# Why are periodic solutions impossible for a differential equation on $\mathbb{R}$?

I am trying to understand a simple property of odes defined on a real line (or a subset of the real line).

Consider an ordinary autonomous differential equation where the dependent variable $x$ is defined on the real line, i.e. $$\dot{x} = f(x),$$ where $x \in \mathbb{R}$, and the dot is differentiation with respect to time $t$. It is known that the corresponding phase portrait has solutions that are equilibria, and all other solutions either move towards $\pm \infty$, or towards / away from the equilibria. Geometrically, if we draw the phase portrait, we only can have equilibria and all other trajectories move right or left. No periodic solutions exist.

So what about nonautonomous ode like $$\dot{x}=\cos t.$$ The general solution is $x(t)=\sin t - \sin t_{0} +x_{0}$. It is clear that all solutions are periodic. Does that contradict the above theory? How would the corresponding phase portrait look like?

• "It is known that the corresponding phase portrait has solutions that are equilibria, and all other solutions either move towards $\mp \infty$, or towards / away from the equilibria." Do you know the reason behind it? I believe if you can prove this statement, you will get your answer. – obareey Oct 18 '15 at 19:53
• Well, if a point $x$ is not equilibrium, then $x(t)$ is monotonically increasing / decreasing function of $t$. Since it is on the real line, it can either move right (mon. increasing), or left (monotonically decreasing) to the equilibrium point, or if there's no equilibrium, to $\pm \infty$. I learnt this fact probably 5 years ago. But untill now, I never thought about a nonautonomous system like the one above - so my original question was, how could one draw the phase portrait in one dimension for a nonautonomous system? – Alex Oct 19 '15 at 12:50
• What I meant in the comment is if you understand why that fact is true, then you can easily see why it fails on non-autonomous case. The phase portrait in non-autonomous case is just the usual plot of $x(t)$ with respect to $t$, i.e. one axis is $t$ and the other one is $x(t)$, since $t$ can be selected as a state. – obareey Oct 19 '15 at 13:07

You can transform every nonautonomous ODE to a autonomous one by introducing a new variable $z$ with $\dot z = 1$ and replacing $t$ by $z$. But this will always add a dimension to the system. Hence it does not contradict the theory, because it only considers the case $$\dot x = f(x)$$ which is an autonomous ODE on $\mathbb R$ and not $\mathbb R^k$ for $k>1$ and not nonautonomous ODEs.
• Yep, I am aware of the trick of replacing time by another variable. I already thought of using a substitution $s=t$ to obtain $\dot{x}=\sin{s}, \quad \dot{s}=1$. But then the resulting phase plane in $(x,s)$ variables does not contain closed curves so in this higher dimension the solution is not periodic anyway? So you are saying that for a nonautonomous ode in one dimension the phase portrait is not defined? – Alex Oct 17 '15 at 21:00
• You are right, the solution $(x,s)$ is not periodic in the phase portrait (which exists in 4-dimensional space). But the main point is that there is a theorem that says A implies B and you are essentially asking "does the fact that C does not imply B contradict the theorem?" – Wauzl Oct 17 '15 at 21:09