Uniqueness of a periodic solution for nonlinear pendulum

Question

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?

Answer 1

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.

Answer 2

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 $F$.

Answer 3

This is nor really an answer, but something to extend the discussion:

I am also trying to understand the solutions of the nonlinear damped pendulum, and since it represent a easy-to-do real world physical model, if this equation is a "right" representation, must consider that the pendulum at some finite-time stops moving, and giving this, you will find yourself with a "situation": since its stops at this ending boundary, you will have a border condition with zero values for a set of points with measure different from zero... situation that could crash the conditions of uniqueness of solutions: as example, for Linear ODEs, this situation makes it impossible from them to support finite duration solutions.

I have found recently these two papers [1] and [2] from V. T. Haimo that develops a theory of continuous-time finite-duration differential equations, and set some conditions to verify if a dynamical system could support finite-duration solutions (actually I get stuck trying to find if the nonlinear damped pendulum equations could support finite-duration solutions, which I am asking here).

But related to the specific discussion of you question, on paper one [1] is stated that: "(...), finite time differential equations cannot be Lipschitz at the origin. As all solutions reach zero in finite time, there is non-uniqueness of solutions through zero in backwards time. This, of course, violates the uniqueness condition for solutions of Lipschitz differential equations."

With that statement (I am assuming is right, and I can´t probe is wrong neither with my knowledge), you will have two possible situations:

• i) Or the solutions to the equations of the nonlinear damped pendulum properly model the physical pendulum so finite-duration solutions are supported, meaning here that they are not unique
• ii) Or the solutions of the equations of the nonlinear damped pendulum are unique, meaning "is not really" an accurate representation of the physical pendulum (as I think everyone in "science" believe it is), so actually finding the answer is quite important, from the philosophical point of view.

Hope it encourage someone to try to apply what the papers said to the nonlinear pendulum equations, and please let me know what you find.