Spring 2016, problem 16

I assume, that the requirement satisfied by $f$ is to be read as:

$f(x + \frac{3 \pi}{2}) ( f''(x) + f(x) ) = 1$ (Eq. 1)

and we can thus infer, that $f(x+ \frac{3 \pi}{2})$ (and therefore $f(x)$) is nonzero everywhere. Since $f$ is in particular assumed continuous, it follows, that either $f >0$ everywhere or or $f \lt 0$ everywhere (otherwise $f$ would have a zero!). We can assume, that the first inequality holds, for if $f \lt 0$ everywhere, then $g := -f > 0$ everywhere and g satisfies the same requirements as $f$.

Since $f$ is even, so is $\frac{1}{f'' + f}$. Also, $f(x + \frac{3 \pi}{2}) = f(x - \frac{\pi}{2})$ from $2\pi$-periodicity. So $f(x - \frac{\pi}{2})$ is even. It is easy to show, that any even function, that stays even when shifted by $\frac{\pi}{2}$, is $\pi$-periodic. And so we have the intermediate result:

$f$ is $\pi$-periodic.

By using the $\pi$ periodicity, it is easily shown, that:

$f(x) = \frac{1}{f''(x - \frac{\pi}{2}) + f(x - \frac{\pi}{2})}$

This equation, taken together with $(Eq. 1)$, yields:

$f(x) f''(x - \frac{\pi}{2}) = f''(x) f(x - \frac{\pi}{2})$.

Define $g(x) := f(x - \frac{\pi}{2})$. Then:

$fg'' = f''g \rightarrow \partial (f'g - f g') = 0 \rightarrow f'g = f g' + C$

with $C$ a constant (the $\partial$ symbol denotes the ordinary derivative operator).

But evaluating the last equation at $y := x + \frac{\pi}{2}$ gives:

$f'(x + \frac{\pi}{2}) f(x) = f(x + \frac{\pi}{2}) f'(x) + C$

And therefore, using $\pi$-periodicity:

$g'(x) f(x) = g(x) f'(x) + C$.

This is only compatible with $f'g = f g' + C$, if $C = 0$. So we have:

$\frac{f'}{f} - \frac{g'}{g} = \partial [ ln(\frac{f}{g}) ] = 0$

So $f = \lambda g$ for some positive constant $\lambda$.

The value of $\lambda$ can't be anything else than 1, for otherwise $f(x)$ would converge to zero for large $x$ (case $ \lambda \lt 1 $ or would be unbounded from above (case $\lambda > 1$). Both cases are excluded for a positive, periodic function. Therefore $\lambda = 1$ and we're finished.

If $f$ is even $f''$ is even and $x \mapsto f\left( {x + 3\frac{\pi }{2}} \right)$ is even. Then $f\left( {x + 3\frac{\pi }{2}} \right) = f\left( { - x + 3\frac{\pi }{2}} \right) = f\left( {x - 3\frac{\pi }{2}} \right)$ and $f$ is $3\pi $ periodic and therefore $\pi $ periodic.

$f''\left( x \right) + f\left( x \right)\left( {1 - \frac{1}{{f\left( x \right)f\left( {x + \frac{\pi }{2}} \right)}}} \right) = 0 $

$f''\left( {x + \frac{\pi }{2}} \right) + f\left( {x + \frac{\pi }{2}} \right)\left( {1 - \frac{1}{{f\left( x \right)f\left( {x + \frac{\pi }{2}} \right)}}} \right) = 0$

$f$ and $g: x \mapsto f\left( {x + \frac{\pi }{2}} \right)$ are two even solutions of the linear équation

$y'' + \left( {1 - \frac{1}{{f\left( x \right)f\left( {x + \frac{\pi }{2}} \right)}}} \right)y = 0$ (1)

$f\left( 0 \right) = a$, $f'\left( 0 \right) = 0$, $a \ne 0$ otherwhise $f=0$

$g\left( 0 \right) = b$, $g'\left( 0 \right) = 0$, $b \ne 0$ otherwhise $g=0$

$h = bf - ag$ is a solution of (1). $h\left( 0 \right) = h'\left( 0 \right) = 0$. Then $h = 0$

$f\left( {x + \frac{\pi }{2}} \right) = \lambda f\left( x \right)$ where $\lambda = \frac{b}{a}$

$f\left( x \right) = f\left( {x + \pi } \right) = {\lambda ^2}f\left( x \right)$

For all $x$, $f\left( x \right) \ne 0$. Then ${\lambda ^2} = 1$

But if $f\left( {x + \frac{\pi }{2}} \right) = - f\left( x \right)$, for some $x \in \mathbb{R}$, $f\left( x \right) = 0$.

$\lambda = 1$ and $f$ is $\frac{\pi }{2}$ periodic.

!!!!!!

All the solutions of $y'' + y = \frac{1}{y}$ seem to be periodic.

They verify $2y'y'' + 2y'y = 2\frac{{y'}}{y}$

and an equation of this type: ${y'^2} + {y^2} = 2\ln y + C$ , $C \ge 1$

The period of this solution is $T = 2\int_{{y_0}}^{{y_1}} {\frac{{dy}}{{\sqrt {C - {y^2} + 2\ln y} }}} $ , ${y_0} \le {y_1}$

where ${y_0}$ and ${y_1}$ are the solutions of ${y^2} - 2\ln y = C$

With $\ln y = \frac{x}{2}$

$T = \int\limits_{{x_0}}^{{x_1}} {\frac{{{e^{\frac{x}{2}}}{\rm{dx}}}}{{\sqrt {C - {e^x} + x} }}} $

where ${x_0}$ and ${x_1}$ are the solutions of ${e^x} = x + C$

With $x = t - \ln \frac{{{\mathop{\rm sh}\nolimits} t}}{t}$

Let $\varphi \left( t \right) = {e^x} - x = \frac{{t{e^t}}}{{{\mathop{\rm sh}\nolimits} t}} - t + \ln \left( {\frac{{{\mathop{\rm sh}\nolimits} t}}{t}} \right) = t\coth t + \ln \left( {\frac{{{\mathop{\rm sh}\nolimits} t}}{t}} \right)$

$\varphi $ is even.

The two solutions of the equation $\varphi \left( t \right) = C$ are opposite.

If $C = \varphi \left( \alpha \right)$ with $\alpha > 0$

With $x = t - \ln \frac{{{\mathop{\rm sh}\nolimits} t}}{t}$

$T = \int\limits_{ - \alpha }^\alpha {\frac{{\sqrt {\frac{{t{e^t}}}{{{\mathop{\rm sh}\nolimits} t}}} \left( {1 + \frac{1}{t} - \coth t} \right)}}{{\sqrt {\varphi \left( \alpha \right) - \varphi \left( t \right)} }}} dt$

$T=\int\limits_{ - \alpha }^\alpha {\frac{{\sqrt {\frac{t}{{{\mathop{\rm sh}\nolimits} t}}} \left( {{e^{\frac{t}{2}}} + \frac{{{e^{\frac{t}{2}}}}}{t} - {e^{\frac{t}{2}}}\coth t} \right)}}{{\sqrt {\varphi \left( \alpha \right) - \varphi \left( t \right)} }}dt}$

By parity,

$T=\int\limits_{ - \alpha }^\alpha {\frac{{\sqrt {\frac{t}{{{\mathop{\rm sh}\nolimits} t}}} \left( {{\mathop{\rm ch}\nolimits} \frac{t}{2} + \frac{{{\mathop{\rm sh}\nolimits} \frac{t}{2}}}{t} - {\mathop{\rm sh}\nolimits} \frac{t}{2}\coth t} \right)}}{{\sqrt {\varphi \left( \alpha \right) - \varphi \left( t \right)} }}dt} $

and finally

$T = 2\int\limits_0^\alpha {\sqrt {\frac{t}{{{\mathop{\rm sh}\nolimits} t}}} \left( {\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{t}{2}}} + \frac{{{\mathop{\rm sh}\nolimits} \frac{t}{2}}}{t}} \right)\frac{{{\mathop{\rm dt}\nolimits} }}{{\sqrt {\varphi \left( \alpha \right) - \varphi \left( t \right)} }}} $

If you want to find the period of the even solution of $y'' + y = \frac{1}{y}$ which verify $y\left( 0 \right) = a$ and $y'\left( 0 \right) = 0$

Calculate $C = {a^2} - 2\ln a$, $C \ge 1$

Find $\alpha > 0$ verifying $\varphi \left( \alpha \right) = C$

$t = \alpha \cos u$

Then $T = 2\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\frac{{\alpha \cos u}}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos u} \right)}}} \left( {\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos u}}{2}}} + \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos u}}{2}}}{{\alpha \cos u}}} \right)}}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos u} \right)}}{{{\alpha ^2}{{\sin }^2}u}}} }}{\mathop{\rm du}\nolimits} } $

I think that

1) as $\alpha $ approaches 0, T approaches $\pi \sqrt 2 $

Near from 0, $\varphi \left( \alpha \right) = 1 + \frac{{{\alpha ^2}}}{2} + o\left( {{\alpha ^2}} \right)$

$\mathop {\lim }\limits_{\alpha \to 0} 2\frac{{\sqrt {\frac{{\alpha \cos u}}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos u} \right)}}} \left( {\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos u}}{2}}} + \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos u}}{2}}}{{\alpha \cos u}}} \right)}}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos u} \right)}}{{{\alpha ^2}{{\sin }^2}u}}} }} = \frac{2}{{\sqrt {\frac{1}{2}\left( {\frac{{1 - {{\cos }^2}u}}{{{\rm{si}}{{\rm{n}}^2}u}}} \right)} }} = 2\sqrt 2 $

$\mathop {\lim }\limits_{\alpha \to 0} T = \int\limits_0^{\frac{\pi }{2}} {2\sqrt 2 {\mathop{\rm dt}\nolimits} } = \pi \sqrt 2 $

(dominated convergence?)

2) T is a decreasing fonction of $\alpha $

??? No proof...

But

$\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos \theta }}{2}}} \le \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos \theta }}{2}}}{{\alpha \cos \theta }}$, $\left( {{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right) \ge \alpha \cos \theta } \right)$

$T \ge 2\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos \theta } \right)}}{{{\alpha ^2}{{\sin }^2}\theta }}} }}\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \left( {\frac{1}{{{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos \theta }}{2}}}} \right){\rm{d\theta }}} $

If $\delta \left( x \right) = \varphi \left( x \right) - 1 - \frac{{{x^2}}}{2}$

$\begin{array}{l}\delta '\left( x \right) & = \varphi '\left( x \right) - x = 2\coth x - \frac{x}{{{{{\mathop{\rm sh}\nolimits} }^2}x}} - \frac{1}{x} - x\\ & = \frac{1}{x}\left( {2x\coth x - {x^2}{{\coth }^2}x - 1} \right)\\ & = - \frac{1}{x}{\left( {1 - x\coth x} \right)^2} \le 0\end{array}$

$\delta$ is a decreasing fonction of $x$ on ${\mathbb{R}^ + }$

and for $x \ge y \ge 0$, $0 \le \varphi \left( x \right) - \varphi \left( y \right) \le \frac{{{x^2} - {y^2}}}{2}$

So $T \ge 2\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{1}{2}\frac{{{\alpha ^2} - {\alpha ^2}{{\cos }^2}\theta }}{{{\alpha ^2}{{\sin }^2}\theta }}} }}\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \left( {\frac{1}{{{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos \theta }}{2}}}} \right){\rm{d\theta }}} $

$T \ge 2\sqrt 2 \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \left( {\frac{1}{{{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos \theta }}{2}}}} \right){\rm{d\theta }}} $ which is a decreasing function of $\alpha$

for $\alpha \le 3$, $T \ge 2\sqrt 2 \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{3\cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {3\cos \theta } \right)}}} \left( {\frac{1}{{{\mathop{\rm ch}\nolimits} \frac{{3\cos \theta }}{2}}}} \right){\rm{d\theta }}} \approx {\rm{2}}{\rm{.341298620}}$

For $x \ge 0$,

$\begin{array}{l}\varphi ''\left( x \right) & = \frac{{2x{\rm{ch}}x}}{{{{{\mathop{\rm sh}\nolimits} }^3}x}} - \frac{2}{{{{{\mathop{\rm sh}\nolimits} }^2}x}} + \frac{1}{{{x^2}}} - \frac{1}{{{{{\mathop{\rm sh}\nolimits} }^2}x}}\\ & = \frac{{2{\mathop{\rm ch}\nolimits} (x)}}{{{{{\mathop{\rm sh}\nolimits} }^3}x}}\left( {x - {\rm{th}}x} \right) + \frac{1}{{{x^2}}} - \frac{1}{{{{{\mathop{\rm sh}\nolimits} }^2}x}} \ge 0\end{array}$

and $\varphi '\left( x \right) \le 2 = \mathop {\lim }\limits_{ + \infty } \varphi '$

For $x \ge y \ge 0$ , $0 \le \varphi \left( x \right) - \varphi \left( y \right)\mathop = \limits_{c \in \left[ {x,y} \right]} \left( {x - y} \right)\varphi '\left( c \right) \le 2\left( {x - y} \right)$

$T \ge 2\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos \theta } \right)}}{{{\alpha ^2}{{\sin }^2}\theta }}} }}\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos \theta }}{2}}}{{\alpha \cos \theta }}{\rm{d\theta }}} $

$T \ge 2\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{{2\left( {\alpha - \alpha \cos \theta } \right)}}{{{\alpha ^2}{{\sin }^2}\theta }}} }}\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos \theta }}{2}}}{{\alpha \cos \theta }}{\rm{d\theta }}} = \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{1 + \cos \theta }}{{\cos \theta }}} \sqrt {{\rm{th}}\frac{{\alpha \cos \theta }}{2}} {\rm{d\theta }}} $ which is an increasing function of $\alpha$.

for $\alpha \ge 3$, $T \ge \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{1 + \cos \theta }}{{\cos \theta }}} \sqrt {{\rm{th}}\frac{{{\rm{3}}\cos \theta }}{2}} {\rm{d\theta }}} \approx {\rm{2}}{\rm{.123182323}}$

For all $\alpha > 0$ , $T \ge 2$

So the solutions of the problem are $y = 1$ and $y = -1$.

3) as $\alpha $ approaches $+ \infty $ , T approaches $\pi$

as $t$ approaches $+ \infty $ $\varphi \left( t \right) \sim 2t$

as $\alpha $ approaches $+ \infty $

$2\frac{{\sqrt {\frac{{\alpha \cos t}}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos t} \right)}}} \left( {\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos t}}{2}}} + \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos t}}{2}}}{{\alpha \cos t}}} \right)}}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos t} \right)}}{{{\alpha ^2}{{\sin }^2}t}}} }} \sim \frac{{2\sqrt {2\alpha \cos t} \,{e^{ - \frac{{\alpha \cos t}}{2}}}\frac{{{e^{\frac{{\alpha \cos t}}{2}}}}}{{2\alpha \cos t}}}}{{\sqrt {\frac{2}{\alpha }\left( {\frac{{1 - \cos t}}{{{{\sin }^2}t}}} \right)} }}$

$\frac{{2\sqrt {2\alpha \cos t} \,{e^{ - \frac{{\alpha \cos t}}{2}}}\frac{{{e^{\frac{{\alpha \cos t}}{2}}}}}{{2\alpha \cos t}}}}{{\sqrt {\frac{2}{\alpha }\left( {\frac{{1 - \cos t}}{{{{\sin }^2}t}}} \right)} }} = \frac{{\sqrt {1 + \cos t} }}{{\sqrt {\cos t} }}$

$\mathop {\lim }\limits_{\alpha \to + \infty } T = \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{1 + \cos t}}{{\cos t}}} {\mathop{\rm dt}\nolimits} } = \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{2}{{1 - 2{{\sin }^2}\frac{t}{2}}}} \cos \frac{t}{2}{\mathop{\rm dt}\nolimits} } $

With ${u = \sqrt 2 \sin \frac{t}{2}}$

$\mathop {\lim }\limits_{\alpha \to + \infty } T = \int\limits_0^1 {\sqrt {\frac{2}{{1 - {u^2}}}} \sqrt 2 {\rm{du}}} = \pi $

(dominated convergence?)

I think that exept for the solutions $y = 1$ and $y = -1$ the minimum period is $\pi$