Prove that the equation of motion of a pendulum has a…

Prove that the equation of motion of a pendulum has a particular solution $y(x)$ that tends to $\pi$ as $x\to\infty$. $y'' + \sin y = 0$.

Step 1: Physical motivation

Here $y$ is the angle of the pendulum from the downward vertical. The equilibria are $y=0$ (stable, bottom) and $y=\pi$ (unstable, top; $\sin\pi=0$, no restoring force). We seek the separatrix: the orbit that leaves the unstable top with vanishing velocity, swings through the bottom, and returns to the top — approaching $y=\pi$ only in the limit $x\to+\infty$.

Step 2: Phase-plane reduction

Since $x$ does not appear explicitly, set $p=y'$ and treat $p$ as a function of $y$, so $y''=p\,dp/dy$: \[p\,\frac{dp}{dy} = -\sin y \ \Rightarrow\ p\,dp = -\sin y\,dy \ \Rightarrow\ \frac{p^2}{2} = \cos y + C. \tag{1}\]

Step 3: Choosing the separatrix energy level

We want $y\to\pi$ with $y'\to0$ as $x\to+\infty$. Setting $p\to0$, $y\to\pi$ in (1) fixes $C=1-\cos\pi$… explicitly $0=\cos\pi+C=-1+C$, so $C=1$: \[\frac{p^2}{2}=1+\cos y = 2\cos^2\frac{y}{2} \ \Rightarrow\ p = y' = \pm2\cos\frac{y}{2}. \tag{2}\]

Step 4: The monotone branch

For $y\in(-\pi,\pi)$, $y/2\in(-\pi/2,\pi/2)$ and $\cos(y/2)>0$ throughout, so the positive branch gives a single smooth, strictly increasing solution across the whole range:

\[ \frac{dy}{dx} = 2\cos\frac{y}{2}. \tag{3} \]

Step 5: Solving the separatrix ODE

Separate variables with $t=y/2$: $\sec t\,dt = dx$. Integrating, $\ln|\sec t+\tan t| = x-x_0$, i.e.\ with $A\equiv e^{x-x_0}$: \[\frac{1+\sin t}{\cos t} = A. \tag{4}\]

Eliminating $\sin t$ via $\sin^2t+\cos^2t=1$ and solving the resulting quadratic in $\cos t$ gives the unique positive root (as $\cos t>0$): \[\cos t = \frac{2A}{A^2+1} = \frac{2}{e^{x-x_0}+e^{-(x-x_0)}} = \operatorname{sech}(x-x_0).\]

Rather than recovering $\sin t$ by a square root — which would lose its sign and produce a kinked, non-differentiable formula — substitute $\cos t$ back into the original, unsquared relation (4):

\[ \sin t = A\cos t - 1 = \frac{2A^2}{A^2+1}-1 = \frac{A^2-1}{A^2+1} = \tanh(x-x_0). \]

This is exact and sign-correct for every $x$, positive or negative of $x_0$. Since $t=y/2\in(-\pi/2,\pi/2)$, inverting with $\arcsin$ (not $\arccos$, which cannot distinguish $t$ from $-t$) gives the single smooth global solution:

$y(x) = 2\arcsin\!\bigl(\tanh(x-x_0)\bigr)$.

Step 6: The limit as $x\to+\infty$

\[ \lim_{x\to+\infty} y(x) = 2\arcsin\!\Bigl(\lim_{x\to+\infty}\tanh(x-x_0)\Bigr) = 2\arcsin(1) = 2\cdot\frac{\pi}{2}. \] $\lim_{x\to+\infty$.

(Symmetrically, $y(x)\to-\pi$ as $x\to-\infty$, and $y(x_0)=0$: the orbit leaves the top, swings through the bottom, and returns to the top — the heteroclinic connection $-\pi\to\pi$.)

Step 7: Existence argument, independent of the closed form

The limit can also be established without solving the ODE explicitly. On the energy curve (2), $y'>0$ for $y\in(-\pi,\pi)$ and $y'=0$ only at $y=\pi$; also $y''=-\sin y<0$ there, so $y$ increases monotonically and is bounded above by $\pi$. Writing $y=\pi-\varepsilon$ near the top, $y'=2\sin(\varepsilon/2)\approx\varepsilon$, so $\varepsilon$ decays like $e^{-x}$ and reaches $0$ only as $x\to\infty$; precisely, $\displaystyle\int_\varepsilon^{\varepsilon_0}\frac{d\varepsilon}{2\sin(\varepsilon/2)}\approx\ln\frac{\varepsilon_0}{\varepsilon}\to\infty$ as $\varepsilon\to0^+$, so $y=\pi$ is never reached at finite $x$. By the monotone convergence theorem, a bounded, monotone increasing function has a limit, and that limit must be the only point where $y'=0$, namely $\pi$.

Step 8: Verification

Differentiate $y=2\arcsin(\tanh u)$, $u=x-x_0$, directly: $y'=\dfrac{2\operatorname{sech}^2u}{\sqrt{1-\tanh^2u}} = \dfrac{2\operatorname{sech}^2u}{\operatorname{sech}u}=2\operatorname{sech}u$, and $y''=-2\operatorname{sech}u\tanh u$. Also $\sin y = 2\sin(y/2)\cos(y/2)=2\tanh u\cdot\sqrt{1-\tanh^2u}=2\tanh u\operatorname{sech}u$. Hence:

\[ y''+\sin y = -2\operatorname{sech}u\tanh u + 2\tanh u\operatorname{sech}u = 0. \ \checkmark \]

As $x\to+\infty$, $y'=2\operatorname{sech}(x-x_0)=2/\cosh(x-x_0)\to0$ exponentially fast, confirming the solution approaches $y=\pi$ tangentially, consistent with $\pi$ being an equilibrium.