Let x ′ ( t ) = f ( t , x ( t ) ) , t ∈ ( 0 , T ) with x ( 0 ) = x 0 f satifies the Lipschitz-condition f ( t , x ) − f ( t , y ) ≤ L | x − y | h ∈ ( 0 , 1 L ) is the step size and the approximation x k for x ( t k ) = h k is given by x k = x k − 1 + h f ( t k , x k ).Now I would be very interested how to derive the error | x k − x ( t k ) | ≤ 1 1 − L h ( | x k − 1 − x ( t k − 1 ) | + h 2 2 max s ∈ [ 0 , T ] | x ″ ( s ) | ) I tried to look up it up in some numerical analysis books but it is always different

Question

Let      x    ′    (  t  )  =  f  (  t  ,  x  (  t  )  )  ,  t  ∈  (  0  ,  T  ) with   x  (  0  )  =      x    0    f satifies the Lipschitz-condition   f  (  t  ,  x  )  −  f  (  t  ,  y  )  ≤  L      |    x  −  y      |    h  ∈  (  0  ,      1    L    ) is the step size and the approximation       x    k   for   x  (      t    k    )  =  h  k is given by       x    k    =      x          k      −      1        +  h  f  (      t    k    ,      x    k    ).Now I would be very interested how to derive the error      |        x    k    −  x  (      t    k    )      |    ≤      1          1      −      L      h            (          |              x              k        −        1              −    x    (          t              k        −        1              )          |        +                  h        2            2              max              s        ∈        [        0        ,        T        ]                    |              x      ″        (    s    )          |        )  I tried to look up it up in some numerical analysis books but it is always different

Edgeriecoereexq · Accepted Answer

First, we get local truncation error.  x  (      t          k      +      1        )  =  x  (      t    k    )  +  h  f  (      t    k    ,  x  (      t    k    )  )  +      τ    k        τ    k    =  x  (      t          k      +      1        )  −  x  (      t    k    )  −  h  f  (      t    k    ,  x  (      t    k    )  )  =            h      2        2        x    ″    (  η  ). Where   η  ∈  (      t    k    ,      t          k      +      1        ).Then we get the bound,      |    x  (      t          k      +      1        )  −      x          k      +      1            |    ≤  (  1  +  h  L  )      |    x  (      t          k        )  −      x          k            |    +      |        τ    k        |    ≤  (  1  +  h  L  )      |    x  (      t          k        )  −      x          k            |    +            h      2        2        max          s      ∈      (      0      ,      T      )            |        x    ″    (  s  )      |    ≤      1          1      −      h      L            |    x  (      t          k        )  −      x          k            |    +            h      2        2        max          s      ∈      (      0      ,      T      )            |        x    ″    (  s  )      |  Where the last step is from geometric series. Maybe it would be helpful if you listed the other results you are talking about, and then we can show that they&#039;re equivalent.

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