10.8 Further Applications of the Divergence Theorem

Transcription

1 10.8 Further Applications of the Divergence Theorem Let D be an open subset of R 3, and let f : D R. Recall that the Laplacian of f, denoted by 2 f, is defined by We say that f is harmonic on D provided 2 f := 2 f x f y f z 2 2 f (x, y, z) = 0 for all (x, y, z) D. Examples of harmonic functions come from electrostatics. 1 / 24

2 Electrostatics Place a unit of electric charge at a fixed point (x 0, y 0, z 0 ). Then the resulting electric field is given by the inverse square law: E = [x x 0, y y 0, z z 0 ] [(x x 0 ) 2 + (y y 0 ) 2 + (z z 0 ) 2 ] 3/2 This vector field is conservative, because a calculation will show that E = f, where the potential f is given by 1 f (x, y, z) = (x x0 ) 2 + (y y 0 ) 2 + (z z 0 ). 2 It s worth singling out this special function of the two variables (x, y, z) and (x 0, y 0, z 0 ). We refer to it as the Green function for R 3. 2 / 24

3 Electrostatics Green function for R 3 Let P = (x, y, z) and Q = (x 0, y 0, z 0 ). The function G(P, Q) = G((x, y, z), (x 0, y 0, z 0 )) = 1 (x x0 ) 2 + (y y 0 ) 2 + (z z 0 ) 2 is called the Green function for R 3. G(P, Q) represents the potential at the point P due to the placement of a point charge at point Q. 1 Note that G(P, Q) = P Q. 3 / 24

4 Exercise. For Q = (x 0, y 0, z 0 ) and P = (x, y, z), we ve defined the Green function G(P, Q) = G((x, y, z), (x 0, y 0, z 0 )) as the potential function associated with the electric field E (x, y, z) that results from placing a point charge at the point (x 0, y 0, z 0 ). Use the above specific formula for the Green function to answer the following questions. a) For fixed Q, how much is lim G(P, Q)? P b) Explain the interpretation of the function P G(P, Q) which relates to work done by the electric field. c) Show that (x, y, z) G((x, y, z), (x 0, y 0, z 0 )) is harmonic at each (x, y, z) (x 0, y 0, z 0 ). 4 / 24

5 More examples of electric fields, potentials, and harmonic functions More generally, suppose we spread electric charge over a bounded surface S, where the charge density is described by means of a function µ(x, y, z) on S, where µ measures charge per unit area on S. Then this sets up an electric field in R 3 and the corresponding potential, denoted by Gµ(x, y, z), is given by Potential due to a charge distribution on a surface S Gµ(x, y, z) = G((x, y, z), (x 0, y 0, z 0 )) µ(x 0, y 0, z 0 )da(x 0, y 0, z 0 ). S This can be written more simply as Gµ(P) = G(P, Q) µ(q) da(q) = S S 1 µ(q) da(q). P Q Note that the corresponding electric field E (x, y, z) is given by E (x, y, z) = (Gµ)(x, y, z). 5 / 24

6 More examples of electric fields, potentials, and harmonic functions Suppose instead we spread electric charge over a bounded solid T, where the charge density is described by means of a function µ(x, y, z) on S, and µ measures charge per unit volume. Then this sets up an electric field in R 3 and the corresponding potential, again denoted by Gµ(x, y, z), is given as follows: Potential due to a charge distribution on a solid T Gµ(x, y, z) = G((x, y, z), (x 0, y 0, z 0 )) µ(x 0, y 0, z 0 )dv (x 0, y 0, z 0 ). T This can be written more simply as Gµ(P) = G(P, Q) µ(q) dv (Q) = T T 1 µ(q) dv (Q). P Q Once again, the corresponding electric field E (x, y, z) is given by E (x, y, z) = (Gµ)(x, y, z). 6 / 24

7 More examples of electric fields, potentials, and harmonic functions Exercise. Explain how you know that the potentials Gµ on the previous two slides are harmonic functions at all points (x, y, z) outside the place where the charge is placed (i.e. outside either S or T ). 7 / 24

8 Poisson Equation Let µ(x, y, z) be a charge density function with domain either a surface S or a solid T. We can view µ(x, y, z) as having domain all of R 3, taking its value to be 0 outside of S or T. Let s refer to the place where µ(x, y, z) is not 0 as the support of µ, i.e. the support of the charge density is the set where we place the charge. So the last exercise indicates that the potentials due to charge distributions are harmonic on the set of points outside of the location of the charge distribution, i.e. if µ(x, y, z) is a charge density function with domain either a surface S or a solid T, then 2 Gµ(x, y, z) = 0 for all (x, y, z) outside the support of µ. Actually these potentials satisfy a more general result known as Poisson s equation: Poisson s Equation 2 Gµ(x, y, z) = µ(x, y, z) for all (x, y, z) 8 / 24

9 Poisson Equation From a mathematical point of view Poisson s equation suggests two mathematically interesting results: 2 Gµ(x, y, z) = µ(x, y, z) 1 If by some means we happen to know the potential Gµ as a function, for example it might be a function which we suspect is a potential, then we can recover the charge density µ by taking the Laplacian of it. 2 For a given function µ(x, y, z) given on some solid T, say we wish to come up with a function F which solves the partial differential equation 2 F (x, y, z) = µ(x, y, z), one such solution F is Gµ, i.e. F (P) = T G(P, Q)µ(Q) dv (Q). 9 / 24

10 Exercise. Let µ(x, y, z) be the charge density of some charge distribution which is spread over some bounded region, either a surface or a solid. Let S be a closed surface in R 3 which entirely contains the region where the charge distribution is located. Let E denote the electric field caused by the above charge distribution. What does the Divergence Theorem tell you about the flux of E out of the region enclosed by S? In order to give the best answer, make use of Poisson s equation described a few slides back. 10 / 24

11 Potential Theory The study of harmonic functions and the functions which arise as potentials due to various charge distributions is known as potential theory. The subject originally arose in connection with certain problems of electrostatics. But in order to put it on a solid mathematical foundation, a lot of mathematics had to be developed. And so as a result the subject took on a mathematical life of its own. 11 / 24

12 Potential Theory For example. how does one give the most general mathematical formulation of a charge distribution? This was answered in the area of mathematics known as measure theory, which was developed in the beginning of the 20th century by Henri Lebesgue. Lebesgue s idea of a measure is precisely what is needed to model the physics idea of a charge distribution. Furthermore, Lebesgue s theory shows that one can integrate functions with respect to measures, so in particular you can define the potential of a measure: Gµ(x, y, z) := G((x, y, z), (x 0, y 0, z 0 ) dµ(x 0, y 0, z 0 ). R 3 But these potentials need not be differentiable at all, yet mathematicians figured out how to make sense of the Poisson equation 2 Gµ = µ i.e. they figured out how to differentiate things which are not differentiable in the usual sense. This was done much later with Laurent Schwartz s theory of distributions. Lebesgue s theory of measure and integration also had a profound impact on the subjects of probability and statistics. 12 / 24

13 The Dirichlet Problem This is one of the fundamental problems of potential theory. We state the problem as follows: Statement of the Dirichlet Problem Let T be a bounded open domain in R 3, with boundary a surface S. Let f : S R. Then the Dirichlet problem consists in extending f continuously and harmonically inside T, i.e. it consists of finding a function F : T S R having the following properties: (i) F is continuous on T S; (ii) F is harmonic on T ; (iii) F = f on S. There are many times one is interested in solving the Dirichlet problem. We next give one problem from electrostatics which is resolved by solving an appropriate Dirichlet problem. 13 / 24

14 Motivation for the Dirichlet Problem: Green functions Recall that the Green function for R 3 1 is given by G(P, Q) = P Q. It represents the potential at P due to the placement of a point charge at Q, where the corresponding electric field is E = G(P, Q) (partial derivatives taken with respect to P). Since G(P, Q) 0 as P, the potential at a point P very far away from Q (we say at the point at ) is 0, so G(P, Q) represents the work done in taking a particle from very far away and bringing it (along any path) to the point P. Using the Green function, as we showed above, we can produce lots of examples of potentials on R 3 by smearing charge over various types of subsets of R 3 and calculating Gµ as a certain integral, where the integral is over the region where the charge is located. 14 / 24

15 Motivation for the Dirichlet Problem: Green functions How can we reformulate these ideas if we wish the entire world to be some bounded open subset T of R 3 rather than all of R 3, and we wish to study potential theory just on T? In order to discuss potentials on T, we should formulate the idea of a Green function for T. And once we formulate it, is there a mathematical way of proving the existence of such a function? 15 / 24

16 Motivation for the Dirichlet Problem: Green functions Let T be a bounded open subset of R 3 with bounding surface S. We want to formulate what is the the Green function for T, denoted by G T (P, Q) where P and Q are points of T. Green function for T The function G T (P, Q) is defined for each P and Q in T as follows: (i) For each Q in T, G T (P, Q) is the potential at P due to the placement of a point charge at Q in T ; (ii) For each Q in T, G T (P, Q) 0 as P goes to any point of the boundary S. Condition (ii) means that the boundary S of T is grounded, i.e. it is held at 0 potential. As a consequence, G T (P, Q) represents the work done in moving a unit charged particle from any boundary point to the point P due to a point charge placed at point Q. We claim that the existence of G T comes down to solving an appropriate Dirichlet problem. 16 / 24

17 Motivation for the Dirichlet Problem: Existence of G T (P, Q) Place a point charge at Q T. Then the potential at any point P R 3 due to this charge at Q is G(P, Q) = 1/ P Q. In order to arrange for S to be grounded, i.e. to force the potential at each P in S to be 0, there must exist a negative charge distribution µ on S with the property that Gµ(P) + Then we can use Gµ(P) for P in T to give us G T : G T (P, Q) = Gµ(P, Q) + 1 = 0 for all P S (1) P Q 1 P Q for all P T. So if we know what is Gµ, then we know the Green function G T (P, Q). Exercise. Which specific Dirichlet problem gives us the desired function Gµ above? 17 / 24

18 Green s Two Identities Let T be a bounded open domain in R 3, with boundary a surface S. Let F (x, y, z) and G(x, y, z) be two scalar fields defined on a neighborhood of S T with continuous second partial derivatives. Then F G is a vector field, namely [FG x, FG y, FG z ]. If we apply the Divergence Theorem to this vector field, we get ( div F ) G dv = F G n da. (2) T In the next exercise we show how to use this to obtain the famous formula known as Green s First Identity. S 18 / 24

19 Green s Two Identities Exercise. 1. Consider the quantity G n on the right side of the above formula. If we evaluate it at a point (x, y, z) on the boundary surface S, it is commonly referred to as the outward pointing normal derivative of G at (x, y, z) and it is denoted by dg (x, y, z). dn Explain why it is described that way. 2. Show why it is the case that div ( F ) G = F G + F 2 G. 3. Use exercises 1 and 2 to rewrite the above statement of the Divergence Theorem in equation (2). The result is known as Green s First Identity. 19 / 24

20 Green s Two Identities T Green s First Identity ( F G + F G) 2 dv = S F dg dn da From Green s First Identity one can deduce the following identity: T Green s Second Identity ( F 2 G G 2 F ) ( dv = F dg dn G df ) da dn S Exercise. Explain how you can deduce Green s Second Identity as a consequence of applying Green s First Identity in the right way. 20 / 24

21 Applications of Green s First Identity Exercise. 1. Suppose we know that G is harmonic on a neighborhood of T S. Explain how we know that dg da = 0. dn S 2. a) Write down what Green s First Identity says in case F = G. b) Show that if F is a solution of the Dirichlet Problem on T S in which F is 0 on S, then F must be identically 0 on T also. c) Show that for any Dirichlet problem on S T, the solution is unique. d) Suppose that f is a function which is harmonic on a neighborhood of T S. Why is it the case that if we merely know the values of f on the boundary surface S, then its values on T are determined? 21 / 24

22 Applications of Green s First Identity Let T be a bounded open subset of R 3 with bounding surface S. Imagine T S is an object which is a conductor of electric charge, i.e. charges are free to move around. Let s distribute charge of total size Q within T S, distributing it in some arbitrary way. The charge is free to move and since like charges repel each other, the charges will move around, until they reach an equilibrium charge distribution, µ(x, y, z). Since the charges repel each other, we expect they will try to get as far from each other as possible, so we would expect the equilibrium charge distribution to have its support on the bounding surface S. Furthermore, physical intuition also suggests that once the charge distribution reaches equilibrium, the total potential at any point of S will be constant. 22 / 24

23 Applications of Green s First Identity Let s denote by µ S the resulting equilibrium charge distribution. As we ve indicated on physical grounds: 1 µ S has its support on S 2 The resulting potential due to this equilibrium charge distribution is constant on S. Exercise. The equilibrium charge distribution sets up an electric field in space. Why is that electric field necessarily equal to 0 on T? 23 / 24

24 Application of Green s Second Identity The following property of harmonic functions was the starting point for a lot of mathematical developments. It is a nice consequence of Green s Second Identity (but the argument is not so obvious). Mean Value Property of Harmonic Functions Let S r (P) denote the sphere of radius r centered at point P R 3. Let F (x, y, z) be harmonic on a neighborhood of this sphere and the ball inside this sphere. Then the average of F over the sphere is equal to its value at the center of the sphere, i.e. 1 F (x, y, z) da = F (P) 4πr 2 S r (P) One can show that the converse of the mean value property also holds, that is, if F is a function with the property that for each sphere (in its domain) the average of F over that sphere is equal to its value at the center of the sphere, then F is harmonic. 24 / 24