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Given the system $$x' = x(1-r)+y(x-r) \\ y' = y(1-r)+x(r-x)\\$$ where $r^2=x^2+y^2$. I don't know how to use the definition of stability of equilibrium to show that the equilibrium $(1,0)$ is unstable.

Definition: An equilibrium $\bf{\bar{x}}$ of an ODE is stable if for all $\epsilon > 0$, there exists $\delta >0$ such that $||\bf{x_0}-\bf{\bar{x}}||<\delta$ implies $||\bf{x(t)}-\bf{\bar{x}}||<\epsilon$ for all $t \geq 0$.

I have evaluated $$r(t)=\frac{r(0)e^t}{1-r(0)+r(0)e^t}$$ and $\\||\bf{x(t)}-\bf{\bar{x}}|| $$=\sqrt{(x(t)-1)^2+(y(t))^2}=\sqrt{(r(t))^2-2x(t)+1}$ and $||\bf{x_0}-\bf{\bar{x}}||$$=\sqrt{(r(0))^2-2x(0)+1}$. But I don't know the next step.

Arctic Char
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Tammy
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  • Is it possible that this problem, to show a particular equilibrium is not stable, was intended to have you apply a proposition that was recently covered, rather than going back to the definition? I didn't follow how you "evaluated $r(t)$", but presumably you reduced the dynamical system for $x,y$ to a single autonomous ODE for $r(t)$. Checking the evolution of that one variable is easier, but even if $r(0) = 1$ is stable, it isn't obvious that guarantees the corresponding stability of $(x(0),y(0)) = (1,0)$. – hardmath Jan 18 '22 at 02:09
  • Thank you for your comment. Professor ask us use Matlab to show that in x-y plane all solution will tend to $(1,0)$ first. But $(1,0)$ is unstable in fact. And the eigenvalues of linearized system at $(1,0)$ is $0$ and $1$. So I can't use Linearized Stability Theory. The professor privided a hint is "examine the definition of stability of equilibrium." – Tammy Jan 18 '22 at 02:34
  • Did you convert to polar coordinates? The radial DE becomes $r' = r(1-r)$. What is the sign of $r'$ when $0 < r < 1, r = 1, r > 1$? – Matthew Cassell Jan 18 '22 at 02:59
  • Thank you all. I think the answer about my question may be found in "Closed orbits of dynamical systems". – Tammy Jan 19 '22 at 02:16

1 Answers1

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You can change the ODE from the cartesian coordinate to the polar cordinate:

$rr'=xx'+yy'$. Therefore, $rr'=(x^2+y^2)(1-r)=r^2(1-r)$, and so $r'=r(1-r)$.

We can conclude that:

For $0<r<1$, $r'>0$ and r is increasing.

For $r>1$, $r'<0$ and r is decreasing.

Hence, we approach the unit circle in this system, and it is wrong to say the equilibrium (1,0) is unstable.

khashayar
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