Approximating y ′ ( t n ) at the relation y ′ ( t n ) = f ( t n , y ( t n ) ) with the difference quotient [ y ( t n + 1 ) − y ( t n ) h ] we get to the Euler method.Approximating the same derivative with the quotient [ y ( t n ) − y ( t n − 1 ) h ] we get to the backward Euler method y n + 1 = y n + h f ( t n + 1 , y n + 1 ) , n = 0 , … , N − 1where y 0 := y 0 .In order to find the formula for the forward Euler method, we use the limit lim h → 0 y ( x 0 + h ) − y ( x 0 ) h for x 0 = t n , h = t n + 1 − t n .In order to find the formula for the backward Euler method, could we pick h = t n − 1 − t n although it is negative?Or how do we get otherwise to the approximation: y ′ ( t n ) ≈ y ( t n ) − y ( t n − 1 ) h

Question

Approximating       y    ′    (      t    n    ) at the relation       y    ′    (      t    n    )  =  f  (      t    n    ,  y  (      t    n    )  ) with the difference quotient       [                  y        (                  t                      n            +            1                          )        −        y        (                  t          n                )            h        ]   we get to the Euler method.Approximating the same derivative with the quotient       [                  y        (                  t                      n                          )        −        y        (                  t                      n            −            1                          )            h        ]   we get to the backward Euler method      y          n      +      1        =      y    n    +  h  f  (      t          n      +      1        ,      y          n      +      1        )  ,  n  =  0  ,  …  ,  N  −  1where       y    0    :=      y    0  .In order to find the formula for the forward Euler method, we use the limit       lim          h      →      0                  y      (              x        0            +      h      )      −      y      (              x        0            )        h   for       x    0    =      t    n    ,  h  =      t          n      +      1        −      t    n  .In order to find the formula for the backward Euler method, could we pick   h  =      t          n      −      1        −      t    n   although it is negative?Or how do we get otherwise to the approximation:      y    ′    (      t    n    )  ≈            y      (              t        n            )      −      y      (              t                  n          −          1                    )        h

iljovskint · Accepted Answer

obviously, you still choose       t          n      +      1        =      t    n    +      h    n  , thus   h  =      h    n    =      t          n      +      1        −      t    n   for the step from       y    n   to       y          n      +      1      .The mean value theorem only tells us that            y      (              t                  n          +          1                    )      −      y      (              t        n            )                      t                  n          +          1                    −              t        n              =      y    ′    (      t    n    +  θ  (      t          n      +      1        −      t    n    )  )where for most of the usual functions       θ    ∈  (  0  ,  1  ) can be found close to 1/2. Setting, for approximation purposes,       θ    =  0 or       θ    =  1 has thus about the same degree of inaccuracy.

Approximating y′(t^n) at the relation y′(t^n)=f(t^n,y(t^n)) with the difference quotient [y(t^(n+1))−y(t^n)/h] we get to the Euler method.

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