Suppose we have a system of ODE's: a a ′ = − a − 2 b and b ′ = 2 a − b with initial conditions a ( 0 ) = 1 and b ( 0 ) = − 1.How can we find the maximum value of the step size such that the norm a solution of the system goes to zero (if we apply the forward Euler formula)?Edit: the main part is to calculate the eigenvalues of the following matrix, based on the Euler method, this becomes ( − 1 − h − 2 − 2 h 2 + 2 h − 1 − h ) The eigenvalues are ( − 1 + 2 i ) ( 1 + h ) and ( − 1 − 2 i ) ( 1 + h )

Question

Suppose we have a system of ODE&#039;s: a      a    ′    =  −  a  −  2  b and       b    ′    =  2  a  −  b with initial conditions   a  (  0  )  =  1 and   b  (  0  )  =  −  1.How can we find the maximum value of the step size such that the norm a solution of the system goes to zero (if we apply the forward Euler formula)?Edit: the main part is to calculate the eigenvalues of the following matrix, based on the Euler method, this becomes      (                            −          1          −          h                          −          2          −          2          h                                      2          +          2          h                          −          1          −          h                      )  The eigenvalues are   (  −  1  +  2  i  )  (  1  +  h  ) and   (  −  1  −  2  i  )  (  1  +  h  )

Madilyn Dunn · Accepted Answer

The Euler discretizations of a differential system      {                                        a            ′                    (          t          )                                    =                          f          (          a          (          t          )          ,          b          (          t          )          )                                                  b            ′                    (          t          )                                    =                          g          (          a          (          t          )          ,          b          (          t          )          )                        are based on the difference systems      {                                        a                          n              +              1                                                          =                                      a                          n                                +          h          ⋅          f          (                      a            n                    ,                      b            n                    )                                                  b                          n              +              1                                                          =                                      b            n                    +          h          ⋅          g          (                      a            n                    ,                      b            n                    )                        for some positive step size   h. In the present case, this blackuces to      (                                        a                          n              +              1                                                                        b                          n              +              1                                            )    =      M    h    ⋅      (                                        a                          n                                                                        b                          n                                            )    ,where      M    h    =      (                            1          −          h                          −          2          h                                      2          h                          1          −          h                      )    .The eigenvalues of       M    h   are  1  −  h  ±  2      i    h  ,hence the square of their common modulus is  (  1  −  h      )    2    +  (  2  h      )    2    =  1  −  h  (  2  −  5  h  )  .When both eigenvalues of       M    h   have modulus less than 1, then   (      a    n    ,      b    n    )  →  (  0  ,  0  ) for every starting point   (      a    0    ,      b    0    ). When this modulus is at least 1, then   (      a    n    ,      b    n    )  →  (  0  ,  0  ) never happens except when   (      a    0    ,      b    0    )  =  (  0  ,  0  ) (this is because in the present situation both eigenvalues have the same modulus).Thus,   (      a    n    ,      b    n    )  →  (  0  ,  0  ) for every starting point   (      a    0    ,      b    0    ) when  0  &amp;lt;  h  &amp;lt;      2    5    .Note that the eigenvalues of the linear differential system are   λ  =  −  1  ±  2      i   such that   ℜ  λ  =  −  1 and       |    λ            |        2    =  5. More generally, for a linear differential system with eigenvalues   λ such that   ℜ  λ  &amp;lt;  0 for every   λ, Euler discretizations yield sequences with limit 0 for every starting point and every positive step size   h such that  h  &amp;lt;      min    λ        (    −    2                  ℜ        λ                              |                λ                              |                    2                      )    .

Nica2t · Accepted Answer

You need to compute the eigenvalues of the system. As a shortcut consider that  (  a  +  i  b      )    ′    =  (  2  i  −  1  )  (  a  +  i  b  )

Suppose we have a system of ODE's: a′=−a−2b and b′=2a−b with initial conditions a(0)=1 and b(0)=−1. How can we find the maximum value of the step size such that the norm a solution of the system goes to zero (if we apply the forward Euler formula)?

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