Consider the following function of $\mathbb{R^+}^2$:

$$f(s_1,s_2)=r_1^2\sin\big(\tfrac{1}{2}(s_2-s_1)\omega_1\big)\sin\big(\tfrac{1}{2}(s_2+s_1)\omega_2\big) + r_2^2\sin\big(\tfrac{1}{2}(s_2+s_1)\omega_1\big)\sin\big(\tfrac{1}{2}(s_2-s_1)\omega_2\big)$$

where $r_1^2+r_2^2=1$. My goal is to find the solution curves of $f$ on $\mathbb{R^+}^2$. There is certainly no closed-form solution, so I'm doing it numerically, for some given values of $(r_1,r_2,\omega_1,\omega_2)$.


Obviously, $f(s_1,s_2)=-f(s_2,s_1)$ so the solution curves of $f$ are symmetric w.r.t. to $s_2=s_1$: it suffices to focus on $s_1\geq s_2$. Note that also $f(-s_1,s_2)=-f(-s_2,s_1)$ so $s_2=-s_1$ is another axis of symmetry for the solutions of the extension of $f$ to $\mathbb{R}^2$, but it is a priori not intersting on $\mathbb{R^+}^2$.

$f$ is a scalar function of two variables so its solution curves are (generically) curves. That's what we can observe to some numeric values of $(r_1,r_2,\omega_1,\omega_2)$:

Mathematica graphics

(Note the axial symmetry mentionned above.)

Calculating some points on the solution curves

For some reason (reduction of computational cost in more complex cases), I'd like to get points on the solution curves of $f$ and then do numerical continuation to compute each curve. By curve, I mean a connex curve, and the solution curves is a family of such connex curves.

The idea is that when $\tfrac{1}{2}(s_2+s_1)\omega_i$ approaches $\pi\mathbb{N}$, only one term is the sum remains. For example, when $s_2+s_1=2\pi/\omega_1$,

$$f(s_1,2\pi/\omega_1-s_1)=r_1^2 \sin(\pi\tfrac{\omega_2}{\omega_1})\sin(s_1\omega_1)$$

and so the roots of $s_1\mapsto f(s_1,2\pi/\omega_1-s_1)$ are the $\pi/\omega_1\mathbb{N}$.

In the, we know that the set of points

$$\mathcal{S}=\Big\{ \frac{\pi}{\omega_i}(p,2p-q),\ p,q\in\mathbb{N}, i\in\{1,2\}\Big\} \cap {\mathbb{R}^+}^2$$

is a subset of the solution curves of $f$, as illustrated below:

Mathematica graphics

or, for another set of parameters $(r_1,r_2,\omega_1,\omega_2)$:

Mathematica graphics

Red and green points correspond to $i=1$ and $i=2$.


On both figures, all the curves pass through the set of points $\mathcal{S}$. Is this always true, or can there be some curves which do not pass through any green or red points?

I am also interested in references or keywords for mathematical tools which could be useful for the study of solution curves of functions of the "type" of $f$ (i.e. sum of products of sines), even though I don't think there are miraculous simplification.

  • $\begingroup$ Nice curves!... A terminology issue: you use the word "roots" instead of (solution) curves. Usually, the term "roots" is reserved for discrete cases. $\endgroup$ – Jean Marie Nov 18 '16 at 17:27
  • $\begingroup$ @JeanMarie Thank you I edited. So is correct to say e.g. that solutions of $g(x,y)=1$ are the solution curves of $g(x,y)-1$? I mean, does "solution curves" imply zeros? $\endgroup$ – anderstood Nov 18 '16 at 18:06
  • 1
    $\begingroup$ Yes, one can say it in this way. $\endgroup$ – Jean Marie Nov 18 '16 at 18:07
  • $\begingroup$ Have you considered defining $t_1 = \frac12(s_2-s_1), t_2 = \frac12(s_2+s_1)$? It would transform $f$ into a sum of two separable functions, which may simplify things a bit. $\endgroup$ – Rahul Nov 18 '16 at 18:15
  • $\begingroup$ @Rahul It's make a slightly simpler indeed (see image) but does not help me find the idea why every curves should pass through the grid of points. That function "becomes": $r_1^2\sin(t_1\omega_1)\sin(t_2\omega_2)+r_2^2\sin(t_1\omega_2)\sin(t_2 \omega_2)$ with $t. $\endgroup$ – anderstood Nov 18 '16 at 18:25

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