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Let $(f,g)$ be a pair of real-valued $C^1$ functions on $\mathbb{R}$ satisfying $$\forall x\in\mathbb{R}\left(f'(x)=g(x)\quad\text{and}\quad g'(x)=-f(x)\right)$$ $$f(0)=0\quad\text{and}\quad g(0)=1$$ Then it is pretty immediate that $f$ and $g$ are both $C^\infty$ on $\mathbb{R}$, this determines a power series which determines uniqueness of the pair. A cute argument (without power series) shows that these functions satisfy $f(x)^2+g(x)^2=1$.

Here's my question: how do we deduce that $f(x)$ and $g(x)$ are $2\pi$ periodic? I understand that we could have started the entire axiomatic system by defining $f(x)$ and $g(x)$ in terms of triangles, but since we already have uniqueness from this setup I wonder how we deduce periodicity from here.

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    $\begingroup$ You show that it is periodic, and then call the smallest period $2\pi$. $\endgroup$ Apr 30, 2020 at 16:32
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    $\begingroup$ See, e.g., Rudin's Principles of Mathematical Analysis, $3$rd edition, page $182 $ $\endgroup$ Apr 30, 2020 at 16:34
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    $\begingroup$ Thanks @J.W.Tanner, what a beautiful argument! $\endgroup$
    – Prototank
    Apr 30, 2020 at 16:36
  • $\begingroup$ using the infinite product expansion periodicity is obvious, doing it this way is just donkey work, no idea why they don't do the infinite product expansion first and then prove periodicity. $\endgroup$
    – jimjim
    May 3, 2020 at 1:31
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    $\begingroup$ @Arjang: That is what Rudin does, I'm not sure I get your point. And periodicity is, a priori, far from obvious from the power series. $\endgroup$
    – copper.hat
    May 3, 2020 at 7:50

2 Answers 2

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Here is a tedious argument that is less slick than Rudin's but perhaps has more geometric content:

The essence of the argument is to show there is an arbitrarily small rotation that is periodic.

Define $c_0 = 0, s_0 = 1, c_{n+1} = \sqrt{{ 1 + c_n \over 2}}, s_{n+1} = \sqrt{{ 1 - c_n \over 2}}$, $Q_n = \begin{bmatrix} c_n & - s_n \\ s_n & c_n \end{bmatrix}$. Note that $c_n, s_n \ge 0$ and $c_n^2+s_n^2 = 1$. Also, $c_n \uparrow 1$ (and hence $s_n \downarrow 0$).

($Q_0$ is a $90^\circ$ rotation, and $Q_{n+1}$ is a rotation through half of the angle of $Q_n$.)

A little work shows that $Q_n Q_n^T = I$, $Q_{n+1}^2 = Q_n$ and $Q_n \to I$. Furthermore, $Q_n^{4n} = I$ and $Q_0 Q_n^T = Q_n^T Q_0$.

Let $J= Q_0$, and $x = (g,f)^T$. Then $x' = Jx$ and $x(0) = e_1$.

Note that $x'(0) = e_2$, hence there is some $T>0$ such that $x(t) > 0$ (coordinate wise) for all $t \in (0,T]$. In particular, there is some $n$ and some $t^* \in (0,T]$ such that $x(t^*) = Q_n x(0) = Q_n e_1$.

Now consider $y(t) = Q_n^T x(t+t^*)$, note that $y(0) = x(0)$ and $y'(t) = Q_n^T J x(t+t^*) = J Q_n^T x(t+t^*)= J y(t)$ and by uniqueness we have $x(t+t^*) = Q_n x(t)$.

In particular, $x(kt^*) = Q_n^kx(0)$ and so $x(4nt) = x(0)$. Hence $x$ is periodic.

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$t \mapsto (g(t), f(t))$ is a movement with speed $1$ on the curve $x^2+y^2=1$. So the movement is periodic with period the length of the curve.

You can cook up an example of a differential equation that gives a movement with unit speed on, say, an ellipse.

In general, consider a system of the form $g'(t) = A(g(t), f(t))$, $f'(t) = B(g(t), f(t))$, where $(A(x,y), B(x,y))= \lambda(x,y) (-\frac{\partial F}{\partial y}, \frac{\partial F}{\partial x})$. Then $F$ is constant on any trajectory of the system. Assume moreover that the initial point lies on a compact connected component of a level curve of $F$ and $\lambda \cdot \nabla F \ne 0$ on it (so it lies on a closed curve). Then again the solution of the system will be periodic.

In the case of a vector $v$ field tangent to a closed curve $C$, the equation $\frac{d\bf{x}}{dt} = v(\bf{x})$ starting on the curve will have a period $$\tau = \int_C \frac{ds}{\|v\|}$$ that can be in principle calculated (approximated).

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    $\begingroup$ The first statement is true but the point is to establish it formally. $\endgroup$
    – copper.hat
    May 3, 2020 at 1:29
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    $\begingroup$ @copper.hat: I agree; while obvious, it is not easy to write a formal proof . $\endgroup$
    – orangeskid
    May 3, 2020 at 7:42
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    $\begingroup$ I agree. I went through 20+ pages trying a length based proof (or other ways of parameterising $S^1$) but either my arguments were circular or the complexity unwarranted. Doesn't mean it can't be done, but it is not entirely trivial. $\endgroup$
    – copper.hat
    May 3, 2020 at 7:46

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