Sketch the phase portrait of a two-dimensional vector field

Question

I am having problem drawing the phase portrait of$$\begin{cases}x'=x-x^3\\y'=-y\end{cases}$$ I have tried many ways but cant grasp my head around it. If anyone could explain this problem or direct me to a website to help me understand.

Thanks

Answer 1

With the help of a MATHEMATICA script

dx = x - x^3; dy = -y; V = x^2 (1 - x^2/2) - y^2; gr1 = StreamPlot[{dx, dy}, {x, -2, 2}, {y, -2, 2}]; sols = Solve[{dx == 0, dy == 0}, {x, y}]; gr2 = Table[Graphics[{Red,Disk[({x, y}/.sols[[k]]),0.06]}],{k,Length[sols]}]; gr2a = ContourPlot[V, {x, -2, 2}, {y, -2, 2}, Contours -> 400, ContourStyle -> Black, ContourShading -> Black]; Show[gr1, gr2,gr2a]

In red we can see the equilibrium points at $(-1,0), (0,0), (1,0)$. The stream goes into the points $(-1,0), (1,0)$ qualifying then as sinks and hence stable equilibrium points. The point $(0,0)$ has a stream which partly enters and partly leaves the point characterizing a saddle point.

In black are shown level surface curves which help us guess a function $z = V(x,y)$ such that it's steepest descent paths would generate the stream plot when projected onto the $z=0$ plane.

NOTE

Here $V(x,y) = x^2(1-\frac{x^2}{2})-y^2$ fits well.