An initial value problem x ′ = A x A = ( 1 1 0 0 3 − 1 0 0 0 0 0 − 2 0 0 2 0 )

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An initial value problem             x        ′    =  A      x    A  =      (                            1                          1                          0                          0                                      3                          −          1                          0                          0                                      0                          0                          0                          −          2                                      0                          0                          2                          0                      )

hoatryearnenizts · Accepted Answer

Observe that the matrix   A has block form      (                            B                          0                                      0                          C                      )  This implies that   (      x    1    ,      x    2    ) is driven by   2  ×  2 matrix   B, while   (      x    3    ,      x    4    ) is driven by matrix   C; there is no interaction between these processes.Since the eigenvalues of   C are   ±  2  i all its orbits are periodic. That is,   (      x    3    ,      x    4    ) are periodic no matter what the initial condition.Since the eigenvalues of   A have nonzero real part, none of its orbits are periodic... except the trivial, zero solution, which is periodic to every period.Hence, the conclusion: solution is periodic if and only if the starting point has the first two coordinates zero.Using the matrix exponential isn&#039;t really necessary. But if you want to:      e          t      A        =      (                                        e                          t              B                                                0                                      0                                      e                          t              C                                            )  where       e          t      C       is sometimes the identity, but       e          t      B       never is, except at   t  =  0.

An initial value problem x′=Ax, A=(1 3 0 0, 1 -1 0 0, 0 0 0 2, 0 0 -2 0)

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