be physically compatible with the observations.

### Static solutions and their perturbation

In the static case, equations (\[second\]) and (\[1st\]) give $$\label{second0}
9M^{6}\left(\frac{\kappa}{a^2}-\frac{\Lambda_{\rm b}}3\right)
+\left[\rho(1+3w)-2\sigma\right] \left(\rho+\sigma -
3m^2\frac{\kappa}{a^2}\right)=0$$ and $$\label{1st0}
m^4\left(\frac{\kappa}{a^2}-\frac{\rho+\sigma}{3m^2}\right)^2=
M^{6}\left(\frac{\kappa}{a^2}-\frac{\Lambda_{\rm b}}{6}-\frac{C}{a^4}\right)\,
,$$ respectively. One should note that Eq. (\[1st0\]) cannot be obtained by integrating Eq. (\[second\]) in the static case because, before integration, Eq. (\[second\]) is multiplied by $\dot{a}$, which, in this case, is identically zero. Nevertheless, it can be obtained from the original equations on the brane (\[eq\_mot2\]) considering the embedding of the brane in the bulk. In fact, this equation determines the value of the black-hole constant $C$ for which a solution of (\[second0\]) is embeddable in the Schwarzschild–(anti)-de Sitter bulk.

Looking for static solutions and investigating their stability, we proceed as in the previous case of general relativity. First, we solve Eq. (\[second\]) with respect to $\ddot a / a$ for $\dot a = 0$: $$\label{1st1}
\frac{\ddot{a}}{a}=-\frac{9M^{6}\left(\frac{\kappa}{a^2}-\frac{\Lambda_{\rm
b}}3\right)+ \left[\rho(1+3w)-2\sigma\right] \left(\rho+\sigma -
3m^2\frac{\kappa}{a^2}\right)}{9M^{6}+6m^2\left(\rho+
\sigma-3m^2\frac{\kappa}{a^2}\right)}\, .$$ Studying homogeneous perturbations of this solution, we set $\dot{a}=0$ as the initial condition. If a solution is stable (unstable) with respect to such perturbations, it will also be stable (unstable) with respect to more generic perturbations with $\dot{a}\neq 0$. Thus, we 