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Let $r(t)=(x(t),y(t),z(t)),t\geq0$ be a parametric curve with $r(0)$ lies on cylinder surface $x^2+2y^2=C$. Let the tangent vector of $r$ is $r'(t)=\left( 2y(t)(z(t)-1), -x(t)(z(t)-1), x(t)y(t)\right)$. Would you help me to show that :

(a) The curve always lies on ylinder surface $x^2+2y^2=C$.

(b) The curve $r(t)$ is periodic (we can find $T_0\neq0$ such that $r(T_0)=r(0)$).If we make the C smaller then the parametric curve $r(t)$ more closer to the origin (We can make a Neighboorhood that contain this parametric curve)

My effort:

(a) Let $V(x,y,z)=x^2+2y^2$. If $V(x,y,z)=C$ then $\frac{d}{dt} V(x,y,z)=0$. Since $\frac{d}{dt} V(x,y,z)=(2x,4y,0)\cdot (x'(t),y'(t),z'(t))=2x(2y(z-1))+4y(-x(z-1))=0$, then $r(0)$ would be parpendicular with normal of cylinder surface. Hence the tangent vector must be on the tangent plane of cylinder. So $r(t)$ must lie on cylinder surface.

(b) From $z'=xy$, I analyze the sign of $z'$ (in 1st quadrant z'>0 so the z component of $r(t)$ increasing and etc.) and conclude that if $r(t)$ never goes unbounded when move to another octan ( But I can't guarante $r(t)$ accros another octan.). I also consider the case when $(x=0, y>0), (x=0, y<0,z>1), (x>0, y=0,z>1$and so on) and draw the vector $r'(t)$.

Thank you so much of your help.

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2 Answers 2

Introducing new coordinates $\bar x$, $\bar y$, $\bar z$ via $$x:=\bar x,\quad y:={1\over\sqrt{2}}\bar y,\quad z:=1+{1\over\sqrt{2}}\bar z$$ our curve now satisfies the differential equations $$\bar x'(t)=\bar y(t)\>\bar z(t),\quad \bar y'(t)=-\bar x(t)\>\bar z(t),\quad \bar z'(t)=\bar x(t)\>\bar y(t)\ ;$$ in other words, it is a solution of the ODE system (I'm omitting the bars now) $$x'=yz,\quad y'=-xz, \quad z'=xy\qquad\bigl((x,y,z)\in{\mathbb R}^3\bigr)\ .\tag{1}$$ After introducing cylindrical coordinates $$x:=r\cos\phi,\quad y:=r\sin\phi$$ the system $(1)$ goes over into $$r'=0,\quad r\>\phi'=-r\>z, \quad z'=r^2\cos\phi\sin\phi\ .\tag{2}$$ Here the first equation says that $r={\rm const.}$ along all solution curves. (This means that in the original setup the solution curves stay on cylinders $x^2+2y^2={\rm const.}$) So from now on the variable $r$ will be considered as a given positive constant, and it is enough to consider the solution curves in the $(\phi,z)$-plane, i.e., on the cylinder of radius $r$ developed into the plane.

Multiplying the two equations $z=-\phi'$ and $z'=r^2\cos\phi\sin\phi$ we obtain that $$0=2zz'+r^2\sin(2\phi)\phi'=\left(z^2-{r^2\over2}\cos(2\phi)\right)'\ .$$ It follows that along the solution curves we have $$\bigl(f(\phi,z):=\bigr)\quad z^2-{r^2\over2}\cos(2\phi)={\rm const.}\quad.$$ enter image description here

The contour plot of $f$ shows that these curves are indeed closed when considered as curves on the cylinder. Some of them go around the cylinder, and some don't. I leave it to you to prove that all these closed orbits are completed in finite time.

But there is a caveat: According to $(1)$ the points with $z=0$ and $x\>y=0$ are stationary. It follows that the trajectories ending in points $(\phi,z)$ with $z=0$ and $\phi$ an odd multiple of ${\pi\over2}$ (see the figure!) are not periodic in time. Rather they take time from $-\infty$ to $+\infty$ connecting two of these special points. In fact one easily checks that $(2)$ with initial conditions $r(0)=1$, $\phi(0)=0$, $z(0)=1$ has the solutions $$r(t)\equiv1,\quad \phi(t)=-\arcsin\bigl(\tanh t\bigr),\quad z(t)={1\over\cosh t}\qquad(-\infty<t<\infty)\ .$$

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We can reparameterize $S=\{(\sqrt{C}\cos u,\frac{\sqrt{C}}{\sqrt{2}}\sin u, v): u,v\in \mathbb{R}\}$ since $(\sqrt{C}\cos u)^2+2\left(\frac{\sqrt{C}}{\sqrt{2}}\sin u\right)^2=C$. Let $r(t)= (x(t),y(t),z(t))$ and $r(0)=(x_0,y_0,z_0)$. Define $V(x,y,z)=x^2+2y^2$. Since $V(x,y,z)=C$ then $\frac{dV}{dt}=0$. But, by chain rule we get $0=\frac{dV}{dt}=\nabla{V}\cdot(x',y',z')$ so the tangent vector of the parametrized curve that intersect $S$ in a point always parpendicular with $\nabla{V}$. Since $r(0)$ be in $S$ and $\nabla{V}$ parpendicular with the tangent plane of $S$ at $r(0)$ , then $r'(0)$ be on the tangent plane of $S$ at $r(0)$. By this argument, we can conclude that $r(t)$ must be on $S$. Since $S=\{(\sqrt{C}\cos u,\frac{\sqrt{C}}{\sqrt{2}}\sin u, v): u,v\in \mathbb{R}\}$ then $x(t)=\sqrt{C}\cos (t-t_0)$ and $y(t)=\frac{\sqrt{C}}{\sqrt{2}}\sin (t-t_0)$ with $t_0$ satisfying $x_0=\sqrt{C}\cos t_0$ and $y_0=-\frac{\sqrt{C}}{\sqrt{2}}\sin t_0$. Since $z'=xy$ then $z'(t)=\frac{C}{2\sqrt{2}}\sin(2t-t_0)$, hence $z(t)=-\frac{C}{4\sqrt{2}}\cos(2t-t_0)$. Since $r(2\pi)=(\sqrt{C}\cos (2\pi-t_0),\frac{\sqrt{C}}{\sqrt{2}}\sin (2\pi-t_0),-\frac{C}{4\sqrt{2}}\cos(2\pi-t_0))=(x_0,y_0,z_0)=r(0)$ then $r(t)$ is periodic.

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