Consider $I:H\rightarrow\mathbb{R}$ defined by $$I(u)=\int_0^R\left\{\dfrac{1}{2}ru_r^2-\gamma ru^2+r\ln(1+u^2)\right\}dr,$$ $\gamma\in(0,1)$, where $H$ is the completion of $$X=\left\{u\in C^1[0,R]:u(0)=0=u(R)]\right\},$$ i.e., the space of differentiable functions on $[0,R]$ vanishing at the boundary, with inner product $$(u,v)=\int_0^R\{ru_rv_r\}dr.$$ Note that we may treat $H$ as an embedded subspace of $W^{1,2}(B_R)$ compose of radially symmetric functions with the property $u(0)=0$.
I am interested in discussing whether the functional $I$ satisfies a Palais-Smale (PS) condition for any values of $\gamma\geq\frac{r_0^2}{2R^2}$? ($r_0\approx2.404$ is the first zero of the Bessel function $J_0$ and note that $R$ should be sufficiently large).
From the definition of the PS condition we may suppose that $\{u_j\}$ is a sequence in $H$ such that
- $\beta\geq|I(u_j)|$
- $\epsilon_j||v||_H\geq|I'(u_j)(v)|$, $\epsilon_j\geq 0$, $v\in H$, $\epsilon_j\rightarrow 0$ as $j\rightarrow\infty$,
and at as a first step aim to show that $\{u_j\}$ is bounded in $H$.
From 2, using Poincare inequality, we may obtain the following: \begin{align} \beta\geq\left(\frac{1}{2}-\gamma \frac{R^2}{r_0^2}\right)\int_0^Rru_{j,r}^2dr. \end{align} Hence, conclude that $\{u_j\}$ is a bounded sequence in $H$ when $\gamma<\frac{r_0^2}{2R^2}$. This also implies that the functional is coercive for such values of $\gamma$. However, this will lead to trivial critical points of the functional.
Hence, in my attempt to use both 1 and 2, I arrive at \begin{align} \beta&\geq \int_0^R\left\{\dfrac{1}{2}ru_r^2-\gamma ru^2+r\ln(1+u^2)\right\}dr\\ &\geq \int_0^R\left\{\dfrac{1}{2}ru_r^2-\gamma ru^2-\xi\dfrac{ru_j^2}{1+u_j^2}\right\}dr, \end{align} where $\xi\in[0,1)$. Then from 2, we may get \begin{align} \int_0^R\dfrac{ru_j^2}{1+u_j^2}\leq \int_0^R\left\{\dfrac{1}{2}ru_r^2-\gamma ru^2\right\}dr-\dfrac{1}{2}\epsilon_j||u_j||_H. \end{align} Inserting above and using a Poincare inequality we may get, \begin{align} \beta\geq (1-\xi)\dfrac{R^2}{r_0^2}\left(\frac{r_0^2}{2R^2}(1-\delta)-\gamma\right) \int_0^R\{ru_r^2\}dr-C\epsilon_j^2, \end{align} where $\delta\in(0,1)$, $C>0$. Hence, to get the boundeness of $\{u_j\}$ again we need $\gamma<\frac{r_0^2}{2R^2}$.
Any discussion is welcomed.