2
$\begingroup$

Consider the system of equations in the four variables $\{x, y, \phi_x, \phi_y \}$: \begin{align} x \cos(\phi_x) + y \sin(\phi_y) &= 0 \\ x \sin(\phi_x) - y \cos(\phi_y) &= 0 \, . \end{align} It's easy to see that if we set $\phi_x=0$ then the two nontrivial solutions are \begin{align} \phi_y &= \pi/2, \quad y = -x \\ \text{and} \quad \phi_y &= -\pi/2, \quad y = x \, . \end{align} Due to the geometric structure of the problem, I suspect that the general solution is \begin{align} \phi_y &= \phi_x + \pi/2, \quad y = -x \\ \text{and} \quad \phi_y &= \phi_x -\pi/2, \quad y = x \tag{$\star$} \, . \end{align} I also noticed that if we square the two equations and add them we get $$ x^2 + y^2 + 2 x y\left(\cos\phi_x \sin\phi_y - \sin\phi_x \cos\phi_y \right) = 0 $$ which admits the same two solution sets as $(\star)$. How can we prove that $(\star)$ is or is not the complete set of solutions?

$\endgroup$
3
$\begingroup$

Note that you can rewrite your system as $$ \begin{bmatrix} \cos(\phi_x) & \sin(\phi_y) \\ -\sin(\phi_x) & \cos(\phi_y) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0. $$ Notice that $$ \det\begin{bmatrix} \cos(\phi_x) & \sin(\phi_y) \\ -\sin(\phi_x) & \cos(\phi_y) \end{bmatrix} = \cos(\phi_x - \phi_y). $$

We consider two cases:

(1) The matrix is invertible, i.e. $\cos(\phi_x - \phi_y) \not= 0$, or equivalently $$ \phi_x - \phi_y \not= \frac{\pi}{2} + \pi n \qquad (*) $$ for an integer $n$. In this case we automatically get $x=y=0$, while $\phi_x, \phi_y$ are arbitrary but satisfy $(*)$.

(2) The matrix is not invertible, so $$ \phi_x = \phi_y + \frac{\pi}{2} + \pi n $$ for an integer $n$. In this case we see that \begin{align*} \cos(\phi_x) &= (-1)^{n+1} \sin(\phi_y),\\ \sin(\phi_x) &= (-1)^n \cos(\phi_y), \end{align*} so that the system reduces to \begin{align*} \sin(\phi_y) ( (-1)^{n+1} x + y ) = 0\\ \cos(\phi_y) ( (-1)^{n+1} x + y ) = 0. \end{align*} Both equations are satisfied if and only if $$ y = (-1)^{n+1}x. $$

$\endgroup$

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.