Fourth Order RK-Method

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1 Fourth Order RK-Method The most commonly used method is Runge-Kutta fourth order method. The fourth order RK-method is y i+1 = y i (k 1 + 2k 2 + 2k 3 + k 4 ), Ordinary Differential Equations (ODE) p.44/89

2 where k 1 = hf(x i, y i ), ( k 2 = hf x i + h 2, y i + k 1 2 ( k 3 = hf x i + h 2, y i + k 2 2 k 4 = hf(x i + h, y i + k 3 ). ) ),, Ordinary Differential Equations (ODE) p.45/89

3 Example 1 Find the approximate solution of the initial value problem dx dt = 1 + x t, 1 t 3 with the initial condition x(1) = 1, using the Runge-Kutta second order and fourth order with step size of h = 1. Ordinary Differential Equations (ODE) p.46/89

4 Solution RK 2nd order method. The formula is y i+1 = y i + h 2 (k 1 + k 2 ), where k 1 = f(x i, t i ), k 2 = f(x i + h, t i + hk 1 ). Here, h = 1 and t 0 = x 0 = 1. Therefore, k 1 = f(t 0, x 0 ) = 1 + x 0 t 0 = 2, k 2 = f(t 0 + h, x 0 + hk 1 ) = f(2, 3) = = 2.5. Ordinary Differential Equations (ODE) p.47/89

5 Thus, y 1 = ( ) = Similary, we calculate y 2. Ordinary Differential Equations (ODE) p.48/89

6 RK 4th order method. The formula is where y i+1 = y i (k 1 + 2k 2 + 2k 3 + k 4 ), k 1 = hf(t i, x i ), ( k 2 = hf t i + h 2, x i + k 1 2 ( k 3 = hf t i + h 2, x i + k 2 2 ) ),, k 4 = hf(t i + h, x i + k 3 ). Ordinary Differential Equations (ODE) p.49/89

7 k 1 = hf(t 0, x 0 ) = f(1, 1) = 2, ( k 2 = hf t 0 + h 2, x 0 + k ) 1 = f(1.5, 2) = , 2 ( k 3 = hf t 0 + h 2, x 0 + k ) 2 = , 2 k 4 = hf(t 0 + h, x 0 + k 3 ) = Ordinary Differential Equations (ODE) p.50/89

8 We find y 1 = ( ) = Similarly, we calculate y 2. Ordinary Differential Equations (ODE) p.51/89

9 System of First Order ODE s We consider the n system of first order equations dy 1 dx dy 2 dx dy n dx = f 1 (x, y 1, y 2,, y n ), y 1 (x 0 ) = y 10, = f 2 (x, y 1, y 2,, y n ), y 2 (x 0 ) = y 20, = f n (x, y 1, y 2,, y n ), y n (x 0 ) = y n0. With the help of single step methods, we find the approximate solution for the system of n equations. Ordinary Differential Equations (ODE) p.52/89

10 We now consider the two first order equations of the form dy 1 dx = f 1(x, y 1, y 2 ), y 1 (x 0 ) = y 10, dy 2 dx = f 2(x, y 1, y 2 ), y 2 (x 0 ) = y 20. We use RK second order and fourth order method to solve the system of equations. Ordinary Differential Equations (ODE) p.53/89

11 2nd Order RK Method The formula for second order RK method with step size h is where y 1,n+1 = = y 1,n (s 1 + k 1 ), y 2,n+1 = y 2,n (s 2 + k 2 ), k 1 = hf 1 (x n, y 1n, y 2n ), k 2 = hf 2 (x n, y 1n, y 2n ), Ordinary Differential Equations (ODE) p.54/89

12 s 1 = hf 1 (x n + h, y 1n + k 1, y 2n + k 2 ), s 2 = hf 2 (x n + h, y 1n + k 1, y 2n + k 2 ). Similarly, the formula for fourth order RK-method is y 1,n+1 = y 1,n (k 1 + 2s 1 + 2l 1 + p 1 ), y 2,n+1 = y 2,n (k 2 + 2s 2 + 2l 2 + p 2 ), Ordinary Differential Equations (ODE) p.55/89

13 where k 1 = hf 1 (x n, y 1n, y 2n ), k 2 = hf 2 (x n, y 1n, y 2n ), s 1 = hf 1 (x n + h 2, y 1n + k 1 2, y 2n + k 2 2 ), s 2 = hf 2 (x n + h 2, y 1n + k 1 2, y 2n + k 2 2 ) Ordinary Differential Equations (ODE) p.56/89

14 l 1 = hf 1 (x n + h 2, y 1n + s 1 2, y 2n + s 2 2 ), l 2 = hf 2 (x n + h 2, y 1n + s 1 2, y 2n + s 2 2 ), p 1 = hf 1 (x n + h, y 1n + l 1, y 2n + l 2 ), p 2 = hf 2 (x n + h, y 1n + l 1, y 2n + l 2 ). Ordinary Differential Equations (ODE) p.57/89

15 Example 2 Use RK-method 2nd order and 4th order to find the approximate solution of y(0.1) and z(0.1) as a solution of pair of equations with the initial conditions dy dx = x + z, dz dx = y x y(0) = 1, z(0) = 1. Take step size h = 0.1. Ordinary Differential Equations (ODE) p.58/89

16 Multistep Methods The general form of a linear m-step multistep method is y n+1 a 1 y n a 2 y n 1 a m y n+1 m h i = b 0 f(x n+1, y n+1 ) + b 1 f(x n, y n ) + + b m f(x n+1 m, y n+1 m ). When b 0 = 0, the method is called explicit, otherwise, it is said to be implicit. In multistep method, we require multiple starting values. Ordinary Differential Equations (ODE) p.64/89

17 Adams-Bashforth Methods We assume a uniform discretization in the x-domain, i.e., we define where h = b a N x i = a + ih,, for some positive integer N. Let the given differential equation is of the form y (x) = f(x, y). We now integrate the given differential equation from x = x i to x = x i+1. Ordinary Differential Equations (ODE) p.65/89

18 This yields the equation y(x i+1 ) y(x i ) = Next, we write xi+1 x i f(x, y(x)) dx. f(x, y(x)) = P m 1 (x) + R m 1 (x), Ordinary Differential Equations (ODE) p.66/89

19 where P m 1 (x) = m j=1 L m 1,j (x)f(x i+1 j, y(x i+1 j )), is the Lagrange form of the polynomial of degree at most m 1 that interpolates f at the m points x i, x i 1, x i 2,, x i+1 m. Ordinary Differential Equations (ODE) p.67/89

20 And R m 1 (x) = f(m) (ξ, y(ξ)) m! Π m j=1 (x x i+1 j) is the corresponding remainder term. Ordinary Differential Equations (ODE) p.68/89

21 Two-Step Adams Bashforth Method Since m = 2, we write f(x, y(x)) = x x i 1 x i x i 1 f(x i, y i ) + x x i x i 1 x i f(x i 1, y i 1 ) + f (ξ, y(ξ)) 2 (x x i )(x x i 1 ). Integrating the Lagrange polynomials yields Ordinary Differential Equations (ODE) p.69/89

22 b 1 = b 2 = xi+1 x i xi+1 x i x x i 1 dx = 3h x i x i 1 2, x x i dx = h x i 1 x i 2. Ordinary Differential Equations (ODE) p.70/89

23 Therefore, the two-step Adams-Bashforth method is y n+1 = y n + h 2 [3f(x n, y n ) f(x n 1, y n 1 )]. Proceeding in a similar manner, the three step Adams-bashforth method is y n+1 = y n + h 12 [23f(x n, y n ) 16f(x n 1, y n 1 ) + 5f(x n 2, y n 2 )]. Ordinary Differential Equations (ODE) p.71/89

24 The four-step Adams-Bashforth method is y n+1 = y n + h 24 [55f(x n, y n ) 59f(x n 1, y n 1 ) + 37f(x n 2, y n 2 ) 9f(x n 3, y n 3 ]. Ordinary Differential Equations (ODE) p.72/89

25 Example 5 Use two-step Adams-Bashforth method to find the approximate solution of dx = 1 + x dt t, x(1) = 1, near x(1.5). Take step size h = 0.5. Ordinary Differential Equations (ODE) p.73/89

26 Solution The two-step Adams-Bashforth formula is x n+1 = x n + h 2 [3f(t n, x n ) f(t n 1, x n 1 )]. We note that for finding x 2, we require x 1 and x 0. We calculate x 1 with the help of second order Taylor s method x 1 = Ordinary Differential Equations (ODE) p.74/89

27 Now, x 2 = x 1 + h 2 [3f(t 1, x 1 ) f(t 0, x 0 )], = Ordinary Differential Equations (ODE) p.75/89

28 Adams-Moulton Methods The derivation of Adams-Moulton methods follows exactly the same procedure as the derivation of the Adams-Bashforth method with one exception. In addition to interpolating f at x i, x i 1, x i 2,, x i+1 m, we also interpolate at x i+1. Ordinary Differential Equations (ODE) p.76/89

29 Hence, we write where f(x, y(x)) = P m (x) + R m (x), P m (x) = m j=0 L m,j (x)f(x i+1 j, y(x i+1 j )), and R m (x) = f(m+1) (ξ, y(ξ)) (m + 1)! Π m j=0(x x i+1 j ). Ordinary Differential Equations (ODE) p.77/89

30 Since P m (x) contains a term involving f(x i+1, y(x i+1 )), the resulting method will be implicit. Furthermore, by using an additional point in the interpolating polynomial the degree of the remainder term is increased by one over the Adams-Bashforth case. Ordinary Differential Equations (ODE) p.78/89

31 2-Step Adams-Moulton Method Since m = 2, we write f(x, y(x)) = (x x i )(x x i 1 ) (x i+1 x i )(x i+1 x i 1 ) f(x i+1, y i+1 ) + (x x i+1)(x x i 1 ) (x i x i+1 )(x i x i 1 ) f(x i, y i ) + (x x i+1)(x x i ) (x i 1 x i+1 )(x i 1 x i ) f(x i 1, y i 1 ) + f (ξ, y(ξ)) 6 (x x i+1 )(x x i )(x x i 1 ). Ordinary Differential Equations (ODE) p.79/89

32 Integrating the Lagrange polynomials yields b 0 = b 1 = b 2 = xi+1 x i (x x i )(x x i 1 ) (x i+1 x i )(x i+1 x i 1 ) xi+1 dt = 5h 12, x i (x x i+1 )(x x i 1 ) (x i x i+1 )(x i x i 1 ) dt = 2h 3, xi+1 x i (x x i+1 )(x x i ) (x i 1 x i+1 )(x i 1 x i ) dt = h 12. Ordinary Differential Equations (ODE) p.80/89

33 Therefore, the two-step Adams-Moulton method is y i+1 y i h = 5 12 f(x i+1, y i+1 ) f(x i, y i ) 1 12 f(x i 1, y i 1 ). To find the approximate solution, we required y 0 and y 1. The initial condition give y 0 and y 1 can be obtained using any third order one-step method. Ordinary Differential Equations (ODE) p.81/89

34 Similarly, we derive the three-step Adams-Moulton method is given by y i+1 = y i + h 24 [9f(x i+1, y i+1 ) +19f(x i, y i ) 5f(x i 1, y i 1 ) +f(x i 2, y i 2 )]. The required starting values for the three step method should be obtained using a fourth-order one-step method. Ordinary Differential Equations (ODE) p.82/89

35 Predictor-Corrector Method The most popular of the predictor-corrector scheme is the Adams fourth-order predictor-corrector method. This uses the four-step, fourth-order Adams-Bashforth method y (0) n+1 = y n + h 24 [55f(x n, y n ) 59f(x n 1, y n 1 ) as a predictor + 37f(x n 2, y n 2 ) 9f(x n 3, y n 3 )], Ordinary Differential Equations (ODE) p.83/89

36 followed by the three-step, fourth-order Adams-Moulton method y (1) n+1 = y n + h 24 [9f(x n+1, y (0) n+1) + 19f(x n, y n ) as a corrector. 5f(x n 1, y n 1 ) + f(x n 2, y n 2 )], Ordinary Differential Equations (ODE) p.84/89

37 Note that the computation of y (0) 4, we require the values (x 0, y 0 ), (x 1, y 1 ), (x 2, y 2 ), (x 3, y 3 ). These values should be given with the given equation or should be computed using single step methods like, Euler s method, Taylor s method or Runge-Kutta methods. To compute the value y (1) 4, we require the values (x 1, y 1 ), (x 2, y 2 ), (x 3, y 3 ) and (x 4, y (0) 4 ). Ordinary Differential Equations (ODE) p.85/89

38 Modifier Let y(x n+1 ) denotes the true solution of y at x n+1. Then y(x n+1 ) = y (0) n h5 f (iv) (ξ 1 ), x n 3 < ξ 1 < x n+1, y(x n+1 ) = y (1) n h5 f (iv) (ξ 2 ), x n 2 < ξ 2 < x n+1. On subtracting, we get 0 = y (0) n+1 y (1) n h5 f (iv) (ξ). Ordinary Differential Equations (ODE) p.86/89

39 This implies that We rewrite as y (1) n+1 y (0) n+1 = h5 f (iv) (ξ). h f(iv) (ξ) = (y(1) n+1 y n+1). (0) Ordinary Differential Equations (ODE) p.87/89

40 Finally, we find the modifier as y(x n+1 ) y (1) n+1 = (y(1) n+1 y (0) n+1) = D n+1. Here, D n+1 is called as modifier. Ordinary Differential Equations (ODE) p.88/89

41 Example Find y(1.4) by Adams Moulton 4th order predictor-corrector pair with modifier as a solution of dy dx = x3 + xy, y(1) = 2, y(1.1) = 1.6, y(1.2) = 0.34 and y(1.3) = with spacing h = 0.1. Ordinary Differential Equations (ODE) p.89/89

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