QG Regularization (Math)

tags: interpolation

In this note, we will construct a QG regularization on sea surface height by using the QG PDE which relates SSH to potential vortity and the stream function. We will convert all variables within the QG PDE to SSH.

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Edits

I document some mistakes I made from the derivation as pointed out by me or others

4-09-2022

• FIXED: Redouane: I made a sign error for the PV in terms of the stream function
  • Wrong:
    $q={\mathrm{\nabla }}^2\psi +c_2\psi$
    $\to$ Correction:
    $q={\mathrm{\nabla }}^2\psi -c_2\psi$
• FIXED: Redouane: I made an error with one of the constants
  • Wrong:
    $c_1=f/g$
    $\to$ Correction:
    $c_1=g/f$
  • Wrong:
    $c_3=f/g{L}_r^2$
    $\to$ Correction:
    $c_3=g/f{L}_r^2$
• FIXED: Redouane: The the
  $det\text{boldsymbol}J(u,u)=0$, duh…
  • $det\text{boldsymbol}J(u,u)={\partial }_{x}u{\partial }_{y}u-{\partial }_{y}u{\partial }_{x}u=0$

22-09-2022

• FIXED: Emmanuel C.: I made another sign error when factoring out the constant,
  $c_1$, which resulted in a wrong final term:
  • Wrong:
    ${\partial }_t{\mathrm{\nabla }}^{2}u\text{color}red+c_2{\partial }_{t}u+c_{1}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)=0$
  • Corrected:
    ${\partial }_t{\mathrm{\nabla }}^{2}u\text{color}green-c_2{\partial }_{t}u+c_{1}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)=0$

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Function

The learned function,

$\text{boldsymbol}{f}_{\theta }$, will map the spatial coordinates,

${\mathbf{x}}_{\varphi }\in {\mathbb{R}}^{D\varphi }$, and time coordinate,

$t\in \mathbb{R}$, to sea surface height,

$u\in \mathbb{R}$.

$$u=\text{boldsymbol}{f}_{\theta }({\mathbf{x}}_{\varphi },t)$$

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Loss

The standard loss term is data-driven

$${\mathcal{L}}_{data}=\text{MSE}(u,\hat{u})=\frac{1}{N}\sum _{n=1}^N{(u-\text{boldsymbol}{f}_{\varphi }({\mathbf{x}}_{\varphi },t))}^2$$

However, there is no penalization to make the field behave the way we would expect. We also want a regularization which makes the field,

$u$, behave how we would expect. This can be achieved by adding a physics-informed loss regularization term to the total loss.

$$\mathcal{L}={\mathcal{L}}_{data}+\lambda {\mathcal{L}}_{phy}$$

This loss term can be minimized by effectively minimizing a PDE function. For example:

$$\text{boldsymbol}{f}_{phy}(\mathbf{x},t):={\partial }_{t}u(\mathbf{x},t)+\mathcal{N}[u(\mathbf{x},t)]=0$$

where

${\partial }_t$ is the derivative of the field,

$u$, wrt to time and

$\mathcal{N}[\cdot ]$ are some partial differential equations. We are interested in minimizing the full PDE, which we denote

$\text{boldsymbol}{f}_{phy}$, st it is 0. So the standard loss function applies.

$${\mathcal{L}}_{phy}=\frac{1}{N}\sum _{n=1}^N(\text{boldsymbol}{f}_{phy}(\mathbf{x},t){)}^2$$

The PDE for this method will be the quasi-geostrophic (QG) equation which the potential vorticity and a stream function. In the following section, we will show how this equation can be modified to act as a physics-informed regularization term on the loss function.

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QG Equation

We have the following PDE for the QG dynamics:

$${\partial }_{t}q+detJ(\psi ,q)=0$$

where

$q(x,t)\in {\mathbb{R}}^2\times \mathbb{R}\to \mathbb{R}$ is the potential vorticity (PV),

$\psi (x,t)\in {\mathbb{R}}^2\times \mathbb{R}\to \mathbb{R}$ is the stream function,

${\partial }_t$ is the partial derivative wrt

$t$,

$\text{boldsymbol}J$, is the Jacobian operator and

$det\text{boldsymbol}J(\cdot ,\cdot )$ is the determinant of the Jacobian.

Objective: We want to convert this PDE in terms of sea surface height (SSH) instead of PV and the stream function.

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QG Equation 4 SSH (TLDR)

Note: For the ease of notation, let's denote

$u$ as the SSH. The above PDE can be written in terms of

$u$

$$c_2{\partial }_{t}u-{\partial }_t{\mathrm{\nabla }}^{2}u-c_{1}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)=0$$

where

$c_1=\frac{g}{f}$ and

$c_2=\frac{1}{L_R^2}$.

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Derivation

Stream Function

Let's define the relationship between the SSH,

$u$, and the stream function

$\psi$.

$$\psi =\frac{g}{f}u=c_{1}u$$

where

$c_1=\frac{g}{f}$. If we plug in the SSH into the PDE, we get:

$${\partial }_{t}q+detJ(c_{1}u,q)=0$$

To simplify the notation, we will factor out the constant,

$c_1$, from the determinant Jacobian term.

$${\partial }_{t}q+c_{1}detJ(u,q)=0$$

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Proof: Constants and determinant Jacobians

$$\begin{array}{rl}detJ(c_{1}u,q)& ={\partial }_x{c}_{1}u{\partial }_{y}q-{\partial }_y{c}_{1}u{\partial }_{x}q\\ detJ(c_{1}u,q)& =c_1{\partial }_{x}u{\partial }_{y}q-c_1{\partial }_{y}u{\partial }_{x}q\\ c_{1}detJ(u,q)& =c_1({\partial }_{x}u{\partial }_{y}q-{\partial }_{y}u{\partial }_{x}q)\end{array}$$

Note: we used the property that

$\partial (cf)=c\partial f$.

QED.

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Potential Vorticity

Now, let's define the relationship between the stream function and the PV. This is given by:

$$\begin{array}{rl}q& ={\mathrm{\nabla }}^2\psi -\frac{1}{L_R^2}\psi \\ & ={\mathrm{\nabla }}^2\psi -c_2\psi \end{array}$$

where

${\mathrm{\nabla }}^2$ is the Laplacian operator and

$c_2=\frac{1}{L_R^2}$. If we plug in SSH,

$u$, into the stream function, as defined above, we get:

$$\begin{array}{rl}q& ={\mathrm{\nabla }}^2(c_{1}u)-c_2(c_{1}u)\\ & =c_1{\mathrm{\nabla }}^{2}u-c_{3}u\end{array}$$

where

$c_3=c_1{c}_2=\frac{g}{f{L}_R^2}$. We can plug in the PV,

$q$, into the PDE in terms of SSH,

$u$.

$${\partial }_t(c_1{\mathrm{\nabla }}^{2}u-c_{3}u)+c_{1}detJ(u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)=0$$

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Now, we can expand this equation but first let's break this up into two terms:

$$\underset{\text{Term I}}{\underbrace{{\partial }_t(c_1{\mathrm{\nabla }}^{2}u-c_{3}u)}}+c_1\underset{\text{Term II}}{\underbrace{detJ(u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)}}=0$$

Now we will tackle both of these terms one at a time.

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Term I

So term I is:

$$f_1:={\partial }_t(c_1{\mathrm{\nabla }}^{2}u-c_{3}u)$$

We can expand this via the partial derivative,

${\partial }_t$.

$$f_1:=c_1{\partial }_t{\mathrm{\nabla }}^{2}u-c_3{\partial }_{t}u$$

So plugging this back into the PDE gives us:

$$c_1{\partial }_t{\mathrm{\nabla }}^{2}u+c_3{\partial }_{t}u+c_1\underset{\text{Term II}}{\underbrace{detJ(u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)}}=0$$

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Term II

We can factorize this determinant Jacobian term.

$$c_{1}detJ(u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)=c_{1}det\text{boldsymbol}J(u,c_1{\mathrm{\nabla }}^{2}u)-det\text{boldsymbol}J(c_{1}u,c_{3}u)$$

And furthermore, we can factor out the constants

$$c_{1}detJ(u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)=c_1^{2}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)-c_1{c}_{3}det\text{boldsymbol}J(u,u)$$

And lastly, we can remove the final term because the determinant Jacobian of the same function is zero. So we have:

$$c_{1}detJ(u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)=c_1^{2}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)$$

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Proof: Determinant Jacobian Expansion

$$det\text{boldsymbol}J(u,u)={\partial }_{x}u{\partial }_{y}u-{\partial }_{y}u{\partial }_{x}u={\partial }_{x}u{\partial }_{y}u-{\partial }_{x}u{\partial }_{y}u=0$$

QED.

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Proof: Determinant Jacobian Expansion

So term II is:

$$f_2:=detJ(c_{1}u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)$$

We know the definition of the determinant of the Jacobian for a vector-valued function,

$\text{boldsymbol}f=[f_1(x,y),f_2(x,y){]}^{\mathrm{\top }}:{\mathbb{R}}^2\to {\mathbb{R}}^2$, as there is an identity.

$$detJ(f_1(x,y),f_2(x,y))={\partial }_x{f}_1{\partial }_y{f}_2-{\partial }_y{f}_1{\partial }_x{f}_2$$

If use this identity with term II, we get:

$$detJ(c_{1}u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)={\partial }_x(c_{1}u){\partial }_y(c_1{\mathrm{\nabla }}^{2}u-c_{3}u)-{\partial }_y(c_{1}u){\partial }_x(c_1{\mathrm{\nabla }}^{2}u-c_{3}u)$$

Again, let's split this up into two subterms and tackle them one-by-one.

$$detJ(c_{1}u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)=\underset{\text{Term IIa}}{\underbrace{{\partial }_x(c_{1}u){\partial }_y(c_1{\mathrm{\nabla }}^{2}u-c_{3}u)}}-\underset{\text{Term IIa}}{\underbrace{{\partial }_y(c_{1}u){\partial }_x(c_1{\mathrm{\nabla }}^{2}u-c_{3}u)}}$$

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Term IIa

We have the following term for Term IIa:

$$f_{2a}:={\partial }_x(c_{1}u){\partial }_y(c_1{\mathrm{\nabla }}^{2}u-c_{3}u)$$

We can expand the terms to get the following:

$$f_{2a}:={\partial }_x(c_{1}u){\partial }_y{\mathrm{\nabla }}^2(c_{1}u)-{\partial }_x(c_{1}u){\partial }_y(c_{3}u)$$

And we can simplify, by factoring out constants, to get:

$$f_{2a}:=c_1^2{\partial }_{x}u{\partial }_y{\mathrm{\nabla }}^{2}u-c_1{c}_3{\partial }_{x}u{\partial }_{y}u$$

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Term IIb

$$f_{2b}:={\partial }_y(c_{1}u){\partial }_x(c_1{\mathrm{\nabla }}^{2}u-c_{3}u)$$

We can expand the terms to get the following:

$$f_{2b}:=c_1^2{\partial }_{y}u{\partial }_x{\mathrm{\nabla }}^{2}u-c_1{c}_3{\partial }_{y}u{\partial }_{x}u$$

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Combined (Term IIa, IIb)

We can substitute all of these two terms into our original expression

$$\begin{gathered} detJ(c_{1}u,c_1{\mathrm{\nabla }}^{2}u+c_{3}u)= \\ =f_{2a}-f_{2b} \\ =(c_1^2{\partial }_{x}u{\partial }_y{\mathrm{\nabla }}^{2}u-c_1{c}_3{\partial }_{x}u{\partial }_{y}u)-(c_1^2{\partial }_{y}u{\partial }_x{\mathrm{\nabla }}^{2}u-c_1{c}_3{\partial }_{y}u{\partial }_{x}u) \\ =c_1^2{\partial }_{x}u{\partial }_y{\mathrm{\nabla }}^{2}u-c_1{c}_3{\partial }_{x}u{\partial }_{y}u-c_1^2{\partial }_{y}u{\partial }_x{\mathrm{\nabla }}^{2}u+c_1{c}_3{\partial }_{y}u{\partial }_{x}u \end{gathered}$$

If we group the terms by operators,

$\partial ,{\mathrm{\nabla }}^2$, then we get:

$$\begin{gathered} detJ(c_{1}u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)= \\ \underset{{\mathrm{\nabla }}^2}{\underbrace{c_1^2{\partial }_{x}u{\partial }_y{\mathrm{\nabla }}^{2}u-c_1^2{\partial }_{y}u{\partial }_x{\mathrm{\nabla }}^{2}u}}-\underset{\partial }{\underbrace{c_1{c}_3{\partial }_{x}u{\partial }_{y}u-c_1{c}_3{\partial }_{y}u{\partial }_{x}u}} \end{gathered}$$

So each of these terms are determinant Jacobian terms,

$det\text{boldsymbol}J$.

$$detJ(c_{1}u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)=c_1^{2}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)-c_1{c}_{3}det\text{boldsymbol}J(u,u)$$

We have the final form for our PDE in terms of SSH,

$u$, which combines terms I and II.

Finally, we can see that the final term is zero because it's the determinant Jacobian of itself. So we are left with:

$$detJ(c_{1}u,c_1{\mathrm{\nabla }}^{2}u-c_{3}u)=c_1^{2}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)$$

QED.

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Final Form

So we have the final form for our PDE in terms of SSH.

$$c_1{\partial }_t{\mathrm{\nabla }}^{2}u-c_3{\partial }_{t}u+c_1^{2}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)=0$$

We will factor out a constant term

$${\partial }_t{\mathrm{\nabla }}^{2}u-c_2{\partial }_{t}u+c_{1}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)=0$$

I will multiply everything by a negative 1 so that we have a PDE of the form:

${\partial }_{t}u+\mathcal{N}()=0$ This give us:

$$c_2{\partial }_{t}u-{\partial }_t{\mathrm{\nabla }}^{2}u-c_{1}det\text{boldsymbol}J(u,{\mathrm{\nabla }}^{2}u)=0$$