lec22 (1) - Introduction to Simulation Lecture 22 Integral...

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Introduction to Simulation - Lecture 22 Thanks to Deepak Ramaswamy, Michal Rewienski, Xin Wang and Karen Veroy Integral Equation Methods Jacob White
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SMA-HPC ©2003 MIT Outline Integral Equation Methods Exterior versus interior problems Start with using point sources Standard Solution Methods in 2-D Galerkin Method Collocation Method Issues in 3-D Panel Integration
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SMA-HPC ©2003 MIT Interior Versus Exterior Problems Temperature known on surface 2 0 T = inside “Temperature in a Tank” 2 0 T = outside Temperature known on surface “Ice Cube in a Bath” Interior Exterior What is the heat flow? Heat Flow surface T n = Thermal conductivity
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SMA-HPC ©2003 MIT v + - 2 0 Outside ∇ Ψ = is given on Surface Ψ potential What is the capacitance? Capacitance surface n ∂Ψ = Dielectric Permitivity Exterior Problem in Electrostatics
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SMA-HPC ©2003 MIT Resonator Discretized Structure Computed Forces Bottom View Computed Forces Top View Drag Force in a Microresonator Courtesy of Werner Hemmert, Ph.D. Used with permission.
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What is common about these problems. Exterior Problems Drag Force in MEMS device - fluid (air) creates drag. Coupling in a Package - Fields in exterior create coupling Capacitance of a Signal Line - Fields in exterior. Quantities of Interest are on the surface MEMS device - Just want surface traction force Package - Just want coupling between conductors Signal Line - Just want surface charge. Exterior Problem is linear and space-invariant MEMS - Exterior Stokes Flow equation (linear). Package - Maxwell’s equations in free space (linear). Signal Line - Laplace’s equation in free space (linear). But problems are geometrically very complex!
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SMA-HPC ©2003 MIT Exterior Problems Why not use Finite-Difference or FEM methods 2-D Heat Flow Example 0 at T = But, must truncate the mesh T Only need on the surface, but T is computed everywhere n Must truncate the mesh, ( ) 0 becomes ( ) 0 T T R ∞ = = Surface
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SMA-HPC ©2003 MIT Laplace’s Equation Green’s Function In 2-D ( ) ( ) 2 2 2 0 0 0 2 2 2 then = 0 for all , , , , u u u x y z x y z x y z + + In 3-D ( ) ( ) ( ) 2 2 2 0 0 0 1 If u x x y y z z = + + Proof: Just differentiate and see! ( ) ( ) 2 2 0 0 2 2 then + = 0 for all , , u u x y x y x y ( ) ( ) ( ) 2 2 0 0 If log u x x y y = +
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SMA-HPC ©2003 MIT Laplace’s Equation in 2-D Simple Idea 2 2 2 2 + = 0 outside u u x y Surface ( ) 0 0 , x y 2 2 2 2 + = 0 outside u u x y Problem Solved Does not match boundary conditions! is given on surface u ( ) ( ) ( ) 2 2 0 0 Let log u x x y y = +
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