1. No category

Approximate Solution of Two-Dimensional Nonlinear Wave
Equation by Optimal Homotopy Asymptotic Method
H. Ullah1, S. Islam1, I. Khan2, L. C. C. Dennis3, M. Fiza1
1
Department of Mathematics, Abdul Wali Khan University
Mardan, 23200 Pakistan.
2
College of Engineering Majmaah University, Majmaah Saudi Arabia
3
Department of Fundamental Applied Sciences, Universiti Teknologi
PETRONAS 31750 Perak Malaysia
Abstract
In this article, the two dimensional nonlinear wave equations are considered. Solution to
the problem is approximated by using Optimal Homotopy Asymptotic Method (OHAM).
The residual and convergence of the proposed method to nonlinear wave equation is
presented through graphs. The resultant analytic series solution of the two dimensional
nonlinear wave equation shows the effectiveness of the proposed method. The
comparison of results has been made with the existing results available in the literature.
Keywords: OHAM, two-dimensional wave equation, Residual Estimate.
1. Introduction
The wave equations play a vital role in diverse areas of engineering, physics and
scientific applications. An enormous amount of research work is already available in the
study of wave equations [1-2]. This article deal with the two-dimensional nonlinear wave
equation of the form
∂ 2u ( x , t )
∂ 2 u ( x, t )
x2 + t 2
−
u
x
,
t
=
1
−
,
0 ≤ x, t ≤ 1
(1)
(
)
∂t 2
∂x 2
2
The differential equations (DEs) can be solved analytically by a number of
perturbation techniques [3-4]. These techniques are fairly simple in calculating the
solutions, but their limitations are based on the assumption of small parameters.
Therefore, the researchers are on the go for some new techniques to overcome these
limitations.
The idea of homotopy was pooled with perturbation. Liao [5] proposed Homotopy
Analysis Method (HAM) in his PhD dissertation and applied it to various nonlinear
engineering problems [6-8] . The Homotopy Perturbation Method (HPM) was initially
introduced by He [9-13]. HPM has been extensively used by several researchers
successfully for physical models [14-16]. Some useful comparisons between HAM and
HPM were done by Domiarrry and Liang [17-18].
Recently Marinca and Herisanu [19-21] introduced OHAM for the solution of
nonlinear problems which made the perturbation methods independent of the assumption
of small parameters, Ullah et al. [22-26] have extended and applied OHAM successfully
for numerous nonlinear phenomena.
1
The motive of this paper is to apply OHAM for the solution of two-dimensional nonlinear
wave equations. In [19-21] OHAM has been proved to be useful for obtaining an
approximate solution of nonlinear differential equations. Here, we have proved that
OHAM is more useful and reliable for the solution of two-dimensional nonlinear wave
equations; hence, showing its validity and greater potential for the solution of transient
physical phenomenon in Science and Engineering.
Section 2 has the basic idea of OHAM formulated for the solution of partial differential
equations. In Section 3, the effectiveness of OHAM for two-dimensional nonlinear wave
equation has been studied.
2. Basic Formulation of OHAM
Consider the partial differential equation of the following form:
A ( u ( x , t ) ) + f ( x , t ) = 0,
x∈Ω
(2)
B ( u, ∂u / ∂x ) = 0,
x∈Γ
(3)
Where A is a differential operator, u ( x, t ) is an unknown function, and x and t denote
spatial and temporal independent variables, respectively, Γ is the boundary of Ω and
f ( x, t ) is a known analytic function. A can be divided into two parts: L and N such
that:
A =L+N
(4)
L is the simpler part of the partial differential equation which is easier to solve, and N
contains the remaining part of A .
According to OHAM, one can construct an optimal homotopy
φ ( x, t; p ) :
Ω × [ 0,1] → ℜ which satisfies:
H (φ ( x, t ; p ) , p ) = (1 − p) {L (φ ( x, t ; p )) + f ( x, t )}
− H ( p) { A (φ ( x, t ; p )) + f ( x, t )} = 0.
(7)
Here the auxiliary function H ( p ) is nonzero for p ≠ 0 and H (0) = 0 . Eq (7) is called
optimal homotopy equation. Clearly, we have:
p = 0 ⇒ H (φ ( x, t;0 ) , 0 ) = L (φ ( x, t;0)) + f ( x, t ) = 0,
(8)
p = 1 ⇒ H (φ ( x, t ;1) ,1) = H (1) { A (φ ( x, t ;1)) + f ( x, t )} = 0,
(9)
Obviously, when p = 0 and p = 1 we obtain
φ ( x, t;0 ) = u0 ( x, t ) and φ ( x, t;1) = u ( x, t ) respectively. Thus, as p varies from 0 to 1 ,
the solution φ ( x, t; p ) approaches from u0 ( x, t ) to u ( x, t ) , where u0 ( x, t ) is obtained
from Eq (7) for p = 0 :
L ( u0 ( x, t ) ) + f ( x, t ) = 0,
B ( u0 , ∂u0 / ∂x ) = 0 .
2
(10)
Next, we choose auxiliary function H ( p ) in the form
H ( p ) = pC1 + p 2C2 + p mCm .
(11)
To get an approximate solution, we expand φ ( x, t; p, Ci ) by Taylor’s series about p in
the following manner,
∞
φ ( x, t ; p, Ci ) = u0 ( x, t ) + ∑ uk ( x, t ; Ci ) p k , i = 1, 2,... .
(12)
k =1
Substituting Eq. (12) into Eq. (7) and equating the coefficient of like powers of p , we
obtain Zeroth order problem, given by Eq. (10), the first and second order problems are
given by Eqs. (13-14) respectively and the general governing equations for uk ( x, t ) are
given by Eq. (15):
L ( u1 ( x, t ) ) = C1N 0 ( u0 ( x, t ) ) , B ( u1 , ∂u1 / ∂x ) = 0
L ( u2 ( x, t ) ) − L ( u1 ( x, t ) ) = C2 N 0 ( u0 ( x, t ) ) +
C1 L ( u1 ( x, t ) ) + N 1 ( u0 ( x, t ), u1 ( x, t ) )  ,
B ( u2 , ∂u2 / ∂x ) = 0,
(13)
(14)
L ( uk ( x, t ) ) − L ( uk −1 ( x, t ) ) = Ck N 0 ( u0 ( x, t ) ) +
k −1
∑ C L ( u
i
k −i
( x, t ) ) + N k −i ( u0 ( x, t ), u1 ( x, t ),..., uk −i ( x, t ) ) ,
(15)
i =1
k = 2,3,...,
B ( uk , ∂uk / ∂x ) = 0,
where N k −i ( u0 ( x, t ), u1 ( x, t ),..., uk −i ( x, t ) ) are the coefficient of p k −i in the expansion of
N (φ ( x, t ; p ) ) about the embedding parameter p.
N (φ ( x, t ; p, Ci ) ) = N 0 ( u0 ( x, t ) ) + ∑ N k ( u0 , u1 , u2 , ... , uk ) p k
(16)
k ≥1
It should be underscored that the u k for k ≥ 0 are governed by the linear equations with
linear boundary conditions that come from the original problem, which can be easily
solved.
It has been observed that the convergence of the series Eq. (12) depends on the auxiliary
constants C1 , C2, ... . If it is convergent at p = 1 , one has:
u ( x, t ; Ci ) = u0 ( x, t ) + ∑ uk ( x, t ; Ci )
(17)
k ≥1
Substituting Eq. (17) into Eq. (1), it results in the following expression for residual:
R ( x, t; Ci ) = L (u ( x, t ; Ci )) + f ( x, t ) + N (u ( x, t; Ci ))
(18)
In actual computation k = 1, 2,3,..., m . If R ( x, t ; Ci ) = 0 then u ( x, t; Ci ) is the exact
solution of the problem. Generally it doesn’t happen, especially in nonlinear problems.
For determining auxiliary constants, Ci , i = 1, 2,..., m , there are a number of methods like
Galerkin’s Method, Ritz Method, Least Squares Method and Collocation Method. The
Method of Least Squares can be applied as under:
3
t
J ( Ci ) = ∫
∫ R ( x, t; C ) dx dt
2
(19)
i
0 Ω
∂J
∂J
∂J
=
= ... =
=0.
∂C1 ∂C2
∂C m
(20)
The m th order approximate solution can be obtained by these optimal constants. The
more general auxiliary function H ( p ) is useful for convergence, which depends on
constants C1 , C2, ...Cm , can be optimally identified by Eq. (20) and is useful in error
minimization.
3. Application of OHAM to Two Dimensional Nonlinear Wave Equations
To demonstrate the effectiveness of the formulation of OHAM, we consider two
dimensional nonlinear wave equations of the form (1) with initial conditions
t2
u ( 0, t ) = ,
2
∂
u ( 0, t ) = 0.
∂x
(21)
Applying the method formulated in section 2, leads to
L=
∂ 2 u ( x, t )
∂x 2
, N = −u ( x , t )
∂ 2u ( x , t )
∂t 2
, f ( x, t ) = 1 −
x2 + t 2
.
2
Zeroth Order Problem
∂u0 ( x, t )
x2 + t 2
,
∂t
2
t2
∂
u0 ( 0, t ) = ,
u0 ( 0, t ) = 0.
2
∂x
= 1−
(22)
Its solution is
u0 ( x, t ) =
x 2 t 2 x 2t 2 x 4
+ −
− .
2 2
4
24
(23)
First Order Problem
∂ 2u1 ( x, t )
∂ 2u0 ( x , t )
∂ 2u0 ( x , t )
 x2 t 2 
= (1 + C1 )
− C1u0 ( x, t )
+ (1 + C1 )  + − 1 ,
∂x 2
∂x 2
∂t 2
 2 2 
∂
u1 ( 0, t ) = 0,
u1 ( 0, t ) = 0.
(24)
∂x
Its solution is
 1
x2 t 2 x4 7 x6
x8 
u1 ( x, t ) =  − t 2 x 2 − +
+
−
(25)
 C1 .
24 24 720 2688 
 4
4
Second Order Problem


∂ 2u1 ( x, t )
∂ 2 u0 ( x , t )
∂ 2u1 ( x, t )
1
+
+
−
,
−
C
C
C
u
x
t
(
)
(
)


1
2
1 0
2
2
2
2
∂ u2 ( x , t ) 
∂x
∂x
∂t
,
=
2
2
2
2
2

∂
u
x
,
t
∂
u
x
,
t
∂x
) − C u x, t
) + C  x + t − 1 
0(
0(
(
)
C1u1 ( x, t )

2
0
2
∂t 2
∂t 2
 2 2


∂
u2 ( 0, t ) = 0,
u2 ( 0, t ) = 0.
∂x
Its solution is
43 12 2
 11

2
10
2
8
 − 5 ( −173 + 56t ) x C1 − 4 x C1 + 66 x (−15C1 + 


2
2
2 2
2
1
−
15
C
+
36
t
C
−
15
C
−
665280
t
x
C
+
C
+
C

.
(
)
(
)
1
1
2
1
2
1
u 2 ( x, t ) =

2661120 
4
2
2
2
2
 +110880 x ((−1 + t )C1 + (−1 + 2t )C1 + (−1 + t )C2 ) 


6
2
2
2
2
 −3696 x ((−7 + 3t )C1 + (−14 + 11t )C1 + (−7 + 3t )C2 ) 
(26)
(27)
Third Order Problem

∂ 2 u 2 ( x, t )
∂ 2u1 ( x, t )
∂ 2 u0 ( x , t )
∂ 2 u2 ( x , t ) 
1
+
+
+
−
,
C
C
C
C
u
x
t
(
)
(
)


1
2
3
1 0
∂x 2
∂x 2
∂x 2
∂t 2



∂ 2u3 ( x, t ) 
∂ 2 u0 ( x , t )
∂ 2u1 ( x, t )
∂ 2u1 ( x, t )
=
−
−
−
C
u
x
,
t
C
u
x
,
t
C
u
x
,
t
(
)
(
)
(
)

,
1 2
1 1
2 0
∂x 2
∂t 2
∂t 2
∂t 2




∂ 2 u0 ( x , t )
∂ 2 u0 ( x , t )
 x2 t 2 
− C3u0 ( x, t )
+ C3  + − 1
 −C2u1 ( x, t )

2
2
∂t
∂t
 2 2



u3 ( 0, t ) = 0,
∂
u3 ( 0, t ) = 0.
∂x
(28)
Its solution is
5
 −12 x 2 (352t 2 (37800 − 12600 x 2 + 2310 x 4 − 135 x 6 + 7 x8 )

 2

2
4
6
8
2
 + x (2217600 − 1034880 x + 108240 x − 7612 x + 215 x ))C1

 + (−13305600 x 4 + 9313920 x 6 − 1591920 x8 + 195800 x10 −





49221
2867
8
12
14
16
2 2
12586 x + 91 x − 240 x − 13 x t (129729600 −



1
u3 ( x , t ) =
64864800 x 4 − 1904760 x 6 + 166452 x8 − 6604 x10 + 231x12 )C13
.
319334400 

2
2
2
4
2
2
4
 −12 x C1 (660(168t (60 − 10 x + x ) + x (1680 − 392 x + 15 x )) 
 + (352t 2 (37800 − 12600 x 2 + 2310 x 4 − 135 x 6 + 7 x8 ) + x 2 (2217600 


 −1034880 x 2 + 108240 x 4 − 7612 x 6 + 215 x8 ))C2 ) − 7920 x 2 (168t 2 ( 


2
4
2
2
4
60 − 10 x + x ) + x (1680 − 392 x + 15 x ))(C2 + C3 )



(29)
Adding equations (23),(25), (27) and (29), we obtain:
u ( x, t , C1 , C2 , C3 ) = u0 ( x, t ) + u1 ( x, t , C1 ) + u2 ( x, t , C1 , C2 )
(30)
The residual can be calculated by using Eq. (18). For calculations of the constants C1 , C2
and C3 , using (30) in (21) and applying the procedure mentioned in (17)-(18), we get
C1 = −1.0639118119872306 , C2 = −9.762978332188523 × 10−4 and
C3 = 1.287603572987478 × 10−4
  0.5 + 3.12815 ×10−7 x 2 − 1.0409 ×10−4 x 4 +  
 

−4 6
−5 8
 2  2.812 × 10 x − 2.52425 × 10 x

 x  +3.37375 × 10 −5 x10 − 2.03977 × 10−6 x12  
 


  +4.50493 × 10−8 x14

.
u ( x, t ) = 
  0.5 + 1.87689 × 10−3 x 2 − 4.4610 × 10−4 x 4  
 
 
−3 6
−3 8
 
 2  +1.53732 × 10 x − 1.38661×10 x
t  +2.28978 ×10−4 x10 − 1.53259 × 10−5 x12  
 
 
 
  +5.3608 ×10−7 x14
 
(31)
The exact solution is [2]
u ( x, t ) =
x2 + t 2
,
2
and HPM solution is [2]
6
(32)
 t 2 x2 x4
x8
x 6t 2 327 x8 
+
−
+
+
−


2
2
24
2688
240
241920 
u ( x, t ) _ HPM =  8 2
.
x10t 2
1038 x10
x12 
 xt
 + 1120 − 21600 + 3628800 − 107520 
(33)
Table 1. Comparison of Exact, HPM and OHAM solutions for different values of x and t
x
0
0.062
0.156
0.225
0.187
0.281
0.343
0.350
0.468
0.369
0.350
0.539
0.780
0.593
0.620
0.656
0.781
0.843
0.968
0.975
1.000
Table 2
t
0
0.162
0.156
0.128
0.275
0.281
0.343
0.263
0.468
0.384
0.870
0.403
0.111
0.593
0.685
0.656
0.781
0.843
0.968
0.692
1.000
Exact solution
0.00000
0.015044
0.024336
0.033504
0.055297
0.078961
0.117649
0.095834
0.219024
0.141801
0.439700
0.226465
0.310360
0.351649
0.426813
0.430336
0.609961
0.710649
0.937024
0.714744
1.000000
HPM solution
0.000000
0.015044
0.024336
0.033505
0.055297
0.078961
0.117650
0.095835
0.219032
0.141810
0.439706
0.226476
0.310263
0.351704
0.426913
0.430462
0.610499
0.711668
0.940295
0.716206
1.004310
OHAM solution
0.00000
0.015044
0.024336
0.033504
0.055297
0.0789608
0.117648
0.095834
0.219022
0.141808
0.439697
0.226463
0.310358
0.351647
0.426808
0.430332
0.609963
0.710652
0.936958
0.714690
0.999877
Comparison of absolute errors of HPM and OHAM solutions for different
values of x and t
x
0
0.062
0.156
0.225
0.187
0.281
0.343
0.350
0.468
t
0
0.162
0.156
0.128
0.275
0.281
0.343
0.263
0.468
E*
0.00000000
6.003 × 10 −12
1.128 × 10 −9
2.613 × 10 −9
1.213 × 10 −8
1.275 × 10 −7
6.370 × 10 −7
3.309 × 10 −7
7.922 × 10 −7
7
E **
0.00000000
1.74057 × 10 −12
6.00903 × 10 −9
2.49951 × 10 −9
3.5136 × 10 −8
1.91487 × 10 −7
5.53487 × 10 −7
4.03006 × 10 −7
2.28841 × 10 −7
0.369
0.350
0.539
0.780
0.593
0.620
0.656
0.781
0.843
0.968
0.975
1.000
0.384
0.870
0.403
0.111
0.593
0.685
0.656
0.781
0.843
0.968
0.692
1.000
1.273 × 10 −6
5.736 × 10 −6
1.122 × 10 −6
9.873 × 10 −5
5.497 × 10 −5
1.010 × 10 −4
1.263 × 10 −4
5.378 × 10 −4
1.019 × 10 −3
3.271 × 10 −3
1.461 × 10 −3
4.310 × 10 −3
8.50245 × 10 −7
3.11566 × 10 −6
2.45830 × 10 −6
4.00013 × 10 −6
4.1745 × 10 −6
4.68453 × 10 −6
3.64289 × 10 −6
2.2877 × 10 −6
2.6668 × 10 −6
6.56401× 10 −5
5.41072 × 10 −5
1.22591× 10 −4
E * = HPM − Exact , E** = OHAM − Exact
Fig. 1, 3D plot for the OHAM solution at t = 1 .
8
Fig. 2, 3D plot for the Exact solution at t = 1 .
1.0
OHAM
Exact
0.9
uHxL
0.8
0.7
0.6
0.5
0.0
0.2
0.4
0.6
0.8
1.0
x
Fig. 3, 2D plot for the OHAM and Exact solutions at t = 1 .
9
0.00002
Residual
0.00001
0
- 0.00001
- 0.00002
- 0.00003
0.0
0.2
0.4
0.6
0.8
1.0
x
Fig. 4, 2D plot for the residual of OHAM at t = 1 .
0.5
Exact
Zeroth Order
First Order
Second Order
Third Order
uHx,tL
0.4
0.3
0.2
0.1
0.0
0.0
0.2
0.4
0.6
0.8
1.0
x
Fig. 5, 2D plot for the convergence of orders of OHAM solution at t = 1 .
4. Results and Discussions
The formulation presented in section 2, provides accurate solutions for the problems
demonstrated in section 3. We have used Mathematica 7 for most of our computational
work. In Table 1 and Figs. 1-3, we have compared the OHAM results with the results
obtained by HPM and Exact for various values of x and t at spatial domain [0,1] for table
1 and at different values of x and fixed value of t = 1 for figures 1-3. In Table 2, we have
presented absolute errors at different values of x and t . Fig. 4 presents the residual at a
spatial domain [0,1] at t = 1 . The convergence of OHAM is presented in Fig. 5 at t = 1 .
5. Conclusion
10
In this paper, we studied the two dimensional nonlinear wave equation. We have used the
OHAM to approximate the solution of the titled problem. It is clear from the present
work that OHAM can successfully be applied to nonlinear phenomena like the one we
had. The obtained results are accurate which shows the effectiveness and validity of the
proposed method. It is observed that OHAM is simpler in applicability, more convenient
to control the convergence and involve less computational work.
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