Let $B \subset \mathbb{R}^N$ be the unit ball and $E = \{u \in H^1_0(B) : u \text{ is radial}\}$. Define the functional $I : E \to \mathbb{R}$ by $$ I(u) = \int_{B} |\nabla u(x)|^2 dx - \frac{1}{p}\int_{B} |x| |u(x)|^{p} dx, $$ for some $p > 2$. Suppose $u \in E$ is a radial critical point of $I$, that is, $$ I'(u) \varphi = \int_{B} \nabla u \nabla \varphi - \int_{B} |x| |u|^{p-2}u \varphi = 0, \quad \forall \varphi \in C^\infty_{0,rad}(B), $$ where $\overline{C^\infty_{0,rad}(B)}^{H^1_0(B)} = E$. How to show $u$ is a weak solution of $$ -\Delta u (x) = |x| u^{p-1}(x), x \in B, \quad u (x) = 0, x \in \partial B, $$ without using the functional $I$ is well defined in $H^1_0(B)$ ? I can't use this fact, because I just know $I$ is well defined in $E$. I know above $PDE$ is equivalent to the $ODE$ $$ -[f''(s) + \frac{N-1}{s} f'(s)] = s |f(s)|^{p-2}f(s), 0 < s < 1, \quad f(0) = \alpha, f'(0) = 0, $$ for some $\alpha \in \mathbb{R}$.
What I tried: Given $\psi \in C^\infty_0(0,1)$ arbitrary, define for $x \neq 0$, $\varphi_0(x) = \psi(|x|)$ and $\varphi_0(0) = \psi(0) = 0$. Notice that $\varphi_0 \in C^\infty_{0,rad}(B)$. So, as $u$ is a critical point of $I$ one has $$ \int_{B} \nabla u \nabla \varphi_0 = \int_{B} |x| |u|^{p-2}u \varphi_0 . $$ If we write $u(x) = f(|x|)$ for some $f \in H^1_0(0,1)$, It is not difficulty to see that $\nabla u(x) \nabla \varphi_0(x) = f'(|x|) \psi'(|x|)$. So, integrating last inequality in polar coordinates one has $$ \sigma_{N}(S^{N-1}) \int_0^1 f'(s) \psi'(s) s^{N-1} ds = \sigma_{N}(S^{N-1}) \int_0^1 s |f(s)|^{p-2} f(s) \psi(s) s^{N-1} ds, $$ that is $$ \int_0^1 f'(s) \psi'(s) s^{N-1} ds = \int_0^1 s |f(s)|^{p-2} f(s) \psi(s) s^{N-1} ds, \quad \forall \psi \in C^\infty_0 (0,1). $$ Is there a way to use my idea to conclude that $$ \int_{B} \nabla u \nabla \varphi - \int_{B} |x| |u|^{p-2}u \varphi = 0, \quad \forall \varphi \in C^\infty_{0}(B) ? $$ Or $$ -\int_0^1 [f''(s) + \frac{N-1}{s} f'(s)] \psi = \int_0^1 s |f(s)|^{p-2}f(s) \psi, \quad \forall \psi \in C^\infty_0(0,1) ? $$
