2
$\begingroup$

Let $1 \le p <2$ be a parameter. Consider the equation

$$ \frac{2^{p/2} (1-\sqrt{s})^p-1}{\sqrt{s}}=-2^{p/2-1}p(1-\sqrt{s})^{p-1}. \tag{1} $$

I am rather certain that for each $1 \le p <2$, there is unique solution $s=s(p)$ in $(\frac{1}{4},1]$.

Question: Is $ p\to s(p)$ monotonically decreasing in $p$? Is it continuous in $p$? How can I prove this rigorously?

Mathematica doesn't give a closed-form formula for $s(p)$.

Motivation:

This question comes from trying to find a "point of contact" when a certain chord between $(0,H(0)), (s,H(s))$ coincides with the tangent to $H$ at $s$, where $H:=F^q$ and $$ F(s) := \begin{cases} 2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4} \\ 1-2s, & \text{ if }\, s \le \frac{1}{4} \end{cases} $$

The equation $\frac{H(s)-H(0)}{s-0}=H'(s)$ is nothing but equation $(1)$ above.


One idea is to assume that $s(p)$ is differentiable, and differentiate equation $(1)$ w.r.t $p$. Doing that, one gets the following (details here) enter image description here

This is a stream line plot of $s(p)$: the function must follow one of these lines, depending on its initial condition. The horizontal-axis is the $p$-variable and the vertical-axis is $s$-variable. $s(p)$ seems monotonically decreasing on the interval, as required. This seems to suggest that there is a unique solution for every initial condition.


Analysis of $p=1,2$:

Let's prove that $s(2)=\frac{1}{4},s(1)=(2-\sqrt 2)^2 \simeq 0.343$.

For $p=1$ the equation reduces to $ \sqrt 2(1-\sqrt s)-1=-\frac{\sqrt s}{\sqrt 2}$. Setting $x=\sqrt s$, we obtain $ 1-\sqrt 2=x(1/\sqrt 2-\sqrt 2) \Rightarrow x=2-\sqrt 2.$

For $p=2$ the equation reduces to

$2(1-\sqrt s)^2-1=-2(1-\sqrt s)\sqrt s \Rightarrow -1=-2(1-\sqrt s)\big((1-\sqrt s)+\sqrt s \big) \Rightarrow 2(1-\sqrt s)=1 \Rightarrow s=\frac{1}{4}.$


$\endgroup$

3 Answers 3

5
$\begingroup$

With $r:=p/2\in[1/2,1)$ and $y:=1-\sqrt s\in[0,1/2)$, rewrite your equation (1) as $$G(r,y):=2^r y^{2 r-1} (r+y-r y)-1=0. \tag{2}$$ For any $r\in(1/2,1)$, $G(r,0)=-1\ne0$, so that $y=0$ is not a solution of equation (2). Also, $y^{2 r-1}$ is undefined for $r=1/2$ and $y=0$. So, in what follows let us assume $r\in(1/2,1)$ and $y\in(0,1/2)$ by default.

Clearly, $G(r,y)$ is strictly and continuously increasing in $y$ from $G(r,0+)=-1<0$ to $G(r,\frac12-)=2^{-r} (1+r - 2^r)>0$, for each $r$; here one may use the concavity of $1+r - 2^r$ in $r$. So, for each $r$, equation (2) has a unique root $$Y:=Y(r)\in(0,1/2). \tag{3}$$ Moreover, $$G'_y(r,y)=2^{r+1} r y^{2 r-2}(r-1/2 + (1-r)y)>0.$$ So, by the implicit function theorem, the function $Y$ is differentiable (and hence continuous). Moreover, $$Y'(r)=-\frac{G'_r(r,y)}{G'_y(r,y)}\Big|_{y=Y} \overset{\text{sign}}=H(r,Y)>H(1/2,Y)\overset{\text{sign}}=h(Y), $$ where $a\overset{\text{sign}}=b$ means $\text{sign}\, a=\text{sign}\,b$, $H(r,y):=-1 + y - (y + r (1 - y)) \ln(2 y^2)$, and $$h(y):= -\frac{1 - y}{1 + y} - \frac12\,\ln(2 y^2).$$ Note that $h(1/2)>0$ and $h'(y)=-\frac{1+y^2}{y (1+y)^2}<0$, whence $h>0$ and hence $Y'>0$.

Thus, $Y(r)$ is continuously increasing in $r$, which means that the root $s$ of your equation (1) is continuously decreasing in $p$, as you conjectured.

$\endgroup$
2
  • $\begingroup$ This seems like a nice solution, thanks. I am sorry to bother you, but if I am not mistaken, the rewriting of the equation should be $G(r,y):=2^r y^{2 r-1} (r+y-ry)-1=0, $ instead of $G(r,y):=2^r y^{2 r-1} (r+y)-1=0. $ (you have a missing term of $ry$ coming from the multiplication of the original equation by $\sqrt s=1-y$). I was able to verify all the steps after this modification, except for the last one which concerns the sign of $G'_r$. $\endgroup$ Sep 14, 2020 at 7:46
  • $\begingroup$ @AsafShachar : I did miss that term. However, now the solution is only a bit simpler. I have also added details. $\endgroup$ Sep 14, 2020 at 14:12
4
$\begingroup$

Put $t=1-\sqrt{s}\in[0,1/2)$ so the equation writes $$ \Big(1-\frac p2\Big)\, t^p+ \frac p2\, t^{p-1}=2^{-\frac p2}$$

Now if we put $u:=t^{p-1}$ the equation takes the form $$u+\Big( \frac2p -1\Big)\,u^q =\frac {2^{1-\frac p 2 }} p$$ with $q=\frac p{p-1} >1$, that can be solved by series (see e.g. here) (this way one covers an interval $1.57<p\le2$ if I'm not wrong. To cover the other values of $p$, close to $1$, one needs to put the equation in other forms).

$\endgroup$
0
0
$\begingroup$

This can be done with Maple and Mathematica as follows. First, let us look at the plot done with Maple

plots:-implicitplot((2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) =
 -2^(p/2 - 1)*p*(1 - sqrt(s))^(p - 1), p = 1 .. 2, s = 1/4 .. 1);

enter image description here

The result suggests that $s(p)$ changes from approximately $0.34$ to approximately $0.25$ as $p$ runs from $1$ to $2$. More exactly, making use of Mathematica, we have

NMaximize[{s, (2^(p/2)*(1 - Sqrt[s])^p - 1)/Sqrt[s] == -2^(p/2 - 1)*
 p*(1 - Sqrt[s])^(p - 1) && p >= 1 && p <= 2}, {p, s}]

$\{0.343146,\{p\to 1.,s\to 0.343146\}\}$ and

NMinimize[{s, (2^(p/2)*(1 - Sqrt[s])^p - 1)/Sqrt[s] == -2^(p/2 - 1)*
 p*(1 - Sqrt[s])^(p - 1) && p >= 1 && p <= 2}, {p, s}]

$\{0.25, \{p -> 2., s -> 0.25\}\}$

We can find the exact values by

solve(eval((2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) = -2^(p/2 - 1)*p*(1 - sqrt(s))^(p - 1), p = 1), s);

$-4\,\sqrt {2}+6$

and

solve(eval((2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) = -2^(p/2 - 1)*p*(1 - sqrt(s))^(p - 1), p = 2), s);

$\frac 1 4$

Now we find the implicit derivative of $s$ with respect to $s$ by

a := implicitdiff((2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) = -2^(p/2 - 1)*p*(1 - sqrt(s))^(p - 1), s, p):

and its maximum value when $p$ runs from $1$ to $2$ by

DirectSearch:-GlobalOptima(a, {(2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) =
 -2^(p/2 - 1)*p*(1-sqrt(s))^(p - 1), 1<= p, p<=2, s<=-4*sqrt(2) + 6, 1/4 <= s}, maximize);

$[-0.0482867952575873, [p = 1.99999990682054, s = 0.250000000105689], 358]$ Because the default absolute error of the GlobalOptima command equals $10^{-6}$, this does the job.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.