4. Method of Green's function: The semi-infinite region z > 0 is free of charge. Its boundary on the left, the z = 0 plane, is kept at a fixed electrostatic potential using some experimental apparatus installed in the region z 0 and a > 0): Vo ( 1 - x 2 + y 2 V (x, y, z = 0) = a 2 if x2+ yz a2. The zero point of the potential is chosen to be at infinity z - too to the right. See Figure 2. free of charges at z > 0 y V(x = 0, y = 0, z> 0)? Z V(x, y, z = 0) given 2 = 0 Figure 2: Electrostatic potential in semi-infinite free space(a) (1 point) To find the potential V(x, y, z) everywhere for z > 0, we would like to use the method of Green's function. Define a suitable Green's function G(r; r') by writing down the Poisson's equation it satisfies and imposing some appropriate boundary conditions. Explain that the Green's function you seek is unique.(b) (0.5 point) Use the method of images to nd 60'; r'). (c) (1 point) Derive an integral solution for V(x, y, z) at z > 0, in terms of 60'; r') and the given potential V(x, y, z = 0) on the z = 0 plane. (d) (1 point) Use the explicit expression for V(x, y, z = 0) given in Eq. (3) to nd the potential along the positive z axis, ie. V(x = 0,y = 0, z) for all z > 0. 1.10 Formal Solution of Electrostatic Boundary- Value Problem with Green Function The solution of the Poisson or Laplace equation in a finite volume V with either Dirichlet or Neumann boundary conditions on the bounding surface S can be obtained by means of Green's theorem (1.35) and so-called Green functions. In obtaining result (1.36)-not a solution-we chose the function y to be 1/ x - x', it being the potential of a unit point source, satisfying the equation: = -478(x - X') (1.31) X - X' The function 1/|x - x' is only one of a class of functions depending on the variables x and x', and called Green functions, which satisfy (1.31). In general, V'2G(x, X' ) = -478(x - x') (1.39) where G(X, X') = + F(x, X' ) (1.40) X - X' with the function F satisfying the Laplace equation inside the volume Vi V'2F(x, X' ) = 0 (1.41)In facing the problem of satisfying the prescribed boundary conditions on (I) or BID/an, we can nd the key by considering result (1.36). As has been pointed out already, this is not a solution satisfying the correct type of boundary condi- tions because both (I) and atbian appear in the surface integral. It is at best an integral relation for (I). With the generalized concept of a Green function and its additional freedom [via the function F(x, x')], there arises the possibility that we can use Green's theorem with up = G(x, x') and choose F (x, x') to eliminate one or the other of the two surface integrals, obtaining a result that involves Only Dirichlet or Neumann boundary conditions. Of course, if the necessary G(x, x') depended in detail on the exact form of the boundary conditions, the method would have little generality. As will be seen immediately, this is not required, and G(x, x') satises rather simple boundary conditions on S. With Green's theorem (1.35), (b = CD, 1,11 = G(x, x'), and the specied prop- erties of G (1.39), it is simple to obtain the generalization of (1.36): me) = 4,; Vp(x')G(x. x') dsx' i' a. .6 1 (1-4 +4? [G(x,n x')a our) (:')] da' The freedom available in the denition of G (1.40) means that we can make the surface integral depend Only on the chosen type of boundary conditions. Thus, for Dirickiet boundary conditions we demand: GD(x, x') = 0 for x' on S (1.43) Then the rst term in the surface integral in (1.42) vanishes and the solution is 1 3 1 66,, = , * ' c1: d 1. @(x) 4% V pt: )Gse, x ) d 4,, S (x), a ( 44) For Neumonn boundary conditions we must be more careful. The obvious choice of boundary condition on G(x, x') seems to be 66\" for x' on S since that makes the second term in the surface integral in (1.42) vanish, as de sired. But an application of Gauss's theorem to (1.39) shows that E do = - 4n 5 an Consequently the simplest allowable boundary condition on GN is 66 4 an" h?" for x' on s (1.45) where S is the total area of the boundary surface. Then the solution is 1 l l 3 I i 532 41760 Vp(X)G~(x.x)dx + GN do (1.46) 4n where ((13),, is the average value of the potential over the whole surface. The customary Neumann problem is the so-called exterior problem in which the vol- (13(3) = (this + 2.1 Method of Images The method of images concerns itself with the problem of one or more point charges in the presence of boundary surfaces, for example, conductors either grounded or held at xed potentials. Under favorable conditions it is possible to infer from the geometry of the situation that a small number of suitably placed charges of appropriate magnitudes, external to the region of interest, can simu- late the required boundary conditions. These charges are called budge charges, and the replacement of the actual problem with boundaries by an enlarged region with image charges but not boundaries is called the method of images. The image charges must be external to the volume of interest, since their potentials must be solutions of the Laplace equation inside the volume; the \"particular integral\" (i.e., solution of the Poisson equation) is provided by the sum of the potentials of the charges inside the volume. A simple example is a point charge located in front of an innite plane con- ductor at zero potential. as shown in Fig. 2.1. It is clear that this is equivalent to the problem of the original charge and an equal and opposite charge located at the mirror-image point behind the plane dened by the position of the conductor. 57 58 Chapter 2 Boundary-Value Problem in Electrostatics: [81 Figure 2.1 Solution by method of images. The original potential problem is on the left, the equivalent-image problem on the right. 2.1 Method of Images The method of images concerns itself with the problem of one or more point charges in the presence of boundary surfaces, for example, conductors either grounded or held at xed potentials. Under favorable conditions it is possible to infer from the geometry of the situation that a small number of suitably placed charges of appropriate magnitudes, external to the region of interest, can simu- late the required boundary conditions. These charges are called budge charges, and the replacement of the actual problem with boundaries by an enlarged region with image charges but not boundaries is called the method of images. The image charges must be external to the volume of interest, since their potentials must be solutions of the Laplace equation inside the volume; the \"particular integral\" (i.e., solution of the Poisson equation) is provided by the sum of the potentials of the charges inside the volume. A simple example is a point charge located in front of an innite plane con- ductor at zero potential. as shown in Fig. 2.1. It is clear that this is equivalent to the problem of the original charge and an equal and opposite charge located at the mirror-image point behind the plane dened by the position of the conductor. 57 58 Chapter 2 Boundary-Value Problem in Electrostatics: [81 Figure 2.1 Solution by method of images. The original potential problem is on the left, the equivalent-image problem on the right. Green's Functions in PDEs - Green's Functions are used to solve nonhomogeneous PDEs . Poisson Equation : V u = f (x, 4, =) - (1) V Boundary Condition ( Dirichlet) u = h (x, y, = ) - (2 ) Cartesian coordinates on S ( boundary of domain D) G : Green's Function which solves Gou as = SSS ( G. Vu) dv + Iff (GV u) av - (3) this PDF Problem: Dirac-delta. function S D VG = 6 ( p-x, q-4, s-2) oh D G ou u as = SSS (GV'u - UV'G) dV - (4) G = O on S - (6) (5) an an S D p, q, s are dummy Variables For integration