# Uniqueness of a periodic solution for nonlinear pendulum

I am working with the system of ODE's or second order differential equation representing the nonlinear pendulum with constant torque and damping.

\begin{equation*} \theta'=v \end{equation*} \begin{equation*} v'=-bv-\sin(\theta)+k \end{equation*} with $b,k>0$ for physics reasons. I determined that we have equilibria at \begin{equation*} \begin{bmatrix}\theta\\v \end{bmatrix}= \begin{bmatrix}\sin^{-1}(k)\\0 \end{bmatrix} \end{equation*}

I have already proved that for $sin^{-1}(k)$ i.e. when there are no equilibria, and for some strip of the cylinder $\mathbb{R}\times S^1$ on which this system is defined, we have a periodic solution (via Poincare-Bendixson)

I need to show that this particular solution is unique. The hint given is to use the energy function for this system: $E(\theta, y)=\frac{1}{2} y^2 -\cos(\theta)+1$ and the fact that $E$ along any periodic solution must have no change.

However, I am getting a bit stuck on how to do this. Previous exercises seem to suggest I should make use of poincare map, $p$ iteration and maybe the contraction mapping to prove existence and uniqueness of a fixed point of the poincare map. My TA suggested that use of the "facts" from physics intuition that if $v_1>v_0$ then $p(v_1)<p(v_0)$ and if $v_1<v_0$ then $p(v_0)<p(v_1)$ as more energy is lost at higher velocity, and therefore we should end up at different vertical spots along the cylinder. I am not comfortable with writing up a solution in which these facts are proved appealing to natural phenomena, is there a better way?

Take the energy function $E(\theta,v) = \frac{1}{2} v^2 - \cos \theta + 1$. If $v$ and $\theta$ are functions of $t$, then the change in time of $E$ is given by \begin{equation} \frac{\text{d}}{\text{d} t} E(v(t),\theta(t)) = \frac{\partial E}{\partial v} \frac{\text{d} v}{\text{d} t} + \frac{\partial E}{\partial \theta} \frac{\text{d} \theta}{\text{d} t} = v \frac{\text{d} v}{\text{d} t} + \sin \theta \frac{\text{d} \theta}{\text{d} t}. \tag{1} \end{equation} Now suppose that $v(t)$ and $\theta(t)$ obey the system of first order differential equations given in your question. That means that we can write $\frac{\text{d} v}{\text{d} t}$ and $\frac{\text{d} \theta}{\text{d} t}$ in terms of $v$ and $\theta$, such that $(1)$ becomes \begin{equation} \frac{\text{d}}{\text{d} t} E = v(-b v-\sin\theta+k)+v \sin \theta = k v-b v^2.\tag{2} \end{equation} So, in general, $E$ is not constant in time, because $v$ is a function of $t$, and in general, $k v - b v^2 \neq 0$. However, we try to pick a constant value of $v$ for which the time derivative of the energy vanishes: that is, we can choose $v$ to be constant and equal to $\frac{k}{b}$. For this special value of $v$, you can now investigate what happens with the system of ODEs.

Furthermore, it is clear that the specific choice $v = \frac{k}{b}$ is the only choice for which $E$ is constant in time. With these ingredients, I'm sure you can finish the argument yourself.

Addition: For a periodic orbit with period $T$, we have $(\theta(0),v(0)) = (\theta(T),v(T))$ and therefore $E(t=0) = E(t=T)$. So, if $E$ has the same value at every point of your solution, and the level set $E=c$ is closed and bounded, this is sufficient to conclude that the orbit is equal to the level set, is closed and bounded, and is therefore periodic (provided there are no equilibria of the system on this level curve $E=c$). I interpret the statement that '$E$ along any periodic solution does not change' as '$E$ has the same value at every point of a periodic solution'. Since there is a unique invariant level set of $E$, characterised by the choice $v = \frac{k}{b}$, there is a unique periodic orbit for which $E$ is constant along that periodic orbit.

That being said, there has to be a reason to assume that for this particular system and this particular choice of energy, all periodic orbits of the system have the property that $E$ is constant along them. This is generally not true: of course, any function of $\theta$ and $v$ has a total zero change over a full period if you consider a periodic orbit, because $(\theta(0),v(0)) = (\theta(T),v(T))$ directly implies $F(\theta(0),v(0)) = F(\theta(T),v(T))$ for any function $F$.

• So I did this on my own but was unclear on how this provided a proof of a unique periodic solution. From my book (Hirsh/Smale/Devaney), for a periodic solution, the total change in E along any solution must be 0 (this is the hint for this problem). Doesn't this mean that technically Energy cannot change over a full period (net change wrt theta is zero) rather than solving for values which cause the time derivative to vanish. – qbert Mar 11 '16 at 9:44
• Good point, I'll update my answer. – Frits Veerman Mar 11 '16 at 10:26
• Still think that by "total change of energy is zero" they mean $\int_{0}^{T} dE(t) = 0$, which is slightly trivial unless you note that this property allows you to tell in which domains it is impossible to find periodic solution. – Evgeny Mar 11 '16 at 15:45
• By domains do you mean values of the parameters? – qbert Mar 11 '16 at 17:18

Below is the answer I settled on. Helpful resources included Hirsch Smale and Devaney as well as Strogatz, nonlinear dynamics and chaos. Previous work included proving the existence of a solution between a strip of $v$ values and $0<\theta<2\pi$, a planar region where I applied Poincare-Bendixson to show there exists a at least a periodic solution.

The hint tells us that the net energy gained for any periodic solution is zero, adding $\int_{0}^{2\pi}\frac{dE}{d\theta}d\theta=0$ Fix $k>1$. Taking the energy function to be: \begin{equation*} E(v,\theta)=\frac{1}{2}v-\cos(\theta)+1 \end{equation*} We have \begin{equation*} \frac{dE}{d\theta}=v\frac{dv}{d\theta}+\sin(\theta)=v\frac{\frac{dv}{dt}}{\frac{d\theta}{dt}} +\sin(\theta) \end{equation*} By the chain rule, yielding: \begin{equation*} \frac{dE}{d\theta}=v\frac{-bv-\sin(\theta)+k}{v}+\sin(\theta)=-bv-\sin(\theta)+k+ \sin(\theta)=-bv+k \end{equation*}

Using our hypothesis that energy along periodic solutions is conserved, if $v$ is a $v$ is the velocity for which $p(v)=v$, i.e. the velocity for a periodic solution, which we know exists in the region $v_{1}<v<v_{2}$ then we have: \begin{equation*} \int_{0}^{2\pi}\frac{dE}{d\theta}d\theta=0\Rightarrow int_{0}^{2\pi}(-bv+k)d\theta =0\Rightarrow \frac{2\pi k}{b}=\int_{0}^{2\pi}v d\theta \end{equation*} But this means that this $v$ is unique. If $v'<v$, then $\int_{0}^{2\pi}v' d\theta <\int_{0}^{2\pi}v d\theta$ and if $v'>v$, then $\int_{0}^{2\pi}v' d\theta >\int_{0}^{2\pi}v d\theta$ by monotonicity of the integral.

Therefore, we can conclude that there is one $v$ for which $p(v)=v$ and $E(\theta,v)=E(\theta,p(v))$ and therefore one periodic solution.

Added: Any tips on how to prove that v is constant in theta would be much appreciated, I believe this is true but a hole in the above proof.