I'm asked to prove in two distinctly different ways, that the following system has no periodic solution in the phase plane:
\begin{alignat*}{2}
\frac{dx}{dt} &= -x+4y \\
\frac{dy}{dt} &= -x-y^{3} \\
\end{alignat*}
For the first proof, I simply apply the Bendixson criterion.
Let
\begin{equation*}
f = \left[
\begin{matrix}
f_{1} \\
f_{2}
\end{matrix}
\right].
\end{equation*}
We have that
\begin{alignat*}{2}
f_{1} &= \dot{x} = -x+4y \\
f_{2} &= \dot{y} = -x-y^{3}
\end{alignat*}
$\therefore$
\begin{alignat*}{2}
\frac{\partial f_{1}}{\partial x} &= -1 \\
\frac{\partial f_{2}}{\partial y} &= -3y^{2}
\end{alignat*}
$\therefore$
\begin{alignat*}{2}
\frac{\partial f_{1}}{\partial x} + \frac{\partial f_{2}}{\partial y} &= -1 -2y^{2} \\
&< 0 &&\forall x,y \in \mathbb{R}.
\end{alignat*}
By the Bendixson criterion, there are no periodic orbits (other than singular points) in a simply connected domain when $f$ is continuously differentiable and $\text{div}(f)$ does not change sign nor does it vanish identically. Thus, the system has no periodic orbits.
I find myself at an impasse for a distinct alternative proof.
First, I tried a change to polar coordinates. The result I got was
\begin{alignat*}{2}
\left[
\begin{matrix}
\dot{r}\\
\dot{\theta}
\end{matrix}
\right]
&=
\left[
\begin{matrix}
3r\cos\theta\sin\theta+r\cos^{2}\theta-r^{3}\sin^{4}\theta \\
\cos\theta\sin\theta – 4\sin^{2}\theta
-\cos^{2}\theta-r^{2}\sin^{3}\theta\cos\theta\end{matrix}
\right].
\end{alignat*}
From here I did not see a clear way of solving for $r$ and $\theta$
or analyzing $\dot{r}$ to rule out periodicity.
Another approach that I considered was to use the theory associated with the Poincare-Bendixson theorem. For instance, if I could show that every solution crosses a transversal an infinite number of times then I could rule out a periodic solution. However, I was not sure how to construct such an argument from the given system.
I also wonder if an eigenvalue analysis of a linearization could be used here as well? I would be happy with any distinctly different approach or a hint towards one that I have tried so far. Thanks in advance.
Best Answer
Hint: For a simple second proof consider Lyapunov function $$ V(x,y)=x^2+4y^2. $$
Actually it could (not always though) because there are results that state what type of an equilibrium can be inside any closed trajectory, but I do not remember them out the top of me head. But my hint is much simpler.