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Question: Use geometric arguments for the model of a negative feedback system,$$\frac{dx}{dt}=\frac{A\theta^2}{\theta^2+x^2}-\gamma x$$where $A, \theta$ and $\gamma$ are positive constants.
$(a)$ How many steady states are there in this system, and what is the stability of these states?
$(b)$ starting from an initial condition of $x=100$ what happens in the limit $t\rightarrow \infty$?

For $(a)$, I try

$$ \begin{align} \frac{A\theta^2}{\theta^2+x^2}-\gamma x&=0\\ \gamma x^3+\gamma \theta^2 x - A\theta^2&=0\\ x^3+\theta^2 x - \frac{A\theta^2}{\gamma}&=0 \end{align} $$

To solve this depressed cubic using closed formula,

$$x=\left\{ \frac{A\theta^2}{2}+\sqrt{\frac{A^2\theta^4}{4}+\frac{\gamma^3\theta^6}{27}} \right\}^{\frac13}+\left\{ \frac{A\theta^2}{2}-\sqrt{\frac{A^2\theta^4}{4}+\frac{\gamma^3\theta^6}{27}} \right\}^{\frac13}$$ With the other two roots found with the cube roots of unity.

It seems very messy to use these steady states to determine the stability of these states. I was wondering what they mean by geometric arguments here. And for $(b)$ I couldn't come up any technique which can help to solve the DE. Any solution or hint will be appreciated. TIA

N00BMaster
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1 Answers1

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To simplify the notation, let's rewrite the ODE as $\frac{dx}{dt}=f(x)-g(x)$, where $f(x):=\frac{A\theta^2}{\theta^2+x^2}$ and $g(x):=\gamma x$.

(a) The question asks how many steady states there are, not to explicitly find them. Now, since $f(x)$ is a positive bell-shaped curve and $g(x)$ is a line with positive slope (see figure), there is only one real solution to the equation $f(x)-g(x)=0$. Let's call this solution $x_*$.

enter image description here

If $x>x_*$, $g(x)>f(x)$, hence $\frac{dx}{dt}<0$ and $x(t)$ decreases with $t$, approaching $x_*$. Similarly, if $x<x_*$, $f(x)>g(x)$, so $\frac{dx}{dt}>0$ and $x(t)$ increases with $t$, again approaching $x_*$. Therefore, $x_*$ is a stable fixed point.

(b) According to the answer to (a), starting from any initial condition (in particular $x=100$), the system will tend to the fixed point $x_*$ as $t\to\infty$.

Gonçalo
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  • I face issued to apply the same logic for the problem $\frac{dx}{dt}=\frac{2x^2}{1+x^3}-x$ then the Equilibria are $x=0,1,\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$. I said to use geometric argument to find the stability of $\frac{-1+\sqrt{5}}{2}$. But if I assume $x_$ then $g(x)=x$ is always bigger then $f(x)=\frac{2x^2}{1+x^3}$ (on $x>0$) implies $x_$ is a stable point but use algebraic way $f'(\frac{-1+\sqrt{5}}{2})>0\implies$ unstable. Where I am wrong? @Goncalo. I know comment shouldn't use for this kind of question but I guess the approach is fit for every scenarios. – N00BMaster Nov 05 '23 at 16:29
  • The fixed points, in increasing order, are $x_1=−(1+\sqrt{5})/2, x_2=0, x_3=(−1+\sqrt{5})/2$, and $x_4=1$. Between $x_2$ and $x_3$, $F(x):=\frac{2x^2}{1+x3}−x<0$, and between $x_3$ and $x_4$, $F(x)>0$ (see figure here), so $x_3$ is an unstable fixed point. A similar consideration shows that $x_1, x_2,$ and $x_4$ are stable fixed points. – Gonçalo Nov 06 '23 at 06:51