## Appendix B Derivation of the Solution
To eliminate the variable \(z_{\rm D}\) in the control equations, the finite Fourier cosine transforms are defined as follows:  
\[\overline{p}_{fD}(u_{n})=\int\limits_{0}^{1}p_{fD}\cos(u_{n}z_{D})dz_{D},\, \overline{p}_{m2D}(u_{n})=\int\limits_{0}^{1}p_{m2D}\cos(u_{n}z_{D})dz_{D} \tag{20}\]  
Thus, the inverse Fourier transforms are:  
\[p_{fD}(z_{D})=\sum\limits_{n=0}^{\infty}\frac{\cos(u_{n}z_{D})}{N(n)} \overline{p}_{fD}(u_{n}),\,p_{m2D}(z_{D})=\sum\limits_{n=0}^{\infty}\frac{\cos (u_{n}z_{D})}{N(n)}\overline{p}_{m2D}(u_{n}) \tag{21}\]  
where,  
\[N(n)=\int\limits_{0}^{1}\cos^{2}(u_{n}z_{D})dz_{D}=\left\{ \begin{array}{l}1, n=0\\ \frac{1}{2},n=1,2,3,\ldots\end{array} \right. \tag{22}\]  
Applying the Fourier transform presented by Equation (20), we can obtain:  
\[\frac{1}{r_{D}}\frac{\partial}{\partial r_{D}}\left(r_{D}\frac{\partial \overline{p}_{fD}}{\partial r_{D}}\right)-\frac{u_{n}^{2}}{h_{1D}^{2}} \overline{p}_{fD}=sf(s)\overline{\overline{p}}_{fD} \tag{23}\]  
\[\frac{1}{r_{D}}\frac{\partial}{\partial r_{D}}\left(r_{D}\frac{\partial \overline{p}_{m2D}}{\partial r_{D}}\right)-\frac{u_{n}^{2}}{h_{2D}^{2}} \overline{p}_{m2D}=\frac{1}{\sigma}\overline{\overline{p}}_{m2D} \tag{24}\]  
and the definite conditions can be written as:  
\[\left.\frac{\partial\overline{\overline{p}}_{m2D}}{\partial r_{D}}\right|_{r _{D}=R_{D}}=0 \tag{25}\]  
\[r_{D}\left.\frac{\partial\overline{\overline{p}}_{fD}}{\partial r_{D}}\right| _{r_{D}=1}=-\frac{1}{sl_{mD}}\int\limits_{z_{wD}-\frac{h_{wD}}{2}}^{z_{wD}+ \frac{h_{wD}}{2}}\cos(u_{n}z_{D})dz_{D}=-\frac{1}{s}\frac{\sin\left(\frac{1}{ 2}u_{n}h_{wD}\right)}{\frac{1}{2}u_{n}h_{wD}}\cos(u_{n}z_{wD}) \tag{26}\]  
\[\overline{\overline{p}}_{fD}=\overline{\overline{p}}_{m2D^{\prime}}\frac{ \partial\overline{\overline{p}}_{fD}}{\partial r_{D}}=M\frac{\partial \overline{\overline{p}}_{m2D}}{\partial r_{D}},\,r_{D}=R_{1D}. \tag{27}\]  
The general solutions of Equations (23) and (24) are:  
\[\overline{\overline{p}}_{fD}=Al_{0}(r_{D}q_{n})+BK_{0}(r_{D}q_{n}),\,q_{n}= \sqrt{sf(s)+\frac{u_{n}^{2}}{h_{1D}^{2}}} \tag{28}\]  
\[\overline{\overline{p}}_{m2D}=Cl_{0}(r_{D}T_{n})+DK_{0}(r_{D}T_{n}),T_{n}= \sqrt{\frac{s}{\sigma}+\frac{u_{n}^{2}}{h_{2D}^{2}}} \tag{29}\]  
Combining the conditions of Equations (25) to (27), the solution in Laplace and Fourier domain is:  
\[\overline{\overline{p}}_{fD}=\frac{1}{s}\frac{\sin\left(\frac{1}{2}u_{n}h_{wD }\right)}{\frac{1}{2}u_{n}h_{wD}}\frac{1}{q_{n}}\Bigg{[}\frac{E_{n}I_{0}(r_{D} q_{n})+F_{n}K_{0}(r_{D}q_{n})}{F_{n}K_{1}(q_{n})-E_{n}I_{1}(q_{n})}\Bigg{]}\cos(u_{n }z_{wD}) \tag{30}\]  
where,  
\[E_{n}=q_{n}G_{n}K_{1}(R_{1D}q_{n})+M\overline{T}_{n}H_{n}K_{0}(R_{1D}q_{n}) \tag{31}\]\[F_{n} = q_{n}G_{n}I_{1}(R_{1D}q_{n})-MT_{n}H_{n}I_{0}(R_{1D}q_{n}) \tag{111}\] \[\frac{1}{r_{D}}\frac{\partial}{\partial r_{D}}\left(r_{D}\frac{\partial \overline{p}_{m2D}}{\partial r_{D}}\right)-\frac{u_{n}^{2}}{h_{2D}^{2}} \overline{p}_{m2D}=\frac{1}{\sigma}\overline{\overline{p}}_{m2D} \tag{24}\]0 (112) \[\frac{1}{r_{D}}\frac{\partial}{\partial r_{D}}\left(r_{D}\frac{\partial \overline{p}_{m2D}}{\partial r_{D}}\right)-\frac{u_{n}^{2}}{h_{2D}^{2}} \overline{p}_{m2D}=\frac{1}{\sigma}\overline{\overline{p}}_{m2D} \tag{24}\]1  
By the utility of inverse Fourier transforms, we can obtain:  
\[ \begin{array}{ll}\widetilde{sp}_{fD}&=\frac{1}{q_{0}}\bigg{[}\frac{E_{0}I_{ 0}(r_{0}q_{0})+F_{0}K_{0}(r_{0}q_{0})}{F_{0}K_{1}(q_{0})-E_{0}I_{0}(q_{0})} \bigg{]}\\ &+2\sum\limits_{n=1}^{\infty}\frac{\sin\left(\frac{1}{2}u_{n}u_{nD}\right)}{ \frac{1}{2}u_{n}u_{nD}}\frac{1}{q_{n}}\bigg{[}\frac{E_{n}I_{0}(r_{0}q_{n})+F_ {n}K_{0}(r_{0}q_{n})}{F_{n}K_{1}(q_{n})-E_{n}I_{0}(q_{n})}\bigg{]}\cos(u_{n}z _{uD})\cos(u_{n}z_{D})\end{array}  \tag{114}\]  
For limited-entry well, the equivalent pressure point can be written as follows by integral-averaging method.