Show that two solutions x n and y n generated by implicit Euler satisfy the inequality | x 0 − y 0 | .Can someone push me in the right direction by explaining to me how exactly implicit Euler's method works? I understand that it is some iteration method where u k + 1 = u k + h f ( t k + 1 , u k + 1 ) .And the u k + 1 is often found by some other iteration method (kind of shaky on this part here). I was wondering if there were some way to manipulate the equation to rewrite x n = x n + 1 − h f ( t k + 1 , x k + 1 ) and y n similarly but I'm not sure if that would get anything done. We are also given that f satisfies the condition where ( x − y ) ( f ( t , x ) − f ( t , y ) ) ≤ 0, which means | x ( t ) − y ( t ) | ≤ | x ( 0 ) − y ( 0 ) | . Any help would be greatly appreciated. Thank you!

Question

Show that two solutions       x    n   and       y    n   generated by implicit Euler satisfy the inequality       |        x    0    −      y    0        |  .Can someone push me in the right direction by explaining to me how exactly implicit Euler&#039;s method works? I understand that it is some iteration method where      u          k      +      1        =      u    k    +  h  f  (      t          k      +      1        ,      u          k      +      1        )  .And the       u          k      +      1       is often found by some other iteration method (kind of shaky on this part here). I was wondering if there were some way to manipulate the equation to rewrite       x    n    =      x          n      +      1        −  h  f  (      t          k      +      1        ,      x          k      +      1        ) and       y    n   similarly but I&#039;m not sure if that would get anything done. We are also given that f satisfies the condition where   (  x  −  y  )  (  f  (  t  ,  x  )  −  f  (  t  ,  y  )  )  ≤  0, which means       |    x  (  t  )  −  y  (  t  )      |    ≤      |    x  (  0  )  −  y  (  0  )      |  . Any help would be greatly appreciated. Thank you!

Tianna Deleon · Accepted Answer

If you just write down the difference of the two iterations and multiply with the right difference of arguments, you find that                    (                  x          +                −                  y          +                          )          2                                    =        (                  x          +                −                  y          +                )        (        x        −        y        )        +        h        (                  x          +                −                  y          +                )        (        f        (                  t          +                ,                  x          +                )        −        f        (                  t          +                ,                  y          +                )        )                                          ≤        (                  x          +                −                  y          +                )        (        x        −        y        )            Now dividing by       |        x    +    −      y    +        |   gives the desired result in one step and thus also in the general case.Using Cauchy-Schwarz, this can also be extended to vector-valued ODE.

Show that two solutions x n </msub> and y n </msub> generate

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