Recommended
More Related Content
Similar to Lane_emden_equation_solved_by_HPM_final
Similar to Lane_emden_equation_solved_by_HPM_final (20)
Recently uploaded
Recently uploaded (20)
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