3
$\begingroup$

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 $\mathbb{R}_+$?

please help me.

$\endgroup$
4
$\begingroup$

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 $y=0$.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.