therefore $v'$ is decreasing. This means that $v'$ is positive in $(0,L)$, since $v'(L)=0$, i.e. $v$ is increasing in $(0,L)$. By multiplying the differential equation of by $v'$ and integrating in $(y,L)$, with $0<y<L$, we get $$d\frac{v'(y)^2}{2}=\int_{v(y)}^{v(L)}f(s)\,ds$$ and, recalling that $v'>0$, we have that any solution of must satisfy $$L=\int_0^L\frac{v'(y)\,dy}{\sqrt{\frac{2}{d}\int_{v(y)}^{v(L)}f(s)\,ds}}=\int_0^1\frac{d\xi}{\sqrt{\frac{2}{d}\int_{\xi}^{1}\frac{f(v(L)\eta)}{v(L)\eta}\eta\,d\eta}}$$ with $0<v(L)<1$. Therefore, if we define the function $$\label{33}
M(\r):=\int_0^1\frac{d\xi}{\sqrt{\frac{2}{d}\int_{\xi}^{1}\frac{f(\r\eta)}{\r\eta}\eta\,d\eta}},$$ which is continuous in $(0,1)$ and measures the length of the interval necessary for a solution of $$\left\{ \begin{array}{l}
  -dV''(y)=f(V(y)) \quad \text{for } y\in(0,y_0)\\
  V(0)=0, \quad V'(y_0)=0 \end{array}\right.$$ to attain its maximum value $\r$ (at $y_0$), the uniqueness of the Cauchy problem associated to the ordinary differential equation of , we have that any solution of $M(\r)=L$ provides a solution of . The function $M$ satisfies $$\lim_{\r\ua 1}M(\r)=+\infty$$ since a maximum equal to $1$ cannot be attained in a finite interval, as seen before. Moreover, thanks to , $$\lim_{\r\da 0}M(\r)=\sqrt{\frac{d}{f'(0)}}\int_0^1\frac{d\xi}{\sqrt{1-\xi^2}}=\sqrt{\frac{d}{f'(0)}}\frac{\pi}{2}<L$$ and, therefore, there exists $\bar{\r}\in(0,1)$ such that $M(\bar{\r})=L$, which provides us with a solution of .

As far as uniqueness, it is easily seen that, under hypothesis , the function $M$ is increasing and therefore there exists a unique value of $\rho$ for which $M(\rho)=L$.

\[re33\] If condition does not 