Introduction
Neural operators approximate the solution operator of a partial differential equation, that is, the map from its input functions, such as initial and boundary data, coefficients and forcing, and its physical parameters to its solution [13 ] . Widely used examples are the Fourier neural operator (FNO) [17 ] and the deep operator network (DeepONet) [21 ] , and both can approximate any continuous operator on a compact set of inputs [12 , 15 ] . Once trained on solutions computed by a conventional solver, a neural operator is evaluated orders of magnitude faster than the solver, which makes it useful whenever a solution operator must be evaluated many times. Fluid dynamics is a main field of application. Fourier neural operators predict turbulent solutions of the two-dimensional Navier–Stokes equations [17 ] , reproduce the long-time statistics of chaotic dissipative systems [18 ] , and forecast the global weather [22 ] . Closest to the present problem, they have been trained on two-dimensional Rayleigh–Bénard convection, a Boussinesq flow driven by buoyancy, where they predict the velocity and temperature fields more accurately than earlier data-driven models [29 ] . Physics-informed neural networks [26 ] and their operator versions [36 , 19 ] add the residual of the equation to the training loss, so that the network learns from the equation as well as from solutions.
Machine learning has also begun to assist mathematical analysis. It has suggested conjectures and guided their proofs in pure mathematics [5 ] , and it has constructed
Funding: J. Wu was partially supported by the National Science Foundation (NSF) under grants DMS-2104682 and DMS-2309748. X. Zheng was partially supported by the NSF under grant DMS-2309747. examples that disprove conjectures or improve known bounds [35 , 28 ] . For the equations of fluid motion, neural networks have found an approximate self-similar blow-up profile of the three-dimensional axisymmetric Euler equations, which also solves the closely related two-dimensional Boussinesq equations [38 ] , and new families of unstable singularities [37 ] , accurate enough to serve as a starting point for computer-assisted proofs. In these works the network proposes candidates and a separate rigorous or high-precision computation decides whether they are valid. The same division of labor underlies multifidelity methods, in which a cheap model explores and an accurate model confirms [24 ] . In the stability theory of fluids, the search for the initial data that maximize the growth of a perturbation over a finite time is an optimization problem, each step of which requires solving the equations forward and backward in time [25 , 10 ] .
Known structure can also be built into the network itself. Conservation laws can be enforced by construction [27 , 8 ] , and asymptotic-preserving networks and neural operators build the limit of a small parameter of the equation into the loss, so that one network stays accurate across the range of the parameter [9 , 39 , 40 ] . Asymptotics-informed operator networks add boundary-layer correctors from singular perturbation theory to the output [16 ] . In these approaches the network learns from solutions computed by a solver, from the residual of the equation, or from both, and the structure built into it comes from the equation or from a formal limit. What rigorous analysis proves about the solutions, such as which quantities are nearly conserved, that is, change only at higher order in the amplitude, how large the various parts of the solution are, and which quantities depend on the data in an ill-conditioned way, is used less systematically in the design of the network. In this paper we design a neural operator from the rigorous long-time analysis of a partial differential equation, the two-dimensional Boussinesq system with anisotropic dissipation, and use it to obtain information that the analysis alone does not give. The analysis decides what the network predicts, how its output is split, and how each part is scaled with the amplitude of the data. The trained network then takes the place of the solver in a search over initial data for a lower bound on the constant of the analytical estimate, and the solver confirms the candidates it finds. The remainder of this introduction describes the physical problem and the analytical result on which the design rests.
Buoyancy-driven flows in the Boussinesq approximation are a canonical model of stratified fluids in the atmosphere and ocean, where the interplay of velocity and temperature (or density) governs the onset of convection, the persistence of shear flows, and the emergence of stratified states [23 , 4 ] . In many geophysical regimes the dissipation is strongly anisotropic : turbulent transport and rotation make the effective diffusion in the horizontal and vertical directions markedly different, and it is natural, both physically and mathematically, to study models in which the dissipation is present in only one direction of each field. The system studied here retains horizontal viscosity in the momentum equation and vertical thermal diffusion in the temperature equation, and no dissipation in the complementary directions.
The stability of such partially dissipated Boussinesq systems near the hydrostatic equilibrium has been established in a sequence of works: for the strip T × R \mathbb{T} \times\mathbb{R} T × R [1 ] , for related partial-dissipation configurations [14 ] , and for the buoyancy-driven case without any thermal diffusion [30 , 6 ] . A recurring theme is that the buoyancy coupling itself is stabilizing: the exchange between the vertical velocity and the temperature converts the momentum equation, which is damped only in the horizontal direction, into a damped-wave-type system, so that small perturbations remain globally regular and uniformly bounded. On the periodic box T 2 \mathbb{T}^2 T 2 H 2 H^2 H 2 [7 ] . The energy method used on the strip in [1 ] does not carry over to the box. On a bounded vertical interval the one-dimensional Sobolev inequality acquires an additional zeroth-order term, which breaks the anisotropic product estimates on which that method rests, and a different mechanism, special to the box, is needed. This estimate is recalled in Appendix A.
What these stability results do not by themselves settle is the long-time limit : given that a small perturbation exists and remains small for all time, what does it converge to, and how much of the limit is determined by the initial data?
If the dissipation is isotropic and present in both fields, as in the full-Laplacian system ∂ t u + u ⋅ ∇ u = − ∇ p + ν Δ u + θ e 2 \partial_t u+u\cdot\nabla u=-\nabla p+\nu\Delta u+\theta e_2 ∂ t u + u ⋅ ∇ u = − ∇ p + ν Δ u + θ e 2 ∂ t θ + u ⋅ ∇ θ + u 2 = η Δ θ \partial_t\theta+u\cdot\nabla\theta+u_2=\eta\Delta\theta ∂ t θ + u ⋅ ∇ θ + u 2 = η Δ θ T 2 \mathbb{T}^2 T 2 L 2 L^2 L 2
d d t ( ∥ u ∥ L 2 2 + ∥ θ ∥ L 2 2 ) = − 2 ν ∥ ∇ u ∥ L 2 2 − 2 η ∥ ∇ θ ∥ L 2 2 . \frac{d}{dt}\left(\lVert u\rVert_{L^2}^2+\lVert\theta\rVert_{L^2}^2\right)=-2\nu\lVert\nabla u\rVert_{L^2}^2-2\eta\lVert\nabla\theta\rVert_{L^2}^2. d t d ( ∥ u ∥ L 2 2 + ∥ θ ∥ L 2 2 ) = − 2 ν ∥ ∇ u ∥ L 2 2 − 2 η ∥ ∇ θ ∥ L 2 2 . and, provided the conserved mean ∫ T 2 u 1 \int_{\mathbb{T}^2}u_1 ∫ T 2 u 1 ν ∂ 11 u \nu\partial_{11}u ν ∂ 11 u η ∂ 22 θ \eta\partial_{22}\theta η ∂ 22 θ
d d t ( ∥ u ∥ L 2 2 + ∥ θ ∥ L 2 2 ) = − 2 ν ∥ ∂ 1 u ∥ L 2 2 − 2 η ∥ ∂ 2 θ ∥ L 2 2 . \frac{d}{dt}\left(\lVert u\rVert_{L^2}^2+\lVert\theta\rVert_{L^2}^2\right)=-2\nu\lVert\partial_1u\rVert_{L^2}^2-2\eta\lVert\partial_2\theta\rVert_{L^2}^2. d t d ( ∥ u ∥ L 2 2 + ∥ θ ∥ L 2 2 ) = − 2 ν ∥ ∂ 1 u ∥ L 2 2 − 2 η ∥ ∂ 2 θ ∥ L 2 2 . The right-hand side vanishes on the velocity modes with k 1 = 0 k_1=0 k 1 = 0 k 2 = 0 k_2=0 k 2 = 0 [30 ] (see also [6 ] ), which treats the two-dimensional Boussinesq system on T 2 \mathbb{T}^2 T 2 full (isotropic) velocity viscosity but no thermal diffusion, and shows that the velocity decays to zero while the temperature tends to an explicit function of x 2 x_2 x 2 [2 ] , and its long-time simulations show the perturbation relaxing exponentially to zero. Our system differs from that of [30 ] in two ways. The velocity dissipation is made anisotropic (horizontal only) and a vertical thermal diffusion is added. The roles are then reversed: it is now the temperature that decays to zero and the vorticity that tends to a stratified limit, that is, a function of x 2 x_2 x 2 Theorem 3.1 ).
The design of our neural operator is guided by the long-time behavior of the system on the periodic box T 2 = [ 0 , 2 π ] 2 \mathbb{T}^2=[0,2\pi]^2 T 2 = [ 0 , 2 π ] 2 Theorem 3.1 . Writing any field as its x 1 x_1 x 1 g ‾ ( x 2 , t ) \overline{g}(x_2,t) g ( x 2 , t ) g ~ = g − g ‾ \widetilde{g}=g-\overline{g} g = g − g x 1 x_1 x 1 u = ( u ‾ 1 ( x 2 ) , 0 ) u=(\overline{u}_1(x_2),0) u = ( u 1 ( x 2 ) , 0 ) ω ‾ = − ∂ 2 u ‾ 1 \overline{\omega}=-\partial_2\overline{u}_1 ω = − ∂ 2 u 1 ∂ 2 2 u ~ 1 u ~ 2 \partial_2^2\widetilde{u}_1\widetilde{u}_2 ∂ 2 2 u 1 u 2 u ~ 1 u ~ 2 \widetilde{u}_1\widetilde{u}_2 u 1 u 2 ε \varepsilon ε ω ‾ ( ⋅ , t ) − ω ‾ 0 \overline{\omega}(\cdot,t)-\overline{\omega}_0 ω ( ⋅ , t ) − ω 0 O ( ε 2 ) O(\varepsilon^2) O ( ε 2 ) ω ‾ 0 \overline{\omega}_0 ω 0 ε 2 \varepsilon^2 ε 2 ε \varepsilon ε
The anisotropy makes the solution operator difficult to learn with a standard neural operator, for three reasons. First, the quantity bounded by the analysis is small. The change of the horizontal average of the vorticity is of order ε 2 \varepsilon^2 ε 2 ε \varepsilon ε ω ˉ 0 \bar{\omega}_0 ω ˉ 0 L 2 L^2 L 2 x 2 x_2 x 2 10 4 10^4 1 0 4 ω ˉ 0 \bar{\omega}_0 ω ˉ 0 ω ( ⋅ , T ) − ω ˉ 0 \omega(\cdot,T)-\bar{\omega}_0 ω ( ⋅ , T ) − ω ˉ 0 ε 2 \varepsilon^2 ε 2 ε \varepsilon ε 2 × 10 3 2 \times10^3 2 × 1 0 3
The remainder of the paper is organized as follows. Section 2 sets up the governing equations, the hydrostatic equilibrium, and the mean/oscillation decomposition. Section 3 states the standing hypotheses, the main result, Theorem 3.1, on the asymptotic stratification and its limit profile, and the linear decay rate, Proposition 3.2. Section 4 introduces the Fourier neural operator (Sec. 4.1), describes the training data (Sec. 4.2), explains why the late-time temperature is not a prediction target (Sec. 4.3), sets out the network and its training (Sec. 4.4), reports the accuracy and cost of the surrogate (Sec. 4.5), tests Theorem 3.1(b) on solver runs and surrogate predictions (Sec. 4.6), compares the surrogate with networks trained without the analysis (Sec. 4.7), and uses the surrogate to search for a lower bound for the constant of Theorem 3.1(b) (Sec. 4.8). Section 5 summarizes the results. Appendix A states the uniform stability estimate and derives from it the bound used in the proof, Appendix B contains the proof of Theorem 3.1, and Appendix C the proof of Proposition 3.2.
Governing equations and the mean/oscillation decomposition
The perturbation system
We consider the two-dimensional anisotropic Boussinesq system from [1 ] , with the domain taken to be the square box Ω = [ 0 , 2 π ] 2 \Omega=[0,2\pi]^2 Ω = [ 0 , 2 π ] 2 U U U P P P Θ \Theta Θ
∂ t U + U ⋅ ∇ U = − ∇ P + ν ∂ 11 U + Θ e 2 , x = ( x 1 , x 2 ) ∈ Ω , t > 0 , ∂ t Θ + U ⋅ ∇ Θ = η ∂ 22 Θ , ∇ ⋅ U = 0. \begin{aligned}
\partial_t U+U\cdot\nabla U&=-\nabla P+\nu\partial_{11}U+\Theta e_2,\qquad x=(x_1,x_2)\in\Omega,\quad t>0,\\
\partial_t\Theta+U\cdot\nabla\Theta&=\eta\partial_{22}\Theta,\\
\nabla\cdot U&=0.
\end{aligned} ∂ t U + U ⋅ ∇ U ∂ t Θ + U ⋅ ∇Θ ∇ ⋅ U = − ∇ P + ν ∂ 11 U + Θ e 2 , x = ( x 1 , x 2 ) ∈ Ω , t > 0 , = η ∂ 22 Θ , = 0. where ν \nu ν η \eta η e 2 e_2 e 2
Denote a hydrostatic equilibrium as ( U h e , P h e , Θ h e ) = ( 0 , 1 2 x 2 2 , x 2 ) (U_{\mathrm{he}},P_{\mathrm{he}},\Theta_{\mathrm{he}})=(0,\frac{1}{2}x_2^2,x_2) ( U he , P he , Θ he ) = ( 0 , 2 1 x 2 2 , x 2 ) ( u , p , θ ) = ( U − U h e , P − P h e , Θ − Θ h e ) (u,p,\theta)=(U-U_{\mathrm{he}},P-P_{\mathrm{he}},\Theta-\Theta_{\mathrm{he}}) ( u , p , θ ) = ( U − U he , P − P he , Θ − Θ he ) 2 π 2\pi 2 π T 2 = [ 0 , 2 π ] 2 \mathbb{T}^2=[0,2\pi]^2 T 2 = [ 0 , 2 π ] 2
The governing equations are, with the notation u = ( u 1 , u 2 ) \mathbf{u}=(u_1,u_2) u = ( u 1 , u 2 )
∂ t u + u ⋅ ∇ u = − ∇ p + ν ∂ 11 u + θ e 2 , (1) \partial_t\mathbf{u}+\mathbf{u}\cdot\nabla\mathbf{u}=-\nabla p+\nu\partial_{11}\mathbf{u}+\theta\mathbf{e}_2, \tag*{(1)} ∂ t u + u ⋅ ∇ u = − ∇ p + ν ∂ 11 u + θ e 2 , (1) ∂ t θ + u ⋅ ∇ θ + u 2 = η ∂ 22 θ , (2) \partial_t\theta+\mathbf{u}\cdot\nabla\theta+u_2=\eta\partial_{22}\theta, \tag*{(2)} ∂ t θ + u ⋅ ∇ θ + u 2 = η ∂ 22 θ , (2) ∇ ⋅ u = 0. (3) \nabla\cdot\mathbf{u}=0. \tag*{(3)} ∇ ⋅ u = 0. (3) An equivalent formulation uses the vorticity ω = ∇ × u = ∂ 1 u 2 − ∂ 2 u 1 \omega=\nabla\times\mathbf{u}=\partial_1u_2-\partial_2u_1 ω = ∇ × u = ∂ 1 u 2 − ∂ 2 u 1 ψ \psi ψ
∂ t ω + u ⋅ ∇ ω = ν ∂ 11 ω + ∂ 1 θ , (4) \partial_t\omega+\mathbf{u}\cdot\nabla\omega=\nu\partial_{11}\omega+\partial_1\theta, \tag*{(4)} ∂ t ω + u ⋅ ∇ ω = ν ∂ 11 ω + ∂ 1 θ , (4) − Δ ψ = ω , (5) -\Delta\psi=\omega, \tag*{(5)} − Δ ψ = ω , (5) u = ∇ ⊥ ψ = ( ∂ 2 ψ , − ∂ 1 ψ ) , (6) \mathbf{u}=\nabla^\perp\psi=(\partial_2\psi,-\partial_1\psi), \tag*{(6)} u = ∇ ⊥ ψ = ( ∂ 2 ψ , − ∂ 1 ψ ) , (6) ∂ t θ + u ⋅ ∇ θ + u 2 = η ∂ 22 θ . (7) \partial_t\theta+\mathbf{u}\cdot\nabla\theta+u_2=\eta\partial_{22}\theta. \tag*{(7)} ∂ t θ + u ⋅ ∇ θ + u 2 = η ∂ 22 θ . (7) Mean/oscillation decomposition
For a field g ( x 1 , x 2 , t ) g(x_1,x_2,t) g ( x 1 , x 2 , t ) x 1 x_1 x 1
g ˉ ( x 2 , t ) : = 1 2 π ∫ 0 2 π g ( x 1 , x 2 , t ) d x 1 , g ~ : = g − g ˉ , (8) \bar{g}(x_2,t):=\frac{1}{2\pi}\int_0^{2\pi}g(x_1,x_2,t)\,dx_1,\qquad\widetilde{g}:=g-\bar{g}, \tag*{(8)} g ˉ ( x 2 , t ) := 2 π 1 ∫ 0 2 π g ( x 1 , x 2 , t ) d x 1 , g := g − g ˉ , (8) so that g ˉ \bar{g} g ˉ x 1 x_1 x 1 g ~ \widetilde{g} g x 1 x_1 x 1 x 2 x_2 x 2 x 1 x_1 x 1 (8) ,
Q 0 g ( x 1 , t ) : = 1 2 π ∫ 0 2 π g ( x 1 , x 2 , t ) d x 2 , g ⊥ : = g − Q 0 g , (9) Q_0g(x_1,t):=\frac{1}{2\pi}\int_0^{2\pi}g(x_1,x_2,t)\,dx_2,\qquad g^\perp:=g-Q_0g, \tag*{(9)} Q 0 g ( x 1 , t ) := 2 π 1 ∫ 0 2 π g ( x 1 , x 2 , t ) d x 2 , g ⊥ := g − Q 0 g , (9) enters only in Appendix A.
The two decompositions separate what each dissipative term can reach. The horizontal viscosity ν ∂ 11 \nu\partial_{11} ν ∂ 11 x 1 x_1 x 1 − ν k 1 2 -\nu k_1^2 − ν k 1 2 g ~ \widetilde{g} g g ˉ \bar{g} g ˉ η ∂ 22 \eta\partial_{22} η ∂ 22 Q 0 g Q_0g Q 0 g g ⊥ g^\perp g ⊥
In the analysis we write the perturbation velocity u \mathbf{u} u (2.1)–(2.3) as u \mathbf{u} u u 0 \mathbf{u}_0 u 0 u \mathbf{u} u
Lemma 2.1 (Mean/oscillation calculus). Let g ˉ \bar{g} g ˉ g ~ = g − g ˉ \widetilde{g}=g-\bar{g} g = g − g ˉ (8) , and let u = ( u 1 , u 2 ) \mathbf{u}=(u_1,u_2) u = ( u 1 , u 2 ) 2 π 2\pi 2 π ω = ∂ 1 u 2 − ∂ 2 u 1 \omega=\partial_1u_2-\partial_2u_1 ω = ∂ 1 u 2 − ∂ 2 u 1 f , g ∈ H 1 ( T 2 ) f,g\in H^1(\mathbb{T}^2) f , g ∈ H 1 ( T 2 ) X X X L 2 L^2 L 2
(a) g ˉ ‾ = g ˉ \overline{\bar{g}}=\bar{g} g ˉ = g ˉ g ~ ~ = 0 \widetilde{\widetilde{g}}=0 g = 0 f ˉ g ~ ‾ = f ˉ g ~ = 0 \overline{\bar{f}\widetilde{g}}=\bar{f}\widetilde{g}=0 f ˉ g = f ˉ g = 0 f g ‾ = f ˉ g ˉ + f ~ g ~ \overline{fg}=\bar{f}\bar{g}+\widetilde{f}\widetilde{g} f g = f ˉ g ˉ + f g
(b) ∂ 2 g ˉ = ∂ 2 g ‾ \partial_2\bar{g}=\overline{\partial_2g} ∂ 2 g ˉ = ∂ 2 g ∂ 2 g ~ = ∂ 2 g ~ \partial_2\widetilde{g}=\widetilde{\partial_2g} ∂ 2 g = ∂ 2 g ∂ 1 g ˉ = 0 \partial_1\bar{g}=0 ∂ 1 g ˉ = 0 ∂ 1 g = ∂ 1 g ~ \partial_1g=\partial_1\widetilde{g} ∂ 1 g = ∂ 1 g ω ˉ = − ∂ 2 u ˉ 1 \bar{\omega}=-\partial_2\bar{u}_1 ω ˉ = − ∂ 2 u ˉ 1 ω ~ = ∂ 1 u ~ 2 − ∂ 2 u ~ 1 \widetilde{\omega}=\partial_1\widetilde{u}_2-\partial_2\widetilde{u}_1 ω = ∂ 1 u 2 − ∂ 2 u 1
(c) ∥ g ∥ X 2 = ∥ g ˉ ∥ X 2 + ∥ g ~ ∥ X 2 \|g\|_X^2=\|\bar{g}\|_X^2+\|\widetilde{g}\|_X^2 ∥ g ∥ X 2 = ∥ g ˉ ∥ X 2 + ∥ g ∥ X 2 ∥ g ˉ ∥ X ≤ ∥ g ∥ X \|\bar{g}\|_X\le\|g\|_X ∥ g ˉ ∥ X ≤ ∥ g ∥ X ∥ g ~ ∥ X ≤ ∥ g ∥ X \|\widetilde{g}\|_X\le\|g\|_X ∥ g ∥ X ≤ ∥ g ∥ X
(d) ∂ 2 u ˉ 2 = 0 \partial_2\bar{u}_2=0 ∂ 2 u ˉ 2 = 0 ∫ T 2 u 2 d x = 0 \int_{\mathbb{T}^2}u_2\,dx=0 ∫ T 2 u 2 d x = 0 u ˉ 2 ≡ 0 \bar{u}_2\equiv0 u ˉ 2 ≡ 0 u 2 = u ~ 2 u_2=\widetilde{u}_2 u 2 = u 2
(e) (Horizontal Poincaré) since the lowest nonzero horizontal frequency on [ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ]
∥ g ~ ∥ L 2 ≤ ∥ ∂ 1 g ~ ∥ L 2 = ∥ ∂ 1 g ∥ L 2 , ∥ ∂ 2 g ~ ∥ L 2 = ∥ ∂ 2 g ~ ∥ L 2 ≤ ∥ ∂ 2 g ∥ L 2 , (10) \|\widetilde{g}\|_{L^2}\le\|\partial_1\widetilde{g}\|_{L^2}=\|\partial_1g\|_{L^2},\qquad\|\partial_2\widetilde{g}\|_{L^2}=\|\widetilde{\partial_2g}\|_{L^2}\le\|\partial_2g\|_{L^2}, \tag*{(10)} ∥ g ∥ L 2 ≤ ∥ ∂ 1 g ∥ L 2 = ∥ ∂ 1 g ∥ L 2 , ∥ ∂ 2 g ∥ L 2 = ∥ ∂ 2 g ∥ L 2 ≤ ∥ ∂ 2 g ∥ L 2 , (10) and hence, the first inequality being applied to g ~ \widetilde{g} g ∂ 2 g ~ = ∂ 2 g ~ \partial_2\widetilde{g}=\widetilde{\partial_2g} ∂ 2 g = ∂ 2 g ∥ g ~ ∥ H 1 ≤ 2 ∥ ∂ 1 g ∥ H 1 \|\widetilde{g}\|_{H^1}\le\sqrt{2}\|\partial_1g\|_{H^1} ∥ g ∥ H 1 ≤ 2 ∥ ∂ 1 g ∥ H 1 ∥ u ~ ∥ H 1 ≤ 2 ∥ ∂ 1 u ∥ H 1 \|\widetilde{\mathbf{u}}\|_{H^1}\le\sqrt{2}\|\partial_1\mathbf{u}\|_{H^1} ∥ u ∥ H 1 ≤ 2 ∥ ∂ 1 u ∥ H 1
(f) (Vertical Poincaré ) if h h h ∫ 0 2 π h ( x 1 , x 2 ) d x 2 = 0 \int_{0}^{2\pi} h(x_{1},x_{2})\,\mathrm{d}x_{2}=0 ∫ 0 2 π h ( x 1 , x 2 ) d x 2 = 0 x 1 x_{1} x 1
∥ h ∥ L 2 ≤ ∥ ∂ 2 h ∥ L 2 . (11) \lVert h\rVert_{L^{2}} \leq\lVert\partial_{2}h\rVert_{L^{2}}. \tag*{(11)} ∥ h ∥ L 2 ≤ ∥ ∂ 2 h ∥ L 2 . (11) (g) ∂ 2 Q 0 g = 0 \partial_{2}Q_{0}g=0 ∂ 2 Q 0 g = 0 ∫ 0 2 π g ⊥ d x 2 = 0 \int_{0}^{2\pi}g^{\perp}\,\mathrm{d}x_{2}=0 ∫ 0 2 π g ⊥ d x 2 = 0 x 1 x_{1} x 1 g = Q 0 g + g ⊥ g=Q_{0}g+g^{\perp} g = Q 0 g + g ⊥ (9) is orthogonal in L 2 ( T 2 ) L^{2}(\mathbb{T}^{2}) L 2 ( T 2 ) ∂ 1 \partial_{1} ∂ 1
Main result: stratification and the limit profile
Standing hypotheses on the initial data
Throughout, the initial data ( u 0 , θ 0 ) ∈ H 2 ( T 2 ) (u_{0},\theta_{0})\in H^{2}(\mathbb{T}^{2}) ( u 0 , θ 0 ) ∈ H 2 ( T 2 ) ∇ ⋅ u 0 = 0 \nabla\cdot u_{0}=0 ∇ ⋅ u 0 = 0
∫ T 2 θ 0 d x = ∫ T 2 u 2 , 0 d x = ∫ T 2 u 1 , 0 d x = 0 , ∥ u 0 ∥ H 2 + ∥ θ 0 ∥ H 2 ≤ ε . (12) \int_{\mathbb{T}^{2}}\theta_{0}\,\mathrm{d}x=\int_{\mathbb{T}^{2}}u_{2,0}\,\mathrm{d}x=\int_{\mathbb{T}^{2}}u_{1,0}\,\mathrm{d}x=0,\qquad\lVert u_{0}\rVert_{H^{2}}+\lVert\theta_{0}\rVert_{H^{2}}\leq\varepsilon. \tag*{(12)} ∫ T 2 θ 0 d x = ∫ T 2 u 2 , 0 d x = ∫ T 2 u 1 , 0 d x = 0 , ∥ u 0 ∥ H 2 + ∥ θ 0 ∥ H 2 ≤ ε . (12) Here and below, with the Fourier coefficients f ^ k = ( 2 π ) − 2 ∫ T 2 f ( x ) e − i k ⋅ x d x \widehat{f}_{k}=(2\pi)^{-2}\int_{\mathbb{T}^{2}}f(x)e^{-\mathrm{i}k\cdot x}\,\mathrm{d}x f k = ( 2 π ) − 2 ∫ T 2 f ( x ) e − i k ⋅ x d x
∥ f ∥ H s 2 = ( 2 π ) 2 ∑ k ∈ Z 2 ( 1 + ∣ k ∣ 2 ) s ∣ f ^ k ∣ 2 , s ∈ R , (13) \lVert f\rVert_{H^{s}}^{2}=(2\pi)^{2}\sum_{k\in\mathbb{Z}^{2}}(1+|k|^{2})^{s}|\widehat{f}_{k}|^{2},\qquad s\in\mathbb{R}, \tag*{(13)} ∥ f ∥ H s 2 = ( 2 π ) 2 k ∈ Z 2 ∑ ( 1 + ∣ k ∣ 2 ) s ∣ f k ∣ 2 , s ∈ R , (13) summed over the components of a vector field.
The mean conditions in (12) are preserved by the flow. Integrating (1) –(2) over T 2 \mathbb{T}^{2} T 2 ∇ ⋅ u = 0 \nabla\cdot u=0 ∇ ⋅ u = 0
d d t ∫ T 2 u 1 d x = 0 , d d t ∫ T 2 θ d x = − ∫ T 2 u 2 d x , d d t ∫ T 2 u 2 d x = ∫ T 2 θ d x . \frac{\mathrm{d}}{\mathrm{d}t}\int_{\mathbb{T}^{2}}u_{1}\,\mathrm{d}x=0,\qquad
\frac{\mathrm{d}}{\mathrm{d}t}\int_{\mathbb{T}^{2}}\theta\,\mathrm{d}x=-\int_{\mathbb{T}^{2}}u_{2}\,\mathrm{d}x,\qquad
\frac{\mathrm{d}}{\mathrm{d}t}\int_{\mathbb{T}^{2}}u_{2}\,\mathrm{d}x=\int_{\mathbb{T}^{2}}\theta\,\mathrm{d}x. d t d ∫ T 2 u 1 d x = 0 , d t d ∫ T 2 θ d x = − ∫ T 2 u 2 d x , d t d ∫ T 2 u 2 d x = ∫ T 2 θ d x . Thus ∫ T 2 u 1 d x \int_{\mathbb{T}^{2}}u_{1}\,\mathrm{d}x ∫ T 2 u 1 d x
∫ T 2 θ d x = cos t ∫ T 2 θ 0 d x − sin t ∫ T 2 u 2 , 0 d x , \int_{\mathbb{T}^{2}}\theta\,\mathrm{d}x=\cos t\int_{\mathbb{T}^{2}}\theta_{0}\,\mathrm{d}x-\sin t\int_{\mathbb{T}^{2}}u_{2,0}\,\mathrm{d}x, ∫ T 2 θ d x = cos t ∫ T 2 θ 0 d x − sin t ∫ T 2 u 2 , 0 d x , ∫ T 2 u 2 d x = sin t ∫ T 2 θ 0 d x + cos t ∫ T 2 u 2 , 0 d x , \int_{\mathbb{T}^{2}}u_{2}\,\mathrm{d}x=\sin t\int_{\mathbb{T}^{2}}\theta_{0}\,\mathrm{d}x+\cos t\int_{\mathbb{T}^{2}}u_{2,0}\,\mathrm{d}x, ∫ T 2 u 2 d x = sin t ∫ T 2 θ 0 d x + cos t ∫ T 2 u 2 , 0 d x , which are periodic in time with period 2 π 2\pi 2 π (12) all three means are zero for all t ≥ 0 t\geq0 t ≥ 0
Statement of the main result
By Theorem A.1 of Appendix A, quoted from [7 ] , there is ε ∗ = ε ∗ ( ν , η ) > 0 \varepsilon_{*}=\varepsilon_{*}(\nu,\eta)>0 ε ∗ = ε ∗ ( ν , η ) > 0 ε ≤ ε ∗ \varepsilon\leq\varepsilon_{*} ε ≤ ε ∗ (1) –(3) with initial data satisfying (12) has a unique global solution. This solution is bounded by C ∗ ε C_{*}\varepsilon C ∗ ε H 2 H^{2} H 2 ∫ 0 ∞ ( ν ∥ ∂ 1 u ∥ H 2 2 + η ∥ ∂ 2 θ ∥ H 2 2 ) d τ \int_{0}^{\infty}(\nu\lVert\partial_{1}u\rVert_{H^{2}}^{2}+\eta\lVert\partial_{2}\theta\rVert_{H^{2}}^{2})\,\mathrm{d}\tau ∫ 0 ∞ ( ν ∥ ∂ 1 u ∥ H 2 2 + η ∥ ∂ 2 θ ∥ H 2 2 ) d τ C ν ε 2 C_{\nu}\varepsilon^{2} C ν ε 2
Averaging the first component of (1) over x 1 x_{1} x 1 − ∂ 1 p -\partial_{1}p − ∂ 1 p ν ∂ 11 u 1 \nu\partial_{11}u_{1} ν ∂ 11 u 1 Lemma 2.1 (b). Since ∇ ⋅ u = 0 \nabla\cdot u=0 ∇ ⋅ u = 0 u ⋅ ∇ u 1 = ∂ 1 ( u 1 2 ) + ∂ 2 ( u 1 u 2 ) u\cdot\nabla u_{1}=\partial_{1}(u_{1}^{2})+\partial_{2}(u_{1}u_{2}) u ⋅ ∇ u 1 = ∂ 1 ( u 1 2 ) + ∂ 2 ( u 1 u 2 ) ∂ t u ‾ 1 = − ∂ 2 u 1 u 2 ‾ \partial_{t}\overline{u}_{1}=-\partial_{2}\overline{u_{1}u_{2}} ∂ t u 1 = − ∂ 2 u 1 u 2 Lemma 2.1 (a), u 1 u 2 ‾ = u ‾ 1 u ‾ 2 + u ~ 1 u ~ 2 \overline{u_{1}u_{2}}=\overline{u}_{1}\overline{u}_{2}+\widetilde{u}_{1}\widetilde{u}_{2} u 1 u 2 = u 1 u 2 + u 1 u 2 (12) , which hold for all t ≥ 0 t\geq0 t ≥ 0 Lemma 2.1 (d), u ‾ 2 = 0 \overline{u}_{2}=0 u 2 = 0 ∂ t u ‾ 1 = − ∂ 2 u ~ 1 u ~ 2 \partial_{t}\overline{u}_{1}=-\partial_{2}\widetilde{u}_{1}\widetilde{u}_{2} ∂ t u 1 = − ∂ 2 u 1 u 2 ω ‾ = − ∂ 2 u ‾ 1 \overline{\omega}=-\partial_{2}\overline{u}_{1} ω = − ∂ 2 u 1 Lemma 2.1 (b),
∂ t ω ‾ = ∂ 22 u ~ 1 u ~ 2 . (14) \partial_{t}\overline{\omega}=\partial_{22}\widetilde{u}_{1}\widetilde{u}_{2}. \tag*{(14)} ∂ t ω = ∂ 22 u 1 u 2 . (14) Theorem 3.1 (Small-data stratification and the limit profile on T 2 \mathbb{T}^{2} T 2 ε ∗ = ε ∗ ( ν , η ) > 0 \varepsilon_{*}=\varepsilon_{*}(\nu,\eta)>0 ε ∗ = ε ∗ ( ν , η ) > 0 Theorem A.1 . Let the initial data satisfy (12) with ε ≤ ε ∗ \varepsilon\leq\varepsilon_{*} ε ≤ ε ∗ ( u , θ ) (u,\theta) ( u , θ ) (1) –(3) . Let ω = ∂ 1 u 2 − ∂ 2 u 1 \omega=\partial_{1}u_{2}-\partial_{2}u_{1} ω = ∂ 1 u 2 − ∂ 2 u 1 ω ‾ \overline{\omega} ω ω ‾ 0 \overline{\omega}_{0} ω 0 (8) of ω \omega ω ω 0 \omega_{0} ω 0 ∫ T 2 θ ( t ) = ∫ T 2 u 2 ( t ) = 0 \int_{\mathbb{T}^{2}}\theta(t)=\int_{\mathbb{T}^{2}}u_{2}(t)=0 ∫ T 2 θ ( t ) = ∫ T 2 u 2 ( t ) = 0 t ≥ 0 t\geq0 t ≥ 0
(a) (Decay of the vertical velocity and the temperature ) As t → ∞ t \to\infty t → ∞ u 2 ( t ) → 0 u_{2}(t) \to0 u 2 ( t ) → 0 θ ( t ) → 0 \theta(t) \to0 θ ( t ) → 0 H s ( T 2 ) H^{s}(\mathbb{T}^{2}) H s ( T 2 ) s < 2 s < 2 s < 2
(b) (Stratification and limit profile of the vorticity ) There is ω ‾ ∞ = ω ‾ ∞ ( x 2 ) ∈ H 1 ( T 2 ) \overline{\omega}_{\infty} = \overline{\omega}_{\infty}(x_{2}) \in H^{1}(\mathbb{T}^{2}) ω ∞ = ω ∞ ( x 2 ) ∈ H 1 ( T 2 ) ω ( ⋅ , t ) → ω ‾ ∞ \omega(\cdot,t) \to\overline{\omega}_{\infty} ω ( ⋅ , t ) → ω ∞ H s ( T 2 ) H^{s}(\mathbb{T}^{2}) H s ( T 2 ) s < 1 s < 1 s < 1 t → ∞ t \to\infty t → ∞
ω ‾ ∞ = ω ‾ 0 + ∫ 0 ∞ ∂ 22 u ~ 1 u ~ 2 ( ⋅ , t ) d t (15) \overline{\omega}_{\infty} = \overline{\omega}_{0} + \int_{0}^{\infty} \partial_{22}\widetilde{u}_{1}\widetilde{u}_{2}(\cdot,t)\,\mathrm{d}t \tag*{(15)} ω ∞ = ω 0 + ∫ 0 ∞ ∂ 22 u 1 u 2 ( ⋅ , t ) d t (15) where the integral converges in H − 2 H^{-2} H − 2
sup t ≥ 0 ∥ ω ‾ ( ⋅ , t ) − ω ‾ 0 ∥ H − 2 ≤ C ε 2 and ∥ ω ‾ ∞ − ω ‾ 0 ∥ H − 2 ≤ C ε 2 , \sup_{t \ge0} \lVert\overline{\omega}(\cdot,t) - \overline{\omega}_{0} \rVert_{H^{-2}} \le C\varepsilon^{2}
\qquad\text{and}\qquad
\lVert\overline{\omega}_{\infty} - \overline{\omega}_{0} \rVert_{H^{-2}} \le C\varepsilon^{2}, t ≥ 0 sup ∥ ω ( ⋅ , t ) − ω 0 ∥ H − 2 ≤ C ε 2 and ∥ ω ∞ − ω 0 ∥ H − 2 ≤ C ε 2 , where C C C ν \nu ν
The proof is given in Appendix B. Since ω ‾ 0 \overline{\omega}_{0} ω 0 u 0 u_{0} u 0 Theorem 3.1 (b) shows that the initial temperature affects the limit only at order ε 2 \varepsilon^{2} ε 2 u 0 u_{0} u 0 θ 0 \theta_{0} θ 0 (12) give limits ω ‾ ∞ \overline{\omega}_{\infty} ω ∞ 2 C ε 2 2C\varepsilon^{2} 2 C ε 2 H − 2 H^{-2} H − 2 (15) , ω ‾ ∞ − ω ‾ 0 \overline{\omega}_{\infty} - \overline{\omega}_{0} ω ∞ − ω 0
The linear decay rate
Proposition 3.2 (Linear decay of ( u 2 , θ ) (u_{2},\theta) ( u 2 , θ ) Consider the linearization of (4) and (2) about the hydrostatic state,
∂ t ω = ν ∂ 11 ω + ∂ 1 θ , ∂ t θ = η ∂ 22 θ − u 2 , (16) \partial_{t}\omega= \nu\partial_{11}\omega+ \partial_{1}\theta,\qquad\partial_{t}\theta= \eta\partial_{22}\theta- u_{2}, \tag*{(16)} ∂ t ω = ν ∂ 11 ω + ∂ 1 θ , ∂ t θ = η ∂ 22 θ − u 2 , (16) with u \mathbf{u} u ω \omega ω (5) and (6) . For each k = ( k 1 , k 2 ) ∈ Z 2 ∖ { 0 } \mathbf{k} = (k_{1},k_{2}) \in\mathbb{Z}^{2} \setminus\{0\} k = ( k 1 , k 2 ) ∈ Z 2 ∖ { 0 }
d u ^ 2 d t = − ν k 1 2 u ^ 2 + k 1 2 ∣ k ∣ 2 θ ^ , d θ ^ d t = − u ^ 2 − η k 2 2 θ ^ . (17) \frac{\mathrm{d}\widehat{u}_{2}}{\mathrm{d}t} = -\nu k_{1}^{2}\widehat{u}_{2} + \frac{k_{1}^{2}}{|\mathbf{k}|^{2}}\widehat{\theta},\qquad
\frac{\mathrm{d}\widehat{\theta}}{\mathrm{d}t} = -\widehat{u}_{2} - \eta k_{2}^{2}\widehat{\theta}. \tag*{(17)} d t d u 2 = − ν k 1 2 u 2 + ∣ k ∣ 2 k 1 2 θ , d t d θ = − u 2 − η k 2 2 θ . (17) (a) If k 1 ≠ 0 k_{1} \ne0 k 1 = 0 k 2 ≠ 0 k_{2} \ne0 k 2 = 0
∣ u ^ 2 ( t ) ∣ 2 + k 1 2 ∣ k ∣ 2 ∣ θ ^ ( t ) ∣ 2 ≤ e − 2 min { ν , η } t ( ∣ u ^ 2 ( 0 ) ∣ 2 + k 1 2 ∣ k ∣ 2 ∣ θ ^ ( 0 ) ∣ 2 ) . |\widehat{u}_{2}(t)|^{2} + \frac{k_{1}^{2}}{|\mathbf{k}|^{2}}|\widehat{\theta}(t)|^{2}
\le e^{-2\min\{\nu,\eta\}t}
\left(|\widehat{u}_{2}(0)|^{2} + \frac{k_{1}^{2}}{|\mathbf{k}|^{2}}|\widehat{\theta}(0)|^{2}\right). ∣ u 2 ( t ) ∣ 2 + ∣ k ∣ 2 k 1 2 ∣ θ ( t ) ∣ 2 ≤ e − 2 m i n { ν , η } t ( ∣ u 2 ( 0 ) ∣ 2 + ∣ k ∣ 2 k 1 2 ∣ θ ( 0 ) ∣ 2 ) . If k 1 = 0 k_{1} = 0 k 1 = 0 k 2 ≠ 0 k_{2} \ne0 k 2 = 0 u ^ 2 ≡ 0 \widehat{u}_{2} \equiv0 u 2 ≡ 0 θ ^ ( t ) = e − η k 2 2 t θ ^ ( 0 ) \widehat{\theta}(t) = e^{-\eta k_{2}^{2}t}\widehat{\theta}(0) θ ( t ) = e − η k 2 2 t θ ( 0 )
(b) If k 2 = 0 k_{2} = 0 k 2 = 0 ν k 1 2 > 2 \nu k_{1}^{2} > 2 ν k 1 2 > 2 θ ^ ( t ) = a + e λ + t + a − e λ − t \widehat{\theta}(t) = a_{+}e^{\lambda_{+}t} + a_{-}e^{\lambda_{-}t} θ ( t ) = a + e λ + t + a − e λ − t a ± a_{\pm} a ± u ^ 2 ( 0 ) \widehat{u}_{2}(0) u 2 ( 0 ) θ ^ ( 0 ) \widehat{\theta}(0) θ ( 0 ) λ − ≤ − ν k 1 2 / 2 < − 1 \lambda_{-} \le-\nu k_{1}^{2}/2 < -1 λ − ≤ − ν k 1 2 /2 < − 1
λ + = − 2 ν k 1 2 + ν 2 k 1 4 − 4 , 1 ν k 1 2 ≤ ∣ λ + ∣ ≤ 2 ν k 1 2 < 1. \lambda_{+} = \frac{-2}{\nu k_{1}^{2} + \sqrt{\nu^{2}k_{1}^{4} - 4}},
\qquad
\frac{1}{\nu k_{1}^{2}} \le|\lambda_{+}| \le\frac{2}{\nu k_{1}^{2}} < 1. λ + = ν k 1 2 + ν 2 k 1 4 − 4 − 2 , ν k 1 2 1 ≤ ∣ λ + ∣ ≤ ν k 1 2 2 < 1. In particular λ + → 0 \lambda_{+} \to0 λ + → 0 ∣ k 1 ∣ → ∞ |k_{1}| \to\infty ∣ k 1 ∣ → ∞ θ \theta θ
The proof is given in Appendix C. The absence of a uniform decay rate in part (b) is consistent with Theorem 3.1 , which treats the nonlinear case and also does not give a decay rate.
A neural-operator surrogate, its training data, and a test of Theorem 1(b)
Neural operators and the Fourier neural operator
A neural operator (NO) [13 ] is a deep-learning architecture that learns a map between two infinite-dimensional function spaces, rather than between finite-dimensional Euclidean vectors as in a standard neural network. Given an input Banach space A \mathcal{A} A U \mathcal{U} U D ⊂ R d \mathcal{D} \subset\mathbb{R}^{d} D ⊂ R d D = T 2 \mathcal{D} = \mathbb{T}^{2} D = T 2 d = 2 d = 2 d = 2 G † : A → U \mathcal{G}^{\dagger} : \mathcal{A} \to\mathcal{U} G † : A → U G ϕ ≈ G † \mathcal{G}_{\phi} \approx\mathcal{G}^{\dagger} G ϕ ≈ G † { ( a ( i ) , z ( i ) ) } \{(a^{(i)}, z^{(i)})\} {( a ( i ) , z ( i ) )} a = ( ω 0 , θ 0 ) a = (\omega_{0}, \theta_{0}) a = ( ω 0 , θ 0 ) ε \varepsilon ε z = ω ( ⋅ , T ) z = \omega(\cdot,T) z = ω ( ⋅ , T ) § 4.2 ).
Architecture. A neural operator is a composition of a lifting map, a sequence of L L L L = 4 L = 4 L = 4
G ϕ = Q ∘ H L ∘ ⋯ ∘ H 1 ∘ P . \mathcal{G}_{\phi} = Q \circ\mathcal{H}_{L} \circ\cdots\circ\mathcal{H}_{1} \circ\mathcal{P}. G ϕ = Q ∘ H L ∘ ⋯ ∘ H 1 ∘ P . The pointwise lifting operator P \mathcal{P} P a a a v 0 v_{0} v 0 projection operator Q Q Q z z z v ℓ : D → R d v ℓ v_{\ell} : \mathcal{D} \to\mathbb{R}^{d_{v_{\ell}}} v ℓ : D → R d v ℓ
v ℓ + 1 ( x ) = ϱ ( W ℓ v ℓ ( x ) + ( K ℓ v ℓ ) ( x ) + b ℓ ( x ) ) , x ∈ D , (18) v_{\ell+1}(x) = \varrho\left(W_{\ell}v_{\ell}(x) + (\mathcal{K}_{\ell}v_{\ell})(x) + b_{\ell}(x)\right), \qquad x \in\mathcal{D}, \tag*{(18)} v ℓ + 1 ( x ) = ϱ ( W ℓ v ℓ ( x ) + ( K ℓ v ℓ ) ( x ) + b ℓ ( x ) ) , x ∈ D , (18) where ϱ \varrho ϱ W ℓ W_{\ell} W ℓ b ℓ b_{\ell} b ℓ K ℓ \mathcal{K}_{\ell} K ℓ integral kernel operator
( K ℓ v ℓ ) ( x ) = ∫ D κ ( ℓ ) ( x , x ′ ) v ℓ ( x ′ ) d x ′ . (19) (\mathcal{K}_{\ell}v_{\ell})(x) = \int_{\mathcal{D}} \kappa^{(\ell)}(x,x')v_{\ell}(x')\,\mathrm{d}x'. \tag*{(19)} ( K ℓ v ℓ ) ( x ) = ∫ D κ ( ℓ ) ( x , x ′ ) v ℓ ( x ′ ) d x ′ . (19) Fourier neural operator. The kernel (4.2) admits several parameterizations, and we use the Fourier neural operator (FNO) [17 ] , which represents the kernel directly in the frequency domain. Assuming a translation-invariant kernel κ ( ℓ ) ( x , x ′ ) = κ ( ℓ ) ( x − x ′ ) \kappa^{(\ell)}(x,x') = \kappa^{(\ell)}(x-x') κ ( ℓ ) ( x , x ′ ) = κ ( ℓ ) ( x − x ′ )
( K ℓ v ℓ ) ( x ) = F − 1 ( S ϕ ( ℓ ) ⋅ F ( v ℓ ) ) ( x ) , (20) (\mathcal{K}_{\ell}v_{\ell})(x) = \mathcal{F}^{-1}\left(S_{\phi}^{(\ell)} \cdot\mathcal{F}(v_{\ell})\right)(x), \tag*{(20)} ( K ℓ v ℓ ) ( x ) = F − 1 ( S ϕ ( ℓ ) ⋅ F ( v ℓ ) ) ( x ) , (20) where F \mathcal{F} F F − 1 \mathcal{F}^{-1} F − 1 S ϕ ( ℓ ) S_{\phi}^{(\ell)} S ϕ ( ℓ ) ∣ k j ∣ \lvert k_{j}\rvert ∣ k j ∣ k max k_{\max} k m a x S ϕ ( ℓ ) S_{\phi}^{(\ell)} S ϕ ( ℓ ) d v ℓ + 1 × d v ℓ d_{v_{\ell+1}} \times d_{v_{\ell}} d v ℓ + 1 × d v ℓ S ( − k ) = S ( k ) ∗ S(-k) = S(k)^{*} S ( − k ) = S ( k ) ∗ F \mathcal{F} F O ( N 2 log N ) \mathcal{O}(N^{2}\log N) O ( N 2 log N ) N × N N \times N N × N
Training data generation
The training pairs are produced by numerical simulation of the perturbation system in its vorticity-stream-function form on the periodic box T 2 \mathbb{T}^{2} T 2 Spatial discretization.. All fields are represented by their truncated Fourier series on an N × N N \times N N × N N = 128 N = 128 N = 128 i k ik ik − Δ ψ = ω -\Delta\psi= \omega − Δ ψ = ω ∣ k ∣ 2 |k|^2 ∣ k ∣ 2 ν ∂ 11 ω \nu\partial_{11}\omega ν ∂ 11 ω η ∂ 22 θ \eta\partial_{22}\theta η ∂ 22 θ − ν k 1 2 ω ^ -\nu k_1^2\widehat{\omega} − ν k 1 2 ω − η k 2 2 θ ^ -\eta k_2^2\widehat{\theta} − η k 2 2 θ u ⋅ ∇ ω = ∇ ⋅ ( u ω ) \mathbf{u}\cdot\nabla\omega= \nabla\cdot(\mathbf{u}\omega) u ⋅ ∇ ω = ∇ ⋅ ( u ω ) u ⋅ ∇ θ = ∇ ⋅ ( u θ ) \mathbf{u}\cdot\nabla\theta= \nabla\cdot(\mathbf{u}\theta) u ⋅ ∇ θ = ∇ ⋅ ( u θ ) ∇ ⋅ u = 0 \nabla\cdot\mathbf{u} = 0 ∇ ⋅ u = 0 u ω \mathbf{u}\omega u ω u θ \mathbf{u}\theta u θ
Time integration.. Writing the two evolution equations as ∂ t ω = N ω + ν ∂ 11 ω \partial_t\omega= \mathcal{N}_{\omega} + \nu\partial_{11}\omega ∂ t ω = N ω + ν ∂ 11 ω ∂ t θ = N θ + η ∂ 22 θ \partial_t\theta= \mathcal{N}_{\theta} + \eta\partial_{22}\theta ∂ t θ = N θ + η ∂ 22 θ
N ω = − u ⋅ ∇ ω + ∂ 1 θ , \mathcal{N}_{\omega} = -\mathbf{u}\cdot\nabla\omega+ \partial_1\theta, N ω = − u ⋅ ∇ ω + ∂ 1 θ , N θ = − u ⋅ ∇ θ − u 2 , \mathcal{N}_{\theta} = -\mathbf{u}\cdot\nabla\theta- u_2, N θ = − u ⋅ ∇ θ − u 2 , the system is advanced with a second-order semi-implicit scheme based on the backward differentiation formula (BDF2): the linear anisotropic dissipation is treated implicitly, while the advection and the buoyancy and coupling terms N ω , N θ \mathcal{N}_{\omega},\mathcal{N}_{\theta} N ω , N θ k = ( k 1 , k 2 ) k = (k_1,k_2) k = ( k 1 , k 2 )
( 3 2 Δ t + ν k 1 2 ) ω ^ n + 1 = 2 ω ^ n − 1 2 ω ^ n − 1 Δ t + N ^ ω ∗ , \left(\frac{3}{2\Delta t}+\nu k_1^2\right)\widehat{\omega}^{n+1}
=
\frac{2\widehat{\omega}^{n}-\frac{1}{2}\widehat{\omega}^{n-1}}{\Delta t}
+\widehat{\mathcal{N}}_{\omega}^{*}, ( 2Δ t 3 + ν k 1 2 ) ω n + 1 = Δ t 2 ω n − 2 1 ω n − 1 + N ω ∗ , ( 3 2 Δ t + η k 2 2 ) θ ^ n + 1 = 2 θ ^ n − 1 2 θ ^ n − 1 Δ t + N ^ θ ∗ , \left(\frac{3}{2\Delta t}+\eta k_2^2\right)\widehat{\theta}^{n+1}
=
\frac{2\widehat{\theta}^{n}-\frac{1}{2}\widehat{\theta}^{n-1}}{\Delta t}
+\widehat{\mathcal{N}}_{\theta}^{*}, ( 2Δ t 3 + η k 2 2 ) θ n + 1 = Δ t 2 θ n − 2 1 θ n − 1 + N θ ∗ , where N ∗ \mathcal{N}^{*} N ∗ ( ω ∗ , θ ∗ ) = ( 2 ω n − ω n − 1 , 2 θ n − θ n − 1 ) (\omega^{*},\theta^{*}) = (2\omega^{n}-\omega^{n-1},2\theta^{n}-\theta^{n-1}) ( ω ∗ , θ ∗ ) = ( 2 ω n − ω n − 1 , 2 θ n − θ n − 1 ) ω ∗ \omega^{*} ω ∗ (5) and (6) . The multistep scheme is initialized with a single first-order step of the same semi-implicit form, backward Euler for the dissipation and forward Euler for N ω , N θ \mathcal{N}_{\omega},\mathcal{N}_{\theta} N ω , N θ Δ t ∥ u ∥ ∞ / h \Delta t\lVert\mathbf{u}\rVert_{\infty}/h Δ t ∥ u ∥ ∞ / h h = 2 π / N h = 2\pi/N h = 2 π / N 1 / 2 1/2 1/2
As is standard for odd-order derivatives on an even grid [34 Ch. 3] , the coefficients of the vorticity at the Nyquist wavenumbers k 1 = N / 2 k_1 = N/2 k 1 = N /2 k 2 = N / 2 k_2 = N/2 k 2 = N /2 ν = η = 1 \nu= \eta= 1 ν = η = 1 Δ t = 10 − 3 \Delta t = 10^{-3} Δ t = 1 0 − 3 T = 100 T = 100 T = 100
Initial data and data sets.. For each sample the initial vorticity ω 0 \omega_0 ω 0 θ 0 \theta_0 θ 0 T 2 \mathbb{T}^2 T 2 ( τ 2 − Δ ) − α (\tau^2-\Delta)^{-\alpha} ( τ 2 − Δ ) − α ( τ 2 + ∣ k ∣ 2 ) − α (\tau^2+|k|^2)^{-\alpha} ( τ 2 + ∣ k ∣ 2 ) − α ω 0 \omega_0 ω 0 α = 2.5 \alpha= 2.5 α = 2.5 τ = 7 \tau= 7 τ = 7 [13 ] , and for θ 0 \theta_0 θ 0 α = 3.5 \alpha= 3.5 α = 3.5 τ = 7 \tau= 7 τ = 7 H s ( T 2 ) H^s(\mathbb{T}^2) H s ( T 2 ) s < α − 1 s < \alpha- 1 s < α − 1 u 0 \mathbf{u}_0 u 0 θ 0 \theta_0 θ 0 H s H^s H s s < 5 / 2 s < 5/2 s < 5/2 H 2 H^2 H 2 ∫ T 2 θ 0 d x = 0 \int_{\mathbb{T}^2}\theta_0\,dx = 0 ∫ T 2 θ 0 d x = 0
u ^ ( k ) = ( i k 2 ∣ k ∣ 2 , − i k 1 ∣ k ∣ 2 ) ω ^ ( k ) , k ≠ 0. \widehat{\mathbf{u}}(k)=\left(\frac{ik_2}{|k|^2},-\frac{ik_1}{|k|^2}\right)\widehat{\omega}(k),\qquad k\ne0. u ( k ) = ( ∣ k ∣ 2 i k 2 , − ∣ k ∣ 2 i k 1 ) ω ( k ) , k = 0. Finally each sample is rescaled so that the smallness hypothesis (3.1) of Theorem 3.1 is realized with equality, the norm being divided evenly between the two fields,
∥ u 0 ∥ H 2 = ∥ θ 0 ∥ H 2 = 1 2 ε . (21) \|u_0\|_{H^2}=\|\theta_0\|_{H^2}=\tfrac{1}{2}\varepsilon. \tag*{(21)} ∥ u 0 ∥ H 2 = ∥ θ 0 ∥ H 2 = 2 1 ε . (21) For each sample we store the initial pair ( ω 0 , θ 0 ) (\omega_0,\theta_0) ( ω 0 , θ 0 ) ( ω ( ⋅ , T ) , θ ( ⋅ , T ) ) (\omega(\cdot,T),\theta(\cdot,T)) ( ω ( ⋅ , T ) , θ ( ⋅ , T )) 128 × 128 128 \times128 128 × 128 ω 0 \omega_0 ω 0 θ 0 \theta_0 θ 0 training set uses ten amplitudes, consecutive ones differing by the factor 2 1 / 2 2^{1/2} 2 1/2
ε j = 0.025 ⋅ 2 ( 2 j − 1 ) / 4 , j = 0 , … , 9 , (22) \varepsilon_j=0.025\cdot2^{(2j-1)/4},\qquad j=0,\ldots,9, \tag*{(22)} ε j = 0.025 ⋅ 2 ( 2 j − 1 ) /4 , j = 0 , … , 9 , (22) with 1000 samples each, 200 initial vorticities each paired with five independent initial temperatures, 10 4 10^4 1 0 4 training-amplitude set uses the same ten amplitudes with 100 samples each, 20 initial vorticities each paired with five independent initial temperatures. The interpolation set uses the five amplitudes ε ∈ { 0.025 , 0.05 , 0.1 , 0.2 , 0.4 } \varepsilon\in\{0.025, 0.05, 0.1, 0.2, 0.4\} ε ∈ { 0.025 , 0.05 , 0.1 , 0.2 , 0.4 } 2 1 / 4 2^{1/4} 2 1/4 1.2 × 10 4 1.2 \times10^4 1.2 × 1 0 4 10 3 10^3 1 0 3
The late-time temperature and why it is not a prediction target
Linear prediction.. By Proposition 3.2, ∣ θ ^ ( k 1 , k 2 , t ) ∣ |\hat{\theta}(k_1,k_2,t)| ∣ θ ^ ( k 1 , k 2 , t ) ∣ k 2 ≠ 0 k_2 \ne0 k 2 = 0 e − min { ν , η } t e^{-\min\{\nu,\eta\}t} e − m i n { ν , η } t e − 100 e^{-100} e − 100 T = 100 T=100 T = 100 ν = η = 1 \nu=\eta=1 ν = η = 1 k 2 = 0 k_2=0 k 2 = 0 ∣ θ ^ ( k 1 , 0 , t ) ∣ |\hat{\theta}(k_1,0,t)| ∣ θ ^ ( k 1 , 0 , t ) ∣ e − t / ( ν k 1 2 ) e^{-t/(\nu k_1^2)} e − t / ( ν k 1 2 ) ∣ λ + ∣ = ( ν k 1 2 ) − 1 + O ( ( ν k 1 2 ) − 3 ) |\lambda_+|=(\nu k_1^2)^{-1}+O((\nu k_1^2)^{-3}) ∣ λ + ∣ = ( ν k 1 2 ) − 1 + O (( ν k 1 2 ) − 3 ) E t ( k 1 ) E_t(k_1) E t ( k 1 ) ∣ θ ^ ( k 1 , 0 , t ) ∣ 2 / ε 2 |\hat{\theta}(k_1,0,t)|^2/\varepsilon^2 ∣ θ ^ ( k 1 , 0 , t ) ∣ 2 / ε 2
E T ( k 1 ) ≈ E 0 ( k 1 ) e − 2 T / ( ν k 1 2 ) , when ν k 1 2 ≫ 1. (23) E_T(k_1)\approx E_0(k_1)e^{-2T/(\nu k_1^2)},\qquad\text{when }\nu k_1^2\gg1. \tag*{(23)} E T ( k 1 ) ≈ E 0 ( k 1 ) e − 2 T / ( ν k 1 2 ) , when ν k 1 2 ≫ 1. (23) For the Gaussian fields of § 4.2 , E 0 ( k 1 ) ∝ ( τ 2 + k 1 2 ) − α E_0(k_1)\propto(\tau^2+k_1^2)^{-\alpha} E 0 ( k 1 ) ∝ ( τ 2 + k 1 2 ) − α α = 3.5 \alpha=3.5 α = 3.5 τ = 7 \tau=7 τ = 7 (23) when T = 100 T=100 T = 100 ν = 1 \nu=1 ν = 1 k 1 ∗ = 9.4 k_1^*=9.4 k 1 ∗ = 9.4
Comparison with the simulations.. The 2000 spectral simulations of the two test sets agree with this analysis. At T = 100 T=100 T = 100 x 1 x_1 x 1 x 1 x_1 x 1 Figure 1 ). The modes with k 2 = 0 k_2=0 k 2 = 0 ∥ θ ( ⋅ , T ) ∥ L 2 2 \|\theta(\cdot,T)\|_{L^2}^2 ∥ θ ( ⋅ , T ) ∥ L 2 2 L 2 L^2 L 2 θ \theta θ ∣ k 1 ∣ |k_1| ∣ k 1 ∣ ∣ θ ^ ( k 1 , 0 , T ) ∣ 2 |\hat{\theta}(k_1,0,T)|^2 ∣ θ ^ ( k 1 , 0 , T ) ∣ 2 k 1 ∗ = 9.4 k_1^*=9.4 k 1 ∗ = 9.4 k 1 k_1 k 1 7 ≤ k 1 ≤ 20 7 \le k_1 \le20 7 ≤ k 1 ≤ 20 ∑ k 1 E T ( k 1 ) \sum_{k_1} E_T(k_1) ∑ k 1 E T ( k 1 ) (23) , with E 0 E_0 E 0 ∣ θ ^ ( k 1 , 0 , T ) ∣ 2 / ε 2 |\hat{\theta}(k_1,0,T)|^2/\varepsilon^2 ∣ θ ^ ( k 1 , 0 , T ) ∣ 2 / ε 2 θ ( ⋅ , T ) \theta(\cdot,T) θ ( ⋅ , T ) x 1 x_1 x 1 θ 0 \theta_0 θ 0 k 2 = 0 k_2=0 k 2 = 0 u 2 u_2 u 2 ∣ k 1 ∣ |k_1| ∣ k 1 ∣ E 0 E_0 E 0 ∣ k 1 ∣ |k_1| ∣ k 1 ∣ ∣ k 1 ∣ |k_1| ∣ k 1 ∣
Figure 1. Top: the initial temperature and the temperature at T = 100 T = 100 T = 100 ε = 0.1 \varepsilon= 0.1 ε = 0.1 Bottom left: ∣ θ ^ ( k 1 , 0 , t ) ∣ 2 |\hat{\theta}(k_1,0,t)|^2 ∣ θ ^ ( k 1 , 0 , t ) ∣ 2 t = 0 t = 0 t = 0 t = T t = T t = T Bottom right: the averages of ∣ θ ^ ( k 1 , 0 , t ) ∣ 2 / ε 2 |\hat{\theta}(k_1,0,t)|^2/\varepsilon^2 ∣ θ ^ ( k 1 , 0 , t ) ∣ 2 / ε 2 t = 0 t = 0 t = 0 t = T t = T t = T E 0 E_0 E 0 E T E_T E T t = 0 t = 0 t = 0 k 1 ∗ = 9.4 k_1^* = 9.4 k 1 ∗ = 9.4 7 ≤ ∣ k 1 ∣ ≤ 20 7 \leq|k_1| \leq20 7 ≤ ∣ k 1 ∣ ≤ 20
Why the temperature is not a prediction target.. By (23) the map θ 0 ↦ θ ( ⋅ , T ) \theta_0 \mapsto\theta(\cdot,T) θ 0 ↦ θ ( ⋅ , T ) k 2 = 0 k_2=0 k 2 = 0 ( E T ( k 1 ) / E 0 ( k 1 ) ) 1 / 2 (E_T(k_1)/E_0(k_1))^{1/2} ( E T ( k 1 ) / E 0 ( k 1 ) ) 1/2 4.8 × 10 − 11 4.8 \times10^{-11} 4.8 × 1 0 − 11 1.4 × 10 − 5 1.4 \times10^{-5} 1.4 × 1 0 − 5 1.9 × 10 − 3 1.9 \times10^{-3} 1.9 × 1 0 − 3 0.368 0.368 0.368 0.902 0.902 0.902 1.00 1.00 1.00 k 1 = 2 , 3 , 4 , 10 , 32 , 64 k_1 = 2, 3, 4, 10, 32, 64 k 1 = 2 , 3 , 4 , 10 , 32 , 64 k 1 = N / 2 = 64 k_1 = N/2 = 64 k 1 = N /2 = 64 10 10 10^{10} 1 0 10 ω ( ⋅ , T ) \omega(\cdot,T) ω ( ⋅ , T ) θ ( ⋅ , T ) \theta(\cdot,T) θ ( ⋅ , T ) Theorem 3.1 (a) the temperature decays to zero, and at T = 100 T = 100 T = 100 L 2 L^2 L 2 1.3 × 10 − 5 1.3 \times10^{-5} 1.3 × 1 0 − 5 θ 0 \theta_0 θ 0
Network structure and training
Network architecture. The Fourier neural operator of § 4.1 is instantiated in two spatial dimensions with L = 4 L = 4 L = 4 k max = 32 k_{\max} = 32 k m a x = 32 − 32 ≤ k 1 < 32 -32 \leq k_1 < 32 − 32 ≤ k 1 < 32 ∣ k 2 ∣ ≤ 32 |k_2| \leq32 ∣ k 2 ∣ ≤ 32 128 × 128 128 \times128 128 × 128
The inputs are the rescaled initial fields ω 0 / ε \omega_0/\varepsilon ω 0 / ε θ 0 / ε \theta_0/\varepsilon θ 0 / ε log ( ε / 0.1 ) \log(\varepsilon/0.1) log ( ε /0.1 )
The outputs are not the final vorticity itself. Theorem 3.1 (b) says that the horizontal average of the vorticity stays within C ε 2 C\varepsilon^2 C ε 2 ω ˉ 0 \bar{\omega}_0 ω ˉ 0
δ ω ˉ : = ω ˉ ( ⋅ , T ) − ω ˉ 0 , (24) \delta\bar{\omega} := \bar{\omega}(\cdot,T) - \bar{\omega}_0, \tag*{(24)} δ ω ˉ := ω ˉ ( ⋅ , T ) − ω ˉ 0 , (24) is a median 5.4 × 10 − 5 5.4 \times10^{-5} 5.4 × 1 0 − 5 1.2 × 10 − 3 1.2 \times10^{-3} 1.2 × 1 0 − 3 ω ˉ 0 \bar{\omega}_0 ω ˉ 0 L 2 L^2 L 2 ε \varepsilon ε ω ( ⋅ , T ) \omega(\cdot,T) ω ( ⋅ , T )
ω N O ( ⋅ , T ) = ω ˉ 0 + s m ( ε ) m N O + s q ( ε ) q N O , (25) \omega^{\mathrm{NO}}(\cdot,T)=\bar{\omega}_0+s_m(\varepsilon)m^{\mathrm{NO}}+s_q(\varepsilon)q^{\mathrm{NO}}, \tag*{(25)} ω NO ( ⋅ , T ) = ω ˉ 0 + s m ( ε ) m NO + s q ( ε ) q NO , (25) where ω ˉ 0 \bar{\omega}_0 ω ˉ 0 ω 0 \omega_0 ω 0 x 1 x_1 x 1 q N O q^{\mathrm{NO}} q NO x 1 x_1 x 1 m N O ( x 2 ) m^{\mathrm{NO}}(x_2) m NO ( x 2 ) (25) . Thus s m m N O s_m m^{\mathrm{NO}} s m m NO s q q N O s_q q^{\mathrm{NO}} s q q NO ω ( ⋅ , T ) − ω ˉ 0 \omega(\cdot,T)-\bar{\omega}_0 ω ( ⋅ , T ) − ω ˉ 0 δ ω ˉ \delta\bar{\omega} δ ω ˉ ω ~ ( ⋅ , T ) \widetilde{\omega}(\cdot,T) ω ( ⋅ , T )
For each training amplitude ε j \varepsilon_j ε j s q ( ε j ) s_q(\varepsilon_j) s q ( ε j ) s m ( ε j ) s_m(\varepsilon_j) s m ( ε j ) ω ~ ( ⋅ , T ) \widetilde{\omega}(\cdot,T) ω ( ⋅ , T ) δ ω ˉ \delta\bar{\omega} δ ω ˉ n j = 1000 n_j=1000 n j = 1000
s q ( ε j ) = ( 1 n j ∑ i = 1 n j ∥ ω ~ i ( ⋅ , T ) ∥ L 2 2 4 π 2 ) 1 / 2 , s m ( ε j ) = ( 1 n j ∑ i = 1 n j ∥ δ ω ˉ i ∥ L 2 2 4 π 2 ) 1 / 2 . (26) s_q(\varepsilon_j)=\left(\frac{1}{n_j}\sum_{i=1}^{n_j}\frac{\lVert\widetilde{\omega}_i(\cdot,T)\rVert_{L^2}^2}{4\pi^2}\right)^{1/2},\qquad
s_m(\varepsilon_j)=\left(\frac{1}{n_j}\sum_{i=1}^{n_j}\frac{\lVert\delta\bar{\omega}_i\rVert_{L^2}^2}{4\pi^2}\right)^{1/2}. \tag*{(26)} s q ( ε j ) = ( n j 1 i = 1 ∑ n j 4 π 2 ∥ ω i ( ⋅ , T ) ∥ L 2 2 ) 1/2 , s m ( ε j ) = ( n j 1 i = 1 ∑ n j 4 π 2 ∥ δ ω ˉ i ∥ L 2 2 ) 1/2 . (26) Between the training amplitudes, log s q \log s_q log s q log s m \log s_m log s m log ε \log\varepsilon log ε ε \varepsilon ε s m s_m s m ε 2.00 \varepsilon^{2.00} ε 2.00 s m / ε 2 s_m/\varepsilon^2 s m / ε 2 5.7 × 10 − 6 5.7\times10^{-6} 5.7 × 1 0 − 6 6.6 × 10 − 6 6.6\times10^{-6} 6.6 × 1 0 − 6 s q s_q s q ε 1.00 \varepsilon^{1.00} ε 1.00 ε \varepsilon ε m N O m^{\mathrm{NO}} m NO q N O q^{\mathrm{NO}} q NO ω 0 / ε \omega_0/\varepsilon ω 0 / ε θ 0 / ε \theta_0/\varepsilon θ 0 / ε
Training process. The training set is the 10 4 10^4 1 0 4 B B B L 2 L^2 L 2 B B B
L = 1 B ∑ i = 1 B ∥ s q q i N O − ω ~ i ( ⋅ , T ) ∥ L 2 ∥ ω ~ i ( ⋅ , T ) ∥ L 2 + γ 1 B ∑ i = 1 B ∥ s m m i N O − δ ω ˉ i ∥ L 2 ∥ δ ω ˉ i ∥ L 2 , γ = 10. (27) \mathcal{L}=\frac{1}{B}\sum_{i=1}^{B}\frac{\lVert s_q q_i^{\mathrm{NO}}-\widetilde{\omega}_i(\cdot,T)\rVert_{L^2}}{\lVert\widetilde{\omega}_i(\cdot,T)\rVert_{L^2}}+\gamma\frac{1}{B}\sum_{i=1}^{B}\frac{\lVert s_m m_i^{\mathrm{NO}}-\delta\bar{\omega}_i\rVert_{L^2}}{\lVert\delta\bar{\omega}_i\rVert_{L^2}},\qquad\gamma=10. \tag*{(27)} L = B 1 i = 1 ∑ B ∥ ω i ( ⋅ , T ) ∥ L 2 ∥ s q q i NO − ω i ( ⋅ , T ) ∥ L 2 + γ B 1 i = 1 ∑ B ∥ δ ω ˉ i ∥ L 2 ∥ s m m i NO − δ ω ˉ i ∥ L 2 , γ = 10. (27) The weight γ \gamma γ δ ω ˉ \delta\bar{\omega} δ ω ˉ Theorem 3.1 (b) bears. Two preliminary runs on 900 samples, with γ = 10 \gamma=10 γ = 10 γ = 100 \gamma=100 γ = 100 δ ω ˉ \delta\bar{\omega} δ ω ˉ γ = 10 \gamma=10 γ = 10 [20 ] with learning rate 3 × 10 − 4 3\times10^{-4} 3 × 1 0 − 4 10 − 4 10^{-4} 1 0 − 4 B = 8 B=8 B = 8 7.5 × 10 5 7.5\times10^5 7.5 × 1 0 5 3.47 × 10 7 3.47\times10^7 3.47 × 1 0 7
Surrogate accuracy and cost
Accuracy. On the 2 × 10 3 2\times10^3 2 × 1 0 3 (25) reproduces the final vorticity ω ( ⋅ , T ) \omega(\cdot,T) ω ( ⋅ , T ) L 2 L^2 L 2 3.8 × 10 − 5 3.8\times10^{-5} 3.8 × 1 0 − 5 2.4 × 10 − 4 2.4\times10^{-4} 2.4 × 1 0 − 4 4.0 × 10 − 5 4.0\times10^{-5} 4.0 × 1 0 − 5 3.3 × 10 − 4 3.3\times10^{-4} 3.3 × 1 0 − 4 Figure 2 shows the training history and Figure 3 three test samples, the best, the median and the worst.
Figure 2. Training history of the Fourier neural operator, versus epoch. Each curve is the average over the batches of each epoch. Black: the loss (4.10). Blue: the first term in (4.10), the relative L 2 L^2 L 2 Orange: the second term in (4.10) without γ \gamma γ L 2 L^2 L 2 δ ω ˉ \delta\bar{\omega} δ ω ˉ
Figure 3. The surrogate on the three samples of the interpolation set with the smallest, median and largest relative L 2 L^2 L 2 ω ( ⋅ , T ) \omega(\cdot,T) ω ( ⋅ , T ) 10 3 10^3 1 0 3 2.7 × 10 − 5 2.7 \times10^{-5} 2.7 × 1 0 − 5 7.0 × 10 − 5 7.0 \times10^{-5} 7.0 × 1 0 − 5 5.4 × 10 − 4 5.4 \times10^{-4} 5.4 × 1 0 − 4 ω 0 \omega_0 ω 0 θ 0 \theta_0 θ 0 ω ( ⋅ , T ) \omega(\cdot,T) ω ( ⋅ , T ) ω N O ( ⋅ , T ) \omega^{\mathrm{NO}}(\cdot,T) ω NO ( ⋅ , T ) (25) , and the pointwise error on its own scale. Fields are drawn with x 1 x_1 x 1 x 1 x_1 x 1 x 1 x_1 x 1
The field is accurate because most of it is ω ˉ 0 \bar{\omega}_0 ω ˉ 0 L 2 L^2 L 2 ω ~ ( ⋅ , T ) \widetilde{\omega}(\cdot,T) ω ( ⋅ , T ) δ ω ˉ \delta\bar{\omega} δ ω ˉ ω ( ⋅ , T ) \omega(\cdot,T) ω ( ⋅ , T )
The error of δ ω ˉ , s m N O − δ ω ˉ \delta\bar{\omega}, s_m^{\mathrm{NO}}-\delta\bar{\omega} δ ω ˉ , s m NO − δ ω ˉ L 2 L^2 L 2 ∣ k 2 ∣ > 16 |k_2| > 16 ∣ k 2 ∣ > 16 δ ω ˉ \delta\bar{\omega} δ ω ˉ δ ω ˉ \delta\bar{\omega} δ ω ˉ 10 − 2 10^{-2} 1 0 − 2 2.4 × 10 − 3 2.4 \times10^{-3} 2.4 × 1 0 − 3 H − 2 H^{-2} H − 2 Theorem 3.1 (b) weights high wavenumbers weakly. In this norm the median relative error of δ ω ˉ \delta\bar{\omega} δ ω ˉ ∥ δ ω ˉ ∥ H − 2 \lVert\delta\bar{\omega}\rVert_{H^{-2}} ∥ δ ω ˉ ∥ H − 2 Theorem 3.1 (b) bounds by C ε 2 C\varepsilon^2 C ε 2 § 4.6 ).
Inference cost. Once trained, the surrogate is far cheaper to evaluate than the spectral solver that generated its data, which costs 10.2 s per simulation on average (§ 4.2 ). On one NVIDIA Tesla V100S (32 GB), a batch of 16 initial fields is mapped to the final vorticity (4.8), including the computation of the inputs and of (4.8), in 21.3 ms, that is, 1.33 ms per field, and a batch of 64 in 73.6 ms, 1.15 ms per field. Against the 10.2 s of the solver, at the batch of 16, the surrogate is about 8 × 10 3 8 \times10^3 8 × 1 0 3
The quantity of Theorem 3.1(b): solver and surrogate
Theorem 3.1 (b) bounds ∥ ω ˉ ( ⋅ , t ) − ω ˉ 0 ∥ H − 2 \lVert\bar{\omega}(\cdot,t)-\bar{\omega}_0\rVert_{H^{-2}} ∥ ω ˉ ( ⋅ , t ) − ω ˉ 0 ∥ H − 2 C ε 2 C\varepsilon^2 C ε 2 C C C ω − ω ˉ 0 = ( ω ˉ − ω ˉ 0 ) + ω ~ \omega-\bar{\omega}_0=(\bar{\omega}-\bar{\omega}_0)+\widetilde{\omega} ω − ω ˉ 0 = ( ω ˉ − ω ˉ 0 ) + ω ∥ ω ~ ∥ H − 2 ≤ ∥ ω ~ ∥ L 2 \lVert\widetilde{\omega}\rVert_{H^{-2}}\leq\lVert\widetilde{\omega}\rVert_{L^2} ∥ ω ∥ H − 2 ≤ ∥ ω ∥ L 2 ∥ ω ~ ∥ L 2 ≤ ∥ ω ∥ L 2 \lVert\widetilde{\omega}\rVert_{L^2}\leq\lVert\omega\rVert_{L^2} ∥ ω ∥ L 2 ≤ ∥ ω ∥ L 2 ∥ ω ∥ L 2 ≤ ∥ u ∥ H 2 \lVert\omega\rVert_{L^2}\leq\lVert u\rVert_{H^2} ∥ ω ∥ L 2 ≤ ∥ u ∥ H 2 Lemma 2.1 (c)) together with (A.2),
∥ ω ( ⋅ , t ) − ω ˉ 0 ∥ H − 2 ≤ C ε 2 + C ∗ ε for every t ≥ 0 , (28) \lVert\omega(\cdot,t)-\bar{\omega}_0\rVert_{H^{-2}}\leq C\varepsilon^2+C_*\varepsilon\qquad\text{for every }t\geq0, \tag*{(28)} ∥ ω ( ⋅ , t ) − ω ˉ 0 ∥ H − 2 ≤ C ε 2 + C ∗ ε for every t ≥ 0 , (28) which is of order ε \varepsilon ε ω ~ \widetilde{\omega} ω i i i T = 100 T=100 T = 100
R ˉ i = ∥ ω ˉ i ( ⋅ , T ) − ω ˉ 0 , i ∥ H − 2 ε i 2 , R i = ∥ ω i ( ⋅ , T ) − ω ˉ 0 , i ∥ H − 2 ε i , (29) \bar{R}_i=\frac{\lVert\bar{\omega}_i(\cdot,T)-\bar{\omega}_{0,i}\rVert_{H^{-2}}}{\varepsilon_i^2},\qquad R_i=\frac{\lVert\omega_i(\cdot,T)-\bar{\omega}_{0,i}\rVert_{H^{-2}}}{\varepsilon_i}, \tag*{(29)} R ˉ i = ε i 2 ∥ ω ˉ i ( ⋅ , T ) − ω ˉ 0 , i ∥ H − 2 , R i = ε i ∥ ω i ( ⋅ , T ) − ω ˉ 0 , i ∥ H − 2 , (29) and by Theorem 3.1 (b), R ˉ i ≤ C \bar{R}_i\leq C R ˉ i ≤ C (28) R i ≤ C ε i + C ∗ R_i\leq C\varepsilon_i+C_* R i ≤ C ε i + C ∗
Table 1 is built as follows. For each of the 2 × 10 3 2\times10^3 2 × 1 0 3 (29) are evaluated with the discrete Fourier coefficients on the 128 × 128 128\times128 128 × 128 ω i ( ⋅ , T ) \omega_i(\cdot,T) ω i ( ⋅ , T ) R ˉ i \bar{R}_i R ˉ i R i R_i R i R ˉ i N O \bar{R}_i^{\mathrm{NO}} R ˉ i NO R i N O R_i^{\mathrm{NO}} R i NO R ˉ \bar{R} R ˉ R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO ∣ R i N O − R ˉ i ∣ / R ˉ i \lvert R_i^{\mathrm{NO}}-\bar{R}_i\rvert/\bar{R}_i ∣ R i NO − R ˉ i ∣ / R ˉ i R R R ∣ R i N O − R i ∣ / R i \lvert R_i^{\mathrm{NO}}-R_i\rvert/R_i ∣ R i NO − R i ∣ / R i Figure 4 shows the same data for R ˉ \bar{R} R ˉ R ˉ i \bar{R}_i R ˉ i R ˉ i N O \bar{R}_i^{\mathrm{NO}} R ˉ i NO R ˉ i \bar{R}_i R ˉ i
Table 1. The quantities of (29) at T = 100 T=100 T = 100 2 × 10 3 2 \times10^3 2 × 1 0 3 R ˉ \bar{R} R ˉ Theorem 3.1 (b) at every time. The interpolation amplitudes (top) were not used in training. The columns ∣ R ˉ N O − R ˉ ∣ / R ˉ \lvert\bar{R}^{\mathrm{NO}}-\bar{R}\rvert/\bar{R} ∣ R ˉ NO − R ˉ ∣ / R ˉ R R R n − 1 n-1 n − 1
Figure 4. (a) R ˉ = ∥ ω ˉ ( ⋅ , T ) − ω ˉ 0 ∥ H − 2 / ε 2 \bar{R}=\lVert\bar{\omega}(\cdot,T)-\bar{\omega}_0\rVert_{H^{-2}}/\varepsilon^2 R ˉ = ∥ ω ˉ ( ⋅ , T ) − ω ˉ 0 ∥ H − 2 / ε 2 T = 100 T=100 T = 100 ε \varepsilon ε 2 × 10 3 2 \times10^3 2 × 1 0 3 2 3 2^3 2 3 ε \varepsilon ε R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO R ˉ \bar{R} R ˉ 2 × 10 3 2 \times10^3 2 × 1 0 3
Table 1 and Figure 4 show four things. First, R ˉ \bar{R} R ˉ ε \varepsilon ε 1.33 × 10 − 6 1.33 \times10^{-6} 1.33 × 1 0 − 6 1.55 × 10 − 6 1.55 \times10^{-6} 1.55 × 1 0 − 6 0.021 0.021 0.021 0.476 0.476 0.476 ∥ ω ˉ ( ⋅ , T ) − ω ˉ 0 ∥ H − 2 \lVert\bar{\omega}(\cdot,T)-\bar{\omega}_0\rVert_{H^{-2}} ∥ ω ˉ ( ⋅ , T ) − ω ˉ 0 ∥ H − 2 c ε p c\varepsilon^p c ε p p = 1.99 p=1.99 p = 1.99 [ 1.95 , 2.03 ] [1.95, 2.03] [ 1.95 , 2.03 ] 2.45 × 10 − 6 2.45 \times10^{-6} 2.45 × 1 0 − 6 4.24 × 10 − 6 4.24 \times10^{-6} 4.24 × 1 0 − 6 2 × 10 3 2 \times10^3 2 × 1 0 3
∥ ω ˉ ( ⋅ , T ) − ω ˉ 0 ∥ H − 2 ≤ 4.25 × 10 − 6 ε 2 \lVert\bar{\omega}(\cdot,T)-\bar{\omega}_0\rVert_{H^{-2}} \le4.25 \times10^{-6}\varepsilon^2 ∥ ω ˉ ( ⋅ , T ) − ω ˉ 0 ∥ H − 2 ≤ 4.25 × 1 0 − 6 ε 2 which is the form of Theorem 3.1 (b) with an empirical constant, and over the range tested nothing in the data suggests that the constant grows as ε \varepsilon ε ε \varepsilon ε § 4.8 shows that initial data of other shapes give a ratio more than four times larger.
Second, at a fixed amplitude R ˉ \bar{R} R ˉ Table 1 is 0.36 to 0.54 times its sample mean. This variation comes from the shapes of the initial data, ω 0 / ε \omega_0/\varepsilon ω 0 / ε θ 0 / ε \theta_0/\varepsilon θ 0 / ε ε \varepsilon ε
Third, the surrogate’s R ˉ i N O \bar{R}^{\mathrm{NO}}_i R ˉ i NO R ˉ i \bar{R}_i R ˉ i R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO R ˉ \bar{R} R ˉ 2 × 10 3 2 \times10^3 2 × 1 0 3 ∣ R ˉ i N O − R ˉ i ∣ / R ˉ i \lvert\bar{R}^{\mathrm{NO}}_i-\bar{R}_i\rvert/\bar{R}_i ∣ R ˉ i NO − R ˉ i ∣ / R ˉ i R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO 4.05 × 10 − 6 4.05 \times10^{-6} 4.05 × 1 0 − 6 R ˉ \bar{R} R ˉ ∣ R ˉ i N O − R ˉ i ∣ / R ˉ i \lvert\bar{R}^{\mathrm{NO}}_i-\bar{R}_i\rvert/\bar{R}_i ∣ R ˉ i NO − R ˉ i ∣ / R ˉ i
Fourth, R R R 1.8 × 10 − 7 1.8 \times10^{-7} 1.8 × 1 0 − 7 2.1 × 10 − 7 2.1 \times10^{-7} 2.1 × 1 0 − 7 ε ≤ 0.06 \varepsilon\le0.06 ε ≤ 0.06 7.4 × 10 − 7 7.4 \times10^{-7} 7.4 × 1 0 − 7 ε = 0.476 \varepsilon= 0.476 ε = 0.476
ω ( ⋅ , T ) − ω ˉ 0 = δ ω ˉ + ω ~ ( ⋅ , T ) , \omega(\cdot,T)-\bar{\omega}_0=\delta\bar{\omega}+\widetilde{\omega}(\cdot,T), ω ( ⋅ , T ) − ω ˉ 0 = δ ω ˉ + ω ( ⋅ , T ) , where ∥ δ ω ˉ ∥ H − 2 = ε 2 R ˉ \lVert\delta\bar{\omega}\rVert_{H^{-2}}=\varepsilon^2\bar{R} ∥ δ ω ˉ ∥ H − 2 = ε 2 R ˉ ε 2 \varepsilon^2 ε 2 Theorem 3.1 (b) drives the oscillation ω ~ \widetilde{\omega} ω T = 100 T=100 T = 100 k 2 = 0 k_2=0 k 2 = 0 § 4.3 . In every run at least 99.99 percent of ∥ ω ~ ( ⋅ , T ) ∥ H − 2 2 \lVert\widetilde{\omega}(\cdot,T)\rVert_{H^{-2}}^2 ∥ ω ( ⋅ , T ) ∥ H − 2 2 k 2 = 0 k_2=0 k 2 = 0 ω ^ ( k 1 , 0 , T ) = i θ ^ ( k 1 , 0 , T ) / ( ν k 1 ) \widehat{\omega}(k_1,0,T)=i\widehat{\theta}(k_1,0,T)/(\nu k_1) ω ( k 1 , 0 , T ) = i θ ( k 1 , 0 , T ) / ( ν k 1 ) H − 2 H^{-2} H − 2 ∥ Q 0 ω ( ⋅ , T ) + ν − 1 ∂ 1 − 1 Q 0 θ ( ⋅ , T ) ∥ H − 2 \lVert Q_0\omega(\cdot,T)+\nu^{-1}\partial_1^{-1}Q_0\theta(\cdot,T)\rVert_{H^{-2}} ∥ Q 0 ω ( ⋅ , T ) + ν − 1 ∂ 1 − 1 Q 0 θ ( ⋅ , T ) ∥ H − 2 0.0015 ∥ Q 0 ω ( ⋅ , T ) ∥ H − 2 0.0015\lVert Q_0\omega(\cdot,T)\rVert_{H^{-2}} 0.0015 ∥ Q 0 ω ( ⋅ , T ) ∥ H − 2 Q 0 Q_0 Q 0 (9) and ∂ 1 − 1 \partial_1^{-1} ∂ 1 − 1 k 1 ≠ 0 k_1\ne0 k 1 = 0 ν ∂ 11 ω + ∂ 1 θ = 0 \nu\partial_{11}\omega+\partial_1\theta=0 ν ∂ 11 ω + ∂ 1 θ = 0 (4) . The H − 2 H^{-2} H − 2 ω ~ ( ⋅ , T ) \widetilde{\omega}(\cdot,T) ω ( ⋅ , T ) 1.8 × 10 − 7 ε 1.8\times10^{-7}\varepsilon 1.8 × 1 0 − 7 ε 2.0 × 10 − 7 ε 2.0\times10^{-7}\varepsilon 2.0 × 1 0 − 7 ε R R R ε \varepsilon ε ε \varepsilon ε δ ω ˉ \delta\bar{\omega} δ ω ˉ ε 2 \varepsilon^2 ε 2 R R R R N O R^{\mathrm{NO}} R NO R R R
Comparison with Fourier neural operators trained without the analysis
To isolate the effect of the design, two further networks were trained that differ from the surrogate only in what the analysis decides. They use the same 10 4 10^4 1 0 4 ω ( ⋅ , T ) \omega(\cdot,T) ω ( ⋅ , T )
L b a s e = 1 B ∑ i = 1 B ∥ ω i N O ( ⋅ , T ) − ω i ( ⋅ , T ) ∥ L 2 ∥ ω i ( ⋅ , T ) ∥ L 2 , (30) \mathcal{L}_{\mathrm{base}}=\frac{1}{B}\sum_{i=1}^{B}\frac{\lVert\omega_i^{\mathrm{NO}}(\cdot,T)-\omega_i(\cdot,T)\rVert_{L^2}}{\lVert\omega_i(\cdot,T)\rVert_{L^2}}, \tag*{(30)} L base = B 1 i = 1 ∑ B ∥ ω i ( ⋅ , T ) ∥ L 2 ∥ ω i NO ( ⋅ , T ) − ω i ( ⋅ , T ) ∥ L 2 , (30) the relative L 2 L^2 L 2 [17 ] . Neither network adds ω ˉ 0 \bar{\omega}_0 ω ˉ 0 δ ω ˉ \delta\bar{\omega} δ ω ˉ ω 0 \omega_0 ω 0 θ 0 \theta_0 θ 0 ω ( ⋅ , T ) \omega(\cdot,T) ω ( ⋅ , T ) ω 0 / ε \omega_0/\varepsilon ω 0 / ε θ 0 / ε \theta_0/\varepsilon θ 0 / ε log ( ε / 0.1 ) \log(\varepsilon/0.1) log ( ε /0.1 ) ω ( ⋅ , T ) / ε \omega(\cdot,T)/\varepsilon ω ( ⋅ , T ) / ε
Table 2 shows the result. Both networks fit the training fields closely, to a relative L 2 L^2 L 2 10 − 3 10^{-3} 1 0 − 3 δ ω ˉ \delta\bar{\omega} δ ω ˉ δ ω ˉ \delta\bar{\omega} δ ω ˉ R ˉ \bar{R} R ˉ ε 0.95 \varepsilon^{0.95} ε 0.95 ε 0.98 \varepsilon^{0.98} ε 0.98 ε 1.17 \varepsilon^{1.17} ε 1.17 ε 1.18 \varepsilon^{1.18} ε 1.18 ω ˉ 0 \bar{\omega}_0 ω ˉ 0 ω ( ⋅ , T ) \omega(\cdot,T) ω ( ⋅ , T )
Table 2. The surrogate and two Fourier neural operators trained on the same data with the loss (4.13), on the 10 3 10^3 1 0 3 ∥ δ ω ˉ ∥ H − 2 \lVert\delta\bar{\omega}\rVert_{H^{-2}} ∥ δ ω ˉ ∥ H − 2 log ϵ \log\epsilon log ϵ
Surrogate investigation of a lower bound for the constant of Theorem 3.1(b)
Theorem 3.1(b) bounds R ˉ \bar{R} R ˉ C C C ν \nu ν C C C 2 × 10 3 2\times10^3 2 × 1 0 3 § 4.6 only show that C ≥ 4.24 × 10 − 6 C\ge4.24\times10^{-6} C ≥ 4.24 × 1 0 − 6 R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO ( ω 0 , θ 0 ) (\omega_0,\theta_0) ( ω 0 , θ 0 ) T T T R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO R ˉ \bar{R} R ˉ ∥ u 0 ∥ H 2 = ∥ θ 0 ∥ H 2 = ϵ / 2 \lVert u_0\rVert_{H^2}=\lVert\theta_0\rVert_{H^2}=\epsilon/2 ∥ u 0 ∥ H 2 = ∥ θ 0 ∥ H 2 = ϵ /2 R ˉ \bar{R} R ˉ ϵ 2 \epsilon^2 ϵ 2 R ˉ \bar{R} R ˉ
First, the surrogate evaluated R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO 10 6 10^6 1 0 6 § 4.2 at each of the five interpolation amplitudes, 5 × 10 6 5\times10^6 5 × 1 0 6 R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO 10 6 10^6 1 0 6 1.36 × 10 − 6 1.36\times10^{-6} 1.36 × 1 0 − 6 1.44 × 10 − 6 1.44\times10^{-6} 1.44 × 1 0 − 6 R ˉ \bar{R} R ˉ Table 1 . However, at each amplitude a few hundred of the 10 6 10^6 1 0 6 R ˉ N O > 4.25 × 10 − 6 \bar{R}^{\mathrm{NO}}>4.25\times10^{-6} R ˉ NO > 4.25 × 1 0 − 6 R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO 5 × 10 6 5\times10^6 5 × 1 0 6 7.34 × 10 − 6 7.34\times10^{-6} 7.34 × 1 0 − 6 R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO R ˉ > 4.25 × 10 − 6 \bar{R}>4.25\times10^{-6} R ˉ > 4.25 × 1 0 − 6 R ˉ = 8.32 × 10 − 6 \bar{R}=8.32\times10^{-6} R ˉ = 8.32 × 1 0 − 6 R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO
Second, starting from the 8 samples with the largest R ˉ N O \bar{R}^{\mathrm{NO}} R ˉ NO log R ˉ N O \log\bar{R}^{\mathrm{NO}} log R ˉ NO ξ k ω \xi_k^\omega ξ k ω ξ k θ \xi_k^\theta ξ k θ
ω ^ 0 ( k ) = c ω ( τ 2 + ∣ k ∣ 2 ) − α ω / 2 ξ k ω , θ ^ 0 ( k ) = c θ ( τ 2 + ∣ k ∣ 2 ) − α θ / 2 ξ k θ \widehat{\omega}_0(k)=c_\omega\left(\tau^2+|k|^2\right)^{-\alpha_\omega/2}\xi_k^\omega,\qquad\widehat{\theta}_0(k)=c_\theta\left(\tau^2+|k|^2\right)^{-\alpha_\theta/2}\xi_k^\theta ω 0 ( k ) = c ω ( τ 2 + ∣ k ∣ 2 ) − α ω /2 ξ k ω , θ 0 ( k ) = c θ ( τ 2 + ∣ k ∣ 2 ) − α θ /2 ξ k θ with α ω = 2.5 \alpha_\omega=2.5 α ω = 2.5 α θ = 3.5 \alpha_\theta=3.5 α θ = 3.5 τ = 7 \tau=7 τ = 7 § 4.2 , where the constants c ω , c θ > 0 c_\omega,c_\theta>0 c ω , c θ > 0 ∥ u 0 ∥ H 2 = ∥ θ 0 ∥ H 2 = ε / 2 \|u_0\|_{H^2}=\|\theta_0\|_{H^2}=\varepsilon/2 ∥ u 0 ∥ H 2 = ∥ θ 0 ∥ H 2 = ε /2 s ≥ 1 s\geq1 s ≥ 1 ∣ ω ^ 0 ( k ) ∣ 2 |\widehat{\omega}_0(k)|^2 ∣ ω 0 ( k ) ∣ 2 ∣ θ ^ 0 ( k ) ∣ 2 |\widehat{\theta}_0(k)|^2 ∣ θ 0 ( k ) ∣ 2 s − 1 2 < ∣ k ∣ ≤ s + 1 2 s-\frac{1}{2}<|k|\leq s+\frac{1}{2} s − 2 1 < ∣ k ∣ ≤ s + 2 1 c ω 2 c_\omega^2 c ω 2 c θ 2 c_\theta^2 c θ 2 [33 ] . Fields with a prescribed power spectrum and modified phases are used in the same way to separate the role of the phases from that of the spectrum [32 ] .
The ascent gives 40 paths in the space of initial data, one from each starting sample. We ran the solver on the 16 paths with the largest R ‾ N O \overline{R}^{\mathrm{NO}} R NO Figure 5 (a)). Up to step 300 the surrogate agrees with the solver to a median relative difference of about 10 percent, and the solver’s R ‾ \overline{R} R R ‾ = 1.87 × 10 − 5 \overline{R}=1.87\times10^{-5} R = 1.87 × 1 0 − 5 256 × 256 256\times256 256 × 256 10 − 5 10^{-5} 1 0 − 5 R ‾ \overline{R} R R ‾ \overline{R} R Figure 5 (b)–(Figure 5 e) shows the case with the largest difference, R ‾ N O = 1.31 × 10 − 5 \overline{R}^{\mathrm{NO}}=1.31\times10^{-5} R NO = 1.31 × 1 0 − 5 R ‾ = 5.9 × 10 − 6 \overline{R}=5.9\times10^{-6} R = 5.9 × 1 0 − 6
Figure 5. Gradient ascent of the surrogate’s R ‾ N O \overline{R}^{\mathrm{NO}} R NO ∣ ω ^ 0 ( k ) ∣ 2 |\widehat{\omega}_0(k)|^2 ∣ ω 0 ( k ) ∣ 2 ∣ θ ^ 0 ( k ) ∣ 2 |\widehat{\theta}_0(k)|^2 ∣ θ 0 ( k ) ∣ 2 R ‾ \overline{R} R R ‾ \overline{R} R 2 × 10 3 2 \times10^3 2 × 1 0 3 θ 0 / ϵ \theta_0/\epsilon θ 0 / ϵ ω 0 / ϵ \omega_0/\epsilon ω 0 / ϵ ϵ = 0.1 \epsilon= 0.1 ϵ = 0.1 ω ( ⋅ , T ) / ϵ \omega(\cdot,T)/\epsilon ω ( ⋅ , T ) / ϵ T = 100 T=100 T = 100
Since Theorem 3.1 (b) bounds R ‾ \overline{R} R R ‾ \overline{R} R ε \varepsilon ε C ≥ 1.87 × 10 − 5 C\geq1.87\times10^{-5} C ≥ 1.87 × 1 0 − 5 ν = η = 1 \nu=\eta=1 ν = η = 1 ε ∗ \varepsilon_* ε ∗
Discussion and conclusions
In this paper we designed a neural operator from a rigorous analysis of the long-time behavior of the two-dimensional anisotropic Boussinesq system on the periodic box T 2 \mathbb{T}^2 T 2 ν \nu ν η \eta η H − 2 H^{-2} H − 2 × 10 3 \times10^3 × 1 0 3 ε 2 \varepsilon^2 ε 2 ε \varepsilon ε 10 4 10^4 1 0 4
The long-time analysis guides the design of the network. First, the network does not learn the final vorticity whole. Its prediction is the initial averaged vorticity ω ‾ 0 \overline{\omega}_0 ω 0 ω ( ⋅ , T ) − ω ‾ 0 \omega(\cdot,T)-\overline{\omega}_0 ω ( ⋅ , T ) − ω 0 ε \varepsilon ε ε 2 \varepsilon^2 ε 2 ε \varepsilon ε T = 100 T=100 T = 100
Departing from the theory leads to failure. When the final temperature was included in the training targets in a preliminary experiment, the training did not converge. When the same network is trained with the standard loss, the error of the whole final vorticity between the network and the solver, it learns the training fields well but misses the quantity bounded by the analysis by a factor of about 30 in the median, and the change of the horizontal average that it predicts grows only about linearly in ϵ \epsilon ϵ
Because the network evaluates an initial datum in milliseconds, against 10.2 s for the solver, and gives the gradient of its prediction with respect to the initial data, it can search the space of initial data far more extensively than the solver. Screening 5 × 10 6 5 \times10^6 5 × 1 0 6 2 × 10 3 2 \times10^3 2 × 1 0 3 C ≥ 1.87 × 10 − 5 C \ge1.87 \times10^{-5} C ≥ 1.87 × 1 0 − 5 ∥ ω ‾ ( ⋅ , t ) − ω ‾ 0 ∥ H − 2 ≤ C ϵ 2 \|\overline{\omega}(\cdot,t)-\overline{\omega}_0\|_{H^{-2}} \le C\epsilon^2 ∥ ω ( ⋅ , t ) − ω 0 ∥ H − 2 ≤ C ϵ 2
The approach extends to other equations for which an analysis identifies a nearly conserved quantity and gives the order in the amplitude of each part of the solution. Such knowledge lets a neural operator learn a small but important quantity accurately, and the trained operator in turn estimates quantities, such as the constants in the estimates, that the analysis does not provide.
Acknowledgments. J. Wu was partially supported by the National Science Foundation (NSF) under grants DMS-2104682 and DMS-2309748. X. Zheng was partially supported by the NSF under grant DMS-2309747. This work was supported in part by computational resources and services provided by the High Performance Computing Center (HPCC) of the Institute for Cyber-Enabled Research at Michigan State University.
We record here the global stability estimate on T 2 \mathbb{T}^{2} T 2 [7 ] , which uses the norms (3.2) and the smallness condition ( ∥ u 0 ∥ H 2 2 + ∥ θ 0 ∥ H 2 2 ) 1 / 2 ≤ ε (\|u_{0}\|_{H^{2}}^{2}+\|\theta_{0}\|_{H^{2}}^{2})^{1/2}\leq\varepsilon ( ∥ u 0 ∥ H 2 2 + ∥ θ 0 ∥ H 2 2 ) 1/2 ≤ ε
Theorem A.1 (Uniform H 2 H^{2} H 2 T 2 \mathbb{T}^{2} T 2 [7 ] ). There is ε ∗ = ε ∗ ( ν , η ) > 0 \varepsilon_{*}=\varepsilon_{*}(\nu,\eta)>0 ε ∗ = ε ∗ ( ν , η ) > 0 0 < ε ≤ ε ∗ 0<\varepsilon\leq\varepsilon_{*} 0 < ε ≤ ε ∗ ( u 0 , θ 0 ) ∈ H 2 ( T 2 ) (u_{0},\theta_{0})\in H^{2}(\mathbb{T}^{2}) ( u 0 , θ 0 ) ∈ H 2 ( T 2 )
( u , θ ) ∈ L ∞ ( 0 , ∞ ; H 2 ( T 2 ) ) ∩ ⋂ 0 ≤ s < 2 C ( [ 0 , ∞ ) ; H s ( T 2 ) ) , (31) (u,\theta)\in L^{\infty}(0,\infty;H^{2}(\mathbb{T}^{2}))\cap\bigcap_{0\leq s<2}C([0,\infty);H^{s}(\mathbb{T}^{2})), \tag*{(31)} ( u , θ ) ∈ L ∞ ( 0 , ∞ ; H 2 ( T 2 )) ∩ 0 ≤ s < 2 ⋂ C ([ 0 , ∞ ) ; H s ( T 2 )) , (31) ∂ t ( u , θ ) ∈ L ∞ ( 0 , ∞ ; L 2 ( T 2 ) ) , \partial_{t}(u,\theta)\in L^{\infty}(0,\infty;L^{2}(\mathbb{T}^{2})), ∂ t ( u , θ ) ∈ L ∞ ( 0 , ∞ ; L 2 ( T 2 )) , and there are an absolute constant C ∗ C_{*} C ∗ C ν C_{\nu} C ν ν \nu ν
sup t ≥ 0 ( ∥ u ( t ) ∥ H 2 + ∥ θ ( t ) ∥ H 2 ) ≤ C ∗ ε , (32) \sup_{t\geq0}\left(\|u(t)\|_{H^{2}}+\|\theta(t)\|_{H^{2}}\right)\leq C_{*}\varepsilon, \tag*{(32)} t ≥ 0 sup ( ∥ u ( t ) ∥ H 2 + ∥ θ ( t ) ∥ H 2 ) ≤ C ∗ ε , (32) \int_{0}^{\infty}\left(\nu|\partial_{1}u|_{H^{2}}^{2}+\eta|\partial_{2}\theta|_{H^{2}}^{2}\right),d\tau\leq C_{\nu}\varepsilon^{2},
{#eq:A.3}
∫ 0 ∞ ∥ ∂ 1 Q 0 θ ∥ L 2 2 d τ ≤ C ν ε 2 . (33) \int_{0}^{\infty}\|\partial_{1}Q_{0}\theta\|_{L^{2}}^{2}\,d\tau\leq C_{\nu}\varepsilon^{2}. \tag*{(33)} ∫ 0 ∞ ∥ ∂ 1 Q 0 θ ∥ L 2 2 d τ ≤ C ν ε 2 . (33) One may take C ∗ = 8 2 C_{*}=8\sqrt{2} C ∗ = 8 2 C ν = 64 ( 2 + ν ) / min { 1 , ν 2 } C_{\nu}=64(2+\nu)/\min\{1,\nu^{2}\} C ν = 64 ( 2 + ν ) / min { 1 , ν 2 } η \eta η ε ∗ \varepsilon_{*} ε ∗
Remark A.2 (The estimate used in the proof). Theorem A.1 yields the following estimate, on which the proof of Theorem 3.1 in Appendix B draws. In addition, the proof of part (b) uses (A.3) directly, so that its constant depends on ν \nu ν C 0 = C 0 ( ν , η ) C_{0}=C_{0}(\nu,\eta) C 0 = C 0 ( ν , η )
sup t ≥ 0 ( ∥ u ( t ) ∥ H 2 2 + ∥ θ ( t ) ∥ H 2 2 ) + 2 ν ∫ 0 ∞ ∥ ∂ 1 u ∥ H 2 2 d τ + 2 η ∫ 0 ∞ ∥ ∂ 2 θ ∥ H 2 2 d τ + ∫ 0 ∞ ∥ ∂ 1 θ ∥ L 2 2 d τ ≤ C 0 ε 2 . (34) \sup_{t\geq0}\left(\|u(t)\|_{H^{2}}^{2}+\|\theta(t)\|_{H^{2}}^{2}\right)+2\nu\int_{0}^{\infty}\|\partial_{1}u\|_{H^{2}}^{2}\,d\tau+2\eta\int_{0}^{\infty}\|\partial_{2}\theta\|_{H^{2}}^{2}\,d\tau+\int_{0}^{\infty}\|\partial_{1}\theta\|_{L^{2}}^{2}\,d\tau\leq C_{0}\varepsilon^{2}. \tag*{(34)} t ≥ 0 sup ( ∥ u ( t ) ∥ H 2 2 + ∥ θ ( t ) ∥ H 2 2 ) + 2 ν ∫ 0 ∞ ∥ ∂ 1 u ∥ H 2 2 d τ + 2 η ∫ 0 ∞ ∥ ∂ 2 θ ∥ H 2 2 d τ + ∫ 0 ∞ ∥ ∂ 1 θ ∥ L 2 2 d τ ≤ C 0 ε 2 . (34) The first three terms are (A.2) and (A.3). Only the last one requires a proof. Decompose ∂ 1 θ = ∂ 1 Q 0 θ + ∂ 1 θ ⊥ \partial_{1}\theta=\partial_{1}Q_{0}\theta+\partial_{1}\theta^{\perp} ∂ 1 θ = ∂ 1 Q 0 θ + ∂ 1 θ ⊥ L 2 ( T 2 ) L^{2}(\mathbb{T}^{2}) L 2 ( T 2 ) x 1 x_{1} x 1 x 2 ↦ ∂ 1 θ ⊥ ( x 1 , x 2 ) x_{2}\mapsto\partial_{1}\theta^{\perp}(x_{1},x_{2}) x 2 ↦ ∂ 1 θ ⊥ ( x 1 , x 2 ) [ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ] ∂ 2 Q 0 θ = 0 \partial_{2}Q_{0}\theta=0 ∂ 2 Q 0 θ = 0
∥ ∂ 1 θ ⊥ ∥ L 2 ≤ ∥ ∂ 2 ∂ 1 θ ⊥ ∥ L 2 = ∥ ∂ 1 ∂ 2 θ ∥ L 2 ≤ ∥ ∂ 2 θ ∥ H 1 . \|\partial_{1}\theta^{\perp}\|_{L^{2}}\leq\|\partial_{2}\partial_{1}\theta^{\perp}\|_{L^{2}}=\|\partial_{1}\partial_{2}\theta\|_{L^{2}}\leq\|\partial_{2}\theta\|_{H^{1}}. ∥ ∂ 1 θ ⊥ ∥ L 2 ≤ ∥ ∂ 2 ∂ 1 θ ⊥ ∥ L 2 = ∥ ∂ 1 ∂ 2 θ ∥ L 2 ≤ ∥ ∂ 2 θ ∥ H 1 . Therefore ∫ 0 ∞ ∥ ∂ 1 θ ⊥ ∥ L 2 2 d τ ≤ η − 1 ∫ 0 ∞ η ∥ ∂ 2 θ ∥ H 2 2 d τ ≤ η − 1 C ν ε 2 \int_{0}^{\infty}\|\partial_{1}\theta^{\perp}\|_{L^{2}}^{2}\,d\tau\leq\eta^{-1}\int_{0}^{\infty}\eta\|\partial_{2}\theta\|_{H^{2}}^{2}\,d\tau\leq\eta^{-1}C_{\nu}\varepsilon^{2} ∫ 0 ∞ ∥ ∂ 1 θ ⊥ ∥ L 2 2 d τ ≤ η − 1 ∫ 0 ∞ η ∥ ∂ 2 θ ∥ H 2 2 d τ ≤ η − 1 C ν ε 2 C 0 = C ∗ 2 + ( 3 + η − 1 ) C ν C_{0}=C_{*}^{2}+(3+\eta^{-1})C_{\nu} C 0 = C ∗ 2 + ( 3 + η − 1 ) C ν ν \nu ν η \eta η
Appendix B. Proof of Theorem 3.1.
We first record Barbalat’s lemma [3 , 11 ] , which turns the finite dissipation integrals of (A.5) into convergence to zero.
Lemma B.1 (Barbalat). Let g : [ 0 , ∞ ) → [ 0 , ∞ ) g:[0,\infty)\to[0,\infty) g : [ 0 , ∞ ) → [ 0 , ∞ ) ∫ 0 ∞ g ( τ ) d τ < ∞ \int_{0}^{\infty}g(\tau)\,d\tau<\infty ∫ 0 ∞ g ( τ ) d τ < ∞ g ( t ) → 0 g(t)\to0 g ( t ) → 0 t → ∞ t\to\infty t → ∞
Proof of part (a)
Proof of Theorem 3.1(a). Global existence, uniqueness and the bound (A.5) are furnished by Theorem A.1 and Remark A.2 of Appendix A. We repeatedly use that, by (A.5), sup t ≥ 0 ( ∥ u ( t ) ∥ H 2 + ∥ θ ( t ) ∥ H 2 ) ≤ C ε \sup_{t\ge0}(\|u(t)\|_{H^2}+\|\theta(t)\|_{H^2})\le C\varepsilon sup t ≥ 0 ( ∥ u ( t ) ∥ H 2 + ∥ θ ( t ) ∥ H 2 ) ≤ C ε C C C ν \nu ν η \eta η
Write v v v u u u θ \theta θ D D D D ∈ { ∂ 1 , ∂ 2 } D\in\{\partial_1,\partial_2\} D ∈ { ∂ 1 , ∂ 2 } ∂ t v \partial_t v ∂ t v L 2 L^2 L 2 ∂ t D v = D ∂ t v \partial_t Dv=D\partial_t v ∂ t D v = D ∂ t v H − 1 H^{-1} H − 1 d d t ∥ D v ∥ L 2 2 = 2 ( ∂ t D v , D v ) L 2 \frac{d}{dt}\|Dv\|_{L^2}^2=2(\partial_t Dv,Dv)_{L^2} d t d ∥ D v ∥ L 2 2 = 2 ( ∂ t D v , D v ) L 2 L 2 L^2 L 2 [31 Ch. III, Lemma 1.2] , in which the pairing is the duality between H − 1 H^{-1} H − 1 H 1 H^1 H 1 w ∈ L l o c 2 ( [ 0 , ∞ ) ; H 1 ( T 2 ) ) w\in L^2_{\mathrm{loc}}([0,\infty);H^1(\mathbb{T}^2)) w ∈ L loc 2 ([ 0 , ∞ ) ; H 1 ( T 2 )) ∂ t w ∈ L l o c 2 ( [ 0 , ∞ ) ; H − 1 ( T 2 ) ) \partial_t w\in L^2_{\mathrm{loc}}([0,\infty);H^{-1}(\mathbb{T}^2)) ∂ t w ∈ L loc 2 ([ 0 , ∞ ) ; H − 1 ( T 2 )) t ↦ ∥ w ( t ) ∥ L 2 2 t\mapsto\|w(t)\|_{L^2}^2 t ↦ ∥ w ( t ) ∥ L 2 2
d d t ∥ w ∥ L 2 2 = 2 ⟨ ∂ t w , w ⟩ H − 1 , H 1 (35) \frac{d}{dt}\|w\|_{L^2}^2=2\langle\partial_t w,w\rangle_{H^{-1},H^1} \tag*{(35)} d t d ∥ w ∥ L 2 2 = 2 ⟨ ∂ t w , w ⟩ H − 1 , H 1 (35) for almost every t t t
Every function g g g g = ∥ w ∥ L 2 2 g=\|w\|_{L^2}^2 g = ∥ w ∥ L 2 2 w = D v w=Dv w = D v w ∈ L ∞ ( 0 , ∞ ; H 1 ) w\in L^\infty(0,\infty;H^1) w ∈ L ∞ ( 0 , ∞ ; H 1 ) ∂ t w = D ∂ t v ∈ L ∞ ( 0 , ∞ ; H − 1 ) \partial_t w=D\partial_t v\in L^\infty(0,\infty;H^{-1}) ∂ t w = D ∂ t v ∈ L ∞ ( 0 , ∞ ; H − 1 ) D D D ∣ d g / d t ∣ = 2 ∣ ∫ T 2 ∂ t v ⋅ D 2 v d x ∣ ≤ 2 ∥ ∂ t v ∥ L 2 ∥ D 2 v ∥ L 2 |dg/dt|=2\left|\int_{\mathbb{T}^2}\partial_t v\cdot D^2v\,dx\right|\le2\|\partial_t v\|_{L^2}\|D^2v\|_{L^2} ∣ d g / d t ∣ = 2 ∫ T 2 ∂ t v ⋅ D 2 v d x ≤ 2∥ ∂ t v ∥ L 2 ∥ D 2 v ∥ L 2 g g g [ 0 , ∞ ) [0,\infty) [ 0 , ∞ )
Step 1: Preservation of the zero means. Write m θ = ∫ T 2 θ m_\theta=\int_{\mathbb{T}^2}\theta m θ = ∫ T 2 θ m u = ∫ T 2 u 2 m_u=\int_{\mathbb{T}^2}u_2 m u = ∫ T 2 u 2 [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) ∂ t ( u , θ ) ∈ L ∞ ( 0 , ∞ ; L 2 ) \partial_t(u,\theta)\in L^\infty(0,\infty;L^2) ∂ t ( u , θ ) ∈ L ∞ ( 0 , ∞ ; L 2 ) T 2 \mathbb{T}^2 T 2 § 3.1 they satisfy m θ ′ = − m u m_\theta'=-m_u m θ ′ = − m u m u ′ = m θ m_u'=m_\theta m u ′ = m θ t t t m θ 2 + m u 2 m_\theta^2+m_u^2 m θ 2 + m u 2 ( m θ 2 + m u 2 ) ′ = − 2 m θ m u + 2 m u m θ = 0 (m_\theta^2+m_u^2)'=-2m_\theta m_u+2m_u m_\theta=0 ( m θ 2 + m u 2 ) ′ = − 2 m θ m u + 2 m u m θ = 0 t = 0 t=0 t = 0 m θ ≡ m u ≡ 0 m_\theta\equiv m_u\equiv0 m θ ≡ m u ≡ 0 u u u ∫ T 2 u 2 = 0 \int_{\mathbb{T}^2}u_2=0 ∫ T 2 u 2 = 0 u ˉ 2 ≡ 0 \bar{u}_2\equiv0 u ˉ 2 ≡ 0 u 2 = u ~ 2 u_2=\widetilde{u}_2 u 2 = u 2
Step 2: Decay of the oscillatory part and of u 2 u_2 u 2 We first show ∥ u ~ ( t ) ∥ H 1 → 0 \|\widetilde{u}(t)\|_{H^1}\to0 ∥ u ( t ) ∥ H 1 → 0 g 1 ( t ) = ∥ ∂ 1 u ( t ) ∥ L 2 2 g_1(t)=\|\partial_1u(t)\|_{L^2}^2 g 1 ( t ) = ∥ ∂ 1 u ( t ) ∥ L 2 2 ∫ 0 ∞ g 1 d τ ≤ ∫ 0 ∞ ∥ ∂ 1 u ∥ H 2 2 d τ < ∞ \int_0^\infty g_1\,d\tau\le\int_0^\infty\|\partial_1u\|_{H^2}^2\,d\tau<\infty ∫ 0 ∞ g 1 d τ ≤ ∫ 0 ∞ ∥ ∂ 1 u ∥ H 2 2 d τ < ∞ g 1 g_1 g 1 g 1 ( t ) → 0 g_1(t)\to0 g 1 ( t ) → 0 ∥ ∂ 1 u ( t ) ∥ L 2 → 0 \|\partial_1u(t)\|_{L^2}\to0 ∥ ∂ 1 u ( t ) ∥ L 2 → 0 ∥ u ~ ∥ L 2 ≤ ∥ ∂ 1 u ∥ L 2 → 0 \|\widetilde{u}\|_{L^2}\le\|\partial_1u\|_{L^2}\to0 ∥ u ∥ L 2 ≤ ∥ ∂ 1 u ∥ L 2 → 0
∥ u ~ ( t ) ∥ H 1 ≤ C ∥ u ~ ∥ L 2 1 / 2 ∥ u ~ ∥ H 2 1 / 2 ≤ C ε 1 / 2 ∥ u ~ ∥ L 2 1 / 2 → 0. \|\widetilde{u}(t)\|_{H^1}\le C\|\widetilde{u}\|_{L^2}^{1/2}\|\widetilde{u}\|_{H^2}^{1/2}\le C\varepsilon^{1/2}\|\widetilde{u}\|_{L^2}^{1/2}\to0. ∥ u ( t ) ∥ H 1 ≤ C ∥ u ∥ L 2 1/2 ∥ u ∥ H 2 1/2 ≤ C ε 1/2 ∥ u ∥ L 2 1/2 → 0. Since u 2 = u ~ 2 u_2=\widetilde{u}_2 u 2 = u 2 ∥ u 2 ( t ) ∥ H 1 ≤ ∥ u ~ ( t ) ∥ H 1 → 0 \|u_2(t)\|_{H^1}\le\|\widetilde{u}(t)\|_{H^1}\to0 ∥ u 2 ( t ) ∥ H 1 ≤ ∥ u ( t ) ∥ H 1 → 0
Step 3: Decay of the temperature. Applying Lemma B.1 to the Lipschitz functions g 2 ( t ) = ∥ ∂ 1 θ ∥ L 2 2 g_2(t)=\|\partial_1\theta\|_{L^2}^2 g 2 ( t ) = ∥ ∂ 1 θ ∥ L 2 2 g 3 ( t ) = ∥ ∂ 2 θ ∥ L 2 2 g_3(t)=\|\partial_2\theta\|_{L^2}^2 g 3 ( t ) = ∥ ∂ 2 θ ∥ L 2 2 ∥ ∂ 1 θ ( t ) ∥ L 2 → 0 \|\partial_1\theta(t)\|_{L^2}\to0 ∥ ∂ 1 θ ( t ) ∥ L 2 → 0 ∥ ∂ 2 θ ( t ) ∥ L 2 → 0 \|\partial_2\theta(t)\|_{L^2}\to0 ∥ ∂ 2 θ ( t ) ∥ L 2 → 0 ∫ T 2 θ ( t ) = 0 \int_{\mathbb{T}^2}\theta(t)=0 ∫ T 2 θ ( t ) = 0 θ ( t ) \theta(t) θ ( t ) k = 0 k=0 k = 0 ∣ k ∣ ≥ 1 |k|\ge1 ∣ k ∣ ≥ 1 T 2 \mathbb{T}^2 T 2
∥ θ ( t ) ∥ L 2 ≤ ∥ ∇ θ ( t ) ∥ L 2 → 0 , \|\theta(t)\|_{L^2}\le\|\nabla\theta(t)\|_{L^2}\to0, ∥ θ ( t ) ∥ L 2 ≤ ∥∇ θ ( t ) ∥ L 2 → 0 , hence ∥ θ ( t ) ∥ H 1 → 0 \|\theta(t)\|_{H^1}\to0 ∥ θ ( t ) ∥ H 1 → 0
Step 4: Decay in H s H^s H s s < 2 s<2 s < 2 For 0 ≤ s ≤ 2 0\le s\le2 0 ≤ s ≤ 2 f ∈ H 2 ( T 2 ) f\in H^2(\mathbb{T}^2) f ∈ H 2 ( T 2 ) ( 1 + ∣ k ∣ 2 ) s = ( ( 1 + ∣ k ∣ 2 ) 2 ) s / 2 (1+|k|^2)^s=((1+|k|^2)^2)^{s/2} ( 1 + ∣ k ∣ 2 ) s = (( 1 + ∣ k ∣ 2 ) 2 ) s /2 ∥ f ∥ H s ≤ ∥ f ∥ L 2 1 − s / 2 ∥ f ∥ H 2 s / 2 \|f\|_{H^s} \le\|f\|_{L^2}^{1-s/2}\|f\|_{H^2}^{s/2} ∥ f ∥ H s ≤ ∥ f ∥ L 2 1 − s /2 ∥ f ∥ H 2 s /2 H 2 H^2 H 2 u 2 ( t ) u_2(t) u 2 ( t ) θ ( t ) \theta(t) θ ( t ) C ε C\varepsilon C ε t t t L 2 L^2 L 2 u 2 ( t ) → 0 u_2(t)\to0 u 2 ( t ) → 0 θ ( t ) → 0 \theta(t)\to0 θ ( t ) → 0 H s H^s H s 0 ≤ s < 2 0\le s<2 0 ≤ s < 2 s < 0 s<0 s < 0 s = 0 s=0 s = 0
Proof of part (b)
Proof of Theorem 3.1(b). Step 1: The vorticity becomes stratified. By Step 2 of the proof of part (a), ∥ u ~ ( t ) ∥ H 1 → 0 \|\widetilde{u}(t)\|_{H^1}\to0 ∥ u ( t ) ∥ H 1 → 0 ω ~ = ω − ω ‾ = ∂ 1 u ~ 2 − ∂ 2 u ~ 1 \widetilde{\omega}=\omega-\overline{\omega}=\partial_1\widetilde{u}_2-\partial_2\widetilde{u}_1 ω = ω − ω = ∂ 1 u 2 − ∂ 2 u 1 Lemma 2.1 (b)), so ∥ ω ( t ) − ω ‾ ( ⋅ , t ) ∥ L 2 = ∥ ω ~ ( t ) ∥ L 2 ≤ C ∥ u ~ ( t ) ∥ H 1 → 0 \|\omega(t)-\overline{\omega}(\cdot,t)\|_{L^2}=\|\widetilde{\omega}(t)\|_{L^2}\le C\|\widetilde{u}(t)\|_{H^1}\to0 ∥ ω ( t ) − ω ( ⋅ , t ) ∥ L 2 = ∥ ω ( t ) ∥ L 2 ≤ C ∥ u ( t ) ∥ H 1 → 0
Step 2: The horizontal average converges. From (14) , for 0 ≤ t 1 ≤ t 2 0\le t_1\le t_2 0 ≤ t 1 ≤ t 2
∥ ω ‾ ( ⋅ , t 2 ) − ω ‾ ( ⋅ , t 1 ) ∥ H − 2 ≤ ∫ t 1 t 2 ∥ ∂ 22 u ~ 1 u ~ 2 ∥ H − 2 d t ≤ ∫ t 1 t 2 ∥ u ~ 1 u ~ 2 ∥ L 2 d t ≤ C ∫ t 1 t 2 ∥ u ~ ( t ) ∥ H 1 2 d t . \begin{aligned}
\|\overline{\omega}(\cdot,t_2)-\overline{\omega}(\cdot,t_1)\|_{H^{-2}}
\le\int_{t_1}^{t_2}\|\partial_{22}\widetilde{u}_1\widetilde{u}_2\|_{H^{-2}}\,dt \\
&\le\int_{t_1}^{t_2}\|\widetilde{u}_1\widetilde{u}_2\|_{L^2}\,dt
\le C\int_{t_1}^{t_2}\|\widetilde{u}(t)\|_{H^1}^2\,dt .
\end{aligned} ∥ ω ( ⋅ , t 2 ) − ω ( ⋅ , t 1 ) ∥ H − 2 ≤ ∫ t 1 t 2 ∥ ∂ 22 u 1 u 2 ∥ H − 2 d t ≤ ∫ t 1 t 2 ∥ u 1 u 2 ∥ L 2 d t ≤ C ∫ t 1 t 2 ∥ u ( t ) ∥ H 1 2 d t . where we used ∥ ∂ 22 g ∥ H − 2 ≤ ∥ g ∥ L 2 \|\partial_{22}g\|_{H^{-2}}\le\|g\|_{L^2} ∥ ∂ 22 g ∥ H − 2 ≤ ∥ g ∥ L 2 ∥ g ~ ∥ L 2 ≤ ∥ g ∥ L 2 \|\widetilde{g}\|_{L^2}\le\|g\|_{L^2} ∥ g ∥ L 2 ≤ ∥ g ∥ L 2 Lemma 2.1 (c)), and the Ladyzhenskaya inequality ∥ u ~ 1 u ~ 2 ∥ L 2 ≤ C ∥ u ~ ∥ H 1 2 \|\widetilde{u}_1\widetilde{u}_2\|_{L^2}\le C\|\widetilde{u}\|_{H^1}^2 ∥ u 1 u 2 ∥ L 2 ≤ C ∥ u ∥ H 1 2 Lemma 2.1 (e), ∥ u ~ ∥ H 1 ≤ 2 ∥ ∂ 1 u ∥ H 1 \|\widetilde{u}\|_{H^1}\le\sqrt{2}\|\partial_1u\|_{H^1} ∥ u ∥ H 1 ≤ 2 ∥ ∂ 1 u ∥ H 1
∫ 0 ∞ ∥ u ~ ∥ H 1 2 d t ≤ 2 ∫ 0 ∞ ∥ ∂ 1 u ∥ H 2 2 d t ≤ 2 ν − 1 C ν ε 2 . \int_0^\infty\|\widetilde{u}\|_{H^1}^2\,dt\le2\int_0^\infty\|\partial_1u\|_{H^2}^2\,dt\le2\nu^{-1}C_\nu\varepsilon^2. ∫ 0 ∞ ∥ u ∥ H 1 2 d t ≤ 2 ∫ 0 ∞ ∥ ∂ 1 u ∥ H 2 2 d t ≤ 2 ν − 1 C ν ε 2 . Therefore the right-hand side of the preceding estimate tends to 0 0 0 t 1 → ∞ t_1\to\infty t 1 → ∞ ω ‾ ( ⋅ , t ) \overline{\omega}(\cdot,t) ω ( ⋅ , t ) H − 2 H^{-2} H − 2 ω ‾ ∞ \overline{\omega}_\infty ω ∞ (14) from 0 0 0 ∞ \infty ∞ (15) . The same estimate with t 1 = 0 t_1=0 t 1 = 0 t 2 = t t_2=t t 2 = t ∥ ω ‾ ( ⋅ , t ) − ω ‾ 0 ∥ H − 2 ≤ C ε 2 \|\overline{\omega}(\cdot,t)-\overline{\omega}_0\|_{H^{-2}}\le C\varepsilon^2 ∥ ω ( ⋅ , t ) − ω 0 ∥ H − 2 ≤ C ε 2 t ≥ 0 t\ge0 t ≥ 0 t 2 = ∞ t_2=\infty t 2 = ∞ ω ‾ ∞ − ω ‾ 0 \overline{\omega}_\infty-\overline{\omega}_0 ω ∞ − ω 0 ν \nu ν
Step 3: The whole field converges to the same limit. By Step 1, ω ( ⋅ , t ) − ω ‾ ( ⋅ , t ) → 0 \omega(\cdot,t)-\overline{\omega}(\cdot,t)\to0 ω ( ⋅ , t ) − ω ( ⋅ , t ) → 0 L 2 ( T 2 ) L^2(\mathbb{T}^2) L 2 ( T 2 ) H − 2 ( T 2 ) H^{-2}(\mathbb{T}^2) H − 2 ( T 2 ) L 2 ↪ H − 2 L^2\hookrightarrow H^{-2} L 2 ↪ H − 2 ω ( ⋅ , t ) → ω ‾ ∞ \omega(\cdot,t)\to\overline{\omega}_\infty ω ( ⋅ , t ) → ω ∞ H − 2 H^{-2} H − 2 ∥ ω ( ⋅ , t ) − ω ‾ 0 ∥ H − 2 → ∥ ω ‾ ∞ − ω ‾ 0 ∥ H − 2 \|\omega(\cdot,t)-\overline{\omega}_0\|_{H^{-2}}\to\|\overline{\omega}_\infty-\overline{\omega}_0\|_{H^{-2}} ∥ ω ( ⋅ , t ) − ω 0 ∥ H − 2 → ∥ ω ∞ − ω 0 ∥ H − 2 C ε 2 C\varepsilon^2 C ε 2
Step 4: Regularity of the limit and convergence in H s H^s H s s < 1 s<1 s < 1 Since ∣ ω ^ k ∣ ≤ ∣ k ∣ ∣ u ^ k ∣ |\widehat{\omega}_k|\le|k||\widehat{u}_k| ∣ ω k ∣ ≤ ∣ k ∣∣ u k ∣ ( 1 + ∣ k ∣ 2 ) ∣ k ∣ 2 ≤ ( 1 + ∣ k ∣ 2 ) 2 (1+|k|^2)|k|^2\le(1+|k|^2)^2 ( 1 + ∣ k ∣ 2 ) ∣ k ∣ 2 ≤ ( 1 + ∣ k ∣ 2 ) 2 ∥ ω ( t ) ∥ H 1 ≤ C ε \|\omega(t)\|_{H^1}\le C\varepsilon ∥ ω ( t ) ∥ H 1 ≤ C ε t t t ω ( t ) \omega(t) ω ( t ) ω ‾ ∞ \overline{\omega}_\infty ω ∞ ∥ ω ‾ ∞ ∥ H 1 ≤ C ε \|\overline{\omega}_\infty\|_{H^1}\le C\varepsilon ∥ ω ∞ ∥ H 1 ≤ C ε ω ‾ ∞ ∈ H 1 ( T 2 ) \overline{\omega}_\infty\in H^1(\mathbb{T}^2) ω ∞ ∈ H 1 ( T 2 ) − 2 ≤ s ≤ 1 -2\le s\le1 − 2 ≤ s ≤ 1 ( 1 + ∣ k ∣ 2 ) s = ( ( 1 + ∣ k ∣ 2 ) − 2 ) ( 1 − s ) / 3 ( 1 + ∣ k ∣ 2 ) ( s + 2 ) / 3 (1+|k|^2)^s=((1+|k|^2)^{-2})^{(1-s)/3}(1+|k|^2)^{(s+2)/3} ( 1 + ∣ k ∣ 2 ) s = (( 1 + ∣ k ∣ 2 ) − 2 ) ( 1 − s ) /3 ( 1 + ∣ k ∣ 2 ) ( s + 2 ) /3
∥ f ∥ H s ≤ ∥ f ∥ H − 2 ( 1 − s ) / 3 ∥ f ∥ H 1 ( s + 2 ) / 3 . \|f\|_{H^s}\le\|f\|_{H^{-2}}^{(1-s)/3}\|f\|_{H^1}^{(s+2)/3}. ∥ f ∥ H s ≤ ∥ f ∥ H − 2 ( 1 − s ) /3 ∥ f ∥ H 1 ( s + 2 ) /3 . With f = ω ( t ) − ω ‾ ∞ f=\omega(t)-\overline{\omega}_\infty f = ω ( t ) − ω ∞ H 1 H^1 H 1 C ε C\varepsilon C ε H − 2 H^{-2} H − 2 ω ( ⋅ , t ) → ω ‾ ∞ \omega(\cdot,t)\to\overline{\omega}_\infty ω ( ⋅ , t ) → ω ∞ H s H^s H s − 2 ≤ s < 1 -2\le s<1 − 2 ≤ s < 1 s < − 2 s<-2 s < − 2 H s H^s H s H − 2 H^{-2} H − 2 s = − 2 s=-2 s = − 2
Appendix C. Proof of Proposition 3.2
Proof. Throughout, k ≠ 0 k\ne0 k = 0 u ^ 2 = − i k 1 ω ^ / ∣ k ∣ 2 \widehat{u}_2=-ik_1\widehat{\omega}/|k|^2 u 2 = − i k 1 ω /∣ k ∣ 2 (16) by − i k 1 / ∣ k ∣ 2 -ik_1/|k|^2 − i k 1 /∣ k ∣ 2 (17) , and the second follows directly. (a) Let k 1 ≠ 0 k_1 \ne0 k 1 = 0 k 2 ≠ 0 k_2 \ne0 k 2 = 0 E k = ∣ u ^ 2 ∣ 2 + ( k 1 2 / ∣ k ∣ 2 ) ∣ θ ^ ∣ 2 E_k = |\hat{u}_2|^2 + (k_1^2/|\mathbf{k}|^2)|\hat{\theta}|^2 E k = ∣ u ^ 2 ∣ 2 + ( k 1 2 /∣ k ∣ 2 ) ∣ θ ^ ∣ 2 (17) ,
d E k d t = − 2 ν k 1 2 ∣ u ^ 2 ∣ 2 − 2 η k 2 2 k 1 2 ∣ k ∣ 2 ∣ θ ^ ∣ 2 ≤ − 2 min { ν k 1 2 , η k 2 2 } E k . \frac{dE_k}{dt} = -2\nu k_1^2|\hat{u}_2|^2 - 2\eta k_2^2\frac{k_1^2}{|\mathbf{k}|^2}|\hat{\theta}|^2 \le-2\min\{\nu k_1^2,\eta k_2^2\}E_k. d t d E k = − 2 ν k 1 2 ∣ u ^ 2 ∣ 2 − 2 η k 2 2 ∣ k ∣ 2 k 1 2 ∣ θ ^ ∣ 2 ≤ − 2 min { ν k 1 2 , η k 2 2 } E k . Since k 1 2 ≥ 1 k_1^2 \ge1 k 1 2 ≥ 1 k 2 2 ≥ 1 k_2^2 \ge1 k 2 2 ≥ 1 min { ν , η } \min\{\nu,\eta\} min { ν , η } k 1 = 0 k_1 = 0 k 1 = 0 k 2 ≠ 0 k_2 \ne0 k 2 = 0 u ^ 2 = − i k 1 ω ^ / ∣ k ∣ 2 = 0 \hat{u}_2 = -ik_1\hat{\omega}/|\mathbf{k}|^2 = 0 u ^ 2 = − i k 1 ω ^ /∣ k ∣ 2 = 0 (17) reduces to d θ ^ / d t = − η k 2 2 θ ^ d\hat{\theta}/dt = -\eta k_2^2\hat{\theta} d θ ^ / d t = − η k 2 2 θ ^
(b) For k 2 = 0 k_2 = 0 k 2 = 0 k 1 ≠ 0 k_1 \ne0 k 1 = 0 (17) is
( − ν k 1 2 1 − 1 0 ) , \begin{pmatrix}
-\nu k_1^2 & 1 \\
-1 & 0
\end{pmatrix}, ( − ν k 1 2 − 1 1 0 ) , with characteristic equation λ 2 + ν k 1 2 λ + 1 = 0 \lambda^2 + \nu k_1^2\lambda+ 1 = 0 λ 2 + ν k 1 2 λ + 1 = 0 ν k 1 2 > 2 \nu k_1^2 > 2 ν k 1 2 > 2 λ ± = ( − ν k 1 2 ± ν 2 k 1 4 − 4 ) / 2 \lambda_\pm= (-\nu k_1^2 \pm\sqrt{\nu^2 k_1^4 - 4})/2 λ ± = ( − ν k 1 2 ± ν 2 k 1 4 − 4 ) /2 θ ^ \hat{\theta} θ ^ λ + \lambda_+ λ + ν k 1 2 + ν 2 k 1 4 − 4 \nu k_1^2 + \sqrt{\nu^2 k_1^4 - 4} ν k 1 2 + ν 2 k 1 4 − 4 0 ≤ ν 2 k 1 4 − 4 ≤ ν k 1 2 0 \le\sqrt{\nu^2 k_1^4 - 4} \le\nu k_1^2 0 ≤ ν 2 k 1 4 − 4 ≤ ν k 1 2 ∣ λ + ∣ |\lambda_+| ∣ λ + ∣ λ − ≤ − ν k 1 2 / 2 \lambda_- \le-\nu k_1^2/2 λ − ≤ − ν k 1 2 /2
