= Solution
Set
$$
g'_0=g\beta(T_1-T_0)^2,
\qquad
g'=g'_0\theta^2,
\qquad
c=\sqrt{g'h}=\theta\sqrt{g'_0h}.
$$
The axisymmetric reduced-gravity equations are
$$
h_t+\frac1r(rhu)_r=0,
$$
$$
u_t+uu_r+g'h_r+\frac h2g'_r=0,
$$
$$
\theta_t+u\theta_r=-\frac K h\theta.
$$
The extra $hg'_r/2$ in momentum is the depth average of the hydrostatic pressure gradient caused by horizontal density variation.
In variables $(u,c,\theta)$ these become
$$
u_t+uu_r+2cc_r-\frac{c^2}{\theta}\theta_r=0,
$$
$$
c_t+uc_r+\frac c2u_r
=-\frac{cu}{2r}-\frac{Kg'_0\theta^2}{c},
$$
$$
\theta_t+u\theta_r
=-\frac{Kg'_0\theta^3}{c^2}.
$$
The three <characteristic speeds> are
$$
\boxed{\lambda_0=u,\qquad\lambda_\pm=u\pm c}.
$$
Along $dr/dt=u$, temperature obeys
$$
\boxed{\frac{d\theta}{dt}=-\frac K h\theta
=-\frac{Kg'_0\theta^3}{c^2}}.
$$
Along $dr/dt=u+s c$, $s=\pm1$, a left-characteristic projection gives
$$
\boxed{
\frac{du}{dt}+2s\frac{dc}{dt}
-s\frac c\theta\frac{d\theta}{dt}
=-s\left(\frac{cu}{r}
+\frac{Kg'_0\theta^2}{c}\right)
}.
$$
Back to article page
