I was just doing some practice questions for a test, but have been stumped by the following for the past couple of hours.I'm given a system such that: d u d t = v &amp; d v d t = − f ( u )with Hamiltonian H = 1 2 ( d u d t ) 2 + ∫ f d u .I have to show that using the forward Euler method leads to a global error for H that grows like n h 2 for step size h and number of steps n.I know that the global error can be calculated via ϵ = | U n − U ( T ) | but I'm not sure how to apply it in this case.Thanks for any help!EDIT: So if I understand correct, I have to calculate | H n + 1 − H n |

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I was just doing some practice questions for a test, but have been stumped by the following for the past couple of hours.I&#039;m given a system such that:            d      u              d      t        =  v        &amp;amp;                  d      v              d      t        =  −  f  (  u  )with Hamiltonian  H  =      1    2              (                        d          u                          d          t                    )        2    +  ∫  f  d  u  .I have to show that using the forward Euler method leads to a global error for   H that grows like   n      h    2   for step size   h and number of steps   n.I know that the global error can be calculated via   ϵ  =      |        U    n    −  U  (  T  )      |   but I&#039;m not sure how to apply it in this case.Thanks for any help!EDIT: So if I understand correct, I have to calculate       |        H          n      +      1        −      H    n        |

Skylar French · Accepted Answer

If you take the example   f  (  u  )  =  u, then you can interpret the Hamilton system in the complex plane as            z      ˙        =            u      ˙        +  i            v      ˙        =  −  i  (  u  +  i  v  )  =  −  i  z  .The Euler forward iteration thus produces elements      z          k      +      1        =      z    k    +  h  (  −  i      z    k    )  =  (  1  −  i  h  )      z    k      ⟹        z    n    =  (  1  −  i  h      )    n        z    0    .The Hamilton function is   H  =      1    2    (      v    2    +      u    2    )  =      1    2        |    z            |        2  , which gives on the Euler solution      H    n    =      1    2        |        z    n              |        2    =      1    2    (  1  +      h    2        )    n        |        z    0              |        2    =      e          n              h        2            +      O      (      n              h        4            )            H    0  so that indeed       H    n    −      H    0    =  n      h    2        e                  1        2            n              h        2            +      O      (      n              h        4            )            H    0  Under more general conditions one gets                              H                      k            +            1                                              =                  1          2                (                  v          k                −        h        f        (                  u          k                )                  )          2                +        F        (                  u          k                +        h                  v          k                )                                          =                  H          k                −        h                  v          k                f        (                  u          k                )        +                  1          2                          h          2                f        (                  u          k                          )          2                         +                 f        (                  u          k                )        (        h                  v          k                )        +                  1          2                          f          ′                (                  u          k                )        (        h                  v          k                          )          2                +        O        (                  h          3                )                                          =                  H          k                +                  1          2                          h          2                                      [                          f        (                  u          k                          )          2                +                  f          ′                (                  u          k                )                  v          k          2                                      ]                          +        O        (                  h          3                )            Now you have to argue that the coefficient of the       h    2   term remains bounded and the claim of the task follows.

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