Let y ′ ( s ) = y 2 ( s ) , y ( 0 ) = 1 be an ode on s ∈ R + ∖ { 1 }. Let h n = s n + 1 − s n . Prove the following inequality with the Euler method ∀ n : s n &lt; 1 the following holds y n ≤ 1 1 − s n The explicit Euler method is: y n + 1 = y n + h n f ( s n , y n )The problem is, I don't know if I have to use somewhere Gronwall's lemma or if it is straight calculation. The solution to the ODE is y ( s ) = 1 1 − s and here is what I've tried so far y n + 1 = y n + h n f ( s n , y n )becomes y n = y n + 1 − h n f ( s n , y n ) y n = y n + 1 − ( s n + 1 − s n ) y n 2 But then I get stucked, since I don't see how I can remove the y n + 1 and s n + 1 terms without getting new recursive terms. Or do I have to use the geometric series?

Question

Let       y    ′    (  s  )  =      y    2    (  s  )  ,  y  (  0  )  =  1 be an ode on   s  ∈            R        +    ∖  {  1  }. Let       h    n    =      s          n      +      1        −      s    n  . Prove the following inequality with the Euler method   ∀  n  :      s    n    &amp;lt;  1 the following holds      y    n    ≤      1          1      −              s        n            The explicit Euler method is:      y          n      +      1        =      y    n    +      h    n    f  (      s    n    ,      y    n    )The problem is, I don&#039;t know if I have to use somewhere Gronwall&#039;s lemma or if it is straight calculation. The solution to the ODE is   y  (  s  )  =      1          1      −      s       and here is what I&#039;ve tried so far      y          n      +      1        =      y    n    +      h    n    f  (      s    n    ,      y    n    )becomes      y    n    =      y          n      +      1        −      h    n    f  (      s    n    ,      y    n    )      y    n    =      y          n      +      1        −  (      s          n      +      1        −      s    n    )      y    n    2  But then I get stucked, since I don&#039;t see how I can remove the       y          n      +      1       and       s          n      +      1       terms without getting new recursive terms. Or do I have to use the geometric series?

Olive Guzman · Accepted Answer

Maybe a proof by induction. You have, since       s    0    =  0,      y    0    =  1  =      1          1      −              s        0              .If      y    n    ≤      1          1      −              s        n            then                              y                      n            +            1                                              =                  y          n                +                  h          n                (                  y          n                          )          2                                                  ≤                  1                      1            −                          s              n                                      +                  h          n                          1                      (            1            −                          s              n                                      )              2                                                                        =                  1                      1            −            (                          s                              n                +                1                                      −                          h              n                        )                          +                  h          n                          1                      (            1            −            (                          s                              n                +                1                                      −                          h              n                        )                          )              2                                                                        =                  1                      1            −                          s                              n                +                1                                                    −                              h            n            2                                (            1            −                          s                              n                +                1                                      )            (            1            −                          s                              n                +                1                                      +                          h              n                                      )              2                                                                        ≤                  1                      1            −                          s                              n                +                1                                                    .            We are done.

retopire07c · Accepted Answer

In recursions of the form       x          n      +      1        =      x    n    +      c    n        x    n    2   you can consider the reciprocal series       u    n    =  1      /        x    n  ,       u    n        x    n    =  1, and use the geometric series/binomial identities to find helpful inequalities like      u          n      +      1        =      1                  x        n            (      1      +              c        n                    x        n            )        =      u    n              1      −              c        n                    x        n                    1      −              c        n        2                    x        n        2              &amp;gt;      u    n    −      c    n  Summation of this gives      u    n    &amp;gt;      u    0    −      ∑          k      =      0              n      −      1            c    k  so that in the original series      x    n    &amp;lt;            x      0              1      −              x        0                    ∑                  k          =          0                          n          −          1                            c        k

Let <msup> y ′ </msup> <mo stretchy="false">( s <mo stretchy="false

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