draft model equations
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@ -7,7 +7,48 @@ Still interesting, leverages analytic derivatives of NNs, but lots of problems
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---
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---
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Some notation from SoL:
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% \newcommand{\pde}{\mathcal{P}} % PDE ops
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% \newcommand{\pdec}{\pde_{s}}
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% \newcommand{\manifsrc}{\mathscr{S}} % coarse / "source"
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% \newcommand{\pder}{\pde_{R}}
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% \newcommand{\manifref}{\mathscr{R}}
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% vc - coarse solutions
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% \renewcommand{\vc}[1]{\vs_{#1}} % plain coarse state at time t
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% \newcommand{\vcN}{\vs} % plain coarse state without time
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% vc - coarse solutions, modified by correction
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% \newcommand{\vct}[1]{\tilde{\vs}_{#1}} % modified / over time at time t
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% \newcommand{\vctN}{\tilde{\vs}} % modified / over time without time
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% vr - fine/reference solutions
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% \renewcommand{\vr}[1]{\mathbf{r}_{#1}} % fine / reference state at time t , never modified
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% \newcommand{\vrN}{\mathbf{r}} % plain coarse state without time
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% \newcommand{\project}{\mathcal{T}} % transfer operator fine <> coarse
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% \newcommand{\loss}{\mathcal{L}} % generic loss function
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% \newcommand{\nn}{f_{\theta}}
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% \newcommand{\dt}{\Delta t} % timestep
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% \newcommand{\corrPre}{\mathcal{C}_{\text{pre}}} % analytic correction , "pre computed"
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% \newcommand{\corr}{\mathcal{C}} % just C for now...
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% \newcommand{\nnfunc}{F} % {\text{NN}}
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Some notation from SoL, move with parts from overview into "appendix"?
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We typically solve a discretized PDE $\mathcal{P}$ by performing discrete time steps of size $\Delta t$.
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Each subsequent step can depend on any number of previous steps,
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$\mathbf{u}(\mathbf{x},t+\Delta t) = \mathcal{P}(\mathbf{u}(\mathbf{x},t), \mathbf{u}(\mathbf{x},t-\Delta t),...)$,
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where
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$\mathbf{x} \in \Omega \subseteq \mathbb{R}^d$ for the domain $\Omega$ in $d$
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dimensions, and $t \in \mathbb{R}^{+}$.
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Numerical methods yield approximations of a smooth function such as $\mathbf{u}$ in a discrete
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setting and invariably introduce errors. These errors can be measured in terms
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of the deviation from the exact analytical solution.
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For discrete simulations of
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PDEs, these errors are typically expressed as a function of the truncation, $O(\Delta t^k)$
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for a given step size $\Delta t$ and an exponent $k$ that is discretization dependent.
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The following PDEs typically work with a continuous
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The following PDEs typically work with a continuous
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velocity field $\mathbf{u}$ with $d$ dimensions and components, i.e.,
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velocity field $\mathbf{u}$ with $d$ dimensions and components, i.e.,
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@ -41,14 +82,49 @@ $\frac{\partial u}{\partial{t}} + u \nabla u = \nu \nabla \cdot \nabla u $ .
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---
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---
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Later on, Navier-Stokes, in 2D:
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Later on, additional equations...
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Navier-Stokes, in 2D:
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$
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$
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\frac{\partial u_x}{\partial{t}} + \mathbf{u} \cdot \nabla u_x =
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\frac{\partial u_x}{\partial{t}} + \mathbf{u} \cdot \nabla u_x =
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- \frac{1}{\rho}\nabla{p} + \nu \nabla\cdot \nabla u_x \\
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- \frac{1}{\rho}\nabla{p} + \nu \nabla\cdot \nabla u_x
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\\
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\frac{\partial u_y}{\partial{t}} + \mathbf{u} \cdot \nabla u_y =
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\frac{\partial u_y}{\partial{t}} + \mathbf{u} \cdot \nabla u_y =
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- \frac{1}{\rho}\nabla{p} + \nu \nabla\cdot \nabla u_y \\
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- \frac{1}{\rho}\nabla{p} + \nu \nabla\cdot \nabla u_y
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\text{subject to} \quad \nabla \cdot \mathbf{u} = 0,
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\\
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\text{subject to} \quad \nabla \cdot \mathbf{u} = 0
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$
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$
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Navier-Stokes, in 2D with Boussinesq:
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%$\frac{\partial u_x}{\partial{t}} + \mathbf{u} \cdot \nabla u_x$
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%$ -\frac{1}{\rho} \nabla p $
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$
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\frac{\partial u_x}{\partial{t}} + \mathbf{u} \cdot \nabla u_x = - \frac{1}{\rho} \nabla p
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\\
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\frac{\partial u_y}{\partial{t}} + \mathbf{u} \cdot \nabla u_y = - \frac{1}{\rho} \nabla p + \eta d
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\\
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\text{subject to} \quad \nabla \cdot \mathbf{u} = 0,
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\\
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\frac{\partial d}{\partial{t}} + \mathbf{u} \cdot \nabla d = 0
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$
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Navier-Stokes, in 3D:
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$
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\frac{\partial u_x}{\partial{t}} + \mathbf{u} \cdot \nabla u_x = - \frac{1}{\rho} \nabla p + \nu \nabla\cdot \nabla u_x
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\\
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\frac{\partial u_y}{\partial{t}} + \mathbf{u} \cdot \nabla u_y = - \frac{1}{\rho} \nabla p + \nu \nabla\cdot \nabla u_y
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\\
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\frac{\partial u_z}{\partial{t}} + \mathbf{u} \cdot \nabla u_z = - \frac{1}{\rho} \nabla p + \nu \nabla\cdot \nabla u_z
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\\
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\text{subject to} \quad \nabla \cdot \mathbf{u} = 0.
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$
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