Consider $I:H\rightarrow\mathbb{R}$ defined by $$I(u)=\int_0^R\left\{\dfrac{1}{2}u_r^2-\xi u^2+\ln(1+u^2)\right\}r\,dr,$$ $\xi\in(0,1)$, where $H$ is the completion of $$X=\left\{u\in C^1[0,R]:u(0)=0=u(R)\right\}.$$ Let $\gamma:H\rightarrow\mathbb{R}$ be defined as $$\gamma(u)=\dfrac{1}{2}I'(u)(u)=\int_0^R\left\{\dfrac{1}{2}u_r^2-\xi u^2+\dfrac{u^2}{1+u^2}\right\}rdr,$$ where $I'(u)(v)$ is the Frechet derivative of $I$ at $u$ in the direction of $v$.
Suppose $M:=\left\{u\in H\backslash\{0\}:\gamma(u)= 0 \right\}$ is nonempty for $\xi$ taking values over some interval.
Claim: Any critical point $u_0\in M$ of $I|_M$ satisfies $I'(u_0)=\mu\gamma'(u_0)$ for some $\mu\in\mathbb{R}$.
$$\gamma'(u)(v)=\int_0^R\left\{u_rv_r-2\xi uv+\dfrac{2uv}{(1+u^2)^2}\right\}rdr$$ and $$I'(u)(v)=\int_0^R\left\{u_rv_r-2\xi uv+\dfrac{2uv}{1+u^2}\right\}rdr.$$
I unable to reached such a conclusion. This may be because I lack an understanding of how the restriction plays a role here. My thoughts are the following: if $u_0$ is a critical point of $I$, then $I'(u_0)=0$ and the claim is trivially true by taking $\mu=0$. However, the claim suggests that $\mu=0$ may not be always the case. I am not really making use of the restriction of $I$ to $M$. Note that in this the problem $\mu=0$ is the case.
A better title may also be suggested for this question.
Finding critical points of $\mathcal{I}|_M$ may be viewed as a constraint minimization problem where the objective function is $\mathcal{I}(u)$ subject to the constraint $\gamma(u)=0$. So, if $u_0$ is a critical point of $\mathcal{I}|_M$, then there is a Lagrange multiplier $\mu$ such that $\mathcal{I}'(u_0)(u)=\mu\gamma'(u_0)(u)$ for every $u\in H$. Morevoer, it can be shown that $\gamma'(u)(u)<0$ and $0=\gamma(u_0)=\dfrac{1}{2}\mathcal{I}'_{\kappa}(u_0)(u_0)=\mu\gamma'(u_0)(u_0). $ So $\mu=0$.