QG Formulations

QG Formulations#

Florian#

(64)#\[\begin{split} \begin{aligned} \frac{\partial q}{\partial t} + \det\boldsymbol{J}(\psi,q) &= 0 \\ \psi &= \frac{g}{f}\eta \\ q &= \nabla^2 \psi - \frac{1}{L_R^2}\psi \end{aligned} \end{split}\]

Hugo#

(65)#\[\begin{split} \begin{aligned} \frac{\partial q}{\partial t} + \det\boldsymbol{J}(\psi,q) &= \nu\nabla^2 q-\mu q -\beta \partial_x\psi + F \\ \psi &= \frac{g}{f}\eta \\ q &= \nabla^2 \psi \\ \boldsymbol{u} &= (-\partial_y\psi,\partial_x\psi) \end{aligned} \end{split}\]

where:

\(\nu\) is the viscosity
\(\mu\) is the linear drag coefficient
\(\beta\) - rossby parameter
\(F\) - source term

Louis#

Here, we have a stacked QG model:

\[\begin{split} \begin{aligned} \mathbf{q} &= [q_1, \ldots, q_N]^\top \\ \mathbf{\psi} &= [\psi_1, \ldots, \psi_N]^\top \end{aligned} \end{split}\]

But we have the same equations as above.

\[\begin{split} \begin{aligned} \frac{\partial \mathbf{q}}{\partial t} + \begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} \cdot \nabla \mathbf{q} &= 0 \\ \begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} &= \nabla^{\perp}\boldsymbol{\psi} \\ \psi &= \frac{g}{f}\eta \\ q &= \nabla^2 \psi - f_0^2\mathbf{A}\psi +\beta y \end{aligned} \end{split}\]

\[\begin{split} \mathbf{A} = \begin{bmatrix} \frac{1}{H_1 g_1'} & \frac{-1}{H_1 g_1'} & \ldots & \ldots & \ldots \\ \frac{-1}{H_1 g_1'} & \frac{1}{H_1}\left(\frac{1}{g_1'} + \frac{1}{g_2'} \right) & \frac{-1}{H_1 g_1'} & \ldots & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ \ldots & \ldots & \frac{-1}{H_{n-1} g_{n-2}'} & \frac{1}{H_{n-1}}\left(\frac{1}{g_{n-2}'} + \frac{1}{g_{n-1}'} \right) & \frac{-1}{H_{n-1} g_{n-2}'} \\ \ldots & \ldots& \ldots & \frac{-1}{H_n g_{n-1}'} & \frac{1}{H_n g_{n-1}'} \\ \end{bmatrix} \end{split}\]

where:

\(f_0 + \beta(y-y_0)\) - Coriolis parameter under the beta-plane approximation with the meridional axis center \(y_0\)
\(\nabla^{\perp}=(-\partial_y,\partial_x)\) - perpendicular gradient
\(\nabla^2 =\partial_{xx}+\partial_{yy}\) - horizontal Laplacian

Solving QG Equations - I#

Given the equations in terms of q and \(\psi\).

\[\begin{split} \begin{aligned} \partial_t q + \det J(\psi, q) &= 0 \\ \psi &= \frac{g}{f}\eta \\ q &= \nabla^2 \psi - \frac{1}{L_R^2}\psi \end{aligned} \end{split}\]

We are stepping through the \(q\) term.

\[ \partial_t q = - \det(\psi, q) \]

Step I: Find \(\psi\)

We need to calculate \(\psi\) from \(q\) using the expression above.

\[\begin{split} \begin{aligned} \nabla^2 \psi - \frac{1}{L_R^2}\psi &= q \\ (\nabla^2 - \frac{1}{L_R^2})\psi &= q \end{aligned} \end{split}\]

which involves solving a linear system of equations:

\[ \psi = (\nabla^2 - \frac{1}{L_R^2})^{-1}q \]

Step II: Find the determinant Jacobian

\[\begin{split} \begin{aligned} -\det J(\psi, q) &= \left(\frac{\partial\psi}{\partial y}\frac{\partial q}{\partial x} - \frac{\partial\psi}{\partial x}\frac{\partial q}{\partial y} \right) \\ \end{aligned} \end{split}\]

Step III: Put Everything together and step

\[ \begin{aligned} \partial_t q = (\nabla^2 - \frac{1}{L_R^2})^{-1}\left(\frac{\partial\psi}{\partial y}\frac{\partial q}{\partial x} - \frac{\partial\psi}{\partial x}\frac{\partial q}{\partial y} \right) \end{aligned} \]

Solving QG Equations for SSH#

\[ \partial_t \left(\nabla^2 - \frac{1}{L_R^2} \right)\eta + \frac{g}{f}\det J(\eta, \nabla^2\eta) = 0 \]

Step I: Calculate Determinant Jacobian#

\[\begin{split} \begin{aligned} -\alpha\beta\det J(\eta, \nabla^2\eta) &= \alpha\beta\left(\frac{\partial\eta}{\partial y}\frac{\partial \nabla^2\eta}{\partial x} - \frac{\partial\eta}{\partial x}\frac{\partial \nabla^2\eta}{\partial y} \right) \\ \end{aligned} \end{split}\]

Step II: Isolate \(\eta\)#

\[\begin{split} \begin{aligned} \partial_t(\nabla^2 - \frac{1}{L_R^2})\eta &= F \\ (\nabla^2 - \frac{1}{L_R^2})\partial_t\eta &= F \\ \end{aligned} \end{split}\]