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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.

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    The [Floquet theorem](http://en.wikipedia.org/wiki/Floquet_theory) might help.2012-12-27
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    I think the whole concept of Floquet theory is about the stability of the periodic solutions not for their existence , right?2012-12-27
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    Have you looked at the Poicare'-Bendixson Theory and Dulac's criteria? For example, see http://people.usd.edu/~jflores/Math735/MMChapter5_5p7.pdf and http://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!2012-12-28
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    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).2012-12-28
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    My suggestion is to add more details to your post. Maybe an example of the kind of problem you are trying to solve.2012-12-29
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    Thank you for your suggestions. I am going through the literature suggested by you but I haven't found anything useful for my problem yet. @PavelM , I am giving you an example of the problem i am having: \frac{dx}{dt}=\mu-d.x2013-01-01
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    I am assuming that the coefficients \mu and d are periodic in time. I want to prove that the existence of a periodic solution of this equation. I am totally confused how to do that.2013-01-01
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    @varu And $d$ and $x$ have the same period, of course? I suggest that you edit the question to add this example to it (you can simply append this example at the end of question). This will give it more visibility and will certainly improve readability.2013-01-01
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    @PavelM I have edited my question. please have a look at it. And yes, d, $\mu$ and x are of the same period. Thanks for your suggestions so far.2013-01-02
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    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 6.8.2013-01-02
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    @PavelM The libraries near my place don't have this book. I have requested this book from some nearby city and i am not sure if i will receive it or not. Please let me know if you have a pdf of the book. I will be grateful.2013-01-04
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    @PavelM Hello again, I went through the Klaus Deimling's book that you suggested me but unfortunately I could not find anything relevant to my problem. Is there anything else you would suggest me? I will be grateful. I have another question that i will be posting shortly. Thanks for your help so far.2013-01-10

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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 $[m,M]$.

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    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.2013-02-10
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    For a linear equation you can find the explicit solution, and then impose the periodicity condition.2013-02-10
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    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.2013-02-11
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    Could you please refer me to some examples online (journal article etc) or in the books where they used this sub solution and super solution argument for the existence? I never saw any example of this kind. I am good in the mechanical stuff but not in the proof things.2013-02-11
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    I got those ideas mainly from [this book](http://books.google.es/books/about/Differential_Equations_A_Dynamical_Syste.html?id=e5K786f9qPkC&redir_esc=y).2013-02-11
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    Thank you very much for the quick response. I am grateful!2013-02-11
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    If i have an equation like $$\dot{x}=\mu g(x)-dx$$ the i have to assume that there exists constants m and M such that $$m\leq \frac{\mu (t) g(x)}{d(t)}\leq M$$, right? Here $g(x)$ is sigmoidal function. I just want to make sure.2013-02-13
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    In that case you want to find constants $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.2013-02-13
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    Thanks a ton for the response.2013-02-13
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    I am applying this technique but the problem is that my subsolution is $m=0$ which is also the solution of my DE. This will give me constant solution which is periodic but this is not what i want. I want positive periodic solution. Is there any particular technique that i can use to omit the possibility of trivial solution.2013-02-13
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    You may try to find anotehr type of subsolution, or do some numerics and try to get some information on the function $x(T,\xi)$.2013-02-14