QG Regularization (Math)

# 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. --- ## 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: $q = \nabla^2\psi + c_2\psi$ $\rightarrow$ Correction: $q = \nabla^2\psi - c_2\psi$ * **FIXED**: *Redouane*: I made an error with one of the constants * Wrong: $c_1 = f/g$ $\rightarrow$ Correction: $c_1 = g/f$ * Wrong: $c_3 = f/gL_r^2$ $\rightarrow$ Correction: $c_3 = g/fL_r^2$ * **FIXED**: *Redouane*: The the $\det\boldsymbol{J}(u,u)=0$, duh... * $\det\boldsymbol{J}(u,u) = \partial_x u \partial_y u - \partial_y u \partial_x u = 0$ **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: $\partial_t \nabla^2 u {\color{red}{+ c_2}} \partial_t u + c_1\det \boldsymbol{J}(u, \nabla^2 u) = 0$ * Corrected: $\partial_t \nabla^2 u {\color{green}{- c_2}} \partial_t u + c_1\det \boldsymbol{J}(u, \nabla^2 u) = 0$ --- ## Function The learned function, $\boldsymbol{f_\theta}$, will map the spatial coordinates, $\mathbf{x}_\phi \in \mathbb{R}^{D\phi}$, and time coordinate, $t \in \mathbb{R}$, to sea surface height, $u \in \mathbb{R}$. $$ u = \boldsymbol{f_\theta}(\mathbf{x}_\phi, t) $$ --- ## Loss The standard loss term is data-driven $$ \mathcal{L}_{data} = \text{MSE}(u, \hat{u}) = \frac{1}{N} \sum_{n=1}^N \left(u - \boldsymbol{f_\phi}(\mathbf{x}_\phi, t) \right)^2 $$ 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: $$ \boldsymbol{f}_{phy}(\mathbf{x},t):= \partial_t u(\mathbf{x},t) + \mathcal{N}[u(\mathbf{x},t)] = 0 $$ 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 $\boldsymbol{f}_{phy}$, st it is 0. So the standard loss function applies. $$ \mathcal{L}_{phy} = \frac{1}{N}\sum_{n=1}^N(\boldsymbol{f}_{phy}(\mathbf{x},t))^2 $$ 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. --- ## QG Equation We have the following PDE for the QG dynamics: $$ \partial_t q + \det J(\psi, q) = 0 $$ where $q(x,t) \in \mathbb{R}^2 \times \mathbb{R} \rightarrow \mathbb{R}$ is the potential vorticity (PV), $\psi(x,t) \in \mathbb{R}^2 \times \mathbb{R} \rightarrow \mathbb{R}$ is the stream function, $\partial_t$ is the partial derivative wrt $t$, $\boldsymbol{J}$, is the Jacobian operator and $\det \boldsymbol{J}(\cdot,\cdot)$ 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. --- ## 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 $u$ $$ c_2 \partial_t u - \partial_t \nabla^2 u - c_1\det \boldsymbol{J}(u, \nabla^2 u)= 0 $$ where $c_1 = \frac{g}{f}$ and $c_2= \frac{1}{L_R^2}$. --- ## 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: $$ \partial_t q + \det J(c_1 u, q) = 0 $$ To simplify the notation, we will factor out the constant, $c_1$, from the determinant Jacobian term. $$ \partial_t q + c_1\det J(u, q) = 0 $$ --- #### **Proof**: Constants and determinant Jacobians $$ \begin{aligned} \det J(c_1 u, q) &= \partial_x c_1 u \partial_y q - \partial_y c_1 u \partial_x q \\ \det J(c_1 u, q) &= c_1 \partial_x u \partial_y q - c_1 \partial_y u \partial_x q \\ c_1\det J(u, q) &= c_1 \left(\partial_x u \partial_y q - \partial_y u \partial_x q\right) \\ \end{aligned} $$ **Note**: we used the property that $\partial (c f) = c \partial f$. **QED**. --- ### Potential Vorticity Now, let's define the relationship between the stream function and the PV. This is given by: $$ \begin{aligned} q &= \nabla^2 \psi - \frac{1}{L_R^2} \psi \\ &= \nabla^2 \psi - c_2 \psi \end{aligned} $$ where $\nabla^2$ 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: $$ \begin{aligned} q &= \nabla^2 (c_1 u) - c_2 (c_1 u) \\ &= c_1 \nabla^2 u - c_3 u \end{aligned} $$ 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, $u$. $$ \partial_t \left(c_1 \nabla^2 u - c_3 u \right) + c_1\det J \left(u, c_1 \nabla^2 u - c_3 u \right) = 0 $$ --- Now, we can expand this equation but first let's break this up into two terms: $$ \underbrace{\partial_t \left(c_1 \nabla^2 u - c_3 u \right)}_{\text{Term I}} + c_1\underbrace{\det J \left( u, c_1 \nabla^2 u - c_3 u \right)}_{\text{Term II}} = 0 $$ Now we will tackle both of these terms one at a time. --- **Term I** So term I is: $$ f_1 := \partial_t \left(c_1 \nabla^2 u - c_3 u \right) $$ We can expand this via the partial derivative, $\partial_t$. $$ f_1 := c_1 \partial_t \nabla^2 u - c_3 \partial_t u $$ So plugging this back into the PDE gives us: $$ c_1 \partial_t \nabla^2 u + c_3 \partial_t u +c_1 \underbrace{\det J \left(u, c_1 \nabla^2 u - c_3 u \right)}_{\text{Term II}} = 0 $$ --- **Term II** We can factorize this determinant Jacobian term. $$ c_1\det J \left(u, c_1 \nabla^2 u - c_3 u \right) = c_1\det \boldsymbol{J}(u, c_1 \nabla^2 u) - \det \boldsymbol{J}(c_1 u, c_3 u) $$ And furthermore, we can factor out the constants $$ c_1\det J \left( u, c_1 \nabla^2 u - c_3 u \right) = c_1^2\det \boldsymbol{J}(u, \nabla^2 u) - c_1c_3 \det \boldsymbol{J}(u, u) $$ And lastly, we can remove the final term because the determinant Jacobian of the same function is zero. So we have: $$ c_1\det J \left( u, c_1 \nabla^2 u - c_3 u \right) = c_1^2\det \boldsymbol{J}(u, \nabla^2 u) $$ --- #### **Proof**: Determinant Jacobian Expansion $$ \det \boldsymbol{J}(u,u) = \partial_x u \partial_y u - \partial_y u \partial_x u = \partial_x u \partial_y u - \partial_x u \partial_y u =0 $$ **QED**. --- #### **Proof**: Determinant Jacobian Expansion So term II is: $$ f_2 := \det J \left(c_1 u, c_1 \nabla^2 u - c_3 u \right) $$ We know the definition of the determinant of the Jacobian for a vector-valued function, $\boldsymbol{f} = [f_1(x,y), f_2(x,y)]^\top: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, as there is an identity. $$ \det J \left( f_1(x,y), f_2(x,y)\right) = \partial_x f_1 \partial_y f_2 - \partial_y f_1 \partial_x f_2 $$ If use this identity with term II, we get: $$ \det J \left(c_1 u, c_1 \nabla^2 u - c_3 u \right) = \partial_x (c_1 u) \partial_y (c_1 \nabla^2u - c_3 u) - \partial_y (c_1 u) \partial_x (c_1 \nabla^2u - c_3 u) $$ Again, let's split this up into two subterms and tackle them one-by-one. $$ \det J \left(c_1 u, c_1 \nabla^2 u - c_3 u \right) = \underbrace{\partial_x (c_1 u) \partial_y (c_1 \nabla^2u - c_3 u)}_{\text{Term IIa}} - \underbrace{\partial_y (c_1 u) \partial_x (c_1 \nabla^2u - c_3 u)}_{\text{Term IIa}} $$ --- **Term IIa** We have the following term for Term IIa: $$ f_{2a} := \partial_x (c_1 u) \partial_y (c_1 \nabla^2u - c_3 u) $$ We can expand the terms to get the following: $$ f_{2a} := \partial_x (c_1 u) \partial_y \nabla^2 (c_1 u) - \partial_x (c_1 u) \partial_y (c_3 u) $$ And we can simplify, by factoring out constants, to get: $$ f_{2a} := c_1^2 \partial_x u \partial_y \nabla^2 u - c_1c_3 \partial_x u \partial_y u $$ --- **Term IIb** $$ f_{2b} := \partial_y (c_1 u) \partial_x (c_1 \nabla^2u - c_3 u) $$ We can expand the terms to get the following: $$ f_{2b} := c_1^2 \partial_y u \partial_x \nabla^2u - c_1 c_3 \partial_y u \partial_x u $$ --- **Combined** (Term IIa, IIb) We can substitute all of these two terms into our original expression $$ \det J \left(c_1 u, c_1 \nabla^2 u + c_3 u \right) = \\ = f_{2a} - f_{2b}\\ = \left(c_1^2 \partial_x u \partial_y \nabla^2 u - c_1c_3 \partial_x u \partial_y u\right) - \left(c_1^2 \partial_y u \partial_x \nabla^2u - c_1 c_3 \partial_y u \partial_x u\right)\\ = c_1^2 \partial_x u \partial_y \nabla^2 u - c_1c_3 \partial_x u \partial_y u - c_1^2 \partial_y u \partial_x \nabla^2u + c_1 c_3 \partial_y u \partial_x u $$ If we group the terms by operators, $\partial, \nabla^2$, then we get: $$ \det J \left(c_1 u, c_1 \nabla^2 u - c_3 u \right) = \\ \underbrace{c_1^2 \partial_x u \partial_y \nabla^2 u - c_1^2 \partial_y u \partial_x \nabla^2u}_{\nabla^2} - \underbrace{c_1c_3 \partial_x u \partial_y u - c_1c_3\partial_y u \partial_x u}_{\partial} $$ So each of these terms are determinant Jacobian terms, $\det \boldsymbol J$. $$ \det J \left(c_1 u, c_1 \nabla^2 u - c_3 u \right) = c_1^2\det \boldsymbol{J}(u, \nabla^2 u) - c_1c_3 \det \boldsymbol{J}(u, u) $$ 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: $$ \det J \left(c_1 u, c_1 \nabla^2 u - c_3 u \right) = c_1^2\det \boldsymbol{J}(u, \nabla^2 u) $$ **QED**. --- ### Final Form So we have the final form for our PDE in terms of SSH. $$ c_1 \partial_t \nabla^2 u - c_3 \partial_t u + c_1^2\det \boldsymbol{J}(u, \nabla^2 u) = 0 $$ We will factor out a constant term $$ \partial_t \nabla^2 u - c_2 \partial_t u + c_1\det \boldsymbol{J}(u, \nabla^2 u)= 0 $$ 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: $$ c_2 \partial_t u - \partial_t \nabla^2 u - c_1\det \boldsymbol{J}(u, \nabla^2 u)= 0 $$