then,

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1 1 1 3 13 32 2 2

1 2 1 2 1 2 2 1 2 1 2 10 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 2 2 1 2 1 2 1

1 1 2 1 2 1 2 1 1

( , , ) ...

( , , ) ( , , )...

( , , ) ( , ,

n n n

n n n n n n



n n n n n n n n n n

V k h m



− − −

− − − − − −= = = = = = = = = = = =

− − − − − − − − −

=

− − −

− − − − −

∑ ∑ ∑ ∑ ∑ ∑ ∑∑∑∑ ∑ ∑

1 ).nm m −−

4. Application

In this section, we apply the three-dimensional differential transform method to the

vibration of a membrane. We demonstrate four examples of the problem under different

conditions. The conditions are (i) without damping term, (ii) with damping term, (iii) with

external force, and (iv) with damping term and external force. The initial and boundary

conditions as defined as follows.

Example 4.1 Considering the equation of the motion for the vibration of a membrane

2 2 2

2 2 2,

w w w

t x y

∂ ∂ ∂= +

∂ ∂ ∂ (15)

with the initial and boundary conditions,

( , ,0) sin sin , 0 1, 0 1,

( , ,0) 0, 0 1, 0 1,

x yw x y x y

a b

wx y x y

t

π π= ≤ ≤ ≤ ≤

∂= ≤ ≤ ≤ ≤

∂

,

( ,0, ) 0, 0 1,

(0, , ) 0, 0 1,

( ,1, ) 0, 0 1,

(1, , ) 0, 0 1, .

w x t x

w y t y

w x t x

w y t y t

= ≤ ≤

= ≤ ≤

= ≤ ≤

= ≤ ≤ ∈�

(16)

Comparing (15) to the general terms of PDEs in Table 1, we have:

( , , ) ( , . ) ( , , ) 1.a x y t b x y t c x y t= = =

Then for all 0, 0i j≥ ≥ and 0,r ≥ we have

( , , ) ( , , ) ( , , ) ( , , ).A i j r B i j r C i j r i j rδ= = =

The following Kronecker symbols can be used:

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

0; ,

x y tx y t x y t x y x y

otherwiseδ δ δ δ δ δ δ

= = == = =

By applying DTM and Kronecker symbols to the given equation of motion for the vibration

of a membrane, we have

]

0 0 0

0 0 0

1( , , 2)

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

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

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

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

( 2)( 1)

k h m

i j r

k h m

i j r

W k h mm m

k i k i A i j r W k i h j m r

h i h i B i j r U k i h j m r

k k h h W k h m W k h mm m

δ

= = =

= = =

+ =+ +

× − + − + − + − −

+ − + − + − − + −

+ + + + + ++ +

∑∑∑

∑∑∑

that is

[ ]( , , 2) ( 2)( 1)( 2)( 1) ( 2, , ) ( , 2, ) /

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

W k h m k k h h W k h m W k h m

m m m m

+ = + + + + + +

+ + + + (17)

Comparing between the coefficient of Taylor’s series of sine and the series in the definition

(14)

2 4 6 8 10 123 5 7 9 112 4 6 8 10 12

sin( )sin( ) ...2! 4! 6! 8! 10! 12!

x y xt x t x t x t x t x tπ π π π π π

π π ≅ − + − + − +

0 0 2

0 0

3 2

2 2

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

(3,1,0) ... (0,1,0) (1,2,0)

(2,2,0) ...

k h

k h

W k h x y W x y W xy W x y

W x y W xy W xy

W x y

∞ ∞

= =

= + +

+ + + +

+ +

∑∑ (18)

Then, we have

0 0,1,2,... 0,2,4,6,...

( )( , ,0) 1,5,9,... 1,3,5,7,...

( )!

( )3,7,11,.. 1,3,5,7,...

( )!

k h

k h

if k and h

k hU k h if k and h

k h

k hif k and h

k h

π

π

+

+

= = +

= = =+

− += =

+

, (19)

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and from the initial condition (16)

( , ,1) 0, , 0,1, 2,3,...W k h k h= = , (20)

and from the boundary condition (16), we have

( ,0, ) 0, , 0,1,2,3,...,

(0, , ) 0, , 0,1,2,3,...,

( ,1, ) 0, , 0,1, 2,3,...,

(1, , ) 0, , 0,1, 2,3,...,

W k m k m

W h m h m

W k m k m

W h m h m

= =

= =

= =

= =

(21)

for each , ,k h m substituting(19)-(21) and by recursive method of (17), we obtain the

coefficients ( , , )W k h m for the series solution.

That is

2 4 2 6 4 8 6 4 3 6 2 31 1( , , ) .

1 1.

6 0 6.

9 6xy t xy t xy t xy x yu x y t t x yπ π π π π π− + −= +− +

As the result in (14), we can not apply the separation of variables technique to find the

analytical solution to the problem with damping term as equation (22). In the following

example, we apply the DTM to find the approximate solution to the problem.

Example 4.2 Considering the equation of motion for the vibration of a membrane with

damping term.

2 2 2

2 2 2,

w w w w

tt x y

∂ ∂ ∂ ∂= + −

∂∂ ∂ ∂ (22)

comparing (22) to general terms of PDEs in table 1, we have

( , , ) ( , . ) ( , , ) 1, ( , , ) 1.a x y t b x y t c x y t d x y t= = = = −


( , , ) ( , , ) ( , , ) ( , , ), ( , , ) ( , , ).A i j r B i j r C i j r i j r D i j r i j rδ δ= = = = −


of a membrane, we have:

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[]

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

( 1) ( , 1, ) / ( 2)( 1).


h W k h m m m

+ = + + + + + +

− + + + + (23)

For each , ,k h m substituting(19)-(21) and by recursive method of (23), we obtain the


That is

2 2 6 4 4 2 3 6 4 3 6 2 5 8 4 55 5 5 5 1 1( , , ) ...

2 6 12 36 48 144w x y t t x t x t x t x t x t xπ π π π π π= − + + − − + +

Example 4.3 Considering the equation of motion for the vibration of a membrane with

external force.

2 2 23

2 2 2.

w w ww

t x y

∂ ∂ ∂= + −

∂ ∂ ∂ (24)

Comparing (24) to general terms of PDEs in table 1, we have:

( , , ) ( , . ) ( , , ) 1, ( , , ) 1.a x y t b x y t c x y t q x y t= = = = −

Then for all 0, 0i j≥ ≥ and 0,r ≥ we have:

( , , ) ( , , ) ( , , ) ( , , ), ( , , ) ( , , ).A i j r B i j r C i j r i j r Q i j r i j rδ δ= = = = −


of a membrane, we have

[

]

2 2 2

2 1 2 1 2 1

1 1 1 1

0 0 0 0 0 0

2 2 1 2 1 2 1 3 2 2 2

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

( , , )

( , , )

, , ) ( , 2, )

( 2)( 1)( / ., , )

k h mk h m

k k h h m m

W k h m

W k k h h m m W


m mk k h h m m

= = = = = =

+ = + + + +

−

+ +

+− − − − +−

−∑∑∑∑∑∑ (25)

For each , ,k h m substituting (19)-(21) and by recursive method of (25), we obtain the


That is:

8 8 102 4 2 6 4 6 4 3 6 2 31 1 1 1

( 3( ))6 30 6 18 36 6 6

( , , ) ...xy t xy t xy t xy x yw x y t x ytπ π π

π π π π π− + + − + − + +− −=

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Example 4.4 Considering the equation of the motion for the vibration of a membrane

with external force and damping term.

2 2 23

2 2 2.

w w w ww

tt x y

∂ ∂ ∂ ∂= + − −

∂∂ ∂ ∂

(26)

Comparing (26) to general terms of PDEs in table 1, we have

( , , ) ( , . ) ( , , ) 1, ( , , ) ( , , ) 1.a x y t b x y t c x y t d x y t q x y t= = = = = −


( , , ) ( , , ) ( , , ) ( , , ), ( , , ) ( , , ) ( , , ).A i j r B i j r C i j r i j r D i j r Q i j r i j rδ δ= = = = = −


of a membrane, we have:

[

]

2 2 2

2 1 2 1 2 1

1 1 1 1 2 2 1 2 1 2 1

0 0 0 0 0 0

3 2 2 2

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

( , , ) ( , , )

( ,

2, )

(

,

1) ( , 1, )

) / ( 2)( 1).

k h mk h m

k k h h m m


h W k h

W k h m W k k h h

m

m

m m

W k k mh h m m

= = = = = =

− − −

− − −

+ = + + + + + +

− + +

+ +

−∑∑∑∑∑∑ (27)

For each , ,k h m substituting (19)-(21) and by recursive method of (27), we obtain the


That is:

6 4

2 2 4 4 6 6 4 2 3 6 4 33 21 1 1 1 1

1/ 6 ( ( ) ( 2 ))2 30 12 12

(3

, , ) .6

..t x t x t xw x y t tt x xπ π π π π π π= +− + + − ++ − −

5. Results and Discussion

Next, we will show the results from examples 4.1 to– 4.4.

Figure 1 Graphical comparison between the amplitude of the analytical solution and the

approximate solution from DTM of the vibration of a membrane.

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In Figure 1, at the initial time, the approximate solution (dashed blue lines) obtained

from DTM in Example 4.1 is close to the analytical solution (solid orange line) calculated

from equation (13). With increasing time, both results continue to lie close to each other.

Figure 2 Graphical result of the amplitude of the approximate solution from DTM of

the vibration of a membrane with a damping term.

The graphical results of the approximate solution from DTM of the vibration of a

vibrating membrane with a damping term (Example 4.2) are shown in Fig.2. We found that

the amplitude of vibration is less than that of the vibrating membrane without a damping

term, as shown by the results in Table (2).

The values for the exact solution, approximate solution with a damping term and error

for the vibration of membrane are shown in Tables 2 below.

Table 2 Data Value at 0.5x = , 0.5y = .


the vibration of a membrane with an external force.

The graphical results of the approximate solution from DTM of the vibration of a

membrane with the external force (Example 4.3) are shown in Fig.3. We found that the

amplitude of vibration is less than that of the vibrating membrane without the external force,

as shown by the results in Table (3).


the vibration of a membrane with a damping term and external force.

The graphical results of the approximate solution from DTM of the vibration a

membrane with a damping term and external force (Example 4.4) are shown in Fig.4. We

found that the amplitude of vibration is less than that of the vibrating membrane without a

damping term and external force as shown by the results in Table (3).

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The values of the approximate solution for the vibration of a membrane with the

external force, with the external force and a damping term, and error for the vibration of a

membrane are shown in Table 3 below.


6. Conclusions

We found that the analytical solutions and approximate solutions are very similar in

case of the problem with no damping and external force. In the case of the problem with a

damping term and external, the analytical solutions are not solvable. Therefore the DTM is

used to find the approximate solutions. The obtained results show that the addition of a

damping term, and addition of external force in the equation of motion of the membrane, can

cause the amplitude of motion for the vibration of a membrane to decrease.

Acknowledgments

This research is supported by Rajamangala University of Technology Tawan-ok

:Chantaburi Campus, Thailand.

References

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differential transform method. Applied Mathematics and computation, 181, 153-162.

Ayaz, F. (2003). On the two-dimentional differential transform method. Applied

Mathematics and computation, 143, 361-374.

Ayaz, F. (2004). Applications of differential transform method to differential-algebraic

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transform method. Applied Mathematics and computation, 147, 547-567.

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Catal, S. (2008), Solution of free vibration equations of beam on elastic soil by using

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solving multidimensional partial differential equations. Indian Journal of Science

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Figure 1 Graphical comparison between the amplitude of the analytical solution and the

approximate solution from DTM of the vibration of a membrane.


the vibration of a membrane with a damping term.

0.2 0.4 0.6 0.8 1.0X

0.2

0.4

0.6

0.8

1.0

W t DTM 0.000

t DTM 0.015

t DTM 0.030

t DTM 0.045

t DTM 0.060

t Analytical 0.000

t Analytical 0.015

t Analytical 0.030

t Analytical 0.045

t Analytical 0.060

Displacement

Position

Displacement

Position

t=0.060

t=0.045

t=0.030

t=0.015

t=0.000

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the vibration of a membrane with an external force.


the vibration of a membrane with a damping term and external force.

Displacement

Position

t=0.060

t=0.045

t=0.000

t=0.015

t=0.030

Displacement

Position

t=0.060

t=0.045

t=0.015

t=0.000

t=0.060

t=0.030

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The fundamental operations of three-dimensional differential transform method

Original function Transformed form

2

2

( , , )3.1 ( , , ) ( , , )

w x y tv x y t a x y t

x

∂=

∂

0 0 0

( , , ) ( 2)( 1)

( , , ) ( 2, , ).

k h m

i j r

V k h m k i k i

A i j r W k i h j m r

= = =

= − + − +

− + − −

∑∑∑

2

2

( , , )3.2 ( , , ) ( , , )

w x y tv x y t b x y t

y

∂=

∂

0 0 0

( , , ) ( 2)( 1)

( , , ) ( , 2, ).

k h m

i j r

V k h m h i h i

B i j r U k i h j m r

= = =

= − + − +

− − + −

∑∑∑

2

2

( , , )3.3 ( , , ) ( , , )

w x y tv x y t c x y t

t

∂=

∂

0 0 0

( , , ) ( 2)( 1)

( , , ) ( , , , 2).

k h m

i j r

V k h m m i m i

C i j r W k i h j m r

= = =

= − + − +

− − − +

∑∑∑

( , , )3.4 ( , , ) ( , , )

w x y tv x y t d x y t

t

∂=

∂ 0 0 0

( , , ) ( 1) ( , , )

( , , ) ( , 1, ).

k h m

i j r

V k h m h i D i j r

D i j r W k i h j m r

= = =

= − +

− − + −

∑∑∑

3.5 ( , , )n l s

v x y t x y t=

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

1 , 1 , 1 ,( ) ( ) ( )

0 , 0 , 0 ,

V k h m k n h l m s k n h l m s where

k n h l m sk n h l m p

k n h l m s

δ δ δ δ

δ δ δ

= − − − = − − −

= = = − = − = − =

≠ ≠ ≠

3.6 ( , , ) ( , , ) ( , , )n

v x y t p x y t w x y t=

1 1 1 3 13 32 2 2

1 2 1 2 1 2 2 1 2 1 2 10 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 2 2 1 2 1 2 1

1 1 2 1 2 1 2 1 1

( , , ) ...

( , , ) ( , , )...

( , , ) ( , ,

n n n

n n n n n n



n n n n n n n n n n

V k h m



− − −

− − − − − −= = = = = = = = = = = =

− − − − − − − − −

=

− − −

− − − − −

∑ ∑ ∑ ∑ ∑ ∑ ∑∑∑∑∑∑

1).nm m−−

33.7 ( , , ) ( , , ) ( , , )v x y t q x y t w x y t=

2 2 2

2 1 2 1 2 1

1 1 1 1

0 0 0 0 0 0

2 2 1 2 1 2 1 3 2 2 2

( , , )( , ,

( , , ) ( , ).

)

,

k h mk h m

k k h h m m

W k h m

W k k h h m m

V k h m

W k k h h m m

= = = = = =

− − − − −

=

−

∑∑∑∑∑∑

Table 1 The fundamental operations of DTM

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t(sec) w(exact

solution)

w(DTM) w(damping) w(exact)- w(DTM)

w(DTM)- w(damping)

0.000t = 1.00000 1.000101 1.0001078 0.000101 0.0007806

0.015t = 0.99113 0.997883 0.9970827 0.000103 0.0008003

0.030t = 0.980081 0.991239 0.9780586 0.000109 0.0131804

0.045t = 0.980081 0.980198 0.9133756 0.000117 0.0668224

0.060t =

0.964679 0.96481 0.7544184

0.000149 0.2103916


t(sec) w(DTM) w(external

force)

w(damping+external

force)

w(DTM)-

w(external force)

w(DTM)-

w(damping+external

force)

0.000t = 1.00000 0.999787 0.999787 0.000313 0.000313

0.015t = 0.99113 0.9810904 0.9834975 0.0167926 0.0143855

0.030t = 0.980081 0.4118883 0.5655184 0.5793507 0.4257206

0.045t = 0.980081 -5.3055571 -3.556495 6.2857551 4.536693

0.060t =

0.902917 -34.028812 -24.203159 34.9936223 25.1679693


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The calculation of the examples 4.2.

2 2 2

2 2 2,

w w w w

tt x y

∂ ∂ ∂ ∂= + −

∂∂ ∂ ∂ (1)

The initial conditions are:

( , ,0) sin sin , 0 , 0 ,

( , ,0) 0, 0 , 0 ,

x yw x y x a y b

a b

wx y x a y b

t

π π= ≤ ≤ ≤ ≤

∂= ≤ ≤ ≤ ≤

∂

(2)

we assumed that 1a = and 1b = .Therefore, the boundary conditions of the equation of

motion of a membrane are definded thus:

( ,0, ) 0, 0 1,

(0, , ) 0, 0 1,

( ,1, ) 0, 0 1,

(1, , ) 0, 0 1, .

w x t x

w y t y

w x t x

w y t y t

= ≤ ≤

= ≤ ≤

= ≤ ≤

= ≤ ≤ ∈ �

(3)

An analytical solution

The following is the derivation of the analytical solution with damping term.

Considering the initial conditions ( , ,0) sin sinw x y x yπ π= and ( , ,0) 0,w

x yt

∂=

∂ the boundary

conditions are shown in (2).The general solution can be derived by the separation of variables

technique:

( , , ) ( ) ( ) ( ),w x y t X x Y y T t= (4)

Subject to the eigenvalue, 2 2,ω α− and

2β , can be written (1)

( ) ( ) ( ) ( ),

( ) ( ) ( ) ( )

T t X x Y y T t

T t X x Y y T t

′′ ′′ ′′ ′= + −

2( ) ( ),

( ) ( )

X x Y y T T

X x Y y T Tω

′′ ′′ ′′ ′+ = + = −

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and

2 2( ) ( ),

( ) ( )

X x Y y

X x Y yω α

′′ ′′− = + =

2( ),

( )

X x

X xα

′′− =

then 2( ) ( ) 0,X x X xα′′ + =

2 2( )( ),

( )

Y y

Y yω α

′′= − −

assumed 2 2 2 ,ω α β− =

then 2( ) ( ) 0,Y y Y yβ′′ + =

then 2( ) ( ) 0,X x X xα′′ + = (5)

2( ) ( ) 0Y y Y yβ′′ + = , (6)

2( ) ( ) ( ) 0.T t T t T tω′′ ′+ + = (7)

We obtain a solution of (5) as 1 2( ) cos sin ,X x C x C xα α= + where 1C and 2C are arbitrary

constants. Subject to the boundary conditions (0) 0X = , we have 1 0.C = Then

2( ) sinX x C xα= full step the boundary condition (1) 0.X = Let 2 0C ≠ give ;m m Iα π= ∈ ,

we have that

( ) sinm mX x C m xπ= . (8)

We obtain a solution of (6) as 3 4( ) cos sinY y C x C yβ β= + .According to boundary

conditions (0) 0Y = , we have 3 0.C = Then the boundary conditions ( ) 0Y y = .

Letting 4 0C ≠ gives ;n n Iβ π= ∈ , then we see that:

( ) sin .n nY y C n yπ= (9)

From

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2 2 2

2 2

,

.m n

ω β α

ω π

= +

= +

2( ) ( ) ( ) 0.T t T t T tω′′ ′+ + =

2 2 0r r ω+ + =

21 1 4,

2 2r

ω−= − ± (10)

We obtain a solution of (7) as

2 2

2, , ,

1 4 1 4( ) ( cos sin )

2 2

t

m n m n m nT t e A t B t

ω ω− − −= + . (11)

According to (8), (9) and (11) can be written

2

2

2

2

( , , ) ( ) ( ) ( ),

1 4sin sin cos

2

1 4sin sin sin ,

2

mn

t

mn m n

t

mn m n

W x y t X x Y y T t

A C C m x n y t e

B C C m x n y t e

ωπ π

ωπ π

−

−

=

−= ⋅

−+ ⋅

(12)

where mn mn m nF A C C= , mn mn m nH B C C= and for all 1,2,3,...mn =

By using the superposition principle (12) become

1 1

2

2

1 1

2

2

( , , ) ( , , ),

1 4sin sin coscos ,

2

1 4sin sin sin cos .

2

mn

m n

t

mn

m n

t

mn

w x y t w x y t

F m x n y t e

H m x n y t e

ωπ π

ωπ π

∞ ∞

= =

−∞ ∞

= =

−

=

−= ⋅

−+ ⋅

∑∑

∑∑ (13)

According to the initial condition ( , ,0) sin sinw x y x yπ π= and ( , ,0) 0.w

x yt

∂=

∂ By using

Fourier series, we have

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2

1 ; 1 1, 2

1.

1 8

mn

mn

F m and n

H

ω π

π

= = = =

=−

Substituting 1mnF = and2

1

1 8mn

Hπ

=−

in (13), we obtain the analytical solution:

2 2

2

2

1 8 1 1 8( , , ) (cos sin )(sin sin )

2 21 8

t

w x y t e t t x yπ π

π ππ

− − −= +

− (14)

Since 2ω π= when substituting to (11), we obtain that 21 4ω− is complex number.

Therefore

( )T t can not be solved.

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