I'm reading Shubin's Invitation to PDE. In section $3.3$. We consider a solution $z\in C^2([a,b]), z\not \equiv 0$ to the Sturm-Liouville problem :
$$z''(x)+q(x)z(x)=0\tag{1}$$
Where $q\in C^{2}([a,b])$, $z(a)=z(b)=0$. It is said that the null points of $z$ in the interval are simple, in the sense that if $z(x_0)=0$ for some $x_0\in [a,b]$, then $z'(x_0)\neq 0$. The author says this fact comes from the existence and uniqueness theorem for order two ODE. I do not see how to deduce this fact from the theorem.
The way I think we could use the theorem is to show that the non-simplicity of a null point would imply that $z'$ is also a solution to the Sturm-Liouville problem on some interval and then deduce by unicity that we must have $z'=z$ on some interval and so $z$ would be an exponential function, which does not solve the problem in general. But I don't see how $z'$ could be a solution.
Any help would be appreciated. Thank you.
We can do this for with uniqueness of first-order systems instead for clarity Let $u = z'$, then construct the following linear system of ODEs: $$ \begin{aligned} z' &= u \\ u' &= -q(x)z. \end{aligned} $$
Suppose that $z(x_0) = z'(x_0) = 0$. In terms of our system, this corresponds to an initial condition $(z_0, u_0) = (0,0)$ and the solution is given by $(z(x), u(z)) = (0,0) \ \forall x> x_0$. One could also reverse the flow by swapping $x\mapsto -x$ and solving the reversed linear ODEs similarly, so one would have $(z(x), u(z)) = (0,0) \ \forall x< x_0$. Uniqueness of solutions to systems of first-order ODEs says that this is the only possibility for $u$ and $z$.
Therefore, if such a point $x_0$ existed for our solution to the Sturm-Liouville problem, then the solution would be zero everywhere in the domain.