Boundary value problem on a circle

Being a physicist, I have never really paid due attention to the little details such as - is the domain on which the differential equation is given an open or closed set? Or, can it be both? Now, my research has led me back to these kinds of questions.

For concreteness, let's say we're solving a simple ODE $y' = im y$, with $m$ a real constant, on a circle. The circle is usually represented as the interval $I = \langle - \pi, +\pi ]$ and so my solution $y(x)$ should be defined for all $x \in I$. However, since the point $x = +\pi$ is a boundary point, there's no way to define the derivative $y'$ at that point (right?) and, therefore, our ODE only makes sense on an open interval $\langle -\pi, +\pi \rangle$. If all this is correct, I would need to define $y(+\pi)$ "by hand" and this definition is nothing more than a boundary condition.

Is the above reasoning correct?

• Is $y(e^{ti})=y(e^{(t+2\pi)i})$ for $-\infty<t<\infty$? – John Wayland Bales Dec 14 '17 at 22:24
• No, $y$ is defined only on the interval $I$, as in the question. – Fizikus Dec 14 '17 at 22:25
• The circle can be defined (topologically) as the quotient $\mathbb{R}/2\pi\mathbb{Z}$. To speak of vector field and ODEs you must give to it a differentiable structure (one usually solves ODEs on manifolds). You can do this exploiting the covering map $p:\mathbb{R}\to \mathbb{R}/2\pi\mathbb{Z}$ (the projection to the quotient). Unfortunately, the vector field you have defined exploiting this map do not pass to the quotient since at the points $-\pi$ and $\pi$ it do not agree. It is not correct to say that $+\pi$ is a boundary point, because the circle is a manifold without boundary. – Warlock of Firetop Mountain Dec 14 '17 at 22:43
• Warlock, so you're saying that the ODE on $I$ is not well defined? – Fizikus Dec 15 '17 at 8:23
• Yes, the vector field that you have defined on the circle is not continuous at $\pm \pi$ because the left and right limits do not agree. Also the ODE $y'(t) = y^2(t)$ for $y(t) \in [-\pi,\pi]$ is not a good example since it defines a continuous vector field on the circle but not a smooth one. What one usually requires is an ODE like $y'(t) = F(y(t))$ with $t\mapsto F(t)$ $2\pi$-periodic and smooth (or at least $C^2$) function. For example $y'(t) = cos^2(y(t))$. – Warlock of Firetop Mountain Dec 15 '17 at 15:07