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1、5 Potential TheoryReference: Introduction to Partial Differential Equations by G. Folland, 1995, Chap. 3.5.1 Problems of Interest.In what follows, we consider an open, bounded subset of Rn with C2 boundary. We letc =Rn (the open complement of ). We are interested in studying the following fourproble
2、ms:(a) Interior Dirichlet Problem. u = 0 x 2 u = g x 2 :(b) Exterior Dirichlet Problem. u = 0 x 2 cu = g x 2 c:(c) Interior Neumann Problem. u = 0 x 2 u” = g x 2 :(d) Exterior Neumann Problem. u = 0 x 2 cu” = g x 2 c:Previously, we have used Greens representation, to show that if u is a C2 solution
3、of theInterior Dirichlet Problem, then u is given byu(x) = Zg(y)G”y(x;y)dS(y);where G(x;y) is the Greens function for . However, in general, it is difficult to calculatean explicit formula for the Greens function. Here, we use a different approach to look forsolutions to the Interior Dirichlet Probl
4、em, as well as to the other three problems above.Again, its difficult to calculate explicit solutions, but we will discuss existence of solutionsand give representations for them.5.2 Definitions and Preliminary Theorems.As usual, let (x) denote the fundamental solution of Laplaces equation. That is,
5、 let(x) 12 lnjxj n = 21n(n2)fi(n) 1jxjn2 n 3:1Let h be a continuous function on . The single layer potential with moment h isdefined asu(x) = Zh(y)(xy)dS(y): (5.1)The double layer potential with moment h is defined asu(x) = Zh(y)”y(xy)dS(y): (5.2)We plan to use these layer potentials to construct so
6、lutions of the problems listed above.Notice that Greens function gives us a solution to the Interior Dirichlet Problem which issimilar to a double layer potential. We will see that for an appropriate choice of h, we canwrite solutions of the Dirichlet problems (a);(b) as double layer potentials and
7、solutions ofthe Neumann problems (c);(d) as single layer potentials.First, we will prove that for a continuous function h, (5.1) and (5.2) are harmonic func-tions for all x =2 .Theorem 1. For h a continuous function on ,1. u and u are defined for all x 2Rn.2. u(x) = u(x) = 0 for all x =2 .Proof.1. W
8、e prove that u is defined for all x 2Rn. A similar proof works for u.First, suppose x =2 . Therefore, ”y(xy) is defined for all y 2 . Consequently,for all x =2 , we haveju(x)jjh(y)jL1()Zflflflfl”y(xy)flflflfl dS(y) C:Next, consider the case when x is in . In this case, the term ”y(x y) in theintegra
9、nd is undefined at x = y. We prove u is defined at this point x by showing thatthe integral in (5.2) still converges.We need to look for a bound onZh(y)”y(xy)dS(y):Recall(xy) = 12 lnjxyj n = 21n(n2)fi(n) 1jxyjn2 n 3:Therefore,yi(xy) = xi yinfi(n)jy xjn;2and,”y(xy) = ry(xy)”(y)= (xy)”(y)nfi(n)jy xjn;
10、where ”(y) is the unit normal to at y.Claim: Fix x 2 . For all y 2 , there exists a constant C 0 such thatj(xy)”(y)j Cjxyj2:Proof of Claim. By assumption, is C2. This means at each point x 2 , thereexists an r 0 and a C2 function f : Rn1 ! R such that - upon relabelling andreorienting if necessary -
11、 we haveB(x;r) = fz 2 B(x;r)jzn f(z1;:;zn1)g:(See Evans - Appendix C.)xrzR -1a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0a1a0rz = f(z ,.,z )1 n-
12、1nnnWithout loss of generality (by reorienting if necessary), we may assume x = 0 and”(x) = (0;:;0;1). Using the fact that our boundary is C2, we know there exists anr 0 and a C2 function f : B(0;r) Rn1 !R such that is given by the graphof the function f near x.First, consider y 2 such that jxyj r.
13、In this case,j(xy)”(y)jjxyj 1rjxyj2 = C(r)jxyj2:Second, consider y 2 such that jxyj r. In this case, we use the fact thatj(xy)”(y)j = j(xy)(”(x)+”(y)”(x)jj(xy)”(x)j+j(xy)(”(y)”(x)j= jynj+j(xy)(”(y)”(x)j:Now,yn = f(y1;:;yn1)3where f 2 C2, f(0) = 0 and rf(0) = 0. Therefore, by Taylors Theorem, we have
14、jynj = jf(y1;:;yn1)j Cj(y1;:;yn1)j2 Cjyj2= Cjxyj2;where the constant C depends only on the bound on the second partial derivativesof f(y1;:;yn1) for j(y1;:;yn1)j r, but this is bounded because by assumptionf 2 C2(B(0;r).Next, we look at j(xy)(”(y)”(x)j. By assumption, is C2 and consequently,” is a C
15、1 function and therefore, there exists a constant C 0 such thatj”(y)”(x)j Cjy xj:Therefore,j(xy)(”(y)”(x)j Cjy xj2:Consequently, our claim is proven. We remark that the constant C will depend on r,but once x is chosen r is fixed. Therefore, we conclude that for x 2 , all y 2 ,flflflfl”y(xy)flflflfl =flflflfl(xy)”(y)nfi(n)jy xjnflflflfl Cjxyj2jxyjn= Cjxyjn2:Therefore,flflflflZh(y)”y(xy)dS(y)flflflfljh(y)jL1()Zflflflfl”y(xy)flflflfl dS(y) CZ1jxyjn2 dS(y) Cusing the fact that is of di
