 Solution

For both parameters the autonomous planar system is $\dot{x}=v$, $\dot{v}=-x(x^2+\lambda x+1)$. It has the conserved energyIndeed $\dot{H}=v\dot{v}+V_{\lambda }^{\prime }(x)\dot{x}=0$. At any equilibrium point $(x_{\ast },0)$ the eigenvalues satisfy

For $\lambda =1$, the quadratic $x^2+x+1$ is strictly positive, so the only equilibrium point is $(0,0)$, with eigenvalues $\pm i$. Since $V_1^{\prime }(x)$ has the sign of $x$, $V_1$ has a unique strict minimum at zero and tends to infinity in both directions. All positive energy levels are closed curves surrounding this stable center equilibrium. They are clockwise: rightward when $v>0$ and leftward when $v<0$.

For $\lambda =5/2$, factor $x^2+(5/2)x+1=(x+2)(x+1/2)$. The three equilibrium points and their classifications areThe two center equilibria are stable but not asymptotically stable; the saddle equilibrium is unstable. At the saddle equilibrium the local unstable and stable directions are $v=(\sqrt{3}/2)(x+1/2)$ and $v=-(\sqrt{3}/2)(x+1/2)$.

The potential values are $V(-2)=-2/3$, $V(0)=0$, and $V(-1/2)=7/192$. Consequently there are closed loops around the left center for $-2/3<H<0$, closed loops in both wells for $0<H<7/192$, and a figure-eight pair of homoclinic orbits at $H=7/192$. At $H=0$ the right center is a stationary point while the left well still has a closed orbit. For $H>7/192$, each closed curve encircles both centers and the saddle.

The two homoclinic orbits satisfy $v=\pm \sqrt{2(7/192-V(x))}$. Their outer turning points can be located exactly fromEvery nonstationary energy curve is oriented to the right in the upper half-plane and to the left in the lower half-plane. The sketches show the closed curves, the separating orbits, the stable and unstable saddle directions and the direction of flow.