A simple DE question on the sign of coefficient and the sign of solutions Hi the following comes from DE notes of my advisor...
If $z_1$ and $z_2$ are two successive zeros of a nontrivial solution $u$ of 
$$
u''+a(x)u=0,
$$
then it is reasonably obvious that $a(x)\ge 0$ for some $x$ between $z_1$ and $z_2$. For otherwise $u''$ and $u$ would have the same sign. This means that $u$ would be convex when $u$ is positive and concave when $u$ is negative.
My thought and mud:
I can see that if $u''=-u$ then $a(x)\ge 0$. But not sure why the rest makes sense.
Could anyone explain to me why 
"For otherwise $u''$ and $u$ would have the same sign... "

The alternative to $a(x) \ge 0$ for some $x \in [z_1, z_2]$ is that $a(x) < 0$ for all $x \in [z_1, z_2]$.  Then, since
$u''(x) = -a(x) u(x) \tag{1}$
with
$-a(x) > 0, \tag{2}$
we see that the signs of $u''(x)$ and $u(x)$ must be everywhere the same.  Note $u(x) \ne 0$ on $[z_1, z_2]$ since $z_1$ and $z_2$ are successive zeroes of the non-trivial solution $u(x)$.
And that, in a nutshell, is why $u(x)$, $u''(x)$ have the same sign unless $a(x) \ge 0$ somewhere in $[z_1, z_2]$.
