## 3 The Proposed Scheme for WSP Minimization  
### Spectrum and Power Allocation Based on Newton's Method for a Matched PL-SL PAIR  
In this subsection, we first discuss the spectrum allocation and power control for one matched PL-SL pair, i.e., \(\rho_{m}\) and \(\beta_{n}\), during one spectrum-sharing subframe, expressed as follows:  
\[\underset{\{p_{\rho,m},p_{s,m},\rho_{m}\}}{min}w_{mn}, \tag{7}\]  
s.t. C1, C2, C3, C4, and C6. When introducing the proposed scheme in this subsection, we omit the subscripts m, n, and mn for clarity and simplicity. From C1 and C3, we can obtain a lower bound on \(p_{\mu}\) as outlined below:  
\[p_{\rho}^{low}=max\Big{\{}p_{min},\Big{(}2^{Q_{\rho}/\theta}-1\Big{)}n_{0}/g_{ \rho}\Big{\}}. \tag{8}\]  
Similarly, a lower bound on \(p_{\lambda}\) can be obtained from C2 and C4 as follows:  
\[p_{\lambda}^{low}=max\Big{\{}p_{min},\Big{[}2^{Q_{\lambda}/(1-\theta)}-1\Big{]} n_{0}/g_{\lambda}\Big{\}}. \tag{9}\]  
It is easy to validate that the objective function in Equation (7) monotonically increases with \(p_{\mu}\) and \(p_{\lambda}\). Their solutions can be expressed as functions of \(\theta\), i.e., \(p_{\mu}^{*}=p_{\rho}^{low}\) and \(p_{\lambda}^{*}=p_{\lambda}^{low}\). Therefore, Equation (7) can be rewritten as presented below:  
\[\underset{\theta}{min}\mu\theta p_{\mu}^{low}+v(1-\theta)p_{\lambda}^{low}. \tag{10}\]  
s.t. \(0<\theta<1\). Next, we will discuss the optimal solution for \(\theta\) based on four possible combinations of \(p_{\rho}^{low}\) and \(p_{\lambda}^{low}\). * **Case 1** If \(p_{\rho}^{low}=\Big{(}2^{Q_{\rho}/\theta}-1\Big{)}n_{0}/g_{\mu}\) and \(p_{\lambda}^{low}=[2^{Q_{\lambda}/(1-\theta)}-1]n_{0}/g_{\lambda}\), defining the following auxiliary functions  
\[f_{1}(x)=\frac{\mu n_{0}}{g_{\mu}}x\Big{(}2^{Q_{\rho}/x}-1\Big{)}, \tag{11}\]  
\[f_{2}(x)=\frac{\
u n_{0}}{g_{\lambda}}(1-x)\Big{[}2^{Q_{\lambda}/(1-x)}-1 \Big{]}, \tag{12}\]  
the objective function in Equation (10) can be rewritten as outlined below:  
\[w(x)=f_{1}(x)+f_{2}(x). \tag{13}\]  
In this case, we can prove that Equation (10) is convex as follows. **Proposition:** optimization problem Equation (10) in case \(1\) is convex. * All that is needed is to check the convexity of Equation (13). The first and second derivatives of \(f_{1}(x)\) and \(f_{2}(x)\) are calculated as follows: \[f_{1}^{\prime}(x)=\frac{\mu n_{0}}{g_{\mu}}\Big{[}2^{Q_{\rho}/x}\Big{(}1- \frac{Q_{\rho}ln2}{x}\Big{)}-1\Big{]},\] (14) \[f_{2}^{\prime}(x)=\frac{\
u n_{0}}{g_{\lambda}}\Big{[}2^{Q_{ \lambda}/(1-x)}\bigg{(}\frac{Q_{\lambda}ln2}{1-x}-1\Big{)}+1\Big{]},\] (15) \[f_{1}^{\prime\prime}(x)=\frac{\mu n_{0}}{g_{\mu}}\frac{(Q_{\rho}ln2)^{2}}{x^ {3}}2^{Q_{\rho}/x},\] (16)\[p_{\rho}^{low}=max\Big{\{}p_{min},\Big{(}2^{Q_{\rho}/\theta}-1\Big{)}n_{0}/g_{ \rho}\Big{\}}. \tag{8}\]0  
Since \(\theta\in(0,1)\), obviously, \(f_{1}^{\prime\prime}(x)>0\) and \(f_{2}^{\prime\prime}(x)>0\). \(w^{\prime\prime}(x)>0\) holds. Therefore, \(w(x)\) is a convex function [34]. It is clear that \(f_{1}^{\prime}(0^{+})=-\infty\) and \(f_{2}^{\prime}(1^{-})=+\infty\), while \(f_{2}^{\prime}(0^{+})\) and \(f_{1}^{\prime}(1^{-})\) are not infinite. Thus, \(w^{\prime}(0^{+})=-\infty\) and \(w^{\prime}(1^{-})=+\infty\). According to the convexity of (13), \(w^{\prime}(x)\) is monotonical so there must exist an \(x^{*}\) satisfying \(w^{\prime}\big{(}x^{*}\big{)}=0\), and the solution of \(\theta\) equals \(x^{*}\). Although the analytical solution of \(x^{*}\) cannot be obtained, we can access its numerical solution through iterative methods such as the Newton-Raphson method [35].