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I am modeling an ellipsoid tumbling in a flow field. I have derived an expression for the Euler angle $\phi(t)$ of the rotation in the $x$-$y$ plane as a function of time, but its range is only $\pi$, from $-\pi/2$ to $\pi/2$. The range should be $2\pi$, and I would like to know if it is possible in this case to identify the angle $-\pi/2$ with $\pi/2$ since the ellipsoid is symmetric, and thus allow this function to represent a full rotation through $2\pi$ of the ellipsoid (over two periods of $\phi$).

Here is what I have found for $\phi$. $$ \phi(t) = \arctan\left( \sqrt{\frac{1+r}{1-r}}\tan \left(\frac{\gamma}{2} \sqrt{1-r^2} t\right)\right), $$

where $r = \frac{{a_1}^2 - {a_2}^2}{{a_1}^2 + {a_2}^2}$, $a_1, a_2$ being the semi-principal axes, and $\gamma \in \mathbb{R}$ is a predetermined constant.

It looks like this:

enter image description here

I believe this is all the information required to answer my question; however, below I have included a derivation of this equation in case this is useful.

Derivation

This derivation follows this publication exactly (I hope!), the results of which I am attempting to reproduce. Represent a linear shear flow field in the laboratory frame as $$ \mathbf{G'} = \left( \begin{matrix} 0&\gamma&0\\ 0&0&0 \\ 0&0&0 \end{matrix} \right) $$ where $\gamma \in \mathbb{R}$, then rotate $\mathbf{G'}$ into the reference frame of the ellipsoid by $\mathbf{G} = \mathbf{RG'R}^T$ where $$ \mathbf{R} = \left( \begin{matrix} \cos \phi& \sin \phi&0\\ -\sin \phi&\cos \phi&0 \\ 0&0&1 \end{matrix} \right) $$ represents a rotation about the $z$ axis (i.e., in the $x$-$y$ plane). Define $\mathbf{E} = \frac{1}{2}(\mathbf{G} + \mathbf{G}^T)$ and $\Omega = \frac{1}{2}(\mathbf{G} - \mathbf{G}^T)$ and then solve the differential equation $$ \phi' = \Omega_{12} + r\,E_{12} $$ with $r$ defined as above. I used a computer to obtain an analytic solution to this ODE.

Possible Issue

The ODE solver (in Mathematica) complained that inverse functions were being used and that thus some solutions may not be found. It is thus possible that I do not have the right solution to the ODE.

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1 Answer 1

up vote 1 down vote accepted

Because $\mathbf{G^\prime}$ assumes only rotation about the $z$-axis, your ODE reduces to $$\phi^\prime=\frac{\gamma}{2}\left(1 + r\cos{2\phi}\right)$$ You see that this governing equation only requires information from $\phi$ in the range of $[0,\pi)$ (or $(-\pi/2,\pi/2]$), because $2\phi$ goes from $0$ to $2\pi$ (or $(-\pi,\pi]$). This is why it admits an $\operatorname{arctan}$ solution. If you were to introduce a small perturbation of the Euler angle $\theta$, the above equation would become more complicated and a solution would require a more detailed, less symmetric solution. So, your interpretation is correct: two of these periods corresponds to one period of rotation for the whole ellipsoid.

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