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Lane - Emden type equation is a nonlinear differential equation appears in many fields such as stellar structure, radioactive cooling and modeling of clusters of galaxies. In this work, this equation is investigated using a semi-analytical method called the Variation of parameters method with an auxiliary parameter. In the applied technique, an unknown auxiliary parameter is inserted in Variation of Parameters Method to solve some special cases of these equations . The used algorithm is easy to implement and very effective. The obtained solutions are also fairly accurate.

Many problems in mathematical physics and astrophysics can be modeled by the so-called Lane-Emden type equation defined in the form

y ″ ( x ) + α x y ′ ( x ) + F ( x , y ) = g ( x ) , α x ≥ 0 , (1)

With the initial conditions:

y ( 0 ) = a , y ′ ( 0 ) = 0 , (2)

where “ ' ” is the differentiation with respect to x, F ( x , y ) is a nonlinear function of x and y, g ( x ) is the non-homogeneous term and α , a are constants. The closed form solution of the Lane-Emden type Equation (1) is always enabled [

Taking α = 2 , g ( x ) = 0 , F ( x , y ) = y k and a = 1 in Equations (1) and (2) respectively, we get

y ″ ( x ) + 2 x y ′ ( x ) + y k = 0 , x ≥ 0. (3)

subject to initial conditions:

y ( 0 ) = 1 , y ′ ( 0 ) = 0. (4)

Equations (3) and (4) are known as the classical Lane-Emden equation.

Choosing α = 2 , g ( x ) = 0 , F ( x , y ) = ( y 2 − C ) 3 / 2 and a = 1 in Equations (1) and (2) respectively, we get the so-called white-dwarf equation that expresses the gravitation potential of the degenerate white-dwarf stars. Isothermal gas spheres [

y ″ ( x ) + 2 x y ′ ( x ) + e y ( x ) = 0 , x ≥ 0 (5)

with the initial conditions:

y ( 0 ) = 0 , y ′ ( 0 ) = 0. (6)

The parameter k in Equation (3) has physical significance in the range 0 ≤ k ≤ 5 so Equation (3) has a closed form solutions for k = 0 , 1 , 5 [

In the present paper, we aim effectively to employ Variation of Parameters Method (VPM) coupling with an unknown auxiliary parameter h to solve the homogeneous as well as non-homogeneous Lane-Emden equation. VPM is free from calculation of the so-called Adomian’s polynomials and is widely applicable because of the reliability of it and the reduction in the size of the computational domain. A comparison with the standard VPM is made to show the reliability and efficiency of the applied algorithm which is simple and is accurately approximates the solution over a large domain.

Variation of Parameters Method (VPM) was first proposed by Ma [

L y ( x ) + R y ( x ) + N y ( x ) + H ( x ) = 0 , x ≥ 0 , (9)

with the initial conditions:

y ( 0 ) = a , y ′ ( 0 ) = 0 , (10)

where is a higher order linear operator that is assumed to be easily invertible, it is equal to ∂ 2 ∂ x 2 for Lane-Emden type Equation, R is a linear operator of order less then N is a nonlinear operator, and H ( x ) is an inhomogeneous term. An unknown auxiliary parameter h can be inserted into Equation (9) so that it can be easily written in the following form

L y ( x ) = L y ( x ) + h ( L y ( x ) + R y ( x ) + N y ( x ) + H ( x ) ) , (11)

or

L y ( x ) = L y ( x ) + h G y ( x ) . (12)

According to the Variation of Parameters Method [

L y ( x ) = 0 is given as

y H ( x ) = c 1 + c 2 x , (13)

For the particular solution, the constants c 1 and c 2 in Equation (13) are replaced by the functions u ( x ) and v ( x ) respectively, thus we have

y p ( x ) = u ( x ) f 1 ( x ) + v ( x ) f 2 ( x ) . (14)

where f 1 ( x ) = 1 and f 2 ( x ) = x . The functions u ( x ) and v ( x ) can be determined as

u ( x ) = ∫ 0 x W 1 W d η and v ( x ) = ∫ 0 x W 1 W d η , (15)

where W is the Wronskian of f 1 ( x ) and f 2 ( x ) i.e.

W = | f 1 f 2 f ′ 1 f ′ 2 | = 1 , (16)

W 1 = | 0 f 2 L y ( x ) + h G y ( x ) f ′ 2 | = − x ( L y ( x ) + h G y ( x ) ) , (17)

and

W 2 = | f 1 0 f ′ 1 L y ( x ) + h G y ( x ) | = L y ( x ) + h G y ( x ) . (18)

Substituting the values of W , W 1 and W 2 in Equation (15), yields

u ( x ) = − ∫ 0 x η ( L y ( η ) + h G y ( η ) ) d η , v ( x ) = ∫ 0 x ( L y ( η ) + h G y ( η ) ) d η . (19)

Hence, the general solution of Equation (12) is

y ( x ) = y H ( x ) + y p ( x ) = c 1 + c 2 x + ∫ 0 x ( x − η ) ( L y ( η ) + h G y ( η ) ) d η . (20)

Applying the initial conditions (10) in Equation (20) and solve for c 1 and c 2 , we get c 1 = a and c 2 = 0 .

Substituting the values of c 1 and c 2 in Equation (20) yields

y ( x ) = a + ∫ 0 x ( x − η ) ( L y ( η ) + h ( L y ( x ) + R y ( x ) + N y ( x ) + H ( x ) ) ) d η . (21)

which can be solved iteratively as

{ y o ( x ) = a y 1 ( x , h ) = y o ( x ) + ∫ 0 x ( x − η ) ( L y 0 ( η ) + h ( L y 0 ( η ) + R y 0 ( η ) + N y 0 ( η ) + H ( η ) ) ) d η , y n + 1 ( x , h ) = y o ( x ) + ∫ 0 x ( x − η ) ( L y n ( η , h ) + h ( L y n ( η , h ) + R y n ( η , h ) + N y n ( η , h ) + H ( η ) ) d η , for n ≥ 1. (22)

Consequently, an exact solution can be obtained when approaches to infinity:

y ( x , h ) = lim n → ∞ y n ( x , h ) . (23)

The valid values of h can be determined by means of the so-called h-curves [

{ y o ( x ) = a , y n + 1 ( x ) = y o ( x ) + ∫ 0 x ( x − η ) ( − R y n ( η ) − N y n ( η ) − H ( η ) ) d η , n ≥ 0. (24)

Now we apply Variation of Parameters Method (VPM) coupled with an unknown auxiliary parameter developed in Section 2 for solving Equation (1). The recursive formulas (22) and (24) for this problem is obtained respectively as

y n + 1 ( x , h ) = y 0 ( x ) + ∫ 0 x ( x − η ) ( ∂ 2 y n ( η , h ) ∂ η 2 + h ( ∂ 2 y n ( η , h ) ∂ η 2 + α η ( ∂ y n ( η , h ) ∂ η ) + f ( η , y n ( η , h ) ) − g ( η ) ) ) d η .

then

y n + 1 ( x , h ) = y 0 ( x ) + ∫ 0 x ( x − η ) ( ( − α η ( ∂ y n ( η , h ) ∂ η ) − f ( η , y n ( η , h ) ) + g ( η ) ) ) d η . (25)

Example 1

Consider the following form of Lane-Emden equation [

y ″ ( x ) + 2 x y ′ ( x ) + y k ( x ) = 0 , x ≥ 0. (26)

where k is a constant. With the initial conditions:

y ( 0 ) = 1 , y ′ ( 0 ) = 0. (27)

Note that the exact solutions for cases of k = 0 , 1 , 5 are known, VPM developed in Section 2 is applied to solve Equations (26) and (27) for some cases of the constant k.

Case (1) for (k = 0):

The closed form solution is: y ( x ) = 1 − x 2 6 . According to standard VPM, we have the following from iterative scheme (24):

{ y 0 ( x ) = y ( 0 ) = 1 , y n + 1 ( x , h ) = 1 + ∫ 0 x ( x − η ) ( ( − 2 η ( ∂ y n ( η , h ) ∂ η ) − 1 ) ) d η . (28)

The result obtained by standard VPM is not valid for large values of x. Now, using the iterative scheme (22) we have:

y 0 ( x ) = y ( 0 ) = 1 ,

y 1 ( x , h ) = 1 + ∫ 0 x ( x − η ) ( ∂ 2 y 0 ( η ) ∂ η 2 + h ( ∂ 2 y 0 ( η ) ∂ η 2 + 2 η ( ∂ y 0 ( η ) ∂ η ) + 1 ) ) d η .

Generally,

y n + 1 ( x , h ) = 1 + ∫ 0 x ( x − η ) ( ∂ 2 y n ( η , h ) ∂ η 2 + h ( ∂ 2 y n ( η , h ) ∂ η 2 + 2 η ( ∂ y n ( η , h ) ∂ η ) + 1 ) ) d η , n ≥ 1. (29)

First few terms of solution are

y 0 ( x ) = y ( 0 ) = 1 , y 1 ( x , h ) = 1 + h x 2 2 , y 2 ( x , h ) = 1 + ( h + 3 h 2 2 ) x 2 , ⋮

It can be seen in

Case (2) for (k = 1):

The closed form solution is y ( x ) = sin x x . According to standard VPM, the absolute error of y 30 ( x ) is shown in

y 0 ( x ) = 1 , y 1 ( x , h ) = 1 + h x 2 2 , y 2 ( x , h ) = 1 + ( h + 3 h 2 2 ) x 2 + h 2 x 4 24 ,

Other terms of the solution can also be obtained in a similar way.

From

y ( x ) = lim n → ∞ y n ( x , h ) ≈ sin x x . (30)

An approximate solution (not exact) was obtained, using VIM, by [

Case (3) for (k = 2):

In this case we analyze the dependence of the convergence regions on the value of a , The h curve for the ninth-order approximation at x = 1 is shown in

Case (4) for (k = 3, k = 4):

For these cases, a comparison is made through

Case (5) for (k = 5):

The closed form solution is y ( x ) = ( 1 + x 2 3 ) − 1 / 2 . Adopting the similar

x | Exact value [ | y 7 ( x ) | Error E 7 | Error by [ |
---|---|---|---|---|

0.0 | 1.0000000 | 1.0000000 | 0.00e+00 | 0.00e+00 |

0.1 | 0.9983358 | 0.9983358 | 1.00e−08 | 1.40e−06 |

0.5 | 0.9598391 | 0.959838 | 1.27e−06 | 2.99e−06 |

1.0 | 0.8550576 | 0.8550575 | 1.29e−07 | 1.99e−06 |

x | Exact value [ | y 7 ( x ) | Error E 7 | Error by [ |
---|---|---|---|---|

0.0 | 1.0000000 | 1.0000000 | 0.00e+00 | 0.00e+00 |

0.1 | 0.9983367 | 0.9983365 | 1.95e−07 | 2.51e−04 |

0.2 | 0.9933862 | 0.9933862 | 9.99e−08 | 2.48e−04 |

0.5 | 0.9603109 | 0.960307 | 3.49e−06 | 2.05e−04 |

1.0 | 0.8608138 | 0.860813 | 1.09e−06 | 1.93e−04 |

x | Exact value [ | Standard VPM | VPM with h |
---|---|---|---|

0.1 | 0.9983367 | 1.1052640 | 0.9983365 (h = −0.3) |

0.2 | 0.9933862 | 1.4239908 | 0.9933862 (h = −0.39) |

0.5 | 0.9603109 | 3.744834 | 0.960307 (h = −0.39) |

1 | 0.8608138 | 9.334329 | 0.860813 (h = −0.36) |

x | Exact value [ | Standard VPM | VPM with h |
---|---|---|---|

0.1 | 0.9983358 | 0.7845924 | 0.9983358 (h = −0.3) |

0.5 | 0.9598391 | −4.8786617 | 0.959838 (h = −0.35) |

1 | 0.8550576 | 66.787236 | 0.8550575 (h = −0.355) |

procedure as in the previous examples,

y 0 ( x ) = 1 , y 1 ( x , h ) = 1 + h x 2 2 , y 2 ( x , h ) = 1 + ( h + 3 h 2 2 ) x 2 + 5 h 2 x 4 24 + h 3 x 6 12 + 5 h 4 x 8 224 + h 5 x 10 288 + h 6 x 12 4224 ,

Other terms of the solution can also be obtained in a similar way, h can be chosen in the range − 0.28 ≤ h ≤ − 0.42 , from

Example 2

For the Isothermal gas spheres equation:

y ″ + 2 x y ′ + e y = 0 , x ≥ 0 , (31)

with the initial conditions ( y ( 0 ) = 0 , y ′ ( 0 ) = 0 ), we can use Taylor series expansion of e y

e y ≅ 1 + y + y 2 2 + y 3 6 + y 4 24 .

y 0 ( x ) = y ( 0 ) = 0 , (32)

y 1 ( x , h ) = h x 2 2 , (33)

y 2 ( x , h ) = ( h + 3 h 2 2 ) x 2 + h 2 x 4 24 + h 3 x 6 240 + h 4 x 8 2688 + h 5 x 10 34560 (34)

Other terms of the solution can also be obtained in a similar way to get:

y ( x ) ≅ − 1 6 x 2 + 1 120 x 4 − 1 1890 x 6 + ⋯

(with h equal to −0.395) which is the same as the solution obtained by [

x | [ | Present method y 6 ( x ) | Error E 6 of present method | Error [ |
---|---|---|---|---|

0.0 | 0.0000000000 | 0.0000000000 | 0.00e+00 | 0.00e+00 |

0.1 | −0.0016658339 | −0.0016658368 | 2.90e−09 | 5.85e−07 |

0.2 | −0.0066533671 | −0.0066534486 | 8.15e−08 | 6.04e−07 |

0.5 | −0.0411539568 | −0.0411526052 | 1.35e−06 | 5.58e−07 |

1.0 | −0.1588273537 | −0.1588280814 | 7.27e−07 | 8.20e−07 |

1.5 | −0.3380131103 | −0.3380148407 | 1.73e−06 | 6.72e−06 |

2 | −0.5599626601 | −0.5599574276 | 5.23e−06 | 1.39e−04 |

2.5 | −0.8100196713 | −0.8104997145 | 4.80e−04 | 3.68e−03 |

x | Wazwaz [ | Standard VPM | VPM with h |
---|---|---|---|

0.1 | −0.0016658339 | 0.1050668 | −0.0016658368 (h = −0.37) |

0.2 | −0.0066533671 | 0.4210945 | −0.0066534486 (h = −0.37) |

0.5 | −0.0411539568 | 2.6614811 | −0.0411526052 (h = −0.4) |

1.0 | −0.1588273537 | 10.748351 | −0.1588280814 (h = −0.4) |

1.5 | −0.3380131103 | −4077070.9 | −0.3380148407 (h = −0.394) |

2.0 | −0.5599626601 | −8413.665 | −0.5599574276 (h = −0.35) |

2.5 | −0.8100196713 | 1.561689 × 10^{70} | −0.8104997145 (h = −0.28) |

Example 3

Consider the following problem [

y ″ ( x ) + 2 x y ′ ( x ) − 2 ( 2 x 2 + 3 ) y ( x ) = 0 , (35)

With the initial conditions

y ( 0 ) = 1 , y ′ ( 0 ) = 0. (36)

having y ( x ) = e x 2 as exact solution. Now, using the iterative scheme (22) we have

y 0 ( x ) = y ( 0 ) = 1 , y 1 ( x , h ) = 1 − 3 h x 2 − h x 4 3 , y 2 ( x , h ) = 1 − 6 h x 2 − 9 h x 2 − 2 h x 4 3 + 17 h 2 x 4 18 + 7 h 2 x 6 5 + h 2 x 8 42

Hence, y ( x ) ≅ e x 2 which is the exact solution of Equation (45) with h = − 0.4 as the range of h is − 0.2 ≤ h ≤ − 0.6 as shown in

In the next examples, the nonhomogeneous Lane-Emden equations are considered.

Example 4

y ″ ( x ) + 2 x y ′ ( x ) + y ( x ) = 6 + 12 x + x 2 + x 3 (37)

With the initial conditions

y ( 0 ) = 0 , y ′ ( 0 ) = 0 , (38)

Which has the following exact solution: y ( x ) = x 2 + x 3 . Now using the iterative scheme (22) we have:

x | Error (Present method) | Error (Parand et al. [ |
---|---|---|

0.01 | 2.32e−14 (h = −0.4) | 2.24e−08 |

0.02 | 9.29e−14 (h = −0.4) | 1.58e−08 |

0.05 | 5.77e−13 (h = −0.4) | 2.12e−08 |

0.1 | 2.20e−12 (h = −0.4) | 1.78e−08 |

0.2 | 3.30e−12 (h = −0.4) | 2.09e−08 |

0.5 | 3.15e−09 (h = −0.4) | 2.62e−08 |

0.7 | 2.07e−09 (h = −0.38) | 3.27e−08 |

0.8 | 5.00e−09 (h = −0.39) | 3.79e−08 |

0.9 | 4.95e−09 (h = −0.399) | 5.48e−08 |

1 | 2.59e−09 (h = −0.399) | 2.51e−09 |

y 0 ( x ) = y ( 0 ) = 0 , y 1 ( x , h ) = − 3 h x 2 − 2 h x 3 − h x 4 12 − h x 5 20 , y 2 ( x , h ) = − 6 h x 2 − 9 h 2 x 2 − 4 h x 3 − 4 h 2 x 3 − h x 4 6 − 7 h 2 x 4 18 − h x 5 10 − 7 h 2 x 5 40 − h 2 x 6 360 − h 2 x 7 840 ,

After number of iterations we get an approximate solution for h in the range of − 0.2 ≤ h ≤ − 0.5 at which y ( x ) ≅ x 2 + x 3 by choosing an auxiliary parameter h = − 0.4 .

Example 5

Consider the equation:

y ″ ( x ) + 8 x y ′ ( x ) + x y ( x ) = x 5 − x 4 + 44 x 2 − 30 x , (39)

with the initial conditions

y ( 0 ) = 0 , y ′ ( 0 ) = 0. (40)

Having y ( x ) = x 4 − x 3 as exact solution, using the iterative scheme (18) we have:

y 0 ( x ) = y ( 0 ) = 0 , y 1 ( x , h ) = 5 h x 3 − 11 h x 4 3 + h x 6 30 − h x 7 42 , y 2 ( x , h ) = 10 h x 3 + 25 h 2 x 3 − 22 h x 4 3 − 121 h 2 x 4 9 + h x 6 15 + 19 h 2 x 6 75 − h x 7 21 − h 2 x 7 7 + h 2 x 9 2160 − h 2 x 10 3780 ,

The exact solution ( x 4 − x 3 ) can be obtained when n approaches to infinity by choosing a suitable value for h in the range − 0.3 ≤ h ≤ − 0.12 which can be obtained from the h curve at x = 1 shown in

x | Error (Present method) | Error [ |
---|---|---|

0.01 | 1.00e−10 | 1.47e−06 |

0.10 | 0.00e+00 | 1.82e−06 |

0.50 | 8.00e−10 | 1.41e−06 |

1.00 | 1.50e−09 | 1.25e−06 |

2.00 | 3.51e−08 | 6.93e−07 |

In this article, Variation of Parameter Method is used coupled with an auxiliary parameter for the Lane-Emden equation. Moreover, the iterative scheme (22) reduces to standard VPM for h = − 1 . The numerical results obtained in this research are very effective, completely reliable and powerful in obtaining analytical solutions of many problems. The approach used in the solution provides a high degree of accuracy only with a few iterations and reduces the size of calculations without using any restrictive assumptions.

The authors declare no conflicts of interest regarding the publication of this paper.

Khalifa, A.S. and Hassan, H.N. (2019) Approximate Solution of Lane-Emden Type Equations Using Variation of Parameters Method with an Auxiliary Parameter. Journal of Applied Mathematics and Physics, 7, 921-941. https://doi.org/10.4236/jamp.2019.74062