Mathematical Methods for Physicists: A concise introduction - Site Map

NUMERICAL SOLUTIONS OF DIFFERENTIAL EQUATIONS
Then
and
k 3 ˆ f …x 0 ‡ h; y 0 ‡ 2hk 2 hk 1 †
ˆ f 0 ‡ h…A 0 ‡ f 0 B 0 †‡…1=2†h 2 fC 0 ‡ 2f 0 D 0 ‡ f 2 0 E 0 ‡ 2B 0 …A 0 ‡ f 0 B 0 †g
‡ O…h 3 †:
…1=6†…k 1 ‡ 4k 2 ‡ k 3 †ˆhf 0 ‡…1=2†h 2 …A 0 ‡ f 0 B 0 †
‡…1=6†h 3 …C 0 ‡ 2f 0 D 0 ‡ f 2 0 E 0 ‡ A 0 B 0 ‡ f 0 B 2 0†‡O…h 4 †:
Comparing this with Eq. (13.26), we see that it agrees with the Taylor series
expansion (up to the term in h 3 ) and the formula is established. Formula (13.25)
can be established in a similar manner by taking one more term of the Taylor
series.
Example 13.6
Using the Runge±Kutta method and h ˆ 0:1, solve
y 0 ˆ x y 2 =10; x 0 ˆ 0; y 0 ˆ 1:
Solution: With h ˆ 0:1, h 4 ˆ 0:0001 and we may use the Runge±Kutta thirdorder
approximation.
First step:
x 0 ˆ 0; y 0 ˆ 0; f 0 ˆ0:1;
k 1 ˆ0:1, y 0 ‡ hk 1 =2 ˆ 0:995;
k 2 ˆ0:049; 2k 2 k 1 ˆ 0:002; k 3 ˆ 0;
y 1 ˆ y 0 ‡ h 6 …k 1 ‡ 4k 2 ‡ k 1 †ˆ0:9951:
Second step: x 1 ˆ x 0 ‡ h ˆ 0:1, y 1 ˆ 0:9951, f 1 ˆ 0:001,
k 1 ˆ 0:001, y 1 ‡ hk 1 =2 ˆ 0:9952;
k 2 ˆ 0:051, 2k 2 k 1 ˆ 0:101, k 3 ˆ 0:099,
y 2 ˆ y 1 ‡ h 6 …k 1 ‡ 4k 2 ‡ k 1 †ˆ1:0002:
Third step: x 2 ˆ x 1 ‡ h ˆ 0:2; y 2 ˆ 1:0002, f 2 ˆ 0:1,
k 1 ˆ 0:1, y 2 ‡ hk 1 =2 ˆ 1:0052,
k 2 ˆ 0:149; 2k 2 k 1 ˆ 0:198; k 3 ˆ 0:196;
y 3 ˆ y 2 ‡ h 6 …k 1 ‡ 4k 2 ‡ k 1 †ˆ1:0151:
475