6. Modified Adomian for linear partial differential equations

INTERNATIONAL JOURNAL OF MATHEMATICS AND SCIENTIFIC COMPUTING (ISSN: 2231-5330), VOL. 3, NO. 2, 2013
17
Modified Adomian for Linear Partial Differential
Equations
Ruchi Nigam
Abstract—In this paper we outline a strategy to use Adomian
decomposition method properly for solving linear partial differential equation with homogeneous and inhomogeneous boundary.
Our fundamental goal in this chapter is to present a further
insight into partial solutions in the decomposition method, and
the resolution of above cases by suitable transformation. The
modifications are necessary in order to make the Adomian
decomposition method efficient for boundary and initial value
problems. The heat equation is chosen as an example and the
cases where improper solution can be obtained are discussed.
Index Terms—Adomian
Laplace Transform.
decomposition;
Heat
while the x equation obtained by inverting the Lx operator
gives zero [1].
In a problem suggested by Professor Y. Cherrault [1],
we use the same boundary conditions but choose u(x, 0) =
x(ℓ − x). Again the x equation gives a zero result. However
the t equation
∞
∑
u = u(x, 0) + L−1
L
un
(2)
x
t
n=0
equation;
u0 = xℓ − x2
u1 = −2t
MSC 2010 Codes – 44A10, 35K05
un≥2 = 0
yields u = xℓ − x + 2t which clearly does not satisfy the
boundary conditions. The terms above, for m > 0, are
I. I NTRODUCTION
L
2
m
um = [L−1
t Lx ] u(x, 0).
ET us look at the possible difficulties in the heat equation
ut = uxx ,
consequently,
0 < x < ℓ, t > 0,
u(0, t) = 0, t > 0
u(ℓ, t) = 0, t > 0
u(x, 0) = sinx, 0 < x < ℓ
u(x, t) =
(1)
n=0
u0 = sinx
u2 = L−1
t Lx
u3 = L−1
t Lx
u4 = L−1
t Lx
∞
∑
n=0
∞
∑
n=0
∞
∑
n=0
∞
∑
n=0
u0 = −tsinx
u1 =
t2
sinx
2!
u2 = −
u3 =
∞
∑
m
[L−1
t Lx ] f (x)
(4)
m=0
Thus, in the usual decomposition notation Lt u = Lx u or
∞
∑
u = u(x, 0) + L−1
L
un ,
x
t
u1 = L−1
t Lx
(3)
t3
sinx
3!
t4
sinx
4!
where u(x, 0) = f (x). In the case f (x) = x(ℓ − x), we see
that the operation L−1
t Lx f (x) annihilates the series in a finite
number of terms and the result does not satisfy the boundary
conditions. This annihilation of the series was not the case with
f (x) = sinx. The resolution of such cases was obtained by
Adomian and Rach [1] by appropriate expansion of the initial
term without making a priori assumptions about the solution.
However, this technique failed to obtain exact solution in the
case of initial-boundary value problems with inhomogeneous
boundary conditions.
A review of previous studies indicates that the difficulty
arise while applying adomian decomposition method to partial
differential equation with homogeneous and inhomogeneous
boundary conditions. Hence we solve (1) using adomian with
Laplace. For Adomian decomposition method one can refer to
[1], [5], [6].
Applying Laplace both sides of equation (1)
d2 u
dx2
d2 u
pu
−
x(ℓ
−
x)
=
∞
∑
dx2
t3
t4
t2
u(x, t) =
un = (1 − t + − + − ....) = e−t sinx Solving by using adomian decomposition method
2! 3! 4!
n=0
d2 u
, Ru = pu, g = −x(ℓ − x),
Lu =
Ruchi Nigam was research scholar at Department of Mathematics, Indian
2
∫dx ∫
Institute of Technology Roorkee, Roorkee-247667 and worked as assistant
pu(x, p) − u(x, 0)
..
.
professor at IJCET, Tuticorin, India. Currently she is on child care leave.
(E-mail: [email protected])
L−1 [.]
x
=
(6)
x1
=
[.]dx2 dx1
0
(5)
0
(7)
INTERNATIONAL JOURNAL OF MATHEMATICS AND SCIENTIFIC COMPUTING (ISSN: 2231-5330), VOL. 3, NO. 2, 2013
∫ x ∫ x1
u0 = u(0, p) + xF (p) −
x(ℓ − x)dx2 dx1
0
0
)
( 3
x4
x ℓ
= F (p)x −
−
3!
3.4
∫ x ∫ x1
u1 =
pu0 dx2 dx1
0
0
( 5
)
x3
x pℓ
x6 p
= p F (p) −
−
3!
5!
3.4.5.6
( 7 2
)
5
x
x p ℓ
x6 p
u2 = p2 F (p) −
−
5!
7!
3.4.5.6.7.8
..
.
Thus,
)
(
p2 x5
px3
+
...
u(x, p) = F (p) x +
3!
5!
( 3
)
x
px5
p2 x7
− ℓ
+
+
.....
3!
5!
7!
( 4
)
6
2 8
x
px
p x
+ 2
+
+
+ ....
4!
6!
8!
18
Applying Laplace with adomian as stated above, we get
∫ x ∫ x1
u = u(0, p) + x.ux (0, p) −
sin xdx2 dx1
0
0
∫ x ∫ x1
+
pudx2 dx1
(12)
0
0
u0 = F (p).x + sin x
px3
− p sin x
u1 = F (p)
3!
p2 x5
+ p2 sin x
u2 = F (p)
5!
p3 x7
u3 = F (p)
− p3 sin x
7!
(13)
Thus,
(8)
The value of F (p) = ux (0, p) are found by applying the
condition at x = ℓ. Substituting the value of F (p) in (8),
we get a series
( 4
)
x
px6
p2 x8
u(x, p) =
+
+
....
12 360 20160
( 3
)
x
px5
p2 x7
+
− −
−
.... ℓ
6
120 5040
)
(
px3
p2 x5
x
+
+
.... ℓ3
+
12
72
1440
(
)
−px p2 x3
p3 x5
+
−
−
.... ℓ5 + ..... (9)
120
720
14400
in closed form above series can be written as
√
1
2
2 cosh{ p(2x − ℓ)/2}
u(x, p) = x(ℓ − x) − 2 + 2
(10)
√
p p
p
cosh( pℓ/2)
(
)
px3
p2 x5
p3 x7
u(x, p) = F (p) x +
+
+
...
3!
5!
7!
(
)
(14)
+ sin x 1 − p + p2 − p3 + ...
The series is clearly of
√
sinh px
sin x
u(x, p) = F (p) √
+
p
p+1
Applying boundary conditions at x = 1, we have
√
sinh p
sin 1
sin 1
F (p) √
+
=
⇒ F (p) = 0
p
p+1
p+1
(15)
Solution in Laplace space is given by
u(x, p) =
sin x
p+1
(16)
Taking Laplace inverse, we get
u(x, t) = sin xe−t
(17)
Although the above example can be solved by Adomian also
because of the presence of the functions sinx and e−t . But
the involvement of all the boundary conditions make laplace
approach more reliable and efficient.
Taking inverse Laplace Transform, we have
Next consider (1) with boundary conditions u(x, 0) =
[
1 (2x − ℓ)2
ℓ2
16 ℓ2 T0 , ux (0, t) = 0, u(ℓ, t) = T1 . It is the case where it is
u(x, t) = x(ℓ − x) − 2t + 2 {
− } + t − 3 . ×difficult to obtain solution by decomposition method. Such
2
4
4
π 4
] a problem can be easily solve by using Laplace approach to
∞
2
2
n
∑
−(2n−1) π t
(−1)
(2n − 1)π(2x − ℓ)/2
2 /4)
4(ℓ
decomposition method. Thus the solution is
e
cosh
3
(2n
−
1)
2.(ℓ/2)
n=1
d2 u
(11)
= pu − T0
dx2
∫ x ∫ x1
x2
Consider equation (1) with inhomogeneous boundary
u0 = −
T0 dx2 dx1 + F (p) = −T0
+ F (p)
2!
conditions u(0, t) = T 1, u(ℓ, t) = T 2, and u(x, 0) = h(x).
(
)
∫ x 0∫ x10
x2
x4
T1 , T2 may be functions or numeric constants. In [1] steady
u1 =
pu0 dx2 dx1 = p F (p) − pT0
2!
4!
state as well as transient solutions are obtained. However, for
(
)
∫0 x ∫0 x1
4
inhomogeneous boundary conditions the result obtained in [1]
x
x6
2
u
=
pu
dx
dx
=
p
F
(p)
−
T
0
2
1
2
1
does not satisfy the boundary conditions. And there was no
4!
6!
(
)
∫0 x ∫0 x1
consideration about the wrong result. We can deal with this dif6
8
x
x
u3 =
pu2 dx2 dx1 = p3 F (p) − T0
ficulty by using Laplace with adomian. For example, consider
6!
8!
0
0
inhomogeneous conditions u(0, t) = 0, u(1, t) = sin 1e−t , and
(18)
u(x, 0) = sin x.
INTERNATIONAL JOURNAL OF MATHEMATICS AND SCIENTIFIC COMPUTING (ISSN: 2231-5330), VOL. 3, NO. 2, 2013
)
(
x4
x6
x2
+ p2 ...
F (p) 1 + p + p2
2!
4!
6!
( 2
)
4
6
x
x
x
x8
− T0
+ p. + p2 . + p3 . ... (19)
2!
4!
6!
8!
u(x, t) =
19
a finite number of terms, we consider the Laplace Transform
method. Thus, we have
d2 u
+ L{g(x, t)}
(29)
dx2
The series is clearly of
where L{g(x, t)} denotes the Laplace transform of the function g(x, t).
T0
T0
√
√
cosh px +
(20) Proceeding as stated above we may obtain the solution.
u(x, p) = F (p) cosh px −
p
p
In case of inhomogeneous heat equation with inhomoThe function F (p) has to be determined by applying the geneous boundary condition the equation can be reduced to
condition at x = ℓ. Substituting the value, we have
homogeneous boundary with the help of the transformation
√
stated earlier and hence can be solve.
cosh px
T0
u(x, p) = (T0 − T1 )
(21) For example, consider the equation (24) where g(x, t) = 2x
√ +
p cosh pℓ
p
and boundary conditions u(0, t) = 0, u(1, t) = 0, u(x, 0) =
Taking inverse Laplace
x − x2 . In this case adomian method provides a solution
u(x, t) = 2xt that does not satisfy the boundary conditions.
∞
4(T1 − T0 ) ∑ (−1)n − (2n−1)22 π2 kt
(2n − 1)πx
4t
While using the Laplace approach we obtain
the series
u(x, t) = T1 +
e
cos
) of the
∑∞ (
n2 π 2 t
π
2n − 1
2ℓ
n=1
sin nπx.
solution u(x, t) = x(1 − x) − π43 n=1 1−en3
(22)
Another way of solving inhomogeneous boundary
condition problem is by reducing the problem to homogeneous
one by applying the transformation introduced in [7], [8].
Let
pu =
III.
CONCLUSION
hence with the technique stated above we can obtain the
solution.
The heat equations are solved Laplace approach to decomposition method. In cases where annihilation of the series
is not the case, solution can be obtained by the usual decomposition method. In cases where Lm
x operator annihilates the
series in a finite number of terms improper solution may be
obtained when consider the initial boundary value problems
involving inhomogeneous boundary conditions. We overcome
the difficulties by using the Laplace approach and the results
are successfully obtained. The transformation used in this
paper is a simple technique for special cases. However, the
powerful Adomian decomposition method can be applied to
much more complicated physical problems [2], [3], [4], [5] in
comparison with other methods. The above examples are used
for clarity and comparison, since they are obviously solvable
by separation of variables. In general, the decomposition is
more widely applicable.
II. I NHOMOGENEOUS HEAT EQUATION
R EFERENCES
u(x, t) = υ(x, t) + g(x), f (x) = h(x) − g(x)
(ℓ − x)
x
g(x) =
T1 + T 2
ℓ
ℓ
substituting this in the problem yields
υt = uxx ,
0 < x < ℓ, t > 0,
υ(0, t) = 0, t > 0
υ(ℓ, t) = 0, t > 0
υ(x, 0) = f (x), 0 < x < ℓ
(23)
Consider the inhomogeneous case
ut = uxx + g(x, t)
(24)
u(x, 0) = u(ℓ, t) = u(0, t) = 0
(25)
write t equation
Lt = Lx + g(x, t)
(26)
According to adomian decomposition, we have
u = u0 + L−1
t Lx
∞
∑
un ,
(27)
n=0
−1
with u0 = L−1
t g + u(x, 0) = Lt g, since initial condition is
zero.
m −1
um = (L−1
(28)
t Lx ) Lt g
are components of u. If annihilation of series is not the case
the adomian method is sufficient to obtain the solution. In the
cases where Lm
x operator acting on g annihilates the series in
[1] G. Adomian and R. Rach, “A further consideration of partial solutions
in the decomposition method” Comput. Math. Appl. vol. 23, pp. 51–64,
1992.
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Physics, Kluwer, Dordrecht, Holland, 1989.
[3] P. Vadasz and S. Olek, “Convergence and accuracy of Adomian’s decomposition method for the solution of Lorenz equations,” Int. J. of Heat and
Mass Transfer, vol. 43, pp. 1715–1734, 2000.
[4] P. Vadasz, “Local and global transitions to chaos and hysteresis in a
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2, pp. 213–245, 1999.
[5] R.C. Mittal and R. Nigam. “Solution of a class of singular boundary value
problems,” Numerical Algorithms, vol. 47, pp. 169–179, 2008.
[6] R.C. Mittal and R.Nigam. Solution of fractional integro-differential equations by adomian decomposition method, International Journal of Applied
Mathematics and Mechanics, vol. 4, no. 2, pp. 87–94, 2008.
[7] T. Myint-U, Partial Differential Equations of Mathematical Physics,
North-Holland, New York, 1980.
[8] R.B. Guenther and J.W. Lee, Partial Differential Equations of Mathematical Physics and Integral Equations, Dover, 1996.

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