JUST THE MATHS SLIDES NUMBER 17.7 NUMERICAL MATHEMATICS 7 - PDF document

“JUST THE MATHS” SLIDES NUMBER 17.7 NUMERICAL MATHEMATICS 7 (Numerical solution) of (ordinary differential equations (B)) by A.J.Hobson 17.7.1 Picard’s method

UNIT 17.7 NUMERICAL MATHEMATICS 7 NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS (B) 17.7.1 PICARD’S METHOD This method of solution is one of successive approxima- tion. It is an iterative method in which the numerical results become more and more accurate, the more times it is used. An approximation for y in terms of x is substituted into the right hand side of the differential equation, d y d x = f ( x, y ) . The equation is then integrated with respect to x , giving y in terms of x as a second approximation. Given numerical values are substituted into the second approximation and the result rounded off to an assigned number of decimal places or significant figures. The iterative process is continued until two consecutive numerical solutions are the same when rounded off to the required number of decimal places. 1

A hint on notation Consider the differential equation d y d x = 3 x 2 , given that y = y 0 = 7 when x = x 0 = 2. This can be solved exactly to give y = x 3 + C, which requires that 7 = 2 3 + C. Hence, y − 7 = x 3 − 2 3 ; or, in more general terms, y − y 0 = x 3 − x 3 0 . Thus, � y � x x 0 3 x 2 d x. y 0 d y = That is, d y � x � x x 0 3 x 2 d x. d x d x = x 0 In future, we shall integrate both sides of the given dif- ferential equation with respect to x , from x 0 to x . 2

EXAMPLES 1. Given that d y d x = x + y 2 , and that y = 0 when x = 0, determine the value of y when x = 0 . 3, correct to four places of decimals. Solution d y � x � x x 0 ( x + y 2 )d x, d x d x = x 0 where x 0 = 0 Hence, � x x 0 ( x + y 2 )d x, y − y 0 = where y 0 = 0. That is, � x 0 ( x + y 2 )d x. y = (a) First Iteration Replace y by y 0 in the function to be integrated. 0 x d x = x 2 � x y 1 = 2 ≃ 0 . 0450 at x = 0 . 3 (b) Second Iteration Now we use 1 = x + x 4 d y d x = x + y 2 4 . 3

Therefore,  x + x 4 d y   � x � x d x d x =  d x,     0 0 4 which gives y − 0 = x 2 2 + x 5 20 . y 2 = x 2 2 + x 5 20 ≃ 0 . 0451 at x = 0 . 3 (c) Third Iteration Now we use d y d x = x + y 2 2 = x + x 4 4 + x 7 20 + x 10 400 . Therefore,  x + x 4 4 + x 7 20 + x 10 d y   � x � x d x d x =  d x,     0 0 400 which gives y − 0 = x 2 2 + x 5 20 + x 8 160 + x 11 4400 . y 3 = x 2 2 + x 5 20 + x 8 160 + x 11 4400 ≃ 0 . 0451 at x = 0 . 3 4

2. If d y d x = 2 − y x, and y = 2 when x = 1, perform three iterations of Picard’s method to estimate a value for y when x = 1 . 2. Work to four places of decimals throughout and state how accurate is the result of the third iteration. (a) First Iteration d y  2 − y � x � x    d x, d x d x = x 0 x 0 x where x 0 = 1. That is,  2 − y � x    d x, y − y 0 = x 0 x where y 0 = 2. Hence,  2 − y � x    d x. y − 2 = 1 x Replacing y by y 0 = 2 in the function being integrated, we have  2 − 2 � x    d x. y − 2 =   1 x Therefore, y = 2 + [2 x − 2 ln x ] x 1 = 2 + 2 x − 2 ln x − 2 + 2 ln 1 = 2( x − ln x ) . y 1 = 2( x − ln x ) ≃ 2 . 0354 when x = 1 . 2 5

(b) Second Iteration d y x = 2 − 2( x − ln x ) = 2 ln x d x = 2 − y 1 . x x Hence, d y 2 lnx � x � x d x d x = x d x. 1 1 That is, � x (ln x ) 2 1 = (ln x ) 2 . � y − 2 = y 2 = 2 + (ln x ) 2 ≃ 2 . 0332 when x = 1 . 2 (c) Third Iteration Finally, we use x − (ln x ) 2 d y x = 2 − 2 d x = 2 − y 2 . x Hence, x − ( lnx ) 2 d y  2 − 2   � x � x d x d x =  d x.     1 1 x That is, x  2 x − 2 ln x − (ln x ) 3   y − 2 = .     3  1 y − 2 = 2 x − 2 ln x − (ln x ) 3 − 2 . 3 6

y 3 = 2 x − 2 ln x − (ln x ) 3 ≃ 2 . 0293 when x = 1 . 2 3 The results of the last two iterations are identical when rounded off to two places of decimals, namely 2.03. 7