Recently I met an ODE problem but after thinking for quite a while I still could not find an answer. Here is the question, which looks very simple:




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Let $y=y(t)$ be a smooth function defined on $[0,\infty)$. If $y(t)$ satisfies the differential inequality
$$(ty')' \le \frac{4t}{(1+t^2)^2}(1-e^{2y})$$with the initial conditions $y(0)=0$ and $y'(0)=0$, then is it true that $y(t) \le 0$ for all $t \ge 0$?




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The motivation is that if the above differential inequality becomes equality with same initial conditions, then $y(t) \equiv 0$. I strongly feel the answer is positive, but so far I don't know how to apply any comparison argument. Any ideas and comments will be fully appreciated.




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Update: For the equality case, we can show $y(t) \equiv 0$ using the following argument:


Let $u(t)=\ln\left(\frac{2}{1+t^2}\right)$, then the equality case becomes 
$$(ty')'= te^{2u}(1-e^{2y})=-(tu')'-te^{2(u+y)}.$$ Let $w=u+y$, then $w$ solves the differential equation $(tw')'+te^{2w}=0$ with the initial condition $w'(0)=0$. Then $w$ must be of the form
$$w(t)=\ln\left(\frac{2c}{1+c^2t^2}\right)$$where $c$ is any positive constant. Hence $$y=w-u=\ln\left(\frac{c(1+t^2)}{1+c^2t^2}\right).$$Using the condition $y(0)=0$, we have $c=1$ and hence $y(t) \equiv 0$.



An ODE comparison problem
$\newcommand{\de}{\delta}
\newcommand{\vp}{\varepsilon}$
The conjecture is false. The previous answer, using numerics, was convincing enough for me, but perhaps not for others. So, here is a rigorous answer, based on the same general idea. 


Let 
\begin{equation\*}
 L(y)(t):=(ty'(t))'-\frac{4t}{(1+t^2)^2}(1-e^{2y(t)}). 
\end{equation\*}
Then 
\begin{equation\*}
 L(y\_1)(t)=0
\end{equation\*}
for real $t>0$, 
where 
\begin{equation\*}
y\_1(t):= \ln \frac{c\_1^2 e^{c\_1 c\_2} t^{c\_1-1}(t^2+1)}
{c\_1^2 e^{2 c\_1 c\_2}+t^{2 c\_1}} 
\end{equation\*}
with $c\_1\ge0$ and any real $c\_2$. In what follows, set 
\begin{equation\*}
 c\_1:=\sqrt{197/199},\quad c\_2:=1/192. 
\end{equation\*}
Then 
\begin{equation\*}
 y\_1(3/10)\approx-0.000101955<0,\quad y\_1'(3/10)\approx-0.0200123<0. 
\end{equation\*}
Let now 
\begin{align\*}
 c&:=-\frac{y\_1'(3/10)^2}{4y\_1(3/10)}\approx0.98203, \tag{1} \\ 
 \de&:=\frac{2y\_1(3/10)}{y\_1'(3/10)}\approx0.0101893, \tag{2} \\ 
 \vp&:=3/10-\de\approx0.289811, \tag{3}
\end{align\*}
so that 
\begin{equation\*}
 c\de^2=-y\_1(3/10),\quad 2c\de=-y\_1'(3/10). \tag{4}
\end{equation\*}
Letting now 
\begin{equation\*}
 y\_2(t):=-c(t-\vp)^2, \tag{5}
\end{equation\*}
we see that 
\begin{equation\*}
 y\_2(\vp)=y\_2'(\vp)=0,\quad y\_2(3/10)=y\_1(3/10),\quad y\_2'(3/10)=y\_1'(3/10), \tag{6}
\end{equation\*}
and
\begin{equation\*}
 L(y\_2)(t)=c(3/5-2 \de-4 t)+
 4 t \,\frac{e^{-2 c (\de+t-3/10)^2}-1}{(t^2+1)^2}
 <c(3/5-2 \de-4 t)<c(3/5-4 \vp)<-0.544<0 \tag{7}
\end{equation\*}
for $t\ge\vp$. 
Letting finally 
\begin{equation\*}
 Y:=y\_1\,1\_{[3/10,\infty)}+y\_2\,1\_{[\vp,3/10)},
\end{equation\*}
we see that 
\begin{equation\*}
 L(Y)\le0
\end{equation\*}
on $[0,\infty)$. However, 
\begin{equation\*}
 Y(6)=y\_1(6)\approx0.00360808>0. 
\end{equation\*}




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The function $Y$ is the solution $y$ to the problem 
$$L(y)=h$$ 
on $(0,\infty)$ with the initial conditions $y(0+)=y'(0+)=0$, where $h:=L(Y)\le0$ and $Y''$ is understood as the right (or left) derivative of $Y'$. Also, $Y=0$ on $(0,\vp]$ and $Y(6)>0.003>0$. 


Let now $\psi$ be a function in $C^\infty((0,\infty))$ such that $\psi=0$ on $(0,\vp/2]$ and $\psi\le0$ on $(0,\infty)$. Let then $Z$ be the solution $z$ to the problem $$L(z)=\psi$$ 
on $(0,\infty)$ with the initial conditions $z(0+)=z'(0+)=0$ (so that $z(\vp/2)=z'(\vp/2)=0$). Choose now $\psi$ to be so close to $h$ in $L^1((0,\infty))$ that $|Z(6)-Y(6)|<0.001$. Then $Z(6)>0.002>0$, as desired. 


