Using Lyapunov function, prove that a critical poin (0,0) is asymptotically stable.Let a linear system x ′ = − 2 x + x y 2 y ′ = − x 3 − y

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Using Lyapunov function, prove that a critical poin (0,0) is asymptotically stable.Let a linear system         x    ′    =  −  2  x  +  x        y      2             y    ′    =  −      x      3    −  y

Superina0xb4i · Accepted Answer

The most obvious try for a Lyapunov function designed to show local stability of a differential system at (0,0) might be  H  (  x  ,  y  )  =      x    2    +      y    2    .In the present case,        d              d      t        H  (  x  ,  y  )  =  −  4      x    2    −  2      y    2    +  2  x  y  (  x  y  −      x    2    )  .Using the bounds   |  x  y  −      x    2    |  ⩽  |  x  y  |  +      x    2   and   2  |  x  y  |  ⩽      x    2    +      y    2  , one sees that        d              d      t        H  (  x  ,  y  )  ⩽  −  4      x    2    −  2      y    2    +  (      x    2    +      y    2    )            1      2        (  3      x    2    +      y    2    )  .Let   D  =  {  (  x  ,  y  )  ∣      x    2    +      y    2    ⩽  2  }. For every   (  x  ,  y  ) in   D,        d              d      t        H  (  x  ,  y  )  ⩽  −  4      x    2    −  2      y    2    +  (  3      x    2    +      y    2    )  =  −  (      x    2    +      y    2    )  ,that is,        d              d      t        H  (  x  ,  y  )  ⩽  −  H  (  x  ,  y  )  .Hence, if   (  x  (  0  )  ,  y  (  0  )  ) is in   D, then   (  x  (  t  )  ,  y  (  t  )  ) stays in   D for every   t  ⩾  0 and, solving the differential inequality       u    ′    ⩽  −  u, one sees that  H  (  x  (  t  )  ,  y  (  t  )  )  ⩽  H  (  x  (  0  )  ,  y  (  0  )  )        e          −      t        .In particular   H  (  x  (  t  )  ,  y  (  t  )  )  →  0 hence   (  x  (  t  )  ,  y  (  t  )  )  →  (  0  ,  0  ) when   t  →  ∞.

Using Lyapunov function, prove that a critical poin (0,0) is asymptotically stable. Let a linear sy

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