Can we obtain a sufficient constraint over the parameters $A$, $B$, and $C$ in the given equation?

Question

I have this equation $$ A\;\big(\cos (x-y)+\cos x+\cos y\big)+B\;\big(\sin (x-y)-\sin x+\sin y\big)+C=0, $$ where $-\pi\leq x,y \leq \pi$, and $A$, $B$, and $C$ are some real constants.

Then, according to the ranges of the trigonometric functions, is there a sufficient constraint over the parameters $A$, $B$, and $C$ to have the real solutions? If so, how can we obtain that condition?

Answer 1

$\def\sf{\sin\phi}\def\cf{\cos\phi}\def\f{\phi}\def\x{\frac x2}\def\y{\frac y2}$ We may assume that at least one of $A,B$ is not zero. Then the solutions of the equation do not change if we divide the whole equation by $\sqrt{A^2+B^2}$. With $$\sin\phi=\frac A{\sqrt{A^2+B^2}},\quad\cos\phi=\frac B{\sqrt{A^2+B^2}}\quad(-\pi<\f\le\pi)\tag1$$ the left hand side of the equation can be rewritten as: $$\begin{align} F(x,y;\phi)&=\sf\big(\cos (x-y)+\cos x+\cos y\big)+\cf\big(\sin (x-y)-\sin x+\sin y\big)\\ \end{align}$$

The function is differentiable, hence it takes the extreme values either on a boundary or at a critical point. It is easy to check that the extreme points on the boundary are $$-\sf\pm 2\cf.\tag2$$

To determine the critical points we compute first the derivatives of the function: $$\begin{align} F_x&=[\cf\cos(x-y)-\sf\sin(x-y)]-[\cf\cos x+\sf\sin x]\\ &=\cos(x-y+\f)-\cos(x-\f)\\ &=2\sin\left(x-\y\right)\sin\left(\y-\f\right);\\ F_y&=[\cf\cos y-\sf\sin y]-[\cf\cos(x-y)-\sf\sin(x-y)]\\ &=\cos(y+\f)-\cos(x-y+\f)\\ &=-2\sin\left(y-\x\right)\sin\left(\x+\f\right);\\ \Delta&= -4\cos\left(x-\y\right)\sin\left(\y-\f\right)\cos\left(y-\x\right)\sin\left(\x+\f\right)-\sin^2(x-y+\f), \end{align}$$ with $\Delta=F_{xx}F_{yy}-F_{xy}F_{yx}$ being the determinant of the Hessian matrix.

The equation of critical points $F_x=F_y=0$ has the following solutions with $m,n\in\mathbb Z$ to be determined from the condition $-\pi\le x,y\le\pi$:

1. $x-\y=\pi m,\quad y-\x=\pi n$;
2. $x-\y=\pi m,\quad \x+\phi=\pi n$;
3. $\y-\phi=\pi m,\quad y-\x=\pi n$;
4. $\y-\phi=\pi m,\quad \x+\phi=\pi n$.

The detailed description of the critical points is given in the following table: $$\begin{array}{l|l|l|l|l} \#& (x,y)& \Delta(x,y) &\text{Type}& F(x,y)\\ \hline 1.& (0,0)& 3\sin^2\f&\text{extreme}& 3\sf\\ &\left(-\frac{2\pi}3,\frac{2\pi}3\right)& 3\sin^2\left(\f+\frac{2\pi}3\right) &\text{extreme}& 3\sin\left(\f+\frac{2\pi}3\right)\\ &\left(\frac{2\pi}3,-\frac{2\pi}3\right)& 3\sin^2\left(\f-\frac{2\pi}3\right) &\text{extreme}& 3\sin\left(\f-\frac{2\pi}3\right)\\ \hline 2.&(-2\f,-4\f)&-\sin^2(3\f)&\text{saddle}&\sin3\phi\\ 3.&(4\f,2\f)&-\sin^2(3\f)&\text{saddle}&\sin3\phi\\ 4.&(-2\f,2\f)&-\sin^2(3\f)&\text{saddle}&\sin3\phi\\ \end{array}\tag3$$

In the cases $2,3,4$ a multiple of $2\pi$ is assumed to be added to corresponding coordinate of the critical point to bring it into the range $[-\pi,\pi]$. This does not affect the values of the Hessian and the function.

The following figure shows the dependence of the local (blue lines) and boundary (red lines) extrema on the parameter $\f$:

One observes that the boundary extrema play no role, so that the sufficient condition in question is: $$ S_\text{min}(\f)\le\frac C{3\sqrt{A^2+B^2}}\le S_\text{max}(\f), $$ where $$ S_\text{min}(\f)=\min\left[\sf,\sin\left(\f+\frac{2\pi}3\right),\sin\left(\f-\frac{2\pi}3\right)\right];\\ S_\text{max}(\f)=\max\left[\sf,\sin\left(\f+\frac{2\pi}3\right),\sin\left(\f-\frac{2\pi}3\right)\right], $$ with $\phi$ defined by $(1)$.

Answer 2

Using Cauchy-Schwarz inequality repeatedly: $C^2 = (A\cos(x-y)+B\sin(x-y) +A\cos x-B\sin x + A\cos y + B\sin y)^2 \le 3(m^2+p^2+q^2), m^2 \le A^2+B^2, n^2 \le A^2+B^2, p^2 \le A^2+B^2\implies C^2 \le 3(A^2+B^2+A^2+B^2+A^2+B^2)= 9(A^2+B^2). $, Thus the necessary condition is: $C^2\le 9A^2+9B^2$.