Lane_emden_equation_solved_by_HPM_final

1. SOLUTION OF THE LANE-
EMDEN EQUATION BY
HOMOTOPY
PERTURBATION METHOD

2. Submitted by-
Soumya Das, Rinku Alam, Aparna Purkait & Murshiul Habib Khandakar
Supervised By-
Dr. Banashree Sen
in partial fulfillment for the award of the degree
of
Master of Science , Applied Mathematics
Academic year : 2022 - 23
Department of Applied Mathematics ,
School of Applied Science and Technology,
Maulana Abul Kalam Azad University of Technology
(formerly WBUT)
Haringhata, Dist- Nadia, West Bengal, India, PIN-
741239

3. CONTENT
 Homotopy Perturbation Methods
 Lane Emden Equation
 Dimensionless Lane Emden Equation
Solve by Homotopy Perturbation Methods
 Conclusion

4. Overview
of
the
project
Introduction
Homotopy Perturbation
 Lane Emden Equation
Outcome Accuracy of HPM Solution

5. X, Y be topological spaces, and f, g : X → Y
continuous maps. A Homotopy from f to g is
a continuous function F : X × [0, 1] → Y
satisfying
F(x, 0) = f(x) and F(x, 1) = g(x), for all x ∈ X.
The two dashed paths shown above
are homotopic relative to their endpoints. The
animation represents one possible homotopy
HOMOTOPY

6. WHY WE PREFER HOMOTOPY
PERTURBATION METHOD ?
 This is relatively a new technic and easy to
handle for solving linear and non-linear partial
differential equation.
 It is simple method compare to another iterative
method.
 For solving this method we get nearest value of
exact solution than another method.
 Error is more less than other method.

7. The Homotopy Perturbation Method:
we consider the following equation
𝐴 𝑢 − 𝑓 𝑟 = 0, 𝑟 ∈ Ω ………………………………….(Equation.1)
with the boundary condition:
𝐵 𝑢,
𝜕𝑢
𝜕𝑛
= 0 , 𝑟∈ Γ,
A = general differential operator,
B =boundary operator,
𝑓 𝑟 = analytical function
G = boundary of the domain Ω.
A = L + N, L is linear and N is nonlinear.
𝐿 𝑢 + 𝑁 𝑢 − 𝑓 𝑟 = 0, 𝑟 ∈ Ω ……………………(Equation.2)

8. Construct a Homotopy
𝐻 𝑣, 𝑝 = 1 − 𝑝 𝐿 𝑣 − 𝐿 𝑢0 + 𝑝 𝐴 𝑣 − 𝑓 𝑟 = 0,where𝑣 ∶ Ω × 0, 1 → ℝ
𝑝 ∈ [0, 1] = embedding parameter
𝑢0 = first approximation , satisfies the boundary conditions.
Satisfies
𝐻 𝑣, 0 = 𝐿 𝑣 − 𝐿 𝑢0 = 0
𝐻 𝑣, 1 = 𝐿 𝑢 + 𝑁 𝑢 − 𝑓 𝑟 = 0
written as a power series in p, as following:
𝑣 = 𝑣0 + 𝑝𝑣1 + 𝑝2 𝑣2 + ⋯
If set 𝑝 = 1
The best approximation is:
𝑢 = lim
𝑝→1
𝑣 = 𝑣0 + 𝑣1 + 𝑣2 + …

9. What is Lane Emden Equation?
The Lane-Emden equations were first developed by the two
astrophysicists Jonathan Homer Lane and Robert Emden .
In mathematics, the Lane – Emden Equation is a second order
singular ordinary differential equation.
In astrophysics, the Lane-Emden equation is essentially a
dimensionless form of Poisson’s equation for the gravitational potential
of self-gravitating and spherically symmetric polytropic fluid.
A polytrope refers to a solution of the Lane – Emden Equation in which
the pressure depends upon the density.

10. For a polytrope we assumes that 𝑃 = 𝑘𝜌𝛾 = 𝑘𝜌(𝑛+1)/𝑛
Where 𝑃 = 𝑔𝑎𝑠 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦,
𝑘 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑓 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑡𝑦, 𝛾 =
𝑛+1
𝑛
, 𝑛 = 𝑝𝑜𝑙𝑦𝑡𝑟𝑜𝑝𝑖𝑐 𝑖𝑛𝑑𝑒𝑥,
𝛾 = 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑖𝑛𝑑𝑒𝑥( a parameter characterizing the behavior of the
specific heat of a gas).
1
𝜉2
𝑑
𝑑𝜉
𝜉2 𝑑𝜃
𝑑𝜉
+ 𝜃𝑛 = 0
The resulting equation is the called the Lane-Emden equation .
Where 𝜉 is a dimensionless radius and 𝜃 is related to the density .The
index 𝑛 is the polytropic index that appears in the polytropic equation of
state.

11. Solution of Lane Emden Equation
For a polytrope 𝑃 = 𝑘𝜌𝛾
………………………… (i)
Where 𝑃 = 𝑔𝑎𝑠 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑘 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑜𝑓 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑡𝑦, 𝛾 =
𝑛+1
𝑛
, 𝑛 =
𝑝𝑜𝑙𝑦𝑡𝑟𝑜𝑝𝑖𝑐 𝑖𝑛𝑑𝑒𝑥, 𝛾 = 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑖𝑛𝑑𝑒𝑥.
Consider The mass continuity equation :
𝑑𝑀
𝑑𝑟
= 4𝜋𝑟2
𝜌(𝑟) ……………………………….. (ii)
Polytrope must be used in hydrostatic equilibrium so,The equation of hydrostatic equilibrium is :
1
𝜌(𝑟)
𝑑𝑃
𝑑𝑟
= −
𝐺𝑀
𝑟2 ……………………………………. (iii)
Differentiating equation (iii) with respect to r :
𝑑
𝑑𝑟
1
𝜌
𝑑𝑃
𝑑𝑟
=
2𝐺𝑀
𝑟3 −
𝐺
𝑟2
𝑑𝑀
𝑑𝑟
Or,
𝑑
𝑑𝑟
1
𝜌
𝑑𝑃
𝑑𝑟
= −
2
𝜌𝑟
𝑑𝑃
𝑑𝑟
− 4𝜋𝐺𝜌 [From equation (ii) and (iii)]

12. Or, 𝑟2 𝑑
𝑑𝑟
1
𝜌
𝑑𝑃
𝑑𝑟
= −
2𝑟
𝜌
𝑑𝑃
𝑑𝑟
− 4𝜋𝐺𝑟2𝜌
Or,
𝑑
𝑑𝑟
𝑟2
𝜌
𝑑𝑃
𝑑𝑟
= −4𝜋𝐺𝑟2𝜌 ……………………………………. (iv)
Now we divide both sides of the equation (iv) by 𝑟2
.
Put, 𝑃 = 𝑘𝜌
𝑛+1
𝑛 and 𝜌 = 𝜌𝑐𝜃𝑛,where 𝜌𝑐 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑎𝑡 𝑐𝑒𝑛𝑡𝑒𝑟.
We get :
1
𝑟2
𝑑
𝑑𝑟
𝑟2
𝜌𝑐𝜃𝑛 𝑘(𝑛 + 1)𝜌𝑐
(1+
1
𝑛
) 𝑑𝜃𝑛+1
𝑑𝑟
= −4𝜋𝐺𝜌𝑐𝜃𝑛
Or, 𝜌𝑐
1
𝑛𝑘 𝑛 + 1
1
𝑟2
𝑑
𝑑𝑟
𝑟2 𝑑𝜃
𝑑𝑟
= −4𝜋𝐺𝜌𝑐𝜃𝑛
Or,
𝜌𝑐
(
1
𝑛
−1)
𝑘 𝑛+1
4𝜋𝐺
1
𝑟2
𝑑
𝑑𝑟
𝑟2 𝑑𝜃
𝑑𝑟
= −𝜃𝑛
………………………………. (v)
Side calculation :
𝑃 = 𝑘𝜌1+
1
𝑛
Or, 𝑃 = 𝑘(𝜌𝑐𝜃𝑛
)(1+
1
𝑛
)
Or, 𝑃 = 𝑘𝜌𝑐
(1+
1
𝑛
)
𝜃𝑛+1
Or,
𝑑𝑃
𝑑𝑟
= 𝑘(𝑛 + 1)𝜌𝑐
(1+
1
𝑛
) 𝑑𝜃𝑛+1
𝑑𝑟

13. Let,
𝜌𝑐
1
𝑛−1
𝑘 𝑛+1
4𝜋𝐺
= 𝛼2
Or,
𝛼2
𝑟2
𝑑
𝑑𝑟
𝑟2 𝑑𝜃
𝑑𝑟
= −𝜃𝑛
Let, 𝑟 = 𝛼𝜉
Or,
1
𝜉2
𝑑
𝑑𝜉
𝜉2 𝑑𝜃
𝑑𝜉
= −𝜃𝑛
Or,
1
𝜉2
𝑑
𝑑𝜉
𝜉2 𝑑𝜃
𝑑𝜉
+ 𝜃𝑛
= 0 ……………………………... (vi)
For n=0
If n=0, then the equation becomes
1
ξ2
d
dξ
ξ2 dθ
dξ
+ 1 = 0
Re-arranging and integrating once gives
ξ2 dθ
dξ
= c1-
1
3
ξ3
dividing both sides by ξ2
and integrating again gives
ϴ(x)= c0-
c1
ξ
-
1
6
ξ2

14. The boundary conditions ϴ(0)=1 and θ′(0) =0 imply that the constants of integration are
c0 = 1 and c1 = 0
Therefore ϴ(ξ)= 1-
1
6
ξ2
For n = 1
If n=1, then the equation become
1
ξ2
d
dξ
ξ2 dθ
dξ
+ θ =0,
d
dξ
ξ2 dθ
dξ
+ θξ2
= 0,
which is the spherical Bessel differential equatin
d
dr
r2 dR
dr
+ k2r2 − n n + 1 R = 0,
with k=1 and n=0, so the solution is,
ϴ(ξ) = Aj0(ξ)+Bn0(ξ)
Applying the boundary condition ϴ(0)=1 gives
ϴ(ξ) = j0(ξ) =
sinξ
ξ
where j0(ξ) is a spherical Bessel function of the first kind

15. For, n = 2,constructing Homotopy structure
𝐿 𝜃 =
𝑑2
𝜃
𝑑𝜉2
+
2
𝜉
𝑑𝜃
𝑑𝜉
𝑁 𝜃 = 𝜃2
𝑑2
𝜃
𝑑𝜉2
+
2
𝜉
𝑑𝜃
𝑑𝜉
1 + 𝜌 𝜃2
− 1 = 0
𝜃 = 𝜃0 + 𝑃𝜃1 + 𝑃2
𝜃2 + 𝑃3
𝜃3 + ⋯
Substituting and equating the power of the 𝑝, we get-
𝑃(0)
:
𝑑2
𝜃0
𝑑𝜉2
+
2
𝜉
𝑑𝜃0
𝑑𝜉
+ 1 = 0
𝑃(1)
:
𝑑2
𝜃1
𝑑𝜉2
+
2
𝜉
𝑑𝜃1
𝑑𝜉
+ 𝜃0
2
− 1 = 0
𝑃(2)
:
𝑑2
𝜃2
𝑑𝜉2
+
2
𝜉
𝑑𝜃2
𝑑𝜉
+ 𝜃1
2
+ 2𝜃 0𝜃1 = 0
Solution gives as
𝜃0 = 1 −
𝜉2
6
n = 2, 𝜃0 plot

16. 𝜃1 = 1 +
𝜉4
60
−
𝜉6
1512
𝜃2 = −
𝜉2
3
+
𝜉4
60
+
𝜉6
1260
+
𝜉8
1512 ∗ 36
+
𝜉8
360 ∗ 36
−
𝜉10
9090 ∗ 55

17. For n = 3,
𝐿 𝜃 =
𝑑2𝜃
𝑑𝜉2 +
2
𝜉
𝑑𝜃
𝑑𝜉
𝑁 𝜃 = 𝜃3
𝐿 𝜃 + 1 + 𝜌 𝑁 𝜃 − 1 = 0
𝑑2𝜃
𝑑𝜉2 +
2
𝜉
𝑑𝜃
𝑑𝜉
1 + 𝜌 𝜃3
− 1 = 0
𝜃 = 𝜃0 + 𝑃𝜃1 + 𝑃2
𝜃2 + 𝑃3
𝜃3 + ⋯
Substituting and equating the power of the 𝑝 , we get-
𝑃(0)
:
𝑑2
𝜃0
𝑑𝜉2 +
2
𝜉
𝑑𝜃0
𝑑𝜉
+ 1 = 0
𝑃(1):
𝑑2
𝜃1
𝑑𝜉2 +
2
𝜉
𝑑𝜃1
𝑑𝜉
+ 𝜃3 − 1 = 0
𝑃(2)
:
𝑑2𝜃2
𝑑𝜉2
+
2
𝜉
𝑑𝜃2
𝑑𝜉
+ 3𝜃0
2
𝜃1 = 0

18. For 𝑃(0),
𝑑2
𝜃0
𝑑𝜉2
+
2
𝜉
𝑑𝜃0
𝑑𝜉
+ 1 = 0
Gives solution as
𝜃0 = 1 −
𝜉2
6
n = 3, 𝜃0 plot

19. For 𝑃(1)
,
𝑑2
𝜃1
𝑑𝜉2
+
2
𝜉
𝑑𝜃1
𝑑𝜉
+ (1 −
𝜉2
6
)3
−1 = 0
Let,
𝑑𝜃1
𝑑𝜉
= 𝑃2
𝑑𝑃2
𝑑𝜉
+
2
𝜉
𝑃2 + (1 −
𝜉2
6
)3
−1 = 0
𝐼. 𝐹 = 𝑒
2
𝜉
𝑑𝜉
= 𝜉2
Multiplying both side with I.F, we get-
𝜉2
𝑃2 = 2.
𝜉3
3
−
𝜉5
2.5
+
𝜉7
12.7
−
𝜉9
216.9
+ 𝐶3
For, 𝜉 = 0, 𝐶3 = 0
𝜃1 =
𝜉2
3
−
𝜉4
2.4.5
+
𝜉6
6.7.12
−
𝜉8
8.9.216
+ 𝐶4
For, 𝜃0 = 1, 𝜉 = 0, 𝐶4 = 1
𝜃1 = 1 +
𝜉2
3
−
𝜉4
2.4.5
+
𝜉6
6.7.12
−
𝜉8
8.9.216

20. For n= 4, Equation reduces to
1
𝜉2
𝑑
𝑑𝜉
𝜉2
𝑑𝜃
𝑑𝜉
= 𝜃4
Or,
𝑑2𝜃
𝑑𝜉2 +
2
𝜉
𝑑𝜃
𝑑𝜉
= 𝜃4
,here linear 𝐿 𝜃 =
𝑑2𝜃
𝑑𝜉2 +
2
𝜉
𝑑𝜃
𝑑𝜉
and non- linera𝑁 𝜃 = 𝜃4
Using homotopy structure,
H(𝜃, 𝑃) = 1 − 𝑃 𝐿 𝜃 + 1 + 𝑃 𝐿 𝜃 + 𝑁 𝜃 = 0
Or, (1-P) + [L 𝜃 − PL 𝜃 ] + PL 𝜃 + 𝑃𝑁 𝜃 = 0
Or, L 𝜃 + 1+P(N 𝜃 − 1) = 0
Or, ,
𝑑2𝜃
𝑑𝜉2 +
2
𝜉
𝑑𝜃
𝑑𝜉
+1+P(𝜃4 − 1) = 0
𝜃 = 𝜃0 + 𝜌𝜃1 + 𝜌2
𝜃2 + 𝜌3
𝜃3 + ⋯
Substituting and equating the power of the 𝑃, we get-
𝑃(0):
𝑑2
𝜃0
𝑑𝜉2
+
2
𝜉
𝑑𝜃0
𝑑𝜉
+ 1 = 0
𝑃(1):
𝑑2𝜃1
𝑑𝜉2
+
2
𝜉
𝑑𝜃1
𝑑𝜉
+ 𝜃4 − 1 = 0

21. 𝑃(2):
𝑑2
𝜃2
𝑑𝜉2
+
2
𝜉
𝑑𝜃2
𝑑𝜉
+ 4𝜃0
2
𝜃1 = 0
𝑃(3):
𝑑2𝜃2
𝑑𝜉2
+
2
𝜉
𝑑𝜃2
𝑑𝜉
+ 6𝜃0
2
𝜃1
2
= 0
……………………..(1)
From equation (1), for 𝑃(0)
𝜃0 = 1 −
𝜉2
6
For 𝑃(1)
For, 𝜉 = 0, 𝐶3 = 0
𝜃1 =
𝜉4
30
−
𝜉6
252
+
𝜉8
388
−
𝜉10
142560
+ 𝐶4
For, 𝜃0 = 1, 𝜉 = 0, 𝐶4 = 1
𝜃1 = 1 +
𝜉4
30
−
𝜉6
252
+
𝜉8
388
−
𝜉10
142560

22. 𝐹𝑜𝑟 𝑛 = 5 𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑎𝑦 𝐼𝑓 𝑊𝑒 𝑐𝑎𝑛 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑒 𝑡ℎ𝑖𝑠 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑤𝑒 𝑔𝑒𝑡 𝑎𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑎𝑠 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 ∶
𝑃(0):
𝑑2𝜃0
𝑑𝜉2
+
2
𝜉
𝑑𝜃0
𝑑𝜉
+ 1 = 0
𝑃(1):
𝑑2𝜃1
𝑑𝜉2
+
2
𝜉
𝑑𝜃1
𝑑𝜉
+ 𝜃5 − 1 = 0
𝑃(2):
𝑑2
𝜃2
𝑑𝜉2
+
2
𝜉
𝑑𝜃2
𝑑𝜉
+ 5𝜃4𝜃1 = 0
𝑃(3):
𝑑2𝜃3
𝑑𝜉2
+
2
𝜉
𝑑𝜃3
𝑑𝜉
+ 5𝜃4𝜃2 + 10𝜃3𝜃1 = 0
And so on
Gives solution as, 𝜃0 = 1 −
𝜉2
6
𝜃1 =
5
6 ∗ 5
.
𝜉4
4
−
10
36 ∗ 7
𝜉6
6
−
10
216 ∗ 9
𝜉8
8
−
5
64 ∗ 11
𝜉10
10
− (
1
65 ∗ 15
)
𝜉12
12

23. For n= 5, Exact Solution Vs HPM Solution ( Upto First Order)
When n =5 , This solution is
finite in mass but infinite in
radial extent, and therefore the
complete polytrope does not
represents a physical solutions.

24. Solution of the Lane-Emden Equation for
n = 0,1,2,3,4,5
 Here, this plot is the solution of Lane-Emden
Equation for the value of n as n = 0, 1, 2, 3, 4
and 5 respectively.
• Lane Emden Equation has analytical solution for
n = 0, 1 ,5
• For n= 0 the density of the solution as a function
of radius is constant. This is the solution for a
constant density incompressible sphere.
 When n =5 , This solution is finite in mass but
infinite in radial extent, and therefore the
complete polytrope does not represents a
physical solutions.

25. Conclusions
• The homotopy perturbation method was used for finding solutions of
linear, nonlinear partial differential equations with initial conditions
and dimensionless Lane-Emden Equation.
• It can be concluded that the homotopy perturbation method is very
powerful and efficient technique in finding exact solutions for wide
classes of problems. In our work we use the MATLAB to calculate the
series obtained from the homotopy perturbation method.
• For solving this method we get nearest value of exact solution than
another method.
• Error is more less than other method.

26. ACKNOWLEDGEMENT
Primarily, we would like to our special thanks of gratitude to our respective
project guider Dr. Banashree Sen who gave this opportunity to work on this
project. We got to learn a lot from this project about “ Using Homotopy
Perturbation Methods in Lane Emden Equation“.
Also we would like to our special thanks to our respective Sir Dr. Abdul
Aziz ,for provided support in completing our Project.
Also, We are thankful for Maulana Abul Kalam Azad University of
Technology(MAKAUT) for providing us this opportunity of Term-Project
program in the curriculum.

27. References
[1] Syed Tauseef Mohyud-Din and Muhammad Aslam Noor, “Homotopy Perturbation Method for
Solving Partial Differential Equations”, Z. Naturforsch. 64a, pp. 157-170, 2009.
[2] He J.H., “Homotopy perturbation technique”, Computer Methods in Applied Mechanics and
Engineering, Volume 178, Issues 3–4, pp. 257-262, 1999.
[3] He J.H., “A coupling method of a homotopy technique and a perturbation technique for non-linear
problems”, International Journal of Non-Linear Mechanics, Volume 35, Issue 1, pp. 37-43, 2000.
[4] He J.H., “Recent development of the homotopy perturbation method”, Topological Methods in
Nonlinear Analysis, Journal of the Juliusz Schauder Center, Volume 31, pp. 205-209, 2008.
[5] Biazar J., Eslami M. and Ghazvini H., "Homotopy Perturbation Method for Systems of Partial
Differential Equations", International Journal of Nonlinear Sciences and Numerical Simulation, vol. 8, no.
3, pp. 413-418, 2007.
[6] Mehdi Ganjiani, “Solution of nonlinear fractional differential equations using homotopy analysis
method”, Applied Mathematical Modelling, Volume 34, Issue 6, pp. 1634-1641, 2010.
[7] He J.H., “Homotopy Perturbation Method with an Auxiliary Term”, Abstract and Applied Analysis,
Volume 2012, Article ID 857612, 2012.
[8] Asian Journal of Science and Applied Technology,ISSN: 2249-0698 Vol.11 No.2, 2022, pp.13-16. The
Research.Publication, www.trp.org.in.DOI: https://doi.org/10.51983/ajsat-2022.11.2.3295

28. Thank You