Numerical ODE Solvers
•
0 likes•138 views
To find Numerical solution of ODE by various methods
Recommended
More Related Content
What's hot
What's hot (20)
Similar to Numerical ODE Solvers
Similar to Numerical ODE Solvers (20)
More from Santhanam Krishnan
More from Santhanam Krishnan (20)
Recently uploaded
Recently uploaded (20)
Numerical ODE Solvers
- 1. NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS
- 2. TAYLOR SERIES METHOD If 𝑑𝑦 𝑑𝑥 = 𝑓 𝑥, 𝑦 , 𝑤𝑖𝑡ℎ 𝑦 𝑥0 = 𝑦0 is the given boundary value problem Then 𝑦 𝑛+1 = 𝑦𝑛 + ℎ𝑦𝑛 ′ + ℎ2 2! 𝑦𝑛 ′′ + ℎ3 3! 𝑦𝑛 ′′′ + ⋯ Where 𝑦 𝑟 = 𝑑 𝑟 𝑦 𝑑𝑥 𝑟
- 3. MERITS AND DEMERITS OF TAYLOR SERIES METHOD 1. Taylor series method is a powerful single step method if we are able to get the successive derivatives easily. This method is useful to give some initial values for powerful numerical methods like Runge-kutta method. 2. The disadvantage is that it became tedious if the higher derivatives are complicated.
- 4. EULER’S METHOD If 𝑑𝑦 𝑑𝑥 = 𝑓 𝑥, 𝑦 , 𝑤𝑖𝑡ℎ 𝑦 𝑥0 = 𝑦0 is the given boundary value problem Then 𝑦 𝑛+1 = 𝑦𝑛 + ℎ𝑓 𝑥 𝑛, 𝑦𝑛 , 𝑛 = 0,1,2 …
- 5. EULER’S ALGORITHM If 𝑥0 is the starting value, 𝑦1 = 𝑦0 + ℎ𝑓(𝑥0, 𝑦0) 𝑦2 = 𝑦1 + ℎ𝑓(𝑥1, 𝑦1). . . 𝑦 𝑛+1 = 𝑦𝑛 + ℎ𝑓(𝑥 𝑛, 𝑦𝑛)
- 6. MODIFIED EULER’S METHOD If 𝑑𝑦 𝑑𝑥 = 𝑓 𝑥, 𝑦 , 𝑤𝑖𝑡ℎ 𝑦 𝑥0 = 𝑦0 is the given boundary value problem Then 𝑦 𝑛+1 = 𝑦𝑛 + ℎ𝑓 𝑥 𝑛 + ℎ 2 , 𝑦𝑛 +
- 7. FOURTH ORDER RUNGE- KUTTA METHOD If 𝑑𝑦 𝑑𝑥 = 𝑓 𝑥, 𝑦 , 𝑤𝑖𝑡ℎ 𝑦 𝑥0 = 𝑦0 is the given boundary value problem 𝑘1 = ℎ𝑓(𝑥 𝑛, 𝑦𝑛) 𝑘2 = ℎ𝑓 𝑥 𝑛 + ℎ 2 , 𝑦𝑛 + 𝑘1 2 𝑘3 = ℎ𝑓 𝑥 𝑛 + ℎ 2 , 𝑦𝑛 + 𝑘2 2 𝑘4 = ℎ𝑓(𝑥 𝑛 + ℎ, 𝑦𝑛 + 𝑘3) ∆𝑦 = 𝑘1+2𝑘2+2𝑘3+𝑘4 6 𝑦 𝑛+1 = 𝑦𝑛 + ∆𝑦 , 𝑛 = 0, 1, 2, 3, . . .
- 8. MERITS OF RK-METHOD Runge- Kutta method is a single step method. This method does not require prior calculation of higher derivatives likes Taylor’s series method. To find the value of y at 𝑥 = 𝑥 𝑟+1 we need the value of y at 𝑥 𝑟 only.
- 9. MULTI STEP METHODS 1. MILNE’S METHOD (MILNE’S PREDICTOR AND CORRECTOR METHOD) 2. ADAM BASHFORTH METHOD(ADAM’S PREDICTOR AND CORRECTOR METHOD)
- 10. MILNE’S METHOD MILNE‘S PREDICTOR FORMULA If 𝑑𝑦 𝑑𝑥 = 𝑓 𝑥, 𝑦 , 𝑤𝑖𝑡ℎ 𝑦 𝑥0 = 𝑦0 is the given boundary value problem 𝑦 𝑛+1 = 𝑦 𝑛 −3 + 4ℎ 3 2𝑦 𝑛−2 ′ − 𝑦 𝑛−1 ′ + 2𝑦 𝑛 ′ MILNE‘S CORRECTOR FORMULA 𝑦 𝑛+1 = 𝑦 𝑛−1 + ℎ 3 𝑦′ 𝑛−1 + 4𝑦𝑛 ′ + 𝑦 𝑛+1 ′ , 𝑤ℎ𝑒𝑟𝑒 𝑦 𝑛+1 ′ = 𝑓 𝑥 𝑛+1, 𝑦 𝑛+1
- 11. In particular, Milne’s predictor formula: 𝑦4 = 𝑦0 + 4ℎ 3 [2𝑦1 ′ − 𝑦2 ′ + 2𝑦3 ′ ] Milne’s Corrector formula: 𝑦4 = 𝑦2 + ℎ 3 [𝑦2 ′ + 4𝑦3 ′ + 𝑦4 ′ ]
- 12. ADAM- BASHFORTH METHOD ADAM- BASHFORTH PREDICTOR FORMULA If 𝑑𝑦 𝑑𝑥 = 𝑓 𝑥, 𝑦 , 𝑤𝑖𝑡ℎ 𝑦 𝑥0 = 𝑦0 is the given boundary value problem 𝑦 𝑛+1 = 𝑦𝑛 + ℎ 24 55 𝑦𝑛 ′ − 59𝑦 𝑛−1 ′ + 37𝑦 𝑛−2 ′ − 9 𝑦 𝑛−3 ′ ADAM- BASHFORTH CORRECTOR FORMULA 𝑦 𝑛+1 = 𝑦𝑛 + ℎ 24 9 𝑦 𝑛+1 ′ + 19𝑦𝑛 ′ − 5𝑦 𝑛−1 ′ + 𝑦 𝑛−2 ′
- 13. In particular, Adam’s predictor formula 𝑦4 = 𝑦3 + ℎ 24 55 𝑦3 ′ − 59𝑦2 ′ + 37𝑦1 ′ − 9 𝑦0 ′ Adam’s corrector formula 𝑦4 = 𝑦3 + ℎ 24 9 𝑦4 ′ + 19𝑦3 ′ − 5𝑦2 ′ + 𝑦1 ′