I am very confused… In a homework task I was asked to draw a phase portrait of the following ODE;
$$\dot{x}=\lambda+x^2$$
I have looked into it online and all the problems I have seen deal with a system of ODEs which you can then turn into a matrix and so on…So far I have determined the critical points of this function, namely;
$$x=\pm\lambda i$$
I have also determined their stability. I have also graphed a bifurcation diagram of this. Would a phase portrait of this ODE simply be vectors converging to the stable solutions and for unstable solutions, arrows will be diverging? Any help will be greatly appreciated.
Phase Portrait of One Dimensional ODE
analysisbifurcationordinary differential equations
Related Solutions
Although there are many ways to do this, I suspect what the problem is guiding you towards doing is to obtain the flow directly by evaluating over every line that intersects the origin in phase space.
So for a sketch, you would draw the line $y = 0.1 x$, and use the expression you found above for $a = 0.1$ to determine the magnitude and direction of the flow on that line. Then try it for a couple of other different lines, and use common sense to fill in the rest.
Your analyses are spot on.
The saddle-node bifurcation requires that $f(x) = 1 + rx + x^2$ has two identical solutions, so the discriminant $∆ = r^2 − 4 = 0$. Therefore, $r = \pm2$ and $x^∗ = \mp1$.
If we draw a phase portrait, it would look like this for $r = -2$
However, I think they want the streamline, which is (compare the phase portrait for stable vs. unstable flow)
For $r = 2$, we have
The streamline is
I also think they are looking for the bifurcation diagram (blue is stable, orange is unstable)
Update I used Mathematica to draw all of the figures.
Best Answer
I guess you are facing with real valued ODE. Therefore, the steady states are those real numbers such that $\dot{x} = 0$. In your case, you get that:
$$\dot{x} = 0 \Rightarrow \lambda + x^2 = 0 \Rightarrow x = \pm \sqrt{-\lambda}.$$
Of course, if $\lambda < 0$, then you don't have any steady state, as $\pm \sqrt{-\lambda}$ are not real numbers. On the other hand, if $\lambda \geq 0$, then $x = \pm \sqrt{-\lambda}$ are real and hence they can be considered as steady states. In particular, for $\lambda = 0$, then there is only one steady state $x=0$.
Regarding the phase portrait, I hope this figure can help you.
The observed phenomenon is also known as "saddle-node" bifurcation.