Small charged Dirichlet conformal branch

By [F6], $X$ and $Y$ are real Banach spaces. The set $U:=\{(u,\epsilon )\in X\times \mathbb{R}:{\min }_{\bar{\mathrm{\Omega }}}(1+u)>0\}$ is open because the minimum changes by at most $\mathrm{\parallel }u-v{\mathrm{\parallel }}_{\mathrm{\infty }}$, which is bounded by the $X$-norm. Define $F:U\to Y$ by $F(u,\epsilon ):=Au-2{\epsilon }^2{N}_S(u)$, with $A=-8\mathrm{\Delta }$ and $N_S(u)=S(1+u{)}^{-3}$. By [A1], the inverse supplier [F2] applies; [F4] gives a continuously differentiable map on $Y$ and the bounded inclusion $X\hookrightarrow Y$, so $F$ is $C^1$. Also $F(0,0)=0$ and $D_{u}F(0,0)=A$, since the $u$-derivative of the nonlinear term is multiplied by ${\epsilon }^2$.

Since $A$ is invertible, [F3] supplies $\rho ,{\epsilon }_0>0$ and a unique local $C^1$ branch $u_{\epsilon }$ in the ball $\mathrm{\parallel }u{\mathrm{\parallel }}_{C^{2,\alpha }}<\rho$ satisfying $F(u_{\epsilon },\epsilon )=0$. Shrink the product neighborhood so that $\mathrm{\parallel }{u}_{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}<1\mathrm{/}2$; then ${\psi }_{\epsilon }=1+u_{\epsilon }>0$, and $F=0$ is equivalent to the defining Dirichlet equation.

For each $\epsilon$, the equation and $S\ge 0$ imply $\mathrm{\Delta }{u}_{\epsilon }\le 0$. Hence $\mathrm{\Delta }(-u_{\epsilon })\ge 0$ and $-u_{\epsilon }=0$ on $\mathrm{\partial }\mathrm{\Omega }$. Applying [F5] to $-u_{\epsilon }$ yields ${\max }_{\bar{\mathrm{\Omega }}}(-u_{\epsilon })=0$, so $u_{\epsilon }\ge 0$ and ${\psi }_{\epsilon }\ge 1$.

Let $v:=A^{-1}(2S)$, which is the unique element of $X$ with $-8\mathrm{\Delta }v=2S$. The fixed-point equation gives $u_{\epsilon }=2{\epsilon }^2{A}^{-1}(S(1+u_{\epsilon }{)}^{-3})$. The branch is continuous and $u_0=0$, so the Hölder norm of $S(1+u_{\epsilon }{)}^{-3}$ is bounded near $\epsilon =0$; boundedness of $A^{-1}$ in [F2] implies $\mathrm{\parallel }{u}_{\epsilon }{\mathrm{\parallel }}_{C^{2,\alpha }}=O({\epsilon }^2)$. If $S=0$, the same identity gives $u_{\epsilon }=0$, so the zero-source case is included.

Subtract $A({\epsilon }^{2}v)=2{\epsilon }^{2}S$ from $A{u}_{\epsilon }=2{\epsilon }^{2}S(1+u_{\epsilon }{)}^{-3}$, obtaining $u_{\epsilon }-{\epsilon }^{2}v=2{\epsilon }^2{A}^{-1}\text{\hspace{0.17em}⁣}\left(S\right((1+u_{\epsilon }{)}^{-3}-1))$. The differentiability estimate in [F4] at $u=0$ gives $\mathrm{\parallel }S((1+u_{\epsilon }{)}^{-3}-1){\mathrm{\parallel }}_{C^{0,\alpha }}\le C_1\mathrm{\parallel }{u}_{\epsilon }{\mathrm{\parallel }}_{C^{0,\alpha }}$ for small $\epsilon$. Combining this with step 3.2, the Hölder product bound, and $\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}_{Y\to X}<\mathrm{\infty }$ proves $\mathrm{\parallel }{u}_{\epsilon }-{\epsilon }^{2}v{\mathrm{\parallel }}_{C^{2,\alpha }}\le C\mathrm{\mid }\epsilon {\mathrm{\mid }}^4$. This is a conditional mathematical consequence of the stated assumptions, not an empirical claim.