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SCIENCE CHINA Information Sciences, Volume 64 , Issue 12 : 222302(2021) https://doi.org/10.1007/s11432-020-3228-8

Differential game-based analysis of multi-attacker multi-defender interaction

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  • ReceivedSep 7, 2020
  • AcceptedMar 23, 2021
  • PublishedSep 18, 2021

Abstract


Acknowledgment

This work was supported in part by Beijing Municipal Natural Science Foundation (Grant No. 4212005), in part by National Natural Science Foundation of China (Grant Nos. 61932005, 61901051).


Supplement

Appendix

Proof of Lemma 2

When $D(t)~-~A(t)~\geqslant~mA(t)$, the Hamiltonian of attack group can be represented as \begin{align} &{H_1}( {t,{C_A}(t),{C_D}(t),{\lambda _{1}}(t),{\lambda _{2}}(t)},m,n ) \\ &= \frac{\gamma }{2}\{ {A^2}(t) - {[D(t) - A(t)]^2}\} - \frac{\alpha }{2}W_1^2(t) + \frac{\beta }{2}S_1^2(t) + {\lambda _{2}}(t)[ {b{W_2}(t) - d{S_2}(t)} ] + {\lambda _{1}}(t)[ {a{W_1}(t) - c{S_1}(t) - hD(t) + hA(t)} ] \\ &= - \frac{\gamma }{2}{D^2}(t) \!+\! \gamma A(t)D(t) \!-\! {\lambda _{1}}(t)hD(t) \!+\! {\lambda _{1}}(t)hA(t) \!-\! \frac{\alpha }{2}W_1^2(t)\! +\! \frac{\beta }{2}S_1^2(t)\! +\! {\lambda _{2}}(t) [ {b{W_2}(t) \!-\! d{S_2}(t)} ] \!+\! {\lambda _{1}}(t)[ {a{W_1}(t) \!-\! c{S_1}(t)} ]. \tag{1} \end{align} The Hamiltonian of defense group can be represented as \begin{align} &{H_2}( {t,{C_A}(t),{C_D}(t),{\mu _{1}}(t),{\mu _{2}}(t)},m,n ) \\ &= \frac{\varphi }{2}{D^2}(t) - \frac{\eta }{2}W_2^2(t) + \frac{\xi }{2}S_2^2(t) + {\mu _{1}}(t)[a{W_1}(t) - c{S_1}(t) - hD(t) + hA(t)] + {\mu _{2}}(t)[ {b{W_2}(t) - d{S_2}(t)} ] \\ &= \frac{\varphi }{2}{D^2}(t) - {\mu _{1}}(t)hD(t) + {\mu _{1}}(t)hA(t) - \frac{\eta }{2}W_2^2(t) + \frac{\xi }{2}S_2^2(t) + {\mu _{1}}(t)[ {a{W_1}(t) - c{S_1}(t)} ] + {\mu _{2}}(t)[ {b{W_2}(t) - d{S_2}(t)} ]. \tag{2} \end{align} Combining (1) and (2), Eq. (Eq:12) can be obtained by solving the partial derivative for group strength.

When $A(t)~-~D(t)~\geqslant~nD(t)$, the Hamiltonian of attack group can be represented as \begin{align} &{H_1}( {t,{C_A}(t),{C_D}(t),{\lambda _{1}}(t),{\lambda _{2}}(t)},m,n ) \\ &= \frac{\gamma }{2}{A^2}(t) - \frac{\alpha }{2}W_1^2(t) + \frac{\beta }{2}S_1^2(t) + {\lambda _{1}}(t)[ {a{W_1}(t) - c{S_1}(t)} ] + {\lambda _{2}}(t)[ {b{W_2}(t) - d{S_2}(t) - fA(t) + fD(t)} ] \\ &= \frac{\gamma }{2}{A^2}(t) - {\lambda _{2}}(t)fA(t) + {\lambda _{2}}(t)fD(t) - \frac{\alpha }{2}W_1^2(t) + \frac{\beta }{2}S_1^2(t) + {\lambda _{1}}(t)[ {a{W_1}(t) - c{S_1}(t)} ] + {\lambda _{2}}(t)[ {b{W_2}(t) - d{S_2}(t)} ]. \tag{3} \end{align} The Hamiltonian of defense group can be represented as \begin{align} &{H_2}( {t,{C_A}(t),{C_D}(t),{\mu _{1}}(t),{\mu _{2}}(t)},m,n ) \\ &= \frac{\varphi }{2}\{ {D^2}(t) - {[A(t) - D(t)]^2}\} - \frac{\eta }{2}W_2^2(t) + \frac{\xi }{2}S_2^2(t) + {\mu _{1}}(t)[ {a{W_1}(t) - c{S_1}(t)} ] + {\mu _{2}}(t)[ {b{W_2}(t) - d{S_2}(t) - fA(t) + fD(t)} ] \\ &= \varphi A(t)D(t) \!-\! \frac{\varphi }{2}{A^2}(t) \!-\! {\mu _{2}}(t)fA(t) \!+\! {\mu _{2}}(t)fD(t) \!-\! \frac{\eta }{2}W_2^2(t) \!+\! \frac{\xi }{2}S_2^2(t) \!+\! {\mu _{1}}(t) [ {a{W_1}(t) \!-\! c{S_1}(t)} ] \!+\! {\mu _{2}}(t)[ {b{W_2}(t) \!-\! d{S_2}(t)} ]. \tag{4} \end{align} Combining (3) and (4), Eq. (Eq:13) can also be obtained by solving the partial derivative for group strength.

With initial boundary conditions $A(0)~=~{A_0}$, $D(0)~=~{D_0}$, the joint state variables need to be fixed in the final states. Thus, the boundary conditions ${\lambda~_{1}}(T)~=~0$, ${\lambda~_{2}}(T)~=~0$, ${\mu~_{1}}(T)~=~0$, ${\mu~_{2}}(T)~=~0$ should be satisfied.


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