Let $u(x)$ and $v(x)$ be solutions of $y'' + p(x)y' + q(x)y = 0$, $p$ and $q$ continuous in $\mathbb{R}$, such that $u(0) = 1$, $u'(0) = 0$, $v(0) = 0$ and $v'(0) = 1$. Prove that, if $x_{1} < x_{2}$ are such that $u(x_1) = u(x_2) = 0$ and $u(x) \neq 0$ for all $x \in (x_1, x_2)$, then there exists a point $c \in (x_1, x_2)$ such that $v(c) = 0$. What's more, prove that such a point is unique.
I don't really know how to start tackling this problem. I know that the Wronskian of $u$ and $v$ is greater than $0$ in the entire interval, and that $v(x_1) \neq 0$ and $v(x_2) \neq 0$, but I couldn't go much farther than that...
Does anyone have a tip to get me going in the right direction?
Thanks in advance!
By the uniqueness of solutions, if $u$ is not identically $0$ there are no points where both $u$ and $u'$ are $0$. Thus if $x_1$ and $x_2$ are adjacent zeros of $u$, $u'(x_1)$ and $u'(x_2)$ are nonzero and have different signs (e.g. if $u > 0$ on $(x_1, x_2)$ we must have $u'(x_1) > 0$ and $u'(x_2) < 0$). Now if $W(u,v)(x_1) = u'(x_1) v(x_1)$ and $W(u,v)(x_2) = u'(x_2) v(x_2)$ have the same sign, $v(x_1)$ and $v(x_2)$ must have different signs. So by the Intermediate Value Theorem...