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5 Potential Theory 5 Potential Theory Reference: Introduction to Partial Differential Equations by G. Folland, 1995, Chap. 3. 5.1 Problems of Interest. In what follows, we consider Ω an open, bounded subset of Rn with C2 boundary. We let Ωc = Rn ¡ Ω (the open complement of Ω). We are interested in studying the following four problems: (a) Interior Dirichlet Problem. ½ ∆u = 0 x 2 Ω u = g x 2 @Ω: (b) Exterior Dirichlet Problem. ½ ∆u = 0 x 2 Ωc u = g x 2 @Ωc: (c) Interior Neumann Problem. ½ ∆u = 0 x 2 Ω @u @º = g x 2 @Ω: (d) Exterior Neumann Problem. ½ ∆u = 0 x 2 Ωc @u c @º = g x 2 @Ω : Previously, we have used Green’s representation, to show that if u is a C2 solution of the Interior Dirichlet Problem, then u is given by Z @G u(x) = ¡ g(y) (x; y) dS(y); @Ω @ºy where G(x; y) is the Green’s function for Ω. However, in general, it is difficult to calculate an explicit formula for the Green’s function. Here, we use a different approach to look for solutions to the Interior Dirichlet Problem, as well as to the other three problems above. Again, it’s difficult to calculate explicit solutions, but we will discuss existence of solutions and give representations for them. 5.2 Definitions and Preliminary Theorems. As usual, let Φ(x) denote the fundamental solution of Laplace’s equation. That is, let ½ 1 ¡ 2¼ ln jxj n = 2 Φ(x) ´ 1 1 n(n¡2)®(n) ¢ jxjn¡2 n ¸ 3: 1 Let h be a continuous function on @Ω. The single layer potential with moment h is defined as Z u(x) = ¡ h(y)Φ(x ¡ y) dS(y): (5.1) @Ω The double layer potential with moment h is defined as Z @Φ u(x) = ¡ h(y) (x ¡ y) dS(y): (5.2) @Ω @ºy We plan to use these layer potentials to construct solutions of the problems listed above. Notice that Green’s function gives us a solution to the Interior Dirichlet Problem which is similar to a double layer potential. We will see that for an appropriate choice of h, we can write solutions of the Dirichlet problems (a); (b) as double layer potentials and solutions of the 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 x2 = @Ω. Theorem 1. For h a continuous function on @Ω, 1. u and u are defined for all x 2 Rn. 2. ∆u(x) = ∆u(x) = 0 for all x2 = @Ω. Proof. 1. We prove that u is defined for all x 2 Rn. A similar proof works for u. First, suppose x2 = @Ω. Therefore, @Φ (x ¡ y) is defined for all y 2 @Ω. Consequently, @ºy for all x2 = @Ω, we have Z ¯ ¯ ¯ ¯ ¯ @Φ ¯ ju(x)j · jh(y)jL1(@Ω) ¯ (x ¡ y)¯ dS(y) · C: @Ω @ºy Next, consider the case when x is in @Ω. In this case, the term @Φ (x ¡ y) in the @ºy integrand is undefined at x = y. We prove u is defined at this point x by showing that the integral in (5.2) still converges. We need to look for a bound on Z @Φ ¡ h(y) (x ¡ y) dS(y): @Ω @ºy Recall ½ 1 ¡ 2¼ ln jx ¡ yj n = 2 Φ(x ¡ y) = 1 1 n(n¡2)®(n) ¢ jx¡yjn¡2 n ¸ 3: Therefore, x ¡ y Φ (x ¡ y) = i i ; yi n®(n)jy ¡ xjn 2 and, @Φ (x ¡ y) = ryΦ(x ¡ y) ¢ º(y) @ºy (x ¡ y) ¢ º(y) = ; n®(n)jy ¡ xjn where º(y) is the unit normal to @Ω at y. Claim: Fix x 2 @Ω. For all y 2 @Ω, there exists a constant C > 0 such that j(x ¡ y) ¢ º(y)j · Cjx ¡ yj2: Proof of Claim. By assumption, @Ω is C2. This means at each point x 2 @Ω, there exists an r > 0 and a C2 function f : Rn¡1 ! R such that - upon relabelling and reorienting if necessary - we have Ω \ B(x; r) = fz 2 B(x; r) j zn > f(z1; : : : ; zn¡1)g: (See Evans - Appendix C.) Ω zn ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ r ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ r z = f(z ,...,z ) ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ n 1 n-1 Rn -1 x Without loss of generality (by reorienting if necessary), we may assume x = 0 and º(x) = (0;:::; 0; 1). Using the fact that our boundary is C2, we know there exists an r > 0 and a C2 function f : B(0; r) ½ Rn¡1 ! R such that @Ω is given by the graph of the function f near x. First, consider y 2 @Ω such that jx ¡ yj ¸ r. In this case, 1 j(x ¡ y) ¢ º(y)j · jx ¡ yj · jx ¡ yj2 = C(r)jx ¡ yj2: r Second, consider y 2 @Ω such that jx ¡ yj · r. In this case, we use the fact that j(x ¡ y) ¢ º(y)j = j(x ¡ y) ¢ (º(x) + º(y) ¡ º(x))j · j(x ¡ y) ¢ º(x)j + j(x ¡ y) ¢ (º(y) ¡ º(x))j = jynj + j(x ¡ y) ¢ (º(y) ¡ º(x))j: Now, yn = f(y1; : : : ; yn¡1) 3 where f 2 C2, f(0) = 0 and rf(0) = 0. Therefore, by Taylor’s Theorem, we have jynj = jf(y1; : : : ; yn¡1)j 2 · Cj(y1; : : : ; yn¡1)j · Cjyj2 = Cjx ¡ yj2; where the constant C depends only on the bound on the second partial derivatives of f(y1; : : : ; yn¡1) for j(y1; : : : ; yn¡1)j · r, but this is bounded because by assumption f 2 C2(B(0; r)). Next, we look at j(x ¡ y) ¢ (º(y) ¡ º(x))j. By assumption, @Ω is C2 and consequently, º is a C1 function and therefore, there exists a constant C > 0 such that jº(y) ¡ º(x)j · Cjy ¡ xj: Therefore, j(x ¡ y) ¢ (º(y) ¡ º(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 @Ω, all y 2 @Ω, ¯ ¯ ¯ ¯ ¯ @Φ ¯ ¯ (x ¡ y) ¢ º(y) ¯ ¯ (x ¡ y)¯ = ¯ ¯ ¯ ¯ ¯ n ¯ @ºy n®(n)jy ¡ xj jx ¡ yj2 · C jx ¡ yjn C = : jx ¡ yjn¡2 Therefore, ¯ Z ¯ Z ¯ ¯ ¯ @Φ ¯ ¯ @Φ ¯ ¯¡ h(y) (x ¡ y) dS(y)¯ · jh(y)j 1 ¯ (x ¡ y)¯ dS(y) ¯ @º ¯ L (@Ω) ¯@º ¯ @Ω y Z @Ω y 1 · C n¡2 dS(y) · C @Ω jx ¡ yj using the fact that @Ω is of dimension n ¡ 1. Therefore, we conclude that u is defined for all x 2 @Ω and consequently for all x 2 Rn as claimed. 2. Next, we will prove that ∆u(x) = 0 for all x 2 Ω. A similar proof works to prove that ∆u(x) = 0. Fix x 2 Ω. We note that for all y 2 @Ω, @Φ (x¡y) is a smooth function. Further, using @ºy the fact that Φ(x ¡ y) is harmonic for all x 6= y, we conclude that ∆ @Φ (x ¡ y) = 0 x @ºy 4 for all y 2 @Ω. Therefore, using the fact that our integral is finite and @Φ (x ¡ y) is @ºy smooth, we conclude that Z @Φ ∆ u(x) = ¡∆ h(y) (x ¡ y) dS(y) x x @º Z @Ω y @Φ = ¡ h(y)∆x (x ¡ y) dS(y) @Ω @ºy = 0: ¤ In the above theorem, we showed that as long as h is a continuous function on @Ω, then u and u, defined in (5.1) and (5.2), respectively, are harmonic functions on Ω. Consequently, if we can choose h appropriately so that our initial condition will be satisfied, then we can find a solution of our particular problem ((a), (b), (c), or (d)). We claim that by choosing h appropriately, u will give us a solution of our interior or exterior Dirichlet problem. Similarly, we will show that by choosing h appropriately, u will give us a solution of our interior or exterior Neumann problems. For a moment, consider the interior Dirichlet problem (a). As proven above, for h a con- tinuous function on @Ω, u defined in (5.2) is harmonic. Now, if we can choose h appropriately, such that for all x0 2 @Ω, lim u(x) = g(x0); x2Ω!x0 then we will have found a solution of the interior Dirichlet problem. Consequently, we are interested in studying the limits of u as we approach the boundary of Ω. In order to study this, we must first prove the following lemma. Lemma 2. (Gauss’ Lemma) Consider the double layer potential, Z @Φ v(x) = ¡ (x ¡ y) dS(y): @Ω @ºy Then, 8 c <> 0 x 2 Ω v(x) = 1 x 2 Ω > : 1=2 x 2 @Ω: Proof. 1. First, for x 2 Ωc, Z @Φ v(x) = ¡ (x ¡ y) dS(y) @º Z@Ω y = ¡ ∆yΦ(x ¡ y) dy Ω = 0 using the Divergence Theorem and the fact that Φ(x ¡ y) is smooth for y 2 Ω, x 2 Ωc. 5 2. Now, for x 2 Ω, Φ(x ¡ y) is not smooth for all y 2 Ω. In order to overcome this problem, we fix ² > 0 sufficiently small such that B(x; ²) is contained within Ω. Then on the region Ω ¡ B(x; ²), Φ(x ¡ y) is smooth, and, consequently, we can say Z 0 = ∆yΦ(x ¡ y) dy ZΩ¡B(x;²) @Φ = (x ¡ y) dS(y) @º Z@(Ω¡B(x;²)) y Z @Φ @Φ = (x ¡ y) dS(y) + (x ¡ y) dS(y) @Ω @ºy @B(x;²) @ºy where º is the outer unit normal to Ω ¡ B(x; ²).
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