$12.$ (a) Determine all critical points of $\dfrac{dx}{dt}=(1+x)\sin y$ , $\dfrac{dy}{dt}=1−x−\cos y$ .
(b) Find the corresponding linear system near each critical point.
(c) Find the eigenvalues of each linear system. What can you conclude about
the nonlinear system?
(d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to to extend them in those cases where the linear system does not provide definite information about the nonlinear system.
12(a) . The critical points are given by the solution set ofthe equations $(1+x)\sin y=0$ , $1-x-\cos y=0$ . If $x=-1$, then we must have $\cos y=2$, which is impossible. Therefore $\sin y=0$ , which implies that $y=n\pi$ , $n=0, \pm 1,2,\ \ldots$. Based on the second equation, $x=1-\cos n\pi. \implies $ The critical points are located at $(0, n_{even} \pi) = (0,2k\pi)$ and $(0, n_{odd} \pi) = (2,\ (2k+1)\pi)$ , where $k=0, \pm 1, \color{red}{ ??? } 2,\ \ldots$.
(1) k is any integer, right? Why didn't solution write $\color{red}{ \pm } 2 $ ?
$(b,\ c)$ . The Jacobian matrix ofthe vector field is $J=\left(\begin{array}{lll} \sin y & (1+ & x)\cos y\\ -1 & \sin y & \end{array}\right). $. Then $J(0,2k\pi)\ =\left(\begin{array}{ll} 0 & 1\\ -1 & 0 \end{array}\right), $ with purely complex eigenvalues $r_{1,2}=\pm i$. The critical points ofthe associated linear systems are centers, which are stable. Note that Theorem 9.3.2 does not provide a definite conclusion regarding the relation between the nature of the critical points of the nonlinear systems and their corresponding linearizations.
(2) This table says a center for a linear system is indetermininate, because for a locally linear system, it can be a center or stable point. But how can you determine that (0,0) is a center? The solution didn't reveal how it determined this? I know the solution graphed with a computer, but what do I do on an exam?
I skip the details for $ J[2,\ (2k+1)\pi]$
The set of critical points is $$ \{(0,2k\pi)\mid k\in\mathbb Z\}\cup\{(2,(2+1)k\pi)\mid k\in\mathbb Z\}. $$ To determine that $(0,0)$ is indeed a center, an approach is to look for some curve $$ y=u(x) $$ invariant by the dynamics. This happens when, for every $t$, $$ y'(t)=u'(x(t))x'(t), $$ This condition holds when, for every $z$, $$ 1-z-\cos u(z)=u'(z)(1+z)\sin u(z), $$ or, equivalently, $$ \frac{\sin u(z)\cdot u'(z)}{1+z}-\frac{1-\cos u(z)}{(1+z)^2}+\frac{z}{(1+z)^2}=0, $$ that is, $$ \frac{\mathrm d}{\mathrm dz}\left(\frac{1-\cos u(z)}{1+z}+\log(1+z)+\frac1{1+z}\right)=0. $$ To sum up, the function $V$ defined by $$ V(x,y)=\frac{2-\cos y}{1+x}+\log(1+x), $$ is invariant by the dynamics. The point $(0,0)$ coincides with the curve $V=1$ and the cycles around $(0,0)$ are the curves $V=1+\varepsilon$ for small positive values of $\varepsilon$.