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Chapter 3: Mathematics of the Poisson Equations

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Chapter 3: Mathematics of the Poisson Equations 3 Mathematics of the Poisson Equation 3.1 Green functions and the Poisson equation (a) The Dirichlet Green function satisfies the Poisson equation with delta-function charge 2 3 − r GD(r; ro) = δ (r − ro) (3.1) and vanishes on the boundary. It is the potential at r due to a point charge (with unit charge) at ro in the presence of grounded (Φ = 0) boundaries The simplest free space green function is just the point charge solution 1 Go = (3.2) 4πjr − roj In two dimensions the Green function is −1 G = log jr − r j (3.3) o 2π o which is the potential from a line of charge with charge density λ = 1 (b) With Dirichlet boundary conditions the Laplacian operator is self-adjoint. The dirichlet Green function is symmetric GD(r; r0) = GD(r0; r). This is known as the Green Reciprocity Theorem, and appears in many clever ways. The intuitive way to understand this is that for grounded boundary b.c. −∇2 is a real self adjoint 2 3 operator (i.e. a real symmetric matrix). Now −∇ GD(r; r0) = δ (r − r0), so in a functional sense 2 GD(r; r0) is the inverse matrix of −∇ . The inverse of a real symmetric matrix is also real and sym- metric. If the Laplace equation with Dirichlet b.c. is discretized for numerical work, these statements become explicitly rigorous. (c) The Poisson equation or the boundary value problem of the Laplace equation can be solved once the Dirichlet Green function is known. Z Z 3 Φ(r) = d xo GD(r; ro)ρ(ro) − dSo no · rro GD(r; ro)Φ(ro) (3.4) V @V where no is the outward directed normal. The first term is a volume integral and is the contribution of the interior charges on the potential. The second term is a surface integral, and is the contribution of the boundary value to the interior. (d) A useful technique to find a Green function is image charges. You should know the image charge green functions i) A plane in 1D and 2D (class) ii) A sphere (homework) iii) A cylinder (homework + recitation) 7 8 CHAPTER 3. MATHEMATICS OF THE POISSON EQUATION (e) The Green function can always be written in the form G(r; ro) = Go(r; ro) +Φind(r; ro) (3.5) | {z } 1 4πjr−roj 2 where the induced potential, Φind(r; ro), is regular and satisfies the homogeneous equation −∇ Φind = 0. The force of a point charge q and the grounded boundaries (i:e: between the charge q and the induced charges on the grounded surfaces) is entirely due to the induced potential1 2 F = − q rrΦind(r; r0) (3.6) r=r0 2 Using the green reciprocity theoem Φind(r; r0) = Φind(r0; r), we can write F = −∇r0 Uint(r0) (3.8) where 1 2 1 2 Uint(ro) = q Φind(ro; ro) = q lim (G(r; ro) − Go(r; ro)) (3.9) 2 2 r!ro (f) Finding the Green function by separation of variables This is best illustrated by example. Pick two 3 dimensions of a surface (say θ; φ). The method is motivated by the fact that δ (r − ro) can be written as a sum 1 1 X δ3(r − r ) = δ(r − r )δ(cos θ − cos θ )δ(φ − φ ) = δ(r − r ) Y (θ; φ)Y ∗ (θ ; φ ) (3.10) o r2 o o o r2 o `m `m o o `m Thus the green function is can also be written as 1 ` X X ∗ G(r; ro) = g`m(r; ro)Y`m(θ; φ)Y`m(θo; φo) (3.11) `=0 −` leading to an equation for g`m(r; ro) 1 @ @ `(` + 1) 1 − r2 + g (r; r ) = δ(r − r ) (3.12) r2 @r @r r2 `m o r2 o This remaining equation in 1D is then solved for the green function following the strategy outlined in Sect. 3.2 (see Eq. (3.37)). This depends on the conditions boundary conditions. Similar expressions can be derived in other coordinates. `+1 ` 2 (g) For free space, the two solutions to Eq. (3.12) are yout(r) = 1=r and yin(r) = r , p(r) = r and p(r)W (r) = 2` + 1. Then the free space Green fcn can be written 1 ` ` 1 X X ∗ 1 r< = [Y`m(θ; φ)Y`m(θo; φo)] `+1 (3.13) 4πjr − roj 2` + 1 `=0 −` r> Some useful identities can be derived from Eq. (3.13): 1 The green function is the potential for a unit charge q = 1. The induced charges are proportional to q. The electro-static field from the induced charges is Eind(r; r0) = −qrrΦind(r; ro) while the force on q is F = qEind(r0; r0). 2 We use h i r Φ (r r ) = 1 r Φ (r r ) + r Φ (r r) = 1 r Φ (r r ) (3.7) r ind ; 0 2 r ind ; 0 r ind 0; 2 r0 ind 0; 0 r=r0 r=r0 3.2. SOLVING THE LAPLACE EQUATION BY SEPARATION 9 i) The generating function of Legendre Polynomials is found by setting ro = z^ and r < 1 with p Y`0 = (2` + 1)=4πP`(cos θ) 1 1 X ` p = r P`(cos θ) (3.14) 2 1 + r − 2r cos θ `=0 ii) The spherical harmonic additionp theorem which we find by writing by setting ro = 1 and r < 1 2 and using 1=jr − roj = 1= 1 + r − 2rr^ · r^o ` 4π X P (r^ · r^ ) = Y (θ; φ)Y ∗ (θ φ ) (3.15) ` o 2` + 1 `m `m o o m=−` where r^ · r^o is the cosine of the angle between the two vectors. iii) The shell structure relation which you find by setting r^ = r^o ` 4π X 1 = Y (θ; φ)Y ∗ (θ; φ) (3.16) 2` + 1 `m `m m=−` This relation is what is responsible for shell structure in the periodic table (h) Similar expansion exists in other coordinates. e.g. in cylindrical coords yout(ρ) = Km(κρ) and yin(ρ) = Im(κρ), leading to 1 1 1 X Z dk h i = eim(φ−φo)eik(z−zo) I (kρ )K (kρ ) (3.17) 4πjr − r j 2π 2π m < m > o m=−∞ 3.2 Solving the Laplace Equation by Separation A summary of separation of variables in different coordinate systems is given in AppendixD. The most important case is spherical and cartesian coordinates. Solving the Laplace equation We use a technique of separation of variables in different coordinate systems. The technique of separation of variables is best illustrated by example. For instance consider a potential in a square geometry. The b a ϕ = 0 on sides z y ϕ (x,y) x o specified on bottom Figure 3.1: A rectangle illustrating separation of vars potential Φ(x; y; z) is specified at z = 0 to be Φo(x; y) and zero on the remaining boundaries (a) We look for solutions of the separated form Φ = Z(z) X(x)Y (y) (3.18) | {z } | {z } ? to surf k to surf 10 CHAPTER 3. MATHEMATICS OF THE POISSON EQUATION Substituting this into the laplace equation, and separating variables gives two equations for X, Y (the parallel directions) d2 − − k2 X(x) =0 ; (3.19) dx2 n d2 − − k2 Y (x) =0 : (3.20) dy2 m and one equation for the perpendicular equation d2 − + k2 Z(z) =0 ; (3.21) dz2 z 2 2 2 where kz = kn + km. The signs of kx; ky; kz are chosen for later convenience, because it will be 2 2 impossible to satisfy the BC for kx < 0 or ky < 0. The first step is always to separate variables and write down the general solutions to the separated equations X(x) =A cos(knx) + B sin(knx) (3.22) Y (y) =A cos(kmy) + B sin(kmy) (3.23) Z(z) =Ae−kz z + Bekz z (3.24) (b) It is best to analyze the parallel equations first which are all of the form of a Sturm Louiville eigen- value equation (see below). These determine the (eigen) functions X(x);Y (y) and the eigenvalues (or separation constants) kx and ky. The general solution for X(x) is X(x) = A cos kxx + B sin kxx ; (3.25) and we are specifying boundary conditions at x = 0 and x = a. In order to satisfy the boundary condition X(0) = X(a) = 0, we must have A = 0 and k = nπ=a, leading to nπ X(x) = B sin(k a) k = n = 1; 2;:::: (3.26) n n a Similarly mπ Y (y) = B sin(k a) k = m = 1; 2;::: (3.27) m m a Thus the parallel directions determine both the functions and the separation constants. The complete eigen functions are nπx mπy (x; y) = sin sin n = 1 ::: 1 m = 1 ::: 1 nm a b (c) Finally we return to the perpendicular direction, Eq. (3.21). This equation does not usually constrain the separation constants. The general solution is Z(z) = Aekz z + Be−kz z (3.28) p 2 2 with kz = kn + km. With Z(z) specified The general solution then is a linear combination 1 1 X X −γnmz +γnmz Anme + Bnme nm(x; y) (3.29) n=1 m=1 3.2. SOLVING THE LAPLACE EQUATION BY SEPARATION 11 Solving the separated equations: After separating variables, all of the equations we wil study can be written in Sturm Louiville form: −d d p(x) + q(x) y(x) = λr(x)y(x) (3.30) dx dx where p(x) and r(x) are postive definite fcns.
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