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I have a flow defined by $\dot{x} = x-x^4+1 :=f(x)$. I need to sketch its phase portrait. Firstly, I have to find its fixed points, these occur at $f(x)=x$. So, $x^4=1 \Rightarrow x= \pm1$. Next, I find the Jacobian, in this case the derivative of $f(x)$, at these points. $f'(x)=1-4x^3 \Rightarrow f'(1)=-3<0, f'(-1)=5>0$. So the points; $x=1$ is stable and $x=-1$ is unstable. Is this the right way to go about obtaining the nature of the fixed points. Further, how do a sketch the phase portrait. I have only seen 2D cases where I solve the DE for $x_1$ and $x_2$ and graph them. Here I only have 1D so I am not sure whether I have to graph the solution of $\dot{x} = x-x^4+1$ and how would I graph the flow lines etc.

Black
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  • @AnthonyQuas sorry. I have flagged it. Hopefully it will be moved there shortly. – Black Oct 15 '14 at 00:09
  • I flagged it for a moderator for possible transfer (I'm not sure if this the right course of action, but I have no doubt I'll be set right if this isn't the right thing...) – anthonyquas Oct 15 '14 at 00:10

1 Answers1

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The fixed points are the constant solutions $x(t)=k$ for some constant $k\in\mathbb R$. Thus we have $x'(t)=0$ or, by the differential equation, $x-x^4+1=0$. (Not $x-x^4+1=x$.) Now, there are two real numbers where $x-x^4+1=0$ and they are approximately at $x=-0.724492$ and $x=1.22074$. The graph is positive between these and negative outside. So the phase line consists of two negatively oriented arrows on the outside and one positively oriented arrow in the middle. You can easily draw it on the $x$-axis together with the graph of $x-x^4+1$.

enter image description here

Mark McClure
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  • Thank you. This was exactly what I was looking for. Just another question, although $x(t)=k$ is the intuitive criterion for a fixed point of a flow, why is $f(x)=x$ the criterion for a 1D map? – Black Oct 15 '14 at 11:25
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    @Black It sounds like you might have a confusion between continuous and discrete dynamics. When studying continuous dynamics of the form $x'=f(x)$ (as in this question), the criteria for a fixed point is not $f(x)=x$, but rather $f(x)=0$. The former condition is the correct criteria in the context of discrete dynamics, i.e. when studying sequences generated by an iteration of the form $x_n=f(x_{n-1})$. I hope that makes sense. – Mark McClure Oct 15 '14 at 11:33
  • thank you. Yes it makes sense now. – Black Oct 15 '14 at 11:46