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!


(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 $x\in(0,\frac\pi2)$.

  • $\begingroup$ 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

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.