Differential equation of period

Question

Suppose that $I=(-\infty,\infty)$ and $p,r$ are periodic of period $\zeta$. Define $P$ on $I$ as; $$ P(x)= \int_{0}^{x} p(t) dt , \forall x\in I $$ Then the following statements are true

1. A solution $\phi$ of the equation $y'+p(x)y=r(x)$ on $I$ is periodic of period $\zeta$ $\leftrightarrow$ $\phi(0)=\phi(\zeta)$.
2. The equation $y'+py=r$ has exactly one periodic solution of period $\zeta \leftrightarrow e^{P(\zeta)} \neq 1$.
3. Every solution of the equation $y'+py=r$ is periodic of period $\zeta \leftrightarrow e^{P(\zeta)} =1$ AND $$ \int_{0}^{\zeta} e^{P(t)} r(t) dt=0 $$

I have already proved statement 1. I let $\phi$ be a solution of the equation $y'+p(x)y=r$. Then I defined another function $\psi$ ...showed that $\psi$ is a solution of it also and after some computation solved that $\phi$ is periodic of period $\zeta$ using the fact that $\chi(x)= ce^{-P(x)}$.

Can anyone assist me in proving parts 2 and 3? Or atleast point me in the right direction. I'm really confused as to how to even begin these parts

Answer 1

For the sake of completeness, I'll start by tackling item (1) even though our OP Jason Moore expresses some satisfaction with his own solution.

1.) Given that $\phi(x)$ is periodic of period $\zeta$, we have by definition that

$\phi(x + \zeta) = \phi(x), \; \forall x \in I = (-\infty, \infty); \tag 1$

thus if we take $x = 0$ we find that

$\phi(\zeta) = \phi(0). \tag{2}$

Next, we note that, if $\phi(x)$ satisfies

$y'(x) + p(x)y(x) = r(x), \tag 3$

that is,

$\phi'(x) + p(x)\phi(x) = r(x), \tag 4$

then we have

$\phi'(x + \zeta) + p(x + \zeta)\phi(x + \zeta) = r(x + \zeta), \tag 5$

which, since $p(x)$ and $r(x)$ are periodic of period $\zeta$, that is, since

$p(x + \zeta) = p(x), \; r(x + \zeta) = r(x), \tag 6$

implies that

$\phi'(x + \zeta) + p(x)\phi(x + \zeta) = r(x); \tag 7$

we see that $\phi(x)$ and $\phi(x + \zeta)$ each satisfy (3)-(4), with initial conditions $\phi(0)$ and $\phi(0 + \zeta) = \phi(\zeta)$ respectively; since we assume that $\phi(\zeta) = \phi(0)$, $\phi(x)$ and $\phi(x + \zeta)$ both satisfy the same differential equation (3)-(4) with identical initial conditions $\phi(0) = \phi(\zeta)$, and thus by uniqueness of solutions we must have

$\phi(x + \zeta) = \phi(x),\; \forall x \in I; \tag 8$

the periodicity of $\phi(x)$ given $\phi(\zeta) = \phi(0)$ is thus established.

2.) With

$P(x) = \displaystyle \int_0^x p(t)\; dt, \tag 9$

the solution $\phi(x)$ to (3)-(4) with initial condition $\phi(0)$ may be expressed in the form

$\phi(x) = e^{-P(x)}(\phi(0) + \displaystyle \int_0^x e^{P(t)}r(t) \; dt), \tag{10}$

which may easily be verified using direct differentiation; from (10),

$\phi'(x) = -P'(x)e^{-P(x)}(\phi(0) + \displaystyle \int_0^x e^{P(t)}r(t) \; dt) + e^{-P(x)}(\phi(0) + \displaystyle \int_0^x e^{P(t)}r(t) \; dt)'$ $= -P'(x)e^{P(x)}(\phi(0) + \displaystyle \int_0^x e^{P(t)}r(t) \; dt) + e^{-P(x)} e^{P(x)}r(x) = -p(x) \phi(x) + r(x), \tag{11}$

since, from (9),

$P'(x) = p(x); \tag{12}$

(11) is easily seen to be equivalent to (3)-(4); we also see that (10) implies

$\phi(0) = e^{-P(0)}(\phi(0) + \displaystyle \int_0^0 e^{P(t)}r(t) \; dt) = e^{P(0)}\phi(0) = \phi(0), \tag{13}$

since, again from (9),

$P(0) = \displaystyle \int_0^0 p(t) \; dt = 0; \tag{14}$

thus (10) is consistent with the initial conditions on (3)-(4) as well.

The formula (10) is in fact exceedingly well-known and its derivation my be found in many textbooks and at many web sites on the internet.

Now suppose

$e^{P(\zeta)} \ne 1; \tag{15}$

then

$e^{-P(\zeta)} \ne 1 \tag{16}$

as well, and if $x = \zeta$, (10) reads

$\phi(\zeta) = e^{-P(\zeta)}(\phi(0) + \displaystyle \int_0^\zeta e^{P(t)}r(t) \; dt), \tag{17}$

which may be re-written as

$e^{P(\zeta)} \phi(\zeta) - \phi(0) = \displaystyle \int_0^\zeta e^{P(t)} r(t) \; dt; \tag{18}$

we have seen in item (1) that $\phi(x)$ is a periodic solution with period $\zeta$ to (3)-(4) if and only if $\phi(\zeta) = \phi(0)$; if we choose $\phi(0)$, based upon (18), so that

$e^{P(\zeta)} \phi(0) - \phi(0) = (e^{P(\zeta)} - 1) \phi(0) = \displaystyle \int_0^\zeta e^{P(t)} r(t) \; dt, \tag{19}$

which is always possible by virtue of (15), then clearly (17) implies

$\phi(\zeta) = e^{-P(\zeta)}(\phi(0) + \displaystyle \int_0^\zeta e^{P(t)}r(t) \; dt) = \phi(0), \tag {20}$

and we may infer from item (1) that the trajectory initialized at $\phi(0)$ is periodic of period $\zeta$. Furthermore, it is now clear from (19)-(20) that $\phi(0)$ is the only possible initial condition such that $\phi(\zeta) = \phi(0)$.

Thus we see that $e^{P(\zeta)} \ne 1$ implies the existence of a unique periodic solution to (2)-(3).

Now suppose there is precisely one periodic trajectory which has initial value $\phi(0)$, and that $e^{P(\zeta)} = 1$. Then we also have $e^{-P(\zeta)} = 1$, and for this solution (20) must hold, whence

$\phi(\zeta) = \phi(0) + \displaystyle \int_0^\zeta e^{P(t)}r(t) \; dt = \phi(0), \tag {21}$

which in turn implies

$\displaystyle \int_0^\zeta e^{P(t)}r(t) \; dt = 0; \tag{22}$

but, via (17) we find

$\phi(\zeta) = \phi(0), \tag{23}$

no matter what value $\phi(0)$ may take. This shows that every solution must be periodic, which contradicts our assumption of a single periodic orbit. Therefore we cannot have $e^{P(\zeta)} = 1$ if the periodic solution is unique; we conclude that $e^{P(\zeta)} \ne 1$ in this case.

Our demonstration of item (2) is thus complete.

3.) We have already seen in item (2) that the combined requirements

$e^{P(\zeta)} = 1,\; \displaystyle \int_0^\zeta e^{P(t)} r(t) \; dt = 0 \tag{24}$

are sufficient to imply every solution to (3)-(4) is periodic; so, supposing every $\phi(x)$ obeying (3)-(4) also satisfies

$\phi(\zeta) = \phi(0),\tag{25}$

then (20), and hence (19), holds for every value of $\phi(0)$; but as we have seen under item (2), $e^{P(\zeta)} \ne 1$ implies that (18) is valid for precisely one value of $\phi(0)$; we thus conclude that

$e^{P(\zeta)} = 1, \tag{26}$

but then (19) immediatly yields

$\displaystyle \int_0^\zeta e^{P(t)} r(t) \; dt = 0, \tag{27}$

and we are done.