I have a nonlinear ODE $$ \dot{x}(t)=a(t)-\mbox{min}(x(t),b) $$ where $b>0$ and the input $a(t)$ is positive, bounded, and periodic with period $1$. How can I show that any periodic equilibrium of the ODE must also have period 1, i.e., the same as the input? (A solution $x^*(t)$ is a periodic equilibrium if there exists $p>0$ such that $x^*(p+t)=x^*(t)$ for all $t\geq0$.)

I know that given $\int_{0}^{1}a(s)ds<b$ then a periodic solution exists and numerical solutions confirm that the period is 1. But I'm not sure how to show this given that it's hard to solve the ODE even after specifiying a $a(t)$ (although a piecewise solution can be constructed by solving it assuming $x(t)>b$ or $x(t)\leq b$ and then attaching the solutions).

  • $\begingroup$ Have you tried to solve by hand to see/guess what happens? $\endgroup$
    – pluton
    Mar 3 '17 at 23:00
  • $\begingroup$ I'm note sure what that means. Also, last paragraph explains what happens. $\endgroup$ Mar 4 '17 at 17:57
  • $\begingroup$ Have some information about regularity of $x^*$? $\endgroup$
    – jJjjJ
    Mar 4 '17 at 18:09
  • $\begingroup$ @jJjjJ yes, given $a(s)$ is right-continious any solution is Lipschitz continuous. $\endgroup$ Mar 4 '17 at 18:12
  • $\begingroup$ Then one can make the following obs. Suppose the solution has a irrational period $p$ and also period $1$. The quotient $p/1$ is irrational so the solution is costant almost everywhere. Thus everywhere for continuity. $\endgroup$
    – jJjjJ
    Mar 4 '17 at 18:17

(For more generality, I will refer to the period of $a(t)$ as $T$ instead of 1).

Allow me to consider the problem in two separate cases. We will justify this later.

First, consider the case that $x < b$. The ODE becomes, $$\dot{x}(t) = a(t) - x(t)$$ which has a steady-state solution (periodic equilibrium) with the same period as the forcing-function $a(t)$. (Proof at the bottom).

Now consider the case that $x>b$. The ODE becomes, $$\dot{x}(t) = a(t) - b$$ which, as you already understand, only has a periodic equilibrium if, $$\frac{1}{T}\int_0^T a(t) dt = b$$ and the period will still be $T$. (Basically, $b$ has to subtract out the mean value of $a(t)$).

However, if we suppose instead that, $$\frac{1}{T}\int_0^T a(t) dt \leq b$$ then $x(t)$ can decrease without bound.

Considering the whole original ODE, this unbounded decrease would just pass smoothly from the $x>b$ case to the $x<b$ case which is not unbounded. I say smoothly because, $$h(t) = \min(f(t),\ g(t))$$ is continuous if both $f$ and $g$ are continuous, and thus for our problem $\dot{x}(t)$ is continuous so $x(t)$ is smooth.

Thus, the periodic equilibrium of $x(t)$ is a smooth function with period $T$ in the case $x \leq b$ and in the case $x \geq b$ as long as $\frac{1}{T}\int_0^T a(t) dt \leq b$. This certifies your claim.

I think all that would remain is to put this reasoning into a more rigorous-looking proof format. Please let me know if you need further clarification, and good luck with your work!

Proof of first lemma:

The equation under consideration is, $$\dot{x}(t) = a(t) - x(t)$$ where a(t) is periodic with period $T$. Note that it is a linear equation in $x(t)$.

Since $a(t)$ is periodic, it can be expressed as a Fourier series; a sum of sinusoids (with rational periods) where $T$ is the least-common-multiple of said periods. Exploiting our ODE's linearity, if we can find the general solution to one sinusoidal input, we can find the solution to a sum of sinusoidal inputs via the superposition principle.

The general solution of a linear ODE with a single sinusoidal inhomogeneity is shown here, and we note that as $t$ becomes large, $x(t)$ becomes completely dominated by a sinusoidal term of the same period as the input. This limiting behavior is the equilibrium the solution settles into, effectively by definition.

For our problem, this means the solution to $\dot{x}(t) + x(t) = a(t)$ settles to an equilibrium that is a sum of sinusoids, each corresponding to the same period as one component of $a(t)$'s fourier series. The overall periodicity of the sum will again be the least-common-multiple of the periods of its constituents, which is $T$ since all the same periods are still present and no new periods were introduced.

Another way to see the periodicity of the equilibrium (steady-state) solution is to consider the actual solution, $$x(t) = x(0)e^{-It} + \int_0^t e^{-I\tau}a(\tau)d\tau$$ where $I$ is the identity matrix, and $e^{A}$ is a matrix exponential. Noting that $e^{-It}$ = $Ie^{-t}$, we see that the right-most term dominates behavior as $t \to \infty$. All that remains is to show that $\int_0^t e^{-I\tau}a(\tau)d\tau$ has period $T$, but in my opinion that is more complicated than the proof I outlined above using Fourier series and linearity.

  • $\begingroup$ Thanks for your answer. It all makes sense and as you said it just needs more rigour. I think however if you prove the first claim (that the ODE in the $x<b$ case has a unique periodic equilibrium with period 1, that could be very helpful and I'll accept your answer.) $\endgroup$ Mar 6 '17 at 21:31
  • $\begingroup$ Added. I know this answer isn't a succinct proof by any means, but I leave that for you to construct. I try my best to instead provide "rigorous insight" in my answers on MSE :) Let me know if anything isn't clear. $\endgroup$
    – jnez71
    Mar 6 '17 at 23:23
  • $\begingroup$ It's very helpful; thanks again. It doesn't let me give you the bounty now but it's yours in 2 hours. $\endgroup$ Mar 6 '17 at 23:26
  • 1
    $\begingroup$ Thanks! Also, I fixed a small typo. The condition is that $\frac{1}{T}\int_0^T a(t)dt \leq b$. I was missing the $\frac{1}{T}$ before because my mind was on the $T=1$ case. You probably knew what I meant. $\endgroup$
    – jnez71
    Mar 7 '17 at 1:03

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