I'm trying to find a closed form solution $(x_0,y_0,v_x,v_y)$ for the following equation, where $a$ and $M(t)$ are known numerically.

$$ \frac{v_x(x_0+tv_x+a)+v_y(y_0+v_yt)}{\sqrt{(x_0+v_xt+a)^2+(y_0+v_yt)^2}}+ \frac{v_x(x_0+tv_x-a)+v_y(y_0+v_yt)}{\sqrt{(x_0+v_xt-a)^2+(y_0+v_yt)^2}} =M(t) \tag1$$

Optimization using gradient descent doesn't yield good results because initilization is realy tedious. Grid search is out of question as I am looking for a fast solution.

It is proved that considering a noise-free measure $M(t)$ and at least $4$ measurments $M(t_1),M(t_2),M(t_3),M(t_4)$ the system will admit $4$ solutions and only one if we resctrict $(x_0,y_0)$ to the first quadrant of the real plane $\mathbb{R}^+\times\mathbb{R}^+$

I would like to solve the system of equations formed by the $4$ equations at $t=t_1,t_2,t_3,t_4$. A closed form solution would be nice but an approximation may suits well.

This feels like it would be accessible with some algebraic geometry dark magic, but my background isn't strong enough to solve this.

Some insights :

  • Having only one measurment $M(t)$ and if we set $v_x,v_y$ the locus of $x_0,y_0$ is an hyperbolae

  • Writting $$ r_1(t)={\sqrt{(x_0+v_xt+a)^2+(y_0+v_yt)^2}}$$ $$r_2(t)={\sqrt{(x_0+v_xt-a)^2+(y_0+v_yt)^2}} $$ We have : $$M(t)=\frac{d}{dt}(r_1(t)+r_2(t))$$

  • Writting $$x(t)=x_0+v_xt$$ $$y(t)=y_0+y_xt$$

    At each time step $t$, solving for $(x,y)$ $$r_1(t)+r_2(t)=2C$$ Yields an ellipse with foci $(-a,0)$ and $(a,0)$ and semi-major axis of length $C$.

  • $\begingroup$ Your eq.1 has the form, $$\frac{A}{\sqrt{B}}+\frac{A}{\sqrt{B}} = M(t)$$ This is easily simplified, or is there a typo in the subscripts somewhere? $\endgroup$ – Tito Piezas III Mar 10 '15 at 4:31
  • $\begingroup$ There was a typo with the sign of $a$ in the two terms. I've corrected it. thanks $\endgroup$ – Antoine Bassoul Mar 10 '15 at 9:25
  • $\begingroup$ Oh, ok. By the way, did you see my answer to your other question? $\endgroup$ – Tito Piezas III Mar 10 '15 at 9:45
  • $\begingroup$ Yes, it looks realy nice, I am working on it at the moment, but with my poor bagage in math, it may takes a bit of time :). As soon as I've understood it and managed to make the needed conclusion, I'll mark it as answered. $\endgroup$ – Antoine Bassoul Mar 10 '15 at 9:49
  • $\begingroup$ One additional insight on the question above : Having only one measurment $M(t)$ and if we set $v_x,v_y$ the locus of $x_0,y_0$ is an hyperbolae. I can give more details but it is cumbersome. $\endgroup$ – Antoine Bassoul Mar 11 '15 at 13:32

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