23 with s 1 as yn yn 1 hb1 f tn 1 c1h y1 331a y1

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ries and substituting the result into the above expression yields ; ; y(tn) = y(tn 1) + hb1 f + c1 hft + ha11 ffy + O(h2)]: ; Comparing the terms of the above series with the Taylor's series h2 (f + ff ) + O(h3) y(tn) = y(tn 1) + hf + 2 t y of the exact solution yields 1 b1 = 1 a11 = c1 = 2 : Substituting these coe cients into (3.3.1), we nd the method to be an implicit midpoint rule ; yn = yn 1 + hf (tn 1 + h=2 Y1) ; ; Y1 = yn 1 + h f (tn 1 + h=2 Y1): 2 The tableau for this method is ; ; 19 (3.3.2a) (3.3.2b) 1 2 1 2 1 The formula has similarities to the midpoint rule predictor-corrector (3.1.11) however, there are important di erences. Here, the backward Euler method (rather than the forward Euler method) may be regarded as furnishing a predictor (3.3.2b) with the midpoint rule providing the corrector (3.3.2a). However, formulas (3.3.2a) and (3.3.2b) are coupled and must be solved simultaneously rather than sequentially. Example 3.3.2. The two-stage method having maximal order four presented in th...
View Full Document

Ask a homework question - tutors are online