# Use linear stability analysis to classify the fixed points of the following system.$f(x)=ax-x^3$ for $a>0$, $a=0$ and $a<0$.

Use linear stability analysis to classify the fixed points of the following system. $f(x)=ax-x^3$ where a can be positive, zero or negative.

I have found that for $a>0$ we have $2$ fixed points For $a=0$ only $x=0$ is a fixed point at which it is stable

• I don't understand how the solution for $a<0.$

• I am not sure why $x=0$ is only the solution

any help will be appreciated thank u!

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The rhs is a 3rd degree polynomial, so expect 3 fixed points in general. Think complex... –  copper.hat Oct 30 '12 at 18:21

A fixed point of $f(x)=ax-x^3$ is a solution of $f(x)=x$, which we transform to $$0 = x^3-ax+x = x\cdot\left(x^2-(a-1)\right).$$ One obvious solution is $x=0$. Let $b=a-1$.

• If $a>1$, i.e. $b>0$, we have two roots of $x^2-b$, namely $x=\pm\sqrt b$. Stability is determined by $f'(x)=a-3x^2$. From $f(\pm\sqrt b) = a-3b=3-2a$, we see that the fixpoints $x=\pm\sqrt {a-1}$ are stable if $1<a<2$ and repelling for $a>2$. Because $f'(0)=a$, the fixpoint $x=0$ is repelling for $a>1$.
• If $a=1$, i.e. $b=0$, we have a double (hence in total a triple) root at $x=0$.
• If $a<1$, i.e. $b<0$, then $x^2-b\ge b>0$ for all $x\in\mathbb R$, hence no additional (real) fixpoint exists. Ans the fixpoint at $x=0$ is stable if $-1<a<1$, repelling if $a<-1$.
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Hagen von Eitzen answered the question perfectly assuming that the system you are looking at is discrete time (i.e., defined by the difference equation $x(n+1) = f(x(n))$). However your comments in the question (e.g., if $a=0$ there is only one fixed point) make me suspect you are looking at a continuous time system (i.e., defined by the ODE $\dot{x}(t) = f(x(t))$. This said, the fact that you added tag chaos-theory might imply that you are actually looking at the discrete time case (my point is that making sure your question is unambiguous helps a lot for someone trying to answer it). Anyway, for completeness I post the continuous time answer as well.

A fixed point of $\dot{x}=ax-x^3$ is a real number that solves $\dot{x} = 0 = ax - x^3 = x(a-x^2)$. Thus $x_{fp}=0$ is always a solution irrelevant of the value of $a$. In addition, if $a>0$ we get two extra solutions $x_{fp}=\pm\sqrt{a}$. We have found the fixed points, next we can linearise around the fixed points to study how they behave locally. The Hartman-Grobman Theorem says that a fixed point $x_{fp}$ locally (in a small neighbourhood of $x_{fp}$) behaves as an asymptotically stable (unstable) fixed point if $\delta_x = 0$ is an asymptotically stable (unstable) fixed point of the `linearised system'

$\dot{\delta}_x = \frac{df(x_{fp})}{dx}\delta_x$.

The linearised system is asymptotically stable if and only if $\frac{df(x_{fp})}{dx}<0$ and unstable if and only if $\frac{df(x_{fp})}{dx}>0$. Notice that the Hartman-Grobman Theorem is not applicable if $\frac{df(x_{fp})}{dx}=0$. Then

$\frac{df(x)}{dx} = a-3x^2$

thus if $a<0$, $\frac{df(0)}{dx} = a <0$ and the unique fixed point $x_{fp}=0$ behaves locally as an asymptotically stable fixed point. If $a=0$, $\frac{df(x)}{dx} =0$ and so we get no information by linearising. Lastly, if $a>0$ then $\frac{df(0)}{dx} = a>0$ and $\frac{df(\pm\sqrt{a})}{dx} = -2a<0$, so $x_{fp}=0$ is unstable and $x_{fp}=\pm\sqrt{a}$ behave locally as an asymptotically stable fixed points.

Just a side note, linearising systems of order one is often a bit pointless given that you can draw conclusions regarding the global behaviour of the system by simply examining the phase line (continuous time) or cob-webbing (discrete time). For more info and a nice introductory book on non-linear systems see

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