I know that there is a typical example: $$\begin{cases}\tag{1} \Delta u-u+u^p=0,\ u>0,\\ \quad\quad u\in H^1(\mathbb{R}^n), \end{cases} $$ where $1<p<2^*-1$. Denote the solution to (1) by $U$ and its linearized operator by $L$, namely $$L(\phi)=\Delta \phi-\phi+pU^{p-1}\phi.$$ Then $U$ is said to be nondegenerate since the kernel of the linearized operator is $span\left\{\frac{\partial U}{\partial x_1},\dots,\frac{\partial U}{\partial x_n}\right\}$. But when we consider the problem: $$\begin{cases}\tag{2} \Delta u+u^{2^*-1}=0,\ u>0,\\ \quad\quad u\in D^{1,2}(\mathbb{R}^n), \end{cases} $$ the kernel of the corresponding linearized operator is $span{\left\{\frac{\partial U_\mu}{\partial \mu}\Big|_{\mu=1},\frac{\partial U_1}{\partial x_1},\dots,\frac{\partial U_1}{\partial x_n}\right\}}$ where $U_\mu$ is the solution to (2) with parameter $\mu$ and thus the definition of nondegeneracy of $U_\mu$ may be changed? Anyway, My question is what is the relation between the nondegeneracy of solution and the kernel of its linearized operator? In other words, what requirements will the kernel of the linearized operator satisfy if the solution to some given PDE is said to be nondegenerate?
By the way, I also want to know the relation between the uniqueness of solution and the nondegeneracy of it since many say that they are two very closely related concepts but do not explain it for details. Thanks!
