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AppliedMathematicsLetters22(2009)1248–1251
Contents lists available at ScienceDirect
AppliedMathematicsLetters
journal homepage: www.elsevier.com/locate/aml
Ontheexactsolutionsforinitialvalueproblemsofsecond-order
differential equations
LazharBougoffa
Al-imamUniversity, Faculty of Science, Department of Mathematics, P.O.Box 90950, Riyadh 11623, Saudi Arabia
a r t i c l e i n f o a b s t r a c t
Article history: In this paper, the solutions of initial value problems for a class of second-order linear
Received24April2008 differentialequationsareobtainedintheexactformbywritingtheequationsinthegeneral
Receivedinrevisedform15December2008 operator form and finding an inverse differential operator for this general operator form.
Accepted26January2009 ©2009ElsevierLtd.Allrightsreserved.
Keywords:
Second-orderdifferential equation
Exact solutions
Inverse differential operator
1. Introduction
Theconsiderationofinitialvalueproblemsforsecond-orderordinarydifferentialequationsismotivatedbyanumberof
physical problems in various fields [1,2].
In recent years, the studies of these types of initial value problems have attracted the attention of many mathematicians
andphysicists.Forexample,Adomian’sdecompositionmethod(ADM)[3,4]whichhasbeenappliedtoawideclassofinitial
andboundaryvalueproblemsfordifferentialequations.
The solution proposed by Adomian [3,4] is to take the differential operator L as the highest-ordered derivative of the
linear part. For example, for the linear (deterministic) ordinary differential equation [4]
2
d u p
dx2 −kx u = f(x) with u(−1) = u(1) = 0.
d2 p −1
Adomian rewrites this equation in the operator form L u = F(u), where L = , F(u) = kx u + f and defines L as
R R xx xx dx2 xx
L−1(.) = x x(.)dxdx, and operates with L−1. Therefore
xx 0 0 xx
′ −1 p −1
u = u(0)+xu(0)+L kx u +L (f (x)),
xx xx
∞
andtheADMconsistsofrepresentingthesolutionuinthedecompositionformgivenbyu = P u .
n
n=0
Manysolutionshavebeenobtainedin[5,6]forfurtherspecificsecond-orderordinarydifferentialequations:Lane–Emden
equation,linearsingularinitialvalueproblemandotherequationsbychoosingadifferentinversedifferentialoperatorsand
using ADM.
In the present paper, we give a novel approach for obtaining the exact solutions of the following initial value problem:
d dy dy
dx p(x)dx +q(x)dx +r(x)y = f(x), x > x0, (1.1)
y(x0) = α, y′(x0) = β, (1.2)
wherep(x) ∈ C1([x0,L]),q(x), r(x) and f(x) are some functions.
E-mail address: bougoffa@hotmail.com.
0893-9659/$–seefrontmatter©2009ElsevierLtd.Allrightsreserved.
doi:10.1016/j.aml.2009.01.038
L. Bougoffa / Applied Mathematics Letters 22 (2009) 1248–1251 1249
The method is based on writing Eq. (1.1), under suitable conditions on the coefficients, in the general operator form
d dz −1
Lxxz = g(x), where Lxxz ≡ dx h(x)dx and we propose an inverse differential operator Lxx of Lxx. Therefore, the exact
solutions of the problem (1.1)–(1.2) can be obtained from operating with L−1.
xx
2. Themethod
Thekeyideaofourmethodisasfollows. R
r(x) dx
Multiplying both sides of Eq. (1.1) by ξ (x) = e q(x) , we get
1
d dy dy
ξ (x) p(x) +ξ (x)q(x) +ξ (x)r(x)y = ξ (x)f(x),
1 dx dx 1 dx 1 1
taking into account ξ′(x)q(x) = ξ (x)r(x), we obtain
1 1
d dy dy ′
ξ (x) p(x) +ξ (x)q(x) +ξ (x)q(x)y = ξ (x)f(x), (2.1)
1 dx dx 1 dx 1 1
d dy d
ξ (x) p(x) +q(x) (ξ (x)y) = ξ (x)f(x), (2.2)
1 dx dx dx 1 1
sothat
d p(x)dy+ q(x) d ξ (x)y = f(x). (2.3)
dx dx ξ (x) dx ( 1 )
1
Let ξ (x)y = z, where ξ (x ) and ξ′(x ) are defined.
1 1 0 1 0
Substituting this into Eq. (2.3), we get
d p(x) 1 ′ z + p(x) dz + q(x) dz = f(x). (2.4)
dx ξ (x) ξ (x) dx ξ (x) dx
1 1 1
HenceEq.(2.4)mayberewrittenas
d p(x) dz+ d p(x) 1 ′z+ q(x) dz = f(x). (2.5)
dx ξ (x) dx dx ξ (x) ξ (x) dx
1 1 1
1 ′ −R r(x)dx
If we choose p(x)(ξ (x)) = c, where c is a constant, that is, cq(x) + p(x)r(x)e q(x) =0.Then,withthischoice,Eq.(2.5)
becomes 1
d s(x)dz+t(x)dz = f(x), (2.6)
dx dx dx
wheres(x) = p(x) andt(x) = q(x) +c.
ξ (x) ξ (x)
1 1 R
t(x) dx
Now,asbefore,multiplyingbothsidesofEq.(2.6)byξ (x) = e s(x) ,weget
2
d dz dz
ξ (x) s(x) +ξ (x)t(x) =ξ (x)f(x). (2.7)
2 dx dx 2 dx 2
Takingintoaccountξ′(x)s(x) = ξ (x)t(x), we obtain
2 2
d dz ′ dz
ξ (x) s(x) +ξ (x)s(x) =ξ (x)f(x), (2.8)
2 dx dx 2 dx 2
sothat
d ξ (x)s(x)dz = ξ (x)f(x). (2.9)
dx 2 dx 2
Now,wewriteEq.(2.9)intheform
Lxxz = g(x), (2.10)
whereL z ≡ d h(x)dz,h(x) = ξ (x)s(x)andg(x) = ξ (x)f(x).
xx dx dx 2 2
1250 L. Bougoffa / Applied Mathematics Letters 22 (2009) 1248–1251
Aformalinverseof(2.10)canbeeasilyfound.Wechooseitas
L−1z(x) = Z x dt Z t z(s)ds,
xx h(t)
x x
0 0
whereL−1L 6= L L−1.ApplyingL−1 toEq.(2.10),weseethat
xx xx xx xx xx
−1 Z x dt Z t ′ ′
(Lxx Lxx)z(x) = x h(t) x h(s)z (s) ds,
Z 0 0
x dt
(L−1Lxx)z(x) = h(t)z′(t) − h(x0)z′(x0) ,
xx h(t)
x0
sothat Z
x dt
(L−1L )z(x) = z(x) − z(x ) − h(x )z′(x ) .
xx xx 0 0 0 h(t)
x
0
Therefore, we obtain
z(x) = z(x ) + h(x )z′(x )Z x dt + L−1(g(x)). (2.11)
0 0 0 h(t) xx
x
0
After z has been found the solution of (1.1)–(1.2) is given by y = 1 z.
ξ (x)
Thus, wehaveprovedthefollowingnewtheorem. 1
Theorem1. Forthegiveninitialvalueproblem(1.1)–(1.2).
If there exists a constant c such that
−R r(x)dx
cq(x) + p(x)r(x)e q(x) =0.
Then, the solution is given by
y = 1 z,
ξ (x)
1
where
z(x) = z(x ) + h(x )z′(x )Z x dt + L−1(g(x)),
0 0 0 h(t) xx
x
0
h(x) = ξ (x)s(x), g(x) = ξ (x)f(x),
2 2
R r(x)dx R t(x)dx
ξ (x) = e q(x) , ξ (x) = e s(x) ,
1 2
s(x) = p(x) , t(x) = q(x) +c,
ξ (x) ξ (x)
1 1
z(x ) = αξ (x ), z′(x ) = αξ′(x ) + βξ (x )
0 1 0 0 1 0 1 0
and
L−1g(x) = Z x dt Z t g(s)ds.
xx h(t)
x x
0 0
In the following we shall apply the above techniques to a few various linear differential equations of mathematical physics.
Example1 (DegenerateHypergeometricEquation).Considerthesingularinitialvalueproblem
2
d y + b−xdy − ay = 0, x > 0,
dx2 x dx x
y(0) = 1, y′(0) = −1.
b
Herep(x) = 1,q(x) = b−x,r(x) = b andf(x) = 0.
x x
If we choose a = −1, then, the conditions of Theorem 1 are fulfilled and straightforward computation yields c = 1,
b 2
1 x −x b −x (b−x)
ξ (x) = , ξ (x) = e ,h(x) = x e ,s(x) = b−xandt(x) = +1.
1 b−x 2 b−x x
Bydirectapplication of Theorem1,wegetz = 1 andtheexactsolutiontothisproblemy(x) = 1 z = b−x.
b ξ (x) b
1
L. Bougoffa / Applied Mathematics Letters 22 (2009) 1248–1251 1251
Example2 (EulerEquation).Considertheinitialvalueproblem
2
x2d y +axdy +by = 3x2, x > 1,
dx2 dx
y(1) = 1, y′(1) = 2.
If we choose a = 1 and b = −1, then the conditions of Theorem 1 are fulfilled and direct calculation produces c = 1,
ξ (x) = 1,ξ (x) = x2,h(x) = x3,t(x) = 2,s(x) = x,g(x) = 3x2 and L−1g(x) = 1 + x − 3.
1 x 2 xx 2 2
2x 1 2
Bydirectapplication of Theorem1,wegetz = x.Therefore,theexactsolutiontothisproblemisy(x) = ξ (x)z = x .
1
Example3 (LegendreEquation).Considertheinitialvalueproblem
d2y 2x dy 2 2
dx2 − 1−x2 dx + 1−x2y = 1−x2, x > −1,
y(−1) = 2, y′(−1) = −1.
Theorem1canbeappliedanddirectcalculationproducesc = 1, ξ (x) = 1, ξ (x) = x(1−x2), h(x) = x2(1−x2), t(x) =
1 x 2
−2x +1, s(x) = x, g(x) = 2xandL−1g(x) = 1 +1.Thusz = 1 −1andy(x) = 1−x.
1−x2 xx x x
3. Conclusion
In conclusion, we have successfully found some exact solutions for a second-order ordinary differential equations by
using a direct method. The idea of this method is to change the problem for solving (1.1) to the general operator form
d dz −1
Lxxz ≡ dx h(x)dx in which the inverse differential operator Lxx of Lxx can be found. Therefore, the exact solutions of such
problem(1.1)–(1.2)areobtainedfromoperatingwithL−1.
xx
References
[1] H.T. Davis, Introduction to Nonlinear Differential and Integral Equations, Dover Publications, New York, 1962.
[2] E. Groswald, Bessel Polynomials, Springer, Berlin, 1978.
[3] G. Adomian, Nonlinear Stochastic Operator Equations, Academic Press, Orlando, FL, 1986.
[4] G. Adomian, Solving Frontier Problems of Physics: The Decomposition Method, Kluwer Academic Publishers, Boston, 1994.
[5] M.M.Hosseini,H.Nasabzadeh,ModifiedAdomiandecompositionmethodforspecificsecondorderordinarydifferentialequations,Appl.Math.Comput.
186(2007)117–123.
[6] A.M. Wazwaz,Anewmethodforsolvingsingularinitialvalueproblemsinthesecond-orderordinarydifferentialequations,Appl.Math.Comput.128
(2002)45–57.
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