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Let $0 < a < 1$. Show that the equation

$$\int_0^x{\left( \sin \left(\frac{\pi \sin\frac{\pi t}{2}}{2} \right)+ \frac{2}{\pi} \sin^{-1} \left( \frac{2}{\pi} \sin^{-1}(t) \right) -2t \right)}dt = \frac{1}{2} \left( a-\frac{2}{\pi}\sin^{-1} \left( \frac{2}{\pi} \sin^{-1}(a) \right) \right) \left( \sin \frac{\pi \sin\frac{\pi a}{2}}{2} -a\right)$$

has at least two solutions on the interval $0 \leq x \leq 1$.

To show an equation $f(x)=g(x)$ has solution, one would usually define $g(x)=f(x)-g(x)$ and find the number of roots of $h(x)$. But in this case, I'm having trouble to find the roots of the equation above.

Can anyone give some hint?

Idonknow
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  • There's a great deal of structure in the integrand; the first thing I'd try is to exploit that structure. Particularly, it seems clear that $\phi(t) = \sin(\pi t/2)$, a homeomorphism of the closed unit interval, plays a starring role. (That's just an idea. I haven't personally made any attempt.) – Andrew D. Hwang May 24 '15 at 10:39
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    I'd wonder about generalization: if the statement is true for $\displaystyle \int\limits_0^x\left(f(t)+f^{-1}(t)-2t\right)dt = \frac{1}{2}\left(a-f^{-1}(a)\right)\left(f(a)-a\right)$ for arbitrary monotonic concave $f: f(0)=0,\ f(1)=1$. – Alexey Burdin May 24 '15 at 10:48

1 Answers1

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Since we are unable to explore $\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}x\right)\right)$ let's consider more general case:
Let $f$ be concave monotonic integrable function such, that $f(0)=0,\ f(1)=1$ and $f^{-1}$ is also integrable.
Consider $\displaystyle g(x)=\int\limits_0^x \left(\left(f(t)-t\right)-\left(t-f^{-1}(t)\right)\right)\,dt$ and $h(x)=\frac{1}{2}\left(f(x)-x\right)\left(x-f^{-1}(x)\right)$
$g(x)$ is the green shaded area minus red shaded area so $g(0)=g(1)=0$ as for $x=1$ these areas are equal. Also $g(x)$ the gray shaded area.
We need to prove that $g(x)=h(a)$ twice. Since $g(x)$ is continuous, $g(0)=g(1)=0$ and $g(x)\ge 0$, we need to find such $x_0$, that $g(x_0)>h(a)>0$ and we've done.
Consider $g(a)$ is the area of the curved gray shaded triangle and $h(a)$ is the area of straight-sided triangle on the same points, thus $g(a)>h(a)$ since $f(x)$ is concave and we've done.
To say, $f(x)$ is increasing and concave as a composition of increasing and concave functions.

Alexey Burdin
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