Using the Lyapunov Function V = 1 2 x 1 2 + 1 2 x 2 2 , prove that the omega(ω)-limit set is non-empty for any initial value given for the dynamical system: x 1 ′ = x 1 + 2 x 2 − 2 x 1 ( x 1 2 + x 2 2 ) 2 x 2 ′ = 4 x 1 + 3 x 2 − 3 x 2 ( x 1 2 + x 2 2 ) 2

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Using the Lyapunov Function   V  =      1    2        x    1    2    +      1    2        x    2    2  , prove that the omega(ω)-limit set is non-empty for any initial value given for the dynamical system:      x    1    ′    =      x    1    +  2      x    2    −  2      x    1    (      x    1    2    +      x    2    2        )    2        x    2    ′    =  4      x    1    +  3      x    2    −  3      x    2    (      x    1    2    +      x    2    2        )    2

priffEmipsy4i37m · Accepted Answer

The Cauchy&#039;s inequality can be rewriten in the form  (      x    1    ⋅  1  +      x    2    ⋅  1      )    2    ≤  (      x    1    2    +      x    2    2    )  ⋅  (      1    2    +      1    2    )or  (      x    1    +      x    2        )    2    ≤  2  (      x    1    2    +      x    2    2    )  .This implies that      V    ˙    ≤  3  (      x    1    +      x    2        )    2    −  2  (      x    1    2    +      x    2    2        )    3    ≤  6  (      x    1    2    +      x    2    2    )  −  2  (      x    1    2    +      x    2    2        )    3  It is easy to check that   ∀  (      x    1    ,      x    2    )  :        x    1    2    +      x    2    2    &amp;gt;      3             V    ˙    &amp;lt;  0, thus the solution for any initial value enters the bounded set      Ω    C    =      {    (          x      1        ,          x      2        )    :              x      1      2        +          x      2      2        &amp;lt;    C    }  in finite time for any   C  &amp;gt;      3   and stays there forever, hence any solution is bounded and, due to the Bolzano–Weierstrass theorem, has a nonempty   ω-limit set.

Using the Lyapunov Function V = 1 2 </mfrac> <msubsup> x 1

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