2.1 NUMERICAL SOLUTION OF SIMULTANEOUS FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS. differential equations with the initial values y(x 0. ; l.

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1 Numerical Methods II UNIT.1 NUMERICAL SOLUTION OF SIMULTANEOUS FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS.1.1 Runge-Kutta Method of Fourth Order 1. Let = f x,y,z, = gx,y,z be the simultaneous first order ordinary differential equations with the initial values yx = y, zx = z. The modified form of expression for computation are as follows. k 1 = hfx,y,z ; l 1 = hgx,y,z k = hf x h,y k 1,z l 1 ; l = hg x h,y k 1,z l 1 k 3 = hf x h,y k,z l ; l 3 = hg x h,y k,z l k 4 = hf x h,y k 3,z l 3 ; l4 = hg x h,y k 3,z l 3 Then according to second order Runge-Kutta method yx h = y k zx h = z l According to fourth order Runge-Kutta method yx h = k1 k 6 k 3 k 4 zx h = l1 l 6 l 3 l 4

2 46 Engineering Mathematics-IV =.1[ =.11 l 4 = hg t h, x k 3, y l 3 =.1[ =.15 xt h = x 1 [ k1 k 6 k 3 k 4 yt h = y 1 6 = 1 1 [ = [ l1 l l 3 l 4 Thus, = 1 1 [ = x.1 = y.1 = Example : equations Use fourth order Runge-Kutta method to solve the system of = 1 xz; = xy given y =, z = 1 at x =.3. Solution: Given simultaneous equations are = 1 xz, = xy y =, z = 1 or, f x,y,z = 1 xz, gx,y,z = xy Here x =, y =, z = 1 given h =.3.

3 Numerical Methods II 47 Now, we shall compute the following. k 1 = hfx, y, z =.3[1 1 =.3 l 1 = hgx, y, z =.3[ = k = hf x h, y k 1, z l 1 =.3[ =.345 l = hg x h, y k 1, z l 1 =.3[ =.675 k 3 = hf x h, y k, z l =.3[ = l 3 = hf yx h = y 1 6 x h, y k, z l =.3[ =.7765 k 4 = hf x h, y k 3, z l 3 =.3[ =.3893 l 4 = hf x h, y k 3, z l 3 =.3[ =.314 [ k1 k k 3 k 4 y.3 = 1 [ =.34483

4 48 Engineering Mathematics-IV zx h = z 1 6 [ l1 l l 3 l 4 z.3 = Picard s Method = 1 1 [ Let the simultaneous differential equations be = f x,y,z = gx,y,z with initial conditions yx = y zx = z. Then = y y = y y 3 = y f x,y,z, = z f x,,, z = z f x,y,z, z 3 = z gx,y,z gx,, gx,y,z so on. Continuing this process, a sequence of functions of x, i.e.,,y,y 3...,z,z 3... are obtained each giving a better approximations of the desired solution. Example 1: Using Picard s method find approximate values of y z corresponding to x =.1 given that y =, z = 1 = x z, = x y. Solution: Given equations are = f x,y,z = x z = gx,y,z = x y y = y f x,y,z z = z gx,y,z x x

5 Numerical Methods II 49 First approximation = y = z x f x,y,z = Second approximation y = y x gx,y,z = 1 x f x,, = x 1 = x x x 4 = 1 4x 1 x x 1 4x 1 x = x 3 x x3 6 z = z Third approximation so on. x gx,, = 1 = 1 4x 3 x x 3 x4 4 x5 y 3 = y x f x,y,z [ x = x 3 x 1 x3 1 4 x4 1 x5 z 3 = z gx,y,z x 1 x 1 1 x6 = 1 4x x 3 x3 1 x4 6 x5 1 x6 5 x7

6 5 Engineering Mathematics-IV when x =.1 =.15, y =.8517 y 3 =.8447 =.65, z = z 3 =.5867 Thus, y.1 =.845, z.1 =.5867 correct to four decimal places..1.3 Taylor s Series Method Let the simultaneous differential equations be with initial conditions yx = y zx = z. If h be the step-size, = yx h = zx h Then Taylor s algorithm for 1 gives = f x,y, = gx,y,z = y hy h! y h3 = z hz h! z h3 3! y 3! z Differentiating 1 successively we get y,z, etc So the values y,y,y... z,z,z... are known. Substituting these in 3 4, we obtain, for the next step. Similarly, y = hy 1 h! y 1 h3 z = hz 1 h! z 1 h3 3! y 1 3! z Since are known, we calculate y, 1 y... 1 z, 1 z... substituting 1 these in 5 6 we get y z. Proceeding further, we can calculate the other values of y z step by step.

7 Numerical Methods II 51 Example 1: Given = z = xz y with y = 1, z =, obtain y z for x =.1,.,.3 by Taylor s series method. Hint: We have y = z z = xz y We use Taylor s series method to find y z.. NUMERICAL SOLUTION OF SECOND ORDER ORDINARY DIFFERENTIAL EQUATIONS..1 Picard s Method Let the second order differential equation be d y = f x,y, 1 with yx = y y x = y. Now, let = z then d y =. Substituting in 1 = f x,y,z with yx = y zx = z The problem reduces to solving the simultaneous equations = z = f 1 x,y,z = f x,y,z subject to yx = y zx = z. By Picard s method we have = y f 1 x,y,z, = z x y = y f 1 x,,, z = z x x x f x,y,z f x,,

8 Numerical Methods II 53 =.5.1. x.5x, z =.1 x.1. x.5x.5.1x =.5.1x. x3 6.5x, =.1.x.x4 4 1x3 3.5x.1 x so on. when x =.1 =.51 y =.574 = y 1 =.49 y = z =.488 y.1 =.574 y.1 =.488 Example : solution of Using Picard s method, obtain the second approximation to the Solution: d y = x3 x3 y so that y = 1, y 1 =. Given the second order differential equation is d y = x3 x3 y with y = 1, y 1 = Let = z so that d y = = z = x3 z x 3 y with y = 1, =. By Picard s methods, = f x,y,z = f 1 x,y,z

9 54 Engineering Mathematics-IV First approximation = y = 1 x f x y z = z = 1 x Second approximation y = y = 1 1 = 1 x = 1 3x4 8 x f x,, z = z 1 3x4 8 x = 1 x = 1 x 3x5 x = 1 f x, y z 1 x3 1x = x3 x f x,, x 3 1 3x4 x 3 1 x 8 x x7 x 3 x4 = 1 x4 8 3x8 64 x4 4 x5 1 Example 3: 1 = 3x4 8 x5 1 3x8 64. Using Picard s method find the solution of the equation y xy y =, with y = 1 y = for x =.1.3. Solution: Given equation is y xy y = with y = 1, y =. Let = y = z so that d y = y = = z = xz y with y = 1, z = = f x,y,z, = f 1 x,y,z.

10 56 Engineering Mathematics-IV when x =.1.1 = 1 y.1 =.955, y 3.1 = =.1 z.1 =.99, z 3.1 =.995 y.1 =.9951 z.1 = y.1 =.995 when x =.3.3 = 1 y.3 =.955 y 3.3 = =.3 z.3 =.91, z 3.3 =.911 y.3 =.9556 z.3 = y.3 = Runge-Kutta Method Let the second order differential equation be d y = f x, y,...1 with yx = y, y x = y. Now, let = z so that d y = = z = f x,y,z = f 1 x,y,z, = f x,y,z with yx = y, zx = z. The problem reduces to solving the simultaneous equations = z = f 1 x,y,z = f x,y,z with yx = y zx = z. Starting at x,y,z taking the step-sizes for x, y, z to be h, k, l respectively, the Runge-Kutta method gives

11 Numerical Methods II 57 K 1 = hf 1 x,y,z l 1 = hf x,y,z K = hf 1 x 1 h, y k 1, z l 1 k 3 = hf 1 x h, y k, z l k 4 = hf 1 x h, y k 3, z l 3 l = hf x h, y k 1, z l 1 l 3 = hf x h, y k, z l l 4 = hf x h, y h 3, z l 3 = y 1 6 k1 k k 3 k 4 = z h 6 l1 l l 3 l 4 Example 1: Using Runge-Kutta method solve y = xy y for x =. correct to 4 decimal places initial conditions are x =, y = 1, y =. Solution: Let Given equation of second order is = z = f 1 x,y,z, then y = xy y with x =, y = 1, y = = xz y = f x,y,z with x =, y = 1, z = we compute k 1, k, k 3, k 4 for f 1 x, y, z l 1, l, l 3, l 4 for f x, y, z. by Runge-Kutta formulae, we have k 1 = hf 1 x,y,z l 1 = hf x,y,z =. = =. 1 =. k = hf 1 x h, y k 1, z l 1 l = hf x h, y k 1, z l 1 =..1 =. =..999 =.1998 k 3 = hf 1 x h, y k, z l l 3 = hf x h, y k, z l =..999 =. = =.1958

12 Numerical Methods II 59 Let = z then d y =. Substituting in 1 = f x,y,z with yx = y zx = z The problem reduces to solving the simultaneous equations. = z = f 1 x,y,z = f x,y,z subject to yx = y, zx = z. The predictor corrector formulae can be written as y P n 1 = y 4h [ n 3 y 3 n y n 1 y n z P n 1 = z 4h [ n 3 z 3 n z n 1 z n y c n 1 = y h [ n 1 y 3 n 1 4y n y n 1 z c n 1 = z h [ n 1 z 3 n 1 4z n z n 1 In particular when n = 3 y P 4 = y 4h 3 z P 4 = z 4h 3 [ y 1 y y 3 [ z 1 z z 3 y c 4 = y h [ y 3 4y 3 y 4 z c 4 = z h [ z 3 4z 3 z 4 3 is called Milne s predictor formula 3 is called Milne s corrector formula.

13 6 Engineering Mathematics-IV Example 1: Given d y x y =, y = 1, y = obtain y for x=,.1,.,.3 by any method. Further compute y.4 by Milne s method. Solution: Let y = z then y = z The given equation is y xy y =...1 with y = 1, y = 1 z xz y = y = z = y = z = f 1 x,y,z with y = 1 = z = f x,y,z = xz y with z = We use Taylor s series method to find y. Differentiating 1 n times we get At x =, y n = n 1y n y n xy n 1 ny n y n = y = 1 gives y = 1, y 4 = gives y 3 = y 5 =... =. Exp yx by Taylor s series = 3, y 6 = 5 3 yx = y x x! y x3 3! y 3... yx = 1 x 3 5 3! 4! x4 x ! From we have zx = y x = x 1 x3 1 8 x5... = xy 3 y.1 = =.995 y. = 1. y.3 = = =.956.

14 6 Engineering Mathematics-IV y.4 = y. h 3 =.98 =.93 [ y. 4y.3 y.4 Hence, y.4 =.93 y.4 = [ Example : Given d y x y = y =.5,y =.1 obtain y for x =,.1,.,.3 by any suitable method. Hence, apply Milne s method to compute y.4. Example 3: Given y = xy y y = 1, y = obtain y for x =,.1,.,.3 by Taylor s series method. Hence, apply Milne s method to compute y.4. Ans: y.4= Taylor s Series Method Example 1: Given y xy y =, y = 1, y =. Obtain y for x =,.1,.,.3 by Taylor s series method. Solution: Let y = z then y = z Given second order differential equation is We use Taylor s series method to find y. y xy y = 1 z xz y = ; y = z Differentiating the given equation n times we get At x =, y n xy n 1 ny n y n = y n = n 1y n y = 1 gives y = 1, y 4 = 3, y 6 = 5 3 gives y 3 = y 5 =... =