y ′ ( t ) = f ( t , y ) , a ≤ t ≤ b , y ( a ) = αapply for the extrapolation techniquewe let h 0 = h 2 and use Euler method with w 0 = α w 1 = w 0 + h 0 f ( a , w 0 )and then midpoint method t i − 1 = a , t i = a + h 0 = a + h 2 to produceapproximation to y ( a + 2 h 0 ) w 2 = w 0 + 2 h 0 f ( a + h 0 , w 1 ) y 1 , 1 = 1 2 [ w 2 + w 1 + h 0 f ( a + 2 h 0 , w 2 ) ]then why this form results O ( h 0 2 ) approximation to y ( t 1 )

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y    ′    (  t  )  =  f  (  t  ,  y  )  ,  a  ≤  t  ≤  b  ,  y  (  a  )  =  αapply for the extrapolation techniquewe let       h    0    =      h    2   and use Euler method with       w    0    =  α      w    1    =      w    0    +      h    0    f  (  a  ,      w    0    )and then midpoint method      t          i      −      1        =  a  ,      t    i    =  a  +      h    0    =  a  +      h    2   to produceapproximation to   y  (  a  +  2      h    0    )      w    2    =      w    0    +  2      h    0    f  (  a  +      h    0    ,      w    1    )      y          1      ,      1        =      1    2    [      w    2    +      w    1    +      h    0    f  (  a  +  2      h    0    ,      w    2    )  ]then why this form results   O  (      h    0    2    ) approximation to   y  (      t    1    )

Michelle Chavez · Accepted Answer

This looks more like an order-blackuced RK3 method. Using the standard Runge-Kutta notation, you get                              k          1                                    =        h        f        (        a        ,                  w          0                )        ,                              w          1                                    =                  w          0                +                              1            2                                    k          1                                              k          2                                    =        h        f        (        a        +                              1            2                          h        ,                  w          0                +                              1            2                                    k          1                )        ,                              w          2                                    =                  w          0                +                  k          2                                              k          3                                    =        h        f        (        a        +        h        ,                  w          0                +                  k          2                )                                      y                      1            ,            1                                              =                              1            2                          (                  w          1                +                  w          2                +                              1            2                                    k          3                )                                          =                  w          0                +                              1            4                                    k          1                +                              1            2                                    k          2                +                              1            4                                    k          3                    This then gives the Butcher tableau                              c          1                                                    c          2                                      a                      21                                                        c          3                                      a                      31                                                a                      32                                                              b          1                                      b          2                                      b          3                      =                    0                                                        1            2                                                            1            2                                              1                    0                    1                                                        1            4                                                            1            2                                                            1            4                              where the first and second order conditions      b    1    +      b    2    +      b    3    =  1        c    1        b    1    +      c    2        b    2    +      c    3        b    3    =      1    2  are both satisfied.

Trevor Rush · Accepted Answer

The solution to the initial value problem has the series  y  (  a  +  h  )  =  α  +  f  (  a  ,  α  )  h  +                    f        t            (      a      ,      α      )      +      f      (      a      ,      α      )              f        y            (      a      ,      α      )        2        h    2    +  O  (      h    3    )where       f    t    =  ∂  f      /    ∂  t and       f    y    =  ∂  f      /    ∂  y. On the other hand, compute the series for       y          11       and you should also find      y          1      ,      1        =  α  +  f  (  a  ,  α  )  h  +                    f        t            (      a      ,      α      )      +      f      (      a      ,      α      )              f        y            (      a      ,      α      )        2        h    2    +  O  (      h    3    )

y′(t)=f(t,y), a<=t<=b, y(a)=alpha

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