# Interlude: Impossibility of periodic oscillations

Being constrained to move on the real line forces solutions to increase or decrease monotonically, or remain constant (fixed point). More precisely, non-constant solutions either approach a fixed point, or diverge to $\pm\infty$ monotonically. To state it more geometrically, a particle flowing on the real line according to $\dot{x}=f(x)$ never reverses direction. Hence there are no periodic solutions to such equations. Let’s prove analytically the absence of periodic solutions. Suppose on the contrary that $x(t)$ is a nontrivial periodic solution, i.e., $x(t)=x(t+T)$ for some $T>0$, and $x(t)\neq x(t+s)$ for all $0 $$\int_{t}^{t+T} f(x) \frac{\text{d}x}{\text{d}t} \ \text{d}t=\int_{x(t)}^{x(t+T)} f(x)\ \text{d}x=0.$$ The first equality follows from the chain rule, and the second from the assumption that$x(t)=x(t+T)$. On the other hand, since$\dot{x}=\text{d}x/\text{d}t=f(x)$, $$\int_{t}^{t+T} f(x) \frac{\text{d}x}{\text{d}t} \ \text{d}t=\int_{t}^{t+T} \left(\frac{\text{d}x}{\text{d}t}\right)^2 \text{d}t\ >0$$ by assumption that$T>0$and$\text{d}x/\text{d}t\$ does not vanish identically. This is a contradiction, thus there is no periodic solution.