# $y''\pm e^ty=0 \implies \mid \cup x_i \mid =? s.t. y(x_i)=0$

I have this question, and i don't know how to solve it:

Show that the solutions of $y''+e^ty=0$ admit an infinite number of zeros.

Also, how to prove that the solutions of $y''-e^ty=0$ admit not more than one zero in $\mathbb{R}_+$?

1) Show that the solutions of $y''+e^ty=0$ admit infinitely many zeros.

Suppose there exists a solution $y$ with finitely many zeros. Therefore, $y$ is positive or negative on $[A,+ \infty)$ for $A$ large enough; if $y<0$ consider $-y$ so that you can suppose $y>0$.

Because $y''(t)=-e^ty(t)<0$ for $t \geq A$, $y'$ is decreasing on $[A,+ \infty)$. Moreover, if $y'$ is not bounded below, then there exist $C<0$ and $t_0>A$ such that $y'(t)<C$ for $t \geq t_0$, hence (by integration) $y(t) < y(t_0)+C(t-t_0) \underset{t\to + \infty}{\longrightarrow} - \infty$: a contradiction with $y>0$. Therefore, $y'$ is bounded below and the limit $\lim\limits_{t \to + \infty} y'(t)=\ell$ exists.

For $\epsilon>0$, there exists $t_1>0$ such that $t \geq t_1$ implies $\ell-\epsilon<y'(t)<\ell+\epsilon$; by integrating, $y(t_1)+ (\ell-\epsilon)(t-t_1)<y(t)<y(t_1)+(\ell+\epsilon)(t-t_1)$. You deduce that $\lim\limits_{t \to + \infty} \frac{y(t)}{t}=\ell$ and $\ell \geq 0$.

For $n \geq 1$, let $c_n \in (n,n+1)$ such that $y'(n+1)-y'(n)=y''(c_n)$. So $c_n \underset{n \to + \infty}{\longrightarrow} + \infty$, and using the above limit, $y''(c_n) \underset{n \to + \infty}{\longrightarrow} 0$.

Because $y'$ is decreasing and $\lim\limits_{t \to + \infty} y'(t)=\ell \geq 0$, you deduce that $y' \geq 0$, so $y$ is nondecreasing. Consequently, $y(t) \geq y(A)>0$ so $y''(t) \leq -e^ty(A) \underset{t \to + \infty}{\longrightarrow} - \infty$: contradiction with $y''(c_n) \underset{n \to + \infty}{\longrightarrow} 0$.

2) Show that the (non zero) solutions of $y''-e^ty=0$ admit at most one zero in $\mathbb{R}_+$.

Let $y$ be a solution of $y''-e^ty=0$ with at least two zeros.

First, suppose there exists an interval $[a,b]$ such that $y(a)=y(b)=0$ and $y(x) \neq 0$ for $x \in (a,b)$. Without loss of generality, suppose $y>0$ on $(a,b)$ (otherwise, consider $-y$). Then $y''=e^ty>0$ on $(a,b)$ and $y'$ is increasing on $(a,b)$. According to Rolle's theorem, there exists $c \in (a,b)$ such that $y'(c)=0$, so $y'(a)<0$.

Because $y'$ is continuous, $y' \leq 0$ on $[a,a+ \epsilon]$ for some $\epsilon >0$ hence $\displaystyle y(t)= \int_a^t y'(s)ds \leq 0$ for $t \in [a,a+\epsilon]$: contradiction with $y>0$ on $(a,b)$.

So there is no such interval $[a,b]$. You deduce that there exists a decreasing sequence $(x_n)$ of zeros. For $n \geq 1$, according to Rolle's theorem, there exists $u_n \in (x_{n+1},x_n)$ such that $y'(u_n)=0$.

We have $0<u_{n+1}<x_{n+1}<u_n<x_n$ for any $n \geq 1$, so $(u_n)$ and $(x_n)$ converge to the same limit $\ell$. We deduce by continuity that $y(\ell)= \lim\limits_{n \to + \infty} y(x_n)=0$ and $y'(\ell)= \lim\limits_{n \to + \infty} y'(u_n)=0$.

Using Cauchy-Lipschitz theorem, you find that the only possibility is $y=0$.

You can solve this equation by noting that putting $$z=e^t$$ then $$\frac{d}{dt}=\frac{d}{dz}\frac{dz}{dt}=e^t\frac{d}{dz}=z\frac{d}{dz}$$ and so the equation becomes $$z\frac{d}{dz}z\frac{dy}{dz}+zy=0$$ or $$z\frac{d^2y}{dz^2}+\frac{dy}{dz}+y=0.$$ taking into account that is always $z\ne 0$. This admits a general solution in terms of Bessel functions as $$y(z)=AJ_0(2\sqrt{z})+BY_0(2\sqrt{z}).$$ being $A$ and $B$ integration constants. This proves the assertion as these Bessel functions have infinite zeros.