# How to prove existence of periodic solutions of ordinary differential equations?

I have a non linear first order ordinary differential equation with periodic coefficients. I am trying to prove that the periodic solution of the differential equation exists. I am giving you an example of the problem I am having:

$$\large\frac{dx}{dt} = \mu - d\cdot x$$

where I assume that $\mu$ and $d$ are periodic in time and have the same period. Now, I have to prove that the solution of the differential equation i.e., $x$ is also periodic in time with the same period.

Is there any particular method I should apply? I need help badly. Your help/suggestion will be greatly appreciated.

• The Floquet theorem might help. – Fabian Dec 27 '12 at 22:26
• I think the whole concept of Floquet theory is about the stability of the periodic solutions not for their existence , right? – math Dec 27 '12 at 22:59
• Have you looked at the Poicare'-Bendixson Theory and Dulac's criteria? For example, see people.usd.edu/~jflores/Math735/MMChapter5_5p7.pdf and people.math.gatech.edu/~bonetto/teaching/6307-fall09/111.pdf. Also, you might have a look at Lyapunov's Second Method. I seem to recall books by Hartman and another by Hale on the matter. Warning, my memory is not serving me well today, so I could be wrong on both counts and leading you astray! – Amzoti Dec 28 '12 at 1:15
• Deimling's book Nonlinear Functional Analysis is a good source for this material; I remember it having a bunch of problems of this particular kind (more than I cared to read, I admit). – user53153 Dec 28 '12 at 5:53
• Certainly not every solution will be periodic. For example, take $\mu\equiv 0$ and $d\equiv 1$; these are periodic function, with any period you wish. The solutions are $x=Ce^{-t}$, only one of which is periodic, namely with $C=0$. So, the best we can hope for is the existence of some periodic solution. The book by Deimling which I mentioned gives a number of methods to obtain such an existence result. I think they are based on fixed point theorems: see Examples 3.2, 11.6, 16.1, 16.2, and Exercises 3.6 and 6.8. – user53153 Jan 2 '13 at 7:55

## 1 Answer

Consider the equation $x'=f(x,t)$ where $f$ satisfies conditions for existence, uniqueness and continuous dependence on initial data of the solution and the periodicity condition $f(x,t+T)=f(x)$. Denote by $x(t,\xi)$ the unique solution such that $x(0)=\xi$. Then $x(t,\xi)$ is periodic of period $T$ if and only if $x(T,\xi)=\xi$.

A strategy to show the existence of periodic solutions is to prove the existence of an initial value $\xi\in\mathbb{R}$ such that $x(T,\xi)=\xi$, that is, that the function $\xi\mapsto x(T,\xi)$ has a fixed point. One possibility is showing the existence of an interval $[a,b]$ such that $x(T,\xi)\in[a,b]$ for all $\xi\in[a,b]$. Sub and supersolutions are a useful tool for this.

Consider the example $x'=\mu-d\,x$ where $m$ and $d$ are periodic of period $T$. Suppose that there exist constants $m$ and $M$ such that $$m\le\frac{\mu(t)}{d(t)}\le M,\quad 0\le t\le T,$$ and hat $\mu/d$ is not constant (so that no constant solutions exist.) Then $v(t)=m$ is a subsolution and $u(t)=M$ is a supersolution. It follows that if $\xi\in[m,M]$, then $x(t,\xi)\in[m,M]$ for all $t\in[0,T]$. By the argument in the previous paragraph there is periodic solution with initial value in $[m,M]$.

• Thank you very much @Julián Aguirre . I will try to follow the same technique. I have another question similar to this. I am trying to find the conditions on $p(t)$ and $q(t)$ for the solution $y(t)$ of the differential equation $$\dot{y}+p(t)y=q(t)$$ to be periodic. Here i assume that the coefficients $p(t)$ and $q(t)$ are periodic in time. I think there must be a rule/condition out in the literature because this is frequently occuring form of ODE but i couldn't find it. – math Feb 10 '13 at 19:47
• For a linear equation you can find the explicit solution, and then impose the periodicity condition. – Julián Aguirre Feb 10 '13 at 22:16
• That's what i did but it is not giving me the condition that i am expecting. So I thought i am doing something wrong. But I will try again now. – math Feb 11 '13 at 16:31
• I got those ideas mainly from this book. – Julián Aguirre Feb 11 '13 at 17:18
• In that case you want to find constants $m<M$ such that $\mu(t)g(m)-d(t)m\ge0$ and $\mu(t)g(M)-d(t)M\le0$ for all $t\in[0,t]$. This implies that $v(t)=m$ is a subsolution and that $u(t)=M$ a supersolution. – Julián Aguirre Feb 13 '13 at 16:59