On the Solution of Abel Differential Equation by Adomian

Applied Mathematical Sciences, Vol. 2, 2008, no. 43, 2105 - 2118

On the Solution of Abel Differential Equation by Adomian Decomposition Method K.I. Al-Dosary Department of Basic Sciences College of Arts and Sciences University of Sharjah, P.O.Box 27272 Sharjah, United Arab Emirates [email protected] N.K. Al-Jubouri Department of Basic Sciences College of Arts and Sciences University of Sharjah, P.O. Box 27272 Sharjah, United Arab Emirates [email protected] H.K. Abdullah Department of Basic Sciences College of Arts and Sciences University of Sharjah, P.O. Box 27272 Sharjah, United Arab Emirates [email protected] Abstract. Abel differential equation with constant coefficients of the form dy = fk y k dt k=0 M

n − is solved using Adomian decomposition method in two formulations An , A polynomials as well as in classical power series. An illustrative example is introduced at the end. Mathematics Subject Classification: 34L30, 34G20, 47J05 Keywords: Abel differential equation; Adomian decomposition; Power series

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K.I. Al-Dosary, N.K. Al-Jubouri and H.K. Abdullah

1. Introduction Consider the Abel differential equation of the first kind of the form dy = fk (t)y k dt k=0 M

This differential equation arose in the context of the studies of N.H. Abel [1] on the theory of elliptic functions. An interesting area of this type of equation also arises in biological systems, see [10], and in many physical, engineering, ecology and economics, see [11]. Thus, methods of solution for these equations are of great importance to engineers and scientists. It represents a natural generalization of the Riccati equation. In this paper we discuss the solution of this equation with constant coefficients fi ’s by applying the Adomian decomposition method. This powerful and extremely efficient method that G. Adomian has developed in the beginning of the 1980’s supplies analytical approximations to the solution of many kinds of equations. The Adomian decomposition method can be an effective procedure for solutions of nonlinear and/or stochastic continuous-time dynamical systems without usual restrictive assumptions. For more details see for instance [2-8], and references therein. We recall briefly some general known notations and formulas of the method that will be used frequently henceforth. Consider the differential equation (1.1)

F y(t) = f0 (t)

where F represents a general non-linear ordinary differential operator involving both linear and non-linear terms. The linear term is decomposed into L+R where L is easily invertible and R is the remainder of the linear operator. Thus the equation may be written in the form Ly + Ry + Ny = f0 (t) where Ny represents the non-linear terms. The solution proposed by Adomian [9] is to take L as the heighest-ordered derivative of the linear part. Solving for Ly, Ly = f0 − Ry − Ny Because L is invertible, an equivalent expression is (1.2)

L−1 Ly = L−1 f0 − L−1 Ry − L−1 Ny

Solving 1.2 for y yields (1.3)

y = F0 (t) + L−1 f0 − L−1 Ry − L−1 Ny

Abel differential equation

The nonlinear term Ny will be equated to

∞

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An , where the An are special

n=0

polynomials of two kinds of formulations to be discussed, and y will be decom∞ yn , with y0 identified as F0 (t) + L−1 f0 : posed into n=0 ∞

∞

−1

yn = y0 − L R

n=0

∞

−1

yn − L

n=0

An

n=0

Consequently, we can write y1 = −L−1 Ry0 − L−1 A0 y2 = −L−1 Ry1 − L−1 A1 In general, we obtain the following recurrence relation yn+1 = −L−1 Ryn − L−1 An If the series y =

∞

yn converges, then the n-term partial sum φn =

n=0

be the approximate solution since lim φn = n→∞

∞

n−1

yi will

i=0

yi = y by definition.

i=0

Two formulations have been developed for the An -polynomials in the method. ∼

One set designated as An and the other by An , see [8]. Either may be used as indicated above. ∼

1.1. The Formulation of An . Consider an equation for which y(t) is the ∞ ∼ ∼ solution containing a non-linear term Ny = f (y) = An . These An polynon=0

mials are defined by ∼

An = Ωn f (y0 ),

n = 0, 1, 2, ...

where Ωn is the operator defined as Ω0 = 1, Ωn = ξn

identity operator n−1

Ων ,

n = 1, 2, 3, ...

ν=0

and ξn is the operator defined as yn dyd

−1 d d2 1 = yn + yn2 2 + ... dy0 2! dy0 ∞ k 1 k d y = k! n dy0k k=1

ξn = e

0

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1.2. The Formulation of An . Consider an equation for which y(t) is the ∞ An . These An polynosolution containing a non-linear term Ny = f (y) = n=0

mials are defined by n

dr f c(r, n) r , An = dy r=1 where c(r, n) =

1 r! α +...+α 1

n ≥ 1,

yα1 yα2 ...yαr r =n

α1 ...αr =0

and A0 = f (y0 ) The aim of this work is to construct the solution of the differential equation of the form dy = fk y k dt k=0 M

(1.4)

where fk ’s are constants coefficients, satisfying initial condition y(t0 ) = α ∼

applying the Adomian’s decomposition method in both An , An polynomials as well as the power series method. We do not loss generality if we assume that the initial condition to be y(0) = 0 just by making use of the shifting ∼ transformation y(t) = y(t + t0 ) − α. Therefore we will use henceforth the initial condition y(0) = 0. The second section of this paper is the analysis of decomposition method in finding the solution. Such analysis indicates some relevant difference in the speed of convergence of the two formulations of decomposition method and the solution obtained by classical power series method. The final section contains illustrative example to obtain approximate solutions by using the decomposition method and classical power series method. 2. Analysis and Implementation of Decomposition Method In this section we will use the Adomian decomposition method to find ap∼

proximation solution of the Abel equation 1.4 in both fomulations An and An as well as in the classical power series method. ∼

2.1. The solution by An . Consider the equation 1.4 with the initial condition d , thus the equation can be written in the form y(0) = 0 and L = dt (2.1)

Ly = f0 + f1 y + f (y)


t fk y k . From the definition of L we have L−1 = [.]dτ , hence

M

where f (y) =

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0

k=2

L−1 Ly = y(t) − y(0) = y(t) Therefore equation 2.1 will be y = L−1 f0 + L−1 f1 y + L−1 f (y)

(2.2)

∞

The function y is decomposed into non-linear term

M

yn with y0 identified as L−1 f0 and the

n=0

∼

An , then equation 2.2 will be

n=0

k=2 ∞

(2.3)

∞

fk y k will be equated to −1

−1

yn = L f0 + L f1

n=0

∞

−1

yn + L

n=0

∞ ∼ An n=0

Consequently, we can write −1

(2.4)

t

y0 = L f0 =

f0 dτ = f0 t 0 ∼

y1 = L−1 f1 y0 + L−1 A0

(2.5)

∼

yn+1 = L−1 f1 yn + L−1 An

(2.6)

n≥1

∼

Thus An polynomials are given by ∼

An = Ωn f (y0) where f (y0 ) =

M k=2

fk y0k and the operator Ωn is defined as Ω0 = I Ωn = ξn

identity operator n−1

Ων ,

ν=0

whereas the operator ξn is given by ξn = eyn D − 1

n≥1

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and D =

d . dy0

Since f is a finite polynomial of order M , then

ξn = eyn D − 1 1 1 1 M M y D + 0 + 0 + ... = yn D + yn2 D 2 + yn3 D 3 + ... + 2! 3! M! n M 1 k k y D = k! n k=1 Therefore ∼

A0 = f (y0) =

M

fk y0k

=

k=2

M

fk f0k tk

k=2

Then from 2.5 we get t y1 =

f1 y0 dτ +

t M

0

0

fk f0k τ k dτ

k=2

Hence

(2.7)

y1 =

M k=1

1 f k fk tk+1 k+1 0

From 2.6 for n = 1, we get ∼

y2 = L−1 f1 y1 + L−1 A1

(2.8)

The first term L−1 f1 y1 of equation 2.8 will be t

−1

L f1 y1 =

f1 0

=

M k=1

M k=1

1 f k fk τ k+1 dτ k+1 0

1 f0k f1 fk tk+2 (k + 1)(k + 2)


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∼

The second term L−1 A1 of equation 2.8 will be −1

t

∼

L A1 =

Ω1 f (y0 )dτ 0

t ξ1 f (y0)dτ

= 0

t M 1 l l = y D f (y0 )dτ l! 1 l=1 0

M M 1 k! l fk y0k−l dτ y1 = l! (k − l)! l=1 k=2 t

l≤k

0

=

M k k=2 l=1

k! fk l!(k − l)!

t y1l y0k−l dτ 0

t M k k k−l y1l τ k−l dτ = fk f0 l k=2 l=1 0

−1

But the formula of the first term L f1 y1 of equation 2.8 can be rewritten in the form t k k L−1 f1 y1 = f or k = 1 fk f0k−l y1l τ k−l dτ l l=1 0

Therefore, equation 2.8 will be in the form M k k k−l y1l τ k−l dτ y2 = fk f0 l k=1 l=1 t

(2.9)

0

k

where the combination notation l is defined classically as k k! = l l!(k − l)! Now the term y3 is found out form 2.6 for n = 2, y3 = = = =

∼

L−1 f1 y2 + L−1 A2 L−1 f1 y2 + L−1 Ω2 f (y0 ) L−1 f1 y2 + L−1 (ξ2 + ξ1 ξ2 )f (y0 ) L−1 f1 y2 + L−1 ξ2 f (y0 ) + L−1 ξ1 ξ2 f (y0)

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The first term L−1 f1 y2 will be t τ M k k k−l y1l ω k−l dωdτ L f1 y2 = f1 fk f0 l k=1 l=1 −1

0

0

The second term L−1 ξ2 f (y0 ) will be t M k k L ξ2 f (y0 ) = y2l y0k−l dτ fk l 0 k=2 l=1 −1

consequently, from the identity The third term L−1 ξ1 ξ2 f (y0 ) canbe written, M M ai bj = ai bj , in the form of summation notation i=1

j=1

i=1,...,M j=1,...,M

M t M l l y mDm y D 1 2 f (y0 )dτ L−1 ξ1 ξ2 f (y0 ) = l! m! 0 m=1 l=1 t M y1l y2m l+m = D f (y0)dτ 0 l,m=1 l!m! M k

k! 1 = fk l!m! (k − l − m)! k=2 l,m=1

t

0

y1l y2my0k−l−mdτ

Therefore (2.10)

y3

t τ M k k = y1l ω k−l dωdτ f1 fk f0k−l l k=1 l=1 +

M k

k fk l

k=2 l=1

0

t

0

0

y2l y0k−l dτ

M k

k! 1 fk + l!m! (k − l − m)! k=2 l,m=1

t 0

y1l y2m y0k−l−mdτ

and so on, one can find the formulas for the next terms y4 ,... 2.2. The Solution by An . In this section we use the formulation of An poly∼

nomials which is equally correct with somewhat slower converging set than An but more convenient. Equation 2.3 will be in the form (2.11)

∞ n=0

−1

−1

yn = L f0 + L f1

∞ n=0

−1

yn + L

∞ n=0

An


2113

with y0 = L−1 f0 = f0 t

(2.12) and

yn+1 = L−1 f1 yn + L−1 An so (2.13)

yn+1 = f1

t

0

yn dτ +

0

t

n≥0

An dτ,

where these An polynomials are defined in the introduction. Consequently, we obtain t t y1 = f1 y0 dτ + f (y0)dτ 0

0

1 f0 f1 t2 + 2

=

t M 0 k=2

fk y0k dτ

Hence, since y0 = f0 t, y1 =

(2.14)

M k=1

1 f k fk tk+1 k+1 0 ∼

Notice that y0 , y1 are of the same formula that obtain from applying An polynomial. The next term y2 is computed as t t y1 dτ + A1 dτ y2 = f1 0

0

so (2.15)

y2 =

M k=1

1 fk f1 f0k tk+2 + (k + 1)(k + 2)

0

t

A1 dτ

where (2.16)

df A1 = y1 dy0 M = k=1

=

1 f k fk tk+1 k+1 0

l=2,...,M k=1,...,M

M

lfl f0l−1 tl−1

l=2

l fk fl f0k+l−1tk+l (k + 1)

From equations 2.15, 2.16, we get, l fk fl f0k+l−1 tk+l+1 (2.17) y2 = (k + 1)(k + l + 1) l=1,...,M k=1,...,M

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Further computation one can obtain (2.18) y3 =

j,k,l=1,...,M

j(j − 1)(k + l + 1) + lj(l + 1) fk fl fj f0k+l+j−2tk+l+j+1 (k + 1)(l + 1)(k + l + 1)(k + l + j + 1)

and so on and so forth. 2.3. The Solution by Power Series. Now we use the classical power series method and compare its output with the Adomian decomposition method in order to observe the fact that the decomposition method is faster and supplies quantitatively reliable results. Equation 1.4 with the initial condition y(0) = 0 implies that y (0) = f0 , thus y in power series is ∞ (2.19) ak tk y= k=1

where ak ’s are constants coefficients. We will use the following notation aα1 aα2 ...aαi , k≥i A(i, k) = α1 +...+αi =k

Remark 1. A(1, k) = ak ,

k = 1, 2, 3, ...

Therefore equation 2.19 will be in the form ∞ (2.20) A(1, k)tk y= k=1

ym =

Remark 2. 2

Proof. y =

∞

∞

A(m, k)tk

k=m

2

k

ak t

k=1 a2 t2

=(a1 t + + ...) (a1 t + a2 t2 + ...) 3 4 =a21 t2 + (a1 a2 + a + (a1 a3 + a2 a2 + 2 a1 ) t a3 a1 )t + ... 2 3 aα1 aα2 t + aα1 aα2 t + aα1 aα2 t4 + ... = α1 +α2 =2

α1 +α2 =3

α1 +α2 =4

=A(2, 2)t2 + A(2, 3)t3 + A(2, 4)t4 + ... ∞ = A(2, k)tk k=2

Assume that the statement is true for m , we show it is true for m + 1. y m+1 = y m y ∞ ∞ =[ A(m, k)tk ][ ai ti ] i=1 k=m ∞ ∞ aα1 aα2 ...aαm tk ][ ai ti ] =[ k=m

α1 +...+αm =k

i=1

∞

= = =


ai

k=m i=1 ∞

k=m+1 ∞

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α1 +...+αm =k

tk+i

aα1 aα2 ...aαm

α1 +...+αm+1 =k

tk

aα1 aα2 ...aαm+1

A(m + 1, k)tk

k=m+1

Therefore by mathematical induction the proof is complete. Hence the right-hand side of equation 1.4 will be M

fk y k

= f0 + f1

k=0

∞

A(1, k)tk + f2

k=1

∞

A(2, k)tk + ... + fM

k=2

∞

A(M, k)tk

k=M 2

= f0 + f1 A(1, 1)t + [f1 A(1, 2) + f2 A(2, 2)]t + [f1 A(1, 3) + f2 A(2, 3) + f3 A(3, 3)]t3 + ...

We define k

Λk =

fi A(i, k) k = 1, 2, 3, ...

i=1

Λ0 = f0 Hence M

k

fk y =

∞

k=0

Λk tk

k=0

On the other hand from equation 2.20, the left-hand side of equation 1.4 will ∞ be kA(1, k)tk−1 . Therefore, equation 1.4 will be k=1 ∞

k−1

kA(1, k)t

k=1

=

∞

Λk tk

k=0

Then ∞

[(k + 1)A(1, k + 1) − Λk ]tk = 0

k=0

This implies the following recurrence relation (k + 1)A(1, k + 1) = Λk ,

k = 0, 1, ...

Then ak+1 = =

1 Λk k+1 k 1 k+1

l=1

fl

α1 +...+αl =k

aα1 aα2 ...aαl

,

k = 1, 2, ...

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Thus a1 = Λ0 = f0 1 1 1 a2 = Λ1 = f1 A(1, 1) = f0 f1 2 2 2 1 1 1 1 Λ2 = [f1 A(1, 2) + f2 A(2, 2)] = [ f0 f12 + f2 f02 ] a3 = 3 3 3 2 1 1 2 a4 = Λ3 = f0 f1 f2 4 4 ...

(2.21)

(2.22)

Therefore 1 1 1 1 y = f0 t + f0 f1 t2 + [ f0 f12 + f2 f02 ]t3 + f02 f1 f2 t4 + ... 2 3 2 4

3. Example We give the following example in order to clarify use of the method and compare their speeds of convergence. Consider the Abel equation dy = 1 + y2 dt with the initial condition y(0) = 0. Here f0 = 1, f1 = 0, f2 = 1, and M = 2. ∼

The solution series based on the An -polynomial has the following n-terms y0

=

y1

=

f0 t = t, 2 1 k=1

k+1

f rom 2.4 f0k fk tk+1 =

1 3 t , 3

f rom 2.7

k 2 k 2 1 f rom 2.9 fk f0k−l y1l τ k−l dτ = 0 + t5 + t7 , l 15 63 k=1 l=1 0 2 4 4 2 4 7 4 13 1 15 9 t + + + t + t , t + t11 + 105 567 405 2475 2079 12285 59535 t

y2

=

y3

=

f rom 2.10

Therefore, if the series converges, then the approximate solution series is given by 1 3 2 5 1 2 4 4 4 2 7 9 y = t+ t + t + + t + + t + + t11 3 15 63 105 567 405 2475 2079 4 13 1 15 t + t + ... + 12285 59535


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The solution series based on the An -polynomial has the following n-terms y0

=

y1

=

y2

=

f0 t = t, 2 1 k+1 k=1 l=1,2 k=1,2

=

f rom 2.12 f0k fk tk+1 =

1 3 t , 3

f rom 2.14

l fk fl f0k+l−1 tk+l+1 , (k + 1)(k + l + 1)

f rom 2.17

[T he term f or l = 1, k = 1] + [T he term f or l = 1, k = 2] + [T he term f or l = 2, k = 1] +

[T he term f or l = 2, k = 2] 1 2 2 1 f1 f1 f0 t3 + f2 f1 f02 t4 + f1 f2 f02 t4 + f2 f2 f03 t5 6 3.4 2.4 3.5 2 5 = t 15 22 7 t, y3 = zeros + f rom 2.18 315 Therefore, if the series converges, then the approximate solution series is given by 2 22 7 1 t + ... y = t + t3 + t5 + 3 15 315 Here we can see that the series based on An converges a little more slowly =

∼

than the series based on An . The solution series using the power series method from 2.21 has the following coefficients a1 = f0 = 1 1 f0 f1 = 0 a2 = 2 1 1 1 [ f0 f12 + f2 f02 ] = a3 = 3 2 3 1 2 f f1 f2 = 0 a4 = 4 0 ... Therefore the series of the solution is 1 y = t + t3 + ... 3 This shows that it is the slowest converging series comparing with that ones obtained by decomposition method. These expansions are exact as one can verify with some effort by expanding the appropriate solution of the differential equation 1.4, namely y = tan t.

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4. References [1] N.H. Abel, Precis d’une theorie des fonctions elliptiques, J. Reine Angew Math., 4, pp. 309-348(1829). [2] G. Adomian, Stochastic Systems, Academic Press, New York, 1983. [3] G. Adomian, The solution of general linear and non-linear stochastic system, in ”Norbert Wiener Memorial Volume,” Modern Trends in Cybernetics and Systems, (J. Rose, Ed.), pp. 160-170, Editura Technica, Bucharest,1976. [4] G. Adomian, Nonlinear stochastic differential equations, J. Math. Anal. Appl. 55, 44-542(1976). [5] G. Adomian, On the modelling and Analysis of nonlinear stochastic systems, in ”proceedings, International Conference on Mathematical modelling” (Avula, Bellman, Luke, and Rigler, Eda.), Vol. 1,pp. 29-40, Univ. of Missouri, Rolla, 1980. [6] G. Adomian, Stochastic systems analysis, in ”Applied Stochastic Processes” (G. Adomian, Ed.), pp. 1-17, Academic Press, New York, 1980. [7] G. Adomian, Solution of nonlinear stochastic physical problems, Rend. Sem. Mat. Univ. Politec. Torino, Numero Speciale, 1-22(1982). [8] G. Adomian, A Review of the Decomposition Method in Applied Mathematics, J. Math. Anal. Appl.135, 501-544(1988). [9] G. Adomian, Solving Frontier Problems of Physics: The Decomposition Method, Kluwer, Dordrecht, 1994. [10] G. Adomian, G.E. Adomian, and R.E. Bellman, Biological system interactions, Proc. Natl. Acad. Sci. USA, Vol. 81, pp. 2938-2940(1984). [11] C.M. Bender, S.A. Orszag, Sec.1.6, in Advanced Mathematical Methods for Scientists and Engineers. New York, Wiley, 1986. Received: January 9, 2008