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The fixed-point equation for $u$ factors asThe cubic derivative is $3u^2+2u+3>0$ for every real $u$; its values at zero and one have opposite signs. Thus the possible first coordinates in the stated domain are exactly $u=0,u_{\ast },1$.

For finite $v$, the second fixed-point equation is $v[v^3-(1+u^2{)}^2]=0$. The renormalization-group fixed points are consequentlyThere are six finite points and three boundary points at infinity. Infinity is interpreted in a compactified domain, not as an ordinary real number. The reciprocal coordinate at infinite coupling $w=1/v$ transforms as $w^{\prime }=(1+u^2{)}^2{w}^4$, making $w=0$ a well-defined boundary fixed point.

At any finite fixed point $(\tilde{u},\tilde{v})$, let $\delta u=u-\tilde{u}$, $\delta v=v-\tilde{v}$. The Jacobian matrix linearization isThe only point with both coordinates strictly interior is $\tilde{u}=u_{\ast }$, $\tilde{v}=(1+u_{\ast }^2{)}^{2/3}$. Using its fixed-point relations, the matrix becomesIts eigenvalues are $\rho$ and $4$. In particular, $u_{\ast }<1/\sqrt{3}$ because the increasing cubic is already positive there, so $\rho >1$ (numerically about $1.68$). The corresponding eigenvectors can be chosen as $(1,c/(\rho -4){)}^T$ and $(0,1{)}^T$. Both discrete multipliers exceed one, henceBoth perturbations are relevant directions of a fixed point under repeated coarse-graining, rather than one stable and one unstable direction. A length-rescaling factor was not specified, so these multipliers should not be assigned numerical critical scaling exponents without additional information. At $a=0,1$, the $u$ multiplier is zero; at $v=0$ or $w=0$ the other multiplier is also zero. Thus the four corner points attract locally within the domain, the two positive finite boundary points at $a=0,1$ are saddles, and the points $(u_{\ast },0),(u_{\ast },\infty )$ have one repulsive and one attractive direction.