Numerical methods for pdes Integral Equation Methods, Lecture I Discretization of boundary Integral equation Notes by Suvranu De and J. White April 23, 2003
Numerical Methods for PDEs Integral Equation Methods, Lecture 1 Discretization of Boundary Integral Equations Notes by Suvranu De and J. White April 23, 2003
1 Outline for this module Overview of Integral Equation Methods Important for many exterior problems (Fluids, Electromagnetics, Acoustics Quadrature and Cubature for computing integrals One and Two dimensional basics Dealing with Singularities 1st and 2nd Kind Integral Equations Collocation, Galerkin and Nystrom theory Alternative Integral Formulations Ansatz approach and Green's theorem Fast Solvers Fast Multipole and FFT-based methods 2 Outline SLIDE 2 Integral Equation Methods Exterior versus interior problems Start with Standard solution methods Collocation Method Galerkin Method Some issues in 3D 3 Interior vs Exterior Problems SLIde 3 Interior Exterior yn on surtace known on surfac Temperature in a tank""Ice cube in a bath What is the heat Heat flow =Thermal conductivity sur face an Note 1 Why use integral equation methods? For both of the heat conduction examples in the above figure, the temperature, T, is a function of the spatial coordinate, z, and satisfies V2T(a)=0. In both
1 Outline for this Module Slide 1 Overview of Integral Equation Methods Important for many exterior problems (Fluids, Electromagnetics, Acoustics) Quadrature and Cubature for computing integrals One and Two dimensional basics Dealing with Singularities 1st and 2nd Kind Integral Equations Collocation, Galerkin and Nystrom theory Alternative Integral Formulations Ansatz approach and Green’s theorem Fast Solvers Fast Multipole and FFT-based methods. 2 Outline Slide 2 Integral Equation Methods Exterior versus interior problems Start with using point sources Standard Solution Methods Collocation Method Galerkin Method Some issues in 3D Singular integrals 3 Interior Vs Exterior Problems Slide 3 Interior Exterior Temperature known on surface 2 ∇ = T 0 inside 2 ∇ = T 0 outside Temperature known on surfac "Temperature in a tank" "Ice cube in a bath" What is the heat flow? Heat flow = Thermal conductivity surface ∂T ∂n Note 1 Why use integral equation methods? For both of the heat conduction examples in the above figure, the temperature, T , is a function of the spatial coordinate, x, and satisfies ∇2T (x)=0. In both 1
problems T(r)is given on the surface, defined by T, and therefore both problems are Dirichlet problems. For the"temperature in a tank"problem, the problem domain, Q2 is the interior of the cube, and for the "ice cube in a bath"problem, the problem domain is the infinitely extending region exterior to the cube. For such an exterior problem, one needs an additional boundary condition to specify what happens sufficiently far away from the cube. Typically, it is assumed there are no heat sources exterior to the cube and therefore limT(x)→0. For the cube problem, we might only be interested in the net heat flow from he surface. That How is given by an integral over the cube surface of the normal derivative of temperature, scaled by a thermal conductivity. It might eem inefficient to use the finite-element or finite-difference methods discussed in previous sections to solve this problem, as such methods will need to compute the temperature everywhere in Q. Indeed, it is possible to write an integral quation which relates the temperature on the surface directly to its surface normal, as we shall see shortly In the four examples below, we try to demonstrate that it is quite common n applications to have exterior problems where the known quantities and the quantities of interest are all on the surface 4 Examples 4.1 Computation of Capacitance SLIDE 4 potential given on s What is the capac Capacitance= Dielectric Permittivity J Example 1: Capacitance problem In the example in the slide, the yellow plates form a parallel-plate capacitor with an applied voltage V. In this 3-D electrostatics problem, the electrostatic potential y satisfies V-y(r)=0 in the region exterior to the plates, and the otential is known on the surface of the plates. In addition, far from the plates. 4-0. What is of interest is the capacitance, C, which satisfies
problems T (x) is given on the surface, defined by Γ, and therefore both problems are Dirichlet problems. For the “temperature in a tank” problem, the problem domain, Ω is the interior of the cube, and for the “ice cube in a bath” problem, the problem domain is the infinitely extending region exterior to the cube. For such an exterior problem, one needs an additional boundary condition to specify what happens sufficiently far away from the cube. Typically, it is assumed there are no heat sources exterior to the cube and therefore lim x→∞ T (x) → 0. For the cube problem, we might only be interested in the net heat flow from the surface. That flow is given by an integral over the cube surface of the normal derivative of temperature, scaled by a thermal conductivity. It might seem inefficient to use the finite-element or finite-difference methods discussed in previous sections to solve this problem, as such methods will need to compute the temperature everywhere in Ω. Indeed, it is possible to write an integral equation which relates the temperature on the surface directly to its surface normal, as we shall see shortly. In the four examples below, we try to demonstrate that it is quite common in applications to have exterior problems where the known quantities and the quantities of interest are all on the surface. 4 Examples 4.1 Computation of Capacitance Slide 4 v + - 2 ∇Ψ= 0 Outsi Ψ is given on S potential What is the capacitance? Capacitance = Dielectric Permittivity ∂Ψ ∂n Note 2 Example 1: Capacitance problem In the example in the slide, the yellow plates form a parallel-plate capacitor with an applied voltage V . In this 3-D electrostatics problem, the electrostatic potential Ψ satisfies ∇2Ψ(x)=0 in the region exterior to the plates, and the potential is known on the surface of the plates. In addition, far from the plates, Ψ → 0. What is of interest is the capacitance, C, which satisfies q = CV 2
where q, the net charge on one of the plates, is given by the surface normal of the potential integrated over one plate and scaled by a dielectric permittivit 4.2 Drag Force in a Microresonator SLIDE 5 tom v⊥ew Note 3 Example 2: Drag force in a MEMS device The example in the slide is a microresonator, it is a structure that can be made vibrate using electrostatic forces. The changing character of those vibra tions can be used to sense rotation. The particulars of how the microresonato operates is not directly relevant to our discussion of integral equations, except for one point. In order to determine how much energy is needed to keep the microresonator vibrating, it is necessary to determine the fluid drag force on comb structures shown in the bottom part of theslide. The Huid is the air sur rounding the structure, and at the micron-scale of these devices, air satisfies the compressible Stokes equation Vu(a)= Vp(a) (1) V·u(x)=0 the fuid velocity and p is the pressure By specifying the comb velocity, and then computing the surface pressure and the normal derivative of velocity one can determine the net drag force on the comb. Once again, this is a problem in which the known quantities and the quantities of interest are on the surface
where q, the net charge on one of the plates, is given by the surface normal of the potential integrated over one plate and scaled by a dielectric permittivity. 4.2 Drag Force in a Microresonator Slide 5 Resonator Discretized Stru Computed Forces Bottom View Computed Forces Top View Note 3 Example 2: Drag force in a MEMS device The example in the slide is a microresonator, it is a structure that can be made to vibrate using electrostatic forces. The changing character of those vibrations can be used to sense rotation. The particulars of how the microresonator operates is not directly relevant to our discussion of integral equations, except for one point. In order to determine how much energy is needed to keep the microresonator vibrating, it is necessary to determine the fluid drag force on comb structures shown in the bottom part of theslide. The fluid is the air surrounding the structure, and at the micron-scale of these devices, air satisfies the incompressible Stokes equation, ∇2u(x) = ∇p(x) (1) ∇ · u(x)=0 where u is the fluid velocity and p is the pressure.By specifying the comb velocity, and then computing the surface pressure and the normal derivative of velocity, one can determine the net drag force on the comb. Once again, this is a problem in which the known quantities and the quantities of interest are on the surface. 3
4.3 Electromagnetic Coupling in a Package SLIDE 6 Picture Thanks to Coventor Note 4 Example 3: Electromagnetic coupling in a package In the 40 lead electronic package pictured in the slide, it is important to de- termine the extent to which signals on different package leads interact. To determine the magnetic interaction between signal currents flowing on different wires, one must solve H(r)=J(a) V·J(x)=0 where h is the magnetic field and J is the signal current density. By specifying he current, and then computing the magnetic field at the surfaces of the leads one can determine the magnetic interaction. Again, this is a problem in which the known quantities and the quantities of interest are on the surface 4.4 Capacitance of Microprocessor Signal Lines LIDE i Note 5 Example 4: Capacitance of microprocessor signal lines This last example in the above slide is a picture of the wiring on a microprocessor integrated circuit. A typical microprocessor has millions of wires, so we are only looking at a small piece of a processor. The critical problem in this example
4.3 Electromagnetic Coupling in a Package Slide 6 Picture Thanks to Coventor. Note 4 Example 3: Electromagnetic coupling in a package In the 40 lead electronic package pictured in the slide, it is important to determine the extent to which signals on different package leads interact. To determine the magnetic interaction between signal currents flowing on different wires, one must solve ∇2H(x) = J(x) (2) ∇ · J(x)=0 where H is the magnetic field and J is the signal current density. By specifying the current, and then computing the magnetic field at the surfaces of the leads, one can determine the magnetic interaction. Again, this is a problem in which the known quantities and the quantities of interest are on the surface. 4.4 Capacitance of Microprocessor Signal Lines Slide 7 Note 5 Example 4: Capacitance of microprocessor signal lines This last example in the above slide is a picture of the wiring on a microprocessor integrated circuit. A typical microprocessor has millions of wires, so we are only looking at a small piece of a processor. The critical problem in this example 4