I am working on this problem \begin{equation} \begin{cases} u''(s)+\frac{2}{u} u'(s)=R^2 f(u) \quad \text{ for } \eta<s<1, \\ u'(\eta)=0, \ u'(1)+\beta R (u(1)-\bar{\sigma})=0, \end{cases} \end{equation} where $R, \beta, \bar{\sigma}>0$, $0 \leq \eta <1$ are constants, $f \in C^1[0,+\infty)$, $f'$ is positive and bounded on $[0,+\infty)$, and $f$ attains $0$ only at the point $\sigma_0 \geq 0$.

It has been proved that this problem has a unique solution $u(s)=U(s,\eta,R)$ with $\sigma_0<u<\bar{\sigma}$ for all $\eta \leq s \leq 1$. What I want to show next is that $U(s,\eta,R)$ is strictly decreasing in $R$ for $\eta \leq s<1$. To do this, denote $v(s)=\frac{\partial U(s,\eta,R)}{\partial R}$, then $v$ solves the following problem \begin{equation} \begin{cases} v''(s)+\frac{2}{s} v'(s)=2Rf(u)+R^2 f'(u)v \quad \text{ for } \eta<s<1, \\ v'(\eta)=0, \ v'(1)+\beta R v(1)=-\beta (u(1)-\bar{\sigma}). \end{cases} \end{equation} Also, it has been proved that $u'(1)>0$, thereby $-\beta (u(1)-\bar{\sigma})=\frac{u'(1)}{R}>0$ follows. With all those prepared, I want to use the maximum principle to prove that $v(s)<0$ for $\eta \leq s<1$, only to see that neither can I find existing results in textbooks or other papers nor conceive a proof by myself.

Could anyone please tell me where to refer to for them, or provide some ideas on how to tackle this kind of problem? Thank you very much.