is determining how long signals take to get from the output of a logical gate to the input of the next gate. To compute that delay, one must determine the capacitance on each of the wires given in the slide picture. To do so requires computing charges given electrostatic potentials as noted above 5 What is common about these problems slide 8 Exterior Problems IEMS device- fluid(air)creates drag Package - Exterior fields create coupling Signal Line- Exterior fields. Quantities of interest are on surface meMS device-Just want surface traction force Signal Line-Just want surface charge Exterior problem is linear and space-invariant mEMS device- Exterior Stoke's flow equation(linear) Package -Maxwell's equations in free space(linear) Signal line- Laplace's equation in free spce(linear) But problems are geometrically very complex 6 Exterior Problems 6.1 Why not use FDM/ FEM? SLidE 9 2-D Heat Flow Example Only need a on the surface, but T is computed everywhere Must truncate the mesh, = T(oo)=0 becomes T(R)=0 Note 6 Heat conduction in 2D In this slide above. we consider a two dimensional exterior heat conduction problem in which the temperature is known on the edges, or surface, of a square Here, the quantity of interest might be the total heat flow out of the square The temperature T satisfies V2T(x)=0x∈9
is determining how long signals take to get from the output of a logical gate to the input of the next gate. To compute that delay, one must determine the capacitance on each of the wires given in the slide picture. To do so requires computing charges given electrostatic potentials as noted above. 5 What is common about these problems? Slide 8 Exterior Problems MEMS device - fluid (air) creates drag Package - Exterior fields create coupling Signal Line - Exterior fields. Quantities of interest are on 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 device - Exterior Stoke’s flow equation (linear) Package - Maxwell’s equations in free space (linear) Signal line - Laplace’s equation in free spce (linear) But problems are geometrically very complex 6 Exterior Problems 6.1 Why not use FDM / FEM? Slide 9 2-D Heat Flow Example T = ∞ 0 at But, must truncate t mesh Surface Only need ∂T ∂n on the surface, but T is computed everywhere. Must truncate the mesh, ⇒ T(∞)=0 becomes T(R)=0. Note 6 Heat conduction in 2D In this slide above, we consider a two dimensional exterior heat conduction problem in which the temperature is known on the edges, or surface, of a square. Here, the quantity of interest might be the total heat flow out of the square. The temperature T satisfies ∇2T (x)=0 x ∈ Ω (3) 5
T(x) giuen a∈r 1imx-∞T(x)=0 where Q is the infinite domain outside the square and r is the region formed by he edges of the square. Using finite-element or finite-difference methods to solve this problem requires introducing an additional approximation beyond discretization error. It is not possible to discretize all of Q, as it is infinite, and therefore the domain must be truncated with an artificial finite boundary. In the slide, the artificial boundary a large ellipse on which we assume the temperature is zero. Clearly, as the radius of the ellipse increases, the truncated problem more accurately represents the domain problem, but the number of unknowns in the discretization increases
T (x) given x ∈ Γ limx→∞T (x)=0 where Ω is the infinite domain outside the square and Γ is the region formed by the edges of the square. Using finite-element or finite-difference methods to solve this problem requires introducing an additional approximation beyond discretization error. It is not possible to discretize all of Ω, as it is infinite, and therefore the domain must be truncated with an artificial finite boundary. In the slide,the artificial boundary is a large ellipse on which we assume the temperature is zero. Clearly, as the radius of the ellipse increases, the truncated problem more accurately represents the domain problem, but the number of unknowns in the discretization increases. 6
7 Laplace’ s Equation 7.1 Green’ s Function SLIDE 10 血n2D If u=log(V(-Io)2+(y-yo)2 hen岩+=0(x,y)≠(xo,) In 3D If hen+分+=0(x,y,2)≠(x0,,20) Proof: Just differentiate and see! Note 7 Greens function for Laplaces equation In the next few slides, we will use an informal semi-numerical approach to deriving the integral form of Laplace's equation. We do this inpart because such a derivation lends insight to the subsequent numerical procedures To start, recall from basic physics that the potential due to a point charge is related only to the distance between the point charge and the evaluation point In 2-D the potential is given by the log of the distance, and in 3-D the potential is inversely proportion to the distance. The precise formulas are given on the slide. A little more formally, direct differentiation reveals that satisfies the 2-D Laplace's equation everywhere except = To, y= yo and u(, y, 2) (x-x0)2+(y-y0)2+(2-20)2 satisfies the 3-D Laplace's equation everywhere except I=Io, y= yo, 2=20 These functions are sometimes referred to as Greens functions for Laplace's equation b Exercise 1 Show by direct differentiation that the functions in(4)and(5) satisfy V-u=0, in the appropriate dimension almost everywhere
7 Laplace’s Equation 7.1 Green’s Function Slide 10 In 2D If u = log (x − x0)2 + (y − y0)2 then ∂2u ∂x2 + ∂2u ∂y2 = 0 ∀ (x, y) = (x0, y0) In 3D If u = √ 1 (x−x0)2+(y−y0)2+(z−z0)2 then ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2 = 0 ∀ (x, y, z) = (x0, y0, z0) Proof: Just differentiate and see! Note 7 Green’s function for Laplace’s equation In the next few slides, we will use an informal semi-numerical approach to deriving the integral form of Laplace’s equation. We do this inpart because such a derivation lends insight to the subsequent numerical procedures. To start, recall from basic physics that the potential due to a point charge is related only to the distance between the point charge and the evaluation point. In 2-D the potential is given by the log of the distance, and in 3-D the potential is inversely proportion to the distance. The precise formulas are given on the slide. A little more formally, direct differentiation reveals that u(x, y) = log (x − x0)2 + (y − y0)2 (4) satisfies the 2-D Laplace’s equation everywhere except x = x0, y = y0 and u(x, y, z) = 1 (x − x0)2 + (y − y0)2 + (z − z0)2 (5) satisfies the 3-D Laplace’s equation everywhere except x = x0, y = y0, z = z0. These functions are sometimes referred to as Green’s functions for Laplace’s equation. Exercise 1 Show by direct differentiation that the functions in (4) and (5) satisfy ∇2u = 0, in the appropriate dimension almost everywhere. 7
8 Laplaces Equation in 2D 8.1 Simple idea SLIDE 11 u is given on surface out fdex-*)+(ry Does not match boundary conditions! Note 8 Simple idea for solving Laplace's equation Here is a simple idea for computing the solution of La s equation o u(r, y)=alog V(a-xo)2+(y-yo where ro, yo is a point inside the square. Clearly such a u will satisfy Vu=0 outside the square, but u may not match the boundary conditions. By adjusting a, it is possible to make sure to match the boundary condition at at least one b Exercise 2 Suppose the potential on the surface of the square is a constant Can you match that constant potential everywhere on the perimeter of the square by judiciously selecting a. I 8.1.1 More points SLIDE 12 u is given on surface a2u du 0 out (x,y)● (., y.) Letu=∑11lg(√(-)2+(-)=∑aG(x-r;y-) Pick the ais to match the boundary conditions
8 Laplace’s Equation in 2D 8.1 Simple idea Slide 11 2 2 2 2 + = 0 outs u u x y ∂ ∂ ∂ ∂ Surface ( ) x y 0 0 , 2 2 2 2 + = 0 outside u u x y ∂ ∂ ∂ ∂ Problem Solved u is given on surface ( ) ( )( ) 2 2 Let log u xx yy = − +− 0 0 Does not match boundary conditions! Note 8 Simple idea for solving Laplace’s equation in 2D Here is a simple idea for computing the solution of Laplace’s equation outside the square. Simply let u(x, y) = α log (x − x0)2 + (y − y0)2 where x0, y0 is a point inside the square. Clearly such a u will satisfy ∇2u = 0 outside the square, but u may not match the boundary conditions. By adjusting α, it is possible to make sure to match the boundary condition at at least one point. Exercise 2 Suppose the potential on the surface of the square is a constant. Can you match that constant potential everywhere on the perimeter of the square by judiciously selecting α. 8.1.1 "More points" Slide 12 2 2 2 2 + = 0 out u u x y ∂ ∂ ∂ ∂ ( ) x y 1 1 , ( ) x2 2 ,y u is given on surface ( ) x y n n , Let u = n i=1 αi log (x − xi)2 + (y − yi)2 = n i=1 αiG(x − xi, y − yi) Pick the αi’s to match the boundary conditions! 8
ote 9 To construct a potential that satisfies Laplace's equation and matches the undary conditions at more points, let u be represented by the potential due to a sum of n weighted point charges in the square's interior. As shown in the slide, we can think of the potential due to a sum of charges as a sum of Greens functions. Of course, we have to determine the weights on the n point charges and the weight on the i n charge is denoted hereby a 8.1.2 More points equations SLIDE 13 Source Strengths selec to give correct potent - testpoints a(x-x,￥-y G(x-x,x-Y Note 10 contd To determine a system of n equations for the n ais, consider selecting a set of test points, as shown in the slide above. Then, by superposition, for each test point ti, yt,, u(xr4,v,)=∑alog√(xt1-xo)2+(vt4-yo) Writing an equation like(6) for each test point yields the matrix equation or the slide The matrix A in the slide has some properties worth notin . A is dense, that is Ai, never equals zero. This is because every charge contributes to every potential If the test points and the charge points are ordered so that the ith test point is nearest the ith charge, then Ai. i will be larger than Ai i for all j Item 2 above seems to suggest that A is diagonally dominant, but this is not the case. Diagonal dominance requires that the absolute sum of the off-diagonal entries is smaller than the magnitude of the diagonal. The matrix above easil violates that condition
Note 9 ...contd To construct a potential that satisfies Laplace’s equation and matches the boundary conditions at more points, let u be represented by the potential due to a sum of n weighted point charges in the square’s interior. As shown in the slide, we can think of the potential due to a sum of charges as a sum of Green’s functions. Of course, we have to determine the weights on the n point charges, and the weight on the i th charge is denoted hereby αi. 8.1.2 "More points equations" Slide 13 ( ) x y 1 1 , ( ) x y 2 2 , ( ) x y n n , ( ) 1 1 x y t t , Source Strengths selec to give correct potent testpoints. ( )( ) ( )( ) ( ) ( ) 1 1 1 1 11 1 1 1 1 1 , ,, , ,, n n n n nn t t t nt n t t n t t t nt n t t Gx x y y Gx x y y x y Gx x y y Gx x y y x y α α −− − − Ψ = −− − − Ψ Note 10 ...contd To determine a system of n equations for the n αi’s,consider selecting a set of n test points, as shown in the slide above. Then, by superposition, for each test point xti , yti , u(xti , yti ) = n i=1 αi log (xti − x0)2 + (yti − y0)2 = n i=1 αiG(xti − x0, yti − y0). (6) Writing an equation like (6) for each test point yields the matrix equation on the slide. The matrix A in the slide has some properties worth noting: • A is dense, that is Ai,j never equals zero. This is because every charge contributes to every potential. • If the test points and the charge points are ordered so that the i th test point is nearest the i th charge, thenAi,i will be larger than Ai,j for all j. Item 2 above seems to suggest that A is diagonally dominant, but this is not the case. Diagonal dominance requires that the absolute sum of the off-diagonal entries is smaller than the magnitude of the diagonal. The matrix above easily violates that condition. 9