# Coupling hydrostatic and nonhydrostatic ocean circulation models. Vertical modes perspective

Pierre Lozano, Eric Blayo, Laurent Debreu

<figure><img src="image_1.png" /></figure>

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Pierre Lozano, Eric Blayo, Laurent Debreu. Coupling hydrostatic and nonhydrostatic ocean circulation models. Vertical modes perspective. EGU 2024 - EGU General Assembly, Apr 2024, Vienna, Austria. hal-05344264

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Distributed under a Creative Commons Attribution 4.0 International License
# Coupling hydrostatic and nonhydrostatic ocean circulation models. Vertical modes perspective.

Pierre Lozano

Eric Blayo

Laurent Debreu

April 16, 2024
<figure><img src="image_3.png" /></figure>

# Motivations

* Hydrostatic when $\delta = \frac{H}{L} \ll 1 \rightarrow$ large scale

* Non Hydrostatic when $\Delta x$ decreases $\rightarrow$ small scale

<figure><img src="image_4.png" /><figcaption>Non hydrostatic zoom in the strait of Gibraltar (CROCO model).</figcaption></figure>

* **Goal:** improve consistency between H and NH solutions in a hybrid framework
<figure><img src="image_5.png" /></figure>

Hydrostatic

$L_H u_H = f_H$

interface: $B_H u_H = B_{NH} u_{NH}$

$L_{NH} u_{NH} = f_{NH}$

update: $u_H = M u_{NH}$

Non Hydrostatic

* Find the interaction operators $B_H$, $B_{NH}$ (one-way) and $M$ (two-way)
<figure><img src="image_6.png" /></figure>

Hydrostatic

$$L_H u_H = f_H$$

interface: $B_H u_H = B_{NH} u_{NH}$

$$L_{NH} u_{NH} = f_{NH}$$

update: $u_H = M u_{NH}$

Non Hydrostatic

* Find the interaction operators $B_H$, $B_{NH}$ (one-way) and $M$ (two-way)

* Physical approach: vertical mode decomposition

* Numerical tool: Perfectly Matched Layers
<figure><img src="image_7.png" /></figure>

* Approximations: Boussinesq and hydrostatic. Linearisation around a state at rest:

$$ 
\left\{ 
\begin{array}{l}
\rho_0 \frac{\partial u}{\partial t} = - \frac{\partial p'}{\partial x} \\
\rho_0 \frac{\partial w}{\partial t} = - \frac{\partial p'}{\partial z} - g \rho' \\
\frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} = 0 \\
\frac{\partial p'}{\partial t} - \frac{N^2}{g} \rho_0 w = 0
\end{array}
\right.
\quad \text{(LNH)}
$$
<figure><img src="image_8.png" /></figure>

* Approximations: Boussinesq and hydrostatic. Linearisation around a state at rest:

$$ \left\{ \begin{array}{l} \rho_0 \frac{\partial u}{\partial t} = - \frac{\partial p'}{\partial x} \\ \frac{\partial w}{\partial t} = - \frac{\partial p'}{\partial z} - g \rho' \\ \frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} = 0 \\ \frac{\partial p'}{\partial t} - \frac{N^2}{g} \rho_0 w = 0 \end{array} \right. \quad (\text{LNH}) $$

Boundary conditions:

$$ \left\{ \begin{array}{l} w(x, -H, t) = 0 \\ \frac{\partial p'}{\partial t}(x, 0, t) = g \rho_0 w(x, 0, t) \end{array} \right. $$
<figure><img src="image_9.png" /></figure>

* Approximations: Boussinesq and hydrostatic. Linearisation around a state at rest:

$$ \left\{ \begin{array}{l} \rho_0 \frac{\partial u}{\partial t} = - \frac{\partial p'}{\partial x} \\ \frac{\partial w}{\partial t} = - \frac{\partial p'}{\partial z} - g \rho' \\ \frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} = 0 \\ \frac{\partial p'}{\partial t} - \frac{N^2}{g} \rho_0 w = 0 \end{array} \right. \quad (\text{LNH}) $$

Boundary conditions:

$$ \left\{ \begin{array}{l} w(x, -H, t) = 0 \\ \frac{\partial p'}{\partial t}(x, 0, t) = g \rho_0 w(x, 0, t) \end{array} \right. $$

* Internal waves dispersion relation:

$$ \Omega^2 = \frac{N^2 k_x^2}{k_x^2 + k_z^2} < N^2 $$

<figure><img src="image_10.png" /><figcaption>Contours of $k_z(k_x, \Omega)$. H (left), NH (right).</figcaption></figure>
<figure><img src="image_11.png" /></figure>

Vertical mode decomposition (LNH)

* (LNH) solutions can be decomposed as:

$$
\left\{
\begin{aligned}
\rho_0 W(x, z, t) &= \int_{\Omega \times N} \sum_n w_n(x, t, \Omega) W_n(z, \Omega) \\
\rho_0 u(x, z, t) &= \int_{\Omega \times N} \sum_n u_n(x, t, \Omega) W'_n(z, \Omega) \\
p'(x, z, t) &= \int_{\Omega \times N} \sum_n g \frac{\rho_n(x, t, \Omega)}{\lambda_n(\Omega)} W'_n(z, \Omega) \\
p'(x, z, t) &= \int_{\Omega \times N} \sum_n \rho_n(x, t, \Omega) (N^2(z) - \Omega^2) W_n(z, \Omega) \\
&\quad - \frac{1}{g} \frac{\partial W_n}{\partial t}(x, t, \Omega) W_n(z, \Omega)
\end{aligned}
\right.
$$
<figure><img src="image_12.png" /></figure>

* (LNH) solutions can be decomposed as:

$$
\left\{
\begin{aligned}
\rho_0 W(x, z, t) &= \int_{\Omega \cap N} \sum_n w_n(x, t, \Omega) W_n(z, \Omega) \\
\rho_0 u(x, z, t) &= \int_{\Omega \cap N} \sum_n u_n(x, t, \Omega) W'_n(z, \Omega) \\
\rho'(x, z, t) &= \int_{\Omega \cap N} \sum_n g \frac{\rho_n(x, t, \Omega)}{\lambda_n(\Omega)} W'_n(z, \Omega) \\
\rho'(x, z, t) &= \int_{\Omega \cap N} \sum_n \rho_n(x, t, \Omega) (N^2(z) - \Omega^2) W_n(z, \Omega) \\
&\quad - \frac{1}{g} \frac{\partial w_n}{\partial t}(x, t, \Omega) W_n(z, \Omega)
\end{aligned}
\right.
$$

* $(W_n(\Omega), \lambda_n(\Omega))_{n \ge 0}$ satisfy a Sturm-Liouville problem at fixed $\Omega < N$:

$$
\left\{
\begin{array}{l}
W'''(z) + \lambda (N^2(z) - \Omega^2) W(z) = 0 \\
W(-H) = 0 \\
W'(0) - g\lambda W(0) = 0
\end{array}
\right.
\qquad (\text{SLNH})
$$
<figure><img src="image_13.png" /></figure>

* (LNH) solutions can be decomposed as:

$$
\left\{
\begin{aligned}
\rho_0 W(x, z, t) &= \int_{\Omega \times N} \sum_n w_n(x, t, \Omega) W_n(z, \Omega) \\
\rho_0 u(x, z, t) &= \int_{\Omega \times N} \sum_n v_n(x, t, \Omega) W'_n(z, \Omega) \\
\rho'(x, z, t) &= \int_{\Omega \times N} \sum_n g \frac{\rho_n(x, t, \Omega)}{\lambda_n(\Omega)} W'_n(z, \Omega) \\
\rho'(x, z, t) &= \int_{\Omega \times N} \sum_n \rho_n(x, t, \Omega) (N^2(z) - \Omega^2) W_n(z, \Omega) \\
&\quad - \frac{1}{g} \frac{\partial w_n}{\partial t}(x, t, \Omega) W_n(z, \Omega)
\end{aligned}
\right.
$$

Example of NH vertical modes.

* $(W_n(\Omega), \lambda_n(\Omega))_{n \ge 0}$ satisfy a Sturm-Liouville problem at fixed $\Omega < N$:

$$
\left\{
\begin{array}{l}
W''(z) + \lambda (N^2(z) - \Omega^2) W(z) = 0 \\
W(-H) = 0 \\
W'(0) - g\lambda W(0) = 0
\end{array}
\right.
\qquad (\text{SLNH})
$$

<figure><img src="image_14.png" /></figure>
<figure><img src="image_15.png" /></figure>

Comparison between H and NH modes

<figure><img src="image_16.png" /></figure>

Vertical mode difference with $\Omega = 0.9N$ (non-hydrostatic case).

<figure><img src="image_17.png" /></figure>

Eigenvalues for $\Omega = 0.1N$ (hydrostatic case) and $\Omega = 0.9N$ (non-hydrostatic case).

Reminder:

$$ \frac{\Omega^2}{N^2} = \frac{k_x^2}{k_x^2 + k_z^2} $$
<figure><img src="image_18.png" /></figure>

Example of basic interface conditions

<figure><img src="image_19.png" /></figure>

H impacts NH

<figure><img src="image_20.png" /></figure>

does not retrieve
input frequency

Höv moller diagram and scalogram at z = 0 for the balanced part of the pressure.
<figure><img src="image_21.png" /></figure>

* Add an **absorbing layer** to attenuate outgoing waves.

<figure><img src="image_22.png" /></figure>
<figure><img src="image_23.png" /></figure>

* Add an **absorbing layer** to attenuate outgoing waves.

<figure><img src="image_24.png" /></figure>

* Example: Korteweg-de Vries equation (similar to (LNH) projected onto the vertical modes)

$$ \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} + \epsilon \frac{\partial^2 u}{\partial x^3} = 0 \quad \text{in } [0, L] $$
<figure><img src="image_25.png" /></figure>

* Add an **absorbing layer** to attenuate outgoing waves.

<figure><img src="image_26.png" /></figure>

* Example: Korteweg-de Vries equation (similar to (LNH) projected onto the vertical modes)

$$ \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} + \epsilon \frac{\partial^2 u}{\partial x^3} = 0 \quad \text{in } [0, L] $$

augmented to:

$$ \left\{ \begin{array}{l} \displaystyle \frac{\partial u}{\partial t} + \sigma u + c \frac{\partial u}{\partial x} + \epsilon \frac{\partial \phi}{\partial x} = 0 \\[1em] \displaystyle \frac{\partial \phi}{\partial t} + \sigma \phi = \frac{\partial}{\partial t} \frac{\partial \psi}{\partial x} \\[1em] \displaystyle \frac{\partial \psi}{\partial t} + \sigma \psi = \frac{\partial}{\partial t} \frac{\partial u}{\partial x} \end{array} \right. \quad \text{in } [L, L + \delta] $$
<figure><img src="image_27.png" /></figure>

• We want to solve:

$$ \frac{\frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} + \varepsilon \frac{\partial^3 u}{\partial x^3} = 0}{\Omega_d} \quad \frac{\frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} = 0}{\Omega_{nd}} $$
<figure><img src="image_28.png" /></figure>

* We want to solve:

$$ \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} + \varepsilon \frac{\partial^3 u}{\partial x^3} = 0 \quad \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} = 0 \qquad \Omega_d \qquad \Omega_{nd} $$

* Let $u = u_d + u_{nd}$ such that:

$$ \begin{cases} \frac{\partial u_{nd}}{\partial t} + c \frac{\partial u_{nd}}{\partial x} = 0 & \text{in } \Omega_d \cup \Omega_{nd} \\ u_d = 0 & \text{in } \Omega_{nd} \end{cases} $$
<figure><img src="image_29.png" /></figure>

* We want to solve:

$$ \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} + \varepsilon \frac{\partial^3 u}{\partial x^3} = 0 \qquad \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} = 0 \qquad \Omega_d \perp \Omega_{nd} $$

* Let $u = u_d + u_{nd}$ such that:

$$ \left\{ \begin{array}{ll} \frac{\partial u_{nd}}{\partial t} + c \frac{\partial u_{nd}}{\partial x} = 0 & \text{in } \Omega_d \cup \Omega_{nd} \\ u_d = 0 & \text{in } \Omega_{nd} \end{array} \right. \Rightarrow \left\{ \begin{array}{ll} \frac{\partial u_d}{\partial t} + c \frac{\partial u_d}{\partial x} + \varepsilon \frac{\partial^3 u_d}{\partial x^3} = -\varepsilon \frac{\partial^3 u_{nd}}{\partial x^3} & \text{in } \Omega_d \\ \frac{\partial u_{nd}}{\partial t} + c \frac{\partial u_{nd}}{\partial x} = 0 & \text{in } \Omega_{nd} \end{array} \right. $$
<figure><img src="image_30.png" /></figure>

* We want to solve:

$$ \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} + \varepsilon \frac{\partial^3 u}{\partial x^3} = 0 \quad \frac{\partial u}{\partial t} + c \frac{\partial u}{\partial x} = 0 \qquad \Omega_d \perp \Omega_{nd} $$

* Let $u = u_d + u_{nd}$ such that:

$$ \left\{ \begin{array}{ll} \frac{\partial u_{nd}}{\partial t} + c \frac{\partial u_{nd}}{\partial x} = 0 & \text{in } \Omega_d \cup \Omega_{nd} \\ u_d = 0 & \text{in } \Omega_{nd} \end{array} \right. \Rightarrow \left\{ \begin{array}{ll} \frac{\partial u_d}{\partial t} + c \frac{\partial u_d}{\partial x} + \varepsilon \frac{\partial^3 u_d}{\partial x^3} = -\varepsilon \frac{\partial^3 u_{nd}}{\partial x^3} & \text{in } \Omega_d \\ \frac{\partial u_{nd}}{\partial t} + c \frac{\partial u_{nd}}{\partial x} = 0 & \text{in } \Omega_{nd} \end{array} \right. $$

* Add a PML to attenuate $u_d$ exiting from $\Omega_d$.

$$ \Omega_d \perp \Omega_{nd} $$
<figure><img src="image_31.png" /></figure>

* **Next step:** derive a PML for the H-NH coupling, using vertical mode decomposition