Phase Plane — Question 7

Question 7 – Solution

Strategy. A decreasing energy supplies a containing region and rules out a closed return without requiring an explicit trajectory.

Step 1: Compute the equilibrium and dissipation. Equilibrium requires $y=0$ and $x+x^3=0$, hence only $(0,0)$. Along solutions, $$E'=y(-x-x^3-cy)+(x+x^3)y=\boxed{-cy^2\le 0}.$$ Thus energy decreases except at instants when $y=0$.

Step 2: Bound the forward solution. Initially $E=3/4$. Since all energy terms are nonnegative, $y^2/2\le 3/4$ gives $|y|\le\sqrt{3/2}$. Also $x^4+2x^2\le 3$, so $(x^2+1)^2\le 4$ and $|x|\le 1$. Therefore $$\boxed{|x(t)|\le 1,\qquad |y(t)|\le\sqrt{3/2}\quad(t\ge 0).}$$ The polynomial field is locally Lipschitz. These bounds keep the state in a compact region, preventing finite-time escape and allowing continuation for all forward times.

Step 3: Interpret zero instantaneous energy loss. At $(1,0)$, the velocity is $(0,-2)$, not zero. Moreover $y(t)=-2t+o(t)$ for small positive $t$, so $y$ becomes nonzero and $E'=-cy^2<0$. The solution enters lower energy levels. A horizontal zero of $E'$ at one instant does not make energy constant on an interval or make its level curve a trajectory.

Step 4: Exclude nonconstant periodic motion. If a solution had period $T>0$, integrating gives $0=E(T)-E(0)=-c\int_0^T y(t)^2\,dt$. Continuity and $c>0$ force $y\equiv 0$, then $x'=0$ and $x+x^3=0$, so the solution is the zero equilibrium. Thus no nonconstant periodic orbit exists. The drawn $E=3/4$ curve bounds a forward-invariant region; it is not a closed orbit. Interior arrows show velocities for the illustrative choice $c=1$.