Consider the IVP d y d t ( t ) = f ( t , y ( t ) ) , t ∈ [ a , b ] , y ( a ) = α ∈ R .One can show that Euler's method, i.e., the scheme y i + 1 = y i + h f ( t i . y i ) , y 0 = α , i = 0 , 1 , 2 , …has local truncation error O ( h 2 ) and global error O ( h ) .For higher order ODEs one can rewrite the ODE as a first order system and then apply Euler as before.Should the local truncation error still be O ( h 2 )?I tried this for a second order ODE y ″ ( t ) = f ( t , y ( t ) , y ′ ( t ) ) , t ∈ [ a , b ] , y ( a ) = α , y ′ ( a ) = βand arrived at y i + 2 = 2 y i + 1 − y i + h 2 f ( t i , y i , f ( t i , y i , y i + 1 − y i h ) ) .Then the local truncation error is y ( t i + 1 ) − y i + 1 = O ( h 3 ) .This of course contradicts my expectation that the error would still be O ( h 2 ).

Question

Consider the IVP            d      y              d      t        (  t  )  =  f  (  t  ,  y  (  t  )  )  ,    t  ∈  [  a  ,  b  ]  ,    y  (  a  )  =  α  ∈      R    .One can show that Euler&#039;s method, i.e., the scheme      y          i      +      1        =      y    i    +  h  f  (      t    i    .      y    i    )  ,        y    0    =  α  ,    i  =  0  ,  1  ,  2  ,  …has local truncation error       O    (      h    2    ) and global error       O    (  h  )  .For higher order ODEs one can rewrite the ODE as a first order system and then apply Euler as before.Should the local truncation error still be       O    (      h    2    )?I tried this for a second order ODE      y    ″    (  t  )  =  f  (  t  ,  y  (  t  )  ,      y    ′    (  t  )  )  ,    t  ∈  [  a  ,  b  ]  ,    y  (  a  )  =  α  ,        y    ′    (  a  )  =  βand arrived at      y          i      +      2        =  2      y          i      +      1        −      y    i    +      h    2    f      (          t      i        ,          y      i        ,    f          (              t        i            ,              y        i            ,                                    y                          i              +              1                                −                      y            i                          h            )        )    .Then the local truncation error is  y  (      t          i      +      1        )  −      y          i      +      1        =      O    (      h    3    )  .This of course contradicts my expectation that the error would still be       O    (      h    2    ).

Rayan Ali · Accepted Answer

Your two-step formula       y          i      +      2        −  2      y          i      +      1        +      y    i    =      h    2    f  (  .  .  .  ) actually requires two summations to get the values of       y    i  . So if the truncation error of it is   O  (      h    3    ), then the first summation gives an error of   O  (      h    2    ) and the second summation an error of   O  (  h  ) for the final result. There is no magical stuff happening.

Consider the IVP dy/dt (t)=f(t,y(t)),t in [a,b],y(a)=alpha in R.

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