Is the solution stable x ′ = − x − 2 y + x 2 y 2 and y ′ = x − 1 2 y − 1 2 x 3 yI have to check whether zero solution (solution for x = y = 0) is stable. Is it Lyapunov stable? Is it asymptotically stable?

Question

Is the solution stable      x    ′    =  −  x  −  2  y  +      x    2        y    2   and       y    ′    =  x  −      1    2    y  −      1    2        x    3    yI have to check whether zero solution (solution for   x  =  y  =  0) is stable. Is it Lyapunov stable? Is it asymptotically stable?

pheniankang · Accepted Answer

{                                                x            ˙                    =          −          x          −          2          y          +          ϕ          (          x          ,          y          )                                                          y            ˙                    =          x          −                      y            2                    +          ψ          (          x          ,          y          )                        The functions   ψ  (  x  ,  y  ) and   ϕ  (  x  ,  y  ) can be ignored (  x  →  0,   y  →  0) at infinitesimal neighborhood of zero. So...      |                            −          1          −          λ                          −          2                                      1                          −          0.5          −          λ                      |        λ    2    +      3    2    λ  +      5    2    =  0      λ    1    =            −      3        4    +                    31            i        4    =  0  ;      λ    2    =            −      3        4    −                    31            i        4    =  0If the all of   λ&#039;s real parts (  R  e  λ) values are below zero, then the zero-solution   y  ≡  0 is asymptotically stable.In this case, it is stable.For the Lyapunov stability investigation the Lyapunov function must be created. It&#039;s look is   V  =  a      x    2    +  b      y    2    ,  a  &amp;gt;  0  ,  b  &amp;gt;  0 - are any parameters one likes.            d      V              d      t        =            d      V              d      x                  d      x              d      t        +            d      V              d      y                  d      y              d      t        =  2  a  x  (  −  x  −  2  y  +      x    2        y    2    )  +  2  b  y  (  x  −      y    2    −      1    2        x    3    y  )  =  −  (  2  a      x    2    +  b      y    2    )  +  (  2  x  y  −      x    3        y    2    )  (  b  −  2  a  )  .Then   b  =  2  a, it easy to see, that             d      V              d      t        =  −  2  a  (      x    2    +      y    2    )  ≤  0.  V  =  a      x    2    +  2  a      y    2    ,  a  &amp;gt;  0- positive definite function and it&#039;s derivative is a negative definite one - so it was a formulation of the Lyapunov theorem about stability

Craig Mendoza · Accepted Answer

I had the same problem, thanks!

Is the solution stable x ′ </msup> = − x &#x2

Answered question

Answer & Explanation

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