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Prove that an autonomous ODE $f(x)=x'$ has no nonconstant periodic solutions.

I guess I could prove it by contradiction by saying $x(t+T) = x(t)$ implies $x(t) =$ constant.

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Can you be more specific, so that we can better help you? –  Matt Jul 21 '11 at 1:46
    
@Derrick: please ask a complete question. Right now, it is really not possible to answer anything useful! –  Mariano Suárez-Alvarez Jul 21 '11 at 2:09

1 Answer 1

I think you are trying to prove that a first-order autonomous ODE $dx/dt = f(x)$ cannot have a periodic solution. Of course it's pretty easy to think of a second-order autonomous ODE that does have a periodic solution, so first-order autonomous systems of differential equations can likewise have periodic solutions.

For the single equation case, think about a periodic function (other than the trivial case of a constant function). If the function is to be differentiable, it must be continuous.

Find two points on a continuous nonconstant periodic function where the function value is the same, but the function is decreasing at one argument and increasing at the other.

This would contradict the autonomous differential equation, right?

Once more, with rigor:

We show that if $x(t)$ is a real periodic solution of the autonomous first-order differential equation:

$$x' = f(x)$$

then $x(t)$ is a constant function.

Suppose not for the sake of contradiction: $x$ has an interval of periodicity $[0,T]$ and attains a bounded range $[X_{min},X_{max}]$. WLOG we assume, by translation in t if necessary, that $x(0) = X_{min} = x(T)$, and for the sake of specificity that $x(t_*) = X_{max}$ for some $t_* \in (0,T)$. Of course those extremal values might also be attained elsewhere in $[0,T]$.

Since $X_{min}, X_{max}$ are extrema of $x(t)$, function $f$ must vanish there:

$$f(X_{min}) = 0 = f(X_{max})$$

Unless $f$ takes a nonzero value somewhere on $[X_{min},X_{max}]$, all solutions $x(t)$ of the differential equation in this range would be constant. So choose $x_* \in (X_{min},X_{max})$ such that $f(x_*)$ is nonzero.

By the Intermediate Value Thm. continuity of $x$ implies $x(t) = x_*$ for some points in $(0,t_*)$ and $(t_*,T)$. In fact there is an open interval around $t_*$ where $x(t) \gt x_*$, so we can pick a greatest $t_{left} \in (0,t_*)$ and a least $t_{right} \in (t_*,T)$ where $x$ takes the value $x_*$.

It follows that $x$ is increasing on the right side of $t_{left}$ and decreasing on the left side of $t_{right}$. But this forces the derivative $x'$ at both points to be zero, contradicting our choice of $x_*$ s.t. $f(x_*)$ is nonzero.

QED

In a sense this is interesting because it assumes nothing about regularity of $f$. Under standard assumptions solutions to initial value problems for the DE are (locally) unique. But as $x(t) = X_{min}$ would be a constant solution of the IVP for $x(0) = X_{min}$, such uniqueness precludes nonconstant solutions.

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Just some slight technical quibbles: 1) your function must be differentiable, not just continuous (you can have a non-constant continuous periodic function that is not increasing or decreasing at any point - but it can't be a solution of a differential equation). 2) a function can be increasing or decreasing at some point but have derivative 0 there (e.g. $t^3$ at $t=0$). Better to specify that the derivative is strictly negative at one point and non-negative at the other. –  Robert Israel Jul 21 '11 at 6:43
    
@Robert Israel: Thanks for the notes. You inspired me to spell out details more fully. –  hardmath Jul 23 '11 at 3:28

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