be the outward unit normal of $\partial\Omega_\varrho$ at $z_i$. We have

    $$\frac{\partial \bar G(z_j,z_i)}{\partial
    n}=-\frac{1}{|z_j-z_i|}\left\langle
    \frac{z_i-z_j}{|z_i-z_j|},n\right\rangle-\frac{\partial
    g(z_j,z_i)}{\partial n},$$ where $n$ is the outward normal unit vector of $\partial\Omega_\varrho$ at $z_i$.

    On the other hand, if $z_j\in\Omega_\varrho$, $j\ne i$, satisfies

    $$\left\langle \frac{z_i-z_j}{|z_i-z_j|},n\right\rangle<0,$$ then,

    $$\left\langle \frac{z_i-z_j}{|z_i-z_j|},n\right\rangle= O(|z_i-z_j|).$$ So, we obtain

    $$\left\langle \frac{z_i-z_j}{|z_i-z_j|},n\right\rangle\ge
    -C|z_i-z_j|, \quad\forall\; j\ne i.$$ As a result, by Lemma \[l43\] and Proposition \[ap2\], we have

    $$\begin{split}
    \frac{\partial K}{\partial n}\ge &
    \frac{4\pi^2{\delta}^2\kappa_i}{|\ln{\varepsilon}||\ln\frac{R}{{\varepsilon}}|}\frac{\partial
    q(z_i)}{\partial n}+
    \frac{2\pi{\delta}^2\kappa_i^2}{(\ln\frac{R}{{\varepsilon}})^2}\frac{\partial
    g(z_i,z_i)}{\partial n}\\
     &+\sum_{j\ne i}^m \frac{2\pi{\delta}^2\kappa_i\kappa_j}{(\ln\frac{R}{{\varepsilon}})^2}\frac{\partial  g(z_j,z_i)}{\partial
     n}-\frac{C
    {\delta}^2}{(\ln\frac{R}{{\varepsilon}})^2}.
    \end{split}$$ On the other hand, we derive from

    $$\frac{\partial g(z_i,z_i)}{\partial n}=\frac{1+o(1)}{2\varrho},$$ and

    $$\frac{\partial g(z_j,z_i)}{\partial
     n}=\frac{1+o(1)}{|\bar z_i-z_j|}\bigl\langle \frac{\bar z_i-z_j}{|\bar
     z_i-z_j|},n\bigr\rangle,$$ where $\bar z_i$ is the reflection point of $z_i$ with respect to $\partial\Omega$.

    It is easy to check that if $|z_j-z_i|\le M\varrho$, where $M>0$ is a fixed large constant, then

    $$\left\langle \frac{\bar z_i-z_j}{|\bar
     z_i-z_j|},n\right\rangle\ge 0.$$ So

    $$\frac{\partial K}{\partial n}\ge \frac{2\pi
    {\delta}^2}{(\ln\frac{R}{{\varepsilon}})^2}\Bigl(
    \frac{\kappa_i^2+o(1)}{2\varrho}-\frac{\kappa_i\kappa_j+o(1)}{M\varrho}-C\Bigr)>0.$$

Therefore, the flow does not leave $D$.

We will prove that $K(Z)$ has a critical point in $K^{c_{{\delta},2}}\setminus K^{c_{{\delta},1}}$.

Suppose that $K(Z)$ has no critical point in $K^{c_{{\delta},2}}\setminus
K^{c_{{\delta},1}}$. Then from Lemma \[l44\] that $K^{c_{{\delta},1}}$ is a deformation retract of $K^{c_{{\delta},2}}$.

It is easy to see that $K^{c_{{\delta},2}}=D$ and

$$\bigl\{Z:  Z\in D,  |z_i-z_j|=\varrho^{\bar L}, \;\text{for some}\;
i\ne 