tailieunhanh - Burden - Numerical Analysis 5e (PWS, 1993) Episode 2 Part 1

Tham khảo tài liệu 'burden - numerical analysis 5e (pws, 1993) episode 2 part 1', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Euler s Method 243 lal Theorem Suppose f is continuous and satisfies a Lipschitz condition with constant L on Z - fi y d t b -K y oo and that a constant M exists with the property that y í I M for all t e a Ỉ . Let y f denote the unique solution to the initial-value problem y1 f f y a t b y a a and w0 vtq . wNbe the approximations generated by Euler s method for some positive integer . Then Ixo - Wi ajr - 1 for each i 0 1 2 . . N. Proof When 0 the result is clearly true since y i0 w0 a. From Eq. we have for i - 0 1 . . N - 1 h2 y ti Xfi WjO 2 and from the equations in Wi I Wf Consequently using the notation yz y q andy i y fi i yz 1 - wi 1 yf - Wj htfit yf - f ĩị wf y i h2 and y 1 - wf 1 yz - wj 4- - f tị y y f . Since f satisfies a Lipschitz condition in the second variable with constant L and ịy 01 we have y Ạ1 - w 1 yf - wf l hf . Referring to Lemma and letting yj - -wyl for each 0 1 . . . N while s hL and t h2M 12 we see that . . .T . . ỉ2m yi 1 - wí 1 s jy0 - w0 Since y0 - vv0 0 and Ì V h ti 1 ĨQ - fi 1 - a we have yi 1 - wi 1 - 1 I t i i T 1 I X for each i 0 1 . N 1. EBE 244 CHAPTER 5 Initial-Value Problems for Ordinary Differential Equations The weakness of Theorem lies in the requirement that a bound be known for the second derivative of the solution. Although this condition often prohibits US from obtain-ing a realistic error bound it should be noted that if df dt and df dy both exist then . J 7 yw Tfl y yw y ty- d dt dt at dy So it is at times possible to obtain an error bound for y t without explicitly knowing y i . EXAMPLE 2 Returning to the initial-value problem ll y y t2 1 0 t 2 y 0 considered in Example 1 we see that since f t y y t2 4- 1 we have ỡ í y ỡy - 1 . for all y so L 1. For this problem we know that the exact solution is y i r l 2 - e so yr t 2 - 4 and id y 0 V 2 for all ÍG 0 2 . Using the inequality in the error bound for Euler s method with h L 1. and M 2 gives the error bound y - .

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