## 4 Optimization Problem Raising and Solving
### Solution to the Optimization Problem  
For the Case 1 that is, \(\alpha\in\left(0,\frac{1}{3}\right)\), combining (30) with (31), the first derivative of the \(C_{Low}\) with respect to variable \(H\) is obtained as  
\[\frac{dC_{Low}}{dH}=\frac{1}{2}\cdot\frac{\varepsilon}{\varepsilon H+2\sigma^{2 }\varsigma^{2}}+\underbrace{\frac{\varepsilon^{2}H+\varepsilon\sigma^{2} \varsigma^{2}}{\sigma^{2}\varsigma^{2}\sqrt{\varepsilon H(\varepsilon H+2 \sigma^{2}\varsigma^{2})}}-\frac{\varepsilon}{\sigma^{2}\varsigma^{2}}}_{ \textcircled{1}}+\]  
\[\frac{\sqrt{\mu}}{\sqrt{\pi}erf\big{(}\sqrt{\mu}\big{)}}\cdot\frac{1}{H\sqrt{ \varsigma^{2}PH}}\cdot tan^{-1}\sqrt{\varsigma^{2}PH+}\]  
\[\underbrace{\frac{\sqrt{\mu}}{\sqrt{\pi}erf\big{(}\sqrt{\mu}\big{)}}\cdot\frac{ 1}{\varsigma^{2}PH^{2}+H}-\frac{\sqrt{\mu}}{\sqrt{\varsigma^{2}PH}\sqrt{\pi} erf\big{(}\sqrt{\mu}\big{)}}\cdot\frac{\varsigma^{2}P}{\sqrt{\varsigma^{2}PH}(1+ \varsigma^{2}PH)}}_{\textcircled{2}=0}. \tag{39}\]  
For expression 1 above, we have  
\[\textcircled{1}=\frac{\big{(}\varepsilon H+\sigma^{2}\varsigma^{2}\big{)}-\sqrt {\varepsilon H(\varepsilon H+2\sigma^{2}\varsigma^{2})}}{\sqrt{\varepsilon H (\varepsilon H+2\sigma^{2}\varsigma^{2})}}.\frac{\varepsilon}{\sigma^{2} \varsigma^{2}}. \tag{40}\]  
In order to prove (40) \(\geq 0\), we just need to simply prove  
\[\varepsilon H+\sigma^{2}\varsigma^{2}\geq\sqrt{\varepsilon H(\varepsilon H+2 \sigma^{2}\varsigma^{2})}, \tag{41}\]  
which is obviously identical to prove that  
\[\varepsilon^{2}H^{2}+2\varepsilon H\sigma^{2}\varsigma^{2}+\left(\sigma^{2} \varsigma^{2}\right)^{2}\geq\varepsilon^{2}H^{2}+2\varepsilon H\sigma^{2} \varsigma^{2}. \tag{42}\]  
Obviously, the inequality (42) is satisfied. Since \(H\) and \(\varepsilon\) are nonnegative, we can easily come to the following conclusion  
\[\frac{dC_{Low}}{dH}\geq 0,\ \ \alpha\in(0,1/3). \tag{43}\]  
As to the Case 2: \(\alpha\in\left[\textcircled{1},1\right]\), similarly, making a derivative of (36) with respect to \(H\), and after some necessary manipulations, we have  
\[\frac{dC_{Low}}{dH}=\frac{1}{2H}+\frac{1}{2}\cdot\frac{\varepsilon}{ \varepsilon H+2\sigma^{2}\varsigma^{2}}+\underbrace{\frac{\varepsilon^{2}H+ \varepsilon\sigma^{2}\varsigma^{2}}{\sigma^{2}\varsigma^{2}\sqrt{\varepsilon H (\varepsilon H+2\sigma^{2}\varsigma^{2})}}-\frac{\varepsilon}{\sigma^{2} \varsigma^{2}}}_{\textcircled{1}}+\]  
\[\frac{1}{2H\sqrt{\varsigma^{2}PH}}\cdot tan^{-1}\sqrt{\varsigma^{2}PH}-\frac{1 }{2}\cdot\frac{\varsigma^{2}P}{\varsigma^{2}PH+1}-\frac{1}{2\varsigma^{2}PH} \cdot\frac{\varsigma^{2}P}{\varsigma^{2}PH+1}\]  
\[=\frac{1}{2}\cdot\frac{\varepsilon}{\varepsilon H+2\sigma^{2}\varsigma^{2}}+ \textcircled{1}+\frac{1}{2H\sqrt{\varsigma^{2}PH}}\cdot tan^{-1}\sqrt{\varsigma ^{2}PH}\geq 0,\ \ \alpha\in[1/3,1]. \tag{44}\]  
The last inequality is satisfied due to the fact that 1\(\textcircled{1}\geq 0\). From both (43) and (44) discussed above, we can see that, no matter what the range of \(\alpha\) is in, \(C_{Low}\) is a monotonically increasing function of parameter \(H\). Thus the optimization problem (38) can be written equivalently as  
\[\frac{\sqrt{\mu}}{\sqrt{\pi}erf\big{(}\sqrt{\mu}\big{)}}\cdot\frac{1}{H\sqrt{ \varsigma^{2}PH}}\cdot tan^{-1}\sqrt{\varsigma^{2}PH+}\]0Due to \(H=GL\tau\), and \(G\) and \(\tau\) are positive constants, we further have  
\[\frac{\sqrt{\mu}}{\sqrt{\pi}erf\big{(}\sqrt{\mu}\big{)}}\cdot\frac{1}{H\sqrt{ \varsigma^{2}PH}}\cdot tan^{-1}\sqrt{\varsigma^{2}PH+}\]1  
Substituting (8) into (6), the expression of \(L\) with the tilting angle \(\theta\) is  
\[\frac{\sqrt{\mu}}{\sqrt{\pi}erf\big{(}\sqrt{\mu}\big{)}}\cdot\frac{1}{H\sqrt{ \varsigma^{2}PH}}\cdot tan^{-1}\sqrt{\varsigma^{2}PH+}\]2 \[\frac{\sqrt{\mu}}{\sqrt{\pi}erf\big{(}\sqrt{\mu}\big{)}}\cdot\frac{1}{H\sqrt{ \varsigma^{2}PH}}\cdot tan^{-1}\sqrt{\varsigma^{2}PH+}\]3  
Further taking the second derivative of \(L\) with respect to \(\theta\) and after simplifying, it is easy to get \(d^{2}L/d\theta^{2}=-L\).