Abstract:
The solution of the Lane-Emden equation is very important to understand the interior
stellar structure, and is of great importance in Mathematics since the equation of
index grater than one represents one class of non-linear differential equations. The
complete solution of the differential equation can be expressed as an infinite Taylor
series of even powers under the boundary conditions to be imposed at the center of
the star.
Department of Mathematics of University of Kelaniya has found an algorithm [I]
which can be used to obtain successive coefficients of the Taylor series and in this
paper the proof of this algorithm will be given.
The Lane – Emden equation of index m [II] has the following form.
0
2
2
2
+ + =
m
y
dx
dy
dx x
d y
(1)
where m is a parameter.
The boundary conditions are
y(0) = 1,
0
1
)0(
=
=
x
dx
dy
y = 0.
Lane- Emden equation is unchanged when –x is substituted for x, the Taylor’s
expansion contains only even powers of x and the derivatives of odd order evaluated
at x = 0 are all zero.
∑
∞
=0
!
)0(
n
n
n
n
y x
= + + + + ... =
6!
)0(
4!
)0(
2!
)0(
)0(
6
6
4
4
2
2
y x y x y x
y ∑
∞
=0
2
2
2( )!
)0(
n
n
n
n
y x
71
By using Leibniz’s formula, and the boundary conditions of (1) we can obtain the
following recurrence relation
( )n
m
n
y
n
n
y
+
+
+ = −
3
1
2
Here
n
n
n
dx
d y
y = (0) and ( ) ).0(
( )
n
n m
n
m
dx
d y
y =
Taylor’s expansion ∑
∞
=0
!
)0(
n
n
n
n
y x
about x = 0 will produce the solution. But
computational difficulties arise in evaluating (y
m
)n .
Therefore we use the following algorithm.
( ) r
r s s s
r
r
m
n
m N(S ,S , ,S )y y y
S !S ! S !
n!
y C ... ...
...
1 2 1 2
1 2
= ∑ ,
where ( , ,........., ) N S1 S2 Sr
is the number of possible arrangements of S S Sr
, ,..., 1 2
such that 1 2 1 1 S + S + ...+ Sr = n , Sr ≥ Sr− ≥ ... ≥ S . The summation is over all the
distinct partitions of n.
The proof of this algorithm is presented in this paper.
The Proof of the Algorithm
It can be shown that 0 yn
)0( = for odd n. Now
( ) r n r s s s
m
y C(S , S ,..., S ) y y ...y 2 ∑ 1 2 1 2
=
Here ( , ,..., ) C S1 S2 Sr
is the coefficient of
r
s s s
y y ...y
1 2
.
If ( ) n ( ) ( ) ( ) m n
m
y y y y 2 1 2 2 = ( ... ) is labeling each y in y
m
and
y DD D y D y
r
r
S
s = ( ... ) =
Now consider the term
1 2 .
....
r
S S S
y y y in ( ) n
m
y 2
.
( ) ( ( ) )
1 1
S 1 S
y = y can be obtained in
1
2
S
nC ways
( ) ( ( ) )
2 2
S 2 S
y = y can be obtained in
2
2 1
S
n S C
−
ways
.
.
.
72
( ) ( ( ) )
r r
S r S
y = y can be obtained in
r
r
S
n S S S 1 2 1C
2 ... − − − − −
ways.
Therefore
1 2 .
...
r
S S S
y y y reduces to
( )
! !... !
2 !
S1 S2 Sr
n
According to the term
1 2 .
...
r
S S S
y y y [= ( ) ( ) ( ) r
S S r S
( y ) ( y ) ...( y )
1 2
1 2
], S S Sr
, ,..., 1 2
can be changed among themselves in N ways, where N is the number of
possible arrangement of S S Sr
, ,..., 1 2
and ... . Sr ≥ Sr−1 ≥ ≥ S1
Therefore
r
C S S S C
m
( 1
, 2
,..., r
)=
( )
! !.... !
2 !
S1 S2 Sr
n
N.
Hence ( )
( )
( , ,... ) .
! !... !
2 !
1 2 ...
1 2
2 S1 S2 Sr
r
r
r
m
n
m N S S S y y y
S S S
n
y = ∑ C
Now we replace 2n by n, to get
( ) S S Sr
r
r
r
m
n
m N S S S y y y
S S S
n
y C 1 2 ...
1 2
1 2
( , ,... )
! !... !
!
= ∑ ,
where
1 2 .
, ,...,
r
S S S
y y y denote , , ,
2 4 6
y y y and so on, ( , ,... ) N S1 S2 Sr denotes the
number of possible arrangement of S S Sr
, ,..., 1 2
, S1 + S2 + ...+ Sr = n and
... . Sr ≥ Sr−1 ≥ ≥ S1
Though the proof applies to positive integers m, the solution obtained by use
of the algorithm holds true for fractional values also.