Analytical Solution of Non-Linear Coupled ODE

Question

I have the following coupled PDE, which describes the trajectory of a non-inertial particle in Taylor-Green Vortex Flow: $$ \frac{\partial}{\partial t} \begin{bmatrix} x_p \\ y_p \end{bmatrix} = e^{-2 \nu t} \cos \left( \beta \begin{bmatrix} x_p \\ y_p \end{bmatrix} \right) \sin \left( \beta \begin{bmatrix} y_p \\ - x_p\end{bmatrix} \right)$$ Assuming some initial condition $x_p(0), y_p(0)$, is it possible to obtain an analytical solution for $x_p(t)$, $y_p(t)$?

I've tried to manipulate this equation with trigonometric identities but was not able to find a simpler form. Additionally, I attempted to change coordinates such that it could be described in terms of polar coordinates with respect to the nearest vortex, (anywhere where the value of $e^{-2 \nu t} \sqrt{\cos^2(\beta x_p) \sin^2 (\beta y_p) - \cos^2(\beta y_p) \sin^2(\beta x_p)} = 0$), but did not find anything productive.

Answer 1

$$\begin{cases} \frac{dx}{dt}=e^{-2\nu t}\cos(\beta x)\sin(\beta y) \\ \frac{dy}{dt}=e^{-2\nu t}\cos(\beta y)\sin(-\beta x) \end{cases}$$ $$\frac{dy}{dx}=\frac{e^{-2\nu t}\cos(\beta y)\sin(-\beta x)}{e^{-2\nu t}\cos(\beta x)\sin(\beta y)}=-\frac{\sin(\beta x)\cos(\beta y)}{\cos(\beta x)\sin(\beta y)} =-\frac{\tan(\beta x)}{\tan(\beta y)}$$ $$\tan(\beta y)dy=-\tan(\beta x)dx$$ $$-\frac1b \ln(\cos(\beta y))=\frac1b \ln(\cos(\beta x))+\text{constant}$$ $$\cos(\beta y)=c\:\cos(\beta x)$$ $$y(x)=\frac{1}{\beta}\cos^{-1}\big(c\:\cos(\beta x)\big)$$ This is the equation of the trajectory of the point $\big((x(t),y(t)\big)$.

In order to find $x(t)$ one put $\quad \sin(\beta y)=\pm\sqrt{1-c^2\cos^2(\beta x)}\quad$ into the firsrt ODE. $$\frac{dx}{dt}=\pm e^{-2\nu t}\cos(\beta x)\sqrt{1-c^2\cos^2(\beta x)}$$ This first order ODE is separable : $$\pm\int e^{-2\nu t}dt=\int \frac{dx}{\cos(\beta x)\sqrt{1-c^2\cos^2(\beta x)}}$$ This gives $t(x)$ and then the inverse function $x(t)$. One can do it with the help of WolframAlpha.

The same to find $y(t)$.

Answer 2

It looks to me.. yes. Continuing from JJacquelin's equation

$$~\frac {\cos \beta y }{\cos \beta x} =c, $$

If we let $\beta x = \psi(t),~ \beta y = \phi(t), $ then we can see that the trajectories can be further split up as hyperbolic asymptotic geodesics.

They obey separable but coupled Sine-Gordon equations on a pseudosphere of another arbitrary Gauss curvature $\dfrac{-1}{a^2}.$

$$ 2 a^2 \psi^{''} =\sin 2 \psi~;$$

$$ 2 c^2 a^2 \phi^{''} =\sin 2 \phi~;$$

The Odes are primed on time $t$. Their solutions are well known.

I suppose this special free? vortex motion arises from non-Euclidean hyperbolic geometry as hyperbolic geodesics.