I have shown that for the given ODE system, that when we apply the forward Euler method to something like y ′ = A y y ( t 0 ) = y 0 t ∈ ( t 0 , T ] where A is diagonalizable; A = P D P − 1 , that the solution given by forward Euler is u n = P ( I + h D ) n P − 1 u 0 where we consider the Forward Euler method to be y k + 1 = y k + h y ′ ( x k , y k ) . NNow, I want to show that if the real parts of λ k (the eigenvalues of A and thus the entries of D) are negative, then h &lt; − 2 R e λ k | λ k | 2 for all k implies that u n → 0.Now, in order to show this, based on what we have obtained for the formula for u n , we need only show that for each | 1 + h λ k | &lt; 1. I am having a really tough time doing this, and have mostly just tried to play around with AM-GM stuff.

Question

I have shown that for the given ODE system, that when we apply the forward Euler method to something like                                          y            ′                                              =        A                  y                                              y                (                  t                      0                          )                            =                  y                      0                                              t                            ∈        (                  t                      0                          ,        T        ]            where   A is diagonalizable;   A  =  P  D      P          −      1      , that the solution given by forward Euler is                              u                      n                          =        P        (        I        +        h        D                  )                      n                                    P                      −            1                                    u                      0                              where we consider the Forward Euler method to be       y          k      +      1        =      y          k        +  h      y    ′    (      x          k        ,      y          k        )  . NNow, I want to show that if the real parts of       λ          k       (the eigenvalues of   A and thus the entries of   D) are negative, then   h  &amp;lt;            −      2      R      e              λ                  k                                    |                    λ                  k                                      |                          2                     for all   k implies that       u          n        →  0.Now, in order to show this, based on what we have obtained for the formula for       u          n      , we need only show that for each       |    1  +  h      λ          k            |    &amp;lt;  1. I am having a really tough time doing this, and have mostly just tried to play around with AM-GM stuff.

beshrewd6g · Accepted Answer

You have to find the set       |    1  +  z      |    &amp;lt;  1 in the complex plane. Squaring this inequality results in  1  +  2    R  e  (  z  )  +      |    z            |        2    &amp;lt;  1    ⟺    R  e  (  z  )  &amp;lt;  −      1    2        |    z            |        2  Now insert   z  =  h            λ        k   and that   h is a real number...

Answered question

Answer & Explanation

New Questions in Integral Calculus