# Estimating number of zeros of solutions of linear 2nd order ode

Let $y''+fy=0$ be a second-order linear ode on $y$, where $f(x)>0$, and $I=\left[ a,b \right)$ be an interval. Suppose we want to estimate the number of zeros of a (not identically zero) solution of the ode on the interval $I$. Note that this question is well defined, because by the Sturm separation theorem, choosing a different solution can change the number of zers by at most 1.
It follows from the Sturm–Picone comparison theorem that if $0<m \leq f(x) \leq M$ on the interval ($m,M$ are numbers), and $N$ is the number of zeros of a solution on $I$, then $$\left \lfloor \frac{(b-a)\sqrt{m}}{\pi} \right \rfloor \leq N \leq \left \lceil \frac{(b-a)\sqrt{M}}{\pi} \right \rceil$$ Now, we can break $I$ into subintervals, and hopefully we get better estimates. A hopeful thought lead to hoping that the number of zeros on $I$ can be approximated by the integral $S=\int_{a}^{b} \frac{\sqrt{f(x)}}{\pi} dx$.

My question is, how good is this approximation? What can be said about the error $\left| N-S \right|$?

• It looks like one always has $N\ge S-1$ but we can easily have $S\approx 0$ for any $N$ without extra assumptions. The second observation is trivial (sawtooth function with smoothed corners). I can try to check the first if it is of any interest for you. – fedja Jun 17 '14 at 21:11
• Your formula is too sensitive to the precise shape of local singularities ($f\approx \delta$, as suggested by fedja). What matters here is $\int_I f$ over small intervals. Also, a very general precise formula would allow you to locate eigenvalues of $-y''+fy$ with high accuracy, so seems unlikely. – Christian Remling Jun 18 '14 at 5:13
• $N\ll S$ is possible too: we can take $f(x)=cx^{-2}$ on $(1,\infty)$, where we must choose $c>0$ small enough so that the equation becomes non-oscillatory. See en.wikipedia.org/wiki/…. So any solution of $y''+fy$ has at most $N$ zeros on $(1,b)$, with $N$ independent of $b$, but of course we can make $S(b)$ arbitrarily large. – Christian Remling Jun 18 '14 at 5:33
• In physics the approximation of the number of zeroes in $I$ by $S$ is known as WKB (or sometimes JWKB) approximation. Google "WKB approximation". The Wikipedia article is a good starting point. – Johannes Trost Jun 18 '14 at 11:23

Here is a specific example with a large $S$ constructed by starting with $y$. Consider $$y=\sqrt[3]{sin(x)}$$ for the interval $[\delta,\pi-\delta]$ with $0 \lt \delta$ quite small.
Actually, an interval like $[\delta,1]$ is adequate.
The point is that $y''+fy=0$ sets $$f=\frac{2+\sin^2(x)}{9\sin^2{x}} \gt\frac{2}{9x^2}$$ so $$\int_{\delta}^1\sqrt{f}dx \gt \frac{\sqrt{2}}{3}(-\ln{\delta})$$ which can be made arbitrarily large by choosing $\delta$ sufficiently small.
I haven't worked out any details, but I'm sure that this could be made into an oscillating example (with arbitrarily large $|S-N|$ ) by some adjustment on the intervals $[k\pi-\delta,k\pi+\delta].$
• The WKB approximation of the number of zeroes in $I$ with (the so called action integral) $S$ is good, if $|df/dx|\ll f^2$ in the interval $I$, which is not the case for your example. See my comment on the question. – Johannes Trost Jun 18 '14 at 11:20