I am trying to show that, for the equation y ′ + α y = 0alternating between a forward Euler method step for y 2 n and a backward Euler step for y 2 n + 1 with time-step h is equivalent to y n + 1 = 1 − α h 1 + α h y n I have that y 1 = y 0 + h ( − α y 0 ) = ( 1 − α h ) y 0 How exactly ( 1 + α h ) works into this I don't see. If I just press forward anyway, using the trapezoid rule for the backward Euler step next, and call y ^ n the approximation of y n using the forward Euler method, I get y 2 = y 1 + h ( y 1 + y ^ 2 2 ) I'm kind of doing this a priori but that seems right to me: the trapezoid rule is to multiply the change in the x-axis by the average of the bases which are y 1 and y 2 . This becomes y 2 = y 1 + h ( y 1 + ( 1 − α h ) y 1 2 ) = ( 1 + h ( 2 − α h 2 ) ) y 1 This doesn't seem to be shaping up to the form I'm supposed to be getting. Am I doing something wrong?

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I am trying to show that, for the equation      y    ′    +  α  y  =  0alternating between a forward Euler method step for       y          2      n       and a backward Euler step for       y          2      n      +      1       with time-step   h is equivalent to      y          n      +      1        =            1      −      α      h              1      +      α      h            y    n  I have that      y    1    =      y    0    +  h  (  −  α      y    0    )  =  (  1  −  α  h  )      y    0  How exactly   (  1  +  α  h  ) works into this I don&#039;t see. If I just press forward anyway, using the trapezoid rule for the backward Euler step next, and call                     y        ^              n   the approximation of       y    n   using the forward Euler method, I get      y    2    =      y    1    +  h      (                            y          1                +                                            y              ^                                2                    2        )  I&#039;m kind of doing this a priori but that seems right to me: the trapezoid rule is to multiply the change in the   x-axis by the average of the bases which are       y    1   and       y    2  . This becomes      y    2    =      y    1    +  h      (                            y          1                +        (        1        −        α        h        )                  y          1                    2        )    =      (    1    +    h          (                        2          −          α          h                2            )        )        y    1  This doesn&#039;t seem to be shaping up to the form I&#039;m supposed to be getting. Am I doing something wrong?

Isaac Molina · Accepted Answer

Let   f  (  z  )  =  −  α  z, so       y    ′    =  f  (  y  ).The forward Euler method is      y          n      +      1        =      y    n    +  h  f  (      y    n    )  =      y    n    −  h  α      y    n  , and the the backwards Euler method is       y          n      +      1        =      y    n    +  h  f  (      y          n      +      1        )  =      y    n    −  h  α      y          n      +      1      So we have       y    2    =      y    1    +  h  f  (      y    2    )  =      y    1    −  h  α      y    2   and       y    1    =      y    0    +  h  f  (      y    0    )  =      y    0    −  h  α      y    0  So putting this to gether we get   (  1  +  h  α  )      y    2    =      y    1    =  (  1  −  h  α  )      y    0  , or      y    2    =            1      −      h      α              1      +      h      α            y    0  Now you can generalize this and fin      y          2      n        =            1      −      h      α              1      +      h      α            y          2      (      n      −      1      )      so in your solution   h probably only refers to half a time step.

Averi Fields · Accepted Answer

Its great, thank you

I am trying to show that, for the equation y′+alpha y=0 alternating between a forward Euler method step for y_2n and a backward Euler step for y_(2n+1) with time-step h is equivalent to y_(n+1)=(1−alpha h)/(1+alpha h)y_n

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