a necessary and sufficient condition for to possess positive bounded solutions.

\[pr32\] Problem admits positive bounded solutions if and only if holds. Moreover, if we assume and , then Problem admits a *unique* positive bounded solution.

We begin with the necessity of . Suppose it does not hold and Problem admits a positive solution $v$. Then, multiplying the differential equation of by $\sin(\frac{\pi}{2L}y)$ and integrating by parts in $(0,L)$, we get $$\begin{gathered}
\int_0^Lf(v(y))\sin\left(\frac{\pi}{2L}y\right)\,dy=d\left(\frac{\pi}{2L}\right)^2\int_0^Lv(y)\sin\left(\frac{\pi}{2L}y\right)\,dy\geq \\
\geq f'(0)\int_0^Lv(y)\sin\left(\frac{\pi}{2L}y\right)\,dy>\int_0^Lf(v(y))\sin\left(\frac{\pi}{2L}y\right)\,dy,\end{gathered}$$ where, for the last relation, we have used the second assumption in . We have reached a contradiction and therefore no positive solution can exist.

Now we pass to the sufficiency. First of all we show that any positive solution of must satisfy $v(y)<1$ in $[0,L]$. Indeed if there was a point $y_0\in(0,L]$ where $v(y_0)=1$, either it would be a maximum of $v$ in a (relative to $(0,L]$) neighborhood of $y_0$, but in such a case $v\equiv 1$ by the uniqueness of the Cauchy problem $-v''=f(v)$ with conditions $v(y_0)=1, v'(y_0)=0$, or there would be $y_1\in(0,L]$ with $v(y_1)=\max v>1$. But in this case, $-dv''(y_1)=f(v(y_1))<0$, which is impossible for a maximum.

As a consequence of this result and , we have that $v''<0$ in $(0,L)$ and 