## 2 [PERSON]'s Function
\begin{array}{cc}\Psi^{2}g(r,\theta,z;r_{0},\theta_{0},z_{0})\\ =\Psi^{2}g_{r}(r,\,r_{0})g_{\theta}(\theta;\theta_{0})g_{z}(z;z_{0})\\ =\frac{\delta g\kappa_{z}}{r}\frac{\partial}{\partial r}\bigg{(}\frac{\partial g _{r}}{\partial r}\bigg{)}+\frac{\delta g\kappa_{z}}{r^{2}}\frac{\partial^{2} \delta g}{\partial r^{2}}+g_{r}g_{\theta}\frac{\partial^{2}g_{z}}{\partial z ^{2}}\\ =\frac{\delta(r-r_{0})\delta(\theta-r_{0})\delta(z-z_{0})}{\partial r}.\end{array}  \tag{29}\]  
First, solve \(\delta(z-z_{0})\) by integrating both sides over \((0,\,l)\)  
\[ \begin{array}{cc}\int_{z=0}^{z=l}\bigg{[}\frac{\delta g\kappa_{z}}{r}\frac {\partial}{\partial r}\bigg{)}+\frac{\delta g\kappa_{z}}{r^{2}}\frac{\partial ^{2}\delta g}{\partial r^{2}}+g_{r}g_{\theta}\frac{\partial^{2}g_{z}}{\partial z ^{2}}\bigg{]}dz\\ =g_{r}g_{\theta}\int_{z=0}^{z=l}\frac{\partial^{2}g_{z}}{\partial z^{2}}dz. \end{array} \]  
Therefore,  
\[g_{r}g_{\theta}\frac{\partial g_{z}}{\partial z}\bigg{|}_{0}^{z=l}=\frac{ \delta(r-r_{0})\delta(\theta-\theta_{0})}{r}. \tag{30}\]  
From Equations (16), (26), and (30), we obtain  
\[ \begin{array}{cc}g_{r}g_{\theta}\frac{\partial g_{z}}{\partial z}\bigg{|}_ {0}^{z=l}=g_{r}g_{\theta}\bigg{[}-\frac{r_{v1}}{a-r_{0}}(z_{0}-z)\bigg{|}_{0} ^{z_{0}}+\frac{1}{2}e^{-\frac{r_{v1}}{a-r_{0}}(z-z_{0})}\bigg{|}_{z_{0}}^{l} \bigg{]}\\ =\frac{\delta g\kappa_{z}}{2}\bigg{[}e^{-\frac{r_{v1}}{a-r_{0}}z_{0}}+e^{- \frac{r_{v1}}{a-r_{0}}(l-z_{0})}-2\bigg{]},\end{array} \]  
and  
\[\frac{g_{r}g_{\theta}}{2}\bigg{[}e^{-\frac{r_{v1}}{a-r_{0}}z_{0}}+e^{-\frac{r_ {v1}}{a-r_{0}}(l-z_{0})}-2\bigg{]}=\frac{\delta(r-r_{0})\delta(\theta-\theta_{ 0})}{r}. \tag{31}\]  
Substituting Equation (21) into the above equation and integrating both sides over \((0,\,2\pi)\) gives  
\[ \begin{array}{cc}\frac{g\kappa_{z}}{2}\bigg{[}e^{-\frac{r_{v1}}{a-r_{0}}z_{0 }}+e^{-\frac{r_{v1}}{a-r_{0}}(l-z_{0})}-2\bigg{]}\sum_{v=-\infty}^{\infty} \int_{0}^{2\pi}\cos[v(\theta-\theta_{0})]d\theta\\ =\frac{\delta(r-r_{0})}{r}\int_{0}^{2\pi}\delta(\theta-\theta_{0})d\theta.\end{array} \]  
Since  
\[\int_{0}^{2\pi}\cos[v(\theta-\theta_{0})]d\theta=\left\{ \begin{array}{cc}2\pi &\text{ if }\theta=\theta_{0}\\ 0&\text{ otherwise}\end{array} \right.,\]the above equation is given by  
\[\pi g_{r}\bigg{[}e^{-\frac{r_{\rm pl}}{a-r_{0}}z_{0}}+e^{-\frac{r_{\rm pl}}{a-r_{0 }}(l-z_{0})}-2\bigg{]}=\frac{\delta(r-r_{0})}{r}. \tag{32}\]  
Substituting Equation (27) into Equation (32) results in  
\[\pi\bigg{[}e^{-\frac{r_{\rm pl}}{a-r_{0}}z_{0}}+e^{-\frac{r_{\rm pl}}{a-r_{0}} (l-z_{0})}(l-z_{0})-2\bigg{]}A_{\rm pl}\ J_{\rm v}\bigg{(}\frac{r_{\rm pl}}{a-r_{0 }}r_{0}\bigg{)}J_{\rm v}\bigg{(}\frac{r_{\rm pl}}{a-r_{0}}r\bigg{)}=\frac{\delta (r-r_{0})}{r}. \tag{33}\]  
Multiplying both sides of Equation (33) by \(rJ_{p}\Big{(}\frac{r_{\rm pl}}{a-r_{0}}r\Big{)}\) and integrating over (0, \(a-r_{0}\)) gives  
\[ \begin{array}{c}\pi\bigg{[}e^{-\frac{r_{\rm pl}}{a-r_{0}}z_{0}}+e^{-\frac{r _{\rm pl}}{a-r_{0}}(l-z_{0})}-2\bigg{]}\\ \times A_{\rm pl}\ J_{\rm v}\ \Big{(}\frac{r_{\rm pl}}{a-r_{0}}r_{0}\Big{)} \int_{0}^{a-r_{0}}J_{\rm v}\ \Big{(}\frac{r_{\rm pl}}{a-r_{0}}r\Big{)}J_{\rm p}\Big{(}\frac{r_{\rm pl}}{a- r_{0}}r\Big{)}rdr\\ =\int_{0}^{a}\delta(r-r_{0})J_{p}\Big{(}\frac{r_{\rm pl}}{a-r_{0}}r\Big{)}\ d. \end{array}  \tag{34}\]  
Given [26]  
\[\int_{0}^{a-r_{0}}J_{\rm\mu}\bigg{(}\frac{\alpha_{\mu}}{a-r_{0}}r\bigg{)}J_{ \rm\
u}\bigg{(}\frac{\alpha_{\
u}}{a-r_{0}}r\bigg{)}rdr=\frac{(a-r_{0})^{2}}{ 2}\big{[}J_{\rm\mu+1}(\alpha_{\mu})\big{]}^{2}\delta_{\mu\
u}\]  
and  
\[\int_{0}^{a-r_{0}}\delta(r-r_{0})J_{\rm\mu}(r)dr=J_{\rm\mu}(r_{0}),\]  
Equation (34) can be rewritten as  
\[ \begin{array}{c}\pi\Big{[}e^{-r_{\rm pl}z_{0}/(a-r_{0})}+e^{-r_{\rm pl}(l-z _{0})/(a-r_{0})}-2\Big{]}A_{\rm pl}\ J_{\rm v}\big{(}\frac{r_{\rm pl}}{a-r_{0}} r_{0}\Big{)}\frac{(a-r_{0})^{2}}{2}[J_{\rm v+1}(r_{\rm pl})]^{2}\\ =J_{\rm v}\big{(}\frac{r_{\rm pl}}{a-r_{0}}r_{0}\big{)}.\end{array}