# Homework Help: Non-autonomous diff eq and periodicity

1. Sep 30, 2014

### BrainHurts

1. The problem statement, all variables and given/known data
Consider the first-order non-autonomous equation $x' = p(t) x$, where $p(t)$ is differentiable and periodic with period $T$. Prove that all solutions of this equation are period with period $T$ if and only if $\int_0^T p(s) ds = 0$.
2. Relevant equations

3. The attempt at a solution

I'm not quite sure, first I got that the general solution to $x' = p(t) x$ is

$x(t) = C e^{\int p(t) dt}$. Now I know that $p(t+T) = p(t)$.

So we have $x' = p(t) x$ $(\star)$, $p$ is periodic with period $T$ and $x(t) = C e^{\int p(t) dt}$

Any hints on how to proceed?

2. Oct 1, 2014

### LCKurtz

I would suggest using a definite integral (integrate from $0$ to $t$) when solving the DE so you can write the solution as$$x(t) = x(0)e^{\int_0^t p(s)~ds}$$Now $x$ is periodic iff $x(t+T) = x(t)$ for all $t$. Write out what that says in terms of your solution. Then see if you can relate that to $p$ being periodic with period $T$.

3. Oct 1, 2014

### BrainHurts

$(\Leftarrow)$

Well $x(T) = x(0) e^{\int_0^T p(s) ds} \Rightarrow x(T) = x(0) \Rightarrow x(0 + T) = x(0)$ so when t = 0 we have that situation covered.

Now we want to show periodicity for all t, $x(t+T) = x(0) e^{\int_T^{t+T} p(s) ds} = x(0) e^{\int_T^{t+T}p(s)ds} = x(0) e^{\int_0^t p(u+T) du} = x(0) e^{\int_0^t p(u) du} = x(t)$

The integral from $T$ to $t+T$ since we want to integrate on an interval with length $t$, and $p(u+T) = p(u)$ since $p$ is $T$ periodic.

The only thing that I'm wondering is I never used the fact that $\int_0^T p(s) ds = 0$.

Does this look OK so far?

Last edited: Oct 1, 2014
4. Oct 1, 2014

### LCKurtz

That would bother me too. In my previous post I showed you$$x(t) = x(0)e^{\int_0^t p(s)~ds}$$Then I asked you what you get if you set $x(t)=x(t+T)$. You haven't done that. Write down exactly what that equation says and see what it says about $p(t)$. You need to do that before you start considering the periodicity of $p(t)$. Once you do that you can try to prove the iff thing about $p$ being periodic.

5. Oct 1, 2014

### BrainHurts

Hmm,

$(\Rightarrow)$

We know that $x(t+T) = x(t)$, and in particular we have that $x(T) = x(0)$.

So $x(T) = x(0) e^{\int_0^T p(s) ds} = x(0) = x(0) e^0$. We see that $e^{\int_0^T p(s) ds} = e^0$, which implies $\int_0^T p(s) ds = 0$.

We already have the fact that $p$ is $T$ - periodic.

I'm not sure what you mean exactly.

6. Oct 1, 2014

### LCKurtz

I am trying to get you to use the fact that $x(t)$ has period $T$ to see what that says about $p(t)$. You need to know the connection between them to show that one being periodic implies the other is. It is premature to show one implies the other until you do that. You already know$$x(t) = x(0) e^{\int_0^t p(s)~ ds}$$Using that formula what is $x(t+T)$? Write that down.$$x(t+T) =~~\text ?$$Now, $x$ is periodic of period $T$ if and only if the left sides are equal which tells you the right sides must be equal. Simplify that as much as you can until you have something about $p$ that must be true.

Then show that "something" happens if and only if $\int_0^Tp(s)ds=0$. That's when you are ready to prove the if and only if. (Remember you are given that $p$ has period $T$.)

Last edited: Oct 1, 2014