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I am trying to figure out how to show that the zero solution to the following is asymptotically stable. $$x'=xy-x^3$$ $$y'=-y+x^3y$$

I was trying to use a liapunov function: $$V(x,y)=\frac{1}{2}[x^2+y^2]$$ Which is always positive. This gives us that $$V'(x,y)=xx'+yy'$$ $$=x(xy-x^3)+y(-y+x^3y)$$ $$=x^2y-x^4-y^2+x^3y^2$$

I know that I need this to be strictly negative for some domain surrounding and including the origin. I am having a hard time showing this. I have tried a couple other liapunov functions as well, but none have been successful either.

Thanks for any help! I am also open to other methods if there is something easier. When I linearized the system though, it seemed to be inconclusive.

I found the linearization to be $$ \begin{bmatrix} x'\\ y'\\ \end{bmatrix}= \begin{bmatrix} y && -\frac{x^3}{y}\\ -\frac{y}{x} && x^3 \\ \end{bmatrix} \begin{bmatrix} x\\ y\\ \end{bmatrix} $$ So I can't plug in $x=y=0$ to find the eigenvalues. Unless I am not doing that right...

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    $\begingroup$ What are the eigenvalues of the fixed point? Note that it is not always possible to find a Lyapunov function approach. You could try changing to polar coordinates and / or examine the system for any symmetries and reversors @MathIsHard $\endgroup$
    – Alex
    May 21 '18 at 23:24
  • $\begingroup$ @Alex I added some more details above about the linarization. Thanks for your comment. $\endgroup$
    – MathIsHard
    May 21 '18 at 23:35
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    $\begingroup$ Your linearisation cannot be correct - the vector field is a polynomial in $x$ and $y$, so its linearisation has to be a polynomial too @MathIsHard $\endgroup$
    – Alex
    May 21 '18 at 23:37
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    $\begingroup$ Let me note two things, which you should be able to use to show that the origin is asymptotically stable: $x=0$ is the stable manifold and $y=0$ is a center manifold (follow right away from the equations). So you only need to look at $x'=-x^3$ for which the origin is asymptotically stable. $\endgroup$
    – John B
    May 21 '18 at 23:44
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    $\begingroup$ Looks to me that your system has two equilibria, with e-values at the origin being $0$ and $-1$, hence you can use center manifold techniques @MathIsHard $\endgroup$
    – Alex
    May 21 '18 at 23:45
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This is an example of the interesting case of a null eigenvalue, aka a non hyperbolic critical point :

The Jacobian (aka the Linearisation matrix) of the given system is :

$$J(x,y) = \begin{bmatrix} y-3x^2 & x \\ 3x^2y & -1+x^3\end{bmatrix}$$

The linearisation matrix for the point $(x,y) = (0,0)$ is :

$$J(0,0) = \begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix}$$

We yield the eigenvalues :

$$\det(J(0,0) - \lambda I) = \begin{vmatrix} -\lambda & 0 \\ 0 & -1-\lambda \end{vmatrix}=0\Rightarrow\lambda(1+\lambda)=0 \Rightarrow \lambda =\begin{cases} 0 \\ -1\end{cases}$$

We notice that we have a null eigenvalue and also $\det(J(0,0)) = 0$, thus the origin $(x,y) = (0,0)$ is a non hyperbolic critical point.

We transform our system on the form :

$$\begin{bmatrix} x \\ y \end{bmatrix}' = \begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + f(x,y)$$

where $f(x,y)$ is a function, such $f(x,y) = (xy-x^3,x^3y) : \mathbb R^2 \to \mathbb R^2$.

Solving the equation :

$$-y(x) + f_2(x,y(x)) = 0 \Rightarrow -y(x) + x^3y(x) = 0 \Rightarrow y(x)(x^3-1)=0 \implies y(x) = 0$$

Finally :

$$f_1(x,y(x)) = f_1(x,0) = -x^3$$

The function $f_1(x,y(x))$ that we yielded is of the form $f_1(x,y(x)) = ax^k + \mathcal{O}(|x|^{k+1})$, since there is a continuous function $g(x)$ such that $g(x) = \mathcal{O}(|x|^{k+1})$ with $k=3$ and $g(x) = 0 \cdot |x|^4$.

Since $f_1(x,y(x)) = -x^3$, the coefficient of the $x$ order is a negative number and the power $k=3$ is an odd number. Thus, the origin $(x,y) = (0,0)$ is an asymptotically stable critical point for the given system.

Drawing a stream plot (phase portrait) for the given dynamical system, we also observe the asymptotic stability of the origin $(x,y) = (0,0)$ :

$\qquad \qquad \qquad \qquad \qquad \quad$ enter image description here

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  • $\begingroup$ In your Jacobian, why is the upper left term just $y$? $\endgroup$
    – imranfat
    May 22 '18 at 0:38
  • $\begingroup$ @imranfat Mistyped. Corrected and thanks. Doesn't change anything though. $\endgroup$
    – Rebellos
    May 22 '18 at 0:42
  • $\begingroup$ @Rebellos Thank you very much for the help. I appreciate your time. When you are talking about the sign and power of the $-x^3$, I got a little lost. Is there a theorem about this idea I could look up? It seems like that could be both negative and positive depending on the value of x, Is that the significance of that line or is it something else? Thanks again! $\endgroup$
    – MathIsHard
    May 22 '18 at 23:52
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    $\begingroup$ @MathIsHard Yes, there is a a theorem-lemma about this. If, by the method I elaborated, you yield an expression of the form : $$f_1(x,y(x)) = ax^k + \mathcal{O}(|x|^{k+1})$$ then if $a<0$ and $k$ is odd, then the origin $(0,0)$ is Asymptotically Stable while on any other case, the origin $(0,0)$ is Unstable. This is the significance of the minus sign, given that $a=-1$ is the coefficient of the $x$ term. $\endgroup$
    – Rebellos
    May 23 '18 at 9:45

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