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Given a cylinder surface $S=\{(x,y,z):x^2+2y^2=C\}$. Let $\gamma(t)=(x(t),y(t),z(t))$ satisfy $\gamma'(t)=(2y(t)(z(t)-1),-x(t)(z(t)-1),x(t)y(t))$. Could we guarante that $\gamma$ always on $S$ and periodic if $\gamma(0)$ on $S$?

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1 Answer 1

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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