Number of zeroes of solution to $y''(x)+e^{x^2}y(x)=0$ in $[0,3π]$

The question is to investigate the number of zeroes of $y''(x)+e^{x^2}y(x)=0$ in $[0,3π]$.

Solving this ODE would not be an easy task as one has to use the power series solution and then investigating the zeroes of the solution will require more analysis. I thought it to compare this ODE with the standard $y''(x)+y(x)=0$ whose solution has three or four zeroes in the interval $[0,3π]$.

Since the coefficient of $y(x)$ is $e^{x^2}\ge 1$ for $x\in [0,3π]$ so the solution of given ODE must have atleast three zeroes in $[0,3π]$. However, what I thought was in the lights of Sturm-Comparison theorem so I am not sure.

Am I correct to interpret this?

For any function $\eta: [0,3\pi] \to \mathbb{R}$, consider functions of the form $$y_\eta(x) = \Im \left[ \exp\left( -\eta(x) + i\int_0^x e^{2\eta(t)} dt + K\right)\right]$$ where $K$ is an arbitrary constant. It is easy to check $y_\eta(x)$ satisfies an ODE

$$y_\eta''(x) + p_\eta(x)y_\eta(x) = 0 \quad\text{ where }\quad p_\eta(x) = e^{4\eta(x)} + \eta''(x) - \eta'(x)^2\tag{*1} $$

When $\eta(x) = \frac{x^2}{4}$, this reduces to $$y_\eta''(x) + \left(e^{x^2} + \frac{2-x^2}{4}\right)y_\eta(x) = 0\tag{*2}$$

Compare this to the ODE at hand

$$y''(x) + p(x)y(x) = 0 \quad\text{ where }\quad p(x) = e^{x^2}\tag{*3}$$

They look "approximately" the same. More precisely, the relative difference between the coefficients of the two ODE, $\frac{p_\eta(x) - p(x)}{p(x)} = \frac{2-x^2}{4}e^{-x^2}$, become tiny as $x$ increases. For example, at $x = 2$, this relative difference already falls below $1\%$. This suggests we can use solutions of $(*2)$ to bound the number of zeros for solutions of $(*3)$.

Notice $p(x) \ge p_\eta(x)$ for $x \in [\sqrt{2},3\pi]$. By Sturm-Picone comparison theorem, we have

$$\#\big\{ x \in [\sqrt{2},3\pi] : y(x) = 0\big\} \ge \#\big\{ x \in [\sqrt{2},3\pi] : y_\eta(x) = 0\big\} - 1$$

In general, if $x_1$ is a zero for $y_\eta(x)$ and $x_2 > x_1$ satisfies $\int_{x_1}^{x_2} e^{2\eta(t)} dt = \pi$, then $x_2$ is also a zero.
When we set $K$ to $-i\int_0^{\sqrt{2}}e^{t^2/2}dt$, $\sqrt{2}$ becomes a zero of $y_\eta(x)$. For this particular choice of $K$,

$$\begin{align} & \#\big\{ x \in [\sqrt{2},3\pi] : y_\eta(x) = 0\big\} = \left\lfloor \frac{1}{\pi}\int_{\sqrt{2}}^{3\pi} e^{t^2/2} dt\right\rfloor + 1\\ \implies & \#\big\{ x \in [0,3\pi] : y(x) = 0\big\} \ge \#\big\{ x \in [\sqrt{2},3\pi] : y(x) = 0\big\} \ge \left\lfloor \frac{1}{\pi}\int_{\sqrt{2}}^{3\pi} e^{t^2/2} dt\right\rfloor \end{align} $$ Together with $\displaystyle\;\frac{1}{\pi}\int_0^{\sqrt{2}}e^{t^2/2}dt \approx 0.658425\;$, we obtain a lower bound for number of zeros. $$\#\big\{ x \in [0,3\pi] : y(x) = 0\big\} - \frac{1}{\pi}\int_0^{3\pi}e^{t^2/2} dt > -1.66$$

To obtain an upper bound, we choose a new $\tilde{\eta} = \frac14\log\left(e^{x^2} + \frac{x^2}{4}\right)$. With help of an CAS, one can verify the corresponding $p_{\tilde{\eta}}(x) \ge p(x)$ for all $x \in [0,3\pi]$. By Sturm-Picone comparison theorem again, we obtain

$$\#\big\{ x \in [0,3\pi] : y_{\tilde{\eta}}(x) = 0\big\} \ge \#\big\{ x \in [0,3\pi] : y(x) = 0\big\} - 1 $$ By choosing a suitable $K$, we can make $y_{\tilde{\eta}}$ satisfy

$$ \#\big\{ x \in [0,3\pi] : y_{\tilde{\eta}}(x) = 0\big\} = \left\lfloor\frac{1}{\pi}\int_0^{3\pi} e^{2\tilde{\eta}(t)}dt\right\rfloor = \left\lfloor\frac{1}{\pi}\int_0^{3\pi} \sqrt{e^{t^2} + \frac{t^2}{4}} dt\right\rfloor $$ Since $\displaystyle\;\sqrt{e^{t^2} + \frac{t^2}{4}} \le e^{t^2/2} + \frac{t^2}{8}e^{-t^2/2}\;$, this leads to an upper bound for number of zeros.

$$\#\big\{ x \in [\sqrt{2},3\pi] : y = 0\big\} -\frac{1}{\pi}\int_0^{3\pi} e^{t^2/2} dt \le 1 + \frac{1}{8\pi}\int_0^{3\pi} t^2 e^{-t^2/2} dt \approx 1.0498678 $$ Combine these, we obtain an estimate of the number of zeros of $y(x)$ in terms of an integral.

$$-1.66 < \#\big\{ x \in [\sqrt{2},3\pi] : y(x) = 0\big\} -\frac{1}{\pi}\int_0^{3\pi} e^{t^2/2} dt < 1.05$$

According to WA, the integral evaluates to $$\frac{1}{\pi}\int_0^{3\pi} e^{t^2/2} dt \approx 663789642044913452.583222...$$ Since number of zeros is always an integer, we can conclude $$ 663789642044913451 \le \#\big\{ x \in [\sqrt{2},3\pi] : y(x) = 0\big\} \le 663789642044913453 $$

_{(Moved from a deleted duplicate question, answered Feb 18 '17 at 8:44, since it contains a more elementary approach)}

See the Sturm-Picone comparison theorem which tells you that you have at least as many roots as $\cos x$ on $[0,3π]$.

You could apply it to the segments $[0,π]$, $[π,2π]$ and $[2π,3π]$ separately to get a better lower bound for the root numbers as you then compare to $y''+e^{(k\pi)^2}y=0$, $k=0,1,2$ so that you get on the respective intervals at least as many roots as $\cos(e^{k^2\pi^2/2}x)$ where the frequencies have numerical values $1,\; 139.045636661,\; 373791533.224$.

With a finer subdivision one can drive this lower bound up to $6.5·10^{17}$ roots inside the interval.

Details on the application of the Sturm-Picone comparison theorem (2/21/17): On $[0,3\pi]$ use $q_1(x)=1$ and $q_2(x)=e^{x^2}$. Then $q_1\le q_2$ and $p_1=1=p_2$, so the theorem applies and any solution $v$ of $v''+q_2v=0$ has at least one root between any two consecutive roots $x_k=k\pi$, $k=0,1,2,3$ of the solution $u(x)=\sin x$ of $u'+q_1u=0$. The roots of $\cos x$ have this property, Which is why one can say that $v$ has at least as many roots as $\cos x$ in that interval.