to the initial time, and it is vanishing.

-   For the null joint term at point $M$ we have $$\label{jointM}
    I_{J}^{null}(M) =  \frac{r_M^{d-1} ~\Omega_{d-1}}{8 \pi G_{N}} \log  \left| \frac{\mathfrak{a}^2}{f(r_M)} \right|$$ It is assumed that $t(M)=0$ at this point, which yields $r ^{\ast}(M)=-\mu^+ = -\frac{\mu+\chi}{2}$. The sign of $f(r_M)$ is negative, and using it is not hard to find $$\label{mtime}
    \frac{d r_M}{d \mu} = -\frac{ f(r_M)}{2}.$$ Using the previous equation we find $$\frac{d I_{J}^\text{null}(M)}{d\mu} = \frac{r_M^{d-2} ~\Omega_{d-1}  }{16 \pi G_{N}} \left( r_M ~f'(r_M)-(d-1)f(r_M) \log \frac{-\mathfrak{a}^2}{f(r_M)}\right).$$ Using we have $$\begin{aligned}
    \label{pointM}
    \frac{d I_{J}^\text{null}(M)}{d\mu} = \frac{(d-2) ~\omega^{d-2}~ \Omega_{d-1}}{16 \pi G_{N}}-\frac{r_M^{d-2} ~\Omega_{d-1} }{8 \pi G_{N}} \left((d-1)~f(r_M) ~\log \frac{-\mathfrak{a}^2}{f(r_M)}\right).\end{aligned}$$ Using , the boundary contribution is given by $$\label{lateboundary}
    \frac{d I_\text{GHY}^\text{sing}(M)}{d\mu} =\frac{\Omega_{d-1} ~d ~\omega^{d-2}}{16 \pi G_{N}}.$$ Finally, the rate of complexity growth in the flat case can be found by adding the last two terms and , $$\label{Scwlate}
    \dot{\mathcal{C}} = \frac{1}{\pi}\frac{d I}{d \mu}=\frac{1}{\pi}\left[ 2 M - \frac{r_M^{d-2} ~\Omega_{d-1} (d-1)}{16 \pi G_{N}  }~f(r_M)\log \frac{-\mathfrak{a}^2}{f(r_M)}\right]$$ This is exactly which is given by taking the flat-space limit.

In the symmetric evolution $r_M$ is always less than the horizon radius and $-\infty<f(r_M)<0$. Using this fact, we find that $\dot{\mathcal{C}}$ given by starts from $-\infty$ and increases to a maximum value 