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Is there an analytical solution satisfying these 3 equations with non-constant z?

$$\frac{dx}{dt}=-z\cdot\cos(\omega t)$$ $$\frac{dy}{dt}=z\cdot\sin(\omega t)$$ $$\frac{dz}{dt}=x\cdot\cos(\omega t) - y\cdot\sin(\omega t)$$

Pick a specific non-zero $\omega$ (e.g., 1 or $\pi$) if you must do so.

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    $\begingroup$ $x=\sin \omega t$, $y=\cos \omega t$, $z=-\omega $ seems to work. $\endgroup$ – abx Oct 16 '18 at 4:42
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    $\begingroup$ Oh, I see. I actually want non-constant z, so will edit my question. $\endgroup$ – bobuhito Oct 16 '18 at 4:53
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    $\begingroup$ $x^2+y^2+z^2$ stays constant for any solution, so you may do dimensional reduction $\endgroup$ – მამუკა ჯიბლაძე Oct 16 '18 at 4:57
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    $\begingroup$ Can the downvoters at least give a reference for how to solve this problem? Sorry, I simply can't see the trick to get the sqrt(1+w^2) sine wave unless I guess a single sine wave from the beginning and solve for parameters. $\endgroup$ – bobuhito Oct 17 '18 at 0:41
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Since $x^2+y^2+z^2$ is conserved, we can switch to the stereographic projection, i. e. substitute $x=\frac{2X}{X^2+Y^2+1}R$, $y=\frac{2Y}{X^2+Y^2+1}R$, $z=\frac{X^2+Y^2-1}{X^2+Y^2+1}R$, with constant $R$. This gives \begin{align*} \frac{dX}{dt}&=\frac12\cos(\omega t)(X^2-Y^2+1)-\sin(\omega t)XY,\\ \\ \frac{dY}{dt}&=\frac12\sin(\omega t)(X^2-Y^2-1)+\cos(\omega t)XY. \end{align*} Denoting $Z=X+iY$, we reduce to a single equation $$ \frac{dZ}{dt}=\frac12\left(e^{i\omega t}Z^2+e^{-i\omega t}\right), $$ solved as $$ Z(t)=e^{-i\omega t}\left(\sqrt{1+\omega^2}\tan\left(\frac12\sqrt{1+\omega^2}t+C_0+iC_1\right)-i\omega\right) $$ (with arbitrary real constants $C_0$, $C_1$).

The latter can be found by e. g. further switching to $F(t)=e^{i\omega t}Z(t)$ which satisfies $$ \frac{dF}{dt}=\frac12\left(1+2i\omega F+F^2\right), $$ giving $$ t=C+\int\frac{2dF}{1+2i\omega F+F^2}=C_0+iC_1+\frac2{\sqrt{1+\omega^2}}\arctan\left(\frac{F+i\omega}{\sqrt{1+\omega^2}}\right) $$

Returning to $(x,y,z)$ it is easy to see that this describes circular motion on a sphere around some axis like this

enter image description here

with the sphere itself at the same time spinning around the $z$ axis (because of the $e^{i\omega t}$ multiplier in $Z=e^{-i\omega t}F$), which means that there must be an easier method to solve the original system. In fact I suspect these are a case of spinning top equations...

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    $\begingroup$ Thanks, it seems logical (though I was wrongly trying to use spherical coordinates from your hint instead of stereographic projection) except for the "solved as" step. I checked that your Z(t) does indeed work but curious about how you found it. If there's a method (i.e., not just a massive lookup) going on there, could you add some intermediate steps? $\endgroup$ – bobuhito Oct 17 '18 at 19:01
  • $\begingroup$ @bobuhito To be honest I just invoked Mathematica. Initially I tried dividing the equation by $Z$, giving $\frac d{dt}(\log Z)=\cos(\omega t-i\log(Z))$ but in the end you get nothing less complicated. If I'll manage to find something more transparent I will add it. Numerical experiments show that going back to $x,y,z$ you get circular trajectories on a sphere which is at the same time itself spinning around the $z$ axis $\endgroup$ – მამუკა ჯიბლაძე Oct 17 '18 at 20:35
  • $\begingroup$ For that you switch to an autonomous system: multiplying the DE by $e^{i\omega t}$ you get for $F(t)=e^{i\omega t}Z(t)$ the equation $F'=\frac12\left(1+2i\omega F+F^2\right)$ or, if you prefer, for $G=\log(F)$, the equation $G'=\cosh(G)+i\omega$ $\endgroup$ – მამუკა ჯიბლაძე Oct 17 '18 at 20:39
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    $\begingroup$ Actually, you can then get a linear equation in a new variable H by defining F=-H'/H/2 (which has simple sine solutions, so the tangent and everything comes naturally now, thanks). $\endgroup$ – bobuhito Oct 18 '18 at 6:41
  • $\begingroup$ @bobuhito You mean $F=-2H'/H$? $\endgroup$ – მამუკა ჯიბლაძე Oct 18 '18 at 7:37
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Here is more direct route. The previous solution suggests that the system might be simplified by imposing rotation around the $z$ axis with frequency $\omega$.

Indeed consider the substitution \begin{align*} x&=\tilde x\cos(\omega t)+\tilde y\sin(\omega t),\\ y&=\tilde y\cos(\omega t)-\tilde x\sin(\omega t),\\ z&=\tilde z, \end{align*} describing rotation around the $z$ axis superimposed on the motion according to $(\tilde x,\tilde y,\tilde z)$. Then for the latter we get \begin{align*} \tilde x'&=-\omega\tilde y-\tilde z,\\ \tilde y'&=\omega\tilde x,\\ \tilde z'&=\tilde x, \end{align*} which can be easily solved and gives \begin{align*} \tilde x&=-A\sqrt{1+\omega^2}\sin\left(\sqrt{1+\omega^2}t\right),\\ \tilde y&=A\omega\cos\left(\sqrt{1+\omega^2}t\right)+B,\\ \tilde z&=A\cos\left(\sqrt{1+\omega^2}t\right)-B\omega. \end{align*} This is rotation around the axis with direction $(0,\frac1{\sqrt{1+\omega^2}},-\frac\omega{\sqrt{1+\omega^2}})$ on a sphere of radius $\sqrt{(A^2+B^2)(1+\omega^2)}$ centered at the origin.

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Maple finds rather complicated closed-form solutions where $z(t) = \cos(\sqrt{1+\omega^2} t)$ or $\sin(\sqrt{1+\omega^2}t)$.

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