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I understand how to find the stability of fixed points by using $f'(x)$ and inserting the values of the fixed points into that. However, I can't figure out how to tell the stability in systems such as

$$\frac{dx}{dt} = a + \sin(x) + \cos(2x)$$

In situations like this, the value of the fixed points varies as '$a$' changes, which I have no trouble finding, its just defining their stability that I struggle with.

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    $\begingroup$ Your question seems too broad. As a starting point, it seems you meant $a$ is time-varying. So you'll enter the realm of nonlinear time-varying systems whose stability analysis requires specific tools such as Lyapunov method. $\endgroup$ – polfosol May 9 '18 at 16:13
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    $\begingroup$ In the case of the question I took the system from, a is a parameter unrelated to time t, but I rushed and didn't clarify, my bad, the stress of exams is getting to me! $\endgroup$ – user6046937 May 9 '18 at 17:14
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Let us assume a system of this type

$$\dot{x}(t)=f(x;a)$$

with the equilibrium point $x_\text{eq}$. In this $a$ is a parameter. In this kind of problem, the equilibrium point is a function of the parameter $a$ (called bifurcation parameter in this context).

In order to obtain the equilibrium point(s) it is necessary to solve the nonlinear equation

$$0 = f(x_\text{eq};a)$$ $$\text{here: } 0 = a+\sin x_\text{eq}+1-2\sin^2x_\text{eq}$$

Then the stability of the equilibrium point can be obtained by Lyapunov's indirect method (also known as linearization).

Hence, you will have to determine

$$\left.\dfrac{\partial f}{\partial x}\right|_{x=x_\text{eq}}.$$

If this value is positive then, the equilibrium point is unstable. If it is negative hen the equilibrium point is at least locally asymptotically stable.

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  • $\begingroup$ The problem is a bit more complicated than that $\endgroup$ – polfosol May 9 '18 at 16:16
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    $\begingroup$ @polfosol: I think the OP is doing a bifurcation analysis. And $a$ is the bifurcation parameter. But let us wait for the OP to explain what he really means. $\endgroup$ – MrYouMath May 9 '18 at 16:18
  • $\begingroup$ MrYouMath is right, a is the bifurcation parameter. My bad, I should have clarified. A typical question would be: a) Does this equation give a well-defined vector field on a circle. (in the question it uses theta instead of x) b) Find all fixed points as a varies and sketch the flux diagram. c) Draw the bifurcation diagram and classify the occurring bifurcations as a varies. $\endgroup$ – user6046937 May 9 '18 at 16:33
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    $\begingroup$ By the way, stability of equilibrium in $\mathbb{R}$ or $\mathbb{S}^1$ can be also simply determined just by looking at signs of $f(x; a)$ near the equilibrium. If only qualitative picture is the matter of interest, this could be enough. $\endgroup$ – Evgeny May 10 '18 at 9:12
  • $\begingroup$ @Evgeny Thank you for the addition:). $\endgroup$ – MrYouMath May 10 '18 at 9:40

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