lec22 (1)

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

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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
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
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!
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 TT R ⇒∞ = = Surface

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SMA-HPC ©2003 MIT Laplace’s Equation Green’s Function In 2-D () ( ) 222 00 0 then = 0 for all , , , , uuu xyz x y z ∂∂∂ ++ In 3-D ( ) 22 2 0 1 If u xx yy zz = −+ Proof: Just differentiate and see! ( ) then + = 0 for all , , uu xy x y ∂∂ ( ) ( ) log ux x y y =− +