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Take the following feedback system:

$\dot{x} = (\theta - k_1) x - k_2 x^3$

Now my book says:

For $\theta > k_1$, the equilibrium $x = 0$ is unstable. I wonder why...

Furthermore my book indicates that it is easy to see that $x(t)$ will converge to one of the two new equilibria $\pm \sqrt{\frac{\theta-k_1}{k_2}}$. Again, how did they obtain this result?

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3 Answers

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Question 1: From stability theory it is known, that

a fixed point $x^*$ of $\dot{x}=f(x)$ is stable $\Leftrightarrow$ all eigenvalues of the jacobian $f'(x^*)$ have negative real parts.

For $f(x)=(\theta - k_1) x - k_2 x^3$ and $x^*=0$ this restricts to $\theta<k_1$

Question 2: You can use the same argument.

For $\theta>k_1$ the new equilibria become stable, as (assuming $k_2>0$) we have

$$f'\left(\pm \sqrt{\frac{\theta-k_1}{k_2}}\right)=\theta-k_1-3k_2\frac{\theta-k_1}{k_2}=-2(\theta-k_1)<0$$

This fundamental change of the dynamics at $\theta=k_1$ is known as a pitchfork bifurcation.

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Starting with:

$$ \dot{x} = (\theta - k_{1})x - k_{2}x^{3} = V(x). $$ The equilibrium occurs at $\dot{x}=0$ i.e.

$$ (\theta - k_{1})x - k_{2}x^{2} = 0 = x\left[(\theta - k_{1}) - k_{2}x^{2}\right] $$ so clearly the stationary points occur, $ x_{0} = 0 $ and when $$ (\theta - k_{1}) - k_{2}x^{2} = 0 $$ To compute the other fixed point we utilise $x^{2} =\frac{(\theta - k_{1})}{k_{2}}$ which you can probably carry from here with.

To prove the stability criteria of the points i.e. the nature of the points you linearise about the stationary points $x_{0}$'s using $x = x_{0} + \delta x$, and inserting into your original ode as follows,

$$ \dot{\delta x} = \frac{\partial V}{\partial x} $$

where the last term is evaluated at the fixed points. This leads to $$ \dot{\delta x} = (\theta - k_{1}) + 3k_{2}x_{0}^{2}\delta x $$ Solving this for $\delta x$ will show how perturbations of the fixed points behave.

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Suppose $\theta - k_1 = k_3 > 0$.

Then $\dot{x} = k_3 x - k_2 x^3.$

If $k_2 \le 0, \dot{x}$ has the same sign as $x$, so $x(t)$ will tend to go further away from 0.

Even if $k_2 > 0,$ for sufficiently small $x, |k_3 x| > |k_2 x^3|,$ so $\dot{x}$ will have the same sign as $x$, so $x(t)$ will tend to go further away from 0.

I.e. 0 is unstable.

You can get the result of where the equilibria are by solving $$\begin{align} 0 = \dot{x} & = (\theta - k_1) x - k_2 x^3 \\ & = x(\theta - k_1 - k_2 x^2) \\ & = x(\sqrt{\theta-k_1} + \sqrt{k_2} x)(\sqrt{\theta-k_1} - \sqrt{k_2} x). \end{align} $$

So the graph of $\dot{x}$ against $x$ is an (anti-symmetric ) cubic. If $k_2 > 0$ the graph slopes down to $-\infty$ as $x \rightarrow + \infty$. I.e. for the equilibrium at $\sqrt{\frac{\theta-k_1}{k_2}}$ the graph is negative for $x$ > equilibrium (leading $x(t)$ down to the equilibrium) and positive for $x$ < equilibrium (leading $x(t)$ up to the equilibrium). I.e. the equilibrium at $\sqrt{\frac{\theta-k_1}{k_2}}$ is stable.

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