Phase Plane — Question 5

Question 5 – Solution

Strategy. Record the time attached to each projected point; the vector field here is not determined by the plane coordinates alone.

Step 1: Verify and eliminate time. Both derivatives are $(1,2t)$, so both candidates solve the system. Integration gives the full family $(x,y)=(t+a,t^2+b)$. The candidate plane curves are respectively $\boxed{y=x^2}$ and $\boxed{y=(x-1)^2}$, each directed toward increasing $x$.

Step 2: Locate the projected crossing and its times. Solving $x^2=(x-1)^2$ gives $(x,y)=(1/2,1/4)$. The first solution reaches it at $t=1/2$, with velocity $(1,1)$; the second reaches it at $t=-1/2$, with velocity $(1,-1)$. The different tangent directions reflect different evaluation times.

Step 3: Check what uniqueness prohibits. At any common time, $V(t)-U(t)=(1,0)$, so the full states never agree. The IVP is unique, but the projected point $(x,y)$ alone does not fix $y'=2t$. There is no single autonomous plane velocity at the projected crossing. Thus the autonomous no-crossing rule is inapplicable. Geometric intersection at different times is not an intersection of these solutions at the same specified time and state.

Step 4: Restore the clock in the state. With new independent variable $\tau$, use $$\boxed{s'=1,\quad x'=1,\quad y'=2s.}$$ The lifted candidates are $(s,x,y)=(t,t,t^2)$ and $(t,t+1,t^2)$. An intersection of lifted curves would first require equal $s$, hence equal $t$, and would then contradict the fixed horizontal difference $1$. Above the plane crossing their clock coordinates are $1/2$ and $-1/2$, so the lifted states are distinct. The figure deliberately labels only projected plane curves and shows their separate rightward directions.