Phase Plane — Question 1

Question 1 – Solution

Strategy. Use a conserved quantity to identify the geometric curve, then use the original equations to recover direction and timing.

Step 1: Locate zero derivatives. The $x$-nullcline is $y=0$, and the $y$-nullcline is $x=0$. Their only common point, $(0,0)$, is the equilibrium. At $(x,0)$ the field is $(0,-4x)$: downward for $x>0$, upward for $x<0$. At $(0,y)$ it is $(y,0)$: rightward for $y>0$, leftward for $y<0$.

Step 2: Determine the conserved curve. Differentiating gives $(4x^2+y^2)'=8xy-8xy=0$. The initial state therefore lies on $$\boxed{4x^2+y^2=4,}$$ an ellipse with horizontal semiaxis $1$ and vertical semiaxis $2$. Equal coordinate scales preserve this actual geometric aspect ratio.

Step 3: Restore orientation and time. Direct substitution verifies $\boxed{x=\cos(2t),\quad y=-2\sin(2t)}$ and the initial data. The least positive period is $\pi$. At the initial point $(1,0)$ the velocity is $(0,-4)$, so the ellipse is traversed clockwise. The conserved equation alone does not specify this orientation or period.

Step 4: Identify all tangent directions. Horizontal tangents occur where $y'=0$, namely $(0,2)$ and $(0,-2)$, where $x'\ne 0$. Vertical tangents occur where $x'=0$, namely $(1,0)$ and $(-1,0)$, where $y'\ne 0$. For $-1<x<1$, the same $x$ has two possible values $y=\pm 2\sqrt{1-x^2}$ on this orbit. A single branch cannot represent the full closed trajectory. The origin is a separate equilibrium, not part of this ellipse. Arrows in the figure indicate increasing time.