クープマンモデル
{ d A d t = − β B d B d t = − α A {\displaystyle {\begin{cases}{\frac {dA}{dt}}=-\beta B\\{\frac {dB}{dt}}=-\alpha A\end{cases}}}
初期条件: A ( 0 ) = A 0 , B ( 0 ) = B 0 {\displaystyle A(0)=A_{0},B(0)=B_{0}}
α {\displaystyle \alpha } は、A軍の武器性能 β {\displaystyle \beta } は、B軍の武器性能 A 0 {\displaystyle A_{0}} はA軍の初期の兵員数 B 0 {\displaystyle B_{0}} はB軍の初期の兵員数
上記の連立微分方程式を解析的に解くと、時間tでのA軍とB軍の残存戦力は以下のようになる
{ υ = a Q − d P a b − c d ω = b P − c Q a b − c d a b − c d = μ 2 − γ 2 γ = c + d 2 κ = c − d 2 μ = κ 2 + a b ε = a b 2 μ ( μ + κ ) [ ( A 0 + d + μ + κ a b − c d P ) − μ + κ b ( B 0 + c + μ − κ a b − c d Q ) ] σ = a b 2 μ ( μ + κ ) [ μ + κ a ( A 0 + d − μ + κ a b − c d P ) + ( B 0 + c − μ − κ a b − c d Q ) ] ( a b ≠ c d ) {\displaystyle {\begin{cases}\upsilon ={\frac {aQ-dP}{ab-cd}}\\\omega ={\frac {bP-cQ}{ab-cd}}\\ab-cd={\mu }^{2}-{\gamma }^{2}\\\gamma ={\frac {c+d}{2}}\\\kappa ={\frac {c-d}{2}}\\\mu ={\sqrt {{\kappa }^{2}+ab}}\\\epsilon ={\frac {ab}{2\mu (\mu +\kappa )}}\left[\left(A_{0}+{\frac {d+\mu +\kappa }{ab-cd}}P\right)-{\frac {\mu +\kappa }{b}}\left(B_{0}+{\frac {c+\mu -\kappa }{ab-cd}}Q\right)\right]\\\sigma ={\frac {ab}{2\mu (\mu +\kappa )}}\left[{\frac {\mu +\kappa }{a}}\left(A_{0}+{\frac {d-\mu +\kappa }{ab-cd}}P\right)+\left(B_{0}+{\frac {c-\mu -\kappa }{ab-cd}}Q\right)\right]\\(ab\neq cd)\end{cases}}}