- When rotation is turned off, we get dispersionless waves. Write an explicit solution for the zonal velocity field.
When we turn rotation off, we get the following equations:
\begin{align*} u_t &= -g \eta_x \\
v_t &= -g\eta_y \\
\eta_t + H(u_x+v_y) &= 0
\end{align*}
Because \(v_0=0\), \(\eta_{y,0} = 0\) and \(v_{y,0} = 0\), this means that \(v=0\) for all time, and so our dependence on \(y\) is removed. Then our new equations are
\begin{align*} u_t &= -g \eta_x \\
\eta_t &= -H u_x
\end{align*}
Upon further manipulation this yields a gravity wave equation:
$$ u_{tt} = gH u_{xx} $$
which generates waves with speed \(\sqrt{gH}\). The solution to the height field, with our initial conditions, is
\begin{align*}
\eta &= -\frac{1}{2}\eta_0\left(\sgn(x+ct) + \sgn(x-ct) \right)
\end{align*}
So if \(-ct < x < ct\), the height is zero, and after an infinite amount of time the height of the entire fluid will be zero.
And we can find the velocity field via the zonal momentum equation:
\begin{align*}
u_t &= -g \eta_x \\
\implies u &= \frac{g}{2c}\eta_0\left(\sgn(x+ct) + \sgn(x-ct) \right)
\end{align*}
So all the potential energy is converted to kinetic energy and radiated away to infinity.
- Now turn rotation on. Use potential vorticity conservation to find a steady state solution.
We come back to our original equations:
\begin{align*} u_t - f_0 v &= -g \eta_x \\
v_t + f_0u &= -g\eta_y \\
\eta_t + H(u_x+v_y) &= 0
\end{align*}
Unlike the non-rotating case, we cannot eliminate the meridional velocity term, as it evolves in the presence of non-zero zonal velocity. However, from our initial conditions there is still no explicit dependence on \(y\), so we may eliminate it:
\begin{align*} u_t - f_0 v &= -g \eta_x \\
v_t + f_0 u &= 0 \\
\eta_t + H u_x &= 0
\end{align*}
In the linearized shallow water equations, the potential vorticity is given by:
\begin{align*}
q &= \zeta - f_0\frac{\eta}{H}
\end{align*}
Because there is no y-dependence, this can be written as \(q = v_x - f_0\frac{\eta}{H}\). Because potential vorticity is conserved, it will have the same value when our steady state is achieved:
$$ q = q_0 = \frac{f_0}{H}\eta_0\sgn(x) $$
We now set \(v_t=\eta_t=u_t=0\) to find a steady state solution. First, we find that \(u = u_x=0\), so zonal velocity is completely eliminated. Combining the zonal momentum equation and PV conservation, we have two equations and two unknowns
\begin{align*}
-f_0v &=-g \eta_x\\
v_x - f_0\frac{\eta}{H} &= \frac{f_0}{H}\eta_0\sgn(x)
\end{align*}
Upon further manipulation, we arrive at the following equation:
\begin{align*}
\eta_{xx} &= \frac{f_0^2}{gH}\left(\eta +\eta_0\sgn(x)\right) = \frac{1}{L_d^2}\left(\eta +\eta_0\sgn(x)\right)
\end{align*}
where \(L_d = \frac{\sqrt{gH}}{f_0}\) is the Rossby radius of deformation. We can solve this ODE using variation of parameters. First consider the homogeneous equation. We will call our solutions to this \(a\):
\begin{align*}
a_{xx} - \frac{1}{L_d^2}a &= 0\\
a &= \left\{\exp\left(\frac{x}{L_d}\right),\exp\left(-\frac{x}{L_d}\right)\right\}
\end{align*}
The Wronskian is
\begin{align*}
W &= a_1a_2' - a_1'a_2 = -\frac{2}{L_d}
\end{align*}
We will now use variation of parameters to find \(b\) such that \(\eta = a_1b_1+a_2b_2\). We enforce the boundary condition that \(\eta \to \eta_0\sgn(x)\) as \(x \to \pm \infty\), and we remove any branches of the solution that cause the solution to blow up:
\begin{align*}
b_1 &= \frac{L_d}{2}\int \exp\left(-\frac{x}{L_d}\right) \frac{\eta_0}{L_d^2}\sgn(x) dx\\\\
&= \begin{cases}
\frac{\eta_0}{2}\exp\left(-\frac{x}{L_d}\right) + C_1 & x<0 \\
-\frac{\eta_0}{2}\exp\left(-\frac{x}{L_d}\right) + C_2 & x>0\\
\end{cases} \\\\
b_2 &= -\frac{L_d}{2}\int \exp\left(\frac{x}{L_d}\right)\frac{\eta_0}{L_d^2}\sgn(x) dx\\\\
&= \begin{cases}
\frac{\eta_0}{2}\exp\left(\frac{x}{L_d}\right) + C_3 & x<0 \\
-\frac{\eta_0}{2}\exp\left(\frac{x}{L_d}\right) + C_4 & x>0
\end{cases}
\end{align*}
So the solution is
\begin{align*}
\eta &= \begin{cases}
\eta_0 + C_1\exp\left(\frac{x}{L_d}\right) + C_3\exp\left(-\frac{x}{L_d}\right) & x<0 \\
-\eta_0+ C_2\exp\left(\frac{x}{L_d}\right) + C_4\exp\left(-\frac{x}{L_d}\right) & x>0
\end{cases}
\end{align*}
To prevent blow-up, we set \(C_2=C_3=0\). At \(x=0\), by symmetry we enforce \(\eta=0\) and ensure continuity there:
\begin{align*}
\eta_0 + C_1\exp\left(\frac{x}{L_d}\right) &= 0 \implies C_1 = -\eta_0\\
-\eta_0+ C_4\exp\left(-\frac{x}{L_d}\right) &=0 \implies C_4 = \eta_0
\end{align*}
Thus
\begin{align*}
\eta &= \begin{cases}
\eta_0\left(1 - \exp\left(\frac{x}{L_d}\right)\right) & x<0 \\
-\eta_0 \left(1 - \exp\left(-\frac{x}{L_d}\right)\right) & x>0
\end{cases}
\end{align*}
And we can solve for the meridional velocity:
\begin{align*}
v &= \frac{g}{f_0}\eta_x = -\eta_0\sqrt{\frac{g}{H}}\exp\left(-\frac{|x|}{L_d}\right)
\end{align*}
