For a given point $p = (u,v) \in \mathcal{R}$, suppose that we can find some $\overline{p} = (\overline{u},v) \in \mathcal{R}$ with $0 \leq \overline{u} < u$ such that the incoming null curve $\overline{p}p$ is in $\mathcal{R}$, then the estimates (<ref>) holds. Here the supremums in (<ref>) are obtained along the outgoing null curve originating from $\Gamma$ which intersects with $\overline{p}$. Even though the dotted line from $\overline{p}$ extends to the past along an incoming null curve may intersect $\mathcal{A}$ and transits into $\mathcal{T}$ before intersecting $C_0^+$, this does not affect our lemma here as it suffices to only find a point $\overline{p}$ along a past-directed incoming null curve from $p$ such that $\overline{p}p$ is in $\mathcal{R}$. If the dotted line from $\overline{p}$ does not intersect $\mathcal{A}\cup\mathcal{T}$, we can set $\overline{u} = 0$.

We use Figure <ref> to illustrate this lemma. For $(u,v) \in \mathcal{R}$, we have $1-\mu > 0$, $\dv r > 0$ and $\du r < 0$. By (<ref>), it implies that $\du \ml \frac{1-\mu}{\dv r}\mr \geq 0$ and $\du \ml \frac{\dv r}{1 - \mu}\mr \leq 0$. Hence, $\frac{\dv r}{1-\mu}$ is decreasing along future-directed incoming null curves, which lies in $\mathcal{R}$. Similarly, via (<ref>), we deduce that $\frac{-\du r}{1 - \mu}$ is increasing along future-directed outgoing null directions. The bound (<ref>) then follows from the monotonicity of $\frac{\dv r}{1-\mu}$ along incoming null directions.

Next, we record a calculus lemma that would be used not only for obtaining estimates of $|\phi|$ and $|Q|$, but also heavily employed throughout the paper.

For any $x \in (0,1)$ and $c > 0$, we have

\begin{equation}\label{xlogx1}
x^c \log \ml \frac{1}{x}\mr \leq \frac{1}{ce}.
\end{equation}

Let $f(x) = x^c \log \ml \frac{1}{x}\mr$. Observe that $f'(x_*) = x_*^{-1+c}\ml -1 + c\log (\frac{1}{x_*})\mr = 0$ when $x_* = e^{-1/c}$. Furthermore, we have $f''(x_*) = x_*^{-2+c}\ml 1-2c-(1-c)c\log(\frac{1}{x_*})\mr = - c e^{-1+2/c} < 0$ for all $c > 0$. This implies that $f$ attains a maximum at $x_* = e^{-1/c} \in (0,1)$, and thus

\begin{equation}\label{xlogx2}
\begin{aligned}
f(x) &\leq \ml e^{-1/c}\mr^c \log \ml e^{1/c}\mr = \frac{1}{ce}
\end{aligned}
\end{equation}

all $x \in (0,1)$.

Lemma <ref> allows us to absorb any logarithmic term $\log \ml \frac{1}{x} \mr$ by $x^c$ for an arbitrarily small $c > 0$ with an explicit upper bound.

With the above preparations, we proceed to derive some novel estimates for $|\phi|$ and $|Q|$, which will be heavily used in subsequent sections.  For notational simplicity, for each $(\overline{u},v) \in \mathcal{R}$, we define

\begin{equation}\label{overlineP}
\overline{P}(\overline{u},v) := \sup_{v'\in(\overline{u},v]} |\phi|(\overline{u},v') < +\infty.
\end{equation}

And we have

For $(u,v) \in \mathcal{R}$, if there exists a $0 \leq \overline{u} < u$ such that $(u',v') \in \mathcal{R}$ for all  $u' \in [\overline{u},u]$ and $v' \in (u',v]$, then for each given $\chi \in (0,1)$,  we have

\begin{equation}\label{Estimates13}
\begin{aligned}
|\phi|(u,v) \leq \overline{P}(\overline{u},v) + \frac{1}{(4\pi)^{\frac{1}{2}}}\ml  \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v)}{\frac{1 - \mu}{ \dv r}(\overline{u},v)}\mr  \mr^{\frac{1}{2}}\ml \log \ml \frac{r(\overline{u},v)}{r(u,v)}\mr \mr^{\frac{1}{2}}
\end{aligned}
\end{equation}

\begin{equation}\label{Estimates14}
\begin{aligned}
&\leq C_1(\overline{u},v;\chi) \cdot \mu^{\oh}(u,v) \cdot r^{\frac{3}{2} - \frac{\chi}{2}}(u,v) \cdot r^{\frac{\chi}{2}}(\overline{u},v) \cdot (v - u)^{\oh}
\end{aligned}
\end{equation}

with constant $C_1(\overline{u},v;\chi)$ given in (<ref>). In particular, we have

\begin{equation}\label{Estimates17}
\frac{Q^2}{r^2}(u,v) \leq C_1^2(\overline{u},v;\chi) \cdot \mu(u,v) \cdot r^{1 - \chi}(u,v) \cdot r^{\chi}(\overline{u},v) \cdot (v - u).
\end{equation}

Furthermore, if $Q$ is paired with $\phi$, we obtain an improved estimate

\begin{equation}\label{Estimates19}
|Q\phi|(u,v) \leq C_5(\overline{u},v;\chi,\xi_1,\xi_2) \cdot \mu(u,v)^{\frac{1}{2\xi_1}} \cdot r^{\frac{3}{2}- \frac{\chi}{2} -\frac{1}{2\xi_1}(1+\xi_2)}(u,v),
\end{equation}

where the constant $C_5$ (depending on $\overline{u},v,\chi,\xi_1$, $\xi_2$) is given in (<ref>),
with any $\xi_1 \geq 1$ and $\xi_2 \in (0,1-\chi)$.

[gray!50] (0,1) to (2,3) to (1.2,3.8) to (0,2.6) to (0,1);
[->] (0,0) node[anchor=north]$\mathcal{O}$ –  (0,6) node[anchor = east]$\Gamma$;
[->] (0,0) – (5,5) node[anchor = west]$C_0^+$;
[thick, dashdotted]  (4.9,5.1) to[in=-120,out= -140] (2.4,2.9) to[out=60,in=0] (2,4) to[out=180,in=-45] (0,4.6);
[thick] (1.2,3.8) node[anchor = east]$p$ – (2,3)node[anchor = south]$\overline{p}$;
[] at (0.5,1) $\mathcal{R}$;
[] at (3.5,4.5) $\mathcal{T}$;
[] at (2,4.3) $\mathcal{A}$;
[] at (-0.3,1) $\overline{p}'$;
[] at (-0.3,3) $p'$;
[thick] (0,1) – (2,3);
[thick] (0,2.6) – (1.2,3.8);
at (1.2,3.8)[circle,fill,inner sep=1.5pt];
at (2,3)[circle,fill,inner sep=1.5pt];

For a given point $p = (u,v) \in \mathcal{R}$, let $\overline{p} = (\overline{u},v)$ with $0 \leq \overline{u} < u$ be also a point in $\mathcal{R}$. Denote $p'p$ and $\overline{p}'\overline{p}$ as outgoing null curves emitting from $p'$ and $\overline{p}'$ along $\Gamma$ respectively. We then have that the shaded region indicated above must be in $\mathcal{R}$. To derive pointwise estimates at $p = (u,v)$, we consider the data along $\overline{p}'\overline{p}$ as the initial data.

A direct consequence of (<ref>) would be that, for $\mu - Q^2/r^2$, it yields

\begin{equation}\label{Estimates18}
\ml \mu - \frac{Q^2}{r^2} \mr(u,v) \geq \mu(u,v) \ml 1 - C_1^2(\overline{u},v;\chi) \cdot r^{1 - \chi}(u,v) \cdot r^{\chi}(\overline{u},v) \cdot (v - u) \mr.
\end{equation}

Hence, we have $\mu - \frac{Q^2}{r^2} \geq 0$ if $C_1^2(\overline{u},v;\chi) \cdot r^{1 - \chi}(u,v) \cdot r^{\chi}(\overline{u},v) \cdot (v - u) \leq 1$, which naturally holds for regions sufficiently close to $\Gamma$. The sign of $\mu - Q^2/r^2$ is essential for this study as it is employed repeatedly throughout the entire paper. Furthermore, from the explicit expression of $C_1$ provided later in (<ref>), we can see that although $C_1(\overline{u},v;\chi)$ is a constant depending on $\overline{u}$ and $v$, it is possible to obtain an upper bound for $C_1$ that is independent of $\overline{u}$ and $v$. This will be addressed in the individual sections on a case-by-case basis.

We begin with estimating $\phi$ as follows. Fix any $(u,v) \in \mathcal{R}$. By the hypothesis, there exists a $\overline{u} \in [0,u)$ such that $(u',v') \in \mathcal{R}$ for all $u' \in [\overline{u},u]$ and $v' \in [u',v']$. (This is illustrated as the shaded region in Figure <ref>.) By the definition of $D_u,$ we have
$\du \phi + \ii \e A_u \phi = D_u \phi$. This implies that

\begin{equation}\label{Estimates16}
\du \ml \phi e^{\int^{u}_{\overline{u}} \ii \e A_u (u',v) \; \D u'}\mr = e^{\int^{u}_{\overline{u}} \ii \e A_u (u',v) \; \D u'} D_u \phi
\end{equation}

for any $\overline{u}$. By conducting integration with respect to $u$ from $\overline{u}$, we obtain a gauge invariant estimate for $\phi$

\begin{equation}\label{Estimates2}
|\phi|(u,v) \leq |\phi| (\overline{u},v) + \int^u_{\overline{u}} |D_u \phi| (u',v) \mathrm{d}u',
\end{equation}

where we have used the fact that $\int^{u}_{\overline{u}} \ii \e A_u (u',v) \; \D u'$ is purely imaginary. Next, notice that (<ref>) is equivalent to the following equation:

\begin{equation}\label{Estimates3}
\begin{aligned}
\du \ml \log \ml \frac{1-\mu}{\dv r}\mr\mr &= \frac{4\pi r |D_u \phi|^2}{-\du r}.
\end{aligned}
\end{equation}

Using (<ref>) and the Cauchy-Schwarz inequality, we deduce

\begin{equation}\label{Estimates5}
\begin{aligned}
|\phi|(u,v) &\leq  |\phi| (\overline{u},v) + \frac{1}{(4\pi)^{\frac{1}{2}}}\ml \int^u_{\overline{u}} \frac{4 \pi r|D_u \phi|^2}{-\du r} (u',v) \mathrm{d}u' \mr^{\frac{1}{2}} \ml \int^u_{\overline{u}} \frac{-\du r}{r} (u',v) \mathrm{d}u' \mr^{\frac{1}{2}} \\
&\leq \overline{P}(\overline{u},v) + \frac{1}{(4\pi)^{\frac{1}{2}}}\ml  \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v)}{\frac{1 - \mu}{ \dv r}(\overline{u},v)}\mr  \mr^{\frac{1}{2}}\ml \log \ml \frac{r(\overline{u},v)}{r(u,v)}\mr \mr^{\frac{1}{2}}.
\end{aligned}
\end{equation}

To estimate $Q$, we shall employ (<ref>). Given any point $(u,v) \in \mathcal{R}$, we perform an integration along an outgoing null curve from $\Gamma$ to $(u,v)$. Together with the fact that $Q|_{\Gamma} = 0$ from (<ref>) and utilizing our bounds in (<ref>), we then derive

\begin{equation}\label{Estimates6}
\begin{aligned}
|Q|(u,v) &\leq \int^v_u 4\pi \e r^{2}|\phi||\dv \phi| (u,v') \mathrm{d}v' \\
&\leq 4\pi \e  \int^v_u r^{2}\ml \overline{P}(\overline{u},v') + \frac{1}{(4\pi)^{\frac{1}{2}}}\ml  \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{ \dv r}(\overline{u},v')}\mr  \mr^{\frac{1}{2}}\ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr^{\frac{1}{2}}\mr|\dv \phi| (u,v') \mathrm{d}v'  \\
&= 4\pi\e (Q_1(u,v) + Q_2(u,v))
\end{aligned}
\end{equation}

\begin{equation}\label{Estimates7}
\begin{aligned}
Q_1(u,v) &:= \overline{P}(\overline{u},v) \int^v_u r^2 |\dv \phi|(u,v') \mathrm{d}v',  \\
Q_2(u,v) &:= \frac{1}{(4\pi)^{\frac{1}{2}}} \int^v_u \ml \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{\dv r}(\overline{u},v')}\mr  \mr^{\frac{1}{2}} \ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr^{\frac{1}{2}} r^2|\dv \phi| (u,v') \mathrm{d}v'.
\end{aligned}
\end{equation}

For $Q_1$, using (<ref>), (<ref>) and $\dv r(u,v) > 0$, we obtain

\begin{equation}\label{Estimates8}
\begin{aligned}
Q_1(u,v) &= \overline{P}(\overline{u},v) \int^v_u r^2 |\dv \phi|(u,v') \mathrm{d}v'  \\
&= \frac{\overline{P}(\overline{u},v)}{(2\pi)^{\frac{1}{2}}}\ml \int^v_u \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r} (u,v') \mathrm{d}v'\mr^{\frac{1}{2}} \ml \int^v_u \frac{ r^2 \dv r}{1 - \mu} (u,v') \mathrm{d}v'\mr^{\frac{1}{2}} \\
&\leq \frac{\overline{P} \bar{B}^{\frac{1}{2}}}{{(2\pi (1-\overline{\mu_*}))^{\oh}}} (\overline{u},v) \cdot r(u,v) \cdot (v-u)^{\oh}  \cdot \ml \int^v_u \dv m (u,v') \mathrm{d}v'\mr^{\oh} \\
&\leq  \frac{\overline{P} \bar{B}^{\frac{1}{2}}}{{(4\pi (1-\overline{\mu_*}))^{\oh}}} (\overline{u},v) \cdot  r^{\frac{3}{2}}(u,v) \cdot \mu^{\oh}(u,v) \cdot (v-u)^{\oh}.
\end{aligned}
\end{equation}

To bound $Q_2$, we shall rely on Lemma <ref> and we will crucially employ the structure of $\frac{\partial_v r}{1-\mu}$-terms after using the equation of $\partial_v m$ and a Hölder's inequality. Also noting that the terms in the arguments of the logarithm functions in (<ref>) are greater than or equal to $1$, we then derive

\begin{equation}\label{Estimates9}
\begin{aligned}
Q_2(u,v) = &\; \frac{1}{(4\pi)^{\frac{1}{2}}} \int^v_u \ml \log \ml \frac{\frac{1 - \mu}{\dv r}(u,v')}{\frac{1 - \mu}{\dv r}(\overline{u},v')}\mr  \mr^{\frac{1}{2}} \ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr^{\frac{1}{2}}r^2|\dv \phi| (u,v') \mathrm{d}v' \\
\leq &\; \frac{1}{(8\pi^2)^{\frac{1}{2}}}\ml \int^v_u \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r} (u,v') \mathrm{d}v'\mr^{\frac{1}{2}} \\
&\times \ml \int^v_u \frac{ r^2 \dv r}{1 - \mu}(u,v') \cdot \ml \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{ \dv r}(\overline{u},v')}\mr  \mr \cdot \ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr \mathrm{d}v'\mr^{\frac{1}{2}} \\
\leq &\; \frac{1}{(8\pi^2)^{\frac{1}{2}}} m^{\oh}(u,v) \\
&\times \ml \int^v_u r^2 (u,v') \cdot \ml \frac{\dv r}{1 - \mu}(\overline{u},v') \mr \cdot
\ml \frac{\frac{\dv r}{1 - \mu}(u,v')}{\frac{\dv r}{1 - \mu}(\overline{u},v')}\mr \cdot \ml \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{ \dv r}(\overline{u},v')}\mr  \mr \cdot \ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr \mathrm{d}v'\mr^{\frac{1}{2}} \\
\leq &\; \frac{\bar{B}^{\frac{1}{2}}}{(16\pi^2 e (1-\overline{\mu_*}))^{\frac{1}{2}}}(\overline{u},v) \cdot \mu^{\oh}(u,v)r^{\oh}(u,v) \cdot \ml \int^v_u r^{2-\chi} (u,v') \cdot r^{\chi}(\overline{u},v') \cdot \ml \frac{r^{\chi}(u,v')}{r^{\chi}(\overline{u},v')}\mr \cdot
\ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr \mathrm{d}v'\mr^{\frac{1}{2}} \\
\leq &\; \frac{\bar{B}^{\frac{1}{2}}}{(16\pi^2 \chi e^2 (1-\overline{\mu_*}))^{\frac{1}{2}}}(\overline{u},v) \cdot \mu^{\oh}(u,v) \cdot r^{\frac{3}{2} - \frac{\chi}{2}}(u,v) \cdot r^{\frac{\chi}{2}}(\overline{u},v) \cdot (v - u)^{\oh}. \\
\end{aligned}
\end{equation}

Here, $\chi \in (0,2)$ is a fixed constant to be chosen.
Combining (<ref>) and (<ref>), we simplify (<ref>) as

\begin{equation}\label{Estimates10}
\begin{aligned}
|Q|(u,v) &\leq 4\pi\e \ml \frac{\overline{P}\bar{B}^{\frac{1}{2}}}{(4\pi (1-\overline{\mu_*}))^{\oh}} + \frac{\bar{B}^{\frac{1}{2}}}{(16\pi^2 \chi e^2 (1-\overline{\mu_*}))^{\oh}}\mr (\overline{u},v) \cdot \mu^{\oh}(u,v) \cdot r^{\frac{3}{2} - \frac{\chi}{2}}(u,v) \cdot r^{\frac{\chi}{2}}(\overline{u},v) \cdot (v - u)^{\oh}   \\
&= C_1(\overline{u},v;\chi) \cdot \mu^{\oh}(u,v) \cdot r^{\frac{3}{2} - \frac{\chi}{2}}(u,v) \cdot r^{\frac{\chi}{2}}(\overline{u},v) \cdot (v - u)^{\oh}
\end{aligned}
\end{equation}

\begin{equation}\label{Estimates11}
C_1(\overline{u},v;\chi) := \frac{4\pi\e \bar{B}^{\oh}}{(1-\overline{\mu_*})^{\oh}} (\overline{u},v) \cdot \ml \frac{\overline{P}(\overline{u},v)}{(4\pi)^{\oh}} + \frac{1}{(16\pi^2 \chi e^2)^{\oh}}\mr.
\end{equation}

In particular, (<ref>) is equivalent to  (<ref>), and (<ref>) follows immediately.

To derive (<ref>), we employ a trick via considering $(Q\phi)^{\xi_1}$ for some $\xi_1 \geq 1$. By the fact that $Q\phi|_\Gamma = 0$ and $m|_{\Gamma} = 0$, via fundamental theorem of calculus and (<ref>), we have

\begin{equation}\label{Estimates20}
\begin{aligned}
|Q\phi|^{\xi_1}(u,v)  &\leq \int^v_u \xi_1 |Q\phi|^{\xi_1 - 1} \dv(Q\phi) (u,v') \D v' \\
&\leq \int^v_u \xi_1  (4\pi \e r^2 |\phi|^{\xi_1 + 1} |Q|^{\xi_1 - 1} + |\phi|^{\xi_1 - 1}|Q|^{\xi_1})|\dv \phi| (u,v') \D v' \\
&\leq \xi_1  \ml \int^v_u \frac{\dv r}{2\pi r^2 (1-\mu)}(4\pi \e r^2 |\phi|^{\xi_1 + 1} |Q|^{\xi_1 - 1} + |\phi|^{\xi_1 - 1}|Q|^{\xi_1})^2  \D v' \mr^{\frac{1}{2}} \ml \int^v_u  \frac{2 \pi r^2(1-\mu)|\dv \phi|^2}{\dv r} \D v'\mr^{\frac{1}{2}}\\
&\leq \xi_1  \ml \int^v_u \frac{\dv r}{2\pi r^2 (1-\mu)}(4\pi \e r^2 |\phi|^{\xi_1 + 1} |Q|^{\xi_1 - 1} + |\phi|^{\xi_1 - 1}|Q|^{\xi_1})^2  (u,v') \D v' \mr^{\frac{1}{2}} m(u,v)^{\frac{1}{2}}\\
&\leq \frac{\xi_1}{\sqrt{2}} \cdot \mu(u,v)^{\frac{1}{2}} \cdot r(u,v)^{\frac{1}{2}} \cdot \ml \int^v_u \frac{\dv r}{2\pi r^2 (1-\mu)}(4\pi \e r^2 |\phi|^{\xi_1 + 1} |Q|^{\xi_1 - 1} + |\phi|^{\xi_1 - 1}|Q|^{\xi_1})^2  (u,v') \D v' \mr^{\frac{1}{2}}.
\end{aligned}
\end{equation}

The first integral on the right can be bounded as follows

\begin{equation}\label{Estimates21}
\begin{aligned}
& \int^v_u \frac{\dv r}{2\pi r^2 (1-\mu)}(4\pi \e r^2 |\phi|^{\xi_1 + 1} |Q|^{\xi_1 - 1} + |\phi|^{\xi_1 - 1}|Q|^{\xi_1})^2  \D v' \\
\leq &\; 16 \pi \e^2 \int^v_u \frac{\dv r}{1-\mu}r^2 |\phi|^{2\xi_1 + 2} |Q|^{2\xi_1 - 2} (u,v') \D v' + \frac{1}{\pi} \int^v_u \frac{\dv r}{1-\mu} \frac{1}{r^2} |\phi|^{2\xi_1 -2} |Q|^{2\xi_1 } (u,v') \D v'.
\end{aligned}
\end{equation}

The first term in the upper bound in (<ref>) can be estimated in the same fashion as in (<ref>) via using (<ref>) and (<ref>). Indeed, for $v' \in (u',v]$, using the fact that $\overline{P}(\overline{u},\cdot)$ and $C_1(\overline{u},\cdot;\chi)$ are non-decreasing, $\dv r > 0$ and $\mu < 1$, we get

\begin{equation}\label{Estimates22}
|\phi|(u,v') \leq \overline{P}(\overline{u},v) + \frac{1}{(4\pi)^{\frac{1}{2}}} \ml  \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{ \dv r}(\overline{u},v')}\mr  \mr^{\frac{1}{2}} \cdot \ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr^{\frac{1}{2}},
\end{equation}

\begin{equation}\label{Estimates23}
|Q|(u,v') \leq C_2(\overline{u},v;\chi) \cdot r^{\frac{3}{2}-\frac{\chi}{2}}(u,v'),
\end{equation}

\begin{equation}\label{Estimates24}
\begin{aligned}
C_{2}(\overline{u},v) &= C_1(\overline{u},v;\chi) \cdot r^{\frac{\chi}{2}}(\overline{u},v) \cdot (v-\overline{u})^{\frac{1}{2}}.
\end{aligned}
\end{equation}

These in turn imply that

\begin{equation}\label{Estimates34}
\begin{aligned}
|\phi|^{2\xi_1 -2}(u,v') &\leq 2^{2\xi_1 -3}(1+\overline{P}(\overline{u},v))^{2\xi_1 -2} + \frac{1}{2\pi^{\xi_1 - 1}}\ml  \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{ \dv r}(\overline{u},v')}\mr  \mr^{\xi_1 - 1} \cdot \ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr^{\xi_1 - 1}, \\
|\phi|^{2\xi_1 +2}(u,v') &\leq 2^{2\xi_1 +1}(1+\overline{P}(\overline{u},v))^{2\xi_1 +2} + \frac{1}{2\pi^{\xi_1 + 1}}\ml  \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{ \dv r}(\overline{u},v')}\mr  \mr^{\xi_1 + 1} \cdot \ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr^{\xi_1 + 1}
\end{aligned}
\end{equation}

by appealing to the inequality $(a+b)^p \leq 2^{p-1}(a^p + b^p)$ for $a,b \geq 0$ and $p \geq 1$.

We start by dealing with the first term on the right of (<ref>) as follows

\begin{equation}\label{Estimates25}
\begin{aligned}
& \int^v_u \frac{\dv r}{1-\mu}r^2 |\phi|^{2\xi_1 + 2} |Q|^{2\xi_1 - 2} (u,v') \D v' \\
\leq &\; C_2^{2\xi_1 - 2} 2^{2\xi_1 + 1}(1+ \overline{P}(\overline{u},v))^{2\xi_1 + 2 }\int^v_u \frac{\dv r}{1-\mu}(u,v')r^{(\xi_1 - 1) \cdot \ml 3 - \chi\mr + 2}(u,v')   \D v' \\
&+ \;  \frac{C_2^{2\xi_1 - 2}}{2 \pi^{\xi_1 + 1}}\int^v_u \frac{\dv r}{1-\mu}(u,v') \cdot r^{(\xi_1 - 1)\ml 3 - \chi\mr + 2}(u,v') \cdot \ml  \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{ \dv r}(\overline{u},v')}\mr  \mr^{\xi_1 + 1} \cdot \ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr^{\xi_1 + 1}  \D v'. \\
\end{aligned}
\end{equation}

By the monotonicity of $\frac{\dv r}{1-\mu}$ along future-directed incoming null directions, for any $\xi_2 \in (0,2)$, we have

\begin{equation}\label{Estimates35}
\begin{aligned}
& C_2^{2\xi_1 - 2} 2^{2\xi_1 + 1}(1+ \overline{P}(\overline{u},v))^{2\xi_1 + 2 }\int^v_u \frac{\dv r}{1-\mu}(u,v') \cdot r^{(\xi_1 - 1)\ml 3 - \chi\mr + 2}(u,v')   \D v' \\
\leq &\; C_2^{2\xi_1 - 2} 2^{2\xi_1 + 1}(1+ \overline{P}(\overline{u},v))^{2\xi_1 + 2 } \frac{\bar{B}(\overline{u},v) r^{\xi_2}(\overline{u},v)}{(1-\overline{\mu_*})(\overline{u},v)} \cdot (v - u) \cdot r^{(\xi_1 - 1)(3-\chi) + 2 - \xi_2}(u,v).
\end{aligned}
\end{equation}

By Lemma <ref>, we further obtain

\begin{equation}\label{Estimates26}
\frac{\dv r}{1-\mu}(u,v') \ml  \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{ \dv r}(\overline{u},v')}\mr  \mr^{\xi_1 + 1} \leq \frac{\dv r}{1-\mu}(\overline{u},v') \ml \frac{\xi_1 + 1}{e}\mr^{\xi_1 + 1},
\end{equation}

\begin{equation}\label{Estimates27}
\ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr^{\xi_1 + 1}  \leq  r^{-\xi_2}(u,v') r^{\xi_2}(\overline{u},v') \ml \frac{\xi_1 + 1}{\xi_2 e}\mr^{\xi_1 + 1}.
\end{equation}

Consequently, the second term in  (<ref>) simplifies to

\begin{equation}\label{Estimates28}
\begin{aligned}
& \frac{C_2^{2\xi_1 - 2}}{2 \pi^{\xi_1 + 1}}\int^v_u \frac{\dv r}{1-\mu}(u,v')r^{(\xi_1 - 1)\ml 3 - \chi\mr + 2}(u,v') \ml  \log \ml \frac{\frac{1 - \mu}{ \dv r}(u,v')}{\frac{1 - \mu}{ \dv r}(\overline{u},v')}\mr  \mr^{\xi_1 + 1}\ml \log \ml \frac{r(\overline{u},v')}{r(u,v')}\mr \mr^{\xi_1 + 1}  \D v' \\
\leq &\; \frac{C_2^{2\xi_1 - 2}}{2\pi^{\xi_1 + 1}}\frac{\bar{B}(\overline{u},v)r^{\xi_2}(\overline{u},v)  }{(1-\overline{\mu_*})(\overline{u},v)} \ml \frac{\xi_1 + 1}{ e}\mr^{\xi_1 + 1}
\ml \frac{\xi_1 + 1}{\xi_2 e}\mr^{\xi_1 + 1} (v-u) r^{(\xi_1-1)(3-\chi) + 2 - \xi_2}(u,v)
\end{aligned}
\end{equation}

for all possible $\xi_1 \geq 1$ and $\chi \in (0,1)$. Thus, we arrive at

\begin{equation}\label{Estimates36}
\begin{aligned}
\int^v_u \frac{\dv r}{1-\mu}r^2 |\phi|^{2\xi_1 + 2} |Q|^{2\xi_1 - 2} (u,v') \D v'
\leq C_3(\overline{u},v;\chi,\xi_1,\xi_2) \cdot r^{(\xi_1-1)(3-\chi) + 2 - \xi_2}(u,v)
\end{aligned}
\end{equation}

\begin{equation}\label{Estimates37}
C_3 :=   C_2^{2\xi_1 - 2}\cdot
(v - \overline{u}) \frac{\bar{B}(\overline{u},v)r^{\xi_2}(\overline{u},v)  }{(1-\overline{\mu_*})(\overline{u},v)}\ml 2^{2\xi_1 + 1}(1+ \overline{P}(\overline{u},v))^{2\xi_1 + 2 }
+ \frac{1}{2\pi^{\xi_1 + 1}}\ml \frac{\xi_1 + 1}{ e}\mr^{\xi_1 + 1}
\ml \frac{\xi_1 + 1}{\xi_2 e}\mr^{\xi_1 + 1} \mr.
\end{equation}

Back to the second term on the right of (<ref>), by an analogous argument, one can show that if $\xi_2 \in (0,1-\chi)$, it then holds

\begin{equation}\label{Estimates29}
\begin{aligned}
\int^v_u \frac{\dv r}{1-\mu} \frac{1}{r^2} |\phi|^{2\xi_1 -2} |Q|^{2\xi_1 } (u,v') \D v'
\leq  C_4(\overline{u},v;\chi,\xi_1,\xi_2) \cdot r^{\xi_1(3-\chi) - 2 - \xi_2}(u,v),
\end{aligned}
\end{equation}

\begin{equation}\label{Estimates38}
C_4 :=  C_2^{2\xi_1}\cdot
(v - \overline{u}) \frac{\bar{B}(\overline{u},v)r^{\xi_2}(\overline{u},v)  }{(1-\overline{\mu_*})(\overline{u},v)}\ml 2^{2\xi_1 - 3}(1+ \overline{P}(\overline{u},v))^{2\xi_1 - 2 }
+ \frac{1}{2\pi^{\xi_1 - 1}}\ml \frac{\xi_1 - 1}{ e}\mr^{\xi_1 - 1}
\ml \frac{\xi_1 - 1}{\xi_2 e}\mr^{\xi_1 - 1} \mr.
\end{equation}

Going back to (<ref>), by applying (<ref>) and (<ref>), we also obtain

\begin{equation}\label{Estimates30}
\begin{aligned}
& \int^v_u \frac{\dv r}{2\pi r^2 (1-\mu)}(4\pi \e r^2 |\phi|^{\xi_1 + 1} |Q|^{\xi_1 - 1} + |\phi|^{\xi_1 - 1}|Q|^{\xi_1})^2  \D v' \\
\leq &\; \ml 16 \pi \e^2 C_3 r^{1+\chi}(\overline{u},v) + \frac{C_4}{\pi} \mr r^{\xi_1(3-\chi)-2-\xi_2}(u,v).
\end{aligned}
\end{equation}

We then apply (<ref>) to (<ref>) and get

\begin{equation}\label{Estimates32}
\begin{aligned}
|Q\phi|^{\xi_1}(u,v) &\leq \frac{\xi_1}{\sqrt{2}}
\ml 16 \pi \e^2 C_3 r^{1+\chi}(\overline{u},v) + \frac{C_4}{\pi} \mr^{\frac{1}{2}}
\mu(u,v)^{\frac{1}{2}} \cdot r^{\xi_1 \ml \frac{3}{2}- \frac{\chi}{2} \mr -\frac{1}{2}(1+\xi_2)}(u,v), \\
\end{aligned}
\end{equation}

\begin{equation}\label{Estimates32a}
\begin{aligned}
|Q\phi|(u,v) &\leq C_5(\overline{u},v;\chi,\xi_1,\xi_2) \cdot \mu(u,v)^{\frac{1}{2\xi_1}} \cdot
r^{\frac{3}{2}- \frac{\chi}{2} -\frac{1}{2\xi_1}(1+\xi_2)}(u,v),
\end{aligned}
\end{equation}

\begin{equation}\label{Estimates33}
C_{5}(\overline{u},v;\chi,\xi_1,\xi_2) :=  \ml \frac{\xi_1}{\sqrt{2}}
\ml 16 \pi \e^2 C_3(\overline{u},v;\chi,\xi_1,\xi_2) \cdot r^{1+\chi}(\overline{u},v) + \frac{C_4 (\overline{u},v;\chi,\xi_1,\xi_2)}{\pi} \mr^{\frac{1}{2}} \mr^{\frac{1}{\xi_1}}.
\end{equation}

§ TRAPPED SURFACE FORMATION

In this section, we shall provide a novel proof and establish a new trapped surface formation criterion for the Einstein-matter fields. In particular, we prove a desired result for the charged case. This will play a crucial role in subsequent sections. If we set $Q\equiv 0$, in the appendix, we also give an approximately 1.5-page new proof of the sharp criterion obtained by Christodoulou for the Einstein-(real) scalar field system. Inspired by [18], for a given coordinate patch $[u_1,u_2]\times[v_1,v_2]$ with $u_1 < u_2$ and $v_1 < v_2$, we denote the dimensionless radius ratio $\delta(u)$ by

\begin{equation}\label{RadiusRatio}
\delta(u) = \frac{r(u,v_2) - r(u,v_1)}{r(u,v_1)}.
\end{equation}

We also define the dimensionless
reduced mass ratio as

\begin{equation}\label{ReducedMassRatio}
\eta^*(u) := \frac{2 \ml m(u,v_2) - m(u,v_1) \mr}{r(u,v_2)} - \frac{2}{r(u,v_2)}\int^{v_2}_{v_1} \frac{Q^2(\dv r)}{2r^2}(u,v')\mathrm{d}v'.
\end{equation}

Note that the additional term containing $Q^2$ is of a special form and is not from the modified Hawking mass. This additional term will play an important role in the later-established instability theorems.

dot/.style = draw, fill = white, circle, inner sep = 0pt, minimum size = 4pt]
[gray!50] (1,4) to (2.5,5.5) to (3,5) to (1.5,3.5) to (1,4);
[gray!80] (0.2,4.8) to (1.2,5.8) to (2,5) to (1,4) to (0.2,4.8);
[thick] (0,0) node[anchor=north]$\mathcal{O}$ –  (0,5);
[] at (-0.3,2.5) $\Gamma$;
[] at (-0.3,5.0) $\mathcal{O}'$;
[->] (0,0) – (6,6) node[anchor = west]$C_0^+$;
[thick, densely dotted] (0,5) to (1.4,6.4);
[thick, loosely dotted] (1.4,6.4) to [out=45,in=175] (4.7,7.0);
[rotate=45] at (0.6,5.9) $\mathcal{B}_0$;
[rotate=0] at (4.3,7.3) $\mathcal{B}\setminus\mathcal{B}_0$;
[fill=white] (1.4,6.4) circle (2pt);
[thick] (0,5) –  (2.5,2.5);
[] at (2.7,2.3)$v_1$;
[thick,dashed] (4,4) –  (1.5,6.5);
[] at (4.2,3.8)$v_3$;
[thick, dashdotted] (0.2,5.2) to [out=30,in=150] (3.5,5.5) to [out=-30,in=150] (3.2,4.2) [out=10] to [in=215] (5,5.1);
[] at (4.15,4.85)$\mathcal{A}$;
[thick] (0.2,4.8) to (1.2,5.8);
[thick] (1.2,5.8) to (2,5);
[thick,dashed] (1,6) to (3.5,3.5);
[] at (3.7,3.3)$v_2$;
[thick] (1,4) to (2.5,5.5) to (3,5) to (1.5,3.5);
[rotate = 45] at (0.75,5.05) $u = u_*$;
[rotate = 45] at (1.4,4.6) $u = u_0$;
[rotate = 45] at (1.9,4.1) $u = \overline{u}$;
[rotate=45] at (1.1,4.9) $R'$;
[rotate=45] at (1.6,4.2) $R$;
[fill=white] (0,5) circle (2pt);

dot/.style = draw, fill = white, circle, inner sep = 0pt, minimum size = 4pt]
[gray!50] (1,4) to (2.5,5.5) to (3,5) to (1.5,3.5) to (1,4);
[gray!80] (0.2,4.8) to (1.2,5.8) to (2,5) to (1,4) to (0.2,4.8);
[thick] (0,0) node[anchor=north]$\mathcal{O}$ –  (0,5);
[] at (-0.3,2.5) $\Gamma$;
[] at (-0.3,5.0) $\mathcal{O}'$;
[->] (0,0) – (6,6) node[anchor = west]$C_0^+$;
[thick, densely dotted] (0,5) to (0.8,5.8);
[thick, loosely dotted] (0.8,5.8) to [out=45,in=175] (4.7,7.0);
[rotate=45] at (0.4,5.7) $\mathcal{B}_0$;
[rotate=0] at (4.3,7.3) $\mathcal{B}\setminus\mathcal{B}_0$;
[fill=white] (0.8,5.8) circle (2pt);
[thick] (0,5) –  (2.5,2.5);
[] at (2.7,2.3)$v_1$;
[thick,dashed] (4,4) –  (1.5,6.5);
[] at (4.2,3.8)$v_3$;
[thick, dashdotted] (0.2,5.2) to [out=30,in=150] (3.5,5.5) to [out=-30,in=150] (3.2,4.2) [out=10] to [in=215] (5,5.1);
[] at (4.15,4.85)$\mathcal{A}$;
[thick] (0.2,4.8) to (1.2,5.8);
[thick] (1.2,5.8) to (2,5);
[thick,dashed] (1,6) to (3.5,3.5);
[] at (3.7,3.3)$v_2$;
[thick] (1,4) to (2.5,5.5) to (3,5) to (1.5,3.5);
[rotate = 45] at (0.75,5.05) $u = u_*$;
[rotate = 45] at (1.4,4.6) $u = u_0$;
[rotate = 45] at (1.9,4.1) $u = \overline{u}$;
[rotate=45] at (1.1,4.9) $R'$;
[rotate=45] at (1.6,4.2) $R$;
[fill=white] (0,5) circle (2pt);

Diagram for the proof of Theorem <ref>. Rectangles $R$ and $R'$ are shaded in a lighter and a darker tone respectively and the apparent horizon $\mathcal{A}$ is indicated as a dotted curve. The scenario on the left corresponds to $|\mathcal{B}_0| \geq v_2 - v_1$, while the scenario on the right corresponds to $|\mathcal{B}_0| < v_2 - v_1$, and $\mathcal{B}_0$ is allowed to be the empty set.

For notational convenience, here we define some commonly used abbreviations. A subscript with $1$ indicates that the quantity is evaluated at $v = v_1$. Similarly, a subscript with $2$ indicates that the quantity is evaluated at $v = v_2$. For a quantity with subscript $0$, this means that it is evaluated at $u = u_0$. For instance, we have $\delta_0 = \delta(u_0)$, $m_1(u) = m(u,v_1)$ and $r_2(u) = r(u,v_2)$. In addition, denote

\begin{equation}\label{TSF43}
\mu^*(u_0) := \sup_{v \in [v_1,v_2]} \mu(u_0,v).
\end{equation}

We are ready to state the below criterion, which is tailored for application in the upcoming instability arguments.

(Trapped Surface Formation Criteria.)
Let $\mathcal{O}' = (v_1,v_1)$ be the future ending point of $\Gamma$ as portrayed in Figure <ref> and set $v_2 >  v_1$ and $u_0 < v_1$ to be sufficiently close to $v_1$. Suppose that there exist $\overline{u} \in [0,u_0)$ and $v_3 > v_2$ such that the coordinate patch $R:= [\overline{u},u_0] \times [v_1,v_3]$ lies in the regular region $\mathcal{R}$. If the reduced mass ratio $\eta^*(u_0)$ satisfies the lower-bound condition

\begin{equation}\label{TSF2}
\eta^*(u_0) > C_{24} \frac{\delta(u_0)}{1+\delta(u_0)} \log \ml \frac{1}{\delta(u_0)}\mr + C_{23} \ml \frac{\delta(u_0)r_1(u_0)}{1-\mu^*(u_0)}\mr,
\end{equation}

with $\delta(u_0) \in (0,1)$ and with constants $C_{23}$ and $C_{24}$ explicitly given in (<ref>) and (<ref>), then a marginally outer trapped surface (MOTS) or a trapped surface is guaranteed to form along the incoming null curve $[u_0, u_*]\times \{v_2\}$ with $u_* \in (u_0,v_0)$.

In the proof of the later instability theorems, by taking into account the blueshift mechanism, for $u_0$ sufficiently close to $v_1$, we will see that $\frac{1}{1-\mu^*(u_0)}$ is of order $\log(\frac{1}{\delta(u_0)})$ and the gained factor $r_1(u_0)$ will play a crucial role.

With $v_2 > v_1$, $u_0 < v_1$ and $u_* \in [u_0,v_1)$ being an arbitrary constant such that $(u_*,v_2) \in \mathcal{Q}$, we denote $R' := [u_0,u_*] \times [v_1,v_2]$. Under condition (<ref>), we explain a mechanism in the regular (untrapped) region, that would lead to MOTS or trapped surface formation.

If there exists a MOTS or a trapped surface within $R'$, then the conclusion of Theorem <ref> is achieved. We then proceed to assume the absence of a MOTS or a trapped surface in $R'$. By Remark <ref>, if $v_2$ and $u_0$ are picked sufficiently close to $v_1$, for each point $(u,v) \in R'$, the hypothesis of Proposition <ref> is satisfied. Consequently, we have that (<ref>) holds for any $\chi \in (0,1)$ and we obtain

\begin{equation}\label{TSF65}
\mu -\frac{Q^2}{r^2}(u,v) \geq \ml 1 - C_1^{2}(\overline{u},v;\chi)r^{1-\chi}(u,v)r^{\chi}(\overline{u},v)(v-u)\mr \mu(u,v).
\end{equation}

Here, $C_1(\overline{u},v;\chi)$ is defined in (<ref>). By the expression there, we can also see that $C_1(\overline{u},v;\chi)$ is monotonically non-decreasing in $v$ for $v \in (u,v_3]$. Hence, we obtain

\begin{equation}\label{TSF66}
\mu -\frac{Q^2}{r^2}(u,v) \geq \ml 1 - C_1^{2}(\overline{u},v_3;\chi)r(\overline{u},v_3)(v-u)\mr \mu(u,v).
\end{equation}

Using $v - u = v - v_1 + v_1 - u \leq (v_2 - v_1) + (v_1 - u_0)$ for all $u \in [u_0,v_1]$ and $v \in [v_1,v_2]$, if we pick $v_2$ and $u_0$ sufficiently close to $v_1$ such that

\begin{equation}\label{TSF67}
(v_2 - v_1) + (v_1 - u_0) \leq \frac{1}{C_1^{2}(\overline{u},v_3;\chi)r(\overline{u},v_3)},
\end{equation}

then we deduce that $\mu - \frac{Q^2}{r^2} \geq 0$ on $R' \subset \mathcal{R}$.

With these preparations, we are ready to prove the main theorem. Introducing a scale-free length $x$ by

\begin{equation}\label{TSF3}
x(u) = \frac{r_2(u)}{r_2(u_0)},
\end{equation}

we then consider the evolution of $\eta^*(x)$ along incoming null curves lying within $R'$. For $x \in (0,1]$, we have

\begin{equation}\label{TSF4}
\begin{aligned}
\frac{\mathrm{d}\eta^*}{\mathrm{d}x} =& \;\frac{\frac{\mathrm{d}\eta^*}{\mathrm{d}u}}{\frac{\mathrm{d}x}{\mathrm{d}u}} = \frac{\du \ml \frac{2(m_2 - m_1)}{r_2} - \frac{2}{r_2}\int^{v_2}_{v_1} \frac{Q^2 \dv r}{r^2} (u,v') \mathrm{d}v'\mr }{\frac{\du r_2}{r_2(u_0)}} \\
&\,\,= \; \frac{r_2(u_0)}{\du r_2} \ml
-\frac{\du r_2}{r_2}\ml\frac{2(m_2 - m_1)}{r_2} +\frac{2}{r_2} \int^{v_2}_{v_1} \frac{Q^2 \dv r}{2r^2} \mathrm{d}v'
\mr  + \frac{2}{r_2}\du(m_2 - m_1)  - \frac{2}{r_2} \int^{v_2}_{v_1} \du \ml \frac{Q^2 \dv r}{2r^2}\mr\mathrm{d}v'\mr \\
&\,\,= \;- \frac{\eta^*}{x} - \frac{2}{x(\du r_2)} \ml \int^{v_2}_{v_1} -\du\dv m + \du \ml \frac{Q^2 \dv r}{2r^2}\mr \mathrm{d}v'\mr.
\end{aligned}
\end{equation}

Note that the term $\du\ml \frac{Q^2 \du r}{2r^2}\mr$ comes from the definition of $\eta^*(u)$ in (<ref>), and it is designed to have a precise cancellation with terms in $-\du \dv m$. This is illustrated using (<ref>), where we see that

\begin{equation}\label{TSF5}
- \du\dv m + \du \ml \frac{Q^2 \dv r}{2r^2} \mr = -\du \ml \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r} \mr.
\end{equation}

Hence, with the help of (<ref>), the integrand in the last line of (<ref>) can be simplified as

\begin{equation}\label{TSF6}
\begin{aligned}
- \du\dv m + \du \ml \frac{Q^2 \dv r}{2r^2} \mr &= -\du \ml \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r} \mr \\
&= - 2 \pi \ml \du\ml \frac{1-\mu}{\dv r}\mr r^2|\dv \phi|^2 + \ml \frac{1-\mu}{\dv r}\mr \du (r^2|\dv \phi|^2) \mr \\
&= 8\pi^2 r^3 (1-\mu)(\du r)(\dv r) \frac{|D_u \phi|^2 |\dv \phi|^2}{(\du r)^2(\dv r)^2} - 2\pi \ml \frac{1-\mu}{\dv r}\mr \du\ml r^2|\dv \phi|^2\mr.
\end{aligned}
\end{equation}

For $\du(r^2|\dv \phi|^2)$, from (<ref>) and the definition of $D_u$, we have

\begin{equation}\label{TSF7}
\du(r\dv \phi) = -\ii \e A_u r \dv \phi - \dv r D_u \phi + \ii\e \frac{Q\phi(-\du r)(\dv r)}{r(1-\mu)}
\end{equation}

and thus

\begin{equation}\label{TSF8}
\begin{aligned}
\du(r^2|\dv \phi|^2) =&\; \du(r \dv \phi) (r\dv \phi)^{\dagger} + \du( (r \dv \phi)^\dagger) (r\dv \phi)
\\ =&\; r (\dv \phi)^\dagger \ml -\ii \e A_u r \dv \phi - \dv r D_u \phi + \ii\e \frac{Q\phi(-\du r)(\dv r)}{r(1-\mu)} \mr \\
&+ r (\dv \phi) \ml -\ii \e A_u r \dv \phi - \dv r D_u \phi + \ii\e \frac{Q\phi(-\du r)(\dv r)}{r(1-\mu)} \mr^{\dagger} \\
=& -2 r \dv r \re{\ml (\dv \phi)^\dagger D_u \phi  \mr} + 2\e Q \frac{(\du r) (\dv r)}{1-\mu} \im{(\phi^\dagger \dv \phi)}.
\end{aligned}
\end{equation}

Combining these calculations, we obtain

\begin{equation}\label{TSF9}
\begin{aligned}
&- \du\dv m + \du \ml \frac{Q^2 \dv r}{2r^2} \mr \\ =& \; (\dv r)(-\du r) \ml -8\pi^2 r^3 (1-\mu) \ml \frac{|D_u \phi| |\dv \phi|}{(-\du r)(\dv r)} \mr^2  + 4\pi r (1-\mu) \frac{\re{\ml (\dv \phi)^\dagger (D_u \phi) \mr}}{\dv r (-\du r)}\mr\\
&\; + 4\pi \e Q (-\du r) \im{(\phi^\dagger (\dv \phi))} \\
\leq&\; 8\pi^2 r^3 (\dv r)(-\du r)(1-\mu) \ml - \ml \frac{|D_u \phi| |\dv \phi|}{(-\du r)(\dv r)} \mr^2  + \frac{1}{2\pi r^2} \frac{|D_u \phi||\dv \phi|}{(-\du r)(\dv r)}\mr + 4\pi \e |Q| (-\du r) |\phi||\dv \phi|. \\
\end{aligned}
\end{equation}

Using the fact that $\du \dv r \leq 0$ in $R'$ and equation (<ref>), the integral in (<ref>) can be simplified to

\begin{equation}\label{TSF10}
\begin{aligned}
-\frac{1}{\du r_2}\int^{v_2}_{v_1} -\du\dv m + \du \ml \frac{Q^2 \dv r}{2r^2}\mr \mathrm{d}v' &\leq \int^{v_2}_{v_1} J_4 (u,v') + J_5 (u,v') \mathrm{d}v'
\end{aligned}
\end{equation}

\begin{equation}\label{TSF11}
J_4 = 8\pi^2 r^3 (\dv r)(1-\mu) \ml - \ml \frac{|D_u \phi| |\dv \phi|}{(-\du r)(\dv r)} \mr^2  + \frac{1}{2\pi r^2} \frac{|D_u \phi||\dv \phi|}{(-\du r)(\dv r)}\mr
\end{equation}

\begin{equation}\label{TSF12}
J_5 = 4 \pi \e |Q||\phi||\dv \phi|.
\end{equation}

For $J_4$, by using an elementary inequality  $-x^2 + ax \leq \frac{a^2}{4}$, we have

\begin{equation}\label{TSF13}
\begin{aligned}
&\int^{v_2}_{v_1} 8\pi^2 r^3 (\dv r)(1-\mu) \ml - \ml \frac{|D_u \phi| |\dv \phi|}{(-\du r)(\dv r)} \mr^2  + \frac{1}{2\pi r^2} \frac{|D_u \phi||\dv \phi|}{(-\du r)(\dv r)}\mr(u,v') \mathrm{d}v' \\
\leq&\; \int^{v_2}_{v_1} 2\pi^2 r^3 (\dv r)(1-\mu) \ml \frac{1}{2\pi r^2}\mr^2(u,v') \mathrm{d}v' \leq\; \frac{1}{2} \int^{v_2}_{v_1}   \frac{\dv r}{r}(u,v') \mathrm{d}v' \\
&\quad\quad=\; \frac{1}{2}\log\ml \frac{r(u,v_2)}{r(u,v_1)}\mr  = \frac{1}{2}\log \ml 1 + \frac{r(u,v_2) - r(u,v_1)}{r(u,v_1)}\mr \leq \frac{1}{2}\delta(u).
\end{aligned}
\end{equation}

Note that via our new approach, $J_4$ is bounded by $\frac{1}{2}\delta(u)$ independent of $\phi$ through a simple computation.

For $J_5$, with (<ref>) and the definition in (<ref>), we proceed to deduce that

\begin{equation}\label{TSF21}
\begin{aligned}
\int^{v_2}_{v_1} J_5 (u,v') \mathrm{d}v'&= \int^{v_2}_{v_1} 4 \pi \e |Q||\phi||\dv \phi| (u,v') \mathrm{d}v' \\
&\leq \sqrt{8\pi} \e \ml \int^{v_2}_{v_1} \frac{2\pi r^2 (1-\mu)|\dv \phi|^2}{\dv r} (u,v') \mathrm{d}v' \mr^{\frac{1}{2}} \ml \int^{v_2}_{v_1} \frac{\dv r}{1-\mu} \frac{|Q|^2|\phi|^2}{r^2}(u,v') \mathrm{d}v'\mr^{\frac{1}{2}} \\
&\leq \sqrt{8\pi} \e \ml \int^{v_2}_{v_1} \dv m - \frac{Q^2 \dv r}{2r^2} (u,v') \mathrm{d}v' \mr^{\frac{1}{2}} \ml \int^{v_2}_{v_1} \frac{\dv r}{1-\mu} \frac{|Q|^2|\phi|^2}{r^2}(u,v') \mathrm{d}v'\mr^{\frac{1}{2}} \\
&\leq \sqrt{4\pi} \e {\eta^*}^{\frac{1}{2}} r_2^{\frac{1}{2}}(u) \ml \int^{v_2}_{v_1} \frac{\dv r}{1-\mu} \frac{|Q|^2|\phi|^2}{r^2}(u,v') \mathrm{d}v'\mr^{\frac{1}{2}}.
\end{aligned}
\end{equation}

Employing (<ref>) of Proposition <ref> with $\xi_1 = 2$ and $\xi_2 = \chi$ for $\chi \in \left(0,\frac{1}{2}\right)$, we obtain the bound

\begin{equation}\label{TSF35}
|Q \phi|^2(u,v) \leq C_{5}(\overline{u},v_2;\chi,2,\chi)^2 r^{\frac{5}{2} - \frac{3\chi}{2}}(u,v),
\end{equation}

where we have used the fact that $\mu < 1$. Returning to (<ref>), with (<ref>) and $\chi \in \left(0, \frac{1}{3}\right]$, we thus have

\begin{equation}\label{TSF37}
\begin{aligned}
\int^{v_2}_{v_1} J_5 (u,v') \mathrm{d}v'
&\leq \sqrt{4\pi} \e {\eta^*}^{\frac{1}{2}} r_2^{\frac{1}{2}}(u) \ml \int^{v_2}_{v_1} \frac{\dv r}{1-\mu} \frac{|Q|^2|\phi|^2}{r^2}(u,v') \mathrm{d}v'\mr^{\frac{1}{2}} \\
&\leq \sqrt{4\pi} \e C_{5} {\eta^*}^{\frac{1}{2}} r_2^{\frac{1}{2}}(u)  \ml \int^{v_2}_{v_1} \frac{\dv r}{1 - \mu}(u_0,v') r^{\frac{1}{2}-\frac{3\chi}{2}}(u,v') \mathrm{d}v'\mr^{\frac{1}{2}} \\
&\leq \sqrt{4\pi} \e C_{5} {\eta^*}^{\frac{1}{2}} r_2^{\frac{3}{4} - \frac{3\chi}{4}}(u) \ml \frac{1}{1 - \mu_0^*}\mr^{\frac{1}{2}} \ml \int^{v_2}_{v_1} \dv r(u_0,v') \mathrm{d}v'\mr^{\frac{1}{2}} \\
&\leq \frac{\sqrt{4\pi} \e C_{5}}{(1-\mu_0^*)^{\frac{1}{2}}} {\eta^*}^{\frac{1}{2}} r_2^{\frac{3}{4}-\frac{3\chi}{4}}(u)  r_1^{\frac{1}{2}}(u_0) \delta(u_0)^{\frac{1}{2}}.
\end{aligned}
\end{equation}

Utilizing (<ref>), (<ref>), (<ref>) and the fact that $x \in (0,1]$, the main inequality (<ref>) simplifies to

\begin{equation}\label{TSF38}
\begin{aligned}
\frac{\mathrm{d}\eta^*}{\mathrm{d}x} &\leq - \frac{\eta^*}{x} - \frac{2}{x(\du r_2)} \ml \int^{v_2}_{v_1} -\du\dv m + \du \ml \frac{Q^2 \dv r}{2r^2}\mr \mathrm{d}v'\mr \\
&\leq - \frac{\eta^*}{x} + \frac{\delta}{x} + \frac{1}{x} \frac{4\sqrt{\pi} \e C_{5}}{(1-\mu_0^*)^{\frac{1}{2}}} {\eta^*}^{\frac{1}{2}} r_2^{\frac{3}{4}-\frac{3\chi}{4}}(u)r_1^{\frac{1}{2}}(u_0) \delta^{\frac{1}{2}}(u_0) \\
&\leq - \frac{\eta^*}{x} + \frac{\delta}{x} + C_{21}x^{-\frac{1}{4}-\frac{3\chi}{4}}\ml \frac{2{\eta^*}^{\frac{1}{2}}\delta_0^{\frac{1}{2}}(r_1)_0^{\frac{1}{2}}}{(1-\mu_0^*)^{\frac{1}{2}}} \mr \\
&\leq - \frac{\eta^*}{x} + \frac{\delta}{x} + \frac{C_{21}}{x^{\frac{1}{2}}} \ml \eta^* + \frac{\delta_0(r_1)_0}{1-\mu_0^*} \mr. \\
\end{aligned}
\end{equation}

Furthermore, the constant $C_{21}$ is defined as

\begin{equation}\label{TSF39}
C_{21} = 2\sqrt{\pi} \e C_{5}r(\overline{u},v_3)^{\frac{3}{4}-\frac{3\chi}{4}}
\end{equation}

with $\chi \in \left(0,\frac{1}{3}\right]$. Meanwhile, for all $(u,v) \in R'$, using $\du \dv r \leq 0$ in $R'$, we get

\begin{equation}\label{TSF40}
\dv r (u,v) \leq \dv r(u_0,v).
\end{equation}

For $u \in [u_0,v_1)$, this implies

\begin{equation}\label{TSF41}
r_2(u) - r_1(u) \leq r_2(u_0) - r_1(u_0).
\end{equation}

Thus, one can establish the following inequality

\begin{equation}\label{TSF42}
\begin{aligned}
\delta(u) = \frac{r_2}{r_1} - 1 &= \frac{r_2 - r_1}{r_2 - (r_2 - r_1)} \leq \frac{r_2(u_0)-r_1(u_0)}{r_2(u) - (r_2(u_0) - r_1(u_0))} \\
&\quad\quad\quad\leq \frac{\delta(u_0)}{\frac{r_2(u)}{r_1(u_0)} - \delta(u_0)} = \frac{\delta_0}{x(u)(1+\delta_0) - \delta_0}.
\end{aligned}
\end{equation}

Rewriting (<ref>), we hence arrive at

\begin{equation}\label{TSF44}
\begin{aligned}
\frac{\mathrm{d}\eta^*}{\mathrm{d}x} +  \ml \frac{1}{x} - \frac{C_{21}}{x^{\frac{1}{2}}}\mr\eta^*&\leq  \delta_0\ml \frac{1}{x(x(1+\delta_0) - \delta_0)} +  \frac{C_{21}(r_1)_0}{x^{\frac{1}{2}}(1-\mu_0^*)} \mr.\\
\end{aligned}
\end{equation}

This can be rewritten as

\begin{equation}\label{TSF44'}
\frac{\mathrm{d}\eta^*}{\mathrm{d}x} + f(x)\eta^* \leq \delta_0 g(x)
\end{equation}

\begin{equation}\label{TSF45}
f(x) = \frac{1}{x} - \frac{C_{21}}{x^{\frac{1}{2}}} \quad \text{ and } \quad g(x) = \frac{1}{x(x(1+\delta_0) - \delta_0)} +  \frac{C_{21}(r_1)_0}{x^{\frac{1}{2}}(1-\mu_0^*)}.
\end{equation}

We then solve the differential inequality (<ref>) and get

$$\frac{\mathrm{d}}{\mathrm{d}x} \ml e^{-\int^1_x f(s) \mathrm{d}s} \eta^* \mr \leq \delta_0 g(x) e^{-\int^1_x f(s) \mathrm{d}s},$$

which implies that, for any $x \in (x(u_*),1)$, we obtain

$$\eta^*_0  - e^{-\int^1_x f(s) \mathrm{d}s} \eta^*(x) \leq \delta_0 \int^1_x g(s) e^{-\int^1_s f(s') \mathrm{d}s'}\mathrm{d}s.$$

Recall that in the regular region definition (<ref>) implies $\eta_*(x) \leq \mu_2(x)$. Rearranging the above inequality, we get

\begin{equation}\label{TSF46}
\mu_2(x) \geq \eta^*(x) \geq e^{\int_x^1 f(s) \D s}\ml \eta_0^* - \delta_0 \int_x^1 g(s) e^{-\int_s^1 f(s') \D s'}\mr.
\end{equation}

We further evaluate the integrals in (<ref>) as follows. We first calculate

\begin{equation}\label{TSF47}
\begin{aligned}
\int^1_x f(s) \mathrm{d}s &= \int^1_x \frac{1}{s} - \frac{C_{21}}{s^{\frac{1}{2}}} \mathrm{d}s = -\log(x) - 2C_{21}\ml 1-x^{\frac{1}{2}} \mr,
\end{aligned}
\end{equation}

which implies

\begin{equation}\label{TSF48}
\begin{aligned}
e^{-\int^1_x f(s) \mathrm{d}s} = xe^{2C_{21}(1-x^{\frac{1}{2}})} \leq xe^{2C_{21}}.
\end{aligned}
\end{equation}

The second integral in (<ref>) obeys

\begin{equation}\label{TSF49}
\begin{aligned}
\int^1_x g(s) e^{-\int^1_s f(s') \mathrm{d}s'}\mathrm{d}s &\leq \int^1_x \ml \frac{1}{s(s(1+\delta_0) - \delta_0)} +  \frac{C_{21}(r_1)_0}{s^{\frac{1}{2}}(1-\mu_0^*)}\mr s e^{2C_{21}} \mathrm{d}s \\
&\leq e^{2C_{21}} \ml \frac{\log\ml \frac{1}{x(1+\delta_0)-\delta_0}\mr}{1+
\delta_0} + \frac{2C_{21} (r_1)_0}{3(1-\mu_0^*)}\ml 1 - x^{\frac{3}{2}}\mr   \mr
\end{aligned}
\end{equation}

if we impose $x > \frac{\delta_0}{1+\delta_0}$. For our following arguments, we restrict ourselves to $x \in \left( \frac{\delta_0}{1+\delta_0}, 1\right]$. Back to (<ref>), with (<ref>) and (<ref>), we now obtain

\begin{equation}\label{TSF50}
\mu_2(x) \geq  \frac{1}{C_{22}x}\ml \eta_0^* - C_{22}\frac{\delta_0}{1+\delta_0} \log\ml\frac{1}{x(1+\delta_0) - \delta_0}\mr - C_{23}\frac{\delta_0 (r_1)_0}{1-\mu_0^*} \mr
\end{equation}

\begin{equation}\label{TSF51}
C_{22} = e^{2C_{21}} \quad \text{ and } \quad C_{23} = \frac{2C_{21}e^{2C_{21}}}{3}.
\end{equation}

Next, we consider the following two cases:

* Suppose that $|\mathcal{B}_0| \geq v_2 - v_1$. This scenario corresponds to the picture on the left of Figure <ref>. As seen from the picture, this allows us to set $u_*$ to be arbitrarily close to $v_1$. Furthermore, observe that at $(u_*, v_2)$, we have

\begin{equation}\label{Scalar21}
x(u_*) \leq \frac{r_1(u_*)}{r_2(u_0)} + \frac{\delta_0}{1 + \frac{r_2(u_0) - r_1(u_0)}{r_1(u_0)}} =  \frac{r_1(u_*)}{r_2(u_0)} + \frac{\delta_0}{1+\delta_0}.
\end{equation}

Henceforth, by (<ref>) and the fact
that $r_1(u_*) \rightarrow 0$ as $u_* \rightarrow v_1^-$, we can have $x(u_*) \leq \frac{p\delta_0}{1+\delta_0}$ for any $p > 1$.

* Assume that $|\mathcal{B}_0| < v_2 - v_1$. We encounter the scenario portrayed in the right picture of Figure <ref>. As observed from the picture, we might not be able to pick $u_*$ to be arbitrarily close to $v_1$ due to the possibly spacelike nature of $\mathcal{B}\setminus \mathcal{B}_0$. Nonetheless,  $r_2(u)\rightarrow 0$ as $(u,v_2)$ approaches any point on $\mathcal{B} \setminus \mathcal{B}_0$. This implies that it suffices to pick $u_*$ along $v = v_2$ to be sufficiently close to $\mathcal{B} \setminus \mathcal{B}_0$ and we conclude $x(u_*) \leq \frac{p\delta_0}{1+\delta_0}$ for any $p > 1$ as in the previous case.

For both scenarios, since $x(u)$ is a decreasing continuous function on $[u_0,u_*]$, there exists $\tilde{u}(k)$ such that $x(\tilde{u}(k)) = \frac{k\delta_0}{1+\delta_0}$ for any $k > 1$. We then evaluate (<ref>) at the point $(\tilde{u}(k), v_2)$. Further applying (<ref>) we then obtain

\begin{equation}\label{TSF60}
\mu(\tilde{u}(k),v_2) \geq \frac{1}{C_{22}k}\ml (C_{24}-C_{22})\log\ml \frac{1}{\delta_0}\mr + C_{22} \log \ml k-1\mr\mr.
\end{equation}

Hence, we would have $\mu(\tilde{u}(k),v_2) \geq 1$ if

\begin{equation}\label{TSF61}
(C_{24}-C_{22})\log\ml \frac{1}{\delta_0}\mr \geq C_{22}\ml k - \log(k-1) \mr.
\end{equation}

Note that with $k>1$ the function $k-\log(k-1)$ is minimized at $k = 2$. Henceforth, we get $\mu(\tilde{u}(2),v_2) \geq 1$ if

\begin{equation}\label{TSF62}
(C_{24}-C_{22})\log\ml \frac{1}{\delta_0}\mr \geq 2C_{22}.
\end{equation}

This requirement is easily satisfied if we set

\begin{equation}\label{TSF54}
C_{24} = 3C_{22},
\end{equation}

since we have $\log\ml\frac{1}{\delta_0}\mr \geq 1$ for $\delta_0 \in (0,1)$. With this choice of $C_{24}$, we then obtain $\mu(\tilde{u}(2),v_2) \geq 1$, which guarantees that a MOTS or a trapped surface is formed in $R'$. This concludes the proof of Theorem <ref>.

§ BV AREA ESTIMATES

In this section, we establish scale-critical BV estimates for $C^1$ solutions with BV initial data. To begin, we first define the concept of area for maps of two Lipschitz functions. Let $\mathcal{U}$ be a domain in the $(u,v)$ plane and let $f,g$ be a pair of Lipschitz functions defined on it. The area $A(\mathcal{U})[f,g]$ for the image of $\mathcal{U}$ lying in a plane by the mapping $(u,v) \rightarrow (f(u),g(v))$ is given by

\begin{equation}\label{Area2}
A(\mathcal{U})[f,g] := \int_{\mathcal{U}} \mlm \frac{\p(f,g)}{\p(u,v)}\mrm \D u \D v
\end{equation}

\begin{equation}\label{Area3}
\frac{\p(f,g)}{\p(u,v)} = \frac{\p f}{\p u}\frac{\p g}{\p v} - \frac{\p f}{\p v}\frac{\p g}{\p u}
\end{equation}

being the Jacobian of the mapping $(u,v) \rightarrow (f(u,v),g(u,v))$.

A major difference between our charged system (<ref>)-(<ref>) and the uncharged scalar-field system investigated in [18] - [21] is that the charged system is not invariant under translations of $\phi$. In other words, for any constant $c$, the following transformation

$$ r\rightarrow r, \quad \Omega \rightarrow \Omega, \quad \phi \rightarrow \phi + c,$$

does not necessarily leaves all equations from (<ref>) - (<ref>) invariant. This means that we will have to modify the area estimates obtained in [19], in which it was assumed without loss of generality that $\phi(0,0) = 0$. In our case, we denote

\begin{equation}\label{tildephi}
\phi_0 := \phi(0,0), \quad \text{ and } \quad  \tp := \phi - \phi_0.
\end{equation}

Correspondingly, we define

\begin{equation}\label{mathfraka}
\mathfrak{a} := \dv(r\phi), \quad \quad \alpha := \frac{\dv(r\phi)}{\dv r} = \frac{\mfa}{\dv r},
\end{equation}

\begin{equation}\label{tildemathfraka}
\tilde{\mathfrak{a}} := \dv(r\tp), \quad  \quad \tilde{\alpha} := \frac{\dv(r\tp)}{\dv r} = \frac{\tmfa}{\dv r}.
\end{equation}

Analogously, we set

\begin{equation}\label{mathfrakb}
\mathfrak{b} := \du(r\phi), \quad \quad \beta := \frac{\du(r\phi)}{\du r} = \frac{\mfb}{\du r},
\end{equation}

\begin{equation}\label{tildemathfrakb}
\tilde{\mathfrak{b}} := \du(r\tp) \quad \quad \tilde{\beta} := \frac{\du(r\tp)}{\du r} = \frac{\tmfb}{\du r}.
\end{equation}

For convenience, we will also denote

\begin{equation}\label{lambda}
\lambda := \dv r, \quad \text{ and } \nu := \du r.
\end{equation}

The above definitions of $\alpha$ and $\beta$ are slightly different from [19], where $\alpha$ and $\beta$ are defined to be just $\dv(r\phi)$ and $\du(r\phi)$. To be consistent with the notations adopted in [21], we will adopt the definitions presented in this paper.

From the above definitions, it is not difficult to see that

\begin{equation}\label{Area39}
\tmfa|_{C_0^+} = \mfa|_{C_0^+} - \frac{1}{2}\phi_0.
\end{equation}

Therefore, the total variations of $\tmfa$ and $\mfa$ are the same. Hence, we define

\begin{equation}\label{Area40}
D := TV_{\{0\} \times (0,v_0)}[\tmfa] = TV_{\{0\} \times (0,v_0)}[\mfa],
\end{equation}

which measures the size of the initial data in the TV (total variation) norm. In addition, we define the following quantities

\begin{equation}\label{Area15}
\begin{aligned}
X(v_0) &= \sup_{u \in (0,v_0)} \int^{v_0}_u
|\dv \phi|(u,v') \D v' = \sup_{u \in (0,v_0)} \int^{v_0}_u
|\dv \tp|(u,v') \D v', \\
Y(v_0) &= \sup_{v \in (0,v_0)} \int^{v}_0
|\du \phi|(u',v) \D u' = \sup_{v \in (0,v_0)} \int^{v}_0
|\du \tp|(u',v) \D u'.
\end{aligned}
\end{equation}

For ease of computation, we pick $v_0$ sufficiently small, such that $\mu - Q^2/r^2 \geq 0$ in $\mathcal{D}(0,v_0)$. We assume that we have a $C^1$ solution on $\mathcal{D}(0,v_0)$ and $\mathcal{D}(0,v_0) \subset \mathcal{R}.$ This means that we can apply Proposition <ref> with $\overline{u} = 0$. In particular, by (<ref>), we have that $\mu - Q^2/r^2 \geq 0$ only if

\begin{equation}\label{Area11}
C_1^2(0,v;\chi)r^{1-\chi}(u,v)r^{\chi}(0,v)(v-u) \leq 1
\end{equation}

for all $(u,v) \in \mathcal{D}(0,v_0).$ By restricting our choice of $v_0$ to stay in $(0,1)$, it suffices to demand that

\begin{equation}\label{Area12}
C_1^2(0,1;\chi)r^{1-\chi}(u,v)r^{\chi}(0,v)(v-u) \leq 1
\end{equation}

since by (<ref>), we know that $C_1(0,\cdot;\chi)$ is a non-decreasing function. In particular, from (<ref>) and the fact that $\du r < 0$, $\dv r > 0$, and $u \geq 0$, if we pick $\chi = \frac{1}{2}$, it then suffices to demand that

\begin{equation}\label{Area13}
v_0^2 \leq \frac{2}{C_1^2\ml 0,1;\frac{1}{2}\mr}.
\end{equation}

For instance, by choosing

\begin{equation}\label{Area14}
v_0 = \min\left\{\oh, \frac{\sqrt{2}}{C_1 \ml 0, 1; \frac{1}{2}\mr}\right\},
\end{equation}

we then have $v_0 < 1$ and (<ref>) are satisfied. Note that such a choice is independent of $D$, which is important in this section. In the arguments that follow, we will refine $v_0$ to be even smaller if necessary, as long as it remains to be independent of $D$.

Last but not least, we work with the following assumptions for solutions in $\mathcal{D}(0,v_0)$:

\begin{equation}\label{Area27}
\inf_{\mathcal{D}(0,v_0)} \dv r \geq \frac{5}{12}, \quad \sup_{\mathcal{D}(0,v_0)} (-\du r) \leq \frac{2}{3}, \quad \sup_{\mathcal{D}(0,v_0)} \mu \leq \frac{1}{3}.
\end{equation}

These assumptions can be verified directly with the same type of arguments as in Section 6 of [19].

With the above preparation, we now state the main theorem of this section.

($BV$ Area Estimates.)
Assume that (<ref>) holds with our choice of $v_0$ to be given in (<ref>). Then there exists a constant $\varepsilon_0 \in (0,1]$ such that if the initial data satisfy

\begin{equation}\label{Area75}
Z(v_0) = \max\{X(v_0),Y(v_0)\} \leq \varepsilon_0, \quad \text{ and } \quad D := TV_{\{0\} \times (0,v_0)}[\tmfa]  \leq \varepsilon_0,
\end{equation}

then we have

\begin{equation}\label{Area76}
Z(v_0) \lesssim D,
\end{equation}

\begin{equation}\label{Area301}
\begin{aligned}
\sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\tmfa] &\lesssim D,\\
\sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\}}[\ttmfb] &\lesssim D,\\
TV_{(0,v_0)}[\tmfa|_{\Gamma}]  &\lesssim D,
\end{aligned} \quad
\begin{aligned}
\sup_{u \in (0,v_0)} TV_{ \{u\} \times (u,v_0) } [\log \lambda] &\lesssim D^2, \\
\sup_{v\in(0,v_0)} TV_{(0,v) \times \{v\}}[\log |\nu|] &\lesssim D^2, \\
TV_{(0,v_0)}[\log(\lambda|_{\Gamma})] &\lesssim D^2, \\
\end{aligned}
\end{equation}

\begin{equation}\label{Area302}
\begin{aligned}
A[\tp^\dagger ,\tmfa/\lambda] &\lesssim D^2, \\
A[\tp^\dagger ,\ttmfb/\nu] &\lesssim D^2, \\
\end{aligned} \quad
\begin{aligned}
A[\lambda, \tp] &\lesssim D^3, \\
A[\nu, \tp] &\lesssim D^3. \\
\end{aligned}
\end{equation}

For a quantity $H$, we write $H \lesssim D$ to mean that there exists a non-negative constant $c$ (possibly depending on $\e$ and $|\phi_0|$) independent of $D$ such that $H \leq cD$. For brevity, we also write $A[f,g]$ to mean $A(\mathcal{D}(0,v_0))[f,g]$ for any pair of Lipschitz functions $f$ and $g$.

To prove this theorem, we first state and prove several lemmas.

For any $(u,v) \in \mathcal{D}(0,v_0)$, we have

\begin{equation}\label{Area16}
\begin{aligned}
\int^{v_0}_u |\dv \tilde{\phi}|(u,v') \D v' &\leq TV_{\{u\} \times (u,v_0)} [\tilde\alpha], \quad \int^{v}_0 |\du \tilde{\phi}|(u',v) \D u' \leq TV_{(0,v) \times \{v\} } [\tilde\beta].
\end{aligned}
\end{equation}

The proof of this lemma is similar to the corresponding one in [19] and we outline the approach.
Using the fact that $r\phi|_{\Gamma} = 0$ and $r\phi \in C^1$, for each $(u,v) \in \mathcal{D}(0,v_0)$, we have

\begin{equation}\label{Area17}
\phi(u,v) = \frac{1}{r(u,v)}
\int^v_u \dv(r\phi) (u,v') \D v'.
\end{equation}

This implies

\begin{equation}\label{Area18}
\begin{aligned}
(\dv \phi)(u,v) &= - \frac{\dv r(u,v)}{r^2(u,v)}\int^{v}_u \dv (r\phi)(u,v') \D v' + \frac{\dv (r\phi)(u,v)}{r(u,v)} \\
&= \frac{(\dv r)(u,v)}{r^2(u,v)} \int^v_u \ml \alpha(u,v) - \alpha(u,v')\mr(\dv r)(u,v') \D v'.
\end{aligned}
\end{equation}

Next, since $\phi$, $r \dv \phi$, $\dv r$ are $C^1$ functions on $\mathcal{D}(0,v_0)$,  this implies that $\alpha \in C^1$ on $\mathcal{D}(0,v_0)$. Consequently, we have

\begin{equation}\label{Area19}
\lim_{v \rightarrow u+}\frac{ \int^v_u \ml \alpha(u,v) - \alpha(u,v')\mr(\dv r)(u,v') \D v'}{r(u,v)} = 0.
\end{equation}

We further denote

\begin{equation}\label{Ialpha}
I_{\alpha}(u,v) := \int^v_u \ml \alpha(u,v) - \alpha(u,v')\mr(\dv r)(u,v') \D v'
\end{equation}

and it satisfies

\begin{equation}\label{Area20}
\dv I_{\alpha} = r \dv \alpha.
\end{equation}

From (<ref>), integrating along $v \in (u,v_0)$ and applying integration by parts, we obtain

\begin{equation}\label{Area21}
\begin{aligned}
\int_u^{v_0} |\dv \phi|(u,v') \D v'
&= - \int_u^{v_0} |I_{\alpha}(u,v')|\D\ml \frac{1}{r(u,v')}\mr = -\frac{|I_{\alpha}|(u,v_0)}{r(u,v_0)} + \int^{v_0}_u \frac{\p |I_{\alpha}(u,v')|}{\p v}\frac{\D v}{r(u,v')} \\
&\quad\quad\leq \int^{v_0}_u \mlm \dv I_{\alpha} (u,v')\mrm \frac{1}{r(u,v')} \D v' = \int^{v_0}_u |\dv \alpha|
(u,v') \D v' = TV_{\{u\} \times (u,v_0)}[\alpha].
\end{aligned}
\end{equation}

Together with the fact that $\dv \phi = \dv \tp$ and $TV_{\{u\} \times (u,v_0)}[\alpha] = TV_{\{u\} \times (u,v_0)}[\tilde{\alpha}]$, inequality (<ref>) implies the first half of (<ref>).

For the second half of (<ref>), we begin by writing

\begin{equation}\label{Area22}
\phi(u,v) = - \frac{1}{r(u,v)}
\int^v_u \du(r\phi) (u',v) \D u'.
\end{equation}

We proceed as above and compute

\begin{equation}\label{Area23}
(\du \phi)(u,v) = \frac{(-\du r)(u,v)}{r^2(u,v)} \int^v_u \ml \beta(u,v) - \beta(u',v)\mr(\du r)(u',v) \D u'.
\end{equation}

Next, we denote

\begin{equation}\label{Ibeta}
I_\beta(u,v) := \int^v_u \ml \beta(u,v) - \beta(u',v)\mr(\du r)(u',v) \D u'
\end{equation}

and observe that

\begin{equation}\label{Area24}
\lim_{u \rightarrow v^-}\frac{I_{\beta}(u,v)}{r(u,v)} = 0 \quad \text{ and } \quad
\du I_\beta = - r \du \beta.
\end{equation}

Integrating (<ref>) along $u \in (0,v)$, we obtain

\begin{equation}\label{Area26}
\begin{aligned}
\int^v_0 |\du \phi|(u',v) \D u' &= -\int^{v}_0 |I_\beta (u',v)| \D \ml \frac{1}{r(u',v)}\mr \\
&\leq \int^{v}_0 |\du \beta|(u',v) \D u' = TV_{(0,v) \times \{v\}}[\beta].
\end{aligned}
\end{equation}

Since $TV_{(0,v) \times \{v\}}[\beta] = TV_{(0,v) \times \{v\}}[\tilde\beta]$, we then prove the second inequality in (<ref>).

We now attempt to derive estimates for the following quantities:

\begin{equation}\label{Area28}
A := \sup_{\mathcal{D}(0,v_0)}|\tmfa|, \quad B := \sup_{\mathcal{D}(0,v_0)}|\tmfb|, \quad P := \sup_{\mathcal{D}(0,v_0)}|\tp|.
\end{equation}

We summarize the estimates for these three quantities in a lemma below.

Suppose that $D \leq 1$. By refining of $v_0$ to be as described in (<ref>), we have

\begin{equation}\label{Area29}
A \leq 4D, \quad B \leq 8D, \quad P \leq 12D.
\end{equation}

We begin by considering the set

\begin{equation}\label{Area70}
\mathcal{\tilde{D}} = \left\{ (\tu,\tv) \in \mathcal{D}(0,v_0): \sup_{\mathcal{D}(0,\tu,\tv)} \mlm \tmfa \mrm \leq 4 D, \quad \sup_{\mathcal{D}(0,\tu,\tv)} \mlm \tp \mrm \leq 12 D \right \}.
\end{equation}

Note that $(0,0) \in \mathcal{\tilde{D}}$ and $\mathcal{\tilde{D}}$ is a closed subset of $\mathcal{D}(0,v_0)$. It remains to show that $\mathcal{\tilde{D}}$ is open.

First, observe that for each $(u,v) \in \mathcal{D}(0,v_0)$, we have

\begin{equation}\label{Area30}
r(u,v) = \int^v_u (\dv r)(u,v') \D v' \geq (v-u) \inf_{\mathcal{D}(0,v_0)} (\dv r) \geq \frac{5}{12}(v-u) \geq \frac{1}{3}(v-u).
\end{equation}

Furthermore, analogous to (<ref>), we get

\begin{equation}\label{Area41}
\tp = \frac{1}{r(u,v)}\int^v_u \tmfa (u,v') \D v'.
\end{equation}

Applying (<ref>) to (<ref>), we hence obtain

\begin{equation}\label{Area31}
\sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \leq \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \sup_{\mathcal{D}(0,\tu,\tv)}\mlm \frac{v - u}{r(u,v)}\mrm \leq 3 \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa|.
\end{equation}

Next, within $\mathcal{D}(0,\tu,\tv)$, we then compute $\du(\dv (r\phi))$ using (<ref>), (<ref>), (<ref>) as follows

\begin{equation}\label{Area32}
\begin{aligned}
\du \ml \dv(r\phi) \mr &= \du r \dv \phi + r \du \dv \phi + \du \phi \dv r + \phi \du \dv r \\
&= \phi \du \dv r - \ii \e  A_u \dv (r\phi) - \ii \e
\frac{\du r \dv r }{1 - \mu}\frac{Q\phi}{r}.
\end{aligned}
\end{equation}

The above implies

\begin{equation}\label{Area42}
\du\ml \dv \ml r\tp \mr \mr = \tp \du \dv r- \ii \e  A_u \dv (r\tp) - \ii \e \phi_0 A_u \dv r  - \ii \e
\frac{\du r \dv r }{1 - \mu}\frac{Q\phi}{r}.
\end{equation}

By viewing (<ref>) as a first order ODE for $\dv(r\tp)$ in the $u$-direction and noting that the integrating factor $e^{\int^{u}_0 \ii \e A_u(u',v) \D u'}$ only contributes to a phase change, we henceforth derive

\begin{equation}\label{Area33}
\begin{aligned}
& e^{\ii \e \int^{u}_0 A_u(u',v) \D u' }\dv(r\tp)(u,v) \\
= &\; \dv(r\tp)(0,v) + \int^u_0 e^{\ii \e \int^{u'}_0 A_u(u'',v) \D u'' }\ml \tp \du \dv r - \ii \e \phi_0 A_u \dv r - \ii \e
\frac{\du r \dv r }{1 - \mu}\frac{Q\phi}{r} \mr(u',v) \D u',
\end{aligned}
\end{equation}

which implies that

\begin{equation}\label{Area34}
\begin{aligned}
|\dv(r\tp)|(u,v) \leq&\; |\dv (r\tp)|(0,v) + \int^u_0 |\tp|(-\du \dv r) (u',v)\D u'   \\
&+ \e \int^u_0 \frac{(-\du r)(\dv r)}{r(1-\mu)}|Q||\phi| (u',v)\D u'  + \e |\phi_0| \int^u_0 |A_u| \dv r (u',v) \D u'.
\end{aligned}
\end{equation}

In the above estimate, we have used the fact that $\mu - Q^2/r^2 \geq 0$ from our choice of $v_0$ in (<ref>) and its implication $-\du \dv r \geq 0$ by (<ref>).

It remains to estimate the integrals in (<ref>). By employing $\dv r\geq \frac{5}{12}$ and (<ref>), the first integral obeys

\begin{equation}\label{Area35}
\begin{aligned}
\int^u_0 |\tp|(-\du \dv r) (u',v)\D u' \leq \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr\ml (\dv r)(0,v) - (\dv r)(u,v) \mr \leq \frac{1}{12}\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr.
\end{aligned}
\end{equation}

For the second integral in (<ref>), since $|\phi| \leq |\tp| + |\phi_0| \leq \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr + |\phi_0|$ for every $(u',v) \in \mathcal{D}(0,\tu,\tv)$, we have

\begin{equation}\label{Area43}
\begin{aligned}
\e \int^u_0 \frac{(-\du r)(\dv r)}{r(1-\mu)}|Q||\phi| (u',v)\D u' \leq & \;  \e \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr \int^u_0 \frac{(-\du r)(\dv r)}{r(1-\mu)}|Q| (u',v)\D u' \\
&+  \e |\phi_0| \int^u_0 \frac{(-\du r)(\dv r)}{r(1-\mu)}|Q| (u',v)\D u'.
\end{aligned}
\end{equation}

For the two integrals in (<ref>), we apply two different estimates for $|Q|$. For the first integral, since $\mathcal{D}(0,v_0) \subset \mathcal{R}$, we have $\mu \leq 1$ and $\mu - Q^2/r^2 \geq 0$, which in turn imply $|Q| \leq \mu^{\frac{1}{2}} r \leq r$. In light of Lemma <ref> and our assumptions in (<ref>), we thus have

\begin{equation}\label{Area36}
\begin{aligned}
\e \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr& \int^u_0 \frac{(-\du r)(\dv r)}{r(1-\mu)}|Q| (u',v)\D u'
\leq \e \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr \int^u_0 \frac{(-\du r)(\dv r)}{1- \mu}(u',v) \D u' \\
&\quad\quad \leq \e \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr \frac{\dv r}{1 - \mu}(0,v)\int^u_0 (-\du r)(u',v) \D u'\\
&\quad\quad \leq \ml \e \frac{3v}{8}\mr \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr \leq \ml \e \frac{3v_0}{8}\mr \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr \leq \frac{1}{12}\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr
\end{aligned}
\end{equation}

as long as we refine $v_0$ from (<ref>) to be

\begin{equation}\label{Area37}
v_0 = \min\left\{\oh, \frac{\sqrt{2}}{C_1 \ml 0, 1; \frac{1}{2}\mr}, \frac{2}{9\e} \right\}.
\end{equation}

For the second integral in (<ref>), we use (<ref>), $Q|_{\Gamma} = 0$ and their implication

\begin{equation}\label{Area44}
\begin{aligned}
|Q|(u,v) \leq 4 \pi \e \int^v_u r^2 |\phi||\dv \phi|(u,v') \D v' \leq 4 \pi \e \ml |\phi_0| + \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr r(u,v) \int^v_u r |\dv \phi|(u,v') \D v'. \\
\end{aligned}
\end{equation}

We further employ (<ref>) and obtain

\begin{equation}\label{Area45}
\begin{aligned}
\int^v_u r|\dv \phi|(u,v') \D v' &= \int^v_u r|\dv \tp|(u,v') \D v' \leq \int^v_u |\dv (r\tp)|(u,v') + |\tp|(\dv r)(u,v') \D v' \\
&\quad \leq \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr (v-u) + \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr r(u,v) \leq 6\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr r(u,v). \\
\end{aligned}
\end{equation}

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

\begin{equation}\label{Area47}
|Q|(u,v) \leq 24\pi\e \ml \mlm \phi_0 \mrm + \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr  r(u,v)^2.
\end{equation}

Analogous to (<ref>), by using (<ref>), we estimate the second integral in (<ref>) and get

\begin{equation}\label{Area48}
\begin{aligned}
\e |\phi_0| \int^u_0 \frac{(-\du r)(\dv r)}{r(1-\mu)}|Q| (u',v)\D u'
&\leq 12 \pi v_0 \e^2 |\phi_0| \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr \ml \mlm \phi_0 \mrm + \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr  \int^u_0 \frac{(-\du r)(\dv r)}{1- \mu}(u',v) \D u' \\
&\leq \frac{9}{2} \pi v_0^2 \e^2 |\phi_0| \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr \ml \mlm \phi_0 \mrm + \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr. \\
\end{aligned}
\end{equation}

For the remaining integral in (<ref>), we begin by estimating $|A_u|$. For $(u',v) \in \tmcd$, employing (<ref>), (<ref>), (<ref>), we deduce

\begin{equation}\label{Area68}
\begin{aligned}
|A_u|(u',v) &\leq \int^{v}_{u'} \frac{2|Q| (-\du r) \dv r}{r^2(1-\mu)} (u',v') \D v' \\
&\leq 48 \pi \e \ml \mlm \phi_0 \mrm + \sup_{\mathcal{D}(0,u',v)}|\tp| \mr  \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr \int^v_{u'} \frac{(-\du r)(\dv r)}{1-\mu}(u',v') \D v'  \\
&\leq 48 \pi \e \ml \mlm \phi_0 \mrm + \sup_{\mathcal{D}(0,u',v)}|\tp| \mr  \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr r(u,v)  \\
&\leq 24 \pi v_0 \e \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr \ml \mlm \phi_0 \mrm + \sup_{\mathcal{D}(0,u',v)}|\tp| \mr.
\end{aligned}
\end{equation}

The above further implies

\begin{equation}\label{Area69}
\e |\phi_0| \int^u_0 |A_u|(\dv r)(u',v) \D u' \leq 12 \pi v_0^2 \e^2 |\phi_0| \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr \ml \mlm \phi_0 \mrm + \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr.
\end{equation}

We now plug (<ref>), (<ref>), (<ref>) into (<ref>). Together with (<ref>), we obtain

\begin{equation}\label{Area50}
\begin{aligned}
|\dv(r\tp)|(u,v) =& \; |\dv (r\tp)|(0,v) + \int^u_0 |\tp|(-\du \dv r) (u',v)\D u' + \e \int^u_0 \frac{(-\du r)(\dv r)}{r(1-\mu)}|Q||\phi| (u',v)\D u' \\
&+ \e |\phi_0| \int^u_0 |A_u| \dv r (u',v) \D u'\\
\leq& \; D + \frac{1}{6}\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr + \frac{33}{2} \pi \e^2 |\phi_0| v_0^2 \ml \mlm \phi_0 \mrm + \sup_{\mathcal{D}(0,\tu,\tv)}|\tp|  \mr \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr.
\end{aligned}
\end{equation}

In (<ref>), we use $D := TV_{\{0\} \times (0,v_0)}[\tmfa]$, $\tmfa(0,0) = 0$ and the implication $|\dv(r\tp)|(0,v) \leq D$. Taking supremum over $\mathcal{D}(0,\tu,\tv)$ and applying (<ref>) and (<ref>), we deduce that

\begin{equation}\label{Area71}
\begin{aligned}
\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr &\leq D + \frac{1}{6}\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \mr + \frac{33}{2} \pi \e^2 |\phi_0| v_0^2 \ml \mlm \phi_0 \mrm + \sup_{\mathcal{D}(0,\tu,\tv)}|\tp|  \mr \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr \\
&\leq D +  \frac{1}{2}\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr + \frac{33}{2} \pi \e^2 |\phi_0| v_0^2 \ml \mlm \phi_0 \mrm + 12 \mr \ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr \\
&\leq D +  \frac{2}{3}\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr,
\end{aligned}
\end{equation}

which gives

\begin{equation}\label{Area52}
\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr \leq 3D.
\end{equation}

Note that in (<ref>) we further restrict our choice of $v_0$ as in (<ref>) to

\begin{equation}\label{Area72}
v_0 = \min\left\{\oh, \frac{\sqrt{2}}{C_1 \ml 0, 1; \frac{1}{2}\mr}, \frac{2}{9\e}, \frac{1}{\sqrt{99\pi\e^2|\phi_0|\ml |\phi_0| + 12 \mr}} \right\}.
\end{equation}

Furthermore, by (<ref>), inequality (<ref>) also leads to

\begin{equation}\label{Area53}
\sup_{\mathcal{D}(0,\tu,\tv)}|\tp| \leq 3\ml \sup_{\mathcal{D}(0,\tu,\tv)}|\tmfa| \mr \leq 9D.
\end{equation}

We thus retrieve our bootstrap estimate in (<ref>) and deduce that $\tmcd$ is an open subset of $\mathcal{D}(0,v_0)$. Moreover, the derived estimates also extend to the entire region $\mathcal{D}(0,v_0)$.

We proceed to estimate $B$. In the following arguments, we take $(u,v) \in \mathcal{D}(0,v_0)$ and set $\tu = \tv = v_0$. By viewing (<ref>) as an ODE in the $v$-direction for $\du \ml r \tp \mr$ and by applying (<ref>), we deduce

\begin{equation}\label{Area54}
\begin{aligned}
\mlm \du \ml r\tp \mr \mrm (u,v)\leq & \; \mlm \dv \ml r\tp \mr \mrm (u,u) +   \int^v_u |\tp|(- \du \dv r)(u,v') \D v'  + \e \int^v_u |A_u| \mlm \dv (r \tp)\mrm  (u,v') \D v' \\
&+ \e |\phi_0|  \int^v_u |A_u|(\dv r) (u,v') \D v' + \e \int^v_u \frac{(-\du r) \dv r}{1 - \mu}\frac{|Q| |\phi|}{r} (u,v') \D v'. \\
\end{aligned}
\end{equation}

Analogous to the previous case, since $\du\dv r \leq 0$ in $\mathcal{D}(0,v_0)$, we bound the first integral in (<ref>) by

\begin{equation}\label{Area60}
\begin{aligned}
\int^v_u |\tp|(-\du \dv r) (u,v')\D v' &\leq P\ml (-\du r)(u,v) - (\dv r)(u,u) \mr \leq \frac{P}{4} \leq 3D.
\end{aligned}
\end{equation}

Furthermore, by (<ref>), we get

\begin{equation}\label{Area55}
\begin{aligned}
|A_u|(u,v') \leq 24 \pi \e (|\phi_0| + P)A v_0 \leq 96 \pi \e (|\phi_0| + 12) v_0 D.
\end{aligned}
\end{equation}

With these, we can see that the second and the third integral in (<ref>) obey

\begin{equation}\label{Area58}
\begin{aligned}
\e \int^v_u |A_u| \mlm \dv (r \tp)\mrm  (u,v') \D v' &\leq 96 \pi\e^2 (|\phi_0| + 12) v_0^2 A D \leq 384 \pi\e^2 (|\phi_0| + 12) v_0^2 D \leq \frac{D}{4}
\end{aligned}
\end{equation}

\begin{equation}\label{Area73}
\begin{aligned}
\e |\phi_0| \int^v_u |A_u| (\dv r) (u,v') \D v' &\leq 96 \pi\e^2 |\phi_0|(|\phi_0| + 12) v_0^2 D \leq \frac{D}{4},
\end{aligned}
\end{equation}

if we refine our choice of $v_0$ from (<ref>) to

\begin{equation}\label{Area59}
v_0 = \min\left\{\oh, \frac{\sqrt{2}}{C_1 \ml 0, 1; \frac{1}{2}\mr}, \frac{2}{9\e}, \frac{1}{\sqrt{384\pi\e^2|\phi_0|\ml |\phi_0| + 12 \mr}}, \frac{1}{\sqrt{1536\pi}\e\sqrt{|\phi_0| + 12}}\right\}.
\end{equation}

To estimate the last integral in (<ref>), by  (<ref>), we first note

\begin{equation}\label{Area62}
|Q\phi| \leq 12 \pi \e \ml \mlm \phi_0 \mrm + 12 \mr^2 A  v_0 r(u,v) \leq 48\pi \e \ml \mlm \phi_0 \mrm + 12 \mr^2 D  v_0 r(u,v).
\end{equation}

Henceforth, the last integral in (<ref>) satisfies

\begin{equation}\label{Area63}
\begin{aligned}
\e \int^v_u \frac{(-\du r) \dv r}{1 - \mu}\frac{|Q| |\phi|}{r} (u,v') \D v' &\leq 48 \pi \e^2 \ml |\phi_0| + 12\mr^2 v_0 D \int^v_u \frac{(-\du r) \dv r}{1 - \mu} (u,v') \D v' \\
&\leq 24 \pi \e^2 \ml |\phi_0| + 12\mr^2 v_0^2 D \leq \frac{D}{2}
\end{aligned}
\end{equation}

if we restrict our choice of $v_0$ from (<ref>) to

\begin{equation}\label{Area64}
v_0 = \min\left\{\oh, \frac{\sqrt{2}}{C_1 \ml 0, 1; \frac{1}{2}\mr}, \frac{2}{9\e}, \frac{1}{\sqrt{384\pi\e^2|\phi_0|\ml |\phi_0| + 12 \mr}}, \frac{1}{\sqrt{1536\pi}\e\sqrt{|\phi_0| + 12}}, \frac{1}{\sqrt{48\pi}\e(|\phi_0| + 12)}\right\}.
\end{equation}

Plugging our bounds from (<ref>), (<ref>), (<ref>) into (<ref>), we deduce that

\begin{equation}\label{Area65}
B \leq A + 3D + \frac{D}{2} + \frac{D}{2} \leq 8D.
\end{equation}

This then completes the proof of Lemma <ref>.

With the above lemma proven, we return to the proof of Theorem <ref>.
By Lemma <ref> we can now write

\begin{equation}\label{Area81}
A \lesssim D, \quad B \lesssim D, \quad P \lesssim D.
\end{equation}

Furthermore, recall that in the proof of Lemma <ref>, we chose $v_0$ such that $\du \dv r \leq 0$. Combined with the fact that $\mathcal{D}(0,v_0) \subset \mathcal{R}$ and $\dv r|_{\Gamma} = -\du r|_{\Gamma}$, for all $(u,v) \in \mathcal{D}(0,v_0)$, (<ref>) then implies

\begin{equation}\label{Area84}
\frac{5}{12} \leq \dv r(u,v) \leq \frac{1}{2}, \quad \text{ and } \quad \frac{1}{2} \leq (-\du r)(u,v) \leq \frac{2}{3}.
\end{equation}

Henceforth, applying

\begin{equation}\label{Area66}
r \dv \tp = \dv \ml r \tp \mr - \tp \dv r, \quad r \du \tp = \du \ml r \tp \mr - \tp \du r, \quad \phi = \phi_0 + \tp,
\end{equation}

and $D \leq \varepsilon_0 \leq 1$, we then get

\begin{equation}\label{Area82}
\sup_{\mathcal{D}(0,v_0)}\mlm r \dv \tp \mrm \lesssim D, \quad \sup_{\mathcal{D}(0,v_0)}\mlm r \du \tp \mrm \lesssim D, \quad \sup_{\mathcal{D}(0,v_0)}\mlm \phi \mrm \lesssim 1.
\end{equation}

Plugging these estimates into (<ref>) and (<ref>), for each $(u,v) \in \mathcal{D}(0,v_0)$, we further have

\begin{equation}\label{Area83}
|Q|(u,v) \lesssim D r^2(u,v) \quad \text{ and } \quad |A_u|(u,v) \lesssim Dr(u,v).
\end{equation}

Note that if needed, we can also use $r(u,v) \leq r(0,v_0) = \frac{1}{2}v_0 \leq 1$. As a consequence, we obtain

\begin{equation}\label{Area114}
|rD_u \phi|(u,v) = |r\du \phi + \ii \e A_u r \phi|(u,v) \lesssim D + Dr(u,v)^2 \lesssim D.
\end{equation}

On top of the estimates in the region $\mathcal{D}(0,v_0)$, we also derive some asymptotics (in $r$) for $C^1$ solutions at points sufficiently close to $\Gamma$. This is to justify that various terms vanish as we take the limit along incoming or (backward) outgoing null curves towards points on $\Gamma$. For a $C^1$ solution, we have that $\phi, r \dv \phi, \text{ and }r \du \phi$ are $C^1$ functions on $\overline{\mathcal{D}(0,v_0)}$. Since $\dv \phi$ and $\du \phi$ are continuous on $\overline{\mathcal{D}(0,v_0)}$, these imply that $r \dv \phi|_{\Gamma} = r \du \phi|_{\Gamma} = 0$.
By the fundamental theorem of calculus and the extreme value theorem, there exists a constant $M(v_0)$ depending on $v_0$ such that

\begin{equation}\label{Area310}
r|\dv \phi|(u,v) \leq M(v_0) (v-u) \quad \text{ and } \quad r|\du \phi| \leq M(v_0)(v-u).
\end{equation}

Under the bootstrap assumption, similar to the proof in (<ref>), we now have $v-u \leq 3r(u,v)$. Hence, we get

\begin{equation}\label{Area311}
|\dv \tilde{\phi}|(u,v) = |\dv \phi|(u,v) \leq 3M(v_0) \quad \text{ and } \quad |\du \tilde{\phi}|(u,v) = |\du \phi| \leq 3M(v_0).
\end{equation}

In addition, since $\dv r$ and $\du r$ are also $C^1$ functions, a similar argument allows us to deduce that there exists a constant $M_1(v_0)$ such that

\begin{equation}\label{Area521}
|\tmfa| (u,v) \leq M_1(v_0)r(u,v) \quad \text{ and } \quad |\tmfb|(u,v) \leq M_1(v_0)r(u,v).
\end{equation}

Additionally, as $\phi$, $\dv r$, and $\du r$ are $C^1$ functions on $\overline{\mathcal{D}(0,v_0)}$, these are bounded functions on $\overline{\mathcal{D}(0,v_0)}$.
Furthermore, since $Q/r$ is yet another $C^1$ function, there exists a constant $M_2(v_0)$ such that

\begin{equation}\label{Area312}
|Q|(u,v) \leq M_2(v_0)r^2(u,v) \quad \text{ and } \quad |A_u|(u,v) \leq M_2(v_0)r(u,v).
\end{equation}

Using (<ref>), together with (<ref>) and $m|_{\Gamma} = 0$, we derive

\begin{equation}\label{Area313}
\begin{aligned}
m(u,v) &\leq \int_u^v \frac{2\pi r^2(1-\mu)|\dv \phi|^2}{\dv r}(u,v') + \frac{Q^2}{r^4}\cdot(r^2 \dv r)(u,v') \D v' \\
&\leq \frac{24}{5}(3M(v_0))^2 \pi r^2 (v - u) + \frac{M_2(v_0)^2}{2} r^2(u,v) (v - u) \\
&\leq \ml \frac{648}{5}\pi^2 M^2(v_0) + \frac{3M_2(v_0)^2}{2}\mr r^3(u,v).
\end{aligned}
\end{equation}

This implies

\begin{equation}\label{Area314}
\mu(u,v) \leq \ml \frac{1296}{5}\pi^2 M^2(v_0) + 3M_2(v_0)^2\mr r^2(u,v).
\end{equation}

We are back to derive estimates for $\tmfa$. Differentiating both sides of (<ref>) with respect to $v$, we get

\begin{equation}
\begin{aligned}
\du \dv \tmfa &= \dv\ml \tp \du \dv r\mr - \ii \e A_u \dv \tmfa - \ii \e \phi_0 (\dv A_u) \dv r - \ii \e \phi_0 A_u \dv \lambda  - \ii \e \dv \ml \frac{\du r \dv r}{1-\mu}\frac{Q \phi}{r}\mr. \\
\end{aligned}
\end{equation}

This can be rewritten as

\begin{equation}\label{Area67}
\begin{aligned}
\du \ml \dv \tmfa \mr &= \du\ml \tp \dv \lambda \mr + G_1 + G_2 + \frac{\p (\lambda, \tp)}{\p(u,v)}
\end{aligned}
\end{equation}

\begin{equation}\label{Area77}
\begin{aligned}
G_1 &= -\ii \e \ml \phi_0 A_u + \frac{\du r }{1-\mu}\frac{Q\phi}{r} \mr \dv \lambda, \\
G_2 &=  -\ii \e \ml (\dv A_u) \tmfa + \phi_0 (\dv A_u)\dv r + A_u \dv \tmfa + \dv r \dv \ml \frac{\du r}{1-\mu}\frac{Q\phi}{r}\mr\mr.
\end{aligned}
\end{equation}

For each $(u,v) \in \mathcal{D}(0,v_0)$, we then have

\begin{equation}\label{Area78}
\begin{aligned}
\mlm \dv \tmfa - \tp \dv \lambda \mrm (u,v) \leq \mlm \dv \tmfa - \tp \dv \lambda \mrm(0,v) + \int^u_0 \mlm G_1 + G_2 + \frac{\p(\lambda,\tp)}{\p(u,v)}\mrm(u',v) \D u'.
\end{aligned}
\end{equation}

Using the fact $\lambda(0,v) = \frac{1}{2}$ and integrating (<ref>) with respect to $v$ from $u$ to $v_0$, we obtain

\begin{equation}\label{Area79}
\begin{aligned}
\int_u^{v_0} \mlm \dv \tmfa - \tp \dv \lambda \mrm (u,v') \D v' \leq \int^{v_0}_u \mlm \dv \tmfa \mrm(0,v') \D v' + \int_u^{v_0}\int^u_0 \ml |G_1| + |G_2| + \mlm\frac{\p(\lambda,\tp)}{\p(u,v)}\mrm \mr(u',v') \D u' \D v'.
\end{aligned}
\end{equation}

By taking the supremum over $u \in (0,v_0)$ in (<ref>) and using (<ref>), (<ref>), we arrive at

\begin{equation}\label{Area80}
\begin{aligned}
\sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\tmfa] \leq& \; D + P \sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\lambda] + A[\lambda,\tp] \\
&+ \sup_{u\in(0,v_0)}\int_u^{v_0}\int^u_0 |G_1| (u',v') \D u' \D v' + \sup_{u\in(0,v_0)}\int_u^{v_0}\int^u_0 |G_2| (u',v') \D u' \D v'.
\end{aligned}
\end{equation}

Next, we turn to estimate the integrals containing $G_1$ and $G_2$ in (<ref>). Observe that, for each $(u,v) \in \mathcal{D}(0,v_0)$, the following inequality holds uniformly in $(u,v)$

\begin{equation}\label{Area85}
\mlm \phi_0 A_u + \frac{\du r}{1- \mu}\frac{Q\phi}{r}\mrm(u,v) \lesssim D r(u,v) \lesssim D.
\end{equation}

This implies that for every $u \in (0,v_0)$, we have

\begin{equation}\label{Area86}
\begin{aligned}
\int_u^{v_0}\int^u_0 |G_1| (u',v') \D u' \D v' &\lesssim D \int_0^{u}\int^{v_0}_u |\dv \lambda| (u',v') \D v' \D u' \lesssim D \sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\lambda],
\end{aligned}
\end{equation}

where we employ the fact $u \leq v_0 \leq 1$ and apply Fubini's theorem to interchange the order of integrations.

To estimate the integrand $G_2$, we start by writing

\begin{equation}\label{Area87}
\begin{aligned}
\dv \ml \frac{\du r}{1-\mu}\frac{Q \phi}{r}\mr &= \frac{\dv \du r}{1-\mu}\frac{Q\phi}{r} + \frac{\du r}{1-\mu}\frac{(\dv Q)\phi}{r} + \frac{\du r}{1-\mu}\frac{Q \dv \phi}{r} - \frac{\du r}{1-\mu}\frac{Q\phi}{r^2}\dv r + \frac{\du r}{(1-\mu)^2}\frac{Q\phi}{r}\dv \mu
\end{aligned}
\end{equation}

and proceed to estimate each of these terms. For the first term, by (<ref>), we have

\begin{equation}\label{Area88}
\begin{aligned}
\mlm \frac{\dv \du r}{1-\mu}\frac{Q\phi}{r} \mrm(u,v) &= \mlm \frac{(\dv r)(\du r)}{(1-\mu)^2}\frac{Q\phi}{r^2}\ml \mu - \frac{Q^2}{r^2}\mr \mrm(u,v) \lesssim D,
\end{aligned}
\end{equation}

where we have used (<ref>), (<ref>), (<ref>), (<ref>), (<ref>) and the fact that $|\mu - Q^2/r^2| = \mu - Q^2/r^2 \leq \mu \leq 1/3$. In the remaining parts of this section, we will not explicitly mention that these are used unless we really need. In a similar fashion, for the second term in (<ref>), by (<ref>), we obtain

\begin{equation}\label{Area89}
\begin{aligned}
\mlm \frac{\du r}{1-\mu}\frac{(\dv Q)\phi}{r}\mrm(u,v) = \mlm \frac{\du r}{1-\mu}\frac{\phi}{r}(-4\pi \e r^2\im(\phi^\dagger \dv \phi))\mrm(u,v) \lesssim D.
\end{aligned}
\end{equation}

The third term in (<ref>) can be estimated directly as

\begin{equation}\label{Area90}
\begin{aligned}
\mlm \frac{\du r}{1-\mu}\frac{Q \dv\phi}{r}\mrm(u,v) \lesssim D^2 \lesssim D.
\end{aligned}
\end{equation}

The fourth term in (<ref>) obeys

\begin{equation}\label{Area91}
\begin{aligned}
\mlm \frac{\du r}{1-\mu}\frac{Q \phi}{r^2}\dv r\mrm(u,v) \lesssim D.
\end{aligned}
\end{equation}

For the fifth term in (<ref>), we appeal to (<ref>) and obtain

\begin{equation}\label{Area92}
\begin{aligned}
\mlm \frac{\du r}{(1-\mu)^2}\frac{Q \phi}{r}\dv \mu\mrm(u,v) &= \mlm \frac{\du r}{(1-\mu)^2}\frac{Q \phi}{r}\ml \frac{4\pi r(1-\mu)|\dv \phi|^2}{\dv r} - \frac{\dv r}{r}\ml \mu - \frac{Q^2}{r^2}\mr \mr\mrm(u,v)  \lesssim D^3 + D \lesssim D.
\end{aligned}
\end{equation}

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

\begin{equation}\label{Area93}
\mlm \dv \ml \frac{\du r}{1-\mu}\frac{Q \phi}{r}\mr\mrm (u,v) \lesssim D.
\end{equation}

For the other terms of $G_2$ in (<ref>), by (<ref>) and $A \lesssim D$, we get

\begin{equation}\label{Area94}
\mlm (\dv A_u)\tmfa\mrm(u,v) = \mlm \frac{2Q\du r \dv r}{r^2(1-\mu)}\tmfa\mrm(u,v) \lesssim D^2 \lesssim D.
\end{equation}

Similarly, we deduce that

\begin{equation}\label{Area95}
\mlm \phi_0(\dv A_u)\dv r\mrm(u,v) = \mlm \frac{2Q\du r \dv r}{r^2(1-\mu)}\phi_0 \dv r\mrm(u,v) \lesssim D.
\end{equation}

For the term $A_u \dv \tmfa$, we first observe from (<ref>) that

\begin{equation}\label{Area96}
|A_u|(u,v) \leq 96v_0 \e (|\phi_0| + 12)D.
\end{equation}

By modifying our choice of $v_0$ from (<ref>) to

\begin{equation}\label{Area97}
v_0 = \min\left\{\oh, \frac{\sqrt{2}}{C_1 \ml 0, 1; \frac{1}{2}\mr}, \frac{2}{9\e}, \frac{1}{\sqrt{384\e^2|\phi_0|\ml |\phi_0| + 12 \mr}}, \frac{1}{\sqrt{1536}\e\sqrt{|\phi_0| + 12}}, \frac{3}{128\e(|\phi_0| + 12)}\right\},
\end{equation}

we obtain

\begin{equation}\label{Area98}
|A_u|(u,v) \leq \frac{9}{4}D \leq \frac{9}{4},
\end{equation}

which in turn implies that

\begin{equation}\label{Area99}
\mlm A_u \dv \tmfa \mrm(u,v) \leq  \frac{9}{4}\mlm \dv \tmfa \mrm(u,v).
\end{equation}

Consequently, we prove

\begin{equation}\label{Area100}
\begin{aligned}
\int^{v_0}_u \int^u_0 |A_u \dv \tmfa|(u',v') \D u' \D v' &= \int^u_0  \int^{v_0}_u |A_u \dv \tmfa|(u',v') \D v' \D u' \leq \frac{1}{2\e} \sup_{u \in (0,v_0)}TV_{\{u\} \times (u,v_0) }[\tmfa].
\end{aligned}
\end{equation}

Combining (<ref>), (<ref>), (<ref>), (<ref>), together with (<ref>), we thus arrive at

\begin{equation}\label{Area101}
\int_u^{v_0}\int^u_0 |G_2| (u',v') \D u' \D v' - \frac{1}{2}\sup_{u \in (0,v_0)}TV_{\{u\} \times (u,v_0) }[\tmfa] \lesssim  \int_u^{v_0}\int^u_0 D \D u' \D v' \lesssim D.
\end{equation}

Since the estimates in (<ref>) and (<ref>) are uniform in $u \in (0,v_0)$, taking supremum over $u \in (0,v_0)$ and substituting them into (<ref>), it yields

\begin{equation}\label{Area103}
\begin{aligned}
\frac{1}{2}\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{aligned}
\end{equation}

Note that in (<ref>), we are allowed to shift the term $\frac{1}{2}\sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\tmfa]$ in the inequality as it is a finite term for a $C^1$ solution.

Next, we move to derive estimates for $\tmfb$. In view of the definition of $\tmfb$ in (<ref>), we define a renormalized quantity

\begin{equation}\label{tildetildemathfrakb}
\ttmfb := \tmfb + \ii \e A_u r \phi = \tp \du r + r \du \tp + \ii \e A_u r \phi = \tp \du r + r D_u \phi.
\end{equation}

By (<ref>), taking derivative of it with respect to $v$, it yields

\begin{equation}\label{Area104}
\begin{aligned}
\dv \ttmfb &= \tp \du \dv r + \ii \e \frac{\du r \dv r}{1 - \mu}\frac{Q\phi}{r}.
\end{aligned}
\end{equation}

Upon taking an additional derivative with respect to $u$, we arrive at

\begin{equation}
\begin{aligned}
\du \dv \ttmfb &= \du \ml \tp \du \dv r \mr + \ii \e \du \ml \frac{\du r \dv r}{1 - \mu}\frac{Q\phi}{r} \mr.
\end{aligned}
\end{equation}

This can be rewritten as

\begin{equation}\label{Area105}
\dv \du \ttmfb = \dv \ml \tp \du \nu \mr + G_3 + G_4 - \frac{\p(\nu,\tp)}{\p(u,v)}
\end{equation}

\begin{equation}\label{Area106}
\begin{aligned}
G_3 = \ii \e \du \nu \ml \frac{\dv r}{1-\mu}\frac{Q\phi}{r}\mr \quad \text{ and } \quad G_4 = \ii \e \du r \du \ml \frac{\dv r}{1-\mu}\frac{Q\phi}{r} \mr.
\end{aligned}
\end{equation}

For each $(u,v) \in \mathcal{D}(0,v_0)$, by integrating along the future-directed outgoing null curve originating from $\Gamma$, we obtain

\begin{equation}\label{Area107}
\begin{aligned}
\mlm \du \ttmfb - \tp \du \nu\mrm(u,v) &\leq \mlm \du \ttmfb - \tp \du \nu\mrm_{\Gamma}(u) + \int^v_u \mlm G_3 + G_4 - \frac{\p(\nu,\tp)}{\p(u,v)} \mrm (u,v') \D v',
\end{aligned}
\end{equation}

where we use (<ref>), (<ref>), (<ref>).
Applying (<ref>), together with $0 \leq \mu - Q^2/r^2 \leq \mu$, (<ref>), (<ref>), (<ref>), we derive

\begin{equation}\label{Area316}
\begin{aligned}
|\dv \nu|(u,v) = \frac{\mu - Q^2/r^2}{1-\mu}\frac{\lambda |\nu|}{r} \lesssim r(u,v).
\end{aligned}
\end{equation}

Employing (<ref>), we also get

\begin{equation}\label{Area317}
\begin{aligned}
|\dv \ttmfb|(u,v) \lesssim D r(u,v) + Dr^2(u,v) \lesssim r(u,v).
\end{aligned}
\end{equation}

Henceforth, we deduce that

\begin{equation}\label{Area122}
\lim_{v \rightarrow u^+} \dv \nu(u,v) = \lim_{v \rightarrow u^+} \dv \ttmfb(u,v) = 0.
\end{equation}

Since by (<ref>) and (<ref>), we also have

\begin{equation}\label{Area116}
\ttmfb|_{\Gamma} = \tmfb|_{\Gamma} = -\tmfa|_{\Gamma} \quad \mbox{ and } \quad \lambda|_{\Gamma} = -\nu|_{\Gamma}.
\end{equation}

By the chain rule, we deduce

\begin{equation}\label{Area120}
\lim_{v \rightarrow u^+} \du \nu(u,v) = \lim_{v \rightarrow u^+} \frac{\D \nu|_{\Gamma}(u)}{\D u} = -\frac{\D \lambda|_{\Gamma}(u)}{\D u}
\end{equation}

\begin{equation}\label{Area121}
\lim_{v \rightarrow u^+} \du \tmfb(u,v) = \lim_{v \rightarrow u^+} \frac{\D \tmfb|_{\Gamma}(u)}{\D u} = -\frac{\D \tmfa|_{\Gamma}(u)}{\D u}.
\end{equation}

In addition, (<ref>) also implies that

\begin{equation}\label{Area118}
\begin{aligned}
TV_{(0,v_0)}[\ttmfb|_\Gamma] = TV_{(0,v_0)}[\tmfa|_\Gamma], \quad TV_{(0,v_0)}[\nu|_\Gamma] = TV_{(0,v_0)}[\lambda|_\Gamma].
\end{aligned}
\end{equation}

Hence, by integrating (<ref>) along a future-directed incoming null curve originating from $C_0^+$ and applying (<ref>) - (<ref>), we obtain

\begin{equation}\label{Area108a}
\begin{aligned}
\int^v_0 \mlm \du \ttmfb - \tp \du \nu\mrm(u',v) \D u' \leq&\; TV_{(0,v_0)}[\ttmfb|_{\Gamma}] + P \cdot TV_{(0,v_0)}[\nu|_{\Gamma}] + A[\nu,\tp] \\
&\; + \int^v_0 \int^v_{u'} |G_3|(u',v') \D v' \D u' + \int^v_0 \int^v_{u'} |G_4|(u',v') \D v' \D u'.
\end{aligned}
\end{equation}

We proceed to estimate the integrals containing $G_3$ and $G_4$ in (<ref>). For the former, for each $(u,v) \in \mathcal{D}(0,v_0)$, we have

\begin{equation}\label{Area109}
|G_3|(u,v) \lesssim D r(u,v) |\du \nu|(u,v) \lesssim D|\du \nu|(u,v)
\end{equation}
and hence

\begin{equation}\label{Area110}
\begin{aligned}
\int^v_0 \int^v_{u'} |G_3|(u',v') \D v' \D u' &= \int^v_0 \int^{v'}_{0} |G_3|(u',v') \D u' \D v' \lesssim D \cdot \sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\}}[\nu].
\end{aligned}
\end{equation}

For the integrand $G_4$, similar to equation (<ref>), we now have

\begin{equation}\label{Area111}
\du \ml \frac{\dv r}{1-\mu}\frac{Q \phi}{r}\mr = \frac{\dv \du r}{1-\mu}\frac{Q\phi}{r} + \frac{\dv r}{1-\mu}\frac{(\du Q)\phi}{r} + \frac{\dv r}{1-\mu}\frac{Q \du \phi}{r} - \frac{\dv r}{1-\mu}\frac{Q\phi}{r^2}\du r + \frac{\dv r}{(1-\mu)^2}\frac{Q\phi}{r}\du \mu.
\end{equation}

Notice that each of the terms above is analogous to the corresponding ones in (<ref>). Their estimates are exactly symmetric (with the use of (<ref>) and (<ref>) instead of (<ref>) and (<ref>) for the second and the fifth terms). Hence, by utilizing (<ref>), we then deduce that

\begin{equation}\label{Area112}
\mlm \du \ml \frac{\dv r}{1-\mu}\frac{Q \phi}{r}\mr \mrm (u,v)  \lesssim D,
\end{equation}

which implies that

\begin{equation}\label{Area113}
\int^v_0 \int^v_{u'} |G_4|(u',v') \D v' \D u' \lesssim D.
\end{equation}

Plugging (<ref>) and (<ref>) into (<ref>), we hence derive

\begin{equation}\label{Area108}
\begin{aligned}
\int^v_0 \mlm \du \ttmfb - \tp \du \nu\mrm(u',v) \D u'
\lesssim & \; TV_{(0,v_0)}[\tmfa|_\Gamma] + P \cdot TV_{(0,v_0)}[\lambda|_\Gamma] + A[\nu,\tp] + D \cdot \sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\}}[\nu] + D.\\
\end{aligned}
\end{equation}

This implies that

\begin{equation}\label{Area318}
\int^v_0 \mlm \du \ttmfb  \mrm (u',v) \D u' \lesssim TV_{(0,v_0)}[\tmfa|_\Gamma] + P \cdot TV_{(0,v_0)}[\lambda|_\Gamma] + A[\nu,\tp] + D \cdot \sup_{v \in (0,v_0)}TV_{(0,v) \times \{v\}}[\nu] + D.
\end{equation}

Taking supremum over $v \in (0,v_0)$, we then conclude that

\begin{equation}\label{Area115}
\begin{aligned}
\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{aligned}
\end{equation}

By scrutinizing the terms in the upper bound in (<ref>), it necessitates estimating $TV_{(0,v_0)}[\tmfa|_\Gamma]$, $TV_{(0,v_0)}[\lambda|_{\Gamma}]$ and $\sup_{v \in (0,v_0)} TV_{(0,v) \times \{v\}}[\nu]$. We now estimate each of them. Starting from $TV_{(0,v_0)}[\tmfa|_\Gamma]$, for a fixed $u \in (0,v_0)$, we integrate (<ref>) along the future-directed incoming null curve $v = u$ starting from $C_0^+$ to $(u,u) \in \Gamma$ and obtain

\begin{equation}\label{Area119}
\begin{aligned}
\dv \tmfa|_{\Gamma}(u) =&\; \tp|_{\Gamma}(u) \dv \lambda|_{\Gamma}(u) + (\dv \tmfa)(0,u) - \ml \tp \dv \lambda \mr (0,u) \\
&+ \int^u_0 G_1(u',u) + G_2(u',u) + \frac{\p(\lambda,\tp)}{\p(u,v)}(u',u) \D u'.
\end{aligned}
\end{equation}

Using similar estimates as in (<ref>) and (<ref>), it is not difficult to see that we have

\begin{equation}\label{Area123}
\lim_{v \rightarrow u^+} \du \lambda(u,v) = \lim_{v \rightarrow u^+} \du \tmfa(u,v) = 0.
\end{equation}

Together with the fact that $\lambda(0,v) = \frac{1}{2}$, this implies that

\begin{equation}\label{Area124}
\begin{aligned}
\frac{\D}{\D u} \tmfa|_{\Gamma}(u) =&\; \tp|_{\Gamma}(u) \frac{\D}{\D u} \lambda|_{\Gamma}(u) + (\dv \tmfa)(0,u)  + \int^u_0 G_1(u',u) + G_2(u',u) + \frac{\p(\lambda,\tp)}{\p(u,v)}(u',u) \D u'.
\end{aligned}
\end{equation}

Integrating (<ref>) with variable $u \in (0,v_0)$, we deduce

\begin{equation}\label{Area125}
\begin{aligned}
TV_{(0,v_0)}[\tmfa|_{\Gamma}] &\leq  P \cdot TV_{(0,v_0)}[\lambda|_{\Gamma}] + D  + A[\lambda,\tp] + \int_0^{v_0}\int^v_0 |G_1|(u',v)  + |G_2|(u',v) \D u' \D v \\
&= P \cdot TV_{(0,v_0)}[\lambda|_{\Gamma}] + D  + A[\lambda,\tp] + \int_0^{v_0}\int^{v_0}_{u'} |G_1|(u',v)  + |G_2|(u',v) \D v \D u'.
\end{aligned}
\end{equation}

For the integral with $|G_1|$, analogous to (<ref>), we deduce

\begin{equation}\label{Area126}
\begin{aligned}
\int^{v_0}_0 \int^{v_0}_{u'} |G_1| (u',v) \D v \D u' &\lesssim D \int^{v_0}_0 \int^{v_0}_{u'} |\dv \lambda| (u',v) \D v \D u'\lesssim D \sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\lambda].  \\
\end{aligned}
\end{equation}

To estimate the integral with $|G_2|$, in a similar fashion as for (<ref>), we obtain

\begin{equation}\label{Area127}
\begin{aligned}\int^{v_0}_0  \int^{v_0}_{u'} |A_u \dv \tmfa|(u',v) \D v \D u' \leq \frac{1}{2\e} \sup_{u \in (0,v_0)}TV_{\{u\} \times (u,v_0) }[\tmfa].
\end{aligned}
\end{equation}

Consequently, by utilizing identical pointwise estimates (<ref>), (<ref>), (<ref>), we derive the below estimate analogous to (<ref>)

\begin{equation}\label{Area128}
\int^{v_0}_0 \int^{v_0}_{u'} |G_2| (u',v) \D v \D u' - \frac{1}{2}\sup_{u \in (0,v_0)}TV_{\{u\} \times (u,v_0) }[\tmfa] \lesssim D.
\end{equation}

Applying (<ref>) and (<ref>) to (<ref>), we then conclude that

\begin{equation}\label{Area129}
\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}

Observe that the upper bound in (<ref>) involves the term $\sup_{v\in(0,v_0)}TV_{\{u\}\times(u,v_0)}[\lambda]$. We now move to bound it. Recall that from (<ref>) we have

\begin{equation}\label{Area130}
\du \log \lambda = \frac{\mu - Q^2/r^2}{1-\mu}\frac{\du r}{r}.
\end{equation}

Upon taking an additional derivative with respect to $v$ and using (<ref>), (<ref>), (<ref>), we arrive at

\begin{equation}\label{Area131}
\begin{aligned}
\dv \du \log \lambda =&\; \frac{\dv\ml \mu -Q^2/r^2\mr}{1-\mu}\frac{\du r}{r} + (\dv \mu) \frac{\mu - Q^2/r^2}{(1-\mu)^2}\frac{\du r}{r} + \frac{\mu - Q^2/r^2}{1-\mu}\frac{\du \dv r}{r} - \frac{\mu - Q^2/r^2}{1-\mu}\frac{(\du r)(\dv r)}{r^2} \\
=&\; \frac{\dv \mu}{1-\mu}\ml 1 + \frac{\mu-Q^2/r^2}{1-\mu}\mr\ml \frac{\du r}{r}\mr - \frac{\dv \ml \frac{Q^2}{r^2}\mr}{1-\mu}\frac{\du r}{r} + \frac{\mu - Q^2/r^2}{1-\mu}\ml \frac{\mu - Q^2/r^2}{1-\mu} - 1\mr\frac{(\du r)(\dv r)}{r^2} \\
=&\; \ml \frac{4\pi r|\dv \phi|^2}{\dv r} - \frac{\dv r}{r}\frac{\mu - Q^2/r^2}{1-\mu}\mr\ml \frac{1-Q^2/r^2}{1-\mu}\mr\ml \frac{\du r}{r}\mr + \ml \frac{8\pi \e Q \im(\phi^{\dagger}\dv \phi) + \frac{2Q^2 \dv r}{r^3}  }{1-\mu}\mr \ml \frac{\du r}{r}\mr \\
&+ \frac{\mu - Q^2/r^2}{1-\mu}\ml \frac{\mu - Q^2/r^2}{1-\mu} - 1\mr\frac{(\du r)(\dv r)}{r^2} \\
=&\; \ml \frac{4\pi r|\dv \phi|^2}{\dv r} \mr \ml \frac{1-Q^2/r^2}{1-\mu}\mr\ml \frac{\du r}{r}\mr + \ml \frac{8\pi \e Q \im(\phi^{\dagger}\dv \phi) + \frac{2Q^2 \dv r}{r^3}  }{1-\mu}\mr \ml \frac{\du r}{r}\mr \\
&-2 \ml \frac{\mu-Q^2/r^2}{1-\mu}\mr \ml \frac{\du r \dv r}{r^2}\mr.
\end{aligned}
\end{equation}

We can also rewrite (<ref>) as

\begin{equation}\label{Area132}
\dv \log (-\nu) = \frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r}.
\end{equation}

By an analogous computation, with the help of (<ref>) and (<ref>), we obtain

\begin{equation}\label{Area133}
\begin{aligned}
\du \dv \log (-\nu) =&\; \ml \frac{4\pi r|D_u \phi|^2}{\du r} \mr \ml \frac{1-Q^2/r^2}{1-\mu}\mr\ml \frac{\dv r}{r}\mr + \ml \frac{-8\pi \e Q \im(\phi^{\dagger}D_u \phi) + \frac{2Q^2 \du r}{r^3}  }{1-\mu}\mr \ml \frac{\dv r}{r}\mr \\
&-2 \ml \frac{\mu-Q^2/r^2}{1-\mu}\mr \ml \frac{\du r \dv r}{r^2}\mr.
\end{aligned}
\end{equation}

Observe that (<ref>) and (<ref>) together imply

\begin{equation}\label{Area134}
\begin{aligned}
\dv \du \log \lambda - \du \dv \log (-\nu) =&\; \frac{4\pi}{r^2}\ml \frac{1-Q^2/r^2}{1-\mu}\mr\ml \frac{(r|\dv \phi|)^2(\du r)}{(\dv r)} - \frac{(r|D_u \phi|)^2(\dv r)}{\du r}\mr \\
&+ \ml \frac{8\pi \e Q}{1-\mu} \mr \ml \frac{1}{1-\mu}\mr\ml (\du r)  \im(\phi^{\dagger}\dv \phi ) + (\dv r)\im(\phi^{\dagger}D_u \phi)\mr.
\end{aligned}
\end{equation}

We then exploit the symmetry in (<ref>). Notice that $\dv \du \log \lambda$ and $\du \dv \log (-\nu)$ have alternative expressions given below

\begin{equation}\label{Area135}
\begin{aligned}
\dv \du \log \lambda =&\; \du\ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r}\mr + \frac{4\pi}{r^2}\ml \frac{1-Q^2/r^2}{1-\mu}\mr\ml \frac{(r|\dv \phi|)^2(\du r)}{(\dv r)} - \frac{(r|D_u \phi|)^2(\dv r)}{\du r}\mr\\
&+ \ml \frac{8\pi \e Q}{1-\mu} \mr \ml \frac{1}{1-\mu}\mr\ml (\du r)  \im(\phi^{\dagger}\dv \phi ) + (\dv r)\im(\phi^{\dagger}D_u \phi)\mr
\end{aligned}
\end{equation}

\begin{equation}\label{Area136}
\begin{aligned}
\du \dv \log(-\nu) =&\; \dv\ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\du r}{r}\mr - \frac{4\pi}{r^2}\ml \frac{1-Q^2/r^2}{1-\mu}\mr\ml \frac{(r|\dv \phi|)^2(\du r)}{(\dv r)} - \frac{(r|D_u \phi|)^2(\dv r)}{\du r}\mr\\
&- \ml \frac{8\pi \e Q}{1-\mu} \mr \ml \frac{1}{1-\mu}\mr\ml (\du r)  \im(\phi^{\dagger}\dv \phi ) + (\dv r)\im(\phi^{\dagger}D_u \phi)\mr.
\end{aligned}
\end{equation}

Next, we define

\begin{equation}\label{Area137}
\tf(u,v) := 4\pi\re\ml \frac{r (\dv \tp)^\dagger}{r}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr\mr(u,v)
\end{equation}

\begin{equation}\label{Area138}
\tfs(u,v) := 4\pi\re\ml \frac{r (\du \tp)^{\dagger}}{r}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \mr(u,v).
\end{equation}

To compute the partial derivative of $f$ with respect to $u$, we first observe from (<ref>) that

\begin{equation}\label{Area139}
\begin{aligned}
\du(r\dv \phi) = -(\dv r)(\du \phi) + E_1 \quad \text{ and } \quad \dv(r\du \phi) = -(\du r)(\dv \phi) + E_1
\end{aligned}
\end{equation}

\begin{equation}\label{Area140}
E_1 := -\ii\e A_u(r\dv \phi + \phi \dv r) - \ii \e \frac{Q\phi}{1-\mu}\frac{(\du r)(\dv r)}{r}
\end{equation}

and $|E_1|(u,v) \lesssim Dr(u,v)$. In addition, we observe that

\begin{equation}\label{Area141}
\frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu} = \frac{r \dv \tp}{\dv r} - \frac{r \du \tp}{\du r} + E_2 \quad \text{ with } \quad E_2 := -\ii \e \ml \frac{A_u r \phi}{\du r}\mr
\end{equation}

and $|E_2|(u,v) \lesssim Dr^2(u,v)$. We then proceed to take the partial derivative with respect to $u$ on both sides of (<ref>). Together with (<ref>) and (<ref>), we obtain

\begin{equation}\label{Area143}
\begin{aligned}
\frac{\p \tf}{\p u} =&\; 4\pi \re \ml -\ml \frac{\du r \dv \tp}{r} + \frac{\dv r\du \tp }{r} - \frac{E_1}{r}\mr^{\dagger} \ml \frac{r\dv \tp}{\dv r} - \frac{r \du \tp}{\du r} + E_2\mr + \ml \frac{r \dv \tp}{r}\mr^{\dagger} \du\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \mr\\
=&\; \frac{4\pi}{r^2}\ml \frac{\dv r}{\du r}(r|\du \tp|)^2 - \frac{\du r}{\dv r}(r|\dv \tp|)^2\mr + 4\pi\re \ml \ml \frac{r \dv \tp}{r}\mr^\dagger \du \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \mr - 4\pi\re\ml (\du \tp)^{\dagger}(\dv \tp) - (\du \tp)(\dv \tp)^\dagger \mr \\
&+ 4\pi\re \ml \ml \frac{E_1^\dagger}{r} \mr \ml \frac{r \dv \tp}{\dv r} - \frac{r \du \tp}{\du r}\mr \mr - 4\pi \re \ml \ml \frac{E_2}{r^2} \mr \ml \du r(r\dv \tp) + \dv r (r \du \tp)\mr \mr + 4\pi\re \ml \frac{E_1^\dagger E_2}{r}\mr \\
=&\; \frac{4\pi}{r^2}\ml \frac{(r|\du \tp|)^2(\dv r)}{\du r} - \frac{ (r|\dv \tp|)^2(\du r)}{\dv r}\mr + 4\pi\re \ml \ml \frac{r \dv \tp}{r}\mr^\dagger \du \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \mr + E_3
\end{aligned}
\end{equation}

\begin{equation}\label{Area144}
E_3 := 4\pi\re \ml \ml \frac{E_1^\dagger}{r} \mr \ml \frac{r \dv \tp}{\dv r} - \frac{r \du \tp}{\du r}\mr \mr - 4\pi \re \ml \ml \frac{E_2}{r^2} \mr \ml \du r(r\dv \tp) + \dv r (r \du \tp)\mr \mr + 4\pi \re \ml \frac{E_1^\dagger E_2}{r}\mr.
\end{equation}

Note that we use the fact that $( \du \tp)^{\dagger}(\dv \tp) - (\du \tp)(\dv \tp)^\dagger$ is purely imaginary. And we have $|E_3|(u,v) \lesssim D^2$. Similarly, we compute the partial derivative of $\tfs$ with respect to $v$ and get

\begin{equation}\label{Area145}
\begin{aligned}
\frac{\p \tfs}{\p v} =&\; 4\pi\re \ml -\ml \frac{\du r \dv \tp}{r} + \frac{\dv r\du \tp }{r} - \frac{E_1}{r}\mr^{\dagger} \ml \frac{r\dv \tp}{\dv r} - \frac{r \du \tp}{\du r} + E_2\mr + \ml \frac{r \du \tp}{r}\mr^{\dagger} \dv\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \mr\\
=&\; \frac{4\pi}{r^2}\ml \frac{(r|\du \tp|)^2(\dv r)}{\du r} - \frac{ (r|\dv \tp|)^2(\du r)}{\dv r}\mr + 4\pi\re \ml \ml \frac{r \du \tp}{r}\mr^\dagger \dv \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \mr + E_3.
\end{aligned}
\end{equation}

The equations for $\frac{\p \tf}{\p u}$ and $\frac{\p \tfs}{\p v}$ will be used in the estimates for $\du \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr$ and $\dv \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr$. To evaluate them, we start by considering (<ref>) and rewrite it as

\begin{equation}\label{Area146}
\begin{aligned}
\du \dv (r\tp) = \dv \du (r\tp) = \tp \du \dv r + E_1.
\end{aligned}
\end{equation}

By the definition of $\tmfa$ from (<ref>) and employing (<ref>), the above equation implies that

\begin{equation}\label{Area147}
\begin{aligned}
\du\ml \frac{\tmfa}{\lambda} \mr = \frac{1}{\dv r}\ml E_1 - \frac{r \dv \tp}{\dv r}\du \dv r\mr = - \frac{\mu - Q^2/r^2}{1-\mu}\ml \frac{r \dv \tp}{\dv r}\mr\ml \frac{\du r}{r}\mr + \frac{E_1}{\dv r}.
\end{aligned}
\end{equation}

Similarly, we also obtain

\begin{equation}\label{Area148}
\begin{aligned}
\dv\ml \frac{\tmfb}{\nu} \mr = - \frac{\mu - Q^2/r^2}{1-\mu}\ml \frac{r \du \tp}{\du r} \mr \ml \frac{\dv r}{r} \mr + \frac{E_1}{\du r}.
\end{aligned}
\end{equation}

By using the definition of $\ttmfb$ in (<ref>), the above further simplifies to

\begin{equation}\label{Area149}
\begin{aligned}
\dv\ml \frac{\ttmfb}{\nu} \mr = - \frac{\mu - Q^2/r^2}{1-\mu}\ml \frac{r \du \tp}{\du r} \mr \ml \frac{\dv r}{r} \mr + \frac{E_4}{\du r}
\end{aligned}
\end{equation}

\begin{equation}\label{Area150}
E_4 := E_1 + (\du r)\dv \ml \frac{\ii \e A_u r \phi}{\du r}\mr.
\end{equation}

Observe that by applying (<ref>) and (<ref>), we get

\begin{equation}\label{Area151}
\begin{aligned}
\dv \ml \frac{A_u r \phi}{\du r}\mr &= - A_u r \phi \ml\frac{\du \dv r}{(\du r)^2}\mr + \frac{1}{\du r}\ml (\dv A_u)(r\phi) + A_u \phi (\dv r) + A_u r(\dv \tp)\mr \\
&= - A_u \phi \ml\frac{(\mu - Q^2/r^2)(\dv r)}{(1-\mu)(\du r)}\mr + \frac{1}{\du r}\ml \ml \frac{2Q(\du r)(\dv r)}{r^2(1-\mu)}\mr(r\phi) + A_u \phi (\dv r) + A_u r(\dv \tp)\mr.
\end{aligned}
\end{equation}

Using the fact that $|E_1|(u,v) \lesssim Dr(u,v)$, we derive the below estimate

\begin{equation}\label{Area152}
\begin{aligned}
|E_4|(u,v) &\leq \mlm E_1\mrm(u,v) + \mlm (\du r)\dv \ml \frac{\ii \e A_u r \phi}{\du r}\mr\mrm \\
&\lesssim D r(u,v) + Dr(u,v) + Dr(u,v) + Dr(u,v) + Dr^2(u,v) \lesssim Dr(u,v).
\end{aligned}
\end{equation}

At the same time, by the expressions in (<ref>) and (<ref>), we compute the following terms

\begin{equation}\label{Area153}
\begin{aligned}
(\dv \tp)^\dagger \du\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu} \mr =&\; (\dv \tp)^\dagger \du \ml \frac{\tmfa}{\lambda}\mr - (\du \tp)^\dagger \dv \ml \frac{\ttmfb}{\nu}\mr + \ml (\du \tp)^\dagger \dv \ml \frac{\ttmfb}{\nu}\mr - (\dv \tp)^\dagger \du \ml \frac{\ttmfb}{\nu}\mr \mr \\
=&\; \frac{1}{r^2}\ml \frac{\mu - Q^2/r^2}{1-\mu} \mr\ml \frac{(r|\du \tp|)^2(\dv r)}{\du r} - \frac{ (r|\dv \tp|)^2(\du r)}{\dv r}\mr + \frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)}\\
&+ \ml \frac{(\dv \tp)^\dagger}{\dv r} \mr E_1 - \ml \frac{(\du \tp)^\dagger}{\du r} \mr E_4,
\end{aligned}
\end{equation}

\begin{equation}\label{Area154}
\begin{aligned}
(\du \tp)^\dagger \dv\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu} \mr =&\; (\dv \tp)^\dagger \du \ml \frac{\tmfa}{\lambda}\mr -(\du \tp)^\dagger \dv \ml \frac{\ttmfb}{\nu} \mr + \ml (\du \tp)^\dagger \dv \ml \frac{\tmfa}{\lambda}\mr - (\dv \tp)^\dagger \du \ml \frac{\tmfa}{\lambda}\mr\mr \\
=&\; \frac{1}{r^2}\ml \frac{\mu - Q^2/r^2}{1-\mu} \mr\ml \frac{(r|\du \tp|)^2(\dv r)}{\du r} - \frac{ (r|\dv \tp|)^2(\du r)}{\dv r}\mr + \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)}\\
&+ \ml \frac{(\dv \tp)^\dagger}{\dv r} \mr E_1 - \ml \frac{(\du \tp)^\dagger}{\du r} \mr E_4.
\end{aligned}
\end{equation}

By substituting (<ref>) into the expression for $\du \tf$ in (<ref>), we obtain

\begin{equation}\label{Area155}
\begin{aligned}
\frac{\p \tf}{\p u} =&\; \frac{4\pi}{r^2}\ml 1 + \frac{\mu - Q^2/r^2}{1-\mu}\mr\ml \frac{(r|\du \tp|)^2(\dv r)}{\du r} - \frac{ (r|\dv \tp|)^2(\du r)}{\dv r}\mr + 4\pi \re \ml \frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)} \mr\\
&+ \re \ml 4\pi\ml \frac{(\dv \tp)^\dagger}{\dv r} \mr E_1 - 4\pi\ml \frac{(\du \tp)^\dagger}{\du r} \mr E_4\mr + E_3 \\
=&\; \frac{4\pi}{r^2}\ml \frac{1 - Q^2/r^2}{1-\mu}\mr\ml \frac{(r|\du \tp|)^2(\dv r)}{\du r} - \frac{ (r|\dv \tp|)^2(\du r)}{\dv r}\mr + 4\pi \re \ml \frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)} \mr + E_5
\end{aligned}
\end{equation}

in which

\begin{equation}\label{Area156}
E_5 := \ml 4\pi\ml \frac{(\dv \tp)^\dagger}{\dv r} \mr E_1 - 4\pi\ml \frac{(\du \tp)^\dagger}{\du r} \mr E_4 \mr + E_3.
\end{equation}

Note that since $|E_1| \lesssim Dr$, $|E_3| \lesssim D^2$, $|E_4| \lesssim Dr$, $|r\dv \tp| \lesssim D$, $|r\du \tp| \lesssim D$, we can deduce that $|E_5| \lesssim D^2$.

By a parallel argument, for $\dv \tfs$, we can utilize (<ref>) to (<ref>) to deduce that

\begin{equation}\label{Area157}
\begin{aligned}
\frac{\p \tfs}{\p v}
=&\; \frac{4\pi}{r^2}\ml \frac{1 - Q^2/r^2}{1-\mu}\mr\ml \frac{(r|\du \tp|)^2(\dv r)}{\du r} - \frac{ (r|\dv \tp|)^2(\du r)}{\dv r}\mr + 4\pi \re \ml \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} \mr + E_5.
\end{aligned}
\end{equation}

Comparing terms in the expressions for $\frac{\p \tf}{\p u}$ and for $\frac{\p \tfs}{\p v}$ above with that for $\dv \du \log \lambda$ and for $\du \dv \log \ml - \nu \mr$ in (<ref>) and (<ref>), it is instructional to consider following decomposition of the term $|D_u \phi|^2$:

\begin{equation}\label{Area158}
\begin{aligned}
|D_u \phi|^2 &= (\du \phi + \ii \e A_u \phi)(\du \phi - \ii \e A_u \phi)^\dagger = |\du \tp|^2 - 2 \e A_u \im(\phi \du \phi^\dagger) + \e^2 A_u^2 |\phi|^2.
\end{aligned}
\end{equation}

This implies that (<ref>) and (<ref>) can be simplified as

\begin{equation}\label{Area159}
\begin{aligned}
\dv \du \log \lambda =&\; \du\ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r}\mr + E_6 \\
&- \frac{4\pi}{r^2}\ml \frac{1-Q^2/r^2}{1-\mu}\mr\ml \frac{(r|\du \tp|)^2(\dv r)}{\du r} - \frac{ (r|\dv \tp|)^2(\du r)}{\dv r}\mr\\
\end{aligned}
\end{equation}

\begin{equation}\label{Area160}
\begin{aligned}
\du \dv \log(-\nu) =&\; \dv\ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\du r}{r}\mr + E_6 \\
&+ \frac{4\pi}{r^2}\ml \frac{1-Q^2/r^2}{1-\mu}\mr\ml \frac{(r|\du \tp|)^2(\dv r)}{\du r} - \frac{ (r|\dv \tp|)^2(\du r)}{\dv r}\mr
\end{aligned}
\end{equation}

\begin{equation}\label{Area161}
\begin{aligned}
E_6 :=&\; \ml \frac{8\pi \e Q}{1-\mu} \mr \ml \frac{1}{1-\mu}\mr\ml (\du r)  \im(\phi^{\dagger}\dv \phi ) + (\dv r)\im(\phi^{\dagger}D_u \phi)\mr \\
&+ 4\pi \ml \frac{1-Q^2/r^2}{1-\mu}\mr\ml \frac{\dv r}{\du r}\mr\ml 2\e A_u \im(\phi \du \phi^\dagger) - \e^2 A_u^2|\phi|^2 \mr
\end{aligned}
\end{equation}

satisfying $|E_6|(u,v) \lesssim D^2$. Substituting the expressions of $\frac{\p \tf}{\p u}$ and $\frac{\p \tfs}{\p v}$ in (<ref>) and in (<ref>) to (<ref>) and to (<ref>), we obtain

\begin{equation}\label{Area162}
\begin{aligned}
\du\ml \dv \log \lambda \mr = \du \ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r} - \tf\mr + 4\pi \re \ml \frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)}\mr + E_5 + E_6
\end{aligned}
\end{equation}

\begin{equation}\label{Area163}
\begin{aligned}
\dv\ml \du \log (-\nu) \mr = \dv \ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\du r}{r} + \tfs\mr - 4\pi \re \ml \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)}\mr - E_5 + E_6.
\end{aligned}
\end{equation}

With the above preparations, we start to estimate the total variations of $\log(\lambda), \log(-\nu)$ and $\log(\lambda|_{\Gamma})$.  Employing (<ref>), for a fixed $(u,v) \in \mathcal{D}(0,v_0)$, we first integrate along a future-directed incoming null curve. Together with the fact that $\dv \lambda (0,v) = 0$, we obtain

\begin{equation}\label{Area164}
\begin{aligned}
\dv \log \lambda (u,v) =&\; \ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r} - \tf \mr(u,v) - \ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r} - \tf \mr(0,v) \\
&+ 4\pi \int^u_0 \re \ml \frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)} \mr (u',v) \D u' + \int^u_0 (E_5 + E_6)(u',v) \D u'.
\end{aligned}
\end{equation}

Further integrating (<ref>) for $v \in (u,v_0)$, we derive

\begin{equation}\label{Area165}
\begin{aligned}
& \int^{v_0}_u \mlm \dv \log \lambda - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r} + \tf\mrm(u,v') \D v'\\
\lesssim &\int^{v_0}_u \mlm \frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r} - \tf\mrm(0,v') \D v' + A[\tp^\dagger,\ttmfb/\nu] + D^2
\end{aligned}
\end{equation}

To obtain the total variation of $\lambda$, it suffices to estimate the integral in the upper bound of (<ref>). To estimate the term $\frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r}$, we appeal to (<ref>) and see that

\begin{equation}\label{Area169}
\dv \log \ml \frac{-\du r}{1-\mu} \mr = \frac{4 \pi r|\dv \phi|^2}{\dv r}.
\end{equation}

By (<ref>), we can rewrite this as

\begin{equation}\label{Area170}
\frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r} - \dv \log \ml 1-\mu \mr = \frac{4 \pi r|\dv \phi|^2}{\dv r}.
\end{equation}

Integrating both sides of (<ref>) with respect to $v'$, we obtain

\begin{equation}\label{Area171}
\begin{aligned}
\int^{v_0}_u \frac{\mu - Q^2/r^2}{1-\mu}\frac{(\dv r)}{r}(u,v') \D v' &= \log(1-\mu(u,v_0)) + 4\pi \int^{v_0}_u \frac{r|\dv \phi|^2}{\dv r}(u,v') \D v' \\
&\lesssim D \int^{v_0}_u |\dv \phi|(u,v') \D v' \lesssim DX
\end{aligned}
\end{equation}

\begin{equation}\label{Area172}
\begin{aligned}
\int^{v_0}_u \frac{\mu - Q^2/r^2}{1-\mu}\frac{(\dv r)}{r}(0,v') \D v' &= \log(1-\mu(u,v_0)) + 4\pi \int^{v_0}_u \frac{r|\dv \phi|^2}{\dv r}(0,v') \D v' \\
&\lesssim D \int^{v_0}_u |\dv \phi|(0,v') \D v' \lesssim D \cdot TV_{\{0\} \times (0,v_0)}[\tilde{\alpha}] \\
&\quad\quad\lesssim D \cdot \ml 2 \cdot TV_{\{0\} \times (0,v_0)}[\tmfa]\mr \lesssim D^2. \\
\end{aligned}
\end{equation}

In the above computations, we have used $\mu|_{\Gamma} = 0$, $\mu < 1$, $\frac{r|\dv \phi|}{\dv r} \lesssim D$, $\lambda(0,v) = \frac{1}{2}$ and Lemma <ref>. Furthermore, from the definition of $\tf$ in (<ref>), we have

\begin{equation}\label{Area173}
|\tf|(u,v) \lesssim D |\dv \phi|(u,v)
\end{equation}

and hence

\begin{equation}\label{Area174}
\begin{aligned}
\int^{v_0}_u |\tf|(u,v') \D v' &\lesssim D \int^{v_0}_u |\dv \phi|(u,v') \D v' \lesssim DX, \\
\int^{v_0}_u |\tf|(0,v') \D v' &\lesssim D \int^{v_0}_u |\dv \phi|(0,v') \D v' \lesssim D^2. \\
\end{aligned}
\end{equation}

Gathering the estimates in (<ref>), (<ref>), (<ref>), by (<ref>), we conclude that

\begin{equation}\label{Area175}
\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}

Next, we move on to estimate the total variation of $\log(-\nu)$. Observe that from (<ref>), (<ref>), and (<ref>) that we have

\begin{equation}\label{Area340}
|\tilde{f}^*|(u,v) \lesssim r(u,v).
\end{equation}

Together with (<ref>), this implies

\begin{equation}\label{Area341}
\tfs|_{\Gamma} = \frac{\mu}{r}|_{\Gamma} = 0.
\end{equation}

Using a symmetric argument as for (<ref>), by (<ref>), we get

\begin{equation}\label{Area176}
\begin{aligned}
\du \log(-\nu)(u,v) =&\; \frac{\D }{\D u}\log(-\nu)(u) + \ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\du r}{r} + 4 \pi \tfs\mr (u,v)  \\
&- 4\pi \int^v_u \re \ml \frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} \mr (u,v') \D v' + \int^v_u (E_6 - E_5)(u,v') \D v'
\end{aligned}
\end{equation}

and hence

\begin{equation}\label{Area177}
\begin{aligned}
&\int^v_0 \mlm \du \log(-\nu) - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\du r}{r} - 4\pi \tfs \mrm(u',v) \D u' \\
\lesssim & \;  TV_{(0,v_0)}[\log(-\nu)|_{\Gamma}] +  A[\tp^\dagger,\tmfa/\lambda] + D^2.
\end{aligned}
\end{equation}

Similar to the arguments made above, we start from (<ref>) and rewrite (<ref>) as

\begin{equation}\label{Area167}
\begin{aligned}
\frac{\mu - Q^2/r^2}{1-\mu}\frac{\du r}{r} - \du \log(1-\mu) &= \frac{4\pi r|D_u \phi|^2}{\du r}.
\end{aligned}
\end{equation}

Integrating with respect to $u'$, recalling the definition of $Y$ in (<ref>) and applying $|D_u \phi| = |\du \phi + \ii \e A_u r \phi| \lesssim |\du \phi| + D$, we deduce that

\begin{equation}\label{Area168}
\begin{aligned}
\int^v_0 \frac{\mu - Q^2/r^2}{1-\mu}\frac{(-\du r)}{r}(u',v) \D u' &\lesssim D \int^v_u |D_u \phi|(u',v) \D u' \lesssim D \int^v_u |\du \tp|(u',v) \D u' + D^2 \lesssim DY + D^2.
\end{aligned}
\end{equation}

In addition, from the expression of $\tfs$ in (<ref>), we see that

\begin{equation}\label{Area178}
|\tfs|(u,v)\lesssim D|\du \phi|(u,v)
\end{equation}

and thus obtain

\begin{equation}\label{Area179}
\begin{aligned}
\int^v_0 |\tfs|(u',v) \D u' &\lesssim D \cdot T.V_{(0,v) \times \{v\}}[\tp]  \lesssim D Y.
\end{aligned}
\end{equation}

By applying (<ref>), (<ref>), (<ref>) to (<ref>), we hence derive

\begin{equation}\label{Area180}
\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}

To estimate $TV_{(0,v_0)}[\log \lambda|_{\Gamma}]$, we appeal to (<ref>), (<ref>), (<ref>). Integrating along future-directed incoming null curve from $C_0^+$ to $(u,u) \in \Gamma$ yields

\begin{equation}\label{Area181}
\begin{aligned}
\frac{\D}{\D u}\log \ml \lambda|_{\Gamma}\mr (u) =&\; - \ml \frac{\mu - Q^2/r^2}{1-\mu}\frac{\dv r}{r} - 4\pi \tf\mr(0,u) \\
&+ 4\pi \int^u_0 \re \ml \frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)}\mr(u',u) \D u' + \int^u_0 (E_5 + E_6) (u',u) \D u'.
\end{aligned}
\end{equation}

By utilizing estimates analogous to (<ref>) and (<ref>), we deduce that

\begin{equation}\label{Area182}
TV_{(0,v_0)}[\log(\lambda|_{\Gamma})] \lesssim D^2 + A[\tp^\dagger,\ttmfb/\nu].
\end{equation}

As the next course of action, we proceed to establish relations between the different area terms that have appeared in the estimates above. Starting from $A[\tp^\dagger,\tmfa/\lambda]$, by (<ref>), we observe that

\begin{equation}\label{Area187}
\du \tmfa = \tp \du \dv r + E_7 \quad \text{ with } \quad E_7 := -\ii \e \ml A_u \dv(r\tp) + \phi_0 A_u \dv r + \frac{\du r \dv r }{1-\mu}\frac{Q \phi}{r}\mr
\end{equation}

and $|E_7|(u,v) \lesssim Dr(u,v)$. This then implies that

\begin{equation}\label{Area183}
\begin{aligned}
\frac{\p(\tp^\dagger,\tmfa/\lambda)}{\p(u,v)} &= \du \tp^\dagger \dv \ml \frac{\tmfa}{\lambda}\mr - \dv \tp^\dagger \du \ml \frac{\tmfa}{\lambda}\mr \\
&= \frac{(\du \tp)^\dagger(\dv \tmfa) - (\dv \tp)^\dagger(\tp \du \dv r + E_7 )}{\lambda} + \frac{(r \dv \tp) + \tp \lambda}{\lambda^2}\frac{\p(\lambda, \tp^\dagger)}{\p(u,v)} \\
&= \frac{(\du \tp)^\dagger}{\lambda}(\dv \tmfa - \tp \dv \lambda) + \frac{(r \dv \tp)}{\lambda^2}\frac{\p(\lambda, \tp^\dagger)}{\p(u,v)}- \frac{(\dv \tp)^\dagger E_7}{\lambda}. \\
\end{aligned}
\end{equation}

Furthermore, applying (<ref>), (<ref>), (<ref>), (<ref>), we get

\begin{equation}\label{Area184}
\mlm \dv \tmfa - \tp \dv \lambda\mrm(u,v) \leq F(v)
\end{equation}

\begin{equation}\label{Area185}
F(v) =  \mlm \dv \tmfa - \tp \dv \lambda \mrm(0,v) + \int^u_0 \mlm G_1 + G_2 + \frac{\p(\lambda,\tp)}{\p(u,v)}\mrm(u',v) \D u'
\end{equation}

\begin{equation}\label{Area186}
\int^{v_0}_0 F(v) \D v \lesssim D\ml 1 + \sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\lambda]\mr  + A[\lambda,\tp^\dagger].
\end{equation}

Note that in (<ref>), we have used $A[\lambda,\tp^\dagger] = A[\lambda,\tp]$. This follows from the fact that, for any complex number $z$, we have $|z| = |z^\dagger|$. Back to (<ref>), we then derive

\begin{equation}\label{Area189}
\begin{aligned}
\int_{D(0,v_0)} \frac{|\du \tp|}{\lambda}|\dv \tmfa - \tp \dv \lambda|(u,v) \D u \D v &\lesssim \ml \sup_{v \in (0,v_0)}\int^v_0 |\du \tp|(u,v) \D u \mr \ml \int^{v_0}_0 F(v) \D v\mr \\
&\lesssim DY\ml 1 + \sup_{u \in (0,v_0)}TV_{\{u\}\times(u,v_0)}[\lambda]\mr  + Y\cdot A[\lambda,\tp^\dagger].
\end{aligned}
\end{equation}

Therefore, equation (<ref>) allows us to deduce that

\begin{equation}\label{Area190}
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^\dagger].
\end{equation}

We conduct a similar strategy to control $A[\tp^\dagger, \ttmfb/\nu]$ as follows. Observe that from (<ref>), we have

\begin{equation}\label{Area191}
\begin{aligned}
\dv \ttmfb &= \tp \du \dv r + E_8 \quad \text{ with } \quad
E_8 := \ii \e \frac{\du r \dv r}{1-\mu}\frac{Q \phi}{r}
\end{aligned}
\end{equation}

and $|E_8|(u,v) \lesssim Dr(u,v)$. Together with (<ref>), we compute $\frac{\p(\tp,\ttmfb/\nu)}{\p(u,v)}$ and obtain

\begin{equation}\label{Area193}
\begin{aligned}
\frac{\p(\tp,\ttmfb/\nu)}{\p(u,v)} &= (\du \tp)^\dagger \dv \ml\frac{\ttmfb}{\nu} \mr - (\dv \tp)^\dagger \du \ml \frac{\ttmfb}{\nu}\mr \\
&= \frac{(\du \tp)^\dagger (\tp \du \dv r + E_8) - (\dv \tp)^\dagger(\du \ttmfb) }{\nu} + \frac{(r \du \tp) + \tp \nu + \ii \e A_u r \phi}{\nu^2}\frac{\p(\nu,\tp^\dagger)}{\p(u,v)} \\
&= -\frac{(\dv \tp)^\dagger}{\nu}\ml \du \ttmfb - \tp \du \nu \mr + \frac{(r \du \tp) + \ii \e A_u r \phi}{\nu^2}\frac{\p(\nu,\tp^\dagger)}{\p(u,v)} + \frac{(\du \tp)^\dagger E_8}{\nu}.
\end{aligned}
\end{equation}

Employing (<ref>), (<ref>), (<ref>), (<ref>), we see that

\begin{equation}\label{Area194}
\mlm \du \ttmfb - \tp \du \nu \mrm(u,v) \leq F^*(u)
\end{equation}

\begin{equation}\label{Area195}
F^*(u) =  \mlm \du \ttmfb - \tp \du \nu\mrm_{\Gamma}(u) + \int^v_u \mlm G_3 + G_4 - \frac{\p(\nu,\tp)}{\p(u,v)} \mrm (u,v') \D v'
\end{equation}

\begin{equation}\label{Area196}
\int_0^{v_0} F^*(u) \D u \lesssim TV_{(0,v_0)}[\ttmfb|_{\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^\dagger].
\end{equation}

Note that in the expression of (<ref>), the definitions of $G_3$ and $G_4$ were given (<ref>). Consequently, we deduce that

\begin{equation}\label{Area197}
\begin{aligned}
&\int_{D(0,v_0)} \frac{| \dv \tp|}{|\nu|}|\du \ttmfb - \tp \du \nu|(u,v) \D u \D v \lesssim \ml \sup_{u \in (0,v_0)}\int^{v_0}_u |\dv \tp|(u,v) \D v\mr \ml \int^{v_0}_0 F^*(u) \D u\mr \\
\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 + X\cdot A[\nu,\tp^\dagger].
\end{aligned}
\end{equation}

Combining (<ref>) and  (<ref>), we now arrive at the following estimate

\begin{equation}\label{Area198}
\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^\dagger].
\end{aligned}
\end{equation}

In view of (<ref>) and (<ref>), we proceed to estimate $A[\lambda,\tp^\dagger]$ and $A[\nu,\tp^\dagger]$. For convenience, we denote

\begin{equation}\label{ttzeta}
\tilde{\theta} := r \dv \tp = r \dv \phi, \quad \quad
\tilde{\zeta} := r \du \tp = r \du \phi, \quad \quad
\ttz := r D_u \phi.
\end{equation}

It is worth noting that, contrary to (<ref>), here $\ttz \neq r D_u \tp$ but rather $\ttz = \tz + \ii \e A_u r \phi.$

Inspired by the definition of $\tf$ in (<ref>), we also define

\begin{equation}\label{Area199}
\tg := 4 \pi \im \ml \frac{\ttheta^\dagger}{r}\ml\frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu} \mr \mr.
\end{equation}

By (<ref>), we have

\begin{equation}\label{Area200}
\tf + \ii \tg = 4\pi \ml \frac{\ttheta^\dagger}{r} \mr\ml  \frac{\ttheta}{\lambda} - \frac{\tz}{\nu} + E_2\mr
\end{equation}

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

\begin{equation}\label{Area201}
-\frac{4\pi}{r^2}\frac{\tz \ttheta^\dagger}{\lambda \nu}(\nu \ttheta - \lambda \tz) = -\frac{\tf \tz}{r} - \ii \frac{\tz \tg}{r} + \frac{4\pi \tz \ttheta^\dagger}{r^2}E_2.
\end{equation}

Together with (<ref>), we now compute $\frac{1}{\lambda}\frac{\p(\lambda,\tp)}{\p(u,v)}$ as follows

\begin{equation}\label{Area202}
\begin{aligned}
\frac{1}{\lambda}\frac{\p(\lambda,\tp)}{\p(u,v)} =&\; \frac{\p \log \lambda}{\p u}\frac{\p \tp}{\p v} - \frac{\p \log \lambda}{\p v}\frac{\p \tp}{\p u} = \frac{\mu - Q^2/r^2}{1-\mu}\frac{\nu \ttheta}{r^2} - \dv \log \lambda \frac{\tz}{r} \\
=&\; \frac{1}{r^2}(\nu \ttheta - \lambda \tz)\ml \frac{\mu - Q^2/r^2}{1-\mu} + 4 \pi \frac{\ttheta^\dagger \tz}{\lambda \nu}\mr - \frac{\tz}{r}\ml \dv \log \lambda - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\lambda}{r}\mr - \frac{4\pi}{r^2}\frac{\ttheta^\dagger \tz}{\lambda \nu}(\nu\ttheta - \lambda \tz) \\
=&\; \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^\dagger \ttz\mr - \frac{\tz}{r}\ml \dv \log \lambda - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\lambda}{r} + \tf \mr \\
&- \ii \frac{\tz \tg}{r} + \frac{4\pi \tz \ttheta^\dagger}{r^2}E_2 - 4\pi \ii \e \frac{\nu \ttheta - \lambda \tz}{\lambda \nu}\frac{\ttheta^\dagger A_u \phi}{r}.  \\
\end{aligned}
\end{equation}

By recalling and re-writing (<ref>), we have

\begin{equation}\label{Area203}
\du \ml \dv \log \lambda - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\lambda}{r} + \tf\mr =  4\pi \re \ml \frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)}\mr + E_5 + E_6,
\end{equation}

with $|E_5|(u,v) \lesssim D^2$ and $|E_6|(u,v) \lesssim D^2$. Henceforth, we obtain

\begin{equation}\label{Area204}
\mlm \dv \log \lambda - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\lambda}{r} + \tf\mrm(u,v) \leq G(v)
\end{equation}

\begin{equation}\label{Area205}
G(v) := \mlm \dv \log \lambda - \frac{\mu - Q^2/r^2}{1-\mu}\frac{\lambda}{r} + \tf\mrm(0,v) + 4\pi \int^v_0 \mlm \frac{\p(\tp^\dagger,\ttmfb/\nu)}{\p(u,v)}\mrm (u',v) \D u' + \int_0^v |E_5| + |E_6| (u',v) \D u'.
\end{equation}

Together with (<ref>) and (<ref>), we have

\begin{equation}\label{Area206}
\int_0^{v_0} G(v) \D v \lesssim D^2 + A[\tp^\dagger, \ttmfb/\nu].
\end{equation}

Next, we define

\begin{equation}\label{Area207}
\tr := r\ml \frac{\mu - Q^2/r^2}{1-\mu}\lambda \nu + 4 \pi \ttheta^\dagger \ttz \mr.
\end{equation}

Back to (<ref>), with $|E_2|(u,v) \lesssim Dr(u,v)^2$, we now deduce that

\begin{equation}\label{Area208a}
\mlm \frac{1}{\lambda}\frac{\p(\lambda,\tp)}{\p(u,v)} \mrm \lesssim D\frac{|\tr|}{r^3} + |\du \tp|G(v) + D \frac{|r^2\tg|}{r^3}  + D^3.
\end{equation}

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

\begin{equation}\label{Area208}
\begin{aligned}
\int_{\mathcal{D}(0,v_0)}\mlm \frac{1}{\lambda}\frac{\p(\lambda,\tp)}{\p(u,v)} \mrm(u,v) \D u \D v \lesssim&\;  D \int_{\mathcal{D}(0,v_0)}\frac{|\tr|}{r^3}(u,v) \D u \D v + \int_{\mathcal{D}(0,v_0)}|\du \tp|(u,v)G(v)\D u \D v \\
&+\; D \int_{\mathcal{D}(0,v_0)}\frac{|r^2\tg|}{r^3}(u,v) \D u \D v  + D^3.
\end{aligned}
\end{equation}

By (<ref>), we note that

\begin{equation}\label{Area209}
\begin{aligned}
\int_{\mathcal{D}(0,v_0)} |\du \tp|(u,v) G(v) \D u \D v &\lesssim Y \int^{v_0}_0 G(v) \D v \lesssim Y (D^2 + A[\tp^\dagger,\ttmfb/\nu]).
\end{aligned}
\end{equation}

Meanwhile, by (<ref>), (<ref>), (<ref>), we have

\begin{equation}\label{Area320}
|\ttheta^\dagger|(u,v) \lesssim r(u,v), \quad |\ttz|(u,v) \lesssim r(u,v) + Dr^3(u,v) \lesssim r(u,v).
\end{equation}

Together with (<ref>) and $0 \leq \mu - Q^2/r^2 \leq \mu$, we prove

\begin{equation}\label{Area321}
|\tr|(u,v) \lesssim r^3(u,v).
\end{equation}

Thus, we deduce that

\begin{equation}\label{Area322}
\frac{\tr}{r^2}|_{\Gamma} = 0.
\end{equation}

Similarly, one can also deduce that $|\tg|(u,v) \lesssim r(u,v)$ and hence it implies

\begin{equation}\label{Area323}
\tg|_{\Gamma} = 0.
\end{equation}

Using (<ref>) and conducting integration by parts, we obtain

\begin{equation}\label{Area210}
\begin{aligned}
\int_{\mathcal{D}(0,v_0)} \frac{|\tr|}{r^3}(u,v) \D u \D v &=
\frac{1}{2}\int^{v_0}_0 \ml \int^v_0 |\tr/\nu| \du\ml \frac{1}{r^2}\mr(u,v) \D u \mr \D v \\
&= - \frac{1}{2}\int^{v_0}_0 \frac{|\tr/\nu|}{r^2}(0,v) \D v - \frac{1}{2}\int^{v_0}_0 \ml \int^v_0 \frac{1}{r^2}\du \mlm \frac{\tr}{\nu}\mrm(u,v) \D u\mr \D v \\
&\leq \frac{1}{2}\int_{\mathcal{D}(0,v_0)}\frac{1}{r^2}\mlm \du\ml \frac{\tr}{\nu}\mr \mrm (u,v) \D u \D v.
\end{aligned}
\end{equation}

In a similar fashion, by (<ref>), we also get

\begin{equation}\label{Area211}
\begin{aligned}
\int_{\mathcal{D}(0,v_0)} \frac{|r^2 \tg|}{r^3}(u,v) \D u \D v \leq \frac{1}{2}\int_{\mathcal{D}(0,v_0)}\frac{1}{r^2}\mlm \du\ml \frac{r^2\tg}{\nu}\mr \mrm (u,v) \D u \D v.
\end{aligned}
\end{equation}

It remains to compute $\du\ml \frac{\tr}{\nu}\mr$ and $\du \ml \frac{r^2 \tg}{\nu}\mr$. For the former, we have

\begin{equation}\label{Area212}
\begin{aligned}
\du \ml \frac{\tr}{\nu}\mr =&\; \du \ml \frac{r}{\nu} \ml \frac{\mu - Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta^\dagger \ttz\mr\mr \\
=&\; \frac{\mu - Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta^\dagger \ttz + r \du \ml \frac{\mu - Q^2/r^2}{1-\mu}\lambda + 4 \pi \frac{\ttheta^\dagger \ttz}{\nu} \mr \\
=&\; \frac{\mu - Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta^\dagger \ttz + r\lambda \du \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr + \frac{\mu-Q^2/r^2}{1-\mu}r \du \lambda \\
&+ 4\pi r \ml \frac{\ttz}{\nu}\mr (D_u \ttheta - \ii \e A_u\ttheta)^\dagger  + 4\pi r \ttheta^\dagger \du \ml \frac{\ttz}{\nu} \mr.
\end{aligned}
\end{equation}

Subsequently, observe that from (<ref>) and  (<ref>), we can write the term $\du \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr$ as

\begin{equation}\label{Area214}
\begin{aligned}
\du \ml \frac{\mu - Q^2/r^2}{1-\mu}\mr =&\; \du \ml \frac{1 -  Q^2/r^2}{1-\mu}\mr = \frac{1-Q^2/r^2}{(1-\mu)^2}\du \mu + \frac{1}{1-\mu}\du \ml -\frac{Q^2}{r^2}\mr \\
&\quad=  \frac{1-Q^2/r^2}{(1-\mu)^2}\ml \frac{4\pi (1-\mu)|\ttz|^2}{r\nu} - \frac{\nu}{r}(\mu - Q^2/r^2)  \mr - \frac{1}{1-\mu}\ml 8\pi \e Q \im(\phi^\dagger D_u \phi) - \frac{2Q^2 \nu}{r^3}\mr.
\end{aligned}
\end{equation}

Furthermore, observe that from (<ref>), we can express the term $D_u \ttheta$ as

\begin{equation}\label{Area215}
D_u \ttheta = D_u(r\dv \phi) = - \lambda \frac{\ttz}{r} - \ii \e \frac{Q \phi \lambda\nu}{r(1-\mu)}.
\end{equation}

In addition, from (<ref>), we have

\begin{equation}\label{Area216}
\du \ml \frac{\ttmfb}{\nu}\mr = \du \ml \frac{\nu \tp + r D_u \phi}{\nu}\mr = \frac{\ttz}{r} - \ii \e A_u \tp + \du \ml \frac{\ttz}{\nu}\mr.
\end{equation}

Collecting (<ref>) and (<ref>) - (<ref>), we can reorganize and simplify (<ref>) as

\begin{equation}\label{Area217}
\begin{aligned}
\du\ml \frac{\tr}{\nu}\mr  =&\; \frac{\mu - Q^2/r^2}{1-\mu}\lambda \nu + 4\pi \ttheta^\dagger \ttz + \ml \frac{\mu-Q^2/r^2}{1-\mu} \mr^2 \lambda \nu \\
&+ r\lambda \ml \frac{1-Q^2/r^2}{(1-\mu)^2}\mr\ml \frac{4\pi (1-\mu)|\ttz|^2}{r\nu} - \frac{\nu}{r}(\mu - Q^2/r^2)  \mr - \frac{r\lambda}{1-\mu}\ml 8\pi \e Q \im(\phi^\dagger D_u \phi) - \frac{2Q^2 \nu}{r^3} \mr  \\
&+ 4\pi r \ml \frac{\ttz}{\nu}\mr \ml - \lambda \frac{\ttz}{r} - \ii \e \frac{Q \phi \lambda\nu}{r(1-\mu)} - \ii \e A_u\ttheta \mr^\dagger + 4\pi r \ttheta^\dagger\ml \du\ml \frac{\ttmfb}{\nu}\mr  - \frac{\ttz}{r} + \ii \e A_u \tp \mr \\
=&\; \frac{4 \pi \lambda(\mu - Q^2/r^2)|\ttz|^2}{\nu(1-\mu)} + 4\pi r \ttheta^\dagger \du \ml \frac{\ttmfb}{\nu}\mr + E_9
\end{aligned}
\end{equation}

\begin{equation}\label{Area218}
E_9 := -\frac{r\lambda}{1-\mu}\ml 8\pi \e Q \im(\tp^\dagger D_u \phi) - \frac{2Q^2\nu}{r^3}\mr + 4 \pi \ii \e r \ml \frac{\ttz}{\nu}\mr\ml \frac{Q\phi \lambda \nu}{r(1-\mu)} + A_u \ttheta\mr^\dagger + 4\pi \ii \e r \ttheta^\dagger A_u \tp.
\end{equation}

Note that in the process of arriving at the last line in (<ref>), numerous cancellations happen.
On top of that, since $|\ttz| \lesssim D$, we then have $|E_9|(u,v) \lesssim D^2 r^2(u,v).$ Moreover, with the help of (<ref>) and (<ref>), we can also write

\begin{equation}\label{Area219}
\begin{aligned}
\ttheta^\dagger \du \ml \frac{\ttmfb}{\nu}\mr =&\; r  \du \ml \frac{\ttmfb}{\nu}\mr \dv \tp^\dagger = r\ml \frac{\p(\ttmfb/\nu,\tp^\dagger)}{\p(u,v)} + \dv \ml \frac{\ttmfb}{\nu}\mr\du \tp^\dagger\mr \\
=&\; r\ml \frac{\p(\ttmfb/\nu,\tp^\dagger)}{\p(u,v)} + \frac{\tz^\dagger}{r} \ml -\frac{\mu - Q^2/r^2}{1-\mu}\frac{\tz}{\nu}\frac{\lambda}{r} + \frac{E_4}{\nu} \mr   \mr \\
=&\; r\ml \frac{\p(\ttmfb/\nu,\tp^\dagger)}{\p(u,v)}\mr - \frac{\lambda(\mu - Q^2/r^2)|\tz|^2}{r\nu(1-\mu)} + \frac{\tz^\dagger}{\nu}E_4 \\
=&\; r\ml \frac{\p(\ttmfb/\nu,\tp^\dagger)}{\p(u,v)}\mr - \frac{\lambda(\mu - Q^2/r^2)|\ttz|^2}{r\nu(1-\mu)} + E_{10}
\end{aligned}
\end{equation}

\begin{equation}\label{Area220}
E_{10} := -\frac{\lambda(\mu-Q^2/r^2)}{r\nu(1-\mu)}(2 \e r^2 A_u \im (\phi \du \tp^\dagger) - r^2 \e^2 A_u^2 |\phi|^2) + \frac{\tz^\dagger}{\nu}E_4
\end{equation}

and $|E_{10}|(u,v) \lesssim D^2 r(u,v)$ since $|E_4|(u,v) \lesssim Dr(u,v)$. Combining (<ref>) and (<ref>), we can now write

\begin{equation}\label{Area221}
\begin{aligned}
\du \ml \frac{\tr}{\nu}\mr &= 4\pi r^2 \frac{\p(\ttmfb/\nu,\tp^\dagger)}{\p(u,v)} + E_{11}
\end{aligned}
\end{equation}

\begin{equation}\label{Area222}
E_{11} := E_9 + 4\pi r E_{10}
\end{equation}

and $|E_{11}|(u,v) \lesssim D^2 r^2(u,v)$. Applying (<ref>), this implies that

\begin{equation}\label{Area223}
\int_{\mathcal{D}(0,v_0)} \frac{|\tr|}{r^3}(u,v) \D u \D v \lesssim A[\tp^\dagger,\ttmfb/\nu] + D^2.
\end{equation}

Next, we continue to estimate $\du\ml \frac{r^2 \tg}{\nu}\mr$ and we first compute

\begin{equation}\label{Area224}
\begin{aligned}
\du\ml \frac{\ttheta^\dagger r}{\nu} \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr\mr &= \ml \ttheta^\dagger
+ r  \du\ml \frac{\ttheta^\dagger}{\nu}\mr    \mr \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr + \frac{r}{\nu}\ttheta^\dagger \du\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr.
\end{aligned}
\end{equation}

Using (<ref>), we can rewrite $\du \ml \frac{\ttheta^\dagger}{\nu}\mr$ as

\begin{equation}\label{Area225}
\begin{aligned}
\du\ml \frac{\ttheta^\dagger}{\nu}\mr &= \frac{\nu \du \ttheta^\dagger - \ttheta^\dagger \du \nu}{\nu^2} = -\frac{\lambda}{r\nu}\tz^\dagger + \frac{E_1^\dagger}{\nu} - \frac{\ttheta^\dagger \du \nu}{\nu^2}.
\end{aligned}
\end{equation}

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

\begin{equation}\label{Area226}
\begin{aligned}
\du \ml \frac{\ttheta^\dagger r}{\nu}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr\mr =&\; \ml \ttheta^\dagger
-\frac{\lambda \tz^\dagger}{\nu} + \frac{rE_1^\dagger}{\nu} - \frac{r\ttheta^\dagger \du \nu}{\nu^2}   \mr \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr \\
&+ \frac{r^2}{\nu}\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,\ttmfb/\nu)}{\p(u,v)} +\frac{\ttheta^\dagger E_1}{r\lambda} - \frac{\tz^\dagger E_4}{r\nu} \mr.
\end{aligned}
\end{equation}

Appealing to (<ref>), we have

\begin{equation}\label{Area227}
\begin{aligned}
\ml \nu\ttheta^\dagger
-\lambda \tz^\dagger - \frac{r\ttheta^\dagger \du \nu}{\nu}   \mr \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr =&\; \ml \nu\ttheta^\dagger
-\lambda \tz^\dagger - \frac{r\ttheta^\dagger \du \nu}{\nu}   \mr \ml \frac{\ttheta}{\lambda} - \frac{\tz}{\nu} + E_2\mr \\
=&\; \frac{\nu|\ttheta|^2}{\lambda} - (\ttheta^\dagger \tz + \tz^\dagger \ttheta) + \frac{\lambda}{\nu}|\tz|^2 + E_2(\nu\ttheta^\dagger - \lambda \tz^\dagger) \\
&- \frac{r}{\nu \lambda}\du \nu|\ttheta|^2 + (\du \nu) \ttheta^\dagger \ml \frac{r}{\nu} \mr\ml \frac{\tz}{\nu} - E_2 \mr.
\end{aligned}
\end{equation}

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

\begin{equation}\label{Area228}
\begin{aligned}
\du \nu \ttheta^\dagger &= r(\du \nu \dv \tp^\dagger) = r \ml \frac{\p(\nu,\tp^\dagger)}{\p(u,v)} + \dv\nu \du \tp^\dagger \mr = r \ml \frac{\p(\nu,\tp^\dagger)}{\p(u,v)} + \frac{\lambda \nu\tz^\dagger}{r^2} \frac{\mu - Q^2/r^2}{1-\mu}\mr. \\
\end{aligned}
\end{equation}

Hence, we obtain

\begin{equation}\label{Area229}
\begin{aligned}
\ml \nu\ttheta^\dagger
-\lambda \tz^\dagger - \frac{r\ttheta^\dagger \du \nu}{\nu}   \mr \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr
=&\; \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} \\
&+  \frac{r^2 \tz}{\nu^2}\frac{\p(\nu,\tp^\dagger)}{\p(u,v)}  + E_2(\nu\ttheta^\dagger - \lambda \tz^\dagger) - E_2 \ml \frac{r^2}{\nu} \mr\ml \frac{\p(\nu,\tp^\dagger)}{\p(u,v)} + \frac{\lambda \nu \tz^\dagger}{r^2} \frac{\mu - Q^2/r^2}{1-\mu}\mr.
\end{aligned}
\end{equation}

Plugging this back to (<ref>), we arrive at

\begin{equation}\label{Area230}
\begin{aligned}
\du\ml\frac{\ttheta^\dagger r}{\nu} \ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr\mr =&\; \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{r^2 \tz}{\nu^3}\frac{\p(\nu,\tp^\dagger)}{\p(u,v)}  \\
&+ \frac{E_2}{\nu}(\nu\ttheta^\dagger - \lambda \tz^\dagger) - E_2 \ml \frac{r^2}{\nu^2} \mr\ml \frac{\p(\nu,\tp^\dagger)}{\p(u,v)} + \frac{\lambda \nu\tz^\dagger}{r^2} \frac{\mu - Q^2/r^2}{1-\mu}\mr+ \frac{rE_1^\dagger}{\nu}\ml \frac{\tmfa}{\lambda} - \frac{\ttmfb}{\nu}\mr\\