$\tilde{\phi}(X,Y;r,s)$ so that $\phi(x,y) dx dy = \tilde{\phi}(X,Y;r,s) dX dY$. Integrals of the density over the domains are tractable if the population fractions are $(r,s)=(\tfrac{1}{3},\tfrac{1}{3})$ so that the peak of the bivariate normal coincides with the vertex where the domains meet (see Figure \[rspDoms\]), which is also the fixed point of the system in the symmetric case. We can exploit this tractability by expressing $\tilde{\phi}(X,Y;r,s)$ as a perturbation of $\tilde{\phi}(X,Y;\tfrac{1}{3},\tfrac{1}{3})$. We first write the population fractions as perturbations about the fixed point $$\begin{aligned}
r &= \frac{1}{3} + \psi_r \\
s &= \frac{1}{3} + \psi_s\end{aligned}$$ where $\psi_r$ and $\psi_s$ are small fluctuations. We then define the following ratio $$h(X,Y;\psi_r,\psi_s) := \frac{\tilde{\phi}(X,Y;\tfrac{1}{3}+\psi_r,\tfrac{1}{3}+\psi_s)}{\tilde{\phi}(X,Y;\tfrac{1}{3},\tfrac{1}{3})}$$ which tends to unity as $\psi_r, \psi_s \rightarrow 0$. The first Taylor polynomial of $h$ about $\psi_r = \psi_s=0$ is $$\begin{gathered}
h_1(X,Y;\psi_r,\psi_s)  = 1 + \frac{6 X \left(m-\sqrt{3} Y\right)}{m} \psi_r \\
 + \frac{6 m \left(X+\sqrt{3} Y\right)-9 X^2-6 \sqrt{3} X Y+9 Y^2}{2 m} \psi_s.\end{gathered}$$ The density $\tilde{\phi}(X,Y;r,s)$ therefore has the following asymptotic behaviour, as $\psi_s, \psi_r \rightarrow 0$ $$\tilde{\phi}(X,Y;r,s) \sim h_1(X,Y;\psi_r,\psi_s) \frac{3 e^{-\frac{3(X^2+Y^2)}{m}}}{m \pi}.$$ We now transform to polar coordinates $X=u \cos \theta, Y= u \sin \theta$, and make use of the fact that in these coordinates the domains are symmetrical 