Finding Equilibrium Solutions & Stability of $(1)$

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In summary, the conversation discusses finding equilibrium solutions and determining their stability for a given set of equations. It is determined that there is only one equilibrium solution at (0,0) and the Jacobian matrix is calculated to determine its stability. It is found to be asymptotically unstable, however upon further analysis, it is determined that the equilibrium is actually asymptotically stable due to the behavior of solutions close to (0,0).
  • #1
evinda
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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)
 
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  • #2
evinda said:
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)
 
  • #3
I like Serena said:
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)
 
  • #4
What can we say about the limits $\lim x(t)$ and $\lim y(t)$ where $(x,y)$ any other solution of the problem? (Thinking)
 
  • #5
evinda said:
... 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)

evinda said:
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.
 

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FAQ: Finding Equilibrium Solutions & Stability of $(1)$

What is an equilibrium solution?

An equilibrium solution is a point in a system where the behavior of the system remains constant over time. In other words, the system is balanced and there is no net change in the system's state.

How do you find equilibrium solutions?

To find equilibrium solutions, you must first set the derivative of the system equal to zero and then solve for the variables. This will give you the values of the variables at which the system is in equilibrium.

What does stability mean in the context of equilibrium solutions?

In the context of equilibrium solutions, stability refers to the behavior of the system when it is perturbed from its equilibrium point. A stable equilibrium solution means that the system will return to its original state after a small disturbance, while an unstable equilibrium solution means that the system will move away from its original state after a small disturbance.

How do you determine the stability of an equilibrium solution?

The stability of an equilibrium solution can be determined by analyzing the derivative of the system at the equilibrium point. If the derivative is positive, the equilibrium point is unstable, and if the derivative is negative, the equilibrium point is stable.

What are some real-life applications of finding equilibrium solutions?

Finding equilibrium solutions is an important tool in fields such as economics, physics, and chemistry. It can be used to model and predict the behavior of systems such as chemical reactions, population dynamics, and market prices. It also helps in designing stable systems and understanding the effects of small disturbances on the overall behavior of a system.

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