Second order Taylor Second order Taylor Method Taylor expansion of y - - PowerPoint PPT Presentation

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Second order Taylor Second order Taylor Method Taylor expansion of y - - PowerPoint PPT Presentation

Apr 29, 2023 604 likes •705 views

Second order Taylor Second order Taylor Method Taylor expansion of y ( t + h ) about y ( t ) yields y ( t ) + hy ( t ) + h 2 2 y ( t ) + h 3 3! y (3) ( c i ) y ( t + h ) = y ( t ) + hf ( t , y ( t )) + h 2 f t + f (

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Second order Taylor

Second order Taylor Method Taylor expansion of y(t + h) about y(t) yields y(t + h) = y(t) + hy′(t) + h2 2 y′′(t) + h3 3! y(3)(ci) = y(t) + hf (t, y(t)) + h2 2 ∂f ∂t + ∂f ∂y f

  • (t, y(t)) + O(h3)

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Second order Taylor

Second order Taylor Method Taylor expansion of y(t + h) about y(t) yields y(t + h) = y(t) + hy′(t) + h2 2 y′′(t) + h3 3! y(3)(ci) = y(t) + hf (t, y(t)) + h2 2 ∂f ∂t + ∂f ∂y f

  • (t, y(t)) + O(h3)

Then the second order Taylor method is: yk+1 − yk h = f (tk, yk) + h 2 ∂f ∂t + ∂f ∂y f

  • (tk, yk)

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Second order Runge-Kutta method

Second order Runge-Kutta method Develop a second order method of the form: yk+1 − yk h = a1f (t, y) + a2f (t + α2, y + δ2f (t, y))

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Second order Runge-Kutta method

Second order Runge-Kutta method Develop a second order method of the form: yk+1 − yk h = a1f (t, y) + a2f (t + α2, y + δ2f (t, y)) We need to solve for a1, a2, α2 and δ2

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Second order Runge-Kutta method

Second order Runge-Kutta method Develop a second order method of the form: yk+1 − yk h = a1f (t, y) + a2f (t + α2, y + δ2f (t, y)) We need to solve for a1, a2, α2 and δ2 To do that, we compare to second order Taylor: yk+1 − yk h = f (tk, yk) + h 2 ∂f ∂t + ∂f ∂y f

  • (tk, yk)

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Second order Runge-Kutta method

Second order Runge-Kutta method Develop a second order method of the form: yk+1 − yk h = a1f (t, y) + a2f (t + α2, y + δ2f (t, y)) We need to solve for a1, a2, α2 and δ2 To do that, we compare to second order Taylor: yk+1 − yk h = f (tk, yk) + h 2 ∂f ∂t + ∂f ∂y f

  • (tk, yk)

So we have to expand a2f (t + α2, y + δ2f (t, y))

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Generalized Taylor (Calc 3 version)

Theorem Let f (t, y) and all first and second order partial derivatives be continuous then f (t + ∆t, y + ∆y) = f (t, y) +

  • ∆t ∂f

∂t (t, y) + ∆y ∂f ∂y (t, y)

  • + R

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Generalized Taylor (Calc 3 version)

Theorem Let f (t, y) and all first and second order partial derivatives be continuous then f (t + ∆t, y + ∆y) = f (t, y) +

  • ∆t ∂f

∂t (t, y) + ∆y ∂f ∂y (t, y)

  • + R

therefore setting ∆t = α2, ∆y = δ2f (t, y) yeilds f (t + α2, y + δ2f (t, y)) = f (t, y) + α2 ∂f ∂t (t, y) + δ2f (t, y)∂f ∂y (t, y) + R

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Comparing apples to apples

RK2

yk+1 − yk h = a1f (t, y) + a2

  • f (t, y) + α2

∂f ∂t (t, y) + δ2f (t, y)∂f ∂y (t, y) + R

  • =

(a1 + a2)f (t, y) + a2α2 ∂f ∂t (t, y) + a2δ2f (t, y)∂f ∂y (t, y) + a2R

Second order Taylor

yk+1 − yk h = f (t, y) + h 2 ∂f ∂t (t, y) + h 2f (t, y)∂f ∂y (t, y) + R

Therefore we need

a1 + a2 = 1, a2α2 = h 2, a2δ2 = h 2

Which has infinitely many solutions of the form a1 + a2 = 1 and α2 = δ2 =

h 2a2

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