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

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 ( t , y ( t )) + O ( h 3 ) = ∂ y f 2 1 / 4

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 ( t , y ( t )) + O ( h 3 ) = ∂ y f 2 Then the second order Taylor method is: � ∂ f � y k +1 − y k = f ( t k , y k ) + h ∂ t + ∂ f ( t k , y k ) ∂ y f h 2 1 / 4

Second order Runge-Kutta method Second order Runge-Kutta method Develop a second order method of the form: y k +1 − y k = a 1 f ( t , y ) + a 2 f ( t + α 2 , y + δ 2 f ( t , y )) h 2 / 4

Second order Runge-Kutta method Second order Runge-Kutta method Develop a second order method of the form: y k +1 − y k = a 1 f ( t , y ) + a 2 f ( t + α 2 , y + δ 2 f ( t , y )) h We need to solve for a 1 , a 2 , α 2 and δ 2 2 / 4

Second order Runge-Kutta method Second order Runge-Kutta method Develop a second order method of the form: y k +1 − y k = a 1 f ( t , y ) + a 2 f ( t + α 2 , y + δ 2 f ( t , y )) h We need to solve for a 1 , a 2 , α 2 and δ 2 To do that, we compare to second order Taylor: � ∂ f � y k +1 − y k = f ( t k , y k ) + h ∂ t + ∂ f ∂ y f ( t k , y k ) 2 h 2 / 4

Second order Runge-Kutta method Second order Runge-Kutta method Develop a second order method of the form: y k +1 − y k = a 1 f ( t , y ) + a 2 f ( t + α 2 , y + δ 2 f ( t , y )) h We need to solve for a 1 , a 2 , α 2 and δ 2 To do that, we compare to second order Taylor: � ∂ f � y k +1 − y k = f ( t k , y k ) + h ∂ t + ∂ f ∂ y f ( t k , y k ) 2 h So we have to expand a 2 f ( t + α 2 , y + δ 2 f ( t , y )) 2 / 4

Generalized Taylor (Calc 3 version) Theorem Let f ( t , y ) and all first and second order partial derivatives be continuous then � � ∆ t ∂ f ∂ t ( t , y ) + ∆ y ∂ f f ( t + ∆ t , y + ∆ y ) = f ( t , y ) + ∂ y ( t , y ) + R 3 / 4

Generalized Taylor (Calc 3 version) Theorem Let f ( t , y ) and all first and second order partial derivatives be continuous then � � ∆ t ∂ f ∂ t ( t , y ) + ∆ y ∂ f f ( t + ∆ t , y + ∆ y ) = f ( t , y ) + ∂ y ( t , y ) + R therefore setting ∆ t = α 2 , ∆ y = δ 2 f ( t , y ) yeilds ∂ f ∂ t ( t , y ) + δ 2 f ( t , y ) ∂ f f ( t + α 2 , y + δ 2 f ( t , y )) = f ( t , y ) + α 2 ∂ y ( t , y ) + R 3 / 4

Comparing apples to apples RK2 ∂ f ∂ t ( t , y ) + δ 2 f ( t , y ) ∂ f y k +1 − y k � � = a 1 f ( t , y ) + a 2 f ( t , y ) + α 2 ∂ y ( t , y ) + R h ∂ f ∂ t ( t , y ) + a 2 δ 2 f ( t , y ) ∂ f = ( a 1 + a 2 ) f ( t , y ) + a 2 α 2 ∂ y ( t , y ) + a 2 R Second order Taylor y k +1 − y k f ( t , y ) + h ∂ f ∂ t ( t , y ) + h 2 f ( t , y ) ∂ f = ∂ y ( t , y ) + R h 2 Therefore we need a 2 α 2 = h a 2 δ 2 = h a 1 + a 2 = 1 , 2 , 2 Which has infinitely many solutions of the form a 1 + a 2 = 1 and h α 2 = δ 2 = 2 a 2 4 / 4