I'm making an exercise in a book from Ravi P. Agarwal, Introduction to Ordinary Differential Equations. I would like to get some help for the following problem:

We are given the differential equation $$ y''+p_1(x)y'+p_2(x)y = 0 $$ where $p_1(x)$ and $p_2(x)$ are continuous functions, and periodic with period $w$ in $\mathbb{R}$.

Show that:

A nontrivial solution $y(x)$ is periodic of period $w$ if and only if $y(0)=y(w)$ and $y'(0)=y'(w)$.

Further if $y_1(x)$ and $y_2(x)$ are two solutions such that $y_1 (0)= 1, y_1'(0)=0, y_2(0)=1, y'_2(0)=0$ then show that:

there exist constants $a, b, c, d$ such that for all $x$ :

$$y_1(x+w)= ay_1(x)+by_2(x)$$ $$y_2(x+w)= cy_1(x)+dy_2(x)$$

I think I have to use a theorem which is in the book. I will give the theorem:

Theorem: Let the matrix $A(x)$ and the function $b(x)$ be continuous and periodic of period $w$ in $\mathbb{R}$. Then the differential system $ u'= A(x)u + b(x)$ has a periodic solution $u(x)$ of period $w$ if and only if $u(0)=u(w)$

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    $\begingroup$ It should be the case that $y_2(0)=0$ and $y_2'(0)=1$? $\endgroup$ – Berkheimer Jan 14 '13 at 13:25
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    $\begingroup$ And you also ask for $y'(0)= y'(w)$ in the first part? $\endgroup$ – Berkheimer Jan 14 '13 at 14:58
  • $\begingroup$ @беркай Yes you are right. I have fixed the question. Could you please give answer the question? :-) $\endgroup$ – Applied mathematician Jan 14 '13 at 22:10

If $y(x)=y(x+w)$, then it is immediate that $y(0)=y(w)$ and $y'(0)=y'(w)$.

For the converse, suppose $y(x)$ be a solution of the given differential equation satisfying $y(0)=y(w)$ and $y'(0)=y'(w)$. Let $v(x)=y(x+w)$. Then,

$$ \tag 1 y''(x+w) + p_1(x+w)y'(x+w) + p_2(x+w)y(x+w)=0 $$

is equivalent to

$$ v''(x) + p_1(x)v'(x) + p_2(x)v(x)=0. $$

Hence, $v(x)$ satisfies the given differential equation. Since, $v(0)=y(w)=y(0)$ and $v'(0)=y'(w)=y'(0)$, by the uniqueness of the initial value problems, it follows that $y(x)=v(x)=y(x+w)$, i.e. $y(x)$ is periodic of period $w$.

Since the Wronskian $W(y_1,y_2)(0)=1 \neq 0$, the solutions $y_1(x)$ and $y_2(x)$ are independent. Moreover, $y_1(x+w)$ and $y_2(x+w)$ satisfy $(1)$. Hence, both can be written as linear combinations of $y_1(x)$ and $y_2(x)$, i.e., $$ y_1(x+w)= a y_1(x) + by_2(x) ; \quad y_2(x+w)= c y_1(x) + d y_2(x). $$

Using the initial conditions, it follows that $$a=y_1(w); \quad b=y_1'(w);\quad c=y_2(w);\quad d=y_2'(w).$$

  • $\begingroup$ Great! Thank you for your help. Can you look at the second part of my question? $\endgroup$ – Applied mathematician Jan 15 '13 at 0:22
  • $\begingroup$ By the way, you should edit the statement: $y_2(0)=0$ and $y_2'(0)=1$. $\endgroup$ – Berkheimer Jan 15 '13 at 0:43
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    $\begingroup$ The Wronskian at the point $0$ must be $1$, this is the idea. We have $W(x)=y_1(x)y_2'(x)-y_1'(x)y_2(x)$ and $W(0)=1$ doesn't hold with your assumption. $\endgroup$ – Berkheimer Jan 15 '13 at 0:57
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    $\begingroup$ I mean we have the Wronskian $1$ with the fundamental solutions satisfying $y_1(0)=1$, $y_1'(0)=0$ etc. Also, the Existence-Uniqueness Theorem assures the existence of these solutions under some conditions, but we are already given these solutions so what I wrote was misleading and I removed this part. $\endgroup$ – Berkheimer Jan 15 '13 at 1:14
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    $\begingroup$ Ok! I will edit my answer. $\endgroup$ – Berkheimer Jan 15 '13 at 1:23

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