my objective contains the forms of $\max(x,0)$ which can be written as { \begin{align} \mathop{\max}\limits_{{\bf{P}},\theta}\quad&R^{sec}_{tot}\\ \textrm{s.t.} \quad\: &C_{c,e} \le R_{c},\\ &\text{tr}\left({\bf{P}}{{\bf{P}}^H}\right) \le {P_T}, \end{align}
where $R_c^{\sec } = {[{R_c} - {C_{c,e}}]^ + }$, $R_{p,k}^{\sec } = {[{R_{p,k}} - {C_{k,e}}]^ + }$,$R_{c}=\min(R_{c,1},R_{c,2})$, $ k=1,2$.and I converted it to { \begin{align} \mathop{\max}\limits_{\boldsymbol{P},\theta,{\boldsymbol{\alpha}}_{p},{\boldsymbol{\alpha}}_{c}} & \mathop{\min} \{ \alpha_{c,1},\alpha_{c,2}\} +\sum\nolimits_{k} \left(\alpha_{p,k}-\alpha_{k,e}\right)-\alpha_{c,e}\\ \textrm{s.t.} \quad\: &R_{p,k} \ge \alpha_{p,k},\\ &R_{c,k} \ge \alpha_{c,k},\\ &C_{k,e} \le \alpha_{k,e},\\ &C_{c,e} \le \alpha_{c,e},\\ &\alpha_{c,e} \le \alpha_{c,1},\\ &\alpha_{c,e} \le \alpha_{c,2},\\ &\text{tr}\left({\bf{P}}{{\bf{P}}^H}\right) \le {P_T}, \end{align} via introducing some non-negative vector, and compute $\max(\min(\alpha_{c,1},\alpha_{c,2}),0)+\max(\alpha_{p,1}-\alpha_{1,e},0)+\max(\alpha_{p,1}-\alpha_{1,e},0)$. But from the simulation, I find the result is not my expectation. Then, what I should do to convert the problem?
If it eventually is $$\max_{x\in \mathcal F_1,y\in \mathcal F_2,z\in \mathcal F_3} \quad x+\max(y,0)+\max(z,0),$$ then you can equivalently rewrite it as $$\max_{x\in \mathcal F_1,y\in \mathcal F_2,z\in \mathcal F_3,\gamma_1,\gamma_2} \quad x+\gamma_1+\gamma_2 \quad \text{ subject to } \quad \max(y,0)\ge \gamma_1 \text { and } \max(z,0)\ge \gamma_2.$$ The latter problem is equivalent to $$\max_{x\in \mathcal F_1,y\in \mathcal F_2,z\in \mathcal F_3, \gamma_1,\gamma_2} \quad x+\gamma_1+\gamma_2 \quad \text{ subject to } \quad y+|y|\ge 2\gamma_1 \text { and } z+|z|\ge 2\gamma_2.$$
So, we need to solve four problems where we play with the sign of $y$ and $z$ and eventually, take the largest value as the optimal value (I do not know an elegant way to remove $|.|$ from the constraints).
By the way, note that in general, $$\max_{x\in \mathbb R^n} \, f(x)$$ is equivalent to $$\max_{x\in \mathbb R^n,t\in \mathbb R} t \quad \text{ subject to } \quad f(x)\ge t.$$ and $\max(z,0)=(z+|z|)/2$. Thus, you can always assume the objective function is linear.