# Proving that simple harmonic motion and pendulum motion are periodic

Consider the equation of motion for simple harmonic motion, \begin{align} f'' + \omega^2 f = 0, \end{align} where $$f: \mathbf{R} \rightarrow \mathbf{R}$$ and $$\omega \in \mathbf{R}$$. Suppose we have some initial conditions $$f(0) = x_0$$, $$f'(0) = v_0$$.

Question: Without explicitly solving this equation, is it possible to prove that any solution is periodic? That is, can we show that if $$f$$ is a solution, then \begin{align} f(t + T) = f(t) \end{align} for some $$T$$?

I also have the same question for the equation of motion of a simple pendulum, \begin{align} f'' + \omega^2 \sin f = 0. \end{align} Feel free to make any assumptions you want about the niceness of the solutions.

• @JohnDouma I don't think that's true. The motion of a pendulum is still periodic for arbitrary amplitude, and we can calculate the period explicitly in terms of an elliptic integral. The difference between the pendulum and the simple harmonic oscillator is that the period of the pendulum is not independent of the initial conditions.
– d_b
Jul 28, 2022 at 23:37
• A possible start suggested by the physics. You can show that $f^2 + f'^2$ is constant (it's the energy). Perhaps ask at physics stackexchange. Jul 28, 2022 at 23:41
• Similar question here
– Sal
Jul 28, 2022 at 23:48
• @Sal In your linked answer you say that the phase portrait is closed and thus we get periodicity. Is that easy to see? Here we have a circle and of course once we return we can easily show that the solution is periodic. However, why does it need to get back? Jul 28, 2022 at 23:57
• @SeverinSchraven If we accept that the level curves of $H(x,p)$ are closed, then periodicty does follow because any one point in phase space uniquely determines the trajectory passing through it, as the equation is of second order. On the other hand, if you are asking how to demonstrate that the level curves are closed: see the answer here
– Sal
Jul 29, 2022 at 0:24

For convenience, we rewrite the equation \begin{align} f'' + \omega^2 \sin f = 0 \end{align} in the form of a system: $$\tag{1} \left\{\begin{array}{lll} \dot x&=&y\\ \dot y&=&-\omega^2 \sin x. \end{array}\right.$$ The system (1) has the first integral/conserved quantity $$V(x,y)= \omega^2(1-\cos x) +\frac{y^2}2.$$ It means that it remains constant along the trajectories of (1). Indeed, let $$(x(t),y(t))$$ be the solution to (1). Then the derivative along the trajectories $$\dot V= \frac{d}{dt}V(x(t),y(t))= \frac{\partial V}{\partial x}\frac{dx}{dt}+ \frac{\partial V}{\partial y}\frac{dy}{dt}= \omega^2\sin x \cdot y + y(-\omega^2 \sin x)\equiv 0.$$ This implies that any solution $$(x(t),y(t))$$ lies on the level curve of $$V$$. By drawing these level curves $$V(x,y)=C$$, one can see that not all of them are closed: One can study the form of the level curves $$\omega^2(1-\cos x) +\frac{y^2}2=C$$ using the fact that they are composed of two graphs of functions: $$\tag{2} y=\pm \sqrt{ 2C-2\omega^2(1-\cos x) }.$$ If $$C>2\omega^2$$, then $$2C-2\omega^2(1-\cos x)>4\omega^2-2\omega^2+2\omega^2\cos x= 2\omega^2(1+\cos x)\ge0$$ and the level curve splits into two separate function graphs, one in the upper, the other in the lower half-plane. Thus, the level curve does not contain a closed curve and therefore the corresponding solution is not periodic.

If $$C<2\omega^2$$, then two graphs, symmetrical about the $$x$$ axis, touch at the points of this axis and form a closed curve. To prove that the solutions corresponding to the values $$C<2\omega^2$$ are periodic, one can use the uniqueness of the solution to the initial value problem. If the solution, having passed along a closed trajectory, returned to the starting point, then the next turn will be identical to the one passed due to the coincidence of the initial conditions. On the other hand, the state cannot stop halfway without going through a full turn along a closed curve, since the velocity of the movement can be bounded from below for any curve, except for the equilibrium points and heteroclinic orbits (see below). As a result, the state travels a finite distance on the phase plane in a finite time.

$$C=2\omega^2$$ (drawn in red) is a special case. It corresponds to the set of heteroclinic orbits.

Update By the way, the representation (2) allows to calculate the period of oscillation. Let us choose some particular constant $$C<2\omega^2$$. The curve (2) near the origin intersects the x-axis at the points $$x=\pm\arccos\left(1-\frac{c}{\omega^2} \right).$$ Denote $$r=\arccos\left(1-\frac{c}{\omega^2} \right)$$.

Since $$x$$ is $$f$$ and $$y$$ is $$f'$$, we have (for definiteness, we choose the upper half of the curve) $$\frac{df}{dt}= \sqrt{ 2C-2\omega^2(1-\cos f) }$$ $$\frac{df}{\sqrt{ 2C-2\omega^2(1-\cos f) }}=dt.$$ Integrating over the upper half of the closed curve, we obtain $$\int_{-r}^{r}\frac{df}{\sqrt{ 2C-2\omega^2(1-\cos f) }}=\int_{0}^{T/2}dt$$ and $$T/2=\int_{-r}^{r}\frac{df}{\sqrt{ 2C-2\omega^2(1-\cos f) }}.$$

• @Lutz Lehmann Thanks, fixed
– AVK
Aug 3, 2022 at 20:40

I don't know if I answered your first question. Let $$g(t)=\omega f(t)+if'(t)$$ where $$i$$ is the imaginary unit. Then $$g'(t)=\omega^2f'(t)+if''(t) =\omega f'(t)-i\omega^2f(t)=-i\omega(\omega f(t)+if'(t))=-i\omega g(t).$$ So $$\frac{g'(t)}{g(t)}=-i\omega$$ which gives $$g(t)=g(0)e^{-i\omega t}.$$ Let $$g(t+T)=g(t)$$ and then one has $$e^{-i\omega T}=1$$. So $$T=\frac{2\pi}{\omega}$$.