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--- moser [2025/11/07 11:54] – spencer
+++ moser [2025/11/07 14:19] (current) – [Inhomogeneous Moser] spencer
@@ Line 59: / Line 59: @@
 (b+1)^2 (f \eta u^b, \eta u^{b+1}) &= (2b+1)\| \nabla (\eta u^{b+1}) \|^2 - \| u^{b+1} \nabla \eta \|^2 - 2b ( \nabla (\eta u^{b+1}), u^{b+1} \nabla \eta) +(b+1)^2(V \eta u^{b+1}, \eta u^{b+1}).
 \end{align*}
+Thus we have
+\begin{align*}
+(2b+1) \| \nabla (\eta u^{b+1})\|^2 &= \| u^{b+1} \nabla \eta \|^2 + 2b (\nabla (\eta u^{b+1}), u^{b+1} \nabla \eta) + (b+1)^2 ((f - Vu) \eta u^b, \eta u^{b+1}) \\
+S^{-1} \| u^{b+1}\|^2_{L^2(B_k)} \le \| \nabla (\eta u^{b+1})\|^2 &\le (2b+1) 4^{k+3} R^{-2} \| u^{b+1}\|^2_{L^2(B_{k-1})} + (b+1)^2 |((f - Vu) \eta u^b, \eta u^{b+1})| \\
+\end{align*}
+To approximate the inhomogeneity, note that at a point $p$ with $R = \frac{|p|}{4}$, then $\| f\|_{L^\infty(B_{2R})} \le C R^{-n}$ by hypothesis, so
+\begin{align*}
+(2b+1) \| \nabla (\eta u^{b+1})\|^2 &= \| u^{b+1} \nabla \eta \|^2 + 2b (\nabla (\eta u^{b+1}), u^{b+1} \nabla \eta) + (b+1)^2 ((f - Vu) \eta u^b, \eta u^{b+1}) \\
+S^{-1} \| u^{b+1}\|^2_{L^{2n/(n-2)}(B_k)} \le \| \nabla (\eta u^{b+1})\|^2 &\le ((2b+1) 4^{k+3} R^{-2} + (b+1)^2 V^\infty)\| u^{b+1}\|^2_{L^2(B_{k-1})} + (b+1)^2 C R^{-n} \|u^{b+\frac{1}{2}}\|^2_{L^2(B_{k-1})} \\
+\end{align*}
+To estimate the $\|u^{b + \frac{1}{2}}\|^2_2 = \| u\|^{2b+1}_{2b+1}$ term, we use the Holder inequality and the previous term in the iteration to bound the $L^{2b+2}$ norm:
+$$
+ \| u\|_{2b+1} \le C\| u \|_{2b+2} (R^n)^{\frac{1}{(2b+1)(2b+2)}}.
+$$
+Thus with $p_k = 2b_k + 2 = 2 \left( \frac{n}{n-2} \right)^k$, it holds that
+\begin{align*}
+(2b+1) \| \nabla (\eta u^{b+1})\|^2 &= \| u^{b+1} \nabla \eta \|^2 + 2b (\nabla (\eta u^{b+1}), u^{b+1} \nabla \eta) + (b+1)^2 ((f - Vu) \eta u^b, \eta u^{b+1}) \\
+S^{-1} \| u\|^{p_k}_{L^{p_{k+1}}(B_k)} &\le ((p_k-1) 4^{k+3} R^{-2} + \frac{p_k^2}{4} V^\infty)\| u\|^{p_k}_{L^{p_k}(B_{k-1})} + C p_k^2 R^{\frac{n}{p_k} - n} \| u\|^{p_k-1}_{L^{p_k}(B_{k-1})}
+\end{align*}