# A system of definitely integral equations involving symmetry and the arc-length formula

The Equations:

Given $\alpha \in (0,1)$. Find a continuous function $f: I_{\alpha} \equiv (0,\alpha) \rightarrow \mathbb{R}$ satisfies all equations below:

I. $$f(x) = f(\alpha-x) \; \; \; \forall x \in I_{\alpha}$$

II. $$1 = \int_0^{\alpha} f(x) \; \mathbb{d}x$$

III. $$\dfrac{1}{\sqrt{1-\alpha^2}} = \int_0^{\alpha} \sqrt{1+f^2(x)} \; \mathbb{d}x$$

$\text{}$

This is not an assignment, but an improvised problem. I've searched, brutely and unsuccessfully, through some elementary candidates, such as functions containing trigonometry and / or inverse trigonometry. However, I might have failed to recognize solutions in those candidates.

• The arc-length formula has a $(f'(x))^2$ and not a $(f(x))^2$. Since you mention arc-length in the title, which one do you want in the integral? – mickep Jan 21 '16 at 14:45
• Sorry, the confusing title is my bad. I mean $(f(x))^2$, and I know the arc-length formula includes $(f'(x))^2$. What I imply in the title is that $f$ plays the role of a derivative. I think the problem is more easily stated this way. – Vincent J. Ruan Jan 21 '16 at 15:31

Edited based on addition of 'continuous' to question Property I implies that $f$ is symmetric around $x = \frac{\alpha}{2}$. Therefore, we can restrain ourselves to the interval $(0,\alpha/2)$ and write the integral conditions as \begin{align} 1 &= 2 \int_0^{\alpha/2} f(x) \text{d} x,\\ \frac{1}{\sqrt{1-\alpha^2}} &= 2\int_0^{\alpha/2} \sqrt{1+f(x)^2} \text{d} x. \end{align} As you don't demand any smoothness of $f$, we could try a piecewise linear function such as \begin{equation} f(x) = \left\{ \begin{array}{lcr} 0 & \text{if} & 0 < x < \frac{\alpha}{2} - \beta -\epsilon\\ \gamma + \frac{\gamma}{\epsilon} (x+\beta - \alpha/2) & \text{if} & \frac{\alpha}{2} - \beta -\epsilon < x < \frac{\alpha}{2} - \beta \\ \gamma & \text{if} & \frac{\alpha}{2} - \beta < x < \frac{\alpha}{2} \end{array}\right. \end{equation} As it turns out, to satisfy the integral constraints, you have to solve \begin{align} 1 &= \gamma(2 \beta + \epsilon), \\ \frac{1}{\sqrt{1-\alpha^2}} &= \alpha + (2 \beta + \epsilon)\sqrt{1+\gamma^2} - 2(\beta+\epsilon) + \frac{\epsilon}{\gamma} \text{arcsinh} \gamma. \end{align}

• Firstly, I forgot about the continuous condition, so it's my bad. Still, thank you for the answer. Secondly, is your formula of $\beta$ derived from the equation $\alpha - 2\beta + 2\beta \sqrt{1+\dfrac{1}{4\beta^2}} = \dfrac{1}{\sqrt{1-\alpha^2}}$ ? If so, there is a condition of $\alpha$ necessary for $\beta$ to be found. Thirdly, based on your idea, one can construct $f$ to be a part-wise linear and continuous function. So, thanks again! – Vincent J. Ruan Jan 21 '16 at 15:52
• Yep, you can smoothen the jump however you like, as long as the function remains symmetric around $x=\alpha/2$. And yes, that's the formula for $\beta$. Feel free to accept the answer! – Frits Veerman Jan 21 '16 at 16:00
• So, one can only find $\beta \in \left [ 0,\dfrac{\alpha}{2} \right ]$ when $\alpha$ is not larger than the real root of $X^3+2X^2-2 = 0$. By the way, the part-wise linear case hasn't been worked out thoroughly, and I doubt we'd not be in a similar situation. Sorry that I can't accept the answer. – Vincent J. Ruan Jan 21 '16 at 16:31
• Absolutely true, no worries. What would you like to have worked out more in the piecewise linear case? – Frits Veerman Jan 21 '16 at 17:00
• Actually, I didn't mean to make you do it, I'd figure it out. Thank you for doing it already. On the other hand, we're in the same position in this case as your previous case: one can only find $\beta, \epsilon \in \left [ 0,\dfrac{\alpha}{2} \right ]$ when $\alpha$ is smaller than the real root of $X^3+2X^2−2=0$. I've found that by analyzing the function $\dfrac{\sqrt{\gamma^2+1}-1- \left ( \dfrac{1}{\sqrt{1-\alpha^2}} - \alpha \right ) \gamma}{\gamma-\ln{(\gamma+\sqrt{1+\gamma^2})}}$. So, I guess piecewise linear cases, in general, provide incomplete solutions. – Vincent J. Ruan Jan 22 '16 at 3:50

This is a 'reverse top-hat' semi-solution (leaving out continuity). In the same manner as this solution, many piecewise smooth solutions can be found. There are more details in the inspiring Frits Veerman's answer and our discussion below that.

The Solution:

$$f(x) = \begin{cases} \gamma & \text{ if } \; 0 < x < \dfrac{\alpha}{2}-\beta \\ -\delta & \text{ if } \; \dfrac{\alpha}{2}-\beta \le x \le \dfrac{\alpha}{2}+\beta \\ \gamma & \text{ if } \; \dfrac{\alpha}{2}+\beta < x < \alpha \end{cases}$$

with $\gamma, \delta > 0$ and $0 < \beta < \dfrac{\alpha}{2}$ and

$$\left\{\begin{matrix} 1 = \gamma(\alpha-2\beta) - \delta(2\beta)\\ \dfrac{1}{\sqrt{1-\alpha^2}} = \sqrt{1+\gamma^2} \cdot (\alpha-2\beta) + \sqrt{1+\delta^2} \cdot (2\beta) \end{matrix}\right.$$.

Now we enter the realm of self-made convenience. Choose $\gamma = \delta$. We have:

$$\left.\begin{matrix} 1 = \gamma(\alpha-4\beta)\\ \dfrac{1}{\sqrt{1-\alpha^2}} = \sqrt{1+\gamma^2} \cdot \alpha \end{matrix}\right\} \Rightarrow \left\{\begin{matrix} \delta = \gamma = \sqrt{\dfrac{1}{\alpha^2(1-\alpha^2)}-1}\\ \beta = \dfrac{\alpha}{4} \left ( 1 - \sqrt{\dfrac{1-\alpha^2}{1-\alpha^2+\alpha^4}} \right ) \end{matrix}\right.$$.