Show that x ∗ = ( 1 , 2 ) is a fixed point of the system x 1 ′ = 2 + 3 x 1 − 2 x 2 − x 1 2 + 2 x 1 x 2 − x 2 2 x 2 ′ = 3 + 4 x 1 − 3 x 2 − x 1 2 + 2 x 1 x 2 − x 2 2 Determine W s ( x ) and W u ( x ), the global stable and unstable manifolds of the fixed point x = ( 1 , 2 ) and give a parametric representation of those manifolds.

Question

Show that       x    ∗    =  (  1  ,  2  ) is a fixed point of the system      x    1    ′    =  2  +  3      x    1    −  2      x    2    −      x    1    2    +  2      x    1        x    2    −      x    2    2        x    2    ′    =  3  +  4      x    1    −  3      x    2    −      x    1    2    +  2      x    1        x    2    −      x    2    2  Determine       W    s    (  x  ) and       W    u    (  x  ), the global stable and unstable manifolds of the fixed point   x  =  (  1  ,  2  ) and give a parametric representation of those manifolds.

Ronald Hickman · Accepted Answer

Plug in   (  1  ,  2  ) to see that you get   0, so the point is an equilibrium.Calculate the jacobian of the right hand side:  D  f  (      x    1    ,      x    2    )  =      (                            3          −          2                      x            1                    +          2                      x            2                                    −          2          +          2                      x            1                    −          2                      x            2                                                4          −          2                      x            1                    +          2                      x            2                                    −          3          +          2                      x            1                    −          2                      x            2                                )  plug in our equilibrium point:  D  f  (  1  ,  2  )  =      (                            3          −          2          +          4                          −          2          +          2          −          4                                      4          −          2          +          4                          −          3          +          2          −          4                      )    =      (                            5                          −          4                                      6                          −          5                      )  Eigenvalues:  (  5  −  λ  )  (  −  5  −  λ  )  +  24  =  −  (  5  −  λ  )  (  5  +  λ  )  +  24  =      λ    2    −  1  =  0Thus   λ  =  ±  1 since our first eigenvalue is negative, constitutes to the stable manifold, and the other is positive, and constitutes to the unstable manifolds .Now our corresponding eigenvectors are   λ  =  1  ⇒  [  1  ,  1  ] and   λ  =  −  1  ⇒  [  1  ,      2    3    ]So our stable subspace:      E    s    = = span  {eigenvector, for   λ  =  −  1  }=  span  ⁡  {  [  1  ,      2    3    ]  }  =  {  (  x  ,  y  )  ∈        R    2       |  y  =      2    3    x  }And unstable subspace:      E    u    =  {eigenvector, for   λ  =  1  } =   span  ⁡  {  [  1  ,  1  ]  }  =  {  (  x  ,  y  )  ∈        R    2       |  y  =  x  }Then the stable and unstable manifolds       W    s    ,      W    u   are then tangential to       E    s    ,      E    u  , at   (  1  ,  2  ), But I am unsure on how to parameterise them, but they must depend and   t. And   l  i      m          t      →      ∞            W    s    (  t  )  =  (  1  ,  2  ), and   l  i      m          t      →      −      ∞            W    u    (  t  )  =  (  1  ,  2  ).