harmonic
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harmonic

Course Number: MATH 517, Fall 2009

College/University: Rutgers

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MATH517: PDE SOME NOTES ON HARMONIC FUNCTIONS Overview. In the next couple lectures, we aim to develop methods to solve boundary value problem of the kind u = f in U , (1) u = g on U . In the beginning lectures, we discussed the separation of variables method to solve (1) in the case U is a round disk in R2 and f 0 in U . The same method can also be made to work on higher dimensional round balls, and the...

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PDE SOME MATH517: NOTES ON HARMONIC FUNCTIONS Overview. In the next couple lectures, we aim to develop methods to solve boundary value problem of the kind u = f in U , (1) u = g on U . In the beginning lectures, we discussed the separation of variables method to solve (1) in the case U is a round disk in R2 and f 0 in U . The same method can also be made to work on higher dimensional round balls, and the eigenfunction expansion method can be used to solve (1) on such domains when f is not identically 0. However these methods leave us no clue as for how to approach (1) on general domains. The first property of solutions to (1) that we will develop is the maximum principle, which states that for a bounded domain U , there exists a constant C > 0 depending only on the domain U such that |u|C 0 (U ) |u|C 0 (U ) + C|u|C 0 (U ) . (2) One immediate consequence of the maximum principle is the uniqueness of solution to (1) in the class C 2 (U ) C 0 (U ): suppose u, v C 2 (U ) C 0 (U ) are solutions to (1), then |u - v|C 0 (U ) = 0, therefore |u - v|C 0 (U ) |u - v|C 0 (U ) + C|(u - v)|C 0 (U ) = 0, thus u v in U . Another consequence of the maximum principle is the following convergence property: suppose that uk C 2 (U ) C 0 (U ) is the unique solution to (1) with fk , gk replacing f, g, respectively, and suppose that there exists f C 0 (U ) and g C 0 (U ) such that fk f 0 0 in C (U ) and gk g in C (U ), then we see through (2) applied to uk - ul that {uk } is Cauchy in C 0 (U ), therefore, there exists a limit u C 0 (U ) with u = g on U . In order for u to solve (1) in the classical sense, we need to find conditions which guarantee that uk u 2 not only in C 0 (U ), but also in Clocal (U ), at least after extracting a subsequence. This can be achieved via Arzela-Ascoli theorem if we can prove that derivatives of uk up to second order are equicontinuous on any compact subset of U . This result can be proved relatively easily for solutions to (1) when f 0 in U , namely, for harmonic functions. See Theorem 4 below. The precise statements on the convergence of solutions to (1) are given in Theorems 6 and 7 below. The equicontinuity of derivatives of solutions of (1) for general f requires some control on the modulus of continuity of f , and can be developed using potential representation. There is a theory, called the Schauder theory, that generalizes such estimates to solutions of elliptic equations with Hlder continuous variable coefficients. We will not have time to o develop this theory in this course. Our main focus will be to find conditions on U and g such that we have a reasonably complete result on the solvability of (1) for the case f 0. Although the convergence result in Theorem 7 looks like a plausible approach, a complete result would require the solvability of (1) for a dense set of data. This can be provided by Poincar's method of balayage. An alternative approach is to use Perron's method which e involves the concept of subharmonic functions. In order to develop the derivatives estimates for harmonic functions in Theorem 4, we first discuss Green's representation and some of 1 2 its consequences, including the regularity properties of harmonic functions, the concept of Green's functions, and the construction of Green's function on round balls and half spaces in any dimension. First we state the Green's identity. Proposition 1. Suppose that U is a bounded domain with piecewise C 1 boundary, and u, v C 2 (U ). Then [u(x)v(x) - v(x)u(x)] dx = U U u(x) u(x) v(x) - v(x) d(x). n(x) n(x) (3) Remark 1. (3) continues to hold when u and v satisfy the weaker condition that u, v C 1 (U ) C 2 (U ) as long as u and v are in L1 (U ). Suppose that u C 2 (U ) is harmonic in U , and BR (x0 ) U , then applying (3) to u and v 1 on Br (x0 ) for any 0 < r < R, we obtain u(x) 0= d(x) Br (x0 ) n(x) = rn-1 B1 (0) u(x0 + r) d r u(x0 + r)d , = rn-1 r B1 (0) from which it follows that B1 (0) u(x0 + r)d is a constant independent of r for 0 < r < R, thus equals |B1 (0)|u(x0 ). We have thus proved Proposition 2. Suppose that u C 2 (U ) is harmonic in U , and BR (x0 ) U . Then for any 0 < r < R, 1 1 u(x0 ) = u(x0 + r)d = u(x)d(x), |B1 (0)| B1 (0) |Br (x0 )| Br (x0 ) and u(x0 ) = 1 |Br (x0 )| u(x)dx. Br (x0 ) As a consequence of Proposition 2, we have Theorem 1. Suppose that u C 2 (U ) C 0 (U ) is harmonic in U . Then min = min u max u = max u. U U U U Furthermore, if U is connected, u can not attain maxU or minU u in U unless u is a constant in U . There are versions of maximum principle for the so called subharmonic (superhamonic) functions. 3 Exercise 1. A C 2 (U ) function is called subharmonic (superhamonic) in U , if u ()0 in U . (i). Prove that if u C 2 (U ) C(U ) is subharmonic in U , then max u = max u. U U (ii). Prove that if u C 2 (U ) C(U ), then max |u| max |u| + C max |u|, U U U where C is a constant depending only on the diameter of U . (Hint: set v = 1 maxU |u| + 2 (d2 - x2 ) maxU |u|, where we may assume that U {x : 0 x1 d}, 1 then (v u) 0 in U , and (v u) 0 on U , then apply (i).) We then introduce the Green's representation formula, which states u(x) = U -u(y) (x - y)dy + U (x - y) u(y) (x - y) - u(y) d(y), ny ny (4) for any C 2 (U ) function u on a piecewise C 1 domain U , where (x) = 1 |x|2-n (n-2)|Sn-1 | 1 1 log |x| 2 if n 3, if n = 2. This is obtained from applying (3) to u(y) and (x - y) on U \ {B (x)} for small > 0 and sending 0, noting that y (x - y) = 0 in U \ {B (x)}, and |(r)|rn-1 0, (r) = -1 as r 0. As a consequence (4), we have Br (0) r Corollary. If u C 2 (U ) is harmonic, then it is smooth in U . Proof. For any proper subdomain V of U with C 1 boundary, we can use the Green's representation (4) on V to express u(x), x V as (x - y) V (x - y) u(y) - u(y) ny ny d(y). Since the integrand is a smooth function of x V for y V , with any derivatives uniformly bounded for y V as long as x V stays away from V , this shows that u is smooth in V. Corollary. If u C 2 (U ) is harmonic, then it is real analytic in U . Proof. This follows from the same representation formula: fix any x0 U , then there exists r0 > 0 such that (x - y) = a (x0 - y)(x - x0 ) , 4 with uniform convergence for |x - x0 | r0 and y U (we should have chosen a subdomain V with C 1 boundary such that x0 V U , as done in the proof of the previous corollary, but will neglect such issues). Thus for any > 0, there is N such that |(x - y) - ||N a (x0 - y)(x - x0 ) | , uniformly for |x - x0 | r0 and y U . Multiplying by u(y) and integrating over y U , ny we obtain u(y) u(y) d(y) - A (x0 )(x - x0 ) | | |d(y), | (x - y) ny ny U U ||N with A (x0 ) = U a (x0 - y) u(y) d(y). Similarly, the other integral is also analytic in ny x U , showing that u is analytic in U . Remark 2. Since for each fixed y, |x-y|2-n is a harmonic function for x Rn (n 3), x = y, we may try to use the superposition principle to construct harmonic functions by |x - y|2-n f (y)dy E over some set E. The proofs of the above two corollaries already used this idea and indicated that it worked when E is taken to be U . When E is taken to be U , however, U |x - y|2-n f (y)dy is not necessarily a harmonic function of x U . This is because xi xj |x - y|2-n is no longer an integrable function of y U , so we can't differentiate twice in x on the integral and pass the differentiation inside the integral to conclude that U |x - y|2-n f (y)dy is harmonic. We do have Proposition 3. If f C 1 (U ), then -x U U (x - y)f (y)dy is a C 2 function of x U , and for x U . (x - y)f (y)dy = f (x), Remark 3. A proof for Proposition 3 will be provided later. With mere continuity of f , (x - y)f (y)dy may not be C 2 function of x U . However, the regularity requirement of U f can be weakened to the following: for some 0 < 1, and C > 0, |f (y1 ) - f (y2 )| C|y1 - y1 | for all y1 , y2 U . (5) o Functions satisfying (5) in U (therefore in U ) with 0 < < 1 are said to be Hlder continuous in U with exponent , and the set of such functions is denoted as C (U ); those functions satisfying (5) in U (therefore in U ) with = 1 are said to be Lipschitz continuous in U , and the set of such functions is denoted as Lip(U ). C (U ) (respectively Lip(U )) usually denotes the set of functions satisfying (5) on any compact subsets of U (the constant C may depend on the compact subset). 5 Remark 4. To solve the Dirichlet problem on U , one can still try to take some E disjoint from U , and use E (x - y)g(y)dy to construct harmonic functions in U . The question becomes: (i) whether such harmonic functions extends continuously to U ? (ii) whether one can achieve all (continuous) boundary value functions on U ? Such harmonic functions indeed extend continuously to U when E = U , U is a piecewise C 1 hypersurface, and g C(U ). However, for x U , in general, xU, x x lim (x - y)g(y)d(y) = g(), x U so even though we can construct a harmonic function x E (x - y)g(y)dy for x U , this harmonic function may not solve (1) for the case f 0. In fact the map g C(U ) (x - y)g(y)d(y) C(U ) is a compact linear map, so U (x - y)g(y)d(y) can not U possibly take on all continuous boundary value functions when g runs through C(U ). It turns out a modification of this idea, using the so called double layer potential, U (x-y) g(y)d(y), ny one can solve the Dirichlet problem this way. Exercise 2. (i). Prove that for a domain U with its boundary being a piecewise C 1 hypersurface and any bounded measurable f defined on U , U (x - y)f (y)d(y) defines a C 2 (U ) C 0 (U ) harmonic function. (ii). Prove that for any bounded measurable f defined in a bounded domain U , U (x - y)f (y)dy defines a C 1, (U ) C 1 (U ) function for every 0 < < 1, and Dxi U (x - y)f (y)dy = U Dxi (x - y)f (y)dy. Exercise 3. For x, y U , x = y, define K(x, y) = (x-y) . Prove that, if U is ny assumed to be piecewise C 2 surface, then for any continuous function f defined on U , (x-y) f (y)d(y) extends continuously to U , and for any x U , ny U xU, x x lim U 1 (x - y) f (y)d(y) = f () + x ny 2 K(, y)f (y)d(y). x U We continue to explore the consequences of (4). Let (y) be a C 2 (U ) C 1 (U ) harmonic function in U , then applying the Green's theorem to and u on U , we have 0= U -u(y)(y)dy + U (y) (y) u(y) - u(y) ny ny d(y). Combining this with (4), we have u(x) = U -u(y) [(x - y) - (y)] dy + U [(x - y) - (y)] u(y) [(x - y) - (y)]) - u(y) ny ny d(y). 6 If, for each x U , we can choose a harmonic function (y) = x (y) such that (x - y) - x (y) 0 for y U and (y) C 1 (U ), then we have u(x) = U -u(y) [(x - y) - x (y)] dy - U u(y) [(x - y) - x (y)] d(y). ny That is, we can represent u in U by its Dirichlet data. When this is possible, we call G(x, y) = (x - y) - x (y) the Green's function (for the Laplace operator with Dirichlet boundary condition) for the region U . We summarize the above discussion as Proposition 4. Suppose that, for each x U , there exists x (y) C 1 (U ) such that y x (y) = 0, (y) = (x - y), x for y U , for y U . Let G(x, y) = (x - y) - x (y). Then y G(x, y) = 0, for y U \ {x}, G(x, y) = 0, for y U , 1 lim |x - y|n-2 G(x, y) = , if n 3, yx (n - 2)|Sn-1 | and for any u C 2 (U ), there holds u(x) = U -u(y) G(x, y)dy - U u(y) G(x, y) d(y). ny For some special domains, we can write out G(x, y) explicitly. First, when U = Rn , for + each x Rn , we define x (y) = (x - y), where x is the mirror image of x in Rn . Then + + (x - y) - x (y) 0 for y Rn , + [(x - y) - x (y)] [(x - y) - x (y)] 2xn := K(x, y), = = n-1 ny yn |S ||x - y|n for x Rn and y Rn . K(x, y) is called the Poisson kernel for Rn . + + + - n n n Theorem 2. Assume g C(R+ ) L (R+ ). Define u(x) for x R+ by and u(x) = 1 |Sn-1 | n R+ 2xn g(y)d(y). |x - y|n Then n n (i). u C (R+ ) L (R+ ), n (ii). x u(x) = 0, in x R+ , n n (iii). limxR+ , xR+ u(x) = g(). x x And there is only one solution u satisfying (i)(iii) above. 7 The last part of the above theorem is proved by maximum principle, and is left as an exercise. Notice that, although the Green's representation was derived under stricter requirement on u: u C 2 (U ), we can apply the representation for any continuous boundary data, which produces a solution which may not be C 2 up to the boundary, but is C 2 in the interior of the domain, and is continuous up to the boundary. Next, when U = BR (0) in Rn . It turns out that for each x = 0 BR (0), if we define R2 x = |x|2 x, then x (y) = ( |x| (x - y)) would work, because |y - x|/|y - x | is independent R of y BR (0), and equals |x|/R, so (x - y) - ( |x| (x - y)) = 0 for all y BR (0). It R x works out that for any y = 0 in B R (0), (y) = ( |x| (x - y)) (R) as x 0, so the R construction of x (y) continues to work in this case, and - [(x - y) - x (y)]) R2 - |x|2 = := K(x, y), ny R|Sn-1 ||x - y|n for x BR (0) and y BR (0). K(x, y) is called the Poisson kernel for BR (0). This generalizes our Poisson kernel on two dimensional discs, which was found based on separation of variables. Theorem 3. Assume g C(BR (0)). Define u(x) for x BR (0) by u(x) = BR (0) R2 - |x|2 g(y)d(y). R|Sn-1 ||x - y|n Then (i). u C (BR (0)), (ii). x u(x) = 0, in x BR (0), (iii). limxBR (0),xBR (0) u(x) = g(). x x And there is only one solution u satisfying (i)(iii) above. Remark 5. Although the Poisson kernels are defined asymmetrically: x in the interior of the domain, and y on the boundary, the Green's function G(x, y) is defined for both x and y inside the domain, x = y, and is a harmonic function in y. A useful property is Proposition 5. For all x, y U , x = y, we have G(x, y) = G(y, x). Thus G(x, y) is also a smooth function in U U \ {(z, z) : z U }, extends to a continuous function in U U \ {(z, z) : z U }, and is harmonic in x U , for x = y. Proposition 6. A C 0 function u in U which satisfies the mean value property for any balls B U is a smooth harmonic function. Proof. Let u be such a C 0 function. For any ball B U , by Theorem 3, there is a harmonic function v C(B) C 2 (B) such that v = u on B. Since both u and v satisfy the maximum principle, we conclude now that v = u in B, i.e., u is harmonic in B. Since this holds for any ball B U , we conclude that u is harmonic in U . 8 Another consequence of the Poisson representation formula on BR (0) is the gradient estimates and Harnack estimates. Theorem 4. If u is harmonic in BR (x0 ), then, for some C = C(n) > 0, | u(x0 )| C max |u|, R BR (x0 ) and for k > 1, | u(x0 )| (Ck)k max |u|, Rk BR (x0 ) for all with || = k. Remark 6. The gradient estimates can be proved in a simpler way: since we already proved that u will be smooth, so uxi (x) is also harmonic. Then by the mean value property, uxi (x0 ) = 1 |Br (x0 )| uxi (x)dx = Br (x0 ) 1 |Br (x0 )| u(x)ni (x)d(x), Br (x0 ) for all 0 < r < R. Thus, | u(x0 )| Sending r R, we obtain | u(x0 )| n max |u|. R BR (x0 ) 1 |Br (x0 )| |u(x)|d(x) Br (x0 ) n max |u|. r Br (x0 ) Theorem 5 (Harnack). (Local Version) If u is a nonnegative harmonic function in BR (0), then R - |x| R + |x| Rn-2 u(0) u(x) Rn-2 u(0), for all x BR (x0 ). n-1 (R + |x|) (R - |x|)n-1 In particular, for any x B R (0), 2 2n-2 u(0) u(x) 2n-2 3u(0). n-1 3 (Global Version) For each connected open set V U , there exists a positive constant C depending on V and U , such that sup u C inf u V V for all nonnegative harmonic function u in U . Corollary (Liouville). A bounded harmonic function on Rn must be a constant. In fact, a harmonic function on Rn bounded from below (or above) must be a constant. The first part can be proved by the gradient estimate, the second part can be proved by the Harnack estimate. The more powerful consequences of the gradient estimates are the convergence theorems. 9 Theorem 6. (i). Uniform limit of a sequence of harmonic functions is harmonic. (ii). Limit of a monotone sequence of harmonic functions is harmonic. (iii). A bounded sequence of harmonic functions on U must have a subsequence that converges on any compact subset of U to a harmonic function. Theorem 7. If, for a sequence gk C(U ), there exists a (unique) solution uk to uk = 0, uk = gk , in U , on U , and gk g uniformly on U , then the Dirichlet problem with g as boundary value has a unique solution. Next we discuss how to use the convergence theorems above to solve the Dirichlet problem on a ball without using the Poisson kernel. This consists of three steps. Let Pk denote polynomials of degree k. Step 1. For any polynomial p Pk-2 , there exists a unique solution v Pk satisfying v = p, v = 0, in BR (0), on BR (0), This follows from considering the linear map T : Pk-2 Pk-2 by T (u) = [(R2 - |x|2 )u(x)]. The maximum principle implies that T has trivial kernel, so must be an isomorphism from Pk-2 to Pk-2 . Step 2. For any p Pk-2 and g Pk , there exists u Pk such that u = p, v = g, v = p - g, v = 0, in BR (0), on BR (0), in BR (0), on BR (0), (6) This follows from Step 1. Since g Pk-2 , by Step 1, there exists v Pk solving The u = v + g is the solution. Step 3. For any g C(BR (0)), take a sequence of polynomials gk such that gk g uniformly as k . Then the maximum principle convergence and theorems provide a limit in C(B R (0)) C 2 (BR (0)) harmonic function with g as boundary value. We can attempt to solve (6) for more general right hand side, as we can already solve it for polynomials. Take f to be continuous on B R (0) for instance. We can take polynomials pk and gk such that pk f uniformly in B R (0), and gk g uniformly on BR (0) as k . Let uk denote the corresponding (polynomial) solution. By Exercise 1, we have uniform bound on ||uk ||C(B R (0) . In fact, {uk } is Cauchy in C(B R (0)). But to prove the convergence of {uk } to a solution of (6), we need higher derivative estimates for {uk }. This can be done using the Green's representations. But a more flexible method is the Bernstein's method. 10 Theorem 8. Let u C 4 (B 1 ) and denote u by f . Then for any r (0, 1), there is a constant N = N (n, r) such that in Br |u| + | u| + | 2 u| N max |f | + max | f | + max | B1 B1 B1 2 f | + max |u| B1 The idea of Bernstein is to verify that some auxiliary function involving u and its derivatives satisfies an appropriate differential inequality (subharmonic, for instance), and thus satisfies the maximum principle. We will illustrate the method by verifying that for any smooth cut-off function supported in B1 and identically equal to 1 on Br , the function w = 2 | u|2 + Cu2 satisfies in B1 , for C large depending on r, ~ w - 2 | f |2 - Cf 2 - Cu2 -N max |f |2 + max | f |2 + max |u|2 B1 B1 B1 ~ for some N depending on C. Thus a generalization of Exercise 2 implies that max w N B1 max |f |2 + max | f |2 + max |u|2 B1 B1 B1 + max w B1 for some N > 0. Then using Exercise 2 again to estimate maxB1 |u|2 in terms of maxB1 |f |2 and maxB1 |u|2 , it follows that max | u|2 N Br max |f |2 + max | f |2 + max |u|2 . B1 B1 B1 A similar construction proves the second derivative estimate. The drawback of this method is the requirement on the higher derivatives of f . The advantage is that it is very flexible and applies even to some nonlinear equations. Next we will use the maximum principle and the convergence theorems to find a harmonic function with prescribed Dirichlet data on fairly general domain. I will describe Poincar's e balayage method, which can be considered as a cousin of the Perron method. The latter is presented in most modern treatment. Poincar's method, as well as Perron's, depends on the maximum principle, the convere gence theorems, and the solvability of Dirichlet problem on balls. The last we obtained by Poisson's kernel or the approximation method. We need to extend the notion of subharmonic functions to C 0 class. Definition. A C 0 (U ) function u is called subharmonic (superharmonic ) in U , if for every ball B U and every harmonic function h in B satisfying u ()h on B, we also have u ()h inside B. The maximum principle extends to these functions in the following way: (i). Suppose U is a bounded domain. If u is subharmonic in U and is continuous up to the boundary of U , then maxU u maxU u. In fact, the strong maximum principle also holds. If v is superharmonic in U , and u v on U , then in any connected component of U either u < v, or u v. 11 (ii). If u is subharmonic in U and B is a ball strictly contained in U , then the harmonic lifting hB (u) of u in B is subharmonic in U , where hB (u) is defined as hB (u) = u(x), for x B , u(x), for x U \ B, with u(x) being the harmonic function in B satisfying u(x) = u(x) on B. Poincar's method consists of several steps. Given a domain U and a continuous boundary e function g. Step 1. Assume first that there exists a subharmonic u0 C(U ) such that u0 |U = g. Step 2. Cover U by a countable number of balls {B1 , B2 , } such that each Bi U . Starting from u0 , we will replace each with its harmonic lifting on successive balls to obtain a sequence of monotone increasing subharmonic functions that are bounded from above. Thus a limiting function u exists. We will prove this u is harmonic in U and will study its boundary behavior. First let B (1) = B1 and define u1 = hB (1) (u0 ). Then (a) u1 (x) u0 , for all x U . (b) u1 is harmonic in B (1) . (c) u1 (x) maxU g, and u1 (x) = g(x) on U . (d) u1 is still subharmonic in U . Order the balls in the following way B1 B2 B1 B2 B3 B1 B2 B3 B4 Define inductively, for each i 2, ui (x) on U by ui = hB (i) (ui-1 ). Then (a) ui (x) ui-1 , for all x U . (b) ui is harmonic in B (i) . (c) ui (x) maxU g, and ui (x) = g(x) on U . (d) ui is still subharmonic in U . Therefore {ui } is a sequence of monotone increasing, subharmonic, continuous functions in U , and is bounded from above. Thus u(x) = limi ui (x) is well-defined for all x U . To prove u is harmonic in U , notice that for each x U , there is a ball Bi(x) such that x Bi(x) . Notice also that because of the ordering of the balls, a subsequence of {ui } is actually monotone, harmonic in Bi(x) . Thus u is harmonic in Bi(x) by the convergence theorem. Step 3. The continuity in U of the solution u in the above step is handled by the concept of barriers. Definition. w C(U ) is called a barrier function at U for the Dirichlet problem on U if 12 a). w is superharmonic in U . b). w() = 0, and w(x) > 0 for x U \ {}. Suppose a barrier function w at U exists. For any given > 0, by the continuity of g at , we can find r0 > 0 such that g(x) - g(x) g() + for all x U with |x - | r0 . There also exists M > 0 depending on w and g such that g(x) - g() M w(x) for all x U with |x - | r0 . Thus g(x) g() + + M w(x), for all x U . Note that g() + + M w(x) is a superharmonic function on U . So in the steps above, we also have u0 ui g() + + M w on U . By the continuity of w and u0 at , we can find r < r0 such that when x U and |x - | r , M w(x) and u0 (x) g() - . Thus g() - u0 (x) u(x) g() + 2 , for all x U with |x - | r , proving the continuity of u at . Step 4. To remove that assumption in Step 1, we first argue that for any polynomial g given, g1 = g + Ax2 and g2 = Ax2 are subharmonic in U for A > 0 sufficiently large. Thus 1 1 the first three steps can be applied to show that the Dirichlet problem has solutions with g1 and g2 as boundary values, thus it also has one with g1 - g2 = g as boundary value. Finally the convergence theorems can be used to prove the existence of solution to the Dirichlet problem with any given continuous boundary value, provided the barrier argument in Step 3 can be carried out. That turns out to depend only on the geometry of the domain U . An easily verified criterion for the existence of a barrier at U is the existence of some exterior ball, i.e., there exists a ball B such that B U = {}. Let x0 be the center of this ball and r be its radius, then w(x) = r2-n - |x - x0 |2-n defines a barrier at . Definition. A boundary point is called regular with respect to the Laplacian if there exists a barrier at that point. We can now summarize our discussion as Theorem 9. The classical Dirichlet problem for the Laplacian in a bounded domain is solvable for arbitrary continuous boundary values if and only if the boundary points are all regular. Points on components of the boundary with codimension 2 or higher are not regular. For instance, we can't solve the Dirichlet boundary value on the domain B \ {0} with prescribed value everywhere on {B \ {0}} = B {0}, as the following theorem shows Theorem 10. Suppose u is harmonic in B \ {0}, and satisfies |u(x)| = o(|x|2-n ) as x 0 (assume n 3). Then u extends to a smooth harmonic function over B. 13 Proof. Let v be the unique solution in B to v = 0, v = u, in B, on B. Then w = u - v is still harmonic in B \ {0}, and satisfies w(x)| = o(|x|2-n ) as x 0. Furthermore w(x) 0 on B. We prove w 0 in B in the following way: for any > 0, we 2-n can find r > 0 such that Br B, and on Br , |w(x)| (|x|2-n - r0 ) (r0 is the radius 2-n of B). Then w(x) + (|x|2-n - r0 ) is a harmonic function on B \ Br , and is nonnegative 2-n on (B \ Br ). Thus by the maximum principle w(x) + (|x|2-n - r0 ) 0 in B \ Br , i.e., 2-n |w(x)| (|x|2-n - r0 ) for all x B \ Br . For any fixed x B \ {0}, x is in B \ Br for all 2-n sufficiently small > 0. Thus |w()| (||2-n - r0 ), and by sending 0, we conclude x x that w() = 0. In conclusion, w 0 in B, and so u v, a smooth harmonic function in x B. The notion of local barrier is sometimes useful. Definition. w is called a local barrier function at U for the Dirichlet problem on U if there is a ball B centered at such that w C(U B) and a). w is superharmonic in U B. b). w() = 0, and w(x) > 0 for x U B \ {}. If a local barrier w at U exists on U B, it is easy to see that we can apply the barrier argument in Step 3 on U B; or alternatively we can define m = minBU w and w(x) = ~ min(w(x), m), m, for x U \ B, for x U B, then w is a (global) barrier function at U . We now provide a proof for Proposition 3 ~ and discuss how to solve the inhomogeneous problem (1). Proof of Proposition 3. We will prove that, assuming (5) on U , the potential integral u(x) = f (y)(x - y) dy is in C 2 (U ), and U u(x) = xi u(x) = x U U (x - y) f (y) dy, xi j (y)yi (x - y)d(y) U (7) (8) xi xj (x - y) [f (y) - f (x)] dy + f (x) 14 for any = (i, j) with || = 2. It then follows that n x u(x) = U x (x - y) [f (y) - f (x)] dy + f (x) i=1 n U i (y)yi (x - y)d(y) y (x - y) dy U \B (x) =f (x) B (x) i=1 i (y)yi (x - y)d(y) + 1-n =f (x) B (x) - |Sn-1 | d(y) = - f (x), which is the stated equality in Proposition 3. The key difficulty in proving (7) and (8) is that the (x - y) in the integrand is singular 2 at y = x; more precisely, x (x - y)f (y) is still integrable in y U , but x (x - y)f (y) is 2 not integrable for y U due to the strength of the singularity in x (x - y) |x - y|-n . It is not hard to prove that x U (x - y)f (y) dy = U x (x - y)f (y) dy, for x U , although one can not seem to apply Lebesgue's dominated convergence theorem directly--see Hints to Exercise 2 below. A direct argument for differentiating twice under the integral sign is not that routine. A streamlined proof for (7) and (8) is to mollify the singularity in (x - y) first: choose a smooth cut-off function (t) such that it is supported in |t| 1 and is identically 1 for |t| 2, and define (t) = (t/ ) for > 0. Then (x - y) (|x - y|) is smooth in x. So if we set u (x) = U f...

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Rutgers - MATH - 517
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Rutgers - MATH - 517
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Rutgers - MATH - 517
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University of Illinois, Urbana Champaign - CS - 105
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University of Illinois, Urbana Champaign - CS - 105
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Cornell - CS - 100
Name(s) &amp; ID(s):_ Assignment P4 CS 100 Summer 1999 Due: Thursday July 29, 1999 at the beginning of class.NOTE: YOU MUST WORK WITH A PARTNER ON THIS ASSIGNMENT. Turn in your assignments according to the instructions given in the first day handout. R
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Cornell - CS - 100
Name &amp; ID:_ Quiz Q1 CS 100 Summer 1999 Due: Wednesday July 7, 1999 5 minutes after the beginning of class Question 1: What is the difference between a public field and a private field in a class definition?Question 2: Compare and contrast: procedu
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Cornell - CS - 100
Name &amp; ID:_ Quiz Q2 CS 100 Summer 1999 Due: Monday July 12, 1999 5 minutes after the beginning of class Question 1: Suppose that a field x in a class is declared thus: public static int x. What do each of the terms before the variable name (public,
Maple Springs - PHYS - 2040
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Cornell - CS - 100
Name &amp; ID:_ Quiz Q3 CS 100 Summer 1999 Due: Wednesday July 14, 1999 5 minutes after the beginning of class Question 1: How would you declare an array of Coordinates where Coordinate is a class as we've defined it in class?Question 2: How would you
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Maple Springs - PHYS - 2040
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Cornell - CS - 100
Name &amp; ID:_ Quiz Q5 CS 100 Summer 1999 Due: Monday July 26, 1999 10 minutes after the beginning of class Question 1: Thinking in Matlab right now, what will the variable a look like after the following statement is executed: a = 10 * [0:9]Question
Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Cornell - CS - 100
Name(s) &amp; ID(s):_ Assignment P2 CS 100 Summer 1999 Due: Thursday July 15, 1999 at the beginning of class.Turn in your assignments according to the instructions given in the first day handout. Remember to include sample output when you turn in your
Maple Springs - PHYS - 2040
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Cornell - CS - 100
Method OverloadingIf two methods of a class (whether both declared in the same class, or bothinherited by a class, or one declared and one inherited) have the same namebut different number or types of parameters, then the method name is said to
Maple Springs - PHYS - 2040
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Cornell - CS - 100
Name(s) &amp; ID(s):_ Assignment P3 CS 100 Summer 1999 Due: Thursday July 22, 1999 at the beginning of class.Turn in your assignments according to the instructions given in the first day handout. Remember to include sample output when you turn in your
Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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Maple Springs - PHYS - 2040
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