solution of ordinary differential equation $x''(t)+e^{t^2} x(t)=0$, for $t\in[0,3\pi]$

Question

Let $x:[0,3\pi]\rightarrow\mathbb{R}$ be a nonzero solution of the ODE $x''(t)+e^{t^2} x(t)=0$, for $t\in[0,3\pi]$. Then the cardinality of the set $\{t\in[0,3\pi]:x(t)=0\}$ is

1. equal to 1

2. greater than or equal to 2

3. equal to 2

4. greater than or equal to 3

Answer 1

See Sturm-Picone comparison theorem which tells you that you have at least as many roots as $\cos t$ 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 $\cos(e^{k^2\pi^2/2}t)$ 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.

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Details on the application of the Sturm-Picone comparison theorem (2/21/17): On $[0,3\pi]$ use $q_1(t)=1$ and $q_2(t)=e^{t^2}$. 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 $t_k=k\pi$, $K=0,1,2,3$ of the solution $u(t)=\sin t$ of $u'+q_1u=0$. The roots of $\cos t$ have this property, Which is why one can say that $v$ has at least as many roots as $\cos t$ in that interval.