0
$\begingroup$

I have to solve the following system of non-linear equations.

  • The variables are $x_i$ and $y_i$ for $i=1,...,n$.
  • For the parameters, we have $a_i\in \mathbb{R}$, $b_i\geq 0$ and $B\geq 0$.

The $n+1$ equations are: $$x_i^2+y_i^2=b_i, \text{ }i=1,...,n,$$ $$\left(\sum_{i=1}^na_ix_i\right)^2+\left(\sum_{i=1}^na_iy_i\right)^2=B.$$

I am wondering what are the conditions on $a_i$, $b_i$ and $B$ so that a solution exists ?

Thank you very much for your help!

$\endgroup$
2
  • 1
    $\begingroup$ No, there is no solution for $b_i=0$ for all $i$ and $B=1$. $\endgroup$ – Dietrich Burde Mar 9 '16 at 23:33
  • $\begingroup$ Ok, I see! I edited my question. Thank you! $\endgroup$ – user321460 Mar 10 '16 at 0:10
0
$\begingroup$

Because $$x_i^2 + y_i^2 = b_i \ge 0$$ you can define $r_i = \sqrt{b_i}$, and switch to polar coordinates: $$\begin{cases} x_i = r_i \cos{\phi_i} \\ y_i = r_i \sin{\phi_i} \end{cases}$$ The first $n$ equations are obviously true in polar coordinates, $$r_i^2 \cos^2 \phi_i + r_i^2 \sin^2 \phi_i = r_i^2 \left ( \cos^2 \phi_i + \sin^2 \phi_i \right ) = r_i^2$$ and the last equation becomes $$\left ( \sum_{i=1}^n a_i r_i \cos \phi_i \right )^2 + \left ( \sum_{i=1}^n a_i r_i \sin \phi_i \right )^2 = B$$ Applying $\left ( \sum_{i=1}^n z_i \right )^2 = \sum_{i=1}^n \sum_{j=1}^n z_i z_j$ we get $$\sum_{i=1}^n \sum_{j=1}^n a_i a_j r_i r_j \cos \phi_i \cos \phi_j + \sum_{i=1}^n \sum_{j=1}^n a_i a_j r_i r_j \sin \phi_i \sin \phi_j = B$$ Because $\cos \phi_i \cos \phi_j + \sin \phi_i \sin \phi_j = \cos(\phi_i - \phi_j)$, we get $$\sum_{i=1}^n \sum_{j=1}^n a_i a_j r_i r_j \cos ( \phi_i - \phi_j ) = B$$ The above means that we are free to choose the orientation of the coordinate system ($\phi_i' = \phi_i + \phi_0$) without affecting the result. If there is a solution, it can be rotated by any angle, and still be a valid solution (so, if there is a solution, there are an infinite number of solutions). For example, we can choose $$\begin{cases}\left ( \sum_{i=1}^n a_i x_i \right )^2 = B \\ \left ( \sum_{i=1}^n a_i y_i \right )^2 = 0 \end{cases}$$ Because $B \ge 0$, we can take the square root on both sides of both equations: $$\begin{cases} \sum_{i=1}^n a_i x_i = \sqrt{B} \\ \sum_{i=1}^n a_i y_i = 0 \end{cases}$$ We are left with two equations to determine $n$ variables, $\phi_i$: $$\begin{cases} \sum_{i=1}^n a_i r_i \cos{\phi_i} = \sqrt{B} \\ \sum_{i=1}^n a_i r_i \sin{\phi_i} = 0 \end{cases}$$ Substituting $r_i = \sqrt{b_i}$ so that $$\begin{cases} x_i = \sqrt{b_i} \cos{\phi_i} \\ y_i = \sqrt{b_i} \sin{\phi_i} \end{cases}$$ we arrive at the form that uses only the known variables $n$, $a_i$, $b_i$, and $B$, plus the $n$ unknown variables $\phi_i$:

$$\begin{cases} \sum_{i=1}^n a_i \sqrt{b_i} \cos{\phi_i} = \sqrt{B} \\ \sum_{i=1}^n a_i \sqrt{b_i} \sin{\phi_i} = 0 \end{cases}$$

I am sure there are lots of patterns for $a_i$ and $b_i$ (constants or sequences) where $\phi_i$ is easy to determine, but I am really not a mathematician, so I'll leave that for those who have the skill and inclination to do so. (I'd continue with numerical methods from this point.)

I do note an interesting property, however. If

$$\sum_{i=1}^n \lvert a_i \rvert \sqrt{b_i} \lt \sqrt{B}$$

there cannot be any solutions, because $\lvert a_i \cos \phi_i \rvert \le \lvert a_i \rvert$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.