Finding Equilibrium Solutions & Stability of $(1)$

[evinda]

Hello! (Wave)

I want to find the equilibrium solutions and determine their stability.

$(1)\left\{\begin{matrix}
\dot{x}=-y-x(1-\sqrt{x^2+y^2})^2\\
\dot{y}=x-y(1-\sqrt{x^2+y^2})^2
\end{matrix}\right.$

I also want to check the behavior of the solutions of $(1)$ when $t \to \infty$. There are four possible answers.

1. there exists exactly one equilibrium solution and it is asymptotically stable. Furthermore, $\forall$ solution $(x,y)$ of $(1)$ we have that $\lim x(t)=x_0$ and $\lim y(t)=y_0$ where $(x_0,y_0)$ equilibrium solution.
   
2. there exists exactly one equilibrium solution and it is asymptotically stable. Furthermore, $\forall$ solution $(x,y)$ of $(1)$ we have that $\lim x(t) \neq x_0$ and $\lim y(t) \neq y_0$ where $(x_0,y_0)$ equilibrium solution.
   
3. $\exists$ exactly one equilibrium solution and it is stable but not asymptotically stable. Also $\forall$ solution $(x,y) \neq (x_0, y_0)$ of $(1)$ we have that $\lim x(t) \neq x_0$ and $\lim y(t) \neq y_0$ where $(x_0,y_0)$ is the equilibrium solution.
   
4. $\exists$ exactly one equilibrium solution and it is stable but not asymptotically stable. Also $\forall$ solution $(x,y) \neq (x_0, y_0)$ of $(1)$ with $x^2(0)+y^2(0)>1$ we have that $x^2(t)+y^2(t) \geq 1$.
   
5. $\exists$ exactly one equilibrium solution and it is unstable. $\forall$ other solution $(x,y)$ of $(1)$ we have that $\lim (x^2(t)+y^2(t))=1$.

I have thought the following so far.In order to find the equilibrium solutions, we set $\dot{x}=0$ and $\dot{y}=0$.

$\dot{x}=0 \Rightarrow -y-x (1-\sqrt{x^2+y^2})^2=0 (\star)$

and

$\dot{y}=0 \Rightarrow x-y (1-\sqrt{x^2+y^2})^2=0 \Rightarrow x=y(1-\sqrt{x^2+y^2})^2$

$(\star): -y-y(1-\sqrt{x^2+y^2})^4=0 \Rightarrow y=0 \text{ or } 1+(1-\sqrt{x^2+y^2})^4=0, \text{ which is rejected}$.

So $y=0$ and $x=0$.So $(0,0)$ is the only equilibrium solution.

If we set $f_1(x,y)=-y-x(1-\sqrt{x^2+y^2})^2$ and $f_2(x)=x-y(1-\sqrt{x^2+y^2})^2$, then we have

$\frac{\partial{f}}{\partial{x}}=-(1-\sqrt{x^2+y^2})^2+\frac{2x^2(1-\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}$.Is this right?

Because in order to determine the stability, we compute the Jacobi matrix at the equilibrium solution, but $\frac{\partial{f}}{\partial{x}}(0,0)$ is not defined.

Am I doing something wrong? (Thinking)

 

[I like Serena]

So $(0,0)$ is the only equilibrium solution.

If we set $f_1(x,y)=-y-x(1-\sqrt{x^2+y^2})^2$ and $f_2(x)=x-y(1-\sqrt{x^2+y^2})^2$, then we have

$\frac{\partial{f}}{\partial{x}}=-(1-\sqrt{x^2+y^2})^2+\frac{2x^2(1-\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}$.

Is this right?

Because in order to determine the stability, we compute the Jacobi matrix at the equilibrium solution, but $\frac{\partial{f}}{\partial{x}}(0,0)$ is not defined.

Am I doing something wrong?

Hey evinda!

$\pd {f_1}x$ is a limit isn't it?
More specifically:
$$\pd {f_1}x(0,0)=\lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h} = \lim_{h\to 0}\frac{-h(1-\sqrt{h^2})^2-0}{h}=-1$$
(Thinking)

 

[evinda]

Hey evinda!

$\pd {f_1}x$ is a limit isn't it?
More specifically:
$$\pd {f_1}x(0,0)=\lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h} = \lim_{h\to 0}\frac{-h(1-\sqrt{h^2})^2-0}{h}=-1$$
(Thinking)

Ah yes... Then we have

$$\pd {f_1}x(0,0)=-1 \\ \pd {f_1}y(0,0)=-1 \\ \pd {f_2}x(0,0)=1 \\ \pd {f_2}y(0,0)=-1$$

and thus the Jacobi matrix at $(0,0)$ gets the following form:

$J=\begin{pmatrix}
-1 & -1\\
1 & -1
\end{pmatrix}$, right?

Then we get that the eigenvalues are these ones: $-1 \pm i$ and so $Re(\lambda_1)=Re(\lambda_2)=-1<0$ which implies that the equilibrium is asymptotically unstable.Is the above correct so far? (Thinking)

 

[evinda]

What can we say about the limits $\lim x(t)$ and $\lim y(t)$ where $(x,y)$ any other solution of the problem? (Thinking)

 

[I like Serena]

... and thus the Jacobi matrix at $(0,0)$ gets the following form:

$J=\begin{pmatrix}
-1 & -1\\
1 & -1
\end{pmatrix}$, right?

Then we get that the eigenvalues are these ones: $-1 \pm i$ and so $Re(\lambda_1)=Re(\lambda_2)=-1<0$ which implies that the equilibrium is asymptotically unstable.

That's the correct Jacobian.
But doesn't $Re(\lambda_1)=Re(\lambda_2)=-1<0$ mean that it's stable? (Shake)
After all, the solutions are of the form $\tilde x=\tilde x_0 e^{\pm it} e^{-t}$, which converges doesn't it? (Wondering)

What can we say about the limits $\lim x(t)$ and $\lim y(t)$ where $(x,y)$ any other solution of the problem? (Thinking)

Let's take a look at a streamplot:
View attachment 8409
What does it show us about those limits? (Wondering)

EDIT: Note in particular that solutions close to (0,0) converge.
Therefore (0,0) is asymptotically stable.