lecture 21

lecture 21 - CE 30125 - Lecture 21 LECTURE 21 SOLUTIONS TO...

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CE 30125 - Lecture 21 p. 21.1 LECTURE 21 SOLUTIONS TO O.D.E.’S Continued • Solve Initial condition • Note that the solution is a function of only while the slope is a function of both and . INSERT FIGURE NO. 96 dy dt ----- fyt , () = y t o y o = y t t , y t y(t) y 0 t f(y,t)=slope
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CE 30125 - Lecture 21 p. 21.2 2nd Order Runge-Kutta Methods • Use two terms for (1) (2) (3) • Expand the 2nd term in Equation (3) as a Taylor series about where and Φ y j 1 + y j Δ t Φ + = Φ a 1 g 1 a 2 g 2 + = Φ a 1 ft j y j , () a 2 j p 1 Δ ty j p 2 Δ t j y j , + , + + = t j y j , j p 1 Δ j , p 2 Δ t j y j , ++ j y j , Δ T f t ---- t j y i , Δ y f y ----- t j y j , = 1 2! Δ T 2 2 f t 2 ------- 2 Δ T Δ y 2 f y t ---------- Δ y 2 2 f y 2 -------- t j y j , Δ Tp 1 Δ t Δ yp 2 Δ t j y j ,
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CE 30125 - Lecture 21 p. 21.3 • Hence the approximation to is expressed as: • We neglect terms of 2nd order and higher. • Hence: • However we recall that: • Substituting in for we obtain an approximating expression for (4) Φ Φ a 1 f j a 2 f j p 1 Δ t f t ---- j p 2 Δ t f j f y ----- j O Δ t () 2 ++ + ⎩⎭ ⎨⎬ ⎧⎫ + = Φ a 1 a 2 + f j a 2 Δ tp 1 f t j p 2 f j f y j + O Δ t 2 = y j 1 + y j Δ t Φ + = Φ y j 1 + y j 1 + y j Δ ta 1 a 2 + f j a 2 Δ t 2 p 1 f t j p 2 f j f y j + O Δ t 3 + =
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CE 30125 - Lecture 21 p. 21.4 • Use Taylor Series to find exact representation for by expanding about (5) • However by definition y j 1 + yt j 1 + () = t j y j 1 + y j Δ t dy dt ----- j Δ t 2 2 ------------ d 2 y 2 -------- j O Δ t 3 ++ + = fty , = d 2 y 2 f t ---- f y ----- + ⎝⎠ ⎛⎞ = d 2 y 2
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lecture 21 - CE 30125 - Lecture 21 LECTURE 21 SOLUTIONS TO...

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