atmospheric dynamics with
play

Atmospheric Dynamics with Polyharmonic Spline RBFs Greg Barnett - PowerPoint PPT Presentation

Photos placed in horizontal position with even amount of white space between photos and header Atmospheric Dynamics with Polyharmonic Spline RBFs Greg Barnett Sandia National Laboratories is a multimission laboratory managed and operated by


  1. Photos placed in horizontal position with even amount of white space between photos and header Atmospheric Dynamics with Polyharmonic Spline RBFs Greg Barnett Sandia National Laboratories is a multimission laboratory managed and operated by National Technology and Engineering Solutions of Sandia, LLC., a wholly owned subsidiary of Honeywell International, Inc., for the U.S. Department of Energy’s National Nuclear Security Administrati on under contract DE-NA0003525. SAND2017-8166 C

  2. Outline ▪ Polyharmonic Spline (PHS) RBFs with Polynomials ▪ 1D Example ▪ Interpolation and Differentiation Weights ▪ Interpolation/Weights in 2D ▪ Weights on the Sphere ▪ Semi-Lagrangian Transport ▪ Governing Equations ▪ Limiter/Fixer ▪ Test Cases ▪ Results ▪ Eulerian Shallow Water Model ▪ Governing Equations ▪ Test Cases ▪ Results ▪ Conclusions and Future Work 2

  3. Table of RBFs 𝜚 𝒚 , 𝒚 ∈ ℝ 𝑒 , 𝜁 ∈ ℝ Name (acronym) Multiquadric (MQ) 2 1 + 𝜁 𝒚 1 Inverse Quadratic (IQ) 2 1 + 𝜁 𝒚 1 Inverse Multiquadric (IMQ) 2 1 + 𝜁 𝒚 2 𝑓 − 𝜁 𝒚 Gaussian (GA) 𝒚 2𝑙+1 Polyharmonic Spline (PHS) 𝒚 2𝑙 log 𝒚 , 𝑙 ∈ ℕ ⋅ = ⋅ 2 3

  4. Some RBFs in 1D Polyharmonic Spline RBFs Infinitely Differentiable RBFs 4

  5. Example: Equi-spaced Interpolation in 1D PHS Basis Functions Approximation 5

  6. Structure of the Linear System 2 1 𝑦 1 𝑦 1 𝑔 1 2 1 𝑦 2 𝑦 2 𝜈 1 𝑔 2 Least Squares Parabola: 2 𝜈 2 1 𝑦 3 𝑦 3 = 𝑔 3 𝜈 1 + 𝜈 2 𝑦 + 𝜈 3 𝑦 2 𝜈 3 2 𝑔 1 𝑦 4 𝑦 4 4 𝑔 2 1 𝑦 5 𝑦 5 5 2 3 4 1 𝑦 1 𝑦 1 𝑦 1 𝑦 1 𝜈 1 𝑔 1 2 3 4 1 𝑦 2 𝑦 2 𝑦 2 𝑦 2 𝜈 2 𝑔 2 Polynomial Interpolant: 2 3 4 𝜈 3 1 𝑦 3 𝑦 3 𝑦 3 𝑦 3 = 𝑔 𝜈 1 + 𝜈 2 𝑦 + 𝜈 3 𝑦 2 + 𝜈 4 𝑦 3 + 𝜈 5 𝑦 4 3 𝜈 4 2 3 4 𝑔 1 𝑦 4 𝑦 4 𝑦 4 𝑦 4 4 𝜈 5 𝑔 2 3 4 1 𝑦 5 𝑦 5 𝑦 5 𝑦 5 5 𝑦 1 − 𝑦 2 3 𝑦 1 − 𝑦 3 3 𝑦 1 − 𝑦 4 3 𝑦 1 − 𝑦 5 3 2 0 1 𝑦 1 𝑦 1 𝜇 1 𝑔 1 𝑦 2 − 𝑦 1 3 𝑦 2 − 𝑦 3 3 𝑦 2 − 𝑦 4 3 𝑦 2 − 𝑦 5 3 2 0 1 𝑦 2 𝑦 2 𝜇 2 𝑔 2 𝑦 3 − 𝑦 1 3 𝑦 3 − 𝑦 2 3 𝑦 3 − 𝑦 4 3 𝑦 3 − 𝑦 5 3 2 0 1 𝑦 3 𝑦 3 𝜇 3 𝑔 PHS Interpolant: 3 𝑦 4 − 𝑦 1 3 𝑦 4 − 𝑦 2 3 𝑦 4 − 𝑦 3 3 𝑦 4 − 𝑦 5 3 2 𝜇 4 0 1 𝑦 4 𝑦 4 𝑔 5 4 3 + 𝜈 1 + 𝜈 2 𝑦 + 𝜈 3 𝑦 2 𝜇 5 = 2 𝑦 5 − 𝑦 1 3 𝑦 5 − 𝑦 2 3 𝑦 5 − 𝑦 3 3 𝑦 5 − 𝑦 4 3 𝑔 0 1 𝑦 5 𝑦 5 5 ෍ 𝜇 𝑘 𝑦 − 𝑦 𝑘 𝜈 1 𝑘=1 0 1 1 1 1 1 0 0 0 𝜈 2 0 𝑦 1 𝑦 2 𝑦 3 𝑦 4 𝑦 5 0 0 0 𝜈 3 0 2 2 2 2 2 𝑦 1 𝑦 2 𝑦 3 𝑦 4 𝑦 5 0 0 0 6

  7. Properties of Polyharmonic Splines ▪ PHS basis includes both RBFs and polynomials ▪ RBFs improve performance and allow the use of irregular nodes ▪ polynomials give convergence to smooth solutions (no saturation error) ▪ The interpolation problem is guaranteed to have a unique solution provided that polynomials are included up to the required degree, and the nodes are unisolvent. For 𝑙 = 1,2,3, … 𝒚 2𝑙 log 𝒚 ▪ 𝜚 𝒚 = ▪ Polynomials up to degree 𝑙 or higher ▪ 𝒚 2𝑙+1 𝜚 𝒚 = ▪ Polynomials up to degree 𝑙 or higher ▪ Rule of thumb for modest polynomial degrees: Twice as many RBFs as polynomials ▪ Condition number of PHS 𝐵 -matrix is invariant under rotation, translation, and uniform scaling ▪ No need to search for optimal shape parameter 7

  8. Interpolation in 2D 𝒚 = 𝑦, 𝑧 𝑜 𝑜 𝑦 𝑗 , 𝑧 𝑗 and corresponding function values 𝑔 Given nodes , find a linear combination of RBF and 𝑗 𝑗=1 𝑗=1 polynomial basis functions that matches the data exactly. 1. Assume the appropriate form of the underlying approximation: 𝑜 𝑛 Φ 𝑦, 𝑧 = ෍ 𝜇 𝑘 𝜚 𝑘 𝑦, 𝑧 + ෍ 𝜈 𝑙 𝑞 𝑙 𝑦, 𝑧 , 𝑘=1 𝑙=1 where 𝜚 𝑘 𝑦, 𝑧 = 𝜚 𝑦 − 𝑦 𝑘 , 𝑧 − 𝑧 𝑘 . 2. Require Φ to match the data at each node: 𝑜 𝑛 Φ 𝑦 𝑗 , 𝑧 𝑗 = ෍ 𝜇 𝑘 𝜚 𝑘 𝑦 𝑗 , 𝑧 𝑗 + ෍ 𝜈 𝑙 𝑞 𝑙 𝑦 𝑗 , 𝑧 𝑗 = 𝑔 𝑗 , 𝑗 = 1,2,3, … , 𝑜. 𝑘=1 𝑙=1 3. Enforce regularity conditions on the coefficients 𝜇 𝑘 : 𝑜 𝜇 𝑘 𝑞 𝑙 𝑦 𝑘 , 𝑧 𝑘 = 0, 𝑙 = 1,2,3, … , 𝑛. ෍ 𝑘=1 4. Solve the symmetric linear system for 𝜇 𝑘 and 𝜈 𝑙 . 8

  9. Differentiation Weights in 2D Interpolation Problem: 𝝁 𝐁 𝐐 𝝂 = 𝒈 𝑷 , 𝐐 𝑈 𝐏 𝑏 𝑗𝑘 = 𝜚 𝑘 𝑦 𝑗 , 𝑧 𝑗 = 𝜚 𝑦 𝑗 − 𝑦 𝑘 , 𝑧 𝑗 − 𝑧 𝑘 , 𝑗, 𝑘 = 1,2,3, … , 𝑜, 𝑞 𝑗𝑙 = 𝑞 𝑙 𝑦 𝑗 , 𝑧 𝑗 , 𝑗 = 1,2,3, … , 𝑜, 𝑙 = 1,2,3, … , 𝑛. Use 𝑀Φ 𝑦, ෤ ෤ 𝑧 to approximate 𝑀𝑔 𝑦, ෤ ෤ 𝑧 : 𝑜 𝑛 𝑀Φ 𝑦, ෤ ෤ 𝑧 = ෍ 𝜇 𝑘 𝑀𝜚 𝑘 𝑦, ෤ ෤ 𝑧 + ෍ 𝜈 𝑙 𝑀𝑞 𝑙 𝑦, ෤ ෤ 𝑧 𝑘=1 𝑙=1 −1 𝝁 𝐁 𝐐 𝒈 = 𝒄 𝝂 = 𝑷 , 𝒅 𝒄 𝒅 𝐐 𝑈 𝐏 weights 𝑐 𝑘 = 𝑀𝜚 𝑘 𝑦, ෤ ෤ 𝑧 , 𝑘 = 1,2,3, … , 𝑜, 𝑑 𝑙 = 𝑀𝑞 𝑙 𝑦, ෤ ෤ 𝑧 , 𝑙 = 1,2,3, … , 𝑛. 9

  10. Derivative Approximation on the Sphere ▪ Given: nodes 𝒚, 𝒛, 𝒜 on the sphere and function values 𝒈 𝜖 𝜖 𝜖 ▪ Find: differentiation matrices (DMs) 𝐗 𝑦 , 𝐗 𝑧 , 𝐗 𝑨 to approximate 𝜖𝑦 , 𝜖𝑧 , 𝜖𝑨 ▪ Method: Use the fact that 𝛼𝑔 ෤ 𝑦, ෤ 𝑧, ǁ 𝑨 is tangent to the sphere at ෤ 𝑦, ෤ 𝑧, ǁ 𝑨 ▪ For each node, get orthogonal unit vectors 𝒇 ො 𝑦 and 𝒇 ො 𝑧 tangent to the sphere 𝜖 𝜖 ▪ Use 2D method to get matrices 𝐗 ො 𝑦 and 𝐗 ො 𝑧 that approximate 𝑦 and 𝜖 ො 𝜖 ො 𝑧 𝜖𝑔 𝜖𝑔 ▪ 𝛼𝑔 = 𝒇 ො 𝑦 + 𝒇 ො 𝑦 𝑧 𝜖 ො 𝜖 ො 𝑧 𝜖𝑔 𝜖𝑔 𝜖𝑔 𝜖 𝜖 ▪ 𝜖𝑦 = 𝛼𝑔 1 = 𝒇 ො 𝑦 + 𝒇 ො 𝑧 = 𝒇 ො 𝑦 + 𝒇 ො 𝑧 𝑔 𝑦 1 𝑧 1 𝑦 1 𝑧 1 𝜖 ො 𝜖 ො 𝜖 ො 𝜖 ො ▪ 𝐗 𝑦 = diag 𝒇 ො 𝑦 1 𝐗 ො 𝑦 + diag 𝒇 ො 𝑧 1 𝐗 ො 𝑧 ▪ 𝐗 𝑧 = diag 𝒇 ො 𝑦 2 𝐗 ො 𝑦 + diag 𝒇 ො 𝑧 2 𝐗 ො 𝑧 ▪ 𝐗 𝑨 = diag 𝒇 ො 𝑦 3 𝐗 ො 𝑦 + diag 𝒇 ො 𝑧 3 𝐗 ො 𝑧 10

  11. Semi-Lagrangian Transport ▪ Governing Equations (velocity 𝒗 is a known function) 𝜖𝜍 ▪ 𝜖𝑢 = −𝒗 ⋅ 𝛼𝜍 − 𝜍𝛼 ⋅ 𝒗, (Eulerian, short time-steps) 𝐸𝑟 𝜖 ▪ 𝐸𝑢 = 𝜖𝑢 + 𝒗 ⋅ 𝛼 𝑟 = 0. (Semi-Lagrangian, long time-steps) ▪ Quasi-Monotone Limiter for 𝑟 (const. along flow trajectories) (𝑜) and 𝑁 𝑙 = max (𝑜) ▪ 𝑛 𝑙 = min 𝑟 𝑙 ℓ 𝑟 𝑙 ℓ ℓ ℓ (𝑜+1) = min 𝑟 𝑙 (𝑜+1) , 𝑁 𝑙 ▪ Set 𝑟 𝑙 (𝑜+1) = max 𝑟 𝑙 (𝑜+1) , 𝑛 𝑙 ▪ Set 𝑟 𝑙 𝑂 ▪ Mass Fixer tracerMass ≡ σ 𝑙=1 𝜍 𝑙 𝑟 𝑙 𝑊 𝑙 ▪ If tracerMass < initialMass, add mass in cells with 𝑟 𝑙 < 𝑁 𝑙 ▪ If tracerMass > initialMass, subtract mass from cells with 𝑟 𝑙 > 𝑛 𝑙 11

  12. Semi-Lagrangian Transport 𝑢 𝑜+1 𝑢 𝑜 12

  13. Pre-Processing ▪ Get DMs 𝐗 𝑦 , 𝐗 𝑧 , and 𝐗 𝑨 for the continuity equation ▪ Find index of the 𝑜 nearest neighbors to each node ▪ For each node 𝒚 𝑙 = 𝑦 𝑙 , 𝑧 𝑙 , 𝑨 𝑙 , write its 𝑜 neighbors in terms of two orthogonal unit vectors 𝒇 ො 𝑦 , 𝒇 ො 𝑧 tangent to the sphere at 𝒚 𝑙 , and one unit vector 𝒇 Ƹ 𝑨 normal to the sphere at 𝒚 𝑙 , so that 𝒚 𝑙 ℓ = ො 𝑦 𝑙 ℓ 𝒇 ො 𝑦 + ො 𝑧 𝑙 ℓ 𝒇 ො 𝑧 + Ƹ 𝑨 𝑙 ℓ 𝒇 Ƹ 𝑨 , ℓ = 1,2,3, … , 𝑜. −1 ▪ Set 𝐃 𝑙 = 𝐁 𝑙 𝐐 𝑙 𝐉 𝑜 𝐏 6×𝑜 , where 𝑈 𝐐 𝑙 𝐏 6×6 𝐁 𝑙 𝑗𝑘 = 𝜚 𝑦 𝑙 𝑗 − ො ො 𝑦 𝑙 𝑘 , ො 𝑧 𝑙 𝑗 − ො 𝑧 𝑙 𝑘 , 𝑗, 𝑘 = 1,2,3, … , 𝑜, 2 . 2 , ෝ 𝐐 𝑙 = 𝟐, ෝ 𝒚 𝑙 , ෝ 𝒛 𝑙 , ෝ 𝒚 𝑙 ෝ 𝒛 𝑙 , ෝ 𝒚 𝑙 𝒛 𝑙 13

  14. Time Stepping 𝜖𝜍 𝜖𝑢 = −𝒗 ⋅ 𝛼𝜍 − 𝜍𝛼 ⋅ 𝒗 from 𝑢 𝑜 to 𝑢 𝑜+1 using several 1. Step explicit Eulerian time steps (RK3) 2. Step 𝒚 ′ = −𝒗 from 𝑢 𝑜+1 to 𝑢 𝑜 to get departure points (RK4) 3. Find the nearest fixed neighbor to each departure point 4. Use the corresponding pre-calculated cardinal coefficients 𝐃 𝑙 and the newly formed row-vectors 𝒄 𝑙 to get rows of the interpolation matrix 𝐗 𝐗 𝑙⋅ = 𝒄 𝑙 𝐃 𝑙 5. Update 𝑟 on fixed nodes using weights 𝐗 6. Cycle quasi-monotone limiter and mass-fixer until the tracer mass is nearly equal to the initial tracer mass (diff<1e-13) 7. Repeat 14

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend