Phase Portrait of One Dimensional ODE

Question

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.

Answer 1

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.