Say we have a function y ( t ), that satisfies the ordinary differential equation d y d t = f ( t , y ) for t ∈ ( t 0 , t max ] ,, where t takes discrete values, t n , with a constant step size h = t n + 1 − t n . We also are given the initial condition, for example, y ( 0 ) = 1.If we were to use the Backward Euler Method: y n + 1 ≈ y n + h f ( t n + 1 , y n + 1 ) ,to approximate a solution for a function, say, f ( y ) = − y, I am told that it is possible to use this information to show that the Backward Euler Scheme is first order, by expanding the global error as a Taylor series in powers of h. (The global error at fixed t is the difference between the approximate solution and the exact solution, y ( t ))Can anyone figure out how this might be possible? I cannot find any resources online that dmeonstrate this idea.

Question

Say we have a function   y  (  t  ), that satisfies the ordinary differential equation                     d            y                      d            t        =  f  (  t  ,  y  )    for   t  ∈  (      t    0    ,      t          max        ]  ,, where   t takes discrete values,       t    n  , with a constant step size   h  =      t          n      +      1        −      t    n  . We also are given the initial condition, for example,   y  (  0  )  =  1.If we were to use the Backward Euler Method:      y          n      +      1        ≈      y    n    +  h  f  (      t          n      +      1        ,      y          n      +      1        )  ,to approximate a solution for a function, say,   f  (  y  )  =  −  y, I am told that it is possible to use this information to show that the Backward Euler Scheme is first order, by expanding the global error as a Taylor series in powers of   h. (The global error at fixed   t is the difference between the approximate solution and the exact solution,   y  (  t  ))Can anyone figure out how this might be possible? I cannot find any resources online that dmeonstrate this idea.

Daniella Cochran · Accepted Answer

You get      y    n    =  (  1  +  h      )          −      n            y    0    =  exp  ⁡  (  −  n  ln  ⁡  (  1  +  h  )  )  =  exp  ⁡  (  −  n  h  +            1      2        n      h    2    +  O  (  n      h    3    )  )  =  exp  ⁡  (  −  t  +            1      2        t  h  +  t  ·  O  (      h    2    )  )so that the first error term is of order 1.

Say we have a function y(t), that satisfies the ordinary differential equation dy/dt=f(t,y) for t in (t_0,t_max], where t takes discrete values, tn, with a constant step size h=t_n+1−t_n. We also are given the initial condition, for example, y(0)=1.

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