# Undamped Pendulum Phase Plane Solution

Given the following ODE which is supposed to represent an undamped pendulum, with x representing the vertical angle:

$$\frac{d^2x}{dt^2}= -2\sin(2x)$$

Make the substitution $$\frac{dx}{dt}= y$$

And find an implicit solution in the x-y phase plane?

Note: I do know how to find the particular solution after substituting $x = A\sin(2x)+B\cos(2x)$ into the first differential equation, and then computing the values of $A$ and $B$ that would lead to the right hand side of the equation. I am completely unsure how to find the implicit solution in the $x$-$y$ phase planes and would greatly appreciate any help in that regard. Expressing the first differential equation in terms of y should lead to the following $$\frac{dy}{dt}= -2\sin(2x)$$

Beyond that, I am completely lost.

• Use conservation of energy – Michael Galuza Jul 20 '15 at 17:12
• mathematicalgarden.wordpress.com/2009/03/29/nonlinear-pendulum – Hans Lundmark Jul 20 '15 at 17:14
• By the way, your comment about $A \sin 2x + B \cos 2x$ is not right. You would get a solution of that form if the left-hand side of the equation were $d^2 y/dx^2$, but here it's $d^2 x/dt^2$, so it's not even a linear equation. – Hans Lundmark Jul 20 '15 at 17:18

Let $$E=\frac{\dot x^2}{2} - \cos{2x}.$$ (it's energy, if $m=1$). Hence $$\frac{dE}{dt} = \dot x \ddot x + 2\dot x\sin{2x}= \dot x(\ddot x + 2\sin2x) = 0$$ So, if we denote $y=\dot x$, we have $$y^2 - 2\cos2x = 2E.$$ It's implicit solution. You may draw it on $xy$-plane. Here is plot for various $E$ ($E=0, 1,\ldots,6$): $\hskip.5in$ • I'm afraid this has not been taught before. What does E represent and how was it derived? – abstract Jul 20 '15 at 18:01
• @abstract $E$ is the total energy of the system. It consists of the kinetic energy ($mv^2/2$, where $v$ is the speed) and the potential energy ($mgh$, where there is a constant gravitational acceleration $g$ and $h$ is the height of the pendulum above the reference height). – Ian Jul 20 '15 at 19:12
• @Ian $mgh$, really? – Michael Galuza Jul 21 '15 at 3:39
• @MichaelGaluza I don't understand your comment. Are you objecting to my presentation, asking whether it really is from gravitational potential in the constant acceleration limit, or what? It is in fact $mgh$, there are just some constants hidden in the choice of units. – Ian Jul 21 '15 at 13:05
• @Ian I talked about that there is no gravitation or smth like that. There is just a non-linear pendulum, and its potentail energy is $-\cos2x$ – Michael Galuza Jul 21 '15 at 13:14

Hint:

Find $\frac{dy}{dx}$ by dividing the two equations, then solve the ODE by separation of variables.

The system of differential equations is: $$\frac{dy}{dt}=-2\sin(2x)\quad\quad(1)\\ \frac{dx}{dt}=y \quad\quad\quad (2)\\$$

Divide (1) by (2) $$\frac{dy}{dx}=\frac{-2\sin(x)}{y}$$

Then you can solve it using separation of variables.

• what will you solve ? you'll just get an equation $x'(t) = f(x(t))$, but beside $x(t) =0$, there is no closed form solution nor for $x(t)$ nor for $x'(t)$, maybe an analytic series or another type or series ? – reuns Jul 20 '15 at 17:50
• @reuns: Solve for $y$ in terms of $x$. I'll explain a little bit more. – KittyL Jul 20 '15 at 18:06
• that what I wrote, $y = x' = f^{-1}(x)$ – reuns Jul 20 '15 at 18:12
• @reuns: I am not sure what you meant. The OP wanted an implicit function. This ODE will give him an explicit function $y=f(x)$ as desired. – KittyL Jul 20 '15 at 18:14
• no what is really desired is an expression for $x(t)$ knowing starting conditions. it's a differential equation, and would be "solving the equation". because it's impossible, he will study the phase space of $(x', x)$ – reuns Jul 20 '15 at 18:14

It is a non-linear pendulum that has to be solved in terms of elliptic integrals. Or else solve it numerically before plotting phase portrait $( \dot x - x)$ for given boundary conditions.

$$\frac{d^2\, (2x) }{dt^2}+ 4\sin(2x) = 0$$