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I want to find the least $3>p>\sqrt{5}$ such that $$ f_p(x)=(p x+x) \sin \left(x-\frac{x}{p}\right)+(p x-x)\sin \left(\frac{x}{p}+x\right)-2 p \cos\left(\frac{2 x}{p}\right)+2 p \cos\left(x-\frac{x}{p}\right)+2 p \cos\left(\frac{x}{p}+x\right)-2 p=0 $$ has a solutions $x\in\left(0,\frac{\pi}{2}\right)$. I use Mathematica to plot $f_p(x)$ for various p. It seems there exists a critical value $p_0$ between $2.24$ and $2.25$ such that if $3>p>p_0$, then $f_p(x)$ has a solution $x\in\left(0,\frac{\pi}{2}\right)$.However, I have no idea to give a rigruous proof and determine $p_0$. Any suggestion, idea, or comment is welcome, thanks!

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(Not a rigorous proof)

If you plot $f_p(x)$ over various $p$ you will find that the solution $x$ is increasing with $p$. Everything follows depends on this observation.

Contour plot of f_p(x)

The minimal $p$ which a solution exists in $x\in(0,\frac\pi2)$ is $p=\sqrt5\approx2.23607$, which could be found by solving $f^{(4)}_p(0)=-\frac{4}{p^3}(p^4-6p^2+5)=0$ (all lower derivatives are identically zero; we are trying to show $x=0$ changes from a quadruple root to quintuple root).

Although this is what you asked, I doubt this is what you really want.

The maximal $p$ can be found by solving $$ 0 = f_p\left(\frac\pi2\right) = p\left( \pi \cos \frac{\pi}{2p} - 2\left(1 + \cos\frac\pi p\right) \right) \implies \cos\frac{\pi}{2p} = \frac\pi4,$$ i.e. $p = \frac{\pi}{2\cos^{-1}(\pi/4)} \approx 2.35340$.

The range of $p$ where solution exists between 0 and $\frac\pi2$ is $2.23607 < p < 2.35340$. Between 2.35340 and 3, there are no solutions $x\in(0,\frac\pi2)$.

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  • $\begingroup$ Thank you for the detailed calculation. Using Mathematica (how did you plot the figure? Mathematica?), I didn't find the solution $x$ is increasing with $p4. So now the question is, how to show this fact rigorously. Do you have any idea? $\endgroup$ – LCH Mar 20 '17 at 17:15
  • $\begingroup$ Yes, you are right. The maximal $p$ is also what I want. $\endgroup$ – LCH Mar 20 '17 at 17:17
  • $\begingroup$ @LCH (1) ContourPlot (2) Maybe you could argue by (a) there is a unique solution in that range and (b) $\frac{\partial f_p(x)}{\partial p} < 0$ for all interesting $p,x$ i.e. $f$ is strictly decreasing with $p$ for the same $x$ $\endgroup$ – kennytm Mar 20 '17 at 17:43
  • $\begingroup$ The minimal $p$ which a solution exists in $x\in(0,\frac\pi2)$ is $p=\sqrt5\approx2.23607$, which could be found by solving $f^{(4)}_p(0)=-\frac{4}{p^3}(p^4-6p^2+5)=0$-> could you please explain this in more details? $\endgroup$ – LCH Mar 20 '17 at 18:58
  • $\begingroup$ @LCH When you decrease $p$, the non-trivial solution will move towards 0. But there already exists a trivial solution at $x=0$ with multiplicity 4 (i.e. $f_p(0) = f_p'(0) = f_p''(0) = f_p^{(3)}(0) = 0$). When we reach the minimal $p$, the non-trivial and trivial solutions will "combine" and have multiplicity 5, i.e. $f_p^{(4)}(0) = 0$. $\endgroup$ – kennytm Mar 20 '17 at 19:08

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