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.
Best 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 $[m,M]$.