Basic state

In the basic steady state there is no fluid motion i.e. $\boldsymbol {U}_0=0$. Now the basic presssure $P_0(z)$ and temperature $T_0(z)$ is given by

$$\frac{d^2T_0}{dz^2}=0$$   and$$\quad \frac{d}{dz}\left(\frac{P_0}{\rho_b}+gz\right) - \alpha g (T_0-T_b) = 0$$

which have solutions given by

$$\boldsymbol {U}=0,\;T_0(z)=T_b-\beta z,\;P_0(z)=p_b-g\rho_b(z+\alpha \beta z^2/2)$$

where $\beta=(T_b-T_u)/d$ is the basic temperature gradient. We anticipate for stability we must have $\beta > 0$.