roots of a solution of second order ODE

Question

How can I prove that if a non trivial solution of the ODE $$u''+(a+b\cos 2x)u=0$$ that has $2n$ roots in the interval $(-0.5 \pi,0.5 \pi)$, then $$(2n-1)^2 \le a+b~?$$

thx in advance

Answer 1

By the Sturm-Picone comparison theorem, this equation is dominated by $$ y''+(a+b)y=0 $$ and has thus at most as many roots as $\sin(\sqrt{a+b}(x+0.5\pi))$. More precisely, between any two consecutive roots of the sine there is at most one root of the solution. The sine has roots $$ \sqrt{a+b}(x_k+0.5\pi)=k\pi\implies x_k=-\frac\pi2+\frac{k\pi}{\sqrt{a+b}}, ~~ k=0,...,m $$ This means that with with $$x_m\le \frac\pi2<x_{m+1}\iff m\le\sqrt{a+b}< m+1,$$ and also including the final interval $(x_m,\frac\pi2)$, there are at most $m+1$ roots of the solution. Thus if $2n=m+1$ in the maximal case then $$ (2n-1)^2\le a+b<4n^2. $$

You will need to consult your sources on the Sturm-Picone theorem and check that its formulation also applies when the coefficient function $q(x)$ in $u''(x)+q(x)u(x)=0$ can become negative.