Plot the solution curves of $\frac {dx} {dt} = - xe^{-x^2}$

Question

Plot the solution curves and phase portrait of $$\frac {dx} {dt} = - xe^{-x^2}$$

It seems that it's increasing for $x < 0$ and decreasing $x > 0$.

Then it seems concave down for $\frac {-1} {\sqrt{2}} < x < \frac {1} {\sqrt{2}}$ and concave up elsewhere.

It doesn't seem like that in Wolfram Alpha. I can't even seem to limit it to $-1 < t < 1$. How do I do that? I don't think I can zoom in without a pro account.

Answer 1

For the concavity part, the second derivative of $x$ with respect to $t$ is $$\frac{d(-xe^{-x^2})}{dx}\frac{dx}{dt}$$ This gives you $$-x(2x^2-1)e^{-x^2}$$

So the curve is concave up at $x<-\frac{1}{\sqrt{2}}$ and $0<x<\frac{1}{\sqrt{2}}$, and concave down elsewhere. This is consistent with the graph Wolfram Alpha gives.

Answer 2

Your sentence "It seems that it's increasing for $x < 0$ and decreasing $x > 0$" says it all. Because of this the phase portrait has an equilibrium at $0$ and the arrows on $(0,+\infty)$ and $(-\infty,0)$ are respectively $\leftarrow$ and $\rightarrow$. Nothing else is needed for the phase portrait.