- #1
Dustinsfl
- 2,281
- 5
Is there a clean may to get the fixed points for
\begin{alignat*}{9}
F - 2B' - cB - \frac{3}{4}AB^2 - \frac{3}{4}A^3 & = & 0 & \quad & \Rightarrow & \quad & B' & = & \frac{1}{2}F - \frac{c}{2}B - \frac{3}{8}AB^2 - \frac{3}{8}A^3\\
2A' + cA - \frac{3}{4}A^2B - \frac{3}{4}B^3 & = & 0 & \quad & \Rightarrow & \quad & A' & = & \frac{3}{8}A^2B + \frac{3}{8}B^3 - \frac{c}{2}A
\end{alignat*}
\begin{alignat*}{9}
F - 2B' - cB - \frac{3}{4}AB^2 - \frac{3}{4}A^3 & = & 0 & \quad & \Rightarrow & \quad & B' & = & \frac{1}{2}F - \frac{c}{2}B - \frac{3}{8}AB^2 - \frac{3}{8}A^3\\
2A' + cA - \frac{3}{4}A^2B - \frac{3}{4}B^3 & = & 0 & \quad & \Rightarrow & \quad & A' & = & \frac{3}{8}A^2B + \frac{3}{8}B^3 - \frac{c}{2}A
\end{alignat*}