&+ \frac{1}{\nu}\ml \frac{\mu-Q^2/r^2}{1-\mu}\mr\ml \frac{|\tz|^2\lambda}{\nu} - \frac{|\ttheta|^2\nu}{\lambda}\mr + \frac{r^2}{\nu}\frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)} + \frac{r^2}{\nu}\ml \frac{\ttheta^\dagger E_1}{r\lambda} - \frac{\tz^\dagger E_4}{r\nu}\mr \\
=&\; r^2 \frac{(\tz/\nu - E_2)}{\nu^2} \frac{\p(\nu,\tp^\dagger)}{\p(u,v)} + r^2\frac{1}{\nu}\frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)} + R_1 + E_{12},
\end{aligned}
\end{equation}

\begin{equation}\label{Area231}
\begin{aligned}
R_1 :=&\;  \frac{1}{\nu}\ml \frac{\nu|\ttheta|^2}{\lambda} - (\ttheta^\dagger \tz + \tz^\dagger \ttheta) + \frac{\lambda}{\nu}|\tz|^2 - \frac{r}{\nu \lambda}\du \nu|\ttheta|^2 + \frac{\lambda |\tz|^2}{\nu} \frac{\mu - Q^2/r^2}{1-\mu}\mr\\
&+ \frac{1}{\nu}\ml \frac{\mu-Q^2/r^2}{1-\mu}\mr\ml \frac{|\tz|^2\lambda}{\nu} - \frac{|\ttheta|^2\nu}{\lambda}\mr, \\
E_{12}:= &\; \frac{E_2}{\nu}(\nu \ttheta^\dagger - \lambda \tz^\dagger) - \frac{E_2 \nu \lambda \tz^\dagger}{\nu^2}\frac{\mu - Q^2/r^2}{1-\mu} + \frac{rE_1^\dagger}{\nu}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr + \frac{r^2}{\nu}\ml \frac{\ttheta^\dagger E_1}{r\lambda} - \frac{\tz^\dagger E_4}{r\nu}\mr.
\end{aligned}
\end{equation}

Note that to arrive at the last equality in (<ref>), numerous cancellations were explored. By recognizing that $\im(R_1) = 0$ and $|E_{12}|(u,v) \lesssim D^2r^2(u,v)$, we can take the imaginary part of (<ref>) and deduce that

\begin{equation}\label{Area232}
\begin{aligned}
\du\ml \frac{r^2 \tg}{\nu}\mr &=   4\pi \im \ml \du \ml \frac{\ttheta^\dagger r}{\nu} \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr\mr \mr \\
&= 4\pi \im \ml r^2 \frac{(\tz/\nu - E_2)}{\nu^2} \frac{\p(\nu,\tp^\dagger)}{\p(u,v)} + r^2\frac{1}{\nu}\frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)}  + E_{12}\mr.
\end{aligned}
\end{equation}

This implies

\begin{equation}\label{Area233}
\mlm \du \ml \frac{r^2 \tg}{\nu}\mr\mrm \lesssim r^2 \ml D\mlm \frac{\p(\nu,\tp^\dagger)}{\p(u,v)}\mrm + \mlm \frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)}\mrm + D^2\mr.
\end{equation}

Plugging (<ref>) back to (<ref>), we thus derive

\begin{equation}\label{Area234}
\int_{\mathcal{D}(0,v_0)} \frac{|r^2 \tg|}{r^3}(u,v) \D u \D v \lesssim D \cdot A[\nu,\tp^\dagger] + A[\tp^\dagger,\ttmfb/\nu] + D^2.
\end{equation}

Henceforth, by utilizing (<ref>), (<ref>), (<ref>), (<ref>) and the fact that $A[\lambda,\tp^\dagger] = A[\lambda,\tp]$, we arrive at

\begin{equation}\label{Area235}
A[\lambda,\tp^\dagger] \lesssim (D+Y)A[\tp^\dagger,\ttmfb/\nu] + D^2\cdot A[\nu,\tp^\dagger] + D^2Y + D^3.
\end{equation}

It remains to estimate $A[\nu,\tp^\dagger]$. Parallel to the above arguments, we start off by defining

\begin{equation}\label{Area236}
\tgs := 4\pi \im\ml \frac{\tz^\dagger}{r}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr\mr
\end{equation}

and we have

\begin{equation}\label{Area237}
\tfs + \ii \tgs = 4\pi \frac{\tz^\dagger}{r}\ml \frac{\ttheta}{\lambda} - \frac{\tz}{\nu} + E_2\mr
\end{equation}

with $E_2$ given in (<ref>). This implies that

\begin{equation}\label{Area238}
\frac{4\pi}{r^2}\frac{\ttheta \tz^\dagger}{\lambda \nu}\ml \nu \ttheta - \lambda \tz\mr = \frac{\tfs \ttheta}{r} + \ii \frac{\tgs \ttheta}{r} - \frac{4\pi \ttheta \tz^\dagger}{r^2}E_2.
\end{equation}

Together with (<ref>), we compute

\begin{equation}\label{Area239}
\begin{aligned}
&\frac{1}{\nu}\frac{\p(\nu,\tp)}{\p(u,v)} = \du \log |\nu| \dv \tp - \dv \log |\nu| \du \tp = \du\log|\nu| \frac{\ttheta}{r} -\frac{\mu - Q^2/r^2}{1-\mu} \frac{\lambda \tz}{r^2} \\
&\quad= \frac{1}{r^2}(\nu \ttheta - \lambda \tz)\ml \frac{\mu - Q^2/r^2}{1-\mu} + 4\pi \frac{\ttheta \tz^\dagger}{\lambda \nu} \mr  + \frac{\ttheta}{r}\ml \du \log|\nu| - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\nu}{r}\mr - 4\pi \frac{\ttheta \tz^\dagger}{r^2}(\nu \ttheta - \lambda \tz) \\
&\quad= \frac{1}{r^2}\ml \frac{\nu \ttheta - \lambda \tz}{\lambda \nu} \mr\ml \frac{\mu - Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta \tz^\dagger \mr  + \frac{\ttheta}{r}\ml \du \log|\nu| - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\nu}{r} - \tfs\mr - \ii \frac{\ttheta\tgs}{r} + 4\pi \frac{\ttheta\tz^\dagger}{r^2}E_2. \\
\end{aligned}
\end{equation}

Note that from (<ref>), we have

\begin{equation}\label{Area240}
\dv \ml \du \log|\nu| - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\nu}{r} - \tfs\mr = -4\pi \re \ml \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)}\mr - E_5 + E_6.
\end{equation}

This implies that

\begin{equation}\label{Area241}
\mlm \du \log|\nu| - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\nu}{r} - \tfs\mrm(u,v) \leq G^*(u)
\end{equation}

\begin{equation}\label{Area242}
G^*(u) := \mlm \frac{\D}{\D u}\log \mlm \nu|_{\Gamma}\mrm \mrm(u) + 4\pi \int^{v_0}_u \mlm \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)}\mrm(u,v) \D v + \int_u^{v_0}|E_5| + |E_6|(u,v) \D v.
\end{equation}

In (<ref>), we have used the fact that both $\frac{\mu - Q^2/r^2}{1-\mu}\frac{\nu}{r}$ and $\tfs$ vanishes at $\Gamma$.
By (<ref>) and (<ref>), we then deduce that

\begin{equation}\label{Area243}
\int_0^{v_0}G^*(u) \D u \lesssim TV_{(0,v_0)}[\lambda|_{\Gamma}] + A[\tp^\dagger,\tmfa/\lambda] + D^2.
\end{equation}

Parallel to (<ref>), we set

\begin{equation}\label{Area244}
\trs := r\ml \frac{\mu - Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta \tz^\dagger\mr.
\end{equation}

Back to (<ref>), we now obtain

\begin{equation}\label{Area245a}
\mlm \frac{1}{\nu}\frac{\p(\nu,\tp)}{\p(u,v)}\mrm \lesssim D\frac{|\trs|}{r^3} + |\dv \tp|G^*(u) + D\frac{|r^2\tgs|}{r^3} + D^3.
\end{equation}

Integrating on the domain $\mathcal{D}(0,v_0)$, we hence derive

\begin{equation}\label{Area245}
\begin{aligned}
\int_{\mathcal{D}(0,v_0)} \mlm \frac{1}{\nu}\frac{\p(\nu,\tp)}{\p(u,v)}\mrm \D u \D v &\lesssim D \int_{\mathcal{D}(0,v_0)}\frac{|\trs|}{r^3}(u,v) \D u \D v + \int_{\mathcal{D}(0,v_0)}|\dv \tp|(u,v)G^*(u) \D u \D v \\
&+ D \int_{\mathcal{D}(0,v_0)}\frac{|r^2\tgs|}{r^3}(u,v) \D u \D v + D^3.
\end{aligned}
\end{equation}

We then estimate the integrals in the right of (<ref>). First, note that by (<ref>), we have

\begin{equation}\label{Area246}
\begin{aligned}
\int_{\mathcal{D}(0,v_0)}|\dv \tp|(u,v) G^*(u) \D u &\lesssim  X \int_0^{v_0}G^*(u) \D u \lesssim X(TV_{(0,v_0)}[\lambda|_{\Gamma}] + A[\tp^\dagger,\tmfa/\lambda] + D^2).
\end{aligned}
\end{equation}

On the other hand, analogously to the above, one can show that

\begin{equation}\label{Area325}
|\trs|(u,v) \lesssim r^3(u,v) \quad \text{ and } \quad  |\tgs|(u,v) \lesssim r(u,v)
\end{equation}

and thus

\begin{equation}\label{Area326}
\frac{\trs}{r^2}|_{\Gamma} = 0 \quad \text{ and } \quad \tgs|_{\Gamma} = 0.
\end{equation}

By employing (<ref>), we perform integration by parts and obtain

\begin{equation}\label{Area247}
\begin{aligned}
\int_{\mathcal{D}(0,v_0)}\frac{|\trs|}{r^3}(u,v) \D u \D v &= \frac{1}{2}\int^{v_0}_0 \ml \int^{v_0}_u \frac{|\trs|}{-
\lambda}\dv \ml \frac{1}{r^2}\mr(u,v) \D v\mr \D u \\
&= - \frac{1}{2}\int^{v_0}_0 \frac{|\trs|/\lambda}{r^2}(u,v_0) \D u + \frac{1}{2}\int^{v_0}_0 \int^{v_0}_u \dv \mlm \frac{\trs}{\lambda}\mrm \cdot \frac{1}{r^2}(u,v) \D v \D u \\
&\leq \frac{1}{2}\int_{\mathcal{D}(0,v_0)}\frac{1}{r^2}\mlm \dv \ml \frac{\trs}{\lambda}\mr\mrm(u,v) \D u \D v.
\end{aligned}
\end{equation}

Meanwhile, a symmetric argument with (<ref>) yields

\begin{equation}\label{Area248}
\begin{aligned}
\int_{\mathcal{D}(0,v_0)}\frac{|r^2 \tgs|}{r^3}(u,v) \D u \D v
&\leq \frac{1}{2}\int_{\mathcal{D}(0,v_0)}\frac{1}{r^2}\mlm \dv \ml \frac{r^2 \tgs}{\lambda}\mr\mrm(u,v) \D u \D v.
\end{aligned}
\end{equation}

It remains to compute $\dv \ml \frac{\trs}{\lambda}\mr$ and $\dv \ml \frac{r^2 \tgs}{\lambda}\mr$. For the former, analogously to (<ref>), we have

\begin{equation}\label{Area249}
\begin{aligned}
\dv \ml \frac{\trs}{\lambda} \mr =&\; \dv \ml \frac{r}{\lambda}\ml \frac{\mu - Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta \tz^\dagger\mr \mr \\
=&\; \frac{\mu- Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta \tz^\dagger +  r \dv \ml \frac{\mu - Q^2/r^2}{1-\mu}\nu + \frac{4\pi \ttheta \tz^\dagger}{\lambda}\mr \\
=&\; \frac{\mu- Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta \tz^\dagger +  r\nu \dv \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr + \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr r\dv \nu  \\
&+ 4\pi r \ml \frac{\ttheta}{\lambda}\mr \dv \tz^\dagger + 4\pi r \tz^\dagger\dv \ml\frac{\ttheta}{\lambda} \mr.
\end{aligned}
\end{equation}

By (<ref>) and (<ref>), we can write $\dv \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr$ as

\begin{equation}\label{Area250}
\begin{aligned}
\dv \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr =&\; \frac{1-Q^2/r^2}{(1-\mu)^2}\dv \mu + \frac{1}{1-\mu}\dv\ml -\frac{Q^2}{r^2}\mr \\
=&\; \frac{1-Q^2/r^2}{(1-\mu)^2}\ml \frac{4\pi (1-\mu)|\ttheta|^2}{r\lambda} - \frac{\lambda}{r}(\mu - Q^2/r^2)\mr + \frac{1}{1-\mu}\ml 8\pi \e Q \im(\phi^\dagger \dv \tp) + \frac{2Q^2\lambda}{r^3}\mr.
\end{aligned}
\end{equation}

In addition, to treat the last term in (<ref>), we observe that

\begin{equation}\label{Area251}
\begin{aligned}
\dv \ml \frac{\tmfa}{\lambda}\mr = \dv\ml \frac{\dv r \tp + r \dv \tp}{\dv r}\mr = \frac{\ttheta}{r} + \dv \ml \frac{\ttheta}{\lambda}\mr.
\end{aligned}
\end{equation}

In conjunction with (<ref>) and (<ref>), the above two expressions imply that (<ref>) simplifies to

\begin{equation}\label{Area252}
\begin{aligned}
\dv \ml \frac{\trs}{\lambda}\mr =&\;\frac{\mu- Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta \tz^\dagger + \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr^2 \lambda \nu  + r\nu\ml \frac{1-Q^2/r^2}{(1-\mu)^2}\ml \frac{4\pi (1-\mu)|\ttheta|^2}{r\lambda} - \frac{\lambda}{r}(\mu - Q^2/r^2)\mr\mr \\
&+ \frac{r\nu}{1-\mu}\ml 8\pi \e Q \im(\phi^\dagger \dv \tp) + \frac{2Q^2\lambda}{r^3}\mr + 4\pi r \ml \frac{\ttheta}{\lambda}\mr \ml -\frac{\nu \ttheta^\dagger}{r} + E_1^\dagger\mr  + 4\pi r \tz^\dagger\ml \dv\ml\frac{\tmfa}{\lambda}\mr - \frac{\ttheta}{r} \mr\\
=&\; \frac{4\pi \nu(\mu - Q^2/r^2)|\ttheta|^2}{\lambda(1-\mu)} + 4\pi r \tz^\dagger \dv \ml \frac{\tmfa}{\lambda} \mr + E_{13}
\end{aligned}
\end{equation}

\begin{equation}\label{Area253}
E_{13} := \frac{r\nu}{1-\mu}\ml 8\pi \e Q \im(\phi^\dagger \dv \tp) + \frac{2Q^2\lambda}{r^3}\mr + \frac{4\pi r \ttheta E_1^\dagger}{\lambda}
\end{equation}

and $|E_{13}|(u,v) \lesssim D^2r^2(u,v).$ Note that to arrive at the last equality in (<ref>), various cancellations were explored. With the help of (<ref>) and (<ref>), we also have

\begin{equation}\label{Area254}
\begin{aligned}
\tz^\dagger \dv \ml \frac{\tmfa}{\lambda}\mr = r\ml \du \tp^\dagger \dv \ml \frac{\tmfa}{\lambda}\mr\mr &= r\ml \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} + \dv \tp^\dagger \du \ml \frac{\tmfa}{\lambda}\mr \mr\\
&= r\frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} - \frac{\nu(\mu-Q^2/r^2)|\ttheta|^2}{r\lambda(1-\mu)} + \frac{\ttheta^\dagger E_1}{\lambda}.
\end{aligned}
\end{equation}

Plugging (<ref>) into (<ref>), we deduce that

\begin{equation}\label{Area255}
\dv \ml \frac{\trs}{\lambda}\mr = 4\pi r^2 \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} + E_{14} \quad \text{ with } \quad E_{14} = E_{13} + \frac{4\pi r\ttheta^\dagger E_1}{\lambda}
\end{equation}

and $|E_{14}|\lesssim D^2r^2(u,v)$. Together with (<ref>), this implies that

\begin{equation}\label{Area257}
\int_{\mathcal{D}(0,v_0)}\frac{|\trs|}{r^3}(u,v) \D u \D v \lesssim A[\tp^\dagger,\tmfa/\lambda] + D^2.
\end{equation}

Next, we proceed to compute $\dv \ml \frac{r^2 \tgs}{\nu}\mr$.  First, observe that

\begin{equation}\label{Area258}
\dv \ml \frac{\tz^\dagger r}{\lambda}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \mr = \ml \tz^\dagger + r \dv\ml \frac{\tz^\dagger}{\lambda}\mr\mr\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr + \frac{r}{\lambda}\tz^\dagger \dv \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr.
\end{equation}

By (<ref>), we can write

\begin{equation}\label{Area259}
\begin{aligned}
\dv \ml \frac{\tz^\dagger}{\lambda}\mr &= \frac{\lambda \dv \tz^\dagger - \tz^\dagger \dv \lambda}{\lambda^2} = -\frac{\nu}{\lambda r}\ttheta^\dagger + \frac{E_1^\dagger}{\lambda} - \frac{\tz^\dagger \dv \lambda }{\lambda^2}.
\end{aligned}
\end{equation}

Combining with (<ref>), we then simplify (<ref>) to

\begin{equation}\label{Area260}
\begin{aligned}
\dv \ml \frac{\tz^\dagger r}{\lambda}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \mr =&\; \ml \tz^\dagger
- \frac{\nu \ttheta^\dagger}{\lambda} + \frac{rE_1^\dagger}{\lambda} - \frac{r\tz^\dagger \dv \lambda}{\lambda^2}  \mr\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \\
&+ \frac{r^2}{\lambda}\ml  \frac{1}{r^2}\ml \frac{\mu - Q^2/r^2}{1-\mu} \mr\ml \frac{|\tz|^2\lambda}{\nu} - \frac{ |\ttheta|^2\nu}{\lambda}\mr + \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} + \frac{\ttheta^\dagger E_1}{r\lambda} - \frac{\tz^\dagger E_4}{r\nu} \mr.
\end{aligned}
\end{equation}

Observe that by (<ref>), we have

\begin{equation}\label{Area261}
\begin{aligned}
\ml \lambda \tz^\dagger - \nu \ttheta^\dagger - \frac{r\tz^\dagger \dv \lambda}{\lambda} \mr\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu} \mr =&\; \ml \lambda \tz^\dagger - \nu \ttheta^\dagger - \frac{r\tz^\dagger \dv \lambda}{\lambda} \mr\ml \frac{\ttheta}{\lambda} - \frac{\tz}{\nu} + E_2 \mr \\
=&\; -\frac{\nu|\ttheta|^2}{\lambda} + (\tz^\dagger \ttheta + \tz \ttheta^\dagger) - \frac{\lambda |\tz|^2}{\nu} + E_2(\lambda \tz^\dagger - \nu \ttheta^\dagger) \\
&+ \frac{r|\tz|^2 \dv \lambda}{\lambda \nu} - (\dv \lambda) \tz^\dagger \ml \frac{r}{\lambda} \mr \ml \frac{\ttheta}{\lambda} + E_2\mr. \\
\end{aligned}
\end{equation}

On the other hand, from (<ref>), we have

\begin{equation}\label{Area262}
\begin{aligned}
(\dv \lambda) \tz^\dagger = r ((\dv \lambda) \du \tp^\dagger) = r\ml \frac{\p(\tp^\dagger,\lambda)}{\p(u,v)} + (\du\lambda) \dv \tp^\dagger\mr = r\ml \frac{\p(\tp^\dagger,\lambda)}{\p(u,v)} + \frac{\lambda \nu \ttheta^\dagger}{r^2}\frac{\mu - Q^2/r^2}{1-\mu}\mr,
\end{aligned}
\end{equation}

and hence the first term in (<ref>) has the following expression

\begin{equation}\label{Area263}
\begin{aligned}
\ml \lambda \tz^\dagger - \nu \ttheta^\dagger - \frac{r\tz^\dagger \dv \lambda}{\lambda} \mr\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu} \mr
=&\; -\frac{\nu|\ttheta|^2}{\lambda} + (\tz^\dagger \ttheta + \tz \ttheta^\dagger) - \frac{\lambda |\tz|^2}{\nu} + \frac{r|\tz|^2 \dv \lambda}{\lambda \nu} - \frac{\nu|\ttheta|^2}{\lambda}\frac{\mu - Q^2/r^2}{1-\mu}  \\
&\; - \frac{r^2 \ttheta}{\lambda^2}\frac{\p(\tp^\dagger,\lambda)}{\p(u,v)} + E_2(\lambda \tz^\dagger - \nu \ttheta^\dagger) - E_2\ml\frac{r^2}{\lambda} \mr \ml \frac{\p(\tp^\dagger,\lambda)}{\p(u,v)} + \frac{\lambda \nu \ttheta^\dagger}{r^2}\frac{\mu - Q^2/r^2}{1-\mu}\mr. \\
\end{aligned}
\end{equation}

Plugging this back to (<ref>), we have therefore obtained

\begin{equation}\label{Area264}
\begin{aligned}
\dv \ml \frac{\tz^\dagger r}{\lambda}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \mr =&\; \frac{1}{\lambda} \ml -\frac{\nu|\ttheta|^2}{\lambda} + (\tz^\dagger \ttheta + \tz \ttheta^\dagger) - \frac{\lambda |\tz|^2}{\nu} + \frac{r|\tz|^2 \dv \lambda}{\lambda \nu} - \frac{\nu|\ttheta|^2}{\lambda}\frac{\mu - Q^2/r^2}{1-\mu} \mr  - \frac{r^2 \ttheta}{\lambda^3}\frac{\p(\tp^\dagger,\lambda)}{\p(u,v)}\\
&+ \frac{E_2}{\lambda}(\lambda \tz^\dagger - \nu \ttheta^\dagger) - E_2\ml\frac{r^2}{\lambda^2} \mr \ml \frac{\p(\tp^\dagger,\lambda)}{\p(u,v)} + \frac{\lambda \nu \ttheta^\dagger}{r^2}\frac{\mu - Q^2/r^2}{1-\mu}\mr + \frac{rE_1^\dagger}{\lambda}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \\
&+  \frac{1}{\lambda}\ml \frac{\mu - Q^2/r^2}{1-\mu} \mr\ml \frac{|\tz|^2\lambda}{\nu} - \frac{ |\ttheta|^2\nu}{\lambda}\mr + \frac{r^2}{\lambda}\frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} + \frac{r^2}{\lambda}\ml \frac{\ttheta^\dagger E_1}{r\lambda} - \frac{\tz^\dagger E_4}{r\nu} \mr  \\
=&\; r^2 \ml \frac{E_2 - \ttheta/\lambda}{\lambda^2} \mr \frac{\p(\tp^\dagger,\lambda)}{\p(u,v)} + r^2 \ml \frac{1}{\lambda} \mr \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} + R_2 + E_{15},
\end{aligned}
\end{equation}

\begin{equation}\label{Area265}
\begin{aligned}
R_2 :=&\;  \frac{1}{\lambda} \ml -\frac{\nu|\ttheta|^2}{\lambda} + (\tz^\dagger \ttheta + \tz \ttheta^\dagger) - \frac{\lambda |\tz|^2}{\nu} + \frac{r|\tz|^2 \dv \lambda}{\lambda \nu} - \frac{\nu|\ttheta|^2}{\lambda}\frac{\mu - Q^2/r^2}{1-\mu} \mr \\
&+ \frac{1}{\lambda}\ml \frac{\mu - Q^2/r^2}{1-\mu} \mr\ml \frac{|\tz|^2\lambda}{\nu} - \frac{ |\ttheta|^2\nu}{\lambda}\mr, \\
E_{15} := &\; \frac{E_2}{\lambda}(\lambda \tz^\dagger - \nu \ttheta^\dagger) - \frac{E_2\nu \ttheta^\dagger}{\lambda}\frac{\mu - Q^2/r^2}{1-\mu}+ \frac{rE_1^\dagger}{\lambda}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr + \frac{r^2}{\lambda}\ml \frac{\ttheta^\dagger E_1}{r\lambda} - \frac{\tz^\dagger E_4}{r\nu} \mr.
\end{aligned}
\end{equation}

Similar to (<ref>), to arrive at the last line in (<ref>), multiple cancellations have to be explored.
Employing $\im(R_2) = 0$ and $|E_{15}|\lesssim D^2r^2(u,v)$, we arrive at

\begin{equation}\label{Area266a}
\begin{aligned}
\dv \ml \frac{r^2 \tgs}{\lambda}\mr &= 4\pi \im \ml \dv \ml \frac{\tz^\dagger r}{\lambda}\ml \frac{\tmfa}{\lambda}- \frac{\ttmfb}{\nu}\mr\mr\mr\\
&= 4\pi \im\ml r^2 \ml \frac{E_2 - \ttheta/\lambda}{\lambda^2} \mr \frac{\p(\tp^\dagger,\lambda)}{\p(u,v)} + r^2 \ml \frac{1}{\lambda} \mr \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} + E_{15} \mr \\
\end{aligned}
\end{equation}

and thus

\begin{equation}\label{Area266}
\mlm \dv \ml \frac{r^2\tgs}{\lambda }\mr\mrm \lesssim r^2\ml
D \mlm \frac{\p(\tp^\dagger,\lambda)}{\p(u,v)}\mrm + \mlm \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)}\mrm + D^2\mr.
\end{equation}

Plugging these back to (<ref>), we then deduce that

\begin{equation}\label{Area267}
\int_{\mathcal{D}(0,v_0)}\frac{|r^2 \tgs|}{r^3}(u,v) \D u \D v \lesssim D\cdot A[\tp^\dagger,\lambda] + A[\tp^\dagger,\tmfa/\lambda] + D^2.
\end{equation}

By incorporating (<ref>), (<ref>), (<ref>) to (<ref>), we arrive at

\begin{equation}\label{Area268}
\begin{aligned}
A[\nu,\tp^\dagger] &\lesssim  X \cdot TV_{(0,v_0)}[\lambda|_{\Gamma}] + (X + D) \cdot A[\tp^\dagger,\tmfa/\lambda] + D^2 \cdot A[\lambda,\tp^\dagger] + D^2X + D^3.\\
\end{aligned}
\end{equation}

Summarizing the area estimates above, we have proven the following:

* (<ref>):

\begin{equation}\label{Area269}
\sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\tmfa] \lesssim  \; D\ml 1 + \sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\lambda]\mr  + A[\lambda,\tp],
\end{equation}

* (<ref>):

\begin{equation}\label{Area270}
\sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\}}[\ttmfb] \lesssim \; TV_{(0,v_0)}[\tmfa|_\Gamma] + D \ml 1 + TV_{(0,v_0)}[\lambda|_{\Gamma}] + \sup_{v \in (0,v_0)} TV_{(0,v) \times \{v\}}[\nu]\mr  + A[\nu,\tp],
\end{equation}

* (<ref>)

\begin{equation}\label{Area271}
\lesssim D\ml 1 + TV_{(0,v_0)}[\lambda|_{\Gamma}] + \sup_{u\in(0,v_0)}TV_{\{u\}\times(u,v_0)}[\lambda]\mr + A[\lambda,\tp],
\end{equation}

* (<ref>):

\begin{equation}\label{Area272}
\sup_{u \in (0,v_0)} TV_{ \{u\} \times (u,v_0) } [\log \lambda] \lesssim D^2 + DX + A[\tp^\dagger, \ttmfb/\nu],
\end{equation}

* (<ref>):

\begin{equation}\label{Area273}
\sup_{v\in(0,v_0)} TV_{(0,v) \times \{v\}}[\log |\nu|] \lesssim D^2 + DY + TV_{(0,v_0)}[\log\lambda|_{\Gamma}] + A[\tp^\dagger,\tmfa/\lambda],
\end{equation}

* (<ref>):

\begin{equation}\label{Area274}
TV_{(0,1)}[\log(\lambda|_{\Gamma})] \lesssim D^2 + A[\tp^\dagger,\ttmfb/\nu],
\end{equation}

* (<ref>):

\begin{equation}\label{Area275}
A[\tp^\dagger, \tmfa/\lambda] \lesssim DY\ml 1 + \sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\lambda]\mr + D^2 + (D+Y)\cdot A[\lambda,\tp],
\end{equation}

* (<ref>):

\begin{equation}\label{Area276}
\begin{aligned}
A[\tp^\dagger, \ttmfb/\nu] \lesssim&\; X \cdot TV_{(0,v_0)}[\ttmfb|_{\Gamma}] + DX\ml 1 +  TV_{(0,v_0)}[\lambda|_{\Gamma}] + \sup_{v \in (0,v_0)} TV_{(0,v) \times \{v\} }[\nu] \mr \\
&+ D^2 + (D + X)\cdot A[\nu,\tp],
\end{aligned}
\end{equation}

* (<ref>):

\begin{equation}\label{Area277}
A[\lambda,\tp] \lesssim (D+Y)A[\tp^\dagger,\ttmfb/\nu] + D^2\cdot A[\nu,\tp] + D^2Y + D^3,
\end{equation}

* (<ref>):

\begin{equation}\label{Area278}
A[\nu,\tp] \lesssim  X \cdot TV_{(0,v_0)}[\lambda|_{\Gamma}] + (X + D) \cdot A[\tp^\dagger,\tmfa/\lambda] + D^2 \cdot A[\lambda,\tp] + D^2X + D^3.
\end{equation}

Note that we have used $A[\nu,\tp^\dagger] = A[\nu,\tp]$ and $A[\lambda,\tp^\dagger] = A[\lambda,\tp]$ to simplify and to rewrite (<ref>) - (<ref>).

Recall that by our hypothesis, we have that (<ref>) holds. Plugging (1) to (6) into (7) to (10), keeping the dominant terms, and employing $TV_{(0,v_0)}[\ttmfb|_{\Gamma}] = TV_{(0,v_0)}[\tmfa|_{\Gamma}]$, we obtain

\begin{equation}\label{Area279}
\begin{aligned}
A[\tp^\dagger,\tmfa/\lambda] &\lesssim DY\ml 1 + D^2 + DX + A[\tp^\dagger,\ttmfb/\nu]\mr + D^2 + (D+Y)\cdot A[\lambda,\tp] \\
&\lesssim DZ + D^2 + DZ\cdot A[\tp^\dagger,\ttmfb/\nu] + (D+Z)A[\lambda,\tp].
\end{aligned}
\end{equation}

\begin{equation}\label{Area280}
\begin{aligned}
A[\tp^\dagger,\ttmfb/\nu] \lesssim&\; X\ml D\ml 1 + TV_{(0,v_0)}[\lambda|_{\Gamma}] + \sup_{v\in(0,v_0)}TV_{(0,v) \times \{v\}}[\lambda]\mr + A[\lambda,\tp] \mr + DX  \\
&+ DX\ml TV_{(0,v_0)}[\lambda|_{\Gamma}]\mr + DX \ml \sup_{v \in (0,v_0)} TV_{(0,v) \times \{v\} }[\nu] \mr + D^2 + (D + X)\cdot A[\nu,\tp], \\
\lesssim&\; DX \ml TV_{(0,v_0)}[\lambda|_{\Gamma}] + \sup_{v\in(0,v_0)}TV_{(0,v) \times \{v\} }[\lambda] +\sup_{v\in(0,v_0)}TV_{(0,v) \times \{v\} }[\nu] \mr \\
&+ X \cdot A[\lambda,\tp] + (D+X) \cdot A[\nu,\tp] + DX + D^2 \\
\lesssim&\; DX \ml A[\tp^\dagger, \ttmfb/\nu] + A[\tp^\dagger,\tmfa/\lambda] \mr + X \cdot A[\lambda,\tp] + (D+X) \cdot A[\nu,\tp] + DX + D^2.
\end{aligned}
\end{equation}

For $\varepsilon_0 \in (0,1)$ being sufficiently small, inequality (8') implies

\begin{equation}\label{Area281}
\begin{aligned}
\lesssim&\; DZ \cdot A[\tp^\dagger,\tmfa/\lambda] + Z \cdot A[\lambda,\tp] + (D+Z) \cdot A[\nu,\tp] + DZ + D^2.
\end{aligned}
\end{equation}

To further simplify (9') and (10'), we consider the following simplifications to (9) and (10)

\begin{equation}\label{Area282}
A[\lambda,\tp] \lesssim (D+Z)A[\tp^\dagger,\ttmfb/\nu] + D^2\cdot A[\nu,\tp] + D^2Z + D^3
\end{equation}

\begin{equation}\label{Area283}
A[\nu,\tp] \lesssim  Z \cdot A[\tp^\dagger,\ttmfb/\nu] + (D + Z) \cdot A[\tp^\dagger,\tmfa/\lambda] + D^2 \cdot A[\lambda,\tp] + D^2Z + D^3.
\end{equation}

Plugging (<ref>) into (<ref>), we get

\begin{equation}\label{Area284}
A[\lambda,\tp] \lesssim (D+Z)A[\tp^\dagger,\ttmfb/\lambda] + D^2(D+Z)A[\tp^\dagger,\tmfa/\nu] + D^4 \cdot A[\lambda,\tp] + D^2Z + D^3.
\end{equation}

For $\varepsilon_0$ sufficiently small, we thus obtain

\begin{equation}\label{Area285}
A[\lambda,\tp] \lesssim (D+Z)A[\tp^\dagger,\ttmfb/\nu] + (D+Z)A[\tp^\dagger,\tmfa/\lambda] + D^2Z + D^3.
\end{equation}

We then plug (<ref>) to (<ref>) and obtain

\begin{equation}\label{Area286}
A[\nu,\tp] \lesssim (D+Z)A[\tp^\dagger,\ttmfb/\nu] + (D+Z)A[\tp^\dagger,\tmfa/\lambda] + D^2Z + D^3.
\end{equation}

Substituting (9') and (10') into (7') and (8'), we then derive

\begin{equation}\label{Area287}
\begin{aligned}
&\lesssim DZ + D^2 + (D+Z)^2 A[\tp^\dagger,\ttmfb/\nu] + (D+Z)^2A[\tp^\dagger,\tmfa/\lambda]
\end{aligned}
\end{equation}

\begin{equation}\label{Area288}
\begin{aligned}
&\lesssim DZ + D^2 + (D+Z)^2 A[\tp^\dagger,\ttmfb/\nu] + (D+Z)^2A[\tp^\dagger,\tmfa/\lambda].
\end{aligned}
\end{equation}

With a sufficiently small $\varepsilon_0$, (<ref>) and (<ref>) imply

\begin{equation}\label{Area294}
\begin{aligned}
&\lesssim DZ + D^2 + (D+Z)^2 A[\tp^\dagger,\ttmfb/\nu]
\end{aligned}
\end{equation}

\begin{equation}\label{Area295}
\begin{aligned}
&\lesssim DZ + D^2  + (D+Z)^2A[\tp^\dagger,\tmfa/\lambda].
\end{aligned}
\end{equation}

By employing (<ref>) into (<ref>), we thus obtain

\begin{equation}\label{Area289}
\begin{aligned}
&\lesssim DZ + D^2 + (D+Z)^4 A[\tp^\dagger,\tmfa/\lambda].
\end{aligned}
\end{equation}

By shrinking $\varepsilon_0$ if necessary, inequality (<ref>) gives

\begin{equation}\label{Area290}
\begin{aligned}
&\lesssim DZ + D^2.
\end{aligned}
\end{equation}

Together with (<ref>), this implies that

\begin{equation}\label{Area291}
\begin{aligned}
&\lesssim DZ + D^2.
\end{aligned}
\end{equation}

Substituting (<ref>) and (<ref>) back to (<ref>) and (<ref>), we hence get

\begin{equation}\label{Area293}
A[\lambda,\tp] \lesssim DZ^2 + D^2Z+ D^3 \quad \text{ and } \quad
A[\nu,\tp] \lesssim DZ^2 + D^2Z+ D^3.
\end{equation}

Plugging (<ref>), (<ref>), (<ref>) into (4), (5), (6), we thus prove

\begin{equation}\label{Area296}
\begin{aligned}
\sup_{u \in (0,v_0)} TV_{ \{u\} \times (u,v_0) } [\log \lambda] &\lesssim DZ + D^2, \\
\sup_{v\in(0,v_0)} TV_{(0,v) \times \{v\}}[\log |\nu|] &\lesssim DZ + D^2, \\
TV_{(0,1)}[\log(\lambda|_{\Gamma})] &\lesssim DZ + D^2. \\
\end{aligned}
\end{equation}

By applying all the above estimates to (1), (2), (3), we now conclude

\begin{equation}\label{Area297}
\begin{aligned}
\sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\tmfa] \lesssim D, \quad \quad
\sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\}}[\ttmfb] \lesssim D, \quad \text{ and } \quad
TV_{(0,v_0)}[\tmfa|_{\Gamma}]  \lesssim D.
\end{aligned}
\end{equation}

In view of Lemma <ref> and (<ref>), we observe that

\begin{equation}\label{Area298}
\begin{aligned}
X &\lesssim \sup_{u \in (0,v_0)}TV_{\{u\} \times (u,v_0)}\left[ \frac{\tmfa}{\lambda}\right] \\
&\lesssim \sup_{u \in (0,v_0)} TV_{\{u\}\times (u,v_0)}[\lambda] + \sup_{u \in (0,v_0)} TV_{\{u\}\times (u,v_0)}[\tmfa] \lesssim D.
\end{aligned}
\end{equation}

A similar argument yields

\begin{equation}\label{Area299}
\begin{aligned}
Y &\lesssim \sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\} }[\tmfb] + \sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\} }[\nu] \\
&\lesssim \sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\} }[\ttmfb] + \sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\} }[\nu] + D \lesssim D.
\end{aligned}
\end{equation}

Henceforth, by employing (<ref>) and (<ref>), we arrive at

\begin{equation}\label{Area300}
Z \lesssim D.
\end{equation}

Substituting (<ref>) back into (<ref>), (<ref>), (<ref>), (<ref>), we thus verify that (<ref>) and (<ref>) holds.

§ C1 EXTENSION THEOREMS

We now establish the $C^1$ extension theorem in this section. We begin by defining

\begin{equation}\label{alpha'}
\alpha' := \frac{1}{\dv r} \dv \left( \frac{\dv (r\phi)}{\dv r} \right) =\frac{\dv \alpha}{\dv r}.
\end{equation}

Recall that the $C^1$-regularity of the solution to the system is equivalent to the boundedness of the term $|\alpha'|$ defined in (<ref>). Furthermore, for the charged scalar field system, the Doppler exponent $\gamma$ is defined in (<ref>) and given by

$$\gamma(u,v) := \int_{0}^u \frac{-\du r}{r}\frac{\mu - Q^2/r^2}{1-\mu}(u',v) \D u'.$$

Our results for this section are summarized in the two theorems below.
($C^1$ Extension Theorem with Doppler Exponent $\gamma$.) Suppose that we have $C^1$ initial data $\alpha$ along $C_0^+$. Let $v_* > 0$ and assume for every $\hat{v} \in (0,v_*)$, a $C^1$ solution exists on $\mathcal{D}(0,\hat{v})$. If we have

\begin{equation}\label{SET4}
\sup_{\mathcal{D}(0,v_*)} \gamma < + \infty,
\end{equation}

then a $C^1$ solution can be extended to $\mathcal{D}(0,\tilde{v})$ with $\tilde{v} > v_*$.

As mentioned in the introduction, analogous to [13], the above theorem implies that the inability to extend the solution past some $v_* > 0$ necessitates the “finite-time" blowup of the Doppler exponent, corresponding to an integration of some physical quantity.

A related but independent $C^1$ extension theorem with mass ratio for the charged system, motivated by  [19], is given as follows.

($C^1$ Extension Theorem with Mass Ratio $\mu$.) Suppose that we have $C^1$ initial data $\alpha$ along $C_0^+$. Let $v_* > 0$ and assume for every $\hat v \in (0,v_*)$, a $C^1$ solution exists on $\mathcal{D}(0,\hat v)$. Then, there exists a constant $\varepsilon' > 0$, such that if we have

\begin{equation}\label{FETsup}
\lim_{u \rightarrow v_*} \sup_{\mathcal{D}(u,v_*)} \mu < \varepsilon',
\end{equation}

then a $C^1$ solution can be extended to $\mathcal{D}(0,\tilde{v})$ for some $\tilde{v} > v_*$.

In this section, we will first prove Theorem <ref>, followed by Theorem <ref>. For both theorems, we will obtain the corresponding extension theorems via two steps. In the first step, under the assumptions in Theorem <ref> or Theorem <ref>, we show that the solution can be extended as a $C^1$ solution up to $\mathcal{D}(0,v_*).$ In the second step, we then extend such a $C^1$ solution to $\mathcal{D}(0,\tilde{v})$ with some $\tilde{v} > v_*$, which completes the proofs.

[gray!30] (0,0) to (1.75,1.75) to (0,3.5) to (0,0);
[gray!60] (0,2) to (1,3) to (0,4) to (0,2);
[->] (0,0) node[anchor=north]$\mathcal{O}$ –  (0,6) node[anchor = east]$\Gamma$;
[->] (0,0) – (5,5) node[anchor = west]$C_0^+$;
(0,3.5) – (1.75,1.75);
(0,4) – (2,2);
(0,5) – (2.5,2.5);
[] at (0.75,1.75) $\mathcal{D}(0,\hat v)$;
[] at (1.9,1.6) $\hat v$;
[] at (2.15,1.85) $v_*$;
[] at (2.65,2.35) $\tilde{v}$;
[anchor = east] at (0,2) $(u,u)$;
at (0,2)[circle,fill,inner sep=1.5pt];
at (0,4)[circle,fill,inner sep=1.5pt];
[anchor = east] at (0,4) $(v_*,v_*)$;
[<-](0.4,3.2) – (2.4,5.2);
[] at (2.7,5.5) $\mathcal{D}(u,v_*)$;

Penrose diagram representing the relative positions of $\hat v, v_*, \tilde{v}$ and the regions $\mathcal{D}(0,\hat{v})$ and $\mathcal{D}(u,v_*)$.

Before we begin with the proof, we list several computations that are commonly used in both proofs.

For any $v_* > 0$ and $u_1 \in (0,v_*)$, assume that a $C^1$ solution exists in $\mathcal{D}(0,v_*)\setminus \mathcal{D}(u_1,v_*)$. For each  $(u,v) \in \mathcal{D}(0,v_*)\setminus \mathcal{D}(u_1,v_*)$, we have

\begin{equation}\label{FET14}
\begin{aligned}
\du \ml \alpha' e^{\int^u_{u_0} \ii \e A_u(u',v) \D u'} \mr &= e^{\int^u_{u_0} \ii \e A_u(u',v) \D u'} \sum_{i=1}^{11}K_i(u,v)
\end{aligned}
\end{equation}

\begin{equation}\label{K}
\begin{aligned}
K_1(u,v) &= -\frac{8 \pi \e Q \ml \im(\phi^\dagger \dv \phi) \mr (\dv \phi)(\du r)}{(\dv r)^2(1-\mu)}, && K_2(u,v)= \frac{4 \pi Q^2|\dv \phi|^2(\dv \phi)(\du r) }{r(\dv r)^3 (1- \mu)}, \\
K_3(u,v) &= -\frac{4 \pi r|\dv \phi|^2(\dv \phi)(\du r) }{(\dv r)^3 (1- \mu)},&& K_4(u,v)= - \frac{2\du r \mu}{r(1-\mu)}\alpha', \\
K_5(u,v) &= \frac{2Q^2 \du r}{r^3(1-\mu)}\alpha', && K_6(u,v)= \frac{5 Q^2\du r \dv \phi}{r^3 \dv r(1-\mu)}, \\
K_7(u,v) &= -\frac{3\du r \dv \phi \mu}{r \dv r(1-\mu)}, && K_8(u,v) = -\frac{4 \pi \ii \e |\dv \phi|^2 Q\phi \du r}{(1-\mu)(\dv r)^2}, \\
K_9(u,v) &= \frac{4 \ii \pi \e^2 \phi\im(\phi^\dagger \dv \phi) r\du r }{ \dv r (1-\mu)}, && K_{10}(u,v) = -\frac{3 \ii\e  Q \du r \dv \phi}{r(1-\mu)(\dv r)}, \\
K_{11}(u,v) &= -\frac{\ii \e Q\phi \du r}{r^2 (1-\mu)}. && &&&\\
\end{aligned}
\end{equation}

Furthermore, if $\gamma(u,v) \leq G$ and $\frac{1}{1-\mu(u,v)} \leq H$ uniformly in $(u,v) \in \mathcal{D}(0,v_*)\setminus \mathcal{D}(u_1,v_*)$ for some $G,H > 0$, then for each of these $(u,v)$ and any $\chi,\iota \in (0,1)$, we have the following estimates:

\begin{equation}\label{C1Lemma2}
\begin{aligned}
|Q|(u,v) &\leq  C_6'(v_*) \mu^{\frac{1}{2}}(u,v)r(u,v), \\
|Q\phi|(u,v) &\leq C_7''(v_*)\mu^{\oh}(u,v)r^{\frac{3}{2}-\frac{\chi}{2}}(u,v) \leq C_7'(v_*)\mu^{\oh}(u,v)r(u,v), \\
|\phi|^2(u,v) &\leq C_8'(v_*)r^{-\frac{3}{2}}(u,v), \\
\end{aligned}
\end{equation}

\begin{equation}\label{SET22}
C_6'(v_*;\chi) := C_6(v_*;\chi)r^{\frac{1}{2}-\frac{\chi}{2}}(0,v_*),
\end{equation}

\begin{equation}\label{SET24}
C_7'(v_*;\chi) := C_7''(v_*;\chi)r^{\frac{1}{2}-\frac{\chi}{2}}(0,v_*),  \quad C_7''(v_*;\chi):= C_7(v_*;\chi)\frac{e^{\oh G}}{(1-\mu_*)^\oh},
\end{equation}

\begin{equation}\label{SET26}
C_8'(v_*;\iota) := \frac{H}{2\pi}r^{\frac{3}{2}}(0,v_*) + 2C_8^2\ml \oh \mr,
\end{equation}

\begin{equation}\label{FET130}
C_6(v_*;\chi) := \frac{2 \pi \e}{1 - \mu^*}\ml \frac{\overline{P}(0,v_*)}{\sqrt{4\pi}} + \frac{1}{\sqrt{16\pi^2 \chi \e^2}} \mr r^{\frac{\chi}{2}}(0,v_*) v_*^{\oh},
\end{equation}

\begin{equation}\label{FET243}
C_7(v_*;\chi) = \ml 8 \pi \e^2 C_{3}(0,v_*;\chi,1,\chi) + \frac{C_6^2}{2(2-\chi)} \mr^{\frac{1}{2}},
\end{equation}

\begin{equation}\label{FET300}
C_{8}(\iota):= \overline{P}(0,v_*)r(0,v_*)^{\oh + \frac{\iota}{2}} +  \frac{v_*^\oh r(0,v_*)^{\frac{\iota}{2}}}{(4\pi e^2 \iota (1-\iota))^{\frac{1}{2}}} \ml \frac{\bar{B}(0,v_*)}{(1-\overline{\mu_*})(0,v_*)}\mr^{\frac{1}{2}}.
\end{equation}

On the other hand, if $\mu(u,v) \leq \varepsilon$ for some $\varepsilon < 1$ for each $(u,v) \in \mathcal{D}(0,v_*)\setminus \mathcal{D}(u_1,v_*)$, we can deduce

\begin{equation}\label{C1Lemma1}
\begin{aligned}
|Q|(u,v) &\leq C_6 \varepsilon^{\frac{1}{2}}r^{\frac{3}{2} - \frac{\chi}{2}}(u,v), \\
|Q\phi|(u,v) &\leq C_7 \frac{\varepsilon^\oh}{(1-\varepsilon)^\oh}r^{\frac{3}{2}-\frac{\chi}{2}}(u,v), \\
|\phi|^2 &\leq \frac{1}{2\pi}\frac{\varepsilon}{1-\varepsilon} + 2C_8^2r^{-1-\iota}(u,v).
\end{aligned}
\end{equation}

Note that $|Q|, |Q\phi|, |\phi|^2$ controlled in Lemma <ref> represent additional terms unique to our charged system. The methodologies employed to derive bounds for these terms are therefore new.

Proof of Lemma <ref>.  Starting from the definition of $\alpha'$ in (<ref>), we compute

\begin{equation}\label{FET2}
\begin{aligned}
\du \alpha' =  \du \ml \frac{1}{\dv r} \dv \ml \frac{\dv (r\phi)}{\dv r}\mr\mr &= -\frac{\du\dv r }{(\dv r)^2}\dv \ml \frac{\dv (r \phi)}{\dv r}\mr + \frac{1}{\dv r}\du \dv \ml \frac{\dv (r\phi)}{\dv r}\mr \\
&= -\frac{\du\dv r }{\dv r}\alpha' + \frac{1}{\dv r}\du \dv \ml \frac{\dv (r\phi)}{\dv r}\mr.
\end{aligned}
\end{equation}

Observe that by (<ref>) we have

\begin{equation}\label{FET3}
-\frac{\du\dv r }{\dv r}\alpha' = -\frac{\du r}{r}\ml \frac{\mu - Q^2/r^2}{1-\mu}\mr \alpha'.
\end{equation}

Employing (<ref>) and (<ref>), we deduce that

\begin{equation}\label{FET4}
\begin{aligned}
\du \ml \frac{\dv (r\phi)}{\dv r}\mr = \du \ml \phi + \frac{r\dv \phi}{\dv r} \mr &= \du \phi + \frac{\dv r \du (r \dv \phi) - (r \dv \phi)\du\dv r }{(\dv r)^2} \\
&= \frac{\du \phi \dv r + \du r \dv \phi + r \du \dv \phi}{\dv r} - \frac{r \dv \phi}{\dv r} \ml \frac{\du \dv r}{\dv r} \mr \\
&= -\frac{\ii \e \Psi(A)}{\dv r} - \frac{ \dv \phi \du r}{ \dv r} \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr
\end{aligned}
\end{equation}

and this implies

\begin{equation}\label{FET5}
\begin{aligned}
\frac{1}{\dv r}\du \dv \ml \frac{\dv (r\phi)}{\dv r}\mr &= - \frac{1}{\dv r}\dv \ml \frac{\ii \e \Psi(A)}{\dv r}\mr - \frac{1}{\dv r} \dv \ml \frac{ \dv \phi \du r}{ \dv r} \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr \mr.
\end{aligned}
\end{equation}

Plugging (<ref>) and (<ref>) into (<ref>), we thus obtain

\begin{equation}\label{FET6}
\begin{aligned}
\du \alpha' &= -\frac{\du r}{r}\ml \frac{\mu - Q^2/r^2}{1-\mu}\mr \alpha' - \frac{1}{\dv r}\dv \ml \frac{\ii \e \Psi(A)}{\dv r}\mr - \frac{1}{\dv r} \dv \ml \frac{ \dv \phi \du r}{ \dv r} \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr \mr \\
&=: J_1(u,v) + J_2(u,v) + J_3(u,v).
\end{aligned}
\end{equation}

For $J_3$, we compute the derivative and get

\begin{equation}\label{FET7}
\begin{aligned}
J_3(u,v) &=  - \frac{1}{\dv r} \dv \ml \frac{ \dv \phi \du r}{ \dv r} \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr \mr \\
&= - \frac{\dv \phi \du r}{(\dv r)^2}\dv \ml \frac{\mu - Q^2/r^2}{1-\mu} \mr - \frac{\mu - Q^2/r^2}{(1-\mu)(\dv r)} \dv \ml \frac{\du r}{r}\frac{r \dv \phi }{\dv r}\mr.
\end{aligned}
\end{equation}

For the term $- \dv \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr$, we have the expression

\begin{equation}\label{FET8.1}
\begin{aligned}
- \dv &\ml \frac{\mu - Q^2/r^2}{1-\mu}\mr = \; \dv \ml 1 + \frac{Q^2/r^2 - 1}{1-\mu}\mr = \; \frac{(1-\mu)\dv (Q^2/r^2) + (Q^2/r^2 -1)\dv \mu}{(1-\mu)^2} \\
=&  -\frac{8 \pi \e Q \im(\phi^\dagger \dv \phi)}{1-\mu}- \frac{2 \dv r Q^2}{r^3(1-\mu)} + \frac{4 \pi r|\dv \phi|^2 (Q^2/r^2-1)}{\dv r (1- \mu)} - \frac{\dv r}{r}\frac{\ml \mu - Q^2/r^2\mr(Q^2/r^2 - 1)}{(1-\mu)^2}.
\end{aligned}
\end{equation}

In the last equality above, we have applied (<ref>) and (<ref>).

For the term $\dv \ml \frac{\du r}{r}\frac{r \dv \phi}{\dv r}\mr$, with the help of (<ref>) and (<ref>), we obtain

\begin{equation}\label{FET8.2}
\begin{aligned}
\dv \ml \frac{\du r}{r}\frac{r \dv \phi }{\dv r}\mr &= \frac{\du r}{r}\dv \ml  \frac{\dv(r\phi)}{\dv r} - \phi\mr + \ml \frac{r \dv \phi}{\dv r} \mr \dv \ml \frac{\du r}{r} \mr \\
&= \frac{\du r \dv r}{r}\alpha' - \frac{\du r \dv\phi}{r} + \ml \frac{r \dv \phi}{\dv r} \mr\ml \frac{r \du \dv r - \dv r \du r}{r^2} \mr \\
&= \frac{\du r \dv r}{r}\alpha' - \frac{\du r \dv\phi}{r} + \frac{\du r \dv \phi}{r}\ml \frac{\mu - Q^2/r^2}{1-\mu} - 1\mr \\
&= \frac{\du r \dv r}{r}\alpha' - \frac{2\du r \dv\phi}{r}  + \frac{\du r \dv \phi}{r}\ml \frac{\mu - Q^2/r^2}{1-\mu}\mr.\\
\end{aligned}
\end{equation}

Substituting both (<ref>) and (<ref>) into (<ref>), $J_3$ can be expanded as

\begin{equation}\label{FET9}
\begin{aligned}
J_3(u,v) =&\; - \frac{\dv \phi \du r}{(\dv r)^2}\dv \ml \frac{\mu - Q^2/r^2}{1-\mu} \mr - \frac{\mu - Q^2/r^2}{(1-\mu)(\dv r)} \dv \ml \frac{\du r}{r}\frac{r \dv \phi }{\dv r}\mr \\
=& \; -\frac{8 \pi \e Q (\im(\phi^\dagger \dv \phi))(\dv \phi)(\du r)}{(\dv r)^2(1-\mu)} - \frac{2 Q^2 (\dv \phi)(\du r)}{r^3 (\dv r)(1-\mu)} + \frac{4 \pi r (Q^2/r^2-1)|\dv \phi|^2(\dv \phi)(\du r) }{(\dv r)^3 (1- \mu)} \\
& \; - \frac{(\dv \phi) (\du r)}{r(\dv r)}\frac{\ml \mu - Q^2/r^2\mr(Q^2/r^2 - 1)}{(1-\mu)^2} -\frac{\du r}{r}\ml \frac{\mu - Q^2/r^2}{1-\mu}\mr \alpha'  \\
&\; + \frac{2 \du r \dv \phi}{r (\dv r)}\ml \frac{\mu - Q^2/r^2}{1-\mu} \mr - \frac{\du r \dv \phi}{r \dv \phi}\ml \frac{\mu - Q^2/r^2}{1-\mu}\mr^2 \\
:= & \sum_{i=1}^7 J_{3,i}(u.v).
\end{aligned}
\end{equation}

Observing that $J_{3,5} \equiv J_1$ and $J_{3,2} + J_{3,4} + J_{3,6} + J_{3,7}$ forms a simpler expression, we can rewrite $J_1 + J_3$ as

\begin{equation}\label{FET10}
\begin{aligned}
J_{1} + J_3 =&\; J_{3,1} + J_{3,3} + 2 J_{3,5} + (J_{3,2} + J_{3,4} + J_{3,6} + J_{3,7}) \\
=&\; -\frac{8 \pi \e Q (\im(\phi^\dagger \dv \phi))(\dv \phi)(\du r)}{(\dv r)^2(1-\mu)}  + \frac{4 \pi r (Q^2/r^2-1)|\dv \phi|^2(\dv \phi)(\du r) }{(\dv r)^3 (1- \mu)} - \frac{2\du r}{r}\ml \frac{\mu - Q^2/r^2}{1-\mu}\mr \alpha' \\
&\; + \frac{\du r \dv \phi}{r \dv r}\ml \frac{5Q^2/r^2 -
3 \mu}{1-\mu}\mr.
\end{aligned}
\end{equation}

With the help of (<ref>), (<ref>), (<ref>), we express the terms in $J_2$ as follows:

\begin{equation}\label{FET11}
\begin{aligned}
J_2(u,v) &=  -\frac{\ii \e}{\dv r}  \dv \ml \frac{A_u \dv (r\phi) - \frac{\Omega^2 Q \phi}{4r}}{\dv r}\mr \\
&= -\frac{\ii \e}{\dv r}  \ml \dv A_u \frac{\dv (r\phi)}{\dv r} + A_u \dv \ml \frac{\dv(r\phi)}{\dv r}\mr - \dv \ml \frac{Q\phi(-\du r)}{r(1-\mu)}\mr\mr \\
&= -\ii \e \ml A_u \alpha' + \dv A_u \frac{\dv (r\phi)}{(\dv r)^2} - \frac{1}{\dv r}\dv \ml \frac{-\du r}{1-\mu}\frac{Q\phi}{r}\mr\mr.
\end{aligned}
\end{equation}

Together with (<ref>), (<ref>), (<ref>),  we arrive at

\begin{equation}\label{FET12}
\begin{aligned}\frac{J_2(u,v)}{-\ii \e} =& \; A_u \alpha' + \ml \frac{2 Q \du r \dv r}{r^2 (1-\mu)}\mr \frac{r \dv \phi + \phi \dv r}{(\dv r)^2} - \frac{1}{\dv r}\ml \frac{Q\phi}{r}\dv \ml \frac{-\du r}{1-\mu}\mr + \frac{-\du r}{1-\mu}\dv \ml \frac{Q\phi}{r}\mr\mr \\
=&\; A_u \alpha' + \frac{2 Q \du r \dv \phi}{r(1-\mu)(\dv r)} + \frac{2Q\phi \du r}{r^2 (1-\mu)} + \frac{1}{\dv r}\frac{Q\phi}{r}\ml \frac{1-\mu}{-\du r}\mr^{-2} \dv \ml \frac{1-\mu}{-\du r}\mr \\
&\; + \frac{\du r}{\dv r(1-\mu)}\ml \frac{\phi \dv Q}{r} + \frac{Q \dv \phi}{r} - \frac{Q\phi}{r^2} \dv r\mr \\
=&\; A_u \alpha' + \frac{2 Q \du r \dv \phi}{r(1-\mu)(\dv r)} + \frac{2Q\phi \du r}{r^2 (1-\mu)} + \frac{4 \pi |\dv \phi|^2 Q\phi \du r}{(1-\mu)(\dv r)^2} \\
&\; - \frac{4  \pi \e \phi\im(\phi^\dagger \dv \phi) r\du r }{ \dv r (1-\mu)} + \frac{Q \du r \dv \phi}{r(1-\mu)(\dv r)} - \frac{Q \phi \du r}{r^2(1-\mu)} \\
=&\; A_u \alpha' + \frac{3 Q \du r \dv \phi}{r(1-\mu)(\dv r)} + \frac{Q\phi \du r}{r^2 (1-\mu)} + \frac{4 \pi |\dv \phi|^2 Q\phi \du r}{(1-\mu)(\dv r)^2} - \frac{4  \pi \e \phi\im(\phi^\dagger \dv \phi) r\du r }{ \dv r (1-\mu)}.
\end{aligned}
\end{equation}

Combining (<ref>) and (<ref>), we hence obtain

\begin{equation}\label{FET13}
\begin{aligned}
\du \alpha' =& \; J_1 + J_3 + J_2 \\
=& \; -\frac{8 \pi \e Q (\im(\phi^\dagger \dv \phi))(\dv \phi)(\du r)}{(\dv r)^2(1-\mu)}  + \frac{4 \pi r (Q^2/r^2-1)|\dv \phi|^2(\dv \phi)(\du r) }{(\dv r)^3 (1- \mu)} \\
&\;- \frac{2\du r}{r}\ml \frac{\mu - Q^2/r^2}{1-\mu}\mr \alpha' + \frac{\du r \dv \phi}{r \dv r}\ml \frac{5Q^2/r^2 -
3 \mu}{1-\mu}\mr \\
&\;  -\ii \e \ml A_u \alpha' + \frac{3 Q \du r \dv \phi}{r(1-\mu)(\dv r)} + \frac{Q\phi \du r}{r^2 (1-\mu)} + \frac{4 \pi |\dv \phi|^2 Q\phi \du r}{(1-\mu)(\dv r)^2} - \frac{4  \pi \e \phi\im(\phi^\dagger \dv \phi) r\du r }{ \dv r (1-\mu)} \mr. \\
\end{aligned}
\end{equation}

Treating $\ii \e A_u$ as a purely imaginary integrating factor, we thus arrive at (<ref>).

We now proceed to obtain estimates for $|Q|, |Q\phi|,$ and $|\phi|^2$. We first prove (<ref>) under the assumption $\mu \leq \varepsilon < 1$ for all $(u,v) \in \mathcal{D}(0,v_*) \setminus \mathcal{D}(u_1,v_*)$.

For $|Q|$, from (<ref>) of Proposition <ref>, by setting $\overline{u} = 0$ and recalling the definitions of $\bar{B}, \overline{\mu_*}, \overline{P}$ in (<ref>) and (<ref>), we have that the $C_1(0,v;\chi)$ in (<ref>) obeys the following upper bound

\begin{equation}\label{FET29}
\begin{aligned}
C_1(0,v;\chi) &\leq \frac{4 \pi \e \ml  \sup_{v' \in [0,v*]} \dv r(0,v') \mr^{\oh}}{(1 - \sup_{v' \in [0,v_*]}\mu(0,v'))^{\frac{1}{2}}}\ml \frac{\overline{P}(0,v_*)}{\sqrt{4\pi}} + \frac{1}{\sqrt{16\pi^2 \chi \e^2}} \mr \\
&\leq \frac{2 \pi \e}{1 - \mu^*}\ml \frac{\overline{P}(0,v_*)}{\sqrt{4\pi}} + \frac{1}{\sqrt{16\pi^2 \chi \e^2}} \mr < + \infty.
\end{aligned}
\end{equation}

Here we use (<ref>) and (<ref>) to bound $\sup_{v' \in [0,v_*]} \dv r(0,v')$ and $\sup_{v' \in [0,v_*]}\mu(0,v')^{\frac{1}{2}}$ respectively. We also appeal to the fact that $\overline{P}(0,v_*) = \sup_{v'\in [0,v_*]} |\phi| < + \infty$ since $\phi \in C^1$ along $C_0^+$ and $\phi|_{\Gamma} < +\infty$ from (<ref>). Plugging these back into (<ref>), we thus obtain

\begin{equation}\label{FET28}
\begin{aligned}
|Q|(u,v) &\leq \frac{2 \pi \e}{1 - \mu^*}\ml \frac{\overline{P}(0,v_*)}{\sqrt{4\pi}} + \frac{1}{\sqrt{16\pi^2 \chi \e^2}} \mr r^{\frac{\chi}{2}}(0,v_*) v_*^{\oh}\varepsilon^{\frac{1}{2}} r^{\frac{3}{2} - \frac{\chi}{2}}(u,v) = C_6 \varepsilon^{\frac{1}{2}}r^{\frac{3}{2} - \frac{\chi}{2}}(u,v)
\end{aligned}
\end{equation}

with $C_6$ defined in (<ref>). Note that the constant $C_6$ defined above depends only on $\chi$ and $v_*$. In addition, for brevity, since $\chi$ and $v_*$ were fixed from the beginning, we can then drop the dependency of $C_6$ and subsequent constants on $v_*$ and $\chi$, if there is no danger of confusion.

For $|Q\phi|$, following a similar strategy as in the proof of Proposition <ref> but with the assumption that $\mu$ is small, we can obtain a better estimate as follows. Since $Q|_\Gamma = 0$ and $|\phi||_{\Gamma} < + \infty$, by (<ref>), we have

\begin{equation}\label{FET64}
\begin{aligned}
Q\phi(u,v) &= \int^v_u (\dv Q) \phi + Q \dv \phi (u,v') \mathrm{d}v',
\end{aligned}
\end{equation}

$$|Q\phi|(u,v) \leq \int^v_u  (4 \pi \e r^2 |\phi|^2 + |Q|) |\dv \phi| (u,v') \mathrm{d}v'.$$

With the help of (<ref>), this implies that

\begin{equation}\label{FET140}
\begin{aligned}
|Q\phi|(u,v) &\leq \ml \int^v_u \frac{\dv r}{2\pi r^2 (1-\mu)}(4 \pi \e r^2 |\phi|^2 + |Q|)^2(u,v')\D v'\mr^{\frac{1}{2}} \ml \int^v_u \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r} (u,v') \D v'\mr^{\frac{1}{2}} \\
&\leq \ml \int^v_u \frac{\dv r}{\pi r^2 (1-\mu)}(16 \pi^2 \e^2 r^4 |\phi|^4 + |Q|^2)(u,v')\D v'\mr^{\frac{1}{2}} \frac{1}{\sqrt{2}}\mu(u,v)^{\frac{1}{2}}r(u,v)^{\frac{1}{2}} \\
&\leq \ml 16 \pi \e^2 \int^v_u \frac{\dv r}{(1-\mu)}r^2 |\phi|^4 (u,v')\D v' + \frac{1}{\pi} \int^v_u \frac{\dv r}{r^2 (1-\mu)}|Q|^2(u,v')\D v'\mr^{\frac{1}{2}} \frac{1}{\sqrt{2}}\varepsilon^{\frac{1}{2}}r(u,v)^{\frac{1}{2}}.
\end{aligned}
\end{equation}

Our strategy to estimate the first integral remains unchanged as in Proposition <ref>. Hence, by setting $\overline{u} = 0$, $\xi_1 = 1$ and $\xi_2 = \chi$ (this is valid since we did not impose $\xi_2 \in (0,1-\chi)$ for this estimate) in (<ref>), we derive the below bound

\begin{equation}\label{FET241}
16 \pi \e^2 \int^v_u \frac{\dv r}{(1-\mu)}r^2 |\phi|^4 (u,v')\D v' \leq 16 \pi \e^2 C_{3}(0,v_*;\chi,1,\chi) r^{2-\chi}(u,v).
\end{equation}

On the other hand, the second integral can be estimated with the help of (<ref>) as follows

\begin{equation}\label{FET242}
\begin{aligned}
\int^v_u \frac{\dv r}{1-\mu} |Q|^2 r^{-2}(u,v') \D v' \leq \int^v_u \frac{\dv r}{1-\varepsilon} C_6^2 \varepsilon r^{1-\chi}(u,v') \D v' = \frac{C_6^2 \varepsilon}{(2-\chi)(1-\varepsilon)}r^{2-\chi}(u,v).
\end{aligned}
\end{equation}

Summarizing (<ref>) and (<ref>), we obtain

\begin{equation}\label{FET69}
\begin{aligned}
|Q\phi|(u,v) &\leq C_7(v_*) \frac{\varepsilon^{\frac{1}{2}}}{(1-\varepsilon)^{\frac{1}{2}}}r^{\frac
{3}{2}-\frac{\chi}{2}} (u,v) \end{aligned}
\end{equation}

with $C_7$ defined in (<ref>).

For $|\phi|^2$, analogous to the proof of Proposition <ref> and (<ref>), we instead consider $\dv(r\phi)$ in an attempt to obtain pointwise estimates for $|\phi|^2$. With $r\phi|_{\Gamma} = 0$, we have

\begin{equation}\label{FET59}
|r\phi|(u,v) \leq \int^v_u r |\dv \phi|(u,v')\D v' + \int^v_u |\phi| \dv r(u,v') \D v'.
\end{equation}

The first term in (<ref>) can be estimated via

\begin{equation}\label{FET70}
\begin{aligned}
\int^v_u r|\dv \phi|(u,v') \D v' &\leq \frac{1}{\sqrt{2\pi}}\ml \int^v_u\frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r}(u,v')\D v'\mr^{\frac{1}{2}}\ml \int^v_u \frac{\dv r}{1-\mu}(u,v')\D v'\mr^{\frac{1}{2}} \\
&\leq \frac{1}{\sqrt{4\pi}}\frac{\varepsilon^{\frac{1}{2}}}{(1-\varepsilon)^{\frac{1}{2}}}r(u,v).
\end{aligned}
\end{equation}

For the second term in (<ref>), by Proposition <ref>, Lemma <ref> and inequality (<ref>), we can exploit its structure and obtain

\begin{equation}\label{FET63}
\begin{aligned}
&\int^v_u|\phi| \dv r(u,v') \D v' \\
\leq &\; \int^v_u (\overline{P}(0,v_*)) \dv r(u,v') \D v' + \int^v_u \frac{1}{(4\pi)^{\frac{1}{2}}} \ml \log \ml\frac{\frac{1-\mu}{\dv r}(u,v')}{\frac{1-\mu}{\dv r}(0,v')} \mr \mr^{\frac{1}{2}}\ml \log \ml \frac{r(0,v')}{r(u,v')} \mr \mr^{\frac{1}{2}}(\dv r)(u,v') \D v' \\
\leq &\;  \overline{P}(0,v_*)r(u,v) + \frac{1}{(4\pi e)^{\frac{1}{2}}} \ml \frac{\bar{B}(0,v_*)}{(1-\overline{\mu_*})(0,v_*)}\mr^{\frac{1}{2}}\int^v_u   (\dv r)^{\frac{1}{2}}(u,v') (1-\mu)^{\frac{1}{2}}(u,v')\ml \log \ml \frac{r(0,v')}{r(u,v')} \mr \mr^{\frac{1}{2}}\D v' \\
\leq &\;  \overline{P}(0,v_*)r(u,v) + \frac{v_*^\oh}{(4\pi e)^{\frac{1}{2}}} \ml \frac{\bar{B}(0,v_*)}{(1-\overline{\mu_*})(0,v_*)}\mr^{\frac{1}{2}} \ml \int^v_u   (\dv r)(u,v') \log \ml \frac{r(0,v')}{r(u,v')} \mr \D v' \mr^{\frac{1}{2}} \\
\leq &\;  \overline{P}(0,v_*) r(u,v) + \frac{v_*^\oh r(0,v_*)^{\frac{\iota}{2}}}{(4\pi e \iota (1-\iota))^{\frac{1}{2}}} \ml \frac{\bar{B}(0,v_*)}{(1-\overline{\mu_*})(0,v_*)}\mr^{\frac{1}{2}} r(u,v)^{\frac{1}{2} -\frac{\iota}{2}} \\
\leq &\; C_8(\iota) r^{\frac{1}{2} - \frac{\iota}{2}}.
\end{aligned}
\end{equation}

In (<ref>) above, the constant $C_8$ is defined in (<ref>) and depends on a parameter $\iota \in (0,1)$.
Combining the two estimates above and using the fact that $(a+b)^2 \leq 2(a^2+b^2)$ for all $a,b \in \mathbb{R}$, we derive

$$r^2|\phi|^2(u,v) \leq \frac{1}{2\pi}\frac{\varepsilon}{1-\varepsilon}r^2(u,v) + 2C_{8}^2(\iota) r^{1-\iota}(u,v),$$

and hence

\begin{equation}\label{FET71}
\begin{aligned}
|\phi|^2(u,v) &\leq \frac{1}{2\pi}\frac{\varepsilon}{1-\varepsilon} + 2C_{8}^2(\iota) r^{-1-\iota}(u,v). \\
\end{aligned}
\end{equation}

We continue to estimate these terms under the assumption that $\gamma(u,v) \leq G$ and $\frac{1}{1-\mu(u,v)} \leq H$ uniformly for $(u,v) \in \mathcal{D}(0,v_*) \setminus \mathcal{D}(u_1,v_*)$.

For $|Q|$, using a similar argument as above with a minor exception of not estimating $\mu(u,v)$ by $\varepsilon$, analogous to (<ref>), we obtain the estimate for $|Q|$ in (<ref>) with $C_6'(v_*;\chi)$ defined in (<ref>).

For $|Q\phi|$, by our assumption $\frac{1}{1-\mu} \leq H$, we have the estimate for $|Q\phi|$ in (<ref>) with $C_7'(v_*;\chi)$ and $C_7''(v_*;\chi)$ defined in (<ref>).

For $|\phi|^2$, employing a similar argument, by (<ref>), $\mu < 1$, and (<ref>), we deduce the estimate for $|\phi|$ in (<ref>) and (<ref>) with  $C_8$ defined in (<ref>) and $\iota$ chosen to be $\oh$.

With Lemma <ref> established, we now start to prove Theorem <ref>.

Proof of Theorem <ref>. By assumption, there is a positive constant $G$ such that

\begin{equation}\label{SET5}
\sup_{\mathcal{D}(0,v_*)}\gamma(u,v) \leq G.
\end{equation}

For each $u \in [0,v_*)$, we define the quantity

\begin{equation}\label{SET6}
B'(u) :=  \sup_{v \in [u,v_*]} |\alpha'|(u,v).
\end{equation}

We will attempt to show that this quantity remains bounded uniformly in $u$, which in turn implies that the constant

\begin{equation}\label{SET7}
B' := \sup_{u \in [0,v_*)} \sup_{v \in [u,v_*]} |\alpha'|(u,v) = \sup_{\mathcal{D}(0,v_*)}|\alpha'| < + \infty
\end{equation}

and the solution extends as a $C^1$ solution up to $\mathcal{D}(0,v_*)$.

Estimates for $B'(u)$. To do so, we proceed as follows. By (<ref>) of Lemma <ref>, we have

\begin{equation}\label{SET8}
\alpha'(u,v)e^{\int_0^u \ii \e A_u(u',v) \D u'} = \alpha'(0,v) + \sum_{i=1}^{11} I_i(u,v)
\end{equation}

with $K_i$ defined in (<ref>) and $I_i := \int_0^u e^{\int^{u'}_{0} \ii \e A_u(u'',v) \D u''} K_i (u',v) \D u'$.
Hence, we obtain

\begin{equation}\label{SET9}
|\alpha'|(u,v) \leq E'(v_*) + \sum_{i=1}^{11} |I_i|(u',v)
\end{equation}

\begin{equation}\label{SET10}
E'(v_*) := \sup_{\{0\} \times [0,v_*]}|\alpha'|.
\end{equation}

and $|I_i|$ satisfying the following estimates:

\begin{equation}\label{I_1'}
|I_1|(u,v) \leq  8\pi \e \int^{u}_{0} (-\du r) \ml \frac{|Q||\phi|}{r^2} \mr  \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr^2 (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_2'}
|I_2|(u,v) \leq 4\pi \int^{u}_{0} (-\du r) \ml\frac{|Q|^2}{r^4}\mr  \ml \frac{1}{1-\mu}\mr \ml \frac{r|\dv \phi|}{\dv r}\mr^3 (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_3'}
|I_3|(u,v) \leq  4\pi \int^{u}_{0} (-\du r)\ml \frac{1}{r^2}\mr\ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr^3 (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_4'}
|I_4|(u,v) \leq  2 \int^{u}_{0} (-\du r)|\alpha'|  \ml \frac{1}{r}\mr \ml \frac{\mu}{1-\mu}\mr(u',v) \mathrm{d}u' ,
\end{equation}

\begin{equation}\label{I_5'}
|I_5|(u,v) \leq  2 \int^{u}_{0} (-\du r)|\alpha'| \ml\frac{|Q|^2}{r^3}\mr \ml \frac{1}{1-\mu}\mr(u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_6'}
|I_6|(u,v) \leq  5 \int^{u}_{0}(-\du r)\ml \frac{|Q|^2}{r^4}\mr\ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_7'}
|I_7|(u,v) \leq 3 \int^{u}_{0} (-\du r)\ml \frac{1}{r^2}\mr\ml \frac{\mu}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_8'}
|I_8|(u,v)  \leq  4\pi \e \int^{u}_{0}(-\du r) \ml\frac{|Q||\phi|}{r^2} \mr \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr^2 (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_9'}
|I_9|(u,v) \leq  4 \pi \e^2  \int^{u}_{0} (-\du r) \ml |\phi|^2 \mr  \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_10'}
|I_{10}|(u,v) \leq  3 \e  \int^{u}_{0} (-\du r) \ml \frac{|Q|}{r^2}\mr  \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_11'}
|I_{11}|(u,v) \leq  \e  \int^{u}_{0} (-\du r) \ml \frac{|Q||\phi|}{r^2}\mr  \ml \frac{1}{1-\mu}\mr (u',v) \mathrm{d}u'.
\end{equation}

Next, we proceed to estimate the following quantities:

$$\frac{1}{1-\mu},\, |\alpha'|,\, \frac{r|\dv \phi|}{\dv r},\, \ml \frac{r|\dv \phi|}{\dv r}\mr^2,\, \ml \frac{r|\dv \phi|}{\dv r}\mr^3,\, |Q|,\, |Q\phi|,\, |\phi|^2 \mbox{ for all } (u,v) \in \mathcal{D}(0,v_*).$$

For these terms listed above, we carry out the following strategy:

$\frac{1}{1-\mu}$: With the help of (<ref>), we compute

$$\du\ml \frac{1}{1-\mu(u,v)}\mr = \frac{\du \mu}{(1 - \mu)^2} = \frac{1}{1-\mu(u,v)} \ml -\frac{4\pi r|D_u \phi|^2}{(-\du r)} + \frac{(-\du r)}{r}\frac{\mu - Q^2/r^2}{1-\mu}\mr.$$

This gives

\begin{equation}\label{SET11}
\begin{aligned}
\frac{1}{1-\mu(u,v)} &= \frac{1}{1-\mu(0,v)}\exp\ml {\int^u_{0} \ml -\frac{4\pi r|D_u \phi|^2}{(-\du r)} + \frac{(-\du r)}{r}\frac{\mu - Q^2/r^2}{1-\mu}\mr (u',v) \mathrm{d}u'} \mr.
\end{aligned}
\end{equation}

Since $\du r < 0$, this implies

\begin{equation}\label{SET12}
\begin{aligned}
\frac{1}{1-\mu(u,v)} &\leq \frac{1}{1-\mu(0,v)}\exp\ml {\int^u_{0} \ml  \frac{(-\du r)}{r}\frac{\mu - Q^2/r^2}{1-\mu}\mr (u',v) \mathrm{d}u'} \mr. \\
\end{aligned}
\end{equation}

Employing the definition of $\gamma$ in (<ref>) and $\mu_*$ in (<ref>), we obtain

\begin{equation}\label{SET13}
\frac{1}{1-\mu}(u,v) \leq \frac{1}{1-\mu_*}e^G.
\end{equation}

$|\alpha'|$: By the definition of $B'(u)$ in (<ref>), for each $(u,v) \in \mathcal{D}(0,v_*)$, we have

\begin{equation}\label{SET14}
|\alpha'|(u,v) \leq B'(u).
\end{equation}

$\frac{r|\dv \phi|}{\dv r}$: Utilizing the definition of $\alpha$ in (<ref>) and $r\phi|_{\Gamma} = 0$, we can rewrite $\phi$ as

\begin{equation}
\phi(u,v) =  \frac{1}{r}\int^v_u (\alpha \dv r)(u,v') \mathrm{d}v',
\end{equation}

which in turn implies that

\begin{equation}
\frac{r \dv \phi}{\dv r}(u,v) = \alpha(u,v) - \frac{1}{r(u,v)}\int^v_u (\alpha \dv r)(u,v') \mathrm{d}v'.
\end{equation}

Using the fact that $r(u,u) = r|_{\Gamma} = 0$ and inequality (<ref>), we obtain

\begin{equation}\label{SET15}
\begin{aligned}
\mlm\frac{r \dv \phi}{\dv r}\mrm(u,v) &\leq
\mlm \alpha(u,v) - \frac{1}{r(u,v)} \int^v_u (\alpha \dv r)(u,v') \mathrm{d}v'\mrm \\
&\leq \frac{1}{r(u,v)}\int^v_u \mlm \alpha(u,v) - \alpha(u,v') \mrm \dv r(u,v') \mathrm{d}v' \\
&\leq \frac{1}{r(u,v)}\int^v_u \ml \int^v_{v'} \mlm \dv \alpha (u,v'')\mrm \mathrm{d}v'' \mr \dv r (u,v') \mathrm{d}v' \\
&\leq \frac{1}{r(u,v)}\int^v_u \ml \int^v_{v'} B'(u) (\dv r)(u,v'') \D v'' \mr \D v'= \frac{B'(u)r(u,v)}{2}.
\end{aligned}
\end{equation}

$\ml \frac{r|\dv \phi|}{\dv r}\mr^3:$ For this term, we proceed with a different strategy. Observe that since

\begin{equation}\label{SET16}
\begin{aligned}
\dv  \ml \frac{r \dv \phi}{\dv r}\mr^3 &= \dv \ml \frac{\dv (r\phi)}{\dv r} - \phi \mr^3 = 3\ml \frac{r \dv \phi}{\dv r}\mr^2 \dv \ml \frac{\dv (r\phi)}{\dv r} - \phi\mr \\
&= 3\ml \frac{r \dv \phi}{\dv r}\mr^2 \ml \alpha ' \dv r - \dv \phi\mr = 3 \alpha' \frac{r^2(\dv \phi)^2}{\dv r} - \frac{3 \dv r}{r}\ml \frac{r \dv \phi}{\dv r}\mr^3,
\end{aligned}
\end{equation}

we then derive

\begin{equation}\label{SET17}
\begin{aligned}
\dv  &\ml r^3\ml \frac{r \dv \phi}{\dv r}\mr^3 \mr = 3r^2 \dv r \ml \frac{r \dv \phi}{\dv r}\mr^3 + r^3\dv \ml \frac{r\dv \phi}{\dv r}\mr^3 \\
&= 3r^2 \dv r \ml \frac{r \dv \phi}{\dv r}\mr^3 + r^3\ml3 \alpha' \frac{r^2(\dv \phi)^2}{\dv r} - \frac{3 \dv r}{r}\ml \frac{r \dv \phi}{\dv r}\mr^3 \mr = 3 \alpha' r^3 \ml \frac{r^2(\dv \phi)^2}{\dv r}\mr.
\end{aligned}
\end{equation}

Since $r^3\ml \frac{r \dv \phi}{\dv r}\mr^3 |_{\Gamma} = 0$, by (<ref>), (<ref>), (<ref>), $\dv r \geq 0, 1-\mu \geq 0, \dv m \geq 0$, we deduce

\begin{equation}\label{SET18}
\begin{aligned}
r^3\ml \frac{r|\dv \phi|}{\dv r} \mr^3& (u,v) \leq \int_u^v \mlm \dv \ml r^3 \ml \frac{r\dv \phi}{\dv r} \mr^3 \mr \mrm(u,v') \D v' \leq 3B'(u)r^3(u,v)\int_u^v \frac{r^2|\dv \phi|^2}{\dv r}(u,v') \D v'\\
&\leq \frac{3e^G}{2\pi(1-\mu_*)} B'(u)r^3(u,v) \int^v_u \dv m(u,v') \D v' \leq \frac{3e^G}{\pi(1-\mu_*)} B'(u)r^4(u,v) \mu(u,v), \\
\end{aligned}
\end{equation}

which in turn implies

\begin{equation}\label{SET19}
\ml \frac{r|\dv \phi|}{\dv r} \mr^3 (u,v) \leq C_3' B'(u)r(u,v) \mu(u,v)
\end{equation}

with the constant $C_1'$ given by

\begin{equation}\label{SET27}
C_3' := \frac{3e^G}{\pi(1-\mu_*)}.
\end{equation}

$\ml \frac{r|\dv\phi|}{\dv r}\mr^2$: Here we apply an interpolation estimate between (<ref>) and (<ref>) to obtain

\begin{equation}\label{SET20}
\ml\frac{r|\partial_v \phi|}{\partial_v r} \mr^2(u,v) = \sqrt{\ml\frac{r|\partial_v \phi|}{\partial_v r} \mr\ml\frac{r|\partial_v \phi|}{\partial_v r} \mr^3} \leq C_2'B'(u)r(u,v) \mu(u,v)^{\oh}
\end{equation}

\begin{equation}\label{SET41}
C_2' := \sqrt{\frac{3e^G}{2\pi(1-\mu_*)}}
\end{equation}

In addition, recall that estimates for $|Q|, |Q\phi|, |\phi|^2$ have been established in Lemma <ref>, with $H = \frac{e^G}{1-\mu_*}$.

With these estimates, we now obtain an upper bound for each of the terms $|I_i|$ as follows. By plugging (<ref>), (<ref>), (<ref>), (<ref>), (<ref>), and (<ref>), using $r \leq r(0,v_*)$ whenever necessary, we have the following estimates:

\begin{equation}\label{SET28}
|I_1|(u,v) \leq 8\pi \e C_7' C_2' r(0,v_*) \int_0^u \frac{- \du r}{r} \frac{\mu}{1-\mu}(u',v) B'(u') \D u',
\end{equation}

\begin{equation}\label{SET29}
|I_2|(u,v) \leq 4\pi (C_6')^2 C_3' \int_0^u \frac{- \du r}{r} \frac{\mu}{1-\mu}(u',v) B'(u') \D u',
\end{equation}

\begin{equation}\label{SET30}
|I_3|(u,v) \leq 4 \pi C_3' \int_0^u \frac{- \du r}{r} \frac{\mu}{1-\mu}(u',v) B'(u') \D u'.
\end{equation}

The next two terms $|I_4|$ and $|I_5|$ involve $|\alpha'|$. By employing (<ref>), we proceed and obtain

\begin{equation}\label{SET31}
|I_4|(u,v) \leq 2 \int_0^u \frac{- \du r}{r} \frac{\mu}{1-\mu}(u',v) B'(u') \D u',
\end{equation}

\begin{equation}\label{SET32}
|I_5|(u,v) \leq 2 (C_6')^2 \int_0^u \frac{- \du r}{r} \frac{\mu}{1-\mu}(u',v) B'(u') \D u'.
\end{equation}

For the next three terms $|I_6|, |I_7|, |I_8|$, we have

\begin{equation}\label{SET33}
|I_6|(u,v) \leq \frac{5 (C_6')^2}{2} \int_0^u \frac{- \du r}{r} \frac{\mu}{1-\mu}(u',v) B'(u') \D u',
\end{equation}

\begin{equation}\label{SET34}
|I_7|(u,v) \leq \frac{3}{2} \int_0^u \frac{- \du r}{r} \frac{\mu}{1-\mu}(u',v) B'(u') \D u',
\end{equation}

\begin{equation}\label{SET35}
|I_8|(u,v) \leq 4\pi \e C_7' C_2' r(0,v_*) \int_0^u \frac{- \du r}{r} \frac{\mu}{1-\mu}(u',v) B'(u') \D u'.
\end{equation}

The integrands in the terms $|I_9|$ and $|I_{10}|$ lack a natural upper bound in $\mu$ to construct the integrand of $\gamma$. Instead, we proceed to show that these integrands have additional powers of $r$ as follows:

\begin{equation}\label{SET36}
|I_9|(u,v) \leq \frac{2 \pi \e^2 C_8' e^G}{1-\mu_*} \int_0^u \frac{- \du r}{r^{\oh}}(u',v) B'(u') \D u',
\end{equation}

\begin{equation}\label{SET37}
|I_{10}|(u,v) \leq \frac{3 \e C_6' e^G r(0,v_*)^\oh}{1-\mu_*} \int_0^u \frac{- \du r}{r^{\oh}}(u',v) B'(u') \D u'.
\end{equation}

For the final term $|I_{11}|$, its integrand lacks a natural upper bound in terms of $B'(u)$. We instead treat this term as an additional forcing term in a Grönwall-type argument that we are about to employ. This implies that we have the following estimate for $|I_{11}|$:

\begin{equation}\label{SET38}
|I_{11}|(u,v) \leq \frac{C_7'' \e e^G}{1-\mu_*} \int_0^u \frac{- \du r}{r^{\oh+\frac{\chi}{2}}}(u',v) \D u' \leq \frac{2C_7'' \e e^G r(0,v_*)^{\oh - \frac{\chi}{2}}}{(1-\mu_*)(1-\chi)}.
\end{equation}

Plugging these estimates for $|I_1|$ to $|I_{11}|$ back into (<ref>), we obtain

\begin{equation}\label{SET39}
|\alpha'|(u,v)\leq E'(v_*) + C_{11} + \int_0^u \ml C_9\frac{- \du r}{r} \frac{\mu}{1-\mu} + C_{10} \frac{- \du r}{r^\oh}\mr(u',v) B'(u') \D u'
\end{equation}

with the constants $C_9$, $C_{10}, C_{11}$ given by

\begin{equation}\label{SET40}
\begin{aligned}
C_9 &:= 12\pi \e C_7' C_2' r(0,v_*) + 4\pi (C_6')^2 C_3' + \frac{9}{2}(C_6')^2 + 4\pi C_3' + \frac{7}{2}, \\
\quad C_{10} &:= \frac{(2\pi \e^2 C_8' + 3\e C_6' r(0,v_*)^\oh)e^G}{1-\mu_*}, \quad C_{11} := \frac{2C_7'' \e e^G r(0,v_*)^{\oh - \frac{\chi}{2}}}{(1-\mu_*)(1-\chi)}. \\
\end{aligned}
\end{equation}

For each $u$, we pick $\tilde{v} \in [u,v_*]$ such that the supremum for (<ref>) is attained. Since (<ref>) is true for all $v \in [u,v_*]$, we thus have

\begin{equation}\label{SET42}
B'(u) \leq E'(v_*) + C_{11} + \int_0^u \ml C_9\frac{- \du r}{r} \frac{\mu}{1-\mu} + C_{10} \frac{- \du r}{r^\oh}\mr(u',\tilde{v}) B'(u') \D u'.
\end{equation}

Applying Grönwall's inequality, we deduce

\begin{equation}\label{SET43}
\begin{aligned}
B'(u) &\leq (E'(v_*)+C_{11})\exp\ml \int_0^u \ml C_9\frac{- \du r}{r} \frac{\mu}{1-\mu} + C_{10} \frac{- \du r}{r^\oh}\mr(u',\tilde{v}) \D u' \mr \\
&\leq (E'(v_*)+C_{11})\exp\ml C_9 \gamma(u,\tilde{v}) + 2C_{10}r(0,\tilde{v})^{\oh} \mr \\
&\leq (E'(v_*)+C_{11})\exp\ml C_9 G + 2C_{10}r(0,v_*)^{\oh} \mr.\\
\end{aligned}
\end{equation}

Observe that the upper bound obtained in (<ref>) is uniform in $u$ and $\tilde{v}$. Hence, from (<ref>), we deduce that

\begin{equation}\label{SET44}
B' \leq (E'(v_*)+C_{11})\exp\ml C_9 G + 2C_{10}r(0,v_*)^{\oh} \mr < + \infty,
\end{equation}

which in turn implies that the solution extends as a $C^1$ solution up to $\mathcal{D}(0,v_*)$.

Extension Beyound Boundary. Next, we will show that there exists some $\xi' > 0$, such that the solution extends as a $C^1$ solution to $\mathcal{D}(0,v_* +  \xi')$. This is done by estimating the change in $\gamma$ across a rectangular region $\{(u,v): u \in [0,v_*], v\in [v_*,v_*+\xi]\}$ for some $\xi > 0$.

[gray!30] (0,0) to (2,2) to (1,3) to (0,2) to (0,1);
[gray!60] (1,3) to (1.5,3.5) to (2.5,2.5) to (2,2) to (1,3);
[->] (0,0) node[anchor=north]$\mathcal{O}$ –  (0,5) node[anchor = east]$\Gamma$;
[->] (0,0) – (4,4) node[anchor = west]$C_0^+$;
(0,4) – (2,2);
[] at (2.15,1.85) $v_*$;
[anchor = east] at (0,2) $(u_1,u_1)$;
at (0,2)[circle,fill,inner sep=1.5pt];
[rotate=45] at (0.85,1.85) $\mathcal{D}(0,u_1,v_*)$;
[dotted] (0,2) – (1.5,3.5);
[] (1.5,3.5) – (2.5,2.5);
[anchor = west] at (2.5,2.5) $v_* + \xi$;
[->] (2,3.5) – (1.5,3);
[anchor = west] at (2,3.5)$U_{\xi}(u_1)$;
[pattern=north east lines, pattern color=black] (0,0) to (2.5,2.5) to (1.5,3.5) to (0,2) to (0,0);

[anchor = west] at (6.5,2) Legend:;
[anchor = west] at (6.9,1.5) $\mathcal{D}(0,u_1,v_*)$;
[gray!30] (6.65,1.6) to (6.85,1.6) to (6.85,1.4) to (6.65,1.4) to (6.65,1.6);
[anchor = west] at (6.9,1) $U_\xi(u_1)$;
[gray!60] (6.65,1.1) to (6.85,1.1) to (6.85,0.9) to (6.65,0.9) to (6.65,1.1);
[anchor = west] at (6.9,0.5) $\mathcal{D}(0,u_1,v_*+\xi)$;
[pattern=north east lines, pattern color=black] (6.65,0.6) to (6.85,0.6) to (6.85,0.4) to (6.65,0.4) to (6.65,0.6);

Illustration of $U_\xi(u_1)$ and $\mathcal{D}(0,v_1,v_* + \xi)$.

For any given $u_1 \in [0,v_*)$, we consider the corresponding rectangular strip given by

\begin{equation}\label{SET45}
U_{\xi}(u_1) = [0,u_1) \times [v_*,v_*+\xi].
\end{equation}

In the spirit of (<ref>), we define

\begin{equation}\label{SET46}
\overline{B'_{\xi}}(u_1) := \sup_{U_{\xi}(u_1)} (|\alpha'|).
\end{equation}

We also note that, under the convention in the previous subsections, we can write

\begin{equation}\label{SET47}
\mathcal{D}(0,u_1,v_* + \xi) = \mathcal{D}(0,u_1,v_*) \cup U_{\xi}(u_1).
\end{equation}

Recall from the first part of the proof that the quantity $B'$ in (<ref>) is bounded, independent of $u_1$, and obeys the following estimate derived in (<ref>):

\begin{equation}\label{SET48}
B'(u) \leq (E'(v_*)+C_{11}(G))\exp\ml C_9(G) G + 2C_{10}(G)r(0,v_*)^{\oh} \mr =: C_{12}(G,v_*).
\end{equation}

Note that in (<ref>), the constants $C_9, C_{10}, C_{11}$ depends on $G$, and we define $C_{12}$ to be a constant depending only on $v_*$ and $G$.

Now, fix some $\hat{\xi} > 0$. We then have the following lemma:

For any $u_1 \in (0,v_*)$ and $\xi \in (0,\hat{\xi}]$, if $\gamma$ satisfies

\begin{equation}\label{SET49}
\sup_{\mathcal{D}(0,u_1,v_* + \xi)} \gamma \leq 3G,
\end{equation}

then we have

\begin{equation}\label{SET50}
\overline{B'_{\xi}}(u_1) \leq \max\{  C_{12}(G,v_*), C_{12}(3G,v_*+\hat{\xi}) \} =: B_{2}.
\end{equation}

The proof of this lemma follows from the fact that in the region $U_\xi(u_1)$ estimates derived as above now rely on the initial data along $\{0\} \times [v_*,v_* + \xi]$ and rely on a new uniform upper bound for $\gamma$ to be $3G$. The remaining details are similar to those outlined in the estimates for $B'(u)$ as above. Note that the upper bound in (<ref>) is independent of $u_1$ and $\xi$.

With the above lemma, we proceed back on track to prove that the $C^1$ solution extends beyond the domain of dependence $\mathcal{D}(0,u_1,v_*)$. Next, we define

\begin{equation}\label{SET51}
u_2 = \sup\left\{ u_1 \in [0,v_*): \sup_{\mathcal{D}(0,u_1,v_*+\xi)} \gamma \leq 3G\right\}.
\end{equation}

By the definition of $u_2$, we then have either $u_2 = v_*$ or

\begin{equation}\label{SET52}
\sup_{\mathcal{D}(0,u_1,v_* + \xi)} \gamma = 3G
\end{equation}

is attained at some $u_2 \in [u_0,v_*)$. We now show that for $\xi$ sufficiently small, the latter will not happen. Suppose for a contradiction that the latter happens. Then, we set $u_1 = u_2$ as in the hypothesis of Lemma <ref>. Consequently, as described in the proof of Lemma <ref>, in the region $U_\xi(u_2)$, we have

\begin{equation}\label{SET54}
\frac{|\dv \phi|}{\dv r}(u,v) \leq \frac{B_2}{2}.
\end{equation}

Note that we can obtain a different estimate for $|Q|$ with a higher power of $r$ albeit without an upper bound in $\mu$. To do so, we leverage on (<ref>), $Q|_\Gamma = 0$ and (<ref>) to obtain

\begin{equation}\label{SET55}
\begin{aligned}
|Q|(u,v) &\leq \int_u^v |\dv Q|(u,v') \D v' \leq 2 \pi \e B_2 C_8' \int_u^v r^{\frac{5}{4}} \dv r(u,v') \D v' \leq \frac{8 \pi \e B_2 \sqrt{C_8'}}{9} r^{\frac{9}{4}}(u,v) \leq C_{13} r^2(u,v),
\end{aligned}
\end{equation}

\begin{equation}\label{SET56}
C_{13} := \frac{8 \pi \e B_2 \sqrt{C_8'} r^{\frac{1}{4}}(0,v_*)}{9}.
\end{equation}

Here, the dependence of $C_8'$ on $v_* + \hat\xi$ is dropped for brevity.

We are now able to estimate $m(u,v)$ as follows. By (<ref>) and $m|_{\Gamma} = 0$, these imply

\begin{equation}\label{SET57}
m(u,v) \leq \int^v_u \ml \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r} + \frac{Q^2 \dv r}{2r^2}\mr(u,v') \D v'.
\end{equation}

Employing (<ref>), we get

\begin{equation}\label{SET58}
\begin{aligned}
\int^v_u \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r}(u,v') \D v'&\leq 2\pi \int^v_u \frac{r^2|\dv \phi|^2}{(\dv r)^2}(\dv r)(u,v') \D v'\\
&\leq \frac{\pi B_2^2 }{2} \int^v_u r^{2}(\dv r)(u,v') \D v' = \frac{\pi B_2^2}{6}r^3(u,v).
\end{aligned}
\end{equation}

Furthermore, by (<ref>), we obtain

\begin{equation}\label{SET59}
\begin{aligned}
\int^v_u \frac{Q^2 \dv r}{2r^2}(u,v') \D v' &\leq \frac{C_{13}^2}{2}\int^v_u r^2(\dv r)(u,v') \D v' = \frac{C_{13}^2}{6}r^3(u,v).
\end{aligned}
\end{equation}

This in turn implies that

\begin{equation}\label{SET60}
\mu(u,v) \leq \ml \frac{\pi B_2^2}{3} + \frac{C_{13}^2}{3}\mr r^2(u,v).
\end{equation}

Meanwhile, from (<ref>) but with $G$ replaced by $3G$, we have

\begin{equation}\label{SET61}
\frac{1}{1-\mu}(u,v) \leq \frac{1}{1-\mu_*}e^{3G}.
\end{equation}

With the above estimates, we can now estimate the change in $\gamma$. Recall the expressions for the derivative of $\gamma$ with respect to $v$ from (<ref>) and (<ref>) as below:

\begin{equation}\label{SET64}
\dv \gamma(u,v) = \int_0^u (T_1 + T_2 + T_3)(u',v) \D u'.
\end{equation}

By utilizing the estimates on relevant quantities above, together with (<ref>), we derive

\begin{equation}\label{SET65}
\begin{aligned}
\int_0^u |T_1|(u',v) \D u' &\leq \frac{8\pi \e e^{3G} }{1-\mu_*}\int_0^u \frac{|Q||\phi||\dv \phi|}{r}(-\du r)(u',v) \D u' \leq \frac{4\pi \e e^{3G} \sqrt{C_8'} C_{13} B_2 }{1-\mu_*}\int_0^u r^{-\frac{1}{4}}(\dv r)(-\du r)(u',v) \D u' \\
&\leq \frac{2\pi \e e^{4G} \sqrt{C_8'} C_{13} B_2 }{1-\mu_*}\int_0^u r^{-\frac{1}{4}}(-\du r)\D u' \leq \frac{(8)^{\frac{3}{4}}\pi \e e^{4G} \sqrt{C_8'} C_{13} B_2 }{3(1-\mu_*)}(v_* + \hat \xi)^{\frac{3}{4}},\\
\end{aligned}
\end{equation}

\begin{equation}\label{SET66}
\begin{aligned}
\int_0^u |T_2|(u',v) \D u' &\leq \frac{4\pi e^{3G}}{1-\mu_*}\ml 1+\frac{C_{13}^2(v_* + \hat \xi)^2}{4}\mr \int_0^u (-\du r)\frac{|\dv \phi|^2}{\dv r}(u',v) \D u' \\
&\leq \frac{\pi e^{3G}B_2^2}{1-\mu_*}\ml 1+\frac{C_{13}^2(v_* + \hat \xi)^2}{4}\mr \int_0^u (-\du r)(\dv r)(u',v) \D u' \\
&\leq \frac{\pi e^{4G}B_2^2}{1-\mu_*}\ml 1+\frac{C_{13}^2(v_* + \hat \xi)^2}{4}\mr \int_0^u (-\du r)(u',v) \D u' \\
&\leq \frac{\pi e^{4G}B_2^2}{2(1-\mu_*)}\ml 1+\frac{C_{13}^2(v_* + \hat \xi)^2}{4}\mr (v_* + \hat \xi), \\
\end{aligned}
\end{equation}

\begin{equation}\label{SET67}
\begin{aligned}
\int_0^u |T_3|(u',v) \D u' &\leq \frac{2e^{3G}}{1-\mu_*}\int_0^u \frac{(-\du r)(\dv r)}{r^2}\ml |\mu| + \frac{2Q^2}{r^2} \mr(u',v) \D u' \\
&\leq \frac{2e^{3G}}{1-\mu_*}\ml \frac{\pi B_2^2}{3} + \frac{7C_{13}^2}{3}\mr\int_0^u (-\du r)(\dv r)(u',v) \D u' \\
&\leq \frac{e^{4G}}{1-\mu_*}\ml \frac{\pi B_2^2}{3} + \frac{7 C_{13}^2}{3}\mr\int_0^u (-\du r)(u',v) \D u' \\
&\leq \frac{e^{4G}}{2(1-\mu_*)}\ml \frac{\pi B_2^2}{3} + \frac{7 C_{13}^2}{3}\mr (v_* + \hat \xi). \\
\end{aligned}
\end{equation}

Note that (<ref>), (<ref>), (<ref>) imply
$\int^u_0 |T_1 + T_2 + T_3|(u',v) \D u' < + \infty$, and hence the expression in (<ref>) is valid by the dominated convergence theorem. Furthermore, for all $(u,v) \in U_\xi(u_2)$, we also obtain

\begin{equation}\label{SET68}
|\dv \gamma|(u,v) \leq C_{14}
\end{equation}

\begin{equation}\label{SET69}
\begin{aligned}
C_{14} :=&\; \frac{(8)^{\frac{3}{4}}\pi \e e^{4G} \sqrt{C_8'} C_{13} B_2 }{3(1-\mu_*)}(v_* + \hat \xi)^{\frac{3}{4}} + \frac{\pi e^{4G}B_2^2}{2(1-\mu_*)}\ml 1+\frac{C_{13}^2(v_* + \hat \xi)^2}{4}\mr(v_* + \hat \xi)\\
&+ \frac{e^{4G}}{2(1-\mu_*)}\ml \frac{\pi B_2^2}{3} + \frac{7 C_{13}^2}{3}\mr(v_* + \hat \xi).
\end{aligned}
\end{equation}

For each $(u,v) \in U_\xi(u_2)$, we thus have

\begin{equation}\label{SET70}
\begin{aligned}
\gamma(u,v) \leq \gamma(u,v_*) + \int_v^{v_*}|\dv \gamma|(u,v') \D v' \leq G + C_{14}\xi.
\end{aligned}
\end{equation}

If we pick

\begin{equation}\label{SET71}
\xi := \min \left\{  \hat{\xi}, \frac{G}{C_{14}}\right\},
\end{equation}

we then deduce that
$\sup_{\mathcal{D}(0,u_1,v_* + \xi)} \gamma \leq 2G$, contradicting (<ref>).

Henceforth, for each $u_1 \in (u_0,v_*)$, the hypothesis of Lemma <ref> is verified. This then implies that (<ref>) holds and hence  $\sup_{\mathcal{D}(u_0,u_1,v_*+\Delta v)}|\alpha'|$ remains bounded as $u_1 \rightarrow v_*$. Therefore, we conclude that the solution extends as a $C^1$ solution up to the closure of $\mathcal{D}(u_0,v_*,v_*+\xi)$. In particular, with $C^1$ initial data prescribed along $\{v_*\} \times [v_*,v_*+\xi]$, we show that the solution further extends as a $C^1$ solution to $\mathcal{D}(v_*,v_*+\xi')$ for some $\xi' \in (0,\xi]$. This concludes the proof of Theorem <ref>.

We continue to prove Theorem <ref>.

Proof of Theorem <ref>. By the assumption, we can find a $u_0 < v_*$ such that

\begin{equation}\label{FET1}
\varepsilon  = \sup_{\mathcal{D}(u_0,v_*)} \mu < \varepsilon'.
\end{equation}

Given any $u_1 \in (u_0,v_*)$, we further define

\begin{equation}\label{D(u0,u1,v*)}
\mathcal{D}(u_0,u_1,v_*) := \mathcal{D}(u_0,v_*)\setminus \overline{\mathcal{D}}(u_1,v_*).
\end{equation}

This region is demonstrated in Figure <ref>. For any given $\delta \in (0,1)$, we set

\begin{equation}\label{Au1}
A'(u_1) := \sup_{\mathcal{D}(u_0,u_1,v_*)} (r^{\delta}|\alpha'|).
\end{equation}

We will show that this quantity remains bounded as $u_1 \rightarrow v_*$. Based on it, we will further prove that

\begin{equation}\label{Bu1}
B'(u_1) := \sup_{\mathcal{D}(u_0,u_1,v_*)}(|\alpha'|)
\end{equation}

remains bounded as $u_1 \rightarrow v_*$. The fact that the solution extends to $\mathcal{D}(0,v_*)$ as a $C^1$ solution then follows.

[gray!30] (0,1) to (1.5,2.5) to (1,3) to (0,2) to (0,1);
[->] (0,0) node[anchor=north]$\mathcal{O}$ –  (0,5) node[anchor = east]$\Gamma$;
[->] (0,0) – (3,3) node[anchor = west]$C_0^+$;
(0,4) – (2,2);
[] at (2.15,1.85) $v_*$;
[anchor = east] at (0,1) $(u_0,u_0)$;
at (0,1)[circle,fill,inner sep=1.5pt];
[anchor = east] at (0,2) $(u_1,u_1)$;
at (0,2)[circle,fill,inner sep=1.5pt];
at (0,4)[circle,fill,inner sep=1.5pt];
[anchor = east] at (0,4) $(v_*,v_*)$;
[rotate=45] at (0.65,2.15) $\mathcal{D}(u_0,u_1,v_*)$;
[dotted] (0,2) – (1,3);
[] (0,1) – (1.5,2.5);

Illustration of $\mathcal{D}(u_0,v_1,v_*)$.

Estimates for $A'(u_1)$.
To bound $A'(u_1)$, we begin by applying Lemma <ref> to obtain (<ref>). From (<ref>), for every $(u,v) \in \mathcal{D}(u_0,u_1,v_*)$, via integration along the $u$ direction, we obtain

\begin{equation}\label{FET114}
\begin{aligned}
\alpha'(u,v) e^{\int^u_{u_0}\ii \e A_u (u',v) \mathrm{d}u'} &= \alpha'(u_0,v) +  \sum_{i=1}^{11} I_i(u,u_0,v). \\
\end{aligned}
\end{equation}

We now define

\begin{equation}\label{D'}
D'(u_0) := \sup_{\{u_0\}\times (u_0,v_*)}(r^{\delta}|\alpha'|)
\end{equation}

and observe that this term is finite as we have $\alpha \in C^1(\{u_0\} \times [u_0,v_*])$.
This implies

\begin{equation}\label{FET15}
\begin{aligned}
|\alpha'(u,v)| &\leq D'(u_0)\ml r(u_0,v) \mr^{-\delta}  + \sum_{i=1}^{11} |I_i|(u,u_0,v)
\end{aligned}
\end{equation}

\begin{equation}\label{FET16‘}
I_i(u,u_0,v) = \int^u_{u_0} e^{\int^{u'}_{u_0}\ii \e A_u (u'',v) \mathrm{d}u''} K_i(u',v) \mathrm{d}u'.
\end{equation}

Since $A_u$ is real-valued, we also obtain

\begin{equation}\label{FET16}
|I_i|(u,u_0,v) \leq \int^u_{u_0} |K_i|(u',v) \mathrm{d}u'.
\end{equation}

Next, we move to derive an upper bound for each of the $I_i$'s. We begin by observing that

\begin{equation}\label{I_1}
|I_1|(u,u_0,v) \leq  8\pi \e \int^{u}_{u_0} (-\du r) \ml \frac{|Q||\phi|}{r^2} \mr  \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr^2 (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_2}
|I_2|(u,u_0,v) \leq 4\pi \int^{u}_{u_0} (-\du r) \ml\frac{|Q|^2}{r^4}\mr  \ml \frac{1}{1-\mu}\mr \ml \frac{r|\dv \phi|}{\dv r}\mr^3 (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_3}
|I_3|(u,u_0,v) \leq  4\pi \int^{u}_{u_0} (-\du r)\ml \frac{1}{r^2}\mr\ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr^3 (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_4}
|I_4|(u,u_0,v) \leq  2 \int^{u}_{u_0} (-\du r)|\alpha'|  \ml \frac{1}{r}\mr \ml \frac{\mu}{1-\mu}\mr(u',v) \mathrm{d}u' ,
\end{equation}

\begin{equation}\label{I_5}
|I_5|(u,u_0,v) \leq  2 \int^{u}_{u_0} (-\du r)|\alpha'| \ml\frac{|Q|^2}{r^3}\mr \ml \frac{1}{1-\mu}\mr(u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_6}
|I_6|(u,u_0,v) \leq  5 \int^{u}_{u_0}(-\du r)\ml \frac{|Q|^2}{r^4}\mr\ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_7}
|I_7|(u,u_0,v) \leq 3 \int^{u}_{u_0} (-\du r)\ml \frac{1}{r^2}\mr\ml \frac{\mu}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_8}
|I_8|(u,u_0,v)  \leq  4\pi \e \int^{u}_{u_0}(-\du r) \ml\frac{|Q||\phi|}{r^2} \mr \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr^2 (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_9}
|I_9|(u,u_0,v) \leq  4 \pi \e^2  \int^{u}_{u_0} (-\du r) \ml |\phi|^2 \mr  \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_10}
|I_{10}|(u,u_0,v) \leq  3 \e  \int^{u}_{u_0} (-\du r) \ml \frac{|Q|}{r^2}\mr  \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u',
\end{equation}

\begin{equation}\label{I_11}
|I_{11}|(u,u_0,v) \leq  \e  \int^{u}_{u_0} (-\du r) \ml \frac{|Q||\phi|}{r^2}\mr  \ml \frac{1}{1-\mu}\mr (u',v) \mathrm{d}u'.
\end{equation}

In order for us to prove desired upper bounds for these $11$ terms, we derive useful estimates for

$$\frac{1}{1-\mu},\, |\alpha'|,\, \frac{r|\dv \phi|}{\dv r},\, \ml \frac{r|\dv \phi|}{\dv r}\mr^2,\, \ml \frac{r|\dv \phi|}{\dv r}\mr^3,\, |Q|,\, |Q\phi|,\, |\phi|^2 \mbox{ for all } (u,v) \in \mathcal{D}(u_0,u_1,v_*).$$

For these listed terms, we carry out the following strategy

$\frac{1}{1-\mu}$: Using the definition of $\varepsilon$ in (<ref>), we have

\begin{equation}\label{FET17}
\frac{1}{1-\mu}(u,v) \leq \frac{1}{1-\varepsilon}.
\end{equation}

$|\alpha'|$: Appealing to the definition of $A(u_1)$ in (<ref>), we get

\begin{equation}\label{FET21}
|\alpha'|(u,v) \leq A'(u_1)r^{-\delta}(u,v).
\end{equation}

$\frac{r|\dv \phi|}{\dv r}$: Recall from the proof of Theorem <ref> that we have

\begin{equation}\label{FET19}
\frac{r \dv \phi}{\dv r}(u,v) = \alpha(u,v) - \frac{1}{r(u,v)}\int^v_u (\alpha \dv r)(u,v') \mathrm{d}v'.
\end{equation}

Using the fact that $r(u,u) = r|_{\Gamma} = 0$, inequality (<ref>) and the definition of $A'(u_1)$ in (<ref>), we obtain

\begin{equation}\label{FET20}
\begin{aligned}
\mlm\frac{r \dv \phi}{\dv r}\mrm(u,v) &\leq
\mlm \alpha(u,v) - \frac{1}{r(u,v)} \int^v_u (\alpha \dv r)(u,v') \mathrm{d}v'\mrm \\
&= \frac{1}{r(u,v)}\int^v_u \mlm \alpha(u,v) - \alpha(u,v') \mrm \dv r(u,v') \mathrm{d}v' \\
&\leq \frac{1}{r(u,v)}\int^v_u \ml \int^v_{v'} \mlm \dv \alpha (u,v'')\mrm \mathrm{d}v'' \mr \dv r (u,v') \mathrm{d}v' \\
&\leq \frac{1}{r(u,v)}\int^v_u \ml \int^v_{v'} A'(u_1) r^{-\delta}(\dv r)(u,v'')
\mathrm{d}v'' \mr \dv r (u,v') \mathrm{d}v' \\
&= \frac{A'(u_1)}{r(u,v)}\ml \frac{r^{2-\delta}}{(1-\delta)} -\frac{r^{2-\delta}}{(1-\delta)(2-\delta)} \mr(u,v)= \frac{A'(u_1)r^{1-\delta}(u,v)}{2-\delta}.
\end{aligned}
\end{equation}

$\ml \frac{r|\dv \phi|}{\dv r}\mr^3$: Here we will obtain an upper bound, which is linear in $A'(u_1)$. We recall a lemma established by Christodoulou in [19] as follows:

Let $f$ be a $C^1$ function on $[0,r_*]$ with $f(0) = 0$ and let

\begin{equation}\label{Lemma2,1}
a := \sup_{r\in(0,r_*)} \ml \frac{g}{r} \mr, \quad b := \sup_{r\in(0,r_*)} \ml r^{\delta} \left| \frac{\mathrm{d}f}{\mathrm{d}r} \right| \mr
\end{equation}

\begin{equation}\label{Lemma2,2}
g(r) = \int^r_0 f^2 \mathrm{d}r.
\end{equation}

Then, for any $\delta \in (0,1)$, we have

\begin{equation}\label{Lemma2,3}
|f(r)|^3 \leq \frac{3ab}{1-\delta}r^{1-\delta}.
\end{equation}

Observe that

\begin{equation}
f(r)^3 = \int^r_0 \frac{\mathrm{d}}{\mathrm{d}r} f(r)^3 \mathrm{d}r
= \int^r_0 3f^2 \ml \frac{\mathrm{d}f}{\mathrm{d}r} \mr \mathrm{d}r.
\end{equation}

This implies

\begin{equation}\label{Lemma2Proof}
\begin{aligned}
|f(r)|^3 &\leq \int^r_0 3f^2 \left| \frac{\mathrm{d}f}{\mathrm{d}r} \right| \mathrm{d}r \leq 3b \int^r_0 r^{-\delta}f^2 \mathrm{d}r = 3b\ml \int^r_0 r^{-\delta} \frac{dg}{dr} \mathrm{d}r \mr \\
&\quad= 3b \ml r^{-\delta}g(r) - r^{-\delta}g(r)|_{r=0} + \delta \int^r_0 gr^{-1-\delta} \mathrm{d}r \mr\\
&\quad\leq 3b \ml ar^{1-\delta} + a\delta \int^r_0 r^{-\delta} \mathrm{d}r \mr = \frac{3ab}{1-\delta}r^{1-\delta}.
\end{aligned}
\end{equation}

Here we use the fact that

\begin{equation}
r^{-\delta}g(r)|_{r=0} = \lim_{r\rightarrow 0}\frac{g(r)}{r^{\delta}} = \lim_{r\rightarrow 0}\frac{f(r)^2}{\delta r^{\delta-1}} = \lim_{r\rightarrow 0} \frac{f(r)^2 r^{1-\delta}}{\delta} = 0.
\end{equation}

We then set $f = \frac{r|\dv \phi|}{\dv r}$ and apply the above lemma. To obtain a bound for $\ml\frac{r|\dv \phi|}{\dv r}\mr^3$, we need to compute $a$ and $b$ as in (<ref>). For each fixed $u \in (u_0,u_1)$ with $r_* = r(u,v_*)$, we have

\begin{equation}\label{FET22}
\begin{aligned}
a(u) &= \sup_{r\in(0,r_*)} \ml \frac{1}{r} \int^r_0 f(r')^2 \mathrm{d}r' \mr = \sup_{v\in(u,v_*)} \ml \frac{1}{r} \int^v_u \ml\frac{(r|\partial_v \phi|)^2}{\partial_v r}\mr(u,v') \mathrm{d}v' \mr.
\end{aligned}
\end{equation}

Noting that (<ref>) implies $
\frac{(r|\partial_v \phi|)^2}{\partial_v r} \leq \frac{\partial_v m}{2\pi (1-\mu)}$, we therefore obtain

\begin{equation}\label{FET24}
\begin{aligned}
a \leq \sup_{v\in(u,v_*)} \ml \frac{1}{2\pi r} \int^v_u \ml\frac{\partial_v m}{1-\mu}\mr \mathrm{d}v' \mr &\leq \frac{1}{4\pi (1-\varepsilon)} \sup_{v\in(u,v_*)}\ml \mu(u,v) - \frac{2m(u,u)}{r(u,v)}\mr \\
&\leq \frac{\varepsilon}{4\pi (1-\varepsilon)}.
\end{aligned}
\end{equation}

For $b$, by applying (<ref>) and (<ref>), we have

\begin{equation}\label{FET25}
\begin{aligned}
b &= \sup_{r\in(0,r_*)} \ml r^{\delta} \left| \frac{\mathrm{d}f}{\mathrm{d}r} \right| \mr =\sup_{v\in(u,v_*)} \ml r^{\delta} \mlm \frac{\dv f}{\dv r} \mrm \mr = \sup_{v\in(u,v_*)} \ml \frac{r^\delta}{\partial_v r} \mlm \dv \ml \frac{\partial_v (r\phi)}{\partial_v r} - \phi\mr \mrm \mr \\
&\quad\quad\leq \sup_{\mathcal{D}(u_0,u_1,v_*)} \ml \frac{r^\delta}{\partial_v r} \mlm \dv \ml \frac{\partial_v (r\phi)}{\partial_v r} - \phi\mr \mrm \mr \\
&\quad\quad\leq \sup_{\mathcal{D}(u_0,u_1,v_*)} \ml r^{\delta}|\alpha'| \mr +
\sup_{\mathcal{D}(u_0,u_1,v_*)} \ml r^{\delta - 1}\ml \frac{r|\partial_v \phi|}{\partial_v r} \mr\mr\\
&\quad\quad\leq A'(u_1) + \frac{1}{2-\delta}A'(u_1) = \frac{3-\delta}{2-\delta}A'(u_1).
\end{aligned}
\end{equation}

Henceforth, combining the lemma together with (<ref>) and (<ref>), we obtain

\begin{equation}\label{FET26}
\begin{aligned}
\ml\frac{r|\partial_v \phi|}{\partial_v r}\mr ^3(u,v) \leq \frac{3ab}{1-\delta}r^{1-\delta} \leq \frac{3(3-\delta)}{4\pi(1-\delta)(2-\delta)}\frac{\varepsilon}{1-\varepsilon}A'(u_1)r^{1-\delta}.
\end{aligned}
\end{equation}

$\ml \frac{r|\dv \phi|}{\dv r}\mr^2:$ A direct interpolation between the upper bounds in (<ref>) and (<ref>) implies that

\begin{equation}\label{FET27}
\begin{aligned}
\ml\frac{r|\partial_v \phi|}{\partial_v r} \mr^2(u,v) = \sqrt{\ml\frac{r|\partial_v \phi|}{\partial_v r} \mr\ml\frac{r|\partial_v \phi|}{\partial_v r} \mr^3} \leq \sqrt{\frac{3(3-\delta)}{4\pi(1-\delta)(2-\delta)^2}\frac{\varepsilon}{1-\varepsilon}}A'(u_1)r^{1-\delta}.
\end{aligned}
\end{equation}

Note that the estimates for $|Q|, |Q\phi|,$ and $|\phi|^2$ have been established in Lemma <ref>.

With these estimates, we now obtain an estimate for each of the $|I_i|$'s as follows. For $I_1$, we have

\begin{equation}\label{FET30}
\begin{aligned}
|I_1|(u,u_0,v) &\leq  8\pi \e \int^{u}_{u_0} (-\du r) \ml \frac{|Q||\phi|}{r^2} \mr  \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr^2 (u',v) \mathrm{d}u' \\
&\leq 8 \pi \e C_{7} A'(u_1) \sqrt{\frac{3(3-\delta)}{4\pi(1-\delta)(2-\delta)^2}} \frac{\varepsilon}{(1-\varepsilon)^{2}}  \int^{u}_{u_0}  (-\du r) r^{\frac{1}{2}-\frac{\chi}{2}-\delta}  (u',v) \mathrm{d}u' \\
&\leq 8 \pi \e C_{7} A'(u_1) \sqrt{\frac{3(3-\delta)}{4\pi(1-\delta)(2-\delta)^2}} \frac{\varepsilon}{(1-\varepsilon)^{2}}  \frac{r^{\frac{3}{2}-\frac{\chi}{2}}(u_0,v)}{\frac{3}{2}-\frac{\chi}{2}} \cdot \frac{1}{r^\delta(u,v)} \\
&\leq \frac{16 \pi \e C_{7}r^{\frac{3}{2}-\frac{\chi}{2}}(0,v_*)}{3-\chi}  \sqrt{\frac{3(3-\delta)}{4\pi(1-\delta)(2-\delta)^2}} \frac{\varepsilon}{(1-\varepsilon)^{2}}   \frac{A'(u_1)}{r^{\delta}(u,v)}.\\
\end{aligned}
\end{equation}

Here we use the fact $\chi\in(0,1)$, $\delta\in(0,1)$ and $\frac12-\frac{\chi}{2}-\delta>-1$.

Following a similar idea, the next term can be bounded as follows:

\begin{equation}\label{FET31}
\begin{aligned}
|I_2|(u,u_0,v) &\leq 4\pi \int^{u}_{u_0} (-\du r) \ml\frac{|Q|^2}{r^4}\mr  \ml \frac{1}{1-\mu}\mr \ml \frac{r|\dv \phi|}{\dv r}\mr^3 (u',v) \mathrm{d}u' \\
&\leq  C_6^2  A'(u_1) \frac{3(3-\delta)}{(1-\delta)(2-\delta)}\frac{\varepsilon^2}{(1-\varepsilon)^2}\int^{u}_{u_0} (-\du r) r^{-\delta - \chi} (u',v) \mathrm{d}u' \\
&\leq  \frac{C_6^2r(0,v_*)^{1-\chi}}{1-\chi} \frac{3(3-\delta)}{(1-\delta)(2-\delta)}\frac{\varepsilon^2}{(1-\varepsilon)^2} \frac{A'(u_1)}{r^{\delta}(u,v)}.
\end{aligned}
\end{equation}

We proceed to derive

\begin{equation}\label{FET33}
\begin{aligned}
|I_3|(u,u_0,v) &\leq 4\pi \int^{u}_{u_0} (-\du r)\ml \frac{1}{r^2}\mr\ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr^3 (u',v) \mathrm{d}u'\\
&\leq  A'(u_1) \frac{3(3-\delta)}{(1-\delta)(2-\delta)}\frac{\varepsilon}{(1-\varepsilon)^2}\int^{u}_{u_0} (-\du r)r^{-1-\delta} (u',v) \mathrm{d}u'  \\
&\leq \frac{3(3-\delta)}{\delta(1-\delta)(2-\delta)}\frac{\varepsilon}{(1-\varepsilon)^2} \frac{A'(u_1)}{r^{\delta}(u,v)}
\end{aligned}
\end{equation}

\begin{equation}\label{FET32}
\begin{aligned}
|I_4|(u,u_0,v) &\leq   2 \int^{u}_{u_0} (-\du r)|\alpha'|  \ml \frac{1}{r}\mr \ml \frac{\mu}{1-\mu}\mr(u',v) \mathrm{d}u'  \\
&\leq 2 A'(u_1) \frac{\varepsilon}{1-\varepsilon}  \int^{u}_{u_0} (-\du r) r^{-1-\delta} (u',v) \mathrm{d}u' \leq \frac{2}{\delta} \frac{\varepsilon}{1-\varepsilon}  \frac{A'(u_1)}{r^{\delta}(u,v)}.
\end{aligned}
\end{equation}

The subsequent three terms obey the following bounds

\begin{equation}\label{FET34}
\begin{aligned}
|I_5|(u,u_0,v) &\leq   2 \int^{u}_{u_0} (-\du r)|\alpha'| \ml\frac{|Q|^2}{r^3}\mr \ml \frac{1}{1-\mu}\mr(u',v) \mathrm{d}u' \\
&\leq 2 C^2_6 A'(u_1) \frac{\varepsilon}{1-\varepsilon} \int^{u}_{u_0} (-\du r) r^{-\delta -\chi} (u',v) \mathrm{d}u' \leq  \frac{2 C^2_6 r^{1-\chi}(0,v_*)}{1-\chi} \frac{\varepsilon}{1-\varepsilon}\frac{A'(u_1)}{r^{\delta}(u,v)},
\end{aligned}
\end{equation}

\begin{equation}\label{FET35}
\begin{aligned}
|I_6|(u,u_0,v) &\leq   5 \int^{u}_{u_0}(-\du r)\ml \frac{|Q|^2}{r^4}\mr\ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u' \\
&\leq  5 C_6^2 A'(u_1) \frac{1}{2-\delta}  \frac{\varepsilon}{1-\varepsilon} \int^{u}_{u_0} (-\du r) r^{-\delta - \chi}(u',v) \mathrm{d}u' \\
&\leq \frac{5 C_6^2 r^{1-\chi}(0,v_*)}{(1-\chi)(2-\delta)} \frac{\varepsilon}{1-\varepsilon} \frac{A'(u_1)}{r^{\delta}(u,v)},
\end{aligned}
\end{equation}

\begin{equation}\label{FET36}
\begin{aligned}
|I_7|(u,u_0,v) &\leq 3 \int^{u}_{u_0} (-\du r)\ml \frac{1}{r^2}\mr\ml \frac{\mu}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u' \\
&\leq  \frac{3A'(u_1)}{2-\delta} \frac{\varepsilon}{1-\varepsilon} \int^{u}_{u_0} (-\du r) r^{-1 -\delta}(u',v) \mathrm{d}u'\leq \frac{3}{\delta(2-\delta)} \frac{\varepsilon}{1-\varepsilon} \frac{A'(u_1)}{r^{\delta}(u,v)}.
\end{aligned}
\end{equation}

For the next term $I_8$, we note that its upper bound is exactly half of that of $I_1$. This implies

\begin{equation}\label{FET38}
\begin{aligned}
|I_8|(u,u_0,v)  &\leq  4\pi \e \int^{u}_{u_0}(-\du r) \ml\frac{|Q||\phi|}{r^2} \mr \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr^2 (u',v) \mathrm{d}u', \\
&\leq \frac{8 \pi \e C_{7}r^{\frac{3}{2}-\frac{\chi}{2}}(0,v_*)}{3-\chi}  \sqrt{\frac{3(3-\delta)}{4\pi(1-\delta)(2-\delta)^2}} \frac{\varepsilon}{(1-\varepsilon)^{2}}   \frac{A'(u_1)}{r^{\delta}(u,v)}. \\
\end{aligned}
\end{equation}

To estimate $|I_9|$, we employ (<ref>) and get

\begin{equation}\label{FET40}
\begin{aligned}
|I_9|(u,u_0,v) &\leq  4 \pi \e^2  \int^{u}_{u_0} (-\du r) \ml |\phi|^2 \mr  \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u' \\
&\leq 4 \pi \e ^2  \frac{A'(u_1)}{2-\delta} \frac{1}{1-\varepsilon} \int^u_{u_0} (-\du r)r^{1-\delta}|\phi|^2(u',v)\mathrm{d}u
&\leq 4 \pi \e ^2  \frac{A'(u_1)}{2-\delta} \frac{1}{1-\varepsilon} \int^u_{u_0} (-\du r)r^{1-\delta} \ml \frac{1}{2\pi}\frac{\varepsilon
}{1-\varepsilon} + 2C_{8}^2(\iota)r^{-1-\iota} \mr (u',v)\mathrm{d}u
&\leq \frac{4 \pi \e ^2 }{2-\delta} \frac{1}{1-\varepsilon} \frac{A'(u_1)}{r^{\delta}(u,v)}  \ml \frac{r^2(0,v_*)}{4\pi}\frac{\varepsilon
}{1-\varepsilon} + \frac{2C_{8}^2(\iota) r^{1-\iota}(u_0,v_*)}{1-\iota}\mr.
\end{aligned}
\end{equation}

The upper bounds for the last two terms $I_{10}$ and $I_{11}$ can be derived as

\begin{equation}\label{FET42}
\begin{aligned}
|I_{10}|(u,u_0,v) &\leq  3 \e  \int^{u}_{u_0} (-\du r) \ml \frac{|Q|}{r^2}\mr  \ml \frac{1}{1-\mu}\mr\ml \frac{r|\dv \phi|}{\dv r}\mr (u',v) \mathrm{d}u' \\
&\leq 3 \e C_6 A'(u_1) \frac{1}{2-\delta}\frac{\varepsilon^{\frac{1}{2}}}{1-\varepsilon}\int^{u}_{u_0} (-\du r) r^{\frac{1}{2} -\frac{\chi}{2}-\delta}(u',v) \mathrm{d}u' \\
&\leq \frac{6 \e C_6 r^{\frac{3}{2}-\frac{\chi}{2}}(u_0,v_*)}{(3 -  \chi)(2-\delta)} \frac{\varepsilon^{\frac{1}{2}}}{1-\varepsilon} \frac{A'(u_1)}{r^{\delta}(u,v)}
\end{aligned}
\end{equation}

\begin{equation}\label{FET43}
\begin{aligned}
|I_{11}|(u,u_0,v) &\leq \e  \int^{u}_{u_0} (-\du r) \ml \frac{|Q||\phi|}{r^2}\mr  \ml \frac{1}{1-\mu}\mr (u',v) \mathrm{d}u'\\
&\leq \e C_{7} \frac{\varepsilon^{\frac{1}{2}}}{(1-\varepsilon)^{\frac{3}{2}}} \int^u_{u_0} (-\du r)r^{-\oh-\frac{\chi}{2}}(u',v)\mathrm{d}u' \\
&\leq \e C_{7} r^{\oh + \delta - \frac{\chi}{2}}(0,v_*) \frac{\varepsilon^{\frac{1}{2}}}{(1-\varepsilon)^{\frac{3}{2}}} \int^u_{u_0} (-\du r)r^{-1-\delta}(u',v)\mathrm{d}u' \\
&\leq \frac{2\e C_{7}r^{\oh + \delta - \frac{\chi}{2}}(0,v_*)}{\delta} \frac{\varepsilon^{\frac{1}{2}}}{(1-\varepsilon)^{\frac{3}{2}}} \frac{1}{r^{\delta}(u,v)}.
\end{aligned}
\end{equation}

Note that in (<ref>), we have restricted our choice for $\chi$ to $\chi < 2\delta$.

Back to (<ref>), multiplying $r^{\delta}(u,v)$ on both sides and taking the supremum over $\mathcal{D}(u_0,u_1,v_*)$, we therefore obtain

\begin{equation}\label{FET45}
A'(u_1) \leq D'(u_0) + \sup_{\mathcal{D}(u_0,u_1,v_*)} \ml \sum_{i=1}^{11} |I_i|r^{\delta}(u,v)\mr
\end{equation}

with $D'(u_0)$ defined in (<ref>). For convenience, we impose the requirement

\begin{equation}\label{FET47}
\varepsilon<\frac12.
\end{equation}

With this restriction, we employ obvious bounds (such as $0 < \iota,\delta,\chi < 1$ and $0 < \varepsilon < \frac{1}{2}$) to simplify the constants appearing in the upper bounds for $|I_1|$ to $|I_{11}|$. We will also apply $1 \leq \frac{1}{(1-\delta)^{\frac{1}{2}}} \leq \frac{1}{1-\delta}\leq \frac{1}{\delta(1-\delta)}$, $\frac{1}{\delta} \leq \frac{1}{\delta(1-\delta)}$, $\frac{1}{(1-\varepsilon)^b} \leq 4$ for any $b \in [1,2]$. Furthermore, for the ease of impending computation, we will lower all powers of $\varepsilon$ that are greater than $\oh$ in the numerator to $\varepsilon^{\frac{1}{2}}$ since $\varepsilon < \frac{1}{2}$. For brevity, we will also define the unified coupling constant $E$ as

\begin{equation}\label{FET49}
E := \max\{1,\e,\e^2\}.
\end{equation}

With the above, we can simplify the constants in our estimates for $|I_i|$. We start by analyzing $|I_1|$ and get

\begin{equation}\label{FET50}
\begin{aligned}
|I_1|(u,u_0,v) &\leq \frac{16 \pi \e C_{7}r^{\frac{3}{2}-\frac{\chi}{2}}(0,v_*)}{3-\chi}  \sqrt{\frac{3(3-\delta)}{4\pi(1-\delta)(2-\delta)^2}} \frac{\varepsilon}{(1-\varepsilon)^{2}}   \frac{A'(u_1)}{r^{\delta}(u,v)} \leq \frac{C_{9,1}\varepsilon^{\frac{1}{2}}}{\delta(1 - \delta)} \frac{A'(u_1)}{r(u,v)^{\delta}}
\end{aligned}
\end{equation}

with constant $C_{9,1} > 0$ depending on $v_*$, $\chi$, $E$. Explicitly for $|I_1|$, this is given by

\begin{equation}
C_{9,1} = 16 \sqrt{6\pi} E C_{7} r^{\frac{3}{2}-\frac{\chi}{2}}(0,v_*).
\end{equation}

One can repeat a similar argument for $|I_i|$ with $i = 2,3,4,5,6,7,8,10$ and get constants $C_{9,i} > 0$ depending on $v_*$, $\chi$, and $E$.[Here, observe that $C_7$ also depends on $v_*, \chi, E$ by its explicit expression in (<ref>).] Note that each of these constants does not depend on $u_0$. For $i = 11$, we employ a similar bound without $A'(u_1)$ and obtain

\begin{equation}\label{FET51}
\begin{aligned}
|I_{11}|(u,u_0,v) &\leq \frac{2\e C_{7}r^{\delta - \frac{\chi}{2}}(0,v_*)}{\delta} \frac{\varepsilon^{\frac{1}{2}}}{1-\varepsilon} \frac{1}{r^{\delta}(u,v)} \leq \frac{C_{10,11}\varepsilon^{\frac{1}{2}}}{\delta(1 - \delta)} \frac{1}{r(u,v)^{\delta}}
\end{aligned}
\end{equation}

with constant $C_{10,11} > 0$ depending on $v_*$, $\chi$, $E$. For $i = 9$, there is a term without $\varepsilon$. Thus, we use the following upper bound

\begin{equation}\label{FET52}
\begin{aligned}
&\leq \ml C_{9,9} \frac{\varepsilon^{\frac{1}{2}}}{\delta(1 - \delta)} + C_{11,9}(u_0;\iota) \mr \frac{A'(u_1)}{r(u,v)^{\delta}}
\end{aligned}
\end{equation}

with $C_{9,9} > 0$ and $C_{11,9}(u_0) > 0$. Note that $C_{9,9}$ depends only on $v_*, \chi, E$, while $C_{11,9}$ depends additionally on $\iota$, $u_0$ and can be expressed as

\begin{equation}\label{FET72}
C_{11,9}(u_0;\iota) = \frac{32 \pi E C_{8}^2(\iota)}{1-\iota}r^{1-\iota}(u_0,v_*).
\end{equation}

Combining these estimates together with (<ref>), we derive

\begin{equation}\label{FET53}
A'(u_1) \leq D'(u_0) + \ml \frac{\sum_{i=1}^{10}C_{9,i}\varepsilon^{\frac{1}{2}} }{\delta(1-\delta)} + C_{11,9}(u_0;\iota)\mr A'(u_1) + \frac{ C_{10,11}}{\delta(1-\delta)}\varepsilon^{\frac{1}{2}}.
\end{equation}

Next, we demand the following constraints on the constants

\begin{equation}\label{FET54}
\frac{\sum_{i=1}^{10}C_{9,i}\varepsilon^{\frac{1}{2}} }{\delta(1-\delta)} < \frac{1}{4},
\end{equation}

\begin{equation}\label{FET73}
C_{11,9}(u_0;\iota) < \frac{1}{8}.\footnote{The choice of $\frac{1}{8}$ here is for simplifying parallel arguments in the next subsection for extension beyond $\mathcal{D}(0,v_*)$.}
\end{equation}

Note that (<ref>) can be rewritten as

\begin{equation}\label{FET74}
\varepsilon < \ml \frac{\delta(1-\delta)}{4\sum_{i=1}^{10}C_{9,i}}\mr^2.
\end{equation}

For (<ref>), this can be attained by picking $u_0$ sufficiently close to $v_*$ with using the explicit expression of $C_{11,9}(u_0)$ in (<ref>), in which for a fixed $\iota \in (0,1)$, the term $r^{1-\iota}(u_0,v_*)$ serves as a small factor which goes to $0$ as $u_0 \rightarrow v_*$. Henceforth, (<ref>) implies

\begin{equation}\label{FET56}
\frac{C_{10,11}}{\delta(1-\delta)}\varepsilon^{\frac{1}{2}} < \frac{C_{10,11}}{4\sum_{i=1}^{10}C_{9,i}} =: C_{12} < + \infty.
\end{equation}

From (<ref>) and (<ref>), if we pick

\begin{equation}\label{FET57}
\varepsilon' = \min\left\{\frac{1}{2},\ml \frac{\delta(1-\delta)}{4\sum_{i=1}^{10}C_{9,i}}\mr^2 \right\},
\end{equation}

then for all $\varepsilon < \varepsilon'$, we have that (<ref>) and (<ref>) hold. Plugging (<ref>), (<ref>), and (<ref>) back into (<ref>), we deduce that

\begin{equation}\label{FET58}
A'(u_1) \leq 2(D'(u_0) + C_{12}) =: C_{13}.
\end{equation}

In other words, $A'(u_1)$ is bounded by a constant depending only on $\delta$, $\iota$, $\chi$. Note that this bound is valid as long as we have $\chi$, $\delta$, $\iota$ satisfying $0< \chi < \delta < 1$ and $0 < \iota < 1$.

Estimates for $B'(u_1)$.
Next, we will further show that $B'(u_1)$ as defined in (<ref>) remains bounded. This is important for concluding the extension of the $C^1$ solution up to the boundary as this in turn implies that the solution in $\mathcal{D}(u_0,u_1,v_*)$ is indeed $C^1$. We start off by observing from (<ref>) and (<ref>) that

\begin{equation}\label{FET107}
\begin{aligned}
\frac{r |\dv \phi|}{\dv r}(u,v) &\leq C_{14}(\delta) r^{1-\delta}(u,v) \mbox{ with } C_{14}(\delta) := \frac{C_{13}}{1-\delta}.
\end{aligned}
\end{equation}

By (<ref>), from $m|_{\Gamma} = 0$, this implies

\begin{equation}\label{FET108}
m(u,v) \leq \int^v_u \ml \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r} + \frac{Q^2 \dv r}{2r^2} \mr(u,v') \D v'.
\end{equation}

The first term in (<ref>) can be estimated with the help of (<ref>) as

\begin{equation}\label{FET109}
\begin{aligned}
\int^v_u \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r}(u,v') \D v'&\leq 2\pi \int^v_u \frac{r^2|\dv \phi|^2}{(\dv r)^2}(\dv r)(u,v') \D v'\\
&\leq 2\pi \int^v_u C_{14}^2 r^{2-2\delta}(\dv r)(u,v') \D v'
= \frac{2\pi C_{14}^2}{3-2\delta}r^{3-2\delta}(u,v).
\end{aligned}
\end{equation}

For the second term, from (<ref>), we have

\begin{equation}\label{FET110}
\begin{aligned}
\int^v_u \frac{Q^2 \dv r}{2r^2}(u,v') \D v' \leq \frac{C_{6}^2}{2}\int^v_u r^{1-\chi}(\dv r)(u,v') \D v' = \frac{C_{6}^2}{2(2-\chi)}r^{2-\chi}(u,v).
\end{aligned}
\end{equation}

If we restrict our choice of $\delta$ to $\left( 0, \frac{1}{2} \right]$, from (<ref>), (<ref>), (<ref>), we observe that

\begin{equation}\label{FET111}
m(u,v) \leq \frac{C_{15}}{2} r^{2-\chi}(u,v)
\end{equation}

\begin{equation}\label{FET112}
C_{15}(\delta,\chi) := \frac{4\pi C_{14}^2}{3-2\delta}r^{1-2\delta+\chi}(0,v_*) + \frac{C_6^2}{(2-\chi)}.
\end{equation}

Consequently, we obtain

\begin{equation}\label{FET113}
\mu(u,v) \leq C_{15}r^{1-\chi}(u,v).
\end{equation}

To see how (<ref>) can help us to obtain a bound for $B'(u_1)$, we first revisit (<ref>) and observe that if we define

\begin{equation}\label{FET115}
D''(u_0) := \sup_{\{u_0\} \times [u_0,v_*]} (|\alpha'|) < +\infty,
\end{equation}

we then have

\begin{equation}\label{FET116}
|\alpha'(u,v)| \leq D''(u_0) + \sum_{i=1}^{11}|I_i|(u,u_0,v)
\end{equation}

with $I_i$ defined in (<ref>) remains unchanged. Analogous to (<ref>), taking supremum over $\mathcal{D}(u_0,u_1,v_*)$ yields

\begin{equation}\label{FET117}
B'(u_1) \leq D''(u_0) + \sup_{\mathcal{D}(u_0,u_1,v_*)} \ml \sum_{i=1}^{11}|I_i|\mr.
\end{equation}

By comparing (<ref>) with  (<ref>), we observe that the supremum is taken without an additional factor of $r^{\delta}(u,v)$ paired with each $|I_i|$.  This is where (<ref>) comes in. By viewing an upper bound for $\mu$ as that for $\varepsilon$, we are able to pick appropriate values of $\delta, \chi, \iota$ such that the term $\varepsilon^{\frac{1}{2}}$ could be replaced by sufficient powers of $r(u,v)$ to counteract the term $r^{-\delta}(u,v)$ in each of the estimates for $|I_{i}|$. The details for relevant terms are provided as follows.

$\ml\frac{r|\dv \phi|}{\dv r}\mr^2$: By (<ref>), we have

\begin{equation}\label{FET119}
\ml \frac{r|\dv \phi|}{\dv r}\mr^2 \leq C_{14}^2 r^{2-2\delta}(u,v).
\end{equation}

$\ml\frac{r|\dv \phi|}{\dv r}\mr^3$: By (<ref>), we obtain

\begin{equation}\label{FET118}
\ml \frac{r|\dv \phi|}{\dv r}\mr^3 \leq C_{14}^3 r^{3-3\delta}(u,v).
\end{equation}

$|Q|$: As (<ref>) was derived from a more general expression in Proposition <ref>, if we instead apply (<ref>), we would obtain

\begin{equation}\label{FET120}
|Q|(u,v) \leq C_6 C_{15}^{\frac{1}{2}} r^{2-\chi}(u,v).
\end{equation}

$|Q\phi|:$ Similar to $|Q|$, if we work through the corresponding arguments and apply (<ref>) instead, we get

\begin{equation}\label{FET121}
|Q\phi| \leq \frac{C_{7}C_{15}^{\frac{1}{2}}}{(1-\varepsilon)^{\frac{1}{2}}}r^{2-\chi}(u,v).
\end{equation}

$|\phi|^2:$ In the same fashion, we deduce

\begin{equation}\label{FET122}
\begin{aligned}
|\phi|^2(u,v) &\leq \frac{C_{15}}{2\pi}\frac{1}{1-\varepsilon} r^{1-\chi}(u,v) + 2C_{8}^2(\iota)r^{-1-\iota}(u,v) \leq \frac{C_{16}}{1-\varepsilon}r^{-1-\iota}(u,v)
\end{aligned}
\end{equation}

with $C_{16}(\chi,\iota,\delta) := \frac{C_{15}(\delta,\chi)}{2\pi}r^{2-\chi+\iota}(0,v_*) + 2 C_{8}^{2}(\iota).$

For the remaining terms with $A'(u_1)$, a direct application of (<ref>) suffices.

With the above preparation, we are now ready to derive improved estimates for $|I_{i}|$ with $i = 1$ to $11$. We summarize the bounds obtained for each $|K_i|(u,v)$ defined in (<ref>) and thus list the bounds for each $|I_i|(u,u_0,v)$ in Table <ref>. Here we use $\lesssim$ to refer to inequalities with constants depending only on $v_*$ and parameters $\delta, \chi, \iota$. Meanwhile, due to (<ref>), the dependence on $A'(u_1)$ will be encoded under $\lesssim$ as $C_{13}$ depends on $\delta, \chi, \iota, v_*$, while being independent of $u_1$. Furthermore, we recall that we require the parameters to satisfy $0 < \chi < \delta \leq \frac{1}{2}$ and $0 < \iota < 1$.

Index $i$     $\frac{|K_i|}{-\du r}(u,v) \lesssim$     $|I_i|(u,u_0,v) \lesssim$

1     $\frac{|Q||\phi|}{r^2} \ml \frac{r|\dv \phi|}{\dv r}\mr \lesssim r^{1-\chi-\delta}$     $r^{2-\chi - \delta}$

2     $\frac{|Q|^2}{r^4}\ml \frac{r|\dv \phi|}{\dv r}\mr^3 \lesssim r^{3-2\chi-3\delta}$     $r^{4-2\chi-3\delta}$

3     $\frac{1}{r^2}\ml \frac{r|\dv \phi|}{\dv r}\mr^3 \lesssim r^{1-3\delta}$     $r^{2-3\delta}$

4     $\frac{|\alpha'|}{r}\mu \lesssim r^{-\chi - \delta}$     $r^{1 - \chi - \delta}$

5     $\frac{|Q|^2}{r^3}|\alpha'|\lesssim r^{1-2\chi - \delta}$     $r^{2-2\chi-\delta}$

6     $\frac{|Q|^2}{r^4}\ml \frac{r|\dv \phi|}{\dv r}\mr \lesssim r^{1-2\chi-\delta}$    $r^{2-2\chi - \delta}$

7     $\frac{\mu}{r^2}\ml \frac{r|\dv \phi|}{\dv r}\mr \lesssim r^{-\chi - \delta}$    $r^{1-\chi-\delta}$

8     $\frac{|Q||\phi|}{r^2}\ml \frac{r|\dv \phi|}{\dv r}\mr^2 \lesssim r^{2-\chi-2\delta}$    $r^{3-\chi -2\delta}$

9     $|\phi|^2\ml \frac{r|\dv \phi|}{\dv r}\mr \lesssim r^{-\delta - \eta}$     $r^{1-\delta - \eta}$

10     $\frac{|Q|^2}{r^2}\ml \frac{r|\dv \phi|}{\dv r}\mr \lesssim r^{3-2\chi-\delta}$    $r^{4-2\chi-\delta}$

11     $\frac{|Q||\phi|}{r^2} \lesssim r^{-\chi}$     $r^{1-\chi}$

Table of improved estimates for $|I_i|$ in terms of powers of $r(u,v)$.

From Table <ref>, we observe the relevant improvement in the estimates for $|I_3|$, $|I_4|$, $|I_7|$, $|I_{11}|$. In addition, we see that

\begin{equation}
\sum_{i=1}^{11}|I_i|(u,u_0,v) \lesssim r^{1-\chi-\delta}(u,v) = r^{\frac{1}{4}}(u,v) \leq r^{\frac{1}{4}}(0,v_*) < +\infty
\end{equation}

if we pick $\delta = \frac{1}{2}$ and $\iota = \chi = \frac{1}{4}$. Henceforth, from (<ref>), we deduce that $B'(u_1)$ uniformly bounded in $u_1$. We then conclude that the solution extends up to $\mathcal{D}(0,v_*)$ as a $C^1$ solution.

Extension Beyond Boundary. Last but not least, we will show that there exists some $\xi' > 0$ such that the solution extends as a $C^1$ solution to $\mathcal{D}(0,v_* +  \xi')$. This is done by estimating the change in $\mu$ across a rectangular region $\{(u,v): u \in [0,v_*], v\in [v_*,v_*+\xi]\}$ for some $\xi > 0$ and through a contradiction argument. Furthermore, to maintain the generality of the arguments and to draw parallel conclusions made in the previous subsections, we will keep $\delta, \chi, \iota$ as parameters to be chosen, and it is implicitly assumed that all constants depend on $\delta, \chi, \iota$ unless stated otherwise. One can observe that the argument below holds if we retain our choice of $\delta = \frac{1}{2}$ and $\chi = \iota = \frac{1}{4}$ as in the previous subsection.

[gray!30] (0,1) to (1.5,2.5) to (1,3) to (0,2) to (0,1);
[gray!60] (1,3) to (1.5,3.5) to (2,3) to (1.5,2.5) to (1,3);
[->] (0,0) node[anchor=north]$\mathcal{O}$ –  (0,5) node[anchor = east]$\Gamma$;
[->] (0,0) – (4,4) node[anchor = west]$C_0^+$;
(0,4) – (2,2);
[] at (2.15,1.85) $v_*$;
[anchor = east] at (0,1) $(u_0,u_0)$;
at (0,1)[circle,fill,inner sep=1.5pt];
[anchor = east] at (0,2) $(u_1,u_1)$;
at (0,2)[circle,fill,inner sep=1.5pt];
[rotate=45] at (0.65,2.15) $\mathcal{D}(u_0,u_1,v_*)$;
[dotted] (0,2) – (1.5,3.5);
[] (0,1) – (2,3);
[] (1.5,3.5) – (2.5,2.5);
[anchor = west] at (2.5,2.5) $v_* + \xi$;
[->] (2,3.5) – (1.5,3);
[anchor = west] at (2,3.5)$U_{\xi}(u_1)$;
[pattern=north east lines, pattern color=black] (0,1) to (2,3) to (1.5,3.5) to (0,2) to (0,1);

[anchor = west] at (6.5,2) Legend:;
[anchor = west] at (6.9,1.5) $\mathcal{D}(u_0,u_1,v_*)$;
[gray!30] (6.65,1.6) to (6.85,1.6) to (6.85,1.4) to (6.65,1.4) to (6.65,1.6);
[anchor = west] at (6.9,1) $U_\xi(u_1)$;
[gray!60] (6.65,1.1) to (6.85,1.1) to (6.85,0.9) to (6.65,0.9) to (6.65,1.1);
[anchor = west] at (6.9,0.5) $\mathcal{D}(u_0,u_1,v_*+\xi)$;
[pattern=north east lines, pattern color=black] (6.65,0.6) to (6.85,0.6) to (6.85,0.4) to (6.65,0.4) to (6.65,0.6);

Illustration of $U_\xi(u_1)$ and $\mathcal{D}(u_0,v_1,v_* + \xi)$.

First, we consider a $u_0 \in [0,v_*)$ sufficiently close to $v_*$ so that (<ref>) holds. Next, we use the continuity of $\dv r$ to deduce that $\dv r \geq 0$ on $\{u_0\} \times [v_*, v_* + \Delta v]$ for some fixed $\Delta v > 0$. Then, for any $0<\xi < \Delta v$, we consider the corresponding rectangular strip given by

\begin{equation}\label{FET75}
U_{\xi}(u_1) = [u_0,u_1) \times [v_*,v_*+\xi].
\end{equation}

In the spirit of (<ref>), we define

\begin{equation}\label{FET76}
\overline{A'_{\xi}}(u_1) := \sup_{U_{\xi}(u_1)} (r^{\delta}|\alpha'|).
\end{equation}

We also note that, under the convention in the previous subsections, we can write

\begin{equation}\label{FET77}
\mathcal{D}(u_0,u_1,v_* + \xi) = \mathcal{D}(u_0,u_1,v_*) \cup U_{\xi}(u_1).
\end{equation}

In below we will show that $\overline{A'_{\xi}}(u_1)$ remains bounded for some $u_1 \in (u_0,v_*)$ sufficiently close to $u_0$. This can be proved by utilizing the continuity of $\mu$ on $\mathcal{Q}$. With $\varepsilon$ and $\varepsilon'$ as in the previous subsections, if we pick $v_* - u_0$ and $\xi$ to be sufficiently small, then we have

\begin{equation}\label{FET78}
\varepsilon''(u_1) := \sup_{\mathcal{D}(u_0,u_1,v_*+\xi)} \mu \leq \frac{1}{2}(\varepsilon + \varepsilon') < \varepsilon'.
\end{equation}

Later, we will show that $\varepsilon''(u_1)$ defined above obeys the same upper bound as in (<ref>) for all $u_1 \in (u_0,v_*)$.

Before we begin detailing the arguments, we will first state and prove two lemmas that would be useful. We first define

\begin{equation}\label{FET83}
A'_* := \sup_{u_1\in(u_0,v_*)}A'(u_1) < +\infty.
\end{equation}

Here $A'_*$ is independent of $u_1$ as we have showed in (<ref>) that $A'(u_1)$ is uniformly bounded in $u_1$ by $C_{13}$.

For any $u_1 \in (u_0,v_*)$, if $\mu$ satisfies

\begin{equation}\label{FET79}
\sup_{\mathcal{D}(u_0,u_1,v_* + \xi)} \mu \leq \frac{1}{2}(\varepsilon + \varepsilon'),
\end{equation}

then we have

\begin{equation}\label{FET80}
\overline{A'_{\xi}}(u_1) \leq \max\left\{A'_*, C_{17}(u_0,v_* + \Delta v) \right\}
\end{equation}

with $C_{17}$ defined in (<ref>).

$ $  Similar to (<ref>), we start by defining

\begin{equation}\label{FET81}
\overline{D'_\xi}(u_0) := \sup_{\{u_0\}\times[v_*,v_*+\xi]}(r^{\delta}|\alpha'|).
\end{equation}

Then, repeating the same derivations for (<ref>) and  (<ref>), but instead with $u_1 \in (u_0,v_*)$ and $v \in [v_*,v_*+\xi]$, we get

\begin{equation}\label{FET82}
|\alpha'|(u,v) \leq \overline{D'_\xi}(u_0)r(u_0,v)^{-\delta} + \sum_{i=1}^{11}|I_i|(u,u_0,v).
\end{equation}

In (<ref>), the expressions for $I_i$ remains unchanged as in (<ref>) to (<ref>). Now, we will obtain an upper bound for each of the terms $|I_1|$ to $|I_{11}|$ as in the previous subsections. A few key differences are listed below.

For $\mu$, we apply (<ref>) and replace each $\varepsilon$ in the estimates obtained above by

\begin{equation}\label{tildeepsilon}
\tilde{\varepsilon} := \frac{1}{2}(\varepsilon + \varepsilon').
\end{equation}

This is justified since instead of $\mu \leq \varepsilon$, we now have $\mu \leq \frac{1}{2}(\varepsilon + \varepsilon')$. Hence, in the region $\mathcal{D}(u_0,u_1,v_*+\xi)$ as the supremum of the quantity $r^{\delta}|\alpha'|$ is bounded above by the maximum of its supremum over $\mathcal{D}(u_0,u_1,v_*)$ and its supremum over $U_\xi(u_1)$ (see Figure <ref>), we obtain

\begin{equation}\label{FET84}
\sup_{\mathcal{D}(u_0,u_1,v_*+\xi)}(r^{\delta}|\alpha'|) \leq \max\{ A'(u_1), \overline{A'_\xi}(u_1)\} \leq \max\{A'_*, \overline{A'_\xi}(u_1)\}.
\end{equation}

For the remaining terms, an analogous logic to the previous sections applies with a slight modification demonstrated as follows. For $(u,v) \in U_\xi(u_1)$, we estimate $\alpha'$ over the entire region $\mathcal{D}(u_0,u_1,v_*+\xi)$ to derive relevant estimates for $\ml \frac{r|\dv \phi|}{\dv r}\mr$, $|Q|$, $|Q\phi|$, and consequently $\ml \frac{r|\dv \phi|}{\dv r}\mr^2$, $\ml \frac{r|\dv \phi|}{\dv r}\mr^3$, $|\phi|^2$. Thus, by (<ref>), we can replace $A'(u_1)$ from (<ref>) to (<ref>) by $\max\{A'_*, \overline{A'_\xi}(u_1)\}$.

With above pointers, we then multiply $r^\delta$ on both sides of (<ref>) and take supremum over $U_{\xi}(u_1)$ to obtain

\begin{equation}\label{FET85}
\begin{aligned}
\overline{A'_\xi}(u_1)  &\leq \overline{D'_\xi}(u_0) + \ml \frac{ \sum_{i=1}^{10}C_{9,i}(v_*+\xi)}{\delta(1-\delta)}\tilde{\varepsilon}^{\frac{1}{2}} + C_{11,9}(u_0;v_*+\xi)\mr \max\{A'_*, \overline{A'_\xi}(u_1)\} + \frac{C_{10,11}(v_*+\xi)}{\delta(1-\delta)} \tilde{\varepsilon}^{\frac{1}{2}} \\
&\leq \overline{D'_\xi}(u_0) + \ml \frac{ \sum_{i=1}^{10}C_{9,i}(v_* + \Delta v)}{\delta(1-\delta)}\tilde{\varepsilon}^{\frac{1}{2}} + C_{11,9}(u_0;v_* + \Delta v)\mr \max\{A'_*, \overline{A'_\xi}(u_1)\} + \frac{C_{10,11}(v_* + \Delta v)}{\delta(1-\delta)} \tilde{\varepsilon}^{\frac{1}{2}},
\end{aligned}
\end{equation}

where we use relevant estimates in the previous subsections, subject to the remarks above. Moreover, as explained in the previous subsections, the constants derived are non-decreasing with increasing $v_*$, which thus allows us to obtain the $\xi-$independent estimates for the constants $C_{9,i}, C_{10,11}, C_{11,9}$. By further refining $\Delta v > 0$ from the start, we could also assume that

\begin{equation}\label{FET89}
C_{11,9}(u_0, v_* + \Delta v) < \frac{1}{4},
\end{equation}

which is analogous to (<ref>). Note that inequality (<ref>) bears resemblance to (<ref>).

With a similar choice of $\varepsilon'$ as in (<ref>) but with constants evaluated at $(u_0,v_*+\Delta v)$, we then deduce that

\begin{equation}\label{FET86}
\varepsilon' \leq \ml \frac{\delta(1-\delta)}{ \sum_{i=1}^{10}C_{9,i}(u_0,v_*+\Delta v)} \mr^2.
\end{equation}

This implies that the counterpart of (<ref>) is still true, if we replace $\varepsilon$ by $\tilde{\varepsilon}$ and it satisfies

\begin{equation}\label{FET87}
\tilde{\varepsilon} = \frac{1}{2}(\varepsilon + \varepsilon') < \varepsilon' \leq \ml \frac{\delta(1-\delta)}{ \sum_{i=1}^{10}C_{9,i}(u_0,v_*+\xi)} \mr^2.
\end{equation}

This implies

\begin{equation}\label{FET88}
\frac{\ml \sum_{i=1}^{10}C_{9,i}(u_0,v_*+\xi)\mr^2}{\delta^2(1-\delta)^2} < \frac{1}{4}.
\end{equation}

Parallel to (<ref>), we also define

\begin{equation}\label{FET124}
C_{17}(u_0,v_*+\Delta v) := 2 \ml \overline{D'_\xi}(u_0) + C_{12}(v_* + \Delta v) \mr.
\end{equation}

By considering two separate cases $A_*' > \overline{A'_{\xi}(u_1)}$ and $A_*' \leq \overline{A'_{\xi}(u_1)}$, following a similar argument as in the previous subsections, analogous to (<ref>), we then deduce that (<ref>) holds. This concludes the proof of this lemma.

We will also equip ourselves with an elementary lemma as follows:

For any $0 < y \leq x$ and $0 \leq p \leq 1$, we have

\begin{equation}\label{FET125}
x^p - y^p \leq (x-y)^p.
\end{equation}

Dividing $y^p$ throughout (<ref>), it is equivalent to show that, for any $r \geq 1$, we have

\begin{equation}\label{FET126}
r^p - 1 \leq (r - 1)^p.
\end{equation}

Let $f(r) = r^p - 1 - (r-1)^p$. For $r \geq 1$, we see that $f(1) = 0$ and $f'(r) = pr^{p-1}\ml 1-\ml 1-\frac{1}{r}\mr^{p-1} \mr \leq 0$. This implies that $f$ is a non-increasing function of $r$ since $1 \leq \ml 1 - \frac{1}{r}\mr^{p-1}$ for $p \in [0,1]$. Thus, we prove $f(r) \leq f(1) = 0$ for all $r \geq 1$ and conclude.

With the above two lemmas, we proceed back on track to prove that the $C^1$ solution extends beyond the domain of dependence $\mathcal{D}(u_0,u_1,v_*)$. Similar to [19], we first define

\begin{equation}\label{FET90}
u_2 = \sup\left\{ u_1 \in (u_0,v_*): \sup_{\mathcal{D}(u_0,u_1,v_*+\xi)} \mu \leq \frac{1}{2}(\varepsilon + \varepsilon')\right\}.
\end{equation}

By the definition of $u_2$, we then have either $u_2 = v_*$ or

\begin{equation}\label{FET91}
\sup_{U_\xi (u_2)} \mu = \frac{1}{2}(\varepsilon + \varepsilon')
\end{equation}

is attained at some $u_2 \in [u_0,v_*)$. We then show that for $\xi$ sufficiently small, the latter will not happen.

Suppose for a contradiction that the latter happens. Then, we set $u_1 = u_2$ as in the hypothesis of Lemma <ref>. Consequently,  by (<ref>), $\sup_{\mathcal{D}(u_0,u_2,v_*+\xi)} \mu$ satisfies (<ref>) and hence by Lemma <ref>, we have that (<ref>) holds. We then estimate $\mu$ in the region $U_\xi(u_2)$ to obtain a contradiction with sufficiently small $\xi$. As described in the proof of Lemma <ref>, in the region $U_\xi(u_2)$, we have

\begin{equation}\label{FET92}
\frac{r|\dv \phi|}{\dv r}(u,v) \leq \frac{r^{1-\delta}(u,v)}{2-\delta} \max\{A'_*, C_{12}\}
\end{equation}

with the argument $(u_0,v_*+\Delta v)$ suppressed when the context is clear. We can then estimate $\mu$ in $U_\xi(u_2)$ as follows. From (<ref>) we first observe that

\begin{equation}\label{FET127}
\frac{Q^2}{r^2}(u,v) \leq \mu(u,v) C_6^2(v_*+\Delta v)r^{1-\chi}(u,v).
\end{equation}

By further requiring $u_0$ to be sufficiently close to $v_*$, we can then demand that

\begin{equation}\label{FET128}
C_6^2(v_*)r^{\frac{1}{2}-\frac{\chi}{2}}(u_0,v_*) \leq \frac{1}{2}.
\end{equation}

Hence, from the expression of $C_6$ in (<ref>), we see that the mapping $v \mapsto C_6^2(v)r^{1-\chi}(u_0,v)$ is continuous. This allows us to refine $\Delta v$ to be sufficiently small (depending on the given $u_0$) such that

\begin{equation}\label{FET132}
C_6(v_* + \Delta)r^{1-\chi}(u_0,v_* + \Delta) \leq \frac{7}{8},
\end{equation}

and consequently,

\begin{equation}\label{FET129}
\frac{Q^2}{r^2}(u,v) \leq \frac{7}{8}\mu(u,v).
\end{equation}

This in turn implies that

\begin{equation}\label{FET131}
\ml \mu - \frac{Q^2}{r^2}\mr(u,v) \geq \frac{\mu}{8}(u,v) \geq 0
\end{equation}

for all $(u,v) \in U_\xi(u_2)$.
With this and the help of (<ref>), we move on to estimate $\mu$ directly via calculating

\begin{equation}\label{FET93}
\begin{aligned}
\dv \mu &= \frac{4\pi r (1-\mu)|\dv \phi|^2}{\dv r} - \frac{\dv r}{r}\ml \mu - \frac{Q^2}{r^2}\mr \leq \frac{4\pi r (1-\mu)|\dv \phi|^2}{\dv r}.
\end{aligned}
\end{equation}

In (<ref>), the inequality follows from the fact that $\dv r > 0$ while $\mu - Q^2/r^2 \geq 0$.
For $(u,v) \in U_\xi(u_2)$, we further perform integration and apply (<ref>) and (<ref>) to obtain

\begin{equation}\label{FET94}
\begin{aligned}
\mu(u,v) &\leq \mu(u,v_*) + 4 \pi \int^v_{v_*} \frac{(1-\mu)}{r} \ml\frac{r|\dv \phi|}{\dv r}\mr^2 (\dv r) (u,v') \; \D v' \\
&\leq \mu(u,v_*) + \frac{4 \pi (\max\{A'_*,C_{17}  \})^2}{(2-\delta)^2} \int^v_{v_*} r^{1 - 2 \delta} (\dv r) (u,v') \; \D v' \\
&= \mu(u,v_*) + \frac{4 \pi (\max\{A'_*,C_{17}\})^2}{2(2-\delta)^2(1-\delta)} \ml r^{2-2\delta}(u,v) - r^{2-2\delta}(u,v_*) \mr. \\
\end{aligned}
\end{equation}

On the other hand, since $\dv r$ is $C^1$ for $(u,v) \in U_\xi(u_2)$, we have $\dv r(u_0,v)$ is $C^1$ on $[v_*,v_*+ \Delta v]$. Thus, there exists $C_{18}(u_0) > 0$, such that $\dv r(u_0,v) \leq C_{18}(u_0)$ for all $v \in [v_*,v_*+\Delta v]$. Hence, for $(u,v) \in U_\xi(u_2)$, using the fact that $\dv \du r \leq 0$ as described above, we obtain

\begin{equation}\label{FET96}
\begin{aligned}
r(u,v) - r(u,v_*) &= r(u_0,v) - r(u_0,v_*) + \int^u_{u_0} \int^v_{v_*} \dv \du r(u',v') \; \D u' \D v' \\
&\leq r(u_0,v) - r(u_0,v_*) = \int_{v_*}^v \dv r (u_0,v') \; \D v' \\
&\leq C_{18}(u_0)(v - v_*) \leq C_{18}\xi.
\end{aligned}
\end{equation}

Now, for $0 < \delta < \frac{1}{2}$ and hence $2 - 2\delta > 1$, we employ (<ref>) and (<ref>) to obtain

\begin{equation}\label{FET95}
\begin{aligned}
\mu(u,v) &\leq \mu(u,v_*) + \frac{4 \pi (\max\{A'_*, C_{17}  \})^2}{2(2-\delta)^2(1-\delta)} (r^{2-2\delta}(u,v) - r^{2-2\delta}(u,v_*)) \\
&\leq \mu(u,v_*) + \frac{4 \pi (\max\{A'_*,C_{17}\})^2}{2(2-\delta)^2(1-\delta)} r^{2-2\delta-1}(u,v_* + \Delta v)(r(u,v) - r(u,v_*)) \\
&\leq \mu(u,v_*) + \frac{4 \pi (\max\{A'_*,C_{17}\})^2 C_{18}}{2(2-\delta)^2(1-\delta)} r^{2-2\delta-1}(u_0,v_*+ \Delta v)\xi.\\
\end{aligned}
\end{equation}

Since $\mu(u,v_*) \leq \varepsilon$, to obtain $\mu(u,v) < \frac{1}{2}(\varepsilon + \varepsilon')$, it suffices to pick $\xi > 0$ sufficiently small, such that

\begin{equation}\label{FET97}
\frac{4 \pi (\max\{A'_*,C_{17}  \})^2 C_{18}}{2(2-\delta)^2(1-\delta)} r^{2-2\delta-1}(u_0,v_*+\Delta v)\xi < \frac{1}{2}(\varepsilon' - \varepsilon).
\end{equation}

In contrast, for the case $\delta \geq \frac{1}{2}$ and hence $2 - 2 \delta \leq 1$, we can apply Lemma <ref> and (<ref>) to obtain

\begin{equation}\label{FET100}
\begin{aligned}
\mu(u,v) &\leq \mu(u,v_*) + \frac{4 \pi (\max\{A'_*,C_{17}  \})^2}{2(2-\delta)^2(1-\delta)} \ml r^{2-2\delta}(u,v) - r^{2-2\delta}(u,v_*) \mr\\
&\leq  \mu(u,v_*) + \frac{4 \pi (\max\{A'_*, C_{17} \})^2}{2(2-\delta)^2(1-\delta)} \ml r(u,v) - r(u,v_*) \mr^{2-2\delta} \\
&\leq  \mu(u,v_*) + \frac{4 \pi (\max\{A'_*, C_{17} \})^2 C_{18}^{2-2\delta}}{2(2-\delta)^2(1-\delta)} \xi^{2-2\delta}.
\end{aligned}
\end{equation}

Similarly to the former case, it suffices to pick $\xi > 0$ sufficiently small such that

\begin{equation}\label{FET101}
\frac{4 \pi (\max\{A'_*,C_{17}  \})^2 C_{18}^{2-2\delta}}{2(2-\delta)^2(1-\delta)} \xi^{2-2\delta} < \frac{1}{2}(\varepsilon' - \varepsilon).
\end{equation}

Combining (<ref>) and (<ref>), while requiring $\xi$  to be less than or equals to $\Delta v$, a valid choice of $\xi$ would be

\begin{equation}\label{FET102}
\xi = \min\left\{ \Delta v, \frac{(2-\delta)^2(1-\delta)(\varepsilon'-\varepsilon)}{8\pi(\max\{A'_*,C_{17} \})^2 C_{18}r^{2-2\delta-1}(u_0,v_*+\xi_1)}, \ml \frac{(2-\delta)^2(1-\delta)(\varepsilon'-\varepsilon)}{8\pi(\max\{A'_*,C_{17}
\})^2 C_{18}^{2-2\delta}}\mr^{\frac{1}{2-2\delta}} \right\}.
\end{equation}

Therefore, for any $\delta \in (0,1)$, such a choice $\xi$ would imply that

\begin{equation}\label{FET103}
\sup_{U_\xi (u_2)} \mu < \frac{1}{2}(\varepsilon + \varepsilon'),
\end{equation}

contradicting (<ref>).

In summary, the above arguments show that $u_2$ in (<ref>) is achieved at $u_2 = v_*$, and hence the bound in (<ref>) continues to hold. This implies that

\begin{equation}\label{FET104}
\sup_{\mathcal{D}(u_0,v_*,v_*+\xi)} \mu \leq \frac{1}{2}(\varepsilon
+ \varepsilon').
\end{equation}

Thus, for each $u_1 \in (u_0,v_*)$, the hypothesis of Lemma <ref> is verified. This then implies that (<ref>) holds. Observe that (<ref>) further implies that  $\overline{A'}(u_1) := \sup_{\mathcal{D}(u_0,u_1,v_*+\Delta v)}(r^{\delta}|\alpha'|)$ remains bounded as $u_1 \rightarrow v_*$. By conducting a similar argument as in previous subsections, we can also obtain improved estimates for $|I_1|$ to $|I_{11}|$. Henceforth, we can deduce that the solution extends as a $C^1$ solution to the closure of $\mathcal{D}(u_0,v_*,v_*+\xi)$. In particular, with $C^1$ initial data along $\{v_*\} \times [v_*,v_*+\xi]$, we show that the solution further extends as a $C^1$ solution to $\mathcal{D}(v_*,v_*+\xi')$ for some $\xi' \in (0,\xi]$. This concludes the proof of Theorem <ref>.

§ FIRST INSTABILITY THEOREM

As previously described, we consider the following picture. We mark $\mathcal{O}' = (v_0,v_0)$ along $\Gamma$ as the start of the singular boundary $\mathcal{B}$ and $\mathcal{J}^-(\mathcal{O}') \subset \mathcal{R}$. Denote $C_0^-$ to be the past-directed incoming null curve emitting from $\mathcal{O}'$. For the remaining parts of this section, we shall focus on $\mathcal{Q}\setminus \mathcal{J}^-(\mathcal{O}')$ lying in the future of both $C_0^+$ and $C_0^-$. Figure <ref> illustrates the above set-up.

dot/.style = draw, fill = white, circle, inner sep = 0pt, minimum size = 4pt, scale=0.8]
[gray!50]  (2.5,2.5) to (0,5) to (1.7,6.7) node[dot] to [out=45,in=175] (4.7,7.0) to (5,5) to (2.5,2.5);
[thick] (0,0) node[anchor=north]$\mathcal{O}$ –  (0,5);
[] at (-0.3,2.5) $\Gamma$;
[] at (-0.3,5.0) $\mathcal{O}'$;
[] at (1,2.5) $\mathcal{J}^-(\mathcal{O}')$;
[->] (0,0) – (5,5) node[anchor = west]$C_0^+$;
[thick, densely dotted] (0,5) to (1.7,6.7);
[thick, loosely dotted](1.7,6.7) to [out=45,in=175] (4.7,7.0);
[] at (0.5,6) $\mathcal{B}_0$;
[rotate=-10] at (4.3,7.3) $\mathcal{B}\setminus\mathcal{B}_0$;
[thick,dashed] (0,5) –  (2.5,2.5);
[] at (2.7,2.3)$v_0$;
[] at (2.7,4.8)$\mathcal{Q} \setminus \mathcal{J}^{-}(\mathcal{O}')$;
[fill=white] (1.7,6.7) circle (2pt);
[fill=white] (0,5) circle (2pt);

Diagram for this section. The shaded region corresponds to the region $\mathcal{Q} \setminus \mathcal{J}^-(\mathcal{O}')$, while the unshaded region corresponds to the region $\mathcal{J}^-(\mathcal{O}')$.

Recall the quantity $\gamma$ from (<ref>) below:

\begin{equation}\label{FIT3}
\gamma(u,v) := \int^u_{0} \frac{(- \du r)}{r} \frac{\mu - Q^2/r^2}{1 - \mu} (u',v) \mathrm{d}u'.
\end{equation}

For notational simplicity, we denote

\begin{equation}\label{FIT4}
\gamma_0(u) := \gamma(u,v_0).
\end{equation}

In lieu of estimates that we will establish, we also define the truncated gamma $\tgam$, with $\overline{u} \in [0,v_0)$ as a parameter, as below

\begin{equation}\label{FIT4a}
\tgam(u,v) := \gamma(u,v;\overline{u}) = \int^u_{\overline{u}} \frac{(- \du r)}{r} \frac{\mu - Q^2/r^2}{1 - \mu} (u',v)\mathrm{d}u'.
\end{equation}

For brevity, the dependence on $\overline{u}$ will be dropped if the given context is clear. In addition, similar to Section <ref>, we use the subscript $()_1$ to indicate that a given quantity is evaluated along $C_0^-$.

We also use $g(u,v)$ and $f(u,v)$ to denote the following functions

\begin{equation}\label{FIT10}
g(u,v) := D_u \phi(u,v) +  \ii \e \frac{Q \phi (\du r)}{r(1-\mu)}(u,v),
\end{equation}

\begin{equation}\label{FIT11}
f(u,v) := \int^u_{0} -\ii \e A_u(u',v) + \frac{(-\du r)}{r} \frac{\mu - Q^2/r^2}{1 - \mu}(u',v) \mathrm{d}u'.
\end{equation}

With them, we define

\begin{equation}\label{FIT14}
I(u) := \int^u_{0} g_0(u')e^{-f_0(u')}\mathrm{d}u'.
\end{equation}

We are now ready to state the main theorem as follows.

(First Instability Theorem.)
Let $\gamma_0(u)$ defined in (<ref>) be unbounded as $u \rightarrow v_0^-$. Suppose that $I$ defined in (<ref>) does not converge to a finite limit as $u \rightarrow v_0^-$ or otherwise

\begin{equation}\label{FITTheorem1}
\ml \frac{r \dv \phi}{\dv r} \mr_0 (0) \neq \lim_{u \rightarrow v_0^-} I(u).
\end{equation}

Then, we can conclude that

* $\mathcal{A}$ is non-empty,

* Either one of the following two scenarios holds:

* $\mathcal{O}' = \mathcal{B}_0$ (and thus both $\mathcal{B}\setminus \mathcal{B}_0$ and $\mathcal{A}$ issue from $\mathcal{O}'$), or

* $\mathcal{O}' \subsetneq \mathcal{B}_0$ and for any sufficiently small open neighborhood $U$ of $\mathcal{O}'$ with $U \subset \mathcal{Q} \setminus \mathcal{J}^-(\mathcal{O}')$, we have $U \cap \mathcal{A} \neq  \emptyset$.

Before we begin with the proof of the theorem, we first establish several useful lemmas.

For the spherically symmetric Einstein-Maxwell-charged scalar field system, if $\mathcal{A} = \emptyset$, then we have $\mathcal{B} \setminus \mathcal{B}_0 = \emptyset$ and $\mathcal{B}_0$ is an outgoing null segment originating from $\mathcal{O}'$ and extends all the way to $v \rightarrow \infty$.

By Proposition <ref>, under the assumption of this lemma, we have that $\mathcal{A} \cup \mathcal{T} = \emptyset$.
For any potential point $p \in \mathcal{B} \setminus \mathcal{B}_0$, we have $r(p) = 0$. From $\mathcal{A} \cup \mathcal{T} = \emptyset$, we also get $\dv r > 0$ everywhere. Henceforth, any outgoing null curve originating from $\Gamma$ will never hit $\mathcal{B} \setminus \mathcal{B}_0$.

Along any incoming null curve with $v \geq v_0$, for all $0 \leq u < v_0$, if $(u',v) \in \mathcal{R}$ for all $u' \in [0,u]$, then we have

\begin{equation}\label{FIT18a}
(\dv r) (u,v) = (\dv r)(0,v) e^{-\gamma(u,v)}.
\end{equation}

In particular, along $v = v_0$, we deduce

\begin{equation}\label{FIT18}
(\dv r)_0 (u) = (\dv r)_0(0) e^{-\gamma_0(u)}
\end{equation}

Furthermore, for any $\overline{u} \in [0,v_0)$, with $\tgam$ defined in (<ref>), we also get

\begin{equation}\label{FIT18b}
(\dv r)(u,v) = (\dv r)(\overline{u},v) e^{-\tgam(u.v)}.
\end{equation}

Using (<ref>), we can rearrange its expression to obtain

\begin{equation}\label{FIT19}
\du \log \ml \dv r\mr = \frac{\du r}{r}\ml \frac{\mu-Q^2/r^2}{1-\mu}\mr.
\end{equation}

Integrating (<ref>), we retrieve (<ref>), (<ref>) and (<ref>).

Along any incoming null curve with $v \geq v_0$, for all $0 \leq \overline{u} \leq u < v_0$, if $(u',v) \in \mathcal{R}$ for all $u' \in [\overline{u},u]$, we have

\begin{equation}\label{FIT5a}
\frac{1}{1-\mu(u,v)} \leq \frac{1}{1-\mu(\overline{u},v)}e^{\tgam(u,v)}.
\end{equation}

In particular, we also get

\begin{equation}\label{FIT5}
\frac{1}{1-\mu(u,v)} \leq \frac{1}{1-\mu(0,v)}e^{\gamma(u,v)}.
\end{equation}

With the help of (<ref>), we compute

$$\du\ml \frac{1}{1-\mu(u,v)}\mr = \frac{\du \mu}{(1 - \mu)^2} = \frac{1}{1-\mu(u,v)} \ml -\frac{4\pi r|D_u \phi|^2}{(-\du r)} + \frac{(-\du r)}{r}\frac{\mu - Q^2/r^2}{1-\mu}\mr.$$

This gives

\begin{equation}\label{FIT6}
\begin{aligned}
\frac{1}{1-\mu(u,v)} &= \frac{1}{1-\mu(\overline{u},v)}\exp\ml {\int^u_{\overline{u}} \ml -\frac{4\pi r|D_u \phi|^2}{(-\du r)} + \frac{(-\du r)}{r}\frac{\mu - Q^2/r^2}{1-\mu}\mr (u',v) \mathrm{d}u'} \mr.
\end{aligned}
\end{equation}

Since $\du r < 0$, this implies

\begin{equation}\label{FIT7}
\begin{aligned}
\frac{1}{1-\mu(u,v)} &\leq \frac{1}{1-\mu(\overline{u},v)}\exp\ml {\int^u_{\overline{u}} \ml  \frac{(-\du r)}{r}\frac{\mu - Q^2/r^2}{1-\mu}\mr (u',v) \mathrm{d}u'} \mr. \\
\end{aligned}
\end{equation}

Employing the definition of $\gamma$ in (<ref>) yields (<ref>) and (<ref>).

Observe that these two lemmas were established when we were proving Theorem <ref>, albeit only for $\overline{u} = 0$.

Next, we remark that the definition of $\gamma$ in (<ref>) naturally arises from the evolution equation for $\frac{r \dv \phi}{\dv r}$ along the $u$-direction. To see this, by (<ref>) and (<ref>), we get

\begin{equation}\label{FIT8}
\begin{aligned}
\du \ml \frac{r \dv \phi}{\dv r} \mr &= \frac{\du (r\dv \phi)}{\dv r} + \frac{r \dv \phi (- \du \dv r)}{(\dv r)^2} \\
&= \frac{- \dv r D_u \phi + \ii \e \frac{Q\phi (-\du r)(\dv r)}{r(1-\mu)}}{\dv r} - \ii \e A_u \frac{r \dv \phi}{\dv r} + \frac{\dv \phi (- \du r)}{(\dv r)} \frac{\mu - Q^2/r^2}{1-\mu} \\
&= -D_u \phi - \ii \e \frac{Q \phi (\du r)}{r(1-\mu)} + \ml \frac{r \dv \phi}{\dv r}\mr \ml -\ii \e A_u + \frac{(-\du r)}{r} \frac{\mu - Q^2/r^2}{1 - \mu} \mr. \\
\end{aligned}
\end{equation}

Using the definition of $f$ and $g$ in (<ref>) and (<ref>), we obtain

\begin{equation}\label{FIT9}
\ml \frac{r \dv \phi}{\dv r} \mr(u,v)  = \ml\ml \frac{r \dv \phi}{\dv r} \mr(0,v) - \int^u_{0} g(u',v)e^{-f(u',v)}\mathrm{d}u' \mr e^{f(u,v)}
\end{equation}

Taking absolute values, we arrive at

\begin{equation}\label{FIT12}
\ml \frac{r |\dv \phi|}{|\dv r|} \mr(u,v)  = \mlm \ml \frac{r \dv \phi}{\dv r} \mr(0,v) - \int^u_{0} g(u',v)e^{-f(u',v)}\mathrm{d}u' \mrm e^{\gamma(u,v)}.
\end{equation}

Employing $(\dv r)_0 > 0$ in lieu of Lemma <ref>, we then derive

\begin{equation}\label{FIT13}
\ml \frac{r |\dv \phi|}{\dv r} \mr_0(u)  = \mlm \ml \frac{r \dv \phi}{\dv r} \mr_0(0) - I(u) \mrm e^{\gamma_0(u)}
\end{equation}

with $I(u)$ defined in (<ref>).

Last but not least, we have the following lemma

Along $v = v_0$, as $u \rightarrow v_0^-$, we have that

\begin{equation}\label{FIT57}
\gamma_0(u) < + \infty \quad \mbox{ implies } \quad \mu_0(u) \rightarrow 0.
\end{equation}

By Proposition <ref>, with $\overline{u} = 0$ and $\chi = \frac{1}{2}$, inequality (<ref>) implies that

\begin{equation}\label{FIT220}
\ml \frac{Q^2}{r^2} \mr_0 (u) \leq C_1^2 \ml 0,v_0;\frac{1}{2}\mr r_0^{\frac{1}{2}}(0)v_0 r_0^{\frac{1}{2}}(u) \mu_0(u).
\end{equation}

We then pick $u_1 \in [0,v_0)$ with $u_1$ sufficiently close to $v_0$, such that $C_1^2 \ml 0,v_0;\frac{1}{2}\mr r_0^{\frac{1}{2}}(0)v_0 r_0^{\frac{1}{2}}(u) \leq \frac{1}{2}$ for all $u \in [u_1,v_0)$. This simplifies (<ref>) to

\begin{equation}\label{FIT221}
0 \leq \ml \frac{Q^2}{r^2} \mr_0 (u) \leq \frac{\mu_0(u)}{2}.
\end{equation}

Recalling the definition of $\gamma_0$ in (<ref>), with a fixed $u_2 \in [u_1,v_0)$, for all $u \in [u_2,v_0)$, we have

\begin{equation}\label{FIT58}
\begin{aligned}
\gamma_0(u) - \gamma_0(u_2) &= \int^u_{u_2} \frac{(-\du r)_0}{r_0} \frac{(\mu - Q^2/r^2)_0}{1-\mu_0}(u')\;\D u' \geq \frac{1}{2}\int^u_{u_2} \frac{(-\du r)_0}{r_0} \frac{\mu_0}{1-\mu_0}(u')\;\D u' \\
&\quad\quad\geq \frac{1}{2}\int^u_{u_2} \frac{(-\du r)_0(u')}{r_0(u')} \frac{\frac{2m_0(u)}{r_0(u')}}{1-\frac{2m_0(u)}{r_0(u')}}\;\D u' = \frac{1}{2}\log \ml \frac{r_0(u')}{r_0(u')-2m_0(u)}\mr|^{u' = u}_{u' = u_2} \\
&\quad\quad\quad\quad= \frac{1}{2}\log \ml \frac{1-\mu_0(u)\frac{r_0(u)}{r_0(u_2)}}{1-\mu_0(u)}\mr.
\end{aligned}
\end{equation}

This in turn implies that

\begin{equation}\label{FIT222a}
\begin{aligned}
e^{2(\gamma_0(u)-\gamma_0(u_2))} - 1 &\geq \frac{\mu_0(u)}{1-\mu_0(u)}\ml 1 - \frac{r_0(u)}{r_0(u_2)} \mr \\
\end{aligned}
\end{equation}

and hence

\begin{equation}\label{FIT222}
0 \leq \mu_0(u) \leq \frac{e^{2 \ml \gamma_0(u) - \gamma_0(u_2)\mr}-1}{1-\frac{r_0(u)}{r_0(u_2)}}.
\end{equation}