Perfect Harmony II

The functional is $I(\psi) \; = \; \int_0^{\frac12\pi} \frac{\psi'(x)^2}{4\psi(x)(1-\psi(x))}\,dx$ By the Cauchy-Schwarz Inequality, $\begin{aligned} \left(\int_0^{\frac12\pi} \frac{\psi'(x)}{\sqrt{\psi(x)(1-\psi(x))}}\,dx\right)^2 & \le \; \left(\int_0^{\frac12\pi}\,dx\right)\left(\int_0^{\frac12\pi} \frac{\psi'(x)^2}{\psi(x)(1-\psi(x))}\,dx\right) \; = \; 2\pi I(\psi) \\ \left(\int_0^1 \frac{du}{\sqrt{u(1-u)}}\right)^2 & \le \; 2\pi I(\psi) \\ \pi^2 & \le \; 2\pi I(\psi) \end{aligned}$ for all suitable $\psi$ . Thus $I(\psi) \ge \tfrac12\pi$ for all suitable $\psi$ (by "suitable" I mean that $\psi'(x)$ has to be sufficiently well-behaved at both $x=0$ and $x=\tfrac12\pi$ for the integral $I(\psi)$ to exist), with equality obtained when $\psi'(x) = k\sqrt{\psi(x)(1-\psi(x))}$ for some constant $k$ . With the boundary conditions $\psi(0) = 0$ and $\psi(\tfrac12\pi) = 1$ , we solve this differential equation to obtain $\psi(x) = \sin^2x$ , and so the minimum value of $\boxed{\tfrac12}\pi$ for $I(\psi)$ can be achieved.

Given the potentially singular nature of the integrand at $x=0$ and $x=\tfrac12\pi$ , the Calculus of Variations approach would have to be established carefully!

Letting the integrand be $L$ , we can simplify it to be $L =\frac { { \psi ' }^{ 2 } }{ 4\psi \left( 1-\psi \right) }$ By the Euler-Lagrange equation, for $I\left( \psi \right)$ to be minimised, $L$ must satisfy the equation $\frac { \partial }{ \partial x } L-\frac { d }{ dx } \left( L-\psi '\frac { \partial }{ \partial \psi ' } L \right) =0$ $L-\psi '\frac { \partial }{ \partial \psi ' } L=c$ , where $c$ is an arbitrary constant. This simplifies to $\psi '=\alpha \sqrt { \psi \left( 1-\psi \right) }$ , where $\alpha =\sqrt{-4c}$ . Now, integrating using the variable seperable method yields $\int { \frac { 1 }{ \sqrt { \psi \left( 1-\psi \right) } } } \frac { \alpha \psi }{ dx } dx=\int { \alpha } dx$ $2\arcsin { \sqrt { \psi } } =\alpha x+\beta$ with $\beta$ as the constant of integration. Making $\psi$ the subject and substituting the limits $\psi \left( 0 \right) =0$ and $\psi \left( \frac { \pi }{ 2 } \right) =1$ gives $\psi =\sin ^{ 2 }{ \left( x \right) }$ and $L=1$ Therefore $I\left( \psi \right) =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ 1 } dx=\boxed{\frac { \pi }{ 2 }}$

The given integral attains an extrimum for the function (sinx)^2 (a standard variational problem). The value of the integral for this function is π/2