## 3 System Model
### Joint Channel and Power Assignment
After solving the channel allocation problem, we continue to analyze how to avoid node interference in the underwater acoustic channel through power control by a non-cooperative game. Two different cognitive users, \(k\) and \(m\), can be taken as two rational and selfish individuals. User \(k\) hopes to increase the transmission power to increase his own utility, but the increase in the transmission power will cause interference to the other user \(m\), which not only reduces the transmission performance of the whole system, but also increases the cost of his own participation in the competition. Therefore, only when the transmitted power of different users achieves a Nash equilibrium in the competition of the game can each user fairly obtain the maximum benefit. In addition, the mutual interference between nodes and the continuous loss of underwater battery energy can be regarded as the costs paid by users in the game, and this part of the costs must be removed to calculate the benefits of users. Therefore, the utility function of cognitive users can be expressed as follows:  
\[ \begin{array}{l}\textit{let }E_{k}=\frac{E-E_{k}}{E}\Rightarrow\\ U_{k}(P_{lk})=\sum\limits_{k=1}^{K}\sum\limits_{l=1}^{N}a_{lk}Blog_{2}\Bigg{(} 1+\frac{P_{lk}h_{lk}}{L_{0}Ba_{lk}+\sum\limits_{m\
eq k,\ m=1}^{M}P_{m}h_{mk}} \Bigg{)}\\ -\Bigg{(}L_{0}Ba_{lk}+\sum\limits_{m\
eq k,\ m=1}^{M}P_{m}h_{mk}+\sum\limits_{ k=1}^{K}P_{lk}h_{km}\Bigg{)}-\frac{1}{E_{k}}\end{array}  \tag{9}\]  
In the above formula, \(\sum\limits_{k=1}^{K}P_{lk}h_{km}\) means that user \(k\) interferes with other nodes. \(\sum\limits_{m\
eq k,\ m=1}^{M}P_{m}h_{mk}\) means that other nodes interfere with user \(k\). \(L_{0}Ba_{lk}\) represents the disturbance by marine noise, including marine mammals. \(\frac{1}{E_{k}}\) represents the increase or decrease in the final yield resulting from the amount of residual energy. The residual energy can be obtained by the detection of the transmitting node. Then, considering the problem of mammal-friendliness, it is necessary to limit the transmitted power in Formula (9), which is expressed as follows:  
\[\left\{ \begin{array}{l}\begin{array}{l}U_{k}(P_{lk})\\ \\ \\ C_{1}:\end{array}&\begin{array}{l}=\sum\limits_{k=1}^{K}\sum\limits_{l=1}^{N}a _{lk}Blog_{2}\Bigg{(}1+\frac{P_{lk}h_{lk}}{L_{0}Ba_{lk}+\sum\limits_{m\
eq k, \ m=1}^{M}P_{m}h_{mk}}\Bigg{)}\\ -\Bigg{(}L_{0}Ba_{lk}+\sum\limits_{m\
eq k,\ m=1}^{M}P_{m}h_{mk}+\sum\limits_{ k=1}^{K}P_{lk}h_{km}\Bigg{)}-\frac{1}{E_{k}}\\ C_{1}:\end{array}\right.\\ P_{k}\leq P_{lim1},P_{m}\leq P_{lim1}\quad b\leq n\leq c\\ C_{2}:\end{array} \right.\\ P_{k}\leq P_{lim2},P_{m}\leq P_{lim2}\quad a\leq n\leq b,\ c\leq n\leq N\end{array} \tag{10}\]The optimal power \(P_{lk}\) in Equation (10) is the Nash equilibrium result of the above game process so the equation can be expressed as a conditional extreme value problem. Then, Formula (10) can be reduced to:  
\[ \begin{array}{ll}U(P_{k},P_{m})&=\sum\limits_{k=1}^{K}\sum\limits_{l=1}^{N}a_ {lk}Blog_{2}\left(1+\frac{P_{lk}h_{lk}}{L_{0}B_{4 lk}+\frac{M}{m+k}\sum\limits_ {n=1}^{M}P_{m}h_{mk}}\right)\\ &-\left(L_{0}Ba_{lk}+\sum\limits_{m+k,\ m=1}^{M}P_{m}h_{mk}+\sum\limits_{k=1}^{ K}P_{k}h_{km}\right)-\frac{1}{E_{k}}\\ &+\sum\limits_{k=1}^{K}\alpha_{k}(P_{nuc}-P_{k})+\sum\limits_{m=1}^{K}\beta_{k }(P_{nuc}-P_{m})\\ &+\sum\limits_{k=1}^{K}\gamma_{k}(P_{nus}-P_{k})+\sum\limits_{m=1}^{K}\delta( P_{nus}-P_{m})\end{array}  \tag{11}\]  
From formula (11), the following can be obtained:  
\[\alpha_{k}(P_{nuc}-P_{k})=0 \tag{12}\]  
\[\gamma_{k}(P_{nus}-P_{k})=0 \tag{13}\]  
\[ \begin{array}{ll}\frac{\partial U(P_{k},P_{m})}{\partial P_{k}^{2}}=0\Rightarrow \\ \left\{\begin{array}{ll}\frac{a_{lk}B_{lk}}{L_{0}B_{4 lk}+\frac{M}{m+k}\sum \limits_{n=1}^{M}P_{n}h_{mk}+P_{k}h_{lk}}-\alpha_{k}=0\\ b\leq n\leq c\\ \frac{a_{lk}B_{lk}}{L_{0}B_{4 lk}+\frac{M}{m+k}\sum\limits_{n=1}^{M}P_{n}h_{ mk}+P_{k}h_{lk}}-\gamma_{k}=0\\ a\leq n\leq b,\ c\leq n\leq N\end{array}\right.\end{array}  \tag{14}\]  
\[P_{lk}=\left\{