Solutions to the L p Mixed Boundary Value Problem in C 1,1 Domains

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1 University of Kentucky UKnowledge Theses and Dissertations--Mathematics Mathematics 206 Solutions to the L p Mixed Boundary Value Problem in C, Domains Laura D. Croyle University of Kentucky, lauradcroyle@gmail.com Digital Object Identifier: Click here to let us know how access to this document benefits you. Recommended Citation Croyle, Laura D., "Solutions to the L p Mixed Boundary Value Problem in C, Domains" (206). Theses and Dissertations--Mathematics This Doctoral Dissertation is brought to you for free and open access by the Mathematics at UKnowledge. It has been accepted for inclusion in Theses and Dissertations--Mathematics by an authorized administrator of UKnowledge. For more information, please contact UKnowledge@lsv.uky.edu.

2 STUDENT AGREEMENT: I represent that my thesis or dissertation and abstract are my original work. Proper attribution has been given to all outside sources. I understand that I am solely responsible for obtaining any needed copyright permissions. I have obtained needed written permission statement(s) from the owner(s) of each thirdparty copyrighted matter to be included in my work, allowing electronic distribution (if such use is not permitted by the fair use doctrine) which will be submitted to UKnowledge as Additional File. I hereby grant to The University of Kentucky and its agents the irrevocable, non-exclusive, and royaltyfree license to archive and make accessible my work in whole or in part in all forms of media, now or hereafter known. I agree that the document mentioned above may be made available immediately for worldwide access unless an embargo applies. I retain all other ownership rights to the copyright of my work. I also retain the right to use in future works (such as articles or books) all or part of my work. I understand that I am free to register the copyright to my work. REVIEW, APPROVAL AND ACCEPTANCE The document mentioned above has been reviewed and accepted by the student s advisor, on behalf of the advisory committee, and by the Director of Graduate Studies (DGS), on behalf of the program; we verify that this is the final, approved version of the student s thesis including all changes required by the advisory committee. The undersigned agree to abide by the statements above. Laura D. Croyle, Student Dr. Russell Brown, Major Professor Dr. Peter Hislop, Director of Graduate Studies

3 Solutions to the L p Mixed Boundary Value Problem in C, Domains DISSERTATION A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in the College of Arts and Sciences at the University of Kentucky By Laura Croyle Lexington, Kentucky Director: Dr. Russell Brown, Professor of Mathematics Lexington, Kentucky 206 Copyright Laura Croyle 206

4 ABSTRACT OF DISSERTATION Solutions to the L p Mixed Boundary Value Problem in C, Domains We look at the mixed boundary value problem for elliptic operators in a bounded C, (R n ) domain. The boundary is decomposed into disjoint parts, D and N, with Dirichlet and Neumann data respectively. Expanding on work done by Ott and Brown, we find a larger range of values of p, < p < n n, for which the Lp mixed problem has a unique solution with the non-tangential maximal function of the gradient in L p ( Ω). KEYWORDS: Boundary value problem, mixed problem, Besov spaces Author s signature: Laura Croyle Date: July 2, 206

5 Solutions to the L p Mixed Boundary Value Problem in C, Domains By Laura Croyle Director of Dissertation: Director of Graduate Studies: Russell Brown Peter Hislop Date: July 2, 206

6 Dedicated to Ralph Oliver Croyle for everything

7 ACKNOWLEDGMENTS First, I would like to express my gratitude and appreciation to my advisor, Dr. Russell Brown, for his continued support and patience over the past 6 years. Thank you for always encouraging me to be brave. Also, thank you to my committee members, Drs. Larry Harris, John Lewis, Constance Wood, and Caicheng Lu, for their guidance. To Sheri Rhine, whose mere presence could bring me to a place of serenity, thank you for your kindness. To the multitude of faculty and staff at UK that supported me and my idiosyncrasies, thank you for your understanding and encouragement; I would not be here without it. The greatest of thanks to my husband, Ralph, for his encouragement and constant faith, for patiently listening to my endless monologues about mathematics, and for always dropping everything whenever I need him. Finally, thank you to my family for always believing in me and cheering me on as I followed my dreams. iii

8 CONTENTS Acknowledgments iii Chapter Introduction Introduction and History Definitions and Preliminaries Main Result Chapter 2 Reverse Hölder Inequality Reverse Hölder Inequality Flattening the Boundary Besov Spaces Proof of Reverse Hölder Inequality Chapter 3 Main Estimate Reverse Hölder Inequality on the Boundary Chapter 4 Final Details Shen s Result Proof of Main Theorem Chapter 5 Future Problems Bibliography Vita iv

9 Chapter Introduction. Introduction and History In this dissertation, we will prove a result for a L p mixed boundary value problem. This and related problems have been studied for decades. Proving existence of solutions to the Dirichlet Boundary Value Problem on Lipschitz domains was done by Dahlberg in 976 [5]. Jerison and Kenig studied the non-tangential maximal function in L 2 for the Neumann Problem for Lipschitz domains in 98 [0]. Optimal results have been given for the L p Neumann Problem on Lipschitz domains by Dahlberg and Kenig in 987 [6]. In this paper we look at the L p mixed problem on a C, domain. The mixed problem has both Neumann and Dirichlet data, given as follows u = 0 u = f D u ν = f N u L p ( Ω) in Ω on D on N where D and N given a Lipschitz dissection of Ω, defined later. Also, u is the non-tangential maximal function of u and is defined in detail in the Section.2. Brown and Ott proved existence of solutions to this problem on Lipschitz domains, but did not get a specific range of p-values []. Brown, Ott, and Taylor, also proved existence to this problem on Lipschitz domains with general decompositions of the boundary, but again did not get a specific range of p-values [6]. By putting stricter conditions on the domain, we are able to prove in this paper existence of solutions for < p < n n. Moreover, when n = 2 this range of p-values is optimal. For our paper, we will prove a Reverse Hölder Inequality for solutions to the weak mixed problem, which will in turn allow us to get these better results for the L p mixed problem. The work of Cafarelli and Peral showed that we could use Reverse Hölder Inequalities to get better L p results [3]. This method was adapted to boundary value problems by Shen, whose work we will use to prove our result [4]. We will prove our Reverse Hölder Inequality by utilizing the work by Savaré on R n + [3]..2 Definitions and Preliminaries We start by introducing notation and problems of interest. Given a domain, Ω, we define for k N and p <, W k,p (Ω) to be the set of functions, u L p (Ω), such that for each multiindex, α, with α k, the weak derivative, α u x, exists and is in α L p (Ω). We define the norm on this space as follows

10 u W k,p (Ω) = α p u α k x α L p (Ω) Note, that for our L p spaces, we use the Lebesgue measure on Ω and on the boundary we use the surface measure, σ. For D Ω, we define CD ( Ω) to be the set of C functions on Ω that are zero in a neighborhood of D and are compactly supported. Moreover, we let W k,p D (Ω) be the closure of C D ( Ω) in the W k,p (Ω) norm. We say that u W k,p loc (Ω), if for each V compactly contained in Ω, u W k,p (V ). We define C, to be the set of Lipschitz functions that have Lipschitz first partial derivatives. We say that Ω is a C, domain if locally the boundary of Ω is the graph of a C, function. We say that a matrix a is elliptic if there is a constant θ > 0, such that ξ T aξ θ ξ 2 for almost every x Ω and all ξ R n. For all of the following we assume that our matrix a is elliptic and symmetric with coefficients that are Lipschitz. Now we introduce the problems of interest in this paper. Given g (W,2 0 (Ω)) and f D tr((w,2 (Ω)) ), the Weak Dirichlet Boundary Value Problem is the problem of finding a function, u in W,2 (Ω), that is a weak solution to /p. diva u = g u = f D in Ω on Ω. A function, u, is a weak solution if u f D W,2 0 (Ω) and it satisfies (.) Ω a u Ψ dx = g, Ψ Ω for all Ψ W,2 0 (Ω). Next, given g (W,2 (Ω)) and f N tr((w,2 (Ω)) ), the Weak Neumann Boundary Value Problem is the problem of finding a function, u W,2 (Ω), that is a weak solution to diva u = g u ν = f N A function, u, is a weak solution if it satisfies in Ω on Ω. (.2) Ω a u Ψ dx = g, Ψ Ω + f N, Ψ Ω for all Ψ W,2 (Ω). One of the boundary value problems we will look at in this paper is the Weak Mixed Boundary Value Problem. First we let D closed and N open give a 2

11 disjoint partition of Ω. Given g (W,2 D (Ω)) and f N tr((w,2 D (Ω)) ), we are searching for u W,2 D (Ω), a weak solution, to diva u = g in Ω u = 0 on D u ν = f N on N. A function, u, is a weak solution if it satisfies (.3) Ω a u Ψ dx = g, Ψ Ω + f N, Ψ Ω for all Ψ W,2 D Ω. Now to introduce a problem closely related to each of these, we need to define the non-tangential maximal function. For a function u defined on Ω, we define the non-tangential maximal function on Ω by u (x) = sup u(y) y Γ α(x) where Γ α (x) is the non-tangential approach region to x given by Γ α (x) = {y Ω x y < ( + α)dist(y, Ω)} for some α > 0. For this paper we use u to stand for ( u) to simplify notation. Now we turn to the main problem of interest for this paper. We say u C 2 (Ω) is a solution to the L p -Mixed Problem, if u = 0 u = 0 u ν = f N u L p ( Ω) in Ω on D on N (.4) where u ν = ν u is the outward normal derivative of u and u is defined in the non-tangential sense, meaning that for x Ω u(x) = y x lim u(y). (.5) y Γ α(x) In the case where N = and we only specify Dirichlet data, we call (.4) the L p - Regularity Problem. When D = and we only specify Neumann data, we call (.4) the L p -Neumann Problem. For our problem we will consider D and N that give a Lipschitz dissection of the boundary. To define what this means, we begin by assuming that our domain, Ω, is R n +. In this case, we say that D and N give a Lipschitz dissection of the boundary if there exists a Lipschitz function Ψ R n 2 R such that D = {(x, x, 0) Ψ(x ) x } 3

12 and N = {(x, x, 0) Ψ(x ) > x } where x R n 2. Moving on to a more general case, suppose that Ω is a C, graph domain, meaning there exists a C, function, ϕ R n R, such that Ω = {(x, x n ) x n > ϕ(x )} where x R n. To have a Lipschitz dissection in this case, we require and D = Ω {Ψ(x ) x } N = Ω {Ψ(x ) > X }. For our general case where Ω is any C, domain, we say that we have a Lipschitz dissection of the boundary if there exists r > 0 small enough, such that for each x Ω, there is a C, graph domain, Ω x, such that Ω B r (x) = Ω x B r (x), and N B r (x) = N x B r (x), D B r (x) = D x B r (x), where D x and N x give a Lipschitz dissection of Ω x. For this paper we refer to bounded C, domains, Ω, that have a Lipschitz dissection of the boundary, given by D and N, as a standard domain. We may fix r 0 > 0 small enough so that for each x Ω, we have a C, function that agrees with the boundary of Ω on B 00r0 (x). Similarly, we choose r 0 small enough so that our Lipschitz function, ϕ, gives a Lipschitz dissection of the boundary of a graph domain Ω x, as in the definition, that agrees with Ω, D, and N on B 00r0 (x). Since Ω is bounded, we have that Ω is compact and we can guarantee existence of such a value for r 0 that works for all x Ω. For simplification of notation, we will use the following notations for our local sets and we define surface balls by Υ r (x) = B r (x) Ω ζ r (x) = B r (x) Ω. 4

13 .3 Main Result The goal of this dissertation is to prove the following theorem. Theorem. (Main Theorem). Let Ω, D, and N be a standard domain, then there exists a unique solution u to the L p Mixed Problem (.4) with f N L p (N), such that for < p < n n. u L p ( Ω) c f N L p (N) In the case where n = 2, we have that our range of p-values is optimal. Consider the following example. Let 0 < α < 2π. Define Ω α = B (0) {(r, θ) 0 < θ < α}. We let D = {(x, 0) 0 < x < } and N = Ω α /D. Consider u(r, θ) = r π 2α sin (θ π 2α ). The main interest of this paper is that we are able to find a larger range of p-values. Straight forward calculation gives that u = 0 on Ω α, u = 0 on D and u ν L on N. When α = π, it is clear that u L p ( Ω), only if p < 2. This example proves that our result is maximal in 2-dimensions. We prove this theorem in multiple steps. We will show that solutions to the weak mixed problem, which satisfy extra conditions will, in fact, be solutions to the L p mixed problem. In Chapter 2, we will work through the proof of a Reverse Hölder Inequality for solutions to the weak mixed problem. This will utilize the work of Savaré [3]. In Chapter 3, we will follow the work of Ott and Brown, [], by using our results from Chapter 2, to prove a related Reverse Hölder Inequality on the boundary for the non-tangential maximal function. Finally, in Chapter 4, we will use the results from Chapter 3 to prove conditions necessary for a theorem by Shen [4] and prove our Main Theorem. Copyright Laura Croyle,

14 Chapter 2 Reverse Hölder Inequality 2. Reverse Hölder Inequality In order to prove the main result we first need to prove a weak Reverse Hölder s Inequality for solutions to weak mixed boundary problems. Recall that Hölder s Inequality implies that for S with positive measure, that ( f p p dx) S ( f q q dx) S for q > p. A Reverse Hölder Inequality is the same result, but in the case where p > q. To prove our main theorem, we will prove a weak reverse Hölder inequality. A weak result refers to a local result which bounds integrals on a neighborhood by integrals on larger neighborhoods. For instance, ( Br(x) f p p dx) ( B2r(x) f q q dx) for p > q. For the duration of this paper, we will be referring to the Reverse Hölder Inequality stated in the following theorem. Theorem 2. (Reverse Hölder Inequality). Let x Ω and 0 < r < r 0. Suppose u is a weak solution to our mixed problem (.3) with g = 0 and f N = κ, a constant, on B r (x) N, then we have for 2 < s < 2n n. ( Υr/2 (x) u s dy) 2.2 Flattening the Boundary s c ( Υr(x) u 2 2 dy) + c κ Before we begin our proof of the Reverse Hölder Inequality, Theorem 2., we need to flatten the boundary, so we can utilize work already done for R n + by Savaré [3]. Although we eventually want to consider the case where g = 0 and a is the identity matrix in (.3) for the proof of Theorem., we are able to prove many of these results in a more general setting. We first need some preliminary results. Lemma 2.2. Given a graph domain, Ω, given by a Lipschitz function, ϕ R n R, we define Φ R n + Ω by Φ(x, x n ) = (x, ϕ(x ) + x n ) where x R n. We can show that Φ is a bi-lipschitz function differentiable almost everywhere with 6

15 detdφ =. Proof. (of Lemma 2.2) We begin by showing that Φ is Lipschitz with a constant of + M, since Φ(x) Φ(y) = (x, ϕ(x ) + x n ) (y, ϕ(y ) + y n ) = (x y) + e n (ϕ(x ) ϕ(y )) x y + ϕ(x ) ϕ(y ) x y + M x y ( + M) x y where we use that ϕ is Lipschitz, with constant M, in the second to last step. It is clear that Φ Ω R n + defined by Φ (y) = (y, y n ϕ(y )) is in fact an inverse for Φ. It follows similarly that Φ is also Lipschitz with constant + M, giving that Φ is a bi-lipschitz function. Since ϕ is Lipschitz, we know that ϕ is differentiable almost everywhere by Rademacher s Theorem. Now, we see that Hence, Φ x i = (0,..., 0,, 0,..., 0, ϕ x i ) if i n (0,..., 0, ) if i = n DΦ = ϕ ϕ x x 2 ϕ x n Therefore, we have that detdφ =. Once we flatten the boundary and have a new mixed problem, the following lemma will allow us to prove that the matrix in our new problem is still elliptic. Lemma 2.3. If a matrix a is elliptic, then for an invertible matrix s, s T as is also elliptic. Proof. (of Lemma 2.3) Let ξ R n. Note that s T asξ ξ = ξ T s T asξ = (sξ) T a(sξ) c sξ 2 7

16 using that a is elliptic. Next note ξ = s sξ s sξ. Together these bounds give the result. Note that the ellipticity constant depends on the norm of the matrix s. The following two lemmas will allow us to show that our functions in the flattened domain lie in the correct spaces. Although we will eventually use Lemma 2.2 and have a function with detdφ =, we are able to prove the following results with weaker conditions. Lemma 2.4. Given domains Ω and Ω, u W, loc (Ω), and Φ Ω Ω, a bi-lipschitz function with m detdφ M, we have the following chain rule and u Φ W, loc (Ω ). x i (u Φ) = n j= u y j Φ Φ j x i (2.) Proof. (of Lemma 2.4) We will prove this result in steps. Step : If u C c (Ω) and Φ C (Ω ), then (2.) holds by the chain rule for smooth functions and the result is obvious. Step 2 : If u C c (Ω) and Φ is as in the statement of the lemma, we can choose a mollifier, ξ, and define Since ϕ is Lipschitz, Φ ε (x) = ξ ε Φ(x). Φ ε (x) Φ(x) cε for all x Ω and Φ ε,j x i converges to Φ j x i test function in Cc (Ω ), we have that pointwise almost everywhere. Given Ψ be a Ω n j= u y j Φ Φ j x i Ψ dx = lim ε 0 + Ω n j= u Φ ε,j Φ ε Ψ dx y j x i by the Lebesgue Dominated Convergence Theorem (LDCT). By the first case and integration by parts we have that, Ω n j= Again using LDCT, we have u Φ ε,j Ψ Φ ε Ψ dx = y j x u Φ ε dx. i Ω x i 8

17 Ψ lim ɛ 0 + u Φ ɛ dx = Ω x u Φ Ψ dx. i Ω x i This result gives (2.). Now, we know that Φ x i that (Φ j,ε ) = ( Φ j. x i x i )ε By the Lebesgue Differentiation Theorem, we have that almost everywhere in Ω. Lastly, since Φ j x i ( Φ j Φ j x i )ε x i exists and is in L (Ω ). We also have L (Ω ) and ( Φ j Φ j x i )ε L (Ω ) x Φ ε dx = Φ j, i L (Ω ) Ω x i L (Ω ) Since u Φ is bounded as is u x i Φ Φ x i, we have u Φ W, loc (Ω ). Step 3 : For this case let u and Φ be as in the statement of the lemma. Let V Ω and set V = Φ(V ). Easily, this gives that V Ω. Choose u k C c (Ω), where u k is supported on V and u k u on V in the W, norm. Now we have V u k Φ u Φ dx = u k (y) u(y) V detdφ dy. Since u k u in L (V ), the above gives that u k Φ u Φ in L (V ). Similarly, since in L and u k y j u y j V n u k Φ Φ j y j x i j= n u y j j= Φ Φ j x i dx n = V ) j= u k y j u y j Φ j x i (Φ (y)) detdφ dy we have that n u k j= y j (Φ) Φ j x i n j= u y j (Φ) Φ j x i in L (V ). Lastly, we get the result as follows by letting Ψ be a test function and getting V n j= u y j Φ Φ j x i Ψ dx = lim k V n u k Φ Φ j Ψ dx j= y j x i = lim k u k Φ Ψ dx V x i = u Φ Ψ dx V x i where we used Step 2 to get the second equality. 9

18 Next, we observe that if u W,2,2 D (Ω), then u Φ WD (Ω ). Lemma 2.5. Let Ω and Ω be C, graph domains and u W,2 D (Ω). Given Φ Ω Ω a bi-lipschitz function with m detdφ M we have that u Φ W,2 D (Ω ) where D = Φ(D). Proof. (of Lemma 2.5) First, since W,2, D (Ω) Wloc (Ω), we have by the chain rule in Lemma 2.4, that u Φ is in W, loc (Ω ). Moreover, we have that Hence, (u Φ) x i = n j= u y j Φ Φ j x i. (2.2) x i (u Φ) L 2 (Ω ) n u = Φ Φ j y j x i M j= n j= L 2 (Ω ) u y j Φ L 2 (Ω ) using that the derivatives of Φ are bounded by the Lipschitz coefficient M and the triangle inequality. Changing variables gives that x i (u Φ) L 2 (Ω ) M n j= M m /2 u y j detdφ /2 L 2 (Ω) n j= c u L 2 (Ω). u y j L 2 (Ω) Since this is finite, we have that can show that u Φ L 2 (Ω ). Therefore we have that u Φ W,2 (Ω ). To show that u Φ has the correct boundary condition, pick {u k } k= C D (Ω) such that u k u in W,2 (Ω). Now we consider u k Φ. We know that u k is zero in a neighborhood of D, so u k Φ is zero in a neighborhood of D. Since u k is bounded and Φ is Lipschitz, we have that u k Φ is also Lipschitz. Since the boundary of Ω is C,, we can extend u k Φ to a Lipschitz function in a neighborhood of Ω using a reflection. We call this extended domain E(Ω ). It is clear that the extension of our function is still zero in a neighborhood of D. Next, choose ε > 0 small enough so that shrinking the x i (u Φ) L 2 (Ω ). Using the same reasoning we extended domain by ε doesn t cut into the original domain. Now we can mollify to give (u k Φ) εk. Choosing ε k = k will suffice, and starting at k K large enough so 0

19 that ε > K. Keep in mind that we are mollifying the extension of our function. Now we know that (u k Φ) εk CD (E(Ω )) and (u k Φ) εk u Φ in W,2 (Ω ). This gives that u Φ W,2 D (Ω ). We are now ready to flatten the boundary, so we will define our new functions on R n +. Given a function u W,2 D (Ω), 0 < r < r 0, and x = (x, x n ) Ω, where x R n, we define v by η(y)(u Φ)(y) on R v(y) = n + B r/m (x, 0) (2.3) 0 else where Φ (y) = (y, y n ϕ(y )) is the function that flattens the boundary of Ω and η C (R n ) satisfies η = η = 0 η c r on B r (x, 0) 2M on R n /B 2r (x, 0) 3M everywhere. When we flatten the boundary, we will want to ensure we still have a Lipschitz dissection of the boundary. As a result, we must choose carefully how we define the dissection of R n +. Given our x Ω, we know there exists a Lipschitz function, Ψ, a graph domain, Ω x, and a Lipschitz dissection of its boundary given by D x and N x that agree with D and N, respectively, on B 00r0 (x). We know D x = {(x, x, x n ) x Ψ(x )}. We define and D = {(x, x, 0) x Ψ(x )} Ñ = {(x, x, 0) x < Ψ(x )}. If we assume that our original function u is a solution to (.3), we will show that v is a weak solution to the following where v ν = b j v x j ν j, div(b v) + v = g + v in R n + v = 0 on D v ν = f N on Ñ (2.4)

20 b lm = g = n detdφ (a ij Φ) ( (Φ ) l Φ) ( (Φ ) m Φ), i,j= x i x j (u Φ)div(b η) 2b (u Φ) η on Υ r (x, 0) M + detdφ (g Φ ) 0 else, and f N = (u Φ)b η ν + + ϕ 2 (f N Φ ) on Υ r (x, 0) M 0 else. Next, we give a careful definition of the operator ( divb + I) in (2.4). For v W,2 D (Rn +), we define a map given by (2.4), denoted B W,2 D (Rn +) (W,2 D (Rn +)), as follows Bv(Ψ)) = R n + (b v Ψ + vψ) dy, Ψ W,2 D (Rn +). In order to prove our Reverse Hölder Inequality, we want to be able to ensure that B has a right inverse. We define B W,2 (R n +) (W,2 0 (R n +)) as Bv(Ψ)) = (b v Ψ + vψ) dy (2.5) R n + for all Ψ W,2 0 (R n +). Also note that the bilinear form Bv(Ψ) is coercive on W,2 0 (R n +) W,2 0 (R n +). Lemma 2.6. Given u W,2 D (Ω), a weak solution to (.3), and a fixed x = (x, x n ) Ω, we have that v defined by (2.3) is in W,2 D (Rn +). Moreover, v is a weak solution to (2.4) and the matrix b is symmetric and elliptic with Lipschitz coefficients. Proof. (of Lemma 2.6) From Lemma 2.3, we have that b is elliptic. That b is symmetric and has Lipschitz coefficients follows from its definition and our assumption that Ω is C,. Now, recall that we flatten a piece of the boundary of Ω using Φ R n + Ω x where Ω x is the region above the graph of a C, function and Ω Φ(R n + B M (x, 0)) = Ω x Φ(B M (x, 0). Define S = Φ(R n + B r M (x, 0)). We know that u W,2 D Using Ω = Ω x and Ω = R n + in Lemma 2.5, we have that v W,2 S (S). By Lemma 2.2, we have that Φ is bi-lipschitz and detdφ =. Ψ W,2 D (Rn +) be a test function. Define η(φ Ψ(x) = (x))ψ(φ (x)) on S 0 else D (Rn +). Now let 2

21 Again, by Lemma 2.5, we have that Ψ W,2 D (Ω). Hence using the weak formulation of our original mixed problem we know that Ω a u Ψ dx = Ω g Ψ dx + N f N Ψ dσ. (2.6) We start by changing variables on the left-hand side of equation (2.6). Ω a u Ψ dx = S n i,j a ij u x i Ψ x j dx. Letting u = u Φ Φ and using the chain rule from Lemma 2.4, we have Ω a u Ψ dx = S Changing variables by letting x = Φ(y), we get a u Ψ dx = Ω Φ (S) n i,j,l n (u Φ) a ij Φ (Φ ) l Ψ dx. i,j,l y l x i x j (u Φ) (a ij Φ) y l ( (Φ ) l x i From the definition of Ψ, straight forward calculation gives that Φ) ( Ψ x j Φ) dy. Ψ n Φ = (Ψ η (Φ ) m Φ + η Ψ (Φ ) m Φ). x j m= y m x j y m x j Using this and the definition we gave for b lm, we have Ω a u Ψ dx = Φ (S)(Ψb (u Φ) η + ηb (u Φ) Ψ) dy. Next, subtracting and adding (u Φ)b Ψ η, we have Ω a u Ψ dx = Φ (S) (Ψb (u Φ) η (u Φ)b Ψ η +(u Φ)b Ψ η + ηb (u Φ) Ψ) dy. Combining the last two terms in this equation gives us the following Ω a u Ψ dx = Φ (S) (Ψb (u Φ) η (u Φ)b Ψ η Recalling our definition of v, we have +b (η(u Ψ)) Ψ) dy. Ω a u Ψ dx = Φ (S) (Ψb (u Φ) η (u Φ)b Ψ η +b v Ψ) dy. (2.7) Now, we need to work through the second term of this equation. Adding and subtracting Ψb (u Φ) η gives 3

22 (u Φ)b Ψ η dy = Φ dy. (S) Φ (S)((u Φ)b Ψ η+ψb (u Φ) η Ψb (u Φ) η) Combining the first two terms of the right-hand side, we have (u Φ)b Ψ η dy = Φ Φ)Ψ) η Ψb (u Φ) η) dy. (S) Φ (S)(b ((u Now using integration by parts on the first term of this equation gives (u Φ)b Ψ η dy = Φ Φ)Ψdiv(b η) dy (S) Φ (S)(u Substituting this into (2.7), gives + (Φ (S))(u Φ)Ψb η ν dσ Φ (S) Ψb (u Φ) η) dy. Ω a u Ψ dx = Φ (S) (2Ψb (u Φ) η + (u Φ)Ψdiv(b η)) dy + b v Ψ dy Φ Φ)Ψb η ν dσ. (S) (Φ (S))(u By definition of η, η is zero on Υ r M (x, 0) R n + and Ψ = 0 on Φ (D S). This gives Ω a u Ψ dx = Φ (S) (2Ψb (u Φ) η + (u Φ)Ψdiv(b η)) dy + Φ (S) b v Ψ dy Ñ B rm (x,0) (u Φ)Ψb η ν dσ. (2.8) Using the same change of variables as earlier, but on the left-hand side of (2.6), we get Ω g Ψ dx + N f N Ψ dσ = Φ (S) detdφ (g Φ )Ψ dy + Φ (N) B r M (x,0) + ϕ 2 (f N Φ )Ψ dσ. (2.9) Now, combining (2.8), (2.9), and (2.6) gives b v Ψ dy = Φ Ñ B ( + ϕ 2 (f N Φ ) + (u Φ)b v ν) Ψ dσ (S) r M (x,0) + Φ (S)( detdφ (g Φ ) 2b (u Φ) η (u Φ)div(b η))ψ dy. 4

23 Finally, recalling the definition of g and f N, along with recalling the support of v, we have R n + b v Ψ dy = Ñ f N Ψ dσ + gψ dy R n + for all Ψ W,2 D (Rn +) as desired. It is obvious from the definition that D and Ñ in (2.4) still give a Lipschitz dissection of the boundary. The following Lemma will allow us to ensure that the difference quotient of our test functions still lie in the same test space. Lemma 2.7. Let Ω = R n + with Ñ and D, a Lipschitz dissection of the boundary, given by one Lipschitz function, Ψ. For x D and h > 0, we have that x + hα D for all α {(α, α, 0) α > M α }, where M is the Lipschitz coefficient for Ψ. Proof. (of Lemma 2.7) Let x D, h > 0, and α be as in the Lemma. First note that x + hα = (x, x, 0) + h(α, α, 0) = (x + hα, x + hα, 0) This gives that x + hα R n +. Next note, Ψ(x + hα ) = Ψ(x + hα ) Ψ(x ) + Ψ(x ) Mh α + Ψ(x ) since Ψ is Lipschitz. Now, since x D Ψ(x + hα ) Mh α + x hα + x by choice of α. Hence we have that x + hα D. Remark: It is clear that we can choose a basis for R n + in {(α, α, 0) α > M α }. We will choose such a basis later in this chapter. 5

24 2.3 Besov Spaces We want to show that our function, v, is in a Besov space. We begin by defining the Besov spaces we are interested in, B s p,q(r n +), as the real interpolation space, B s p,q(r n +) = (L p (R n +), W,p (R n +)) s,q. For a definition of interpolation spaces see Bergh and Löfström [2]. We also define some related spaces found in Savaré [3]. For θ > 0, define D +θ (Rn +) = w W,2 (R n +) sup h>0 k=,...,n For this paper, we are interested in B /2 2, w( + he k ) w L 2 (R n +) h θ <. and B3/2 2,. In order to prove our function is in one of these Besov spaces, we need an alternate definition for B 3/2 2, (Rn +). We start by proving some necessary results. Lemma 2.8. Let {β k } n k= give a basis for Rn +, then u D β +θ is an equivalent norm on D +θ. = u W,2 (R n + sup { u( + hβ k) u( ) L 2 (R n +) } +) h>0 h θ k=,...,n Proof. (of Lemma 2.8) First, since we know that {e,..., e n } is a basis for R n +, we have that β i = n j= c ij e j, so n ( + hβ i ) u( ) L 2 (R n = u( + h +) j= n u( + h j= c ij e j ) u( ) L 2 (R n +) c ij e j ) u( + hc i,n e n ) L 2 (R n +) + ( + hc i,n e n ) u( ) L 2 (R n +) where we subtracted and added u( + hc i,n e n ) and used the triangle inequality. Note that we can change variables on the first term using y = x + hc i,n e n. If c i,n < 0, we can translate the last term as well to get the following result n 2 u( + hβ i ) u( ) L 2 (R n u( + h +) Continuing in this matter for each e j, we get j= +h θ sup k>0 c ij e j ) u( ) L 2 (R n +) u( + ke n ) u( ) L 2 (R n +) k θ. 6

25 ( + hβ i ) u( ) L 2 (R n h +) θ n From this, it follows easily that The other direction follows analogously. j= sup k>0 u D β c u D. +θ +θ u( + ke j ) u( ) L 2 (R n +) k θ. In order to prove several of the following results, we need to utilize a basic result for difference quotients found in Evans [7, p ]. First, we define for ξ R n {0}/{0}, and τ ξ u = u( + ξ) (2.0) ξ u = u( + ξ) u( ). (2.) ξ Note that this notation is a little confusing, ξ is not the Laplacian. Lemma 2.9. Let v W,p (R n +) for p < be supported on a unit ball, then for ξ R n with ξ n = 0, we have ξ v L p (R n +) c v L p (R n +). Lemma 2.0. For D 2 defined earlier (θ = ), we have that where B defined in (2.5). W 2,2 (R n +) = {u D 2(R n +) Bu L 2 (R n +)} Proof. (of Lemma 2.0) If u W 2,2 (R n +), easily we have that u D 2 (Rn +) and integration by parts gives that ( Bu)(Ψ) = R n + l(z)ψ(z)dz for some l L 2 (R n +). This is what we mean for Bu L 2 (R n +), which gives the first direction. For the other direction, suppose that u D 2 (Rn +) and Bu L 2 (R n +). We will begin by proving that 2 u x 2 L2 (R n +) will follow from showing that 2 u n x i x j L 2 (R+) n for (i, j) (n, n). By Theorem on interior H 2 -regularity on page 327 of Evans [7], we have that u W 2,2 loc (Rn +), since Bu L 2 (R n +). Moreover, if l L 2 (R n +) is the function satisfying ( Bu)(Ψ) = R n + l(z)ψ(z)dz, 7

26 then by the remarks following Theorem on page 328 of Evans [7], we have that div(b u) + u = l almost everywhere. Writing this in non-divergence form gives n b ij i,j= 2 n u + x i x j almost everywhere. Rearranging, we have b nn 2 u x 2 n = Since b is elliptic, we know that n b ij i,j= i+j 2n 0 < θ θ e n 2 k= 2 n u + x i x j c k u x k + u = l k= c k u x k + u l. n b ij (e n ) i (e n ) j = b nn. i,j= This gives that b nn > 0 and therefore we can divide to obtain, 2 u x 2 n = n i,j= i+j 2n b ij b nn 2 u + x i x j n k= c k u + u l b nn x k b nn b nn almost everywhere. Now, c k and b ij are in L (R n +), b nn ε, and we know that u, u x k, and l are in L 2 (R n +). If we know that 2 u x i x j L 2 (R n +) for (i, j) (n, n), then we would have that 2 u L2 (R n +) and thus u W 2,2 (R n +) as desired. We only need to prove that 2 u x i x j x 2 n L 2 (R n +) for (i, j) (n, n). Since u D 2 (Rn +), we have that for i n u( + he i ) u( ) L 2 (R n +) c h for all h > 0. Thus we have a sequence of {h k } k= for which u x j ( + h k e i ) u x j ( ) converges to v ij L 2 (R n +) weakly. Given Φ C c (R n +), we have that h k R n + u Φ u Φ(y h k e i ) Φ(y) dy = x j x lim dy i R n + x j k h k u = lim k ( Φ(y h ke i ) Φ(y) ) dy R n + x j h k by the Dominated Convergence Theorem. Now changing variables gives 8

27 R n + u Φ dy = lim x j x i k R n + u x j (y + h k e i ) u = R n + v ij (y)φ(y) dy h k x j (y) Φ(y) dy by weak convergence of the difference quotient. Hence the weak derivative of u namely and so 2 u x 2 n 2 u x j x i, is equal to v ij. Hence we have for (i, j) (n, n) that L2 (R n +). This gives us the result. Remark: It is obvious that W,2 (R n +) = {u W,2 (R n +) Bu (W,2 0 (R n +)) } x j, 2 u x j x i L 2 (R n +) since for u W,2 (R n +), we know Bu (W,2 0 (R n +)). Now, we need a basic interpolation result before we can continue, but first we take the following definition of compatible couples of Banach Spaces from Bergh and Löfström [2]. We say that X 0 and X is a compatible couple if there is a normed Hausdorff topological vector space X such that X 0 and X are subspaces of X. Then, X 0 + X and X 0 X are also subspaces of X. We denote compatible couples by (X 0, X ). The following result can be found in Theorem 3..2 in Bergh and Löfström [2]. Lemma 2.. Suppose there exists an operator T, such that for compatible couples (X 0, X ) and (Y 0, Y ) we have T X 0 Y 0 T Y Y then for 0 < θ < and p, we have that T (X 0, X ) θ,p (Y 0, Y ) θ,p. We also need the following Lemma from Baiocchi []. Lemma 2.2. Let (X 0, X ) and (Z 0, Z ) be compatible couples of Banach Spaces and A X 0 + X Z 0 + Z. Suppose a right inverse, Â, to A exists such that  Z 0 X 0  Z X  A X 0 X 0  A X X A  = id. 9

28 Let Y 0 = {u X 0 Au Z 0 } Y = {u X Au Z }, then we have for 0 < θ < and q that (Y 0, Y ) θ,q = {u (X 0, X ) θ,q Au (Z 0, Z ) θ,q }. For completeness, we include the proof of this Lemma. Proof. (of Lemma 2.2) First, to prove the forward direction of the containment, note that we have where i is the natural inclusion and i Y k X k A Y k Z k for k = 0, easily. Lemma 2. gives for 0 < θ < and q, that i (Y 0, Y ) θ,q (X 0, X ) θ,q and A (Y 0, Y ) θ,q (Z 0, Z ) θ,q. This means that if u (Y 0, Y ) θ,q, then u (X 0, X ) θ,q and Au (Z 0, Z ) θ,q. Therefore, (Y 0, Y ) θ,q {u (X 0, X ) θ,q Au (Z 0, Z ) θ,q }. Now to get the other direction of the containment, start by defining P(u, f) = u Â(Au f). If u X 0 and f Z 0, then Au f Z 0. We know by assumption that Â(Au f) X 0, hence Similarly, we have that Next note, P(u, f) X 0 Z 0 X 0. P(u, f) X Z X. 20

29 A(P(u, f)) = A(u Â(Au f)) = Au AÂ(Au f) = Au (Au f) since AÂ = id. Hence A(P(u, f)) = f. In either case we have that, P X k Z k Y k for k = 0,. Again by Lemma 2., we know that P (X 0 Z 0, X Z ) θ,q (Y 0, Y ) θ,q for 0 < θ < and q. Let u {u (X 0, X ) θ,q Au (Z 0, Z ) θ,q }, then (u, Au) (X 0 Z 0, X Z ) θ,q. This means that P(u, Au) (Y 0, Y ) θ,q. Now note that P(u, Au) = u. Therefore we have that u (Y 0, Y ) θ,q as desired. Lemma 2.3. Our operator B, defined in (2.5) satisfies the conditions from Lemma 2.2. Meaning there exists a one-sided inverse, ˆB, such that. ˆB (W,2 0 (R n +)) W,2 0 (R n +) W,2 (R n +) 2. ˆB L2 (R F n + ) D 2 (Rn +) 3. B ˆB = id 4. ˆB B W,2 (R n +) W,2 (R n +) 5. ˆB B D 2 (Rn +) D 2 (Rn +) Proof. (of Lemma 2.3) First, given f (W,2 0 (R n +)), Lax-Milgram gives that there exists w W,2 0 (R n +) such that ( Bw)(Ψ) = f(ψ) for all Ψ W,2 0 (R n +), since B is coercive. Define ˆB(f) = w. Note that is clear directly from Lax-Milgram. Result 4 follows directly from and the fact that B W,2 (R n +) W,2 (R n +). Next we have that, B( ˆB(f))(Ψ) = ( Bw)(Ψ) = f(ψ) which gives 3. Now to show 5, let v = ˆB Bu with u D 2 (Rn +). We want to show that v D 2 (Rn +). By 3, we know that Bv = Bu. Hence Bu(Ψ) = Bv(Ψ) = (b v Ψ + vψ) dy. R n + 2

30 Also, we know that v W,2 0 (R n +) by definition of ˆB. Easily, we have ξ v W,2 0 (R n +), so we can use it as a test function. Hence Now define, and Changing variables gives, ( Bu)( ξ v) = R n + (b v ( ξ v) + v ξ v) dy. (2.2) τ ξ ( Bv)(Ψ) = R n + (τ ξ (b v) Ψ + τ ξ vψ) dy ξ ( Bv)(Ψ) = τ ξ( Bv)(Ψ) Bv)(Ψ). ξ τ ξ ( Bv)(Ψ) = (b v (τ R n ξ Ψ) + vτ ξ Ψ) dy + = ( Bv)(τ ξ Ψ) = ( Bu)(τ ξ Ψ) = τ ξ ( Bu)(Ψ) where we use the same reasoning for the last step as on the first two, but in the reverse direction. Hence letting Ψ = ξ v we have that (τ ξ Bu)( ξ v) = R n + (τ ξ (b v) ( ξ v) + τ ξ v ξ v) dy. (2.3) Now subtracting (2.2) from (2.3) and dividing by ξ gives that ( ξ Bu)( ξ v) = ξ ((τ ξ (b v) b v) ( ξ v) + (τ ξ v v) ξ v) dy. R n + Adding and subtracting (τ ξ b) v to the integrand gives τ ξ (b v) b v = (τ ξ b)τ ξ ( v) (τ ξ b) v + (τ ξ b) v b v = (τ ξ b)(τ ξ v v) + (τ ξ b b) v. Hence, ( ξ Bu)( ξ v) = R n + ((τ ξ b) ( ξ v) ( ξ v) + ( ξ b) v ( ξ v) + ( ξ v)( ξ v)) dy. Since τ ξ b is elliptic, we have 22

31 min(λ, ) R n + ( ( ξ v) 2 + ξ v 2 ) dy ( ξ Bu)( ξ v) R n + ξ b ( ξ v) dy ( ξ Bu)( ξ v) +M b v L 2 (R n +) ( ξ v) L 2 (R n +) where we use that b is Lipschitz with constant M b and Cauchy-Schwarz s Inequality. Next use Cauchy s Inequality with an ε on the last term to get min(λ, ) R n + ( ( ξ v) 2 + ξ v 2 ) dy ( ξ Bu)( ξ v) + c v 2 L 2 (R n +) Combining the similar terms, we have + min(λ, ) ( ξ v) 2 L 2 2 (R n +). R n + ( ( ξ v) 2 + ξ v 2 ) dy c ( ξ Bu)( ξ v) + c v 2 L 2 (R n +). (2.4) Now, ( ξ ˆBu)( ξ v) = R n + ( ξ (b u) ( ξ v) + ξ u ξ v) dy = R n + ((τ ξ b ξ u + ξ b u) ( ξ v) + ξ u ξ v) dy using our work from earlier. Next, ( ξ Bu)( ξ v) ( b L (R n +) ξ u L 2 (R n + M +) b u L 2 (R n +)) ( ξ v) L 2 (R n +) + ξ u L 2 (R n +) ξ v L 2 (R n +). Again Cauchy s Inequality with an ε, used twice, gives ( ξ Bu)( ξ v) c( ξ u 2 L 2 (R n +) + u 2 L 2 (R n +) ) + (/2) ( ξv) 2 L 2 (R n +) + ξ u L 2 (R n +) ξ v L 2 (R n +). 2 Hence, substituting this into (2.4) gives R n + ( ξ v 2 + ξ v 2 ) dy c( ξ u 2 L 2 (R n +) + u 2 L 2 (R n +) + ξ u L 2 (R n +) ξ v L 2 (R n +) + v 2 L 2 (R n +) ). 23

32 Since u D 2 (Rn +) and v W,2 (R n +), setting ξ = he i for i =,..., n, we easily get v D 2 (Rn +). Hence we have 5. To show 2, let f L 2 (R n +). We know that for u = ˆBf, that u W,2 0 (R n +) and R n + (b u Ψ + uψ) dy = R n + fψ dy for all Ψ W,2 (R n +). Using ξ u for a test function gives R n + (b u ( ξ u) + u ξ u) dy = R n + f ξ u dy. (2.5) Letting τ ξ ξ u be a test function and translating gives that R n + (τ ξ (b u) ( ξ u)τ ξ u ξ u) dy = R n + fτ ξ ( ξ u) dy. (2.6) Subtracting (2.5) and (2.6) gives R n + ((τ ξ (b u) b u) ξ u + (τ ξ u u) ξ u) dy = R n + f(τ ξ ( ξ u) ξ u) dy. Now the same work as before gives that R n + ( ξ u 2 + ξ u 2 ) dy R n + f ξ ( ξ u) dy + c u 2 L 2 (R n +). Fix w = ξ u and use Cauchy-Schwarz Inequality to get R n + ( ξ u 2 + ξ u 2 ) dy f L 2 (R n +) ξ (w) L 2 (R n +) + c u 2 L 2 (R n +). Next, using Lemma 2.9 we have that R n + ( ξ u 2 + ξ u 2 ) dy f L 2 (R n +) ξ u L 2 (R n +) + c u L 2 (R n +). Using Cauchy s Inequality with an ε on the first term on the right-hand side and rearranging gives R n + ( ξ u 2 + ξ u 2 ) dy c f L 2 (R n +) + c u L 2 (R n +). Since f L 2 (R n +) and u L 2 (R n +), we have that u D 2 by using ξ = he i for i =,..., n. Hence we have the lemma. Lemma 2.4. D 3/2 (Rn +) = (W,2 (R n +), D 2(R n +)) /2, Proof. (of Lemma 2.4) As noted in Savaré, the proof can be carried out using semigroup theory (See Bergh and Löfstrom) [2, p. 56]. 24

33 Lemma 2.5. We have the following alternate characterization of the Besov Space, B 3/2 2, (Rn +), B 3/2 2, (Rn +) = {w D 3/2 (Rn +) Bw (W,2 (R n +), L 2 (R n +)) /2, } where B is defined in (2.5). Proof. (of Lemma 2.5) This alternate definition for our Besov Space follows from Lemma 2.2, Lemma 2.3, and Lemma 2.4. From this definition, the following gives a norm for B 3/2 2, (Rn +) v 3/2 B = v W 2, (Rn +),2 (R n + sup { v( + hβ k v( )) L 2 (R n +) } (2.7) +) h>0 h k=,2,...,n + Bv (W,2 (R n +),L 2 (R n +)) /2, where {β k } is a basis of R n {0}. Next, we need a couple of basic results about Besov spaces. Lemma 2.6. If v B 3/2 (R n +), then v B /2 2, (Rn +). Moreover, v B /2 2, (Rn +) c v B 3/2 2, (Rn +). The Lemma above is obvious and follows directly from the definition of interpolation spaces. Lemma 2.7. If v B /2 2, (Ω) for a C domain, Ω, then for 2 < q < 2n n, we have v L q (Ω) c v B /2 2, (Ω) if v is supported on a ball. The constant, c, depends on this support. Proof. (of Lemma 2.7) Theorem 2 from pg 279 of Evans [7] gives that W,2 (Ω) L 2n n 2 (Ω), meaning for u W,2 (Ω) we have u L 2n n 2 c u W,2 (Ω). We also have that L 2 (Ω) L 2 (Ω). Hence, by Lemma 2. we have that (L 2 (Ω), W,2 (Ω)) θ,q (L 2 (Ω), L 2n n 2 (Ω)) θ,q. Recall that B θ 2,q = (L2 (Ω), W,2 (Ω)) θ,q. We define the Lorentz space by where 0 < p 0 < p,, and L p,q (Ω) = (L p 0 (Ω), L p (Ω)) θ,q 25

34 p = θ + θ p 0 p with 0 < θ <. Note that this is not the standard definition of a Lorentz space, but rather an equivalent definition given by Theorem 5.2. on page 09 of Interpolation Spaces by Bergh and Löfström [2]. In our particular case, p 0 = 2, p = 2n n 2, θ = 2, and q =. This gives us that p = 2n n. Hence we have that This means for u B /2 2, (Ω), we have that B /2 2, L 2n n, (Ω). u 2n c u L n, (Ω) B /2 (Ω). (2.8) Next we have another definition of the Lorentz Space which holds only when q = given by 2, We define L p, (Ω) = {f sup (t p m({ f > t})) < }. t>0 W p (f) = {sup t>0 (t p p m({ f > t}))} and we have that W p (f) c f L p, (Ω). Fix f L p, (Ω). Let E Ω be a set with finite measure that contains the support of f. Let 2 < s < p, then E f s dy = E s 0 f t s dt dy. Next using Fubini to change the order of integration, we have E f s dy = = 0 0 = s 0 st s {y E f(y) >t} dy dt st s m({ f > t}) dt A t s m({ f > t})dt + s ca s m(e) + cw p p (f) A A t s p dt t s m({ f > t}) dt where we use that t p m({ f > t}) is bounded by the pth-power of of W p (f). Hence, we have E f s dy ca s m(e) + cw p p (f)a s p. 26

35 Setting A = W p (f) gives f L s (Ω) cw p (f). Combining this with (2.8) gives the result. 2.4 Proof of Reverse Hölder Inequality In this section, we will use a difference quotient method to prove our Reverse Hölder Inequality. To prove Theorem 2., we begin by assuming that r =. Lemma 2.8. Assuming that our Neumann data f N has a constant value of κ on B (x), then for our function v, defined in (2.3), we have v B 3/2 2, (Rn +). Moreover, recalling u in the definition of v, we have v B 3/2 2, (Rn +) c u L 2 (Υ (x)) + c u L 2 (Υ (x)) + c g L 2 (Υ (x)) + c κ. Proof. (of Lemma 2.8) Recalling Lemma 2.5, we will begin by getting a bound on the D 3/2 (Rn +) norm of v and also the (W,2 (R n +), L 2 (R n +)) /2, norm of Bv. Let {β i } n i= be a basis for Rn + chosen so that each β {(α, α, 0) α > M α } from Lemma 2.7. Let h > 0. From this Lemma we easily see that v( + hβ i ) W,2 D (Rn +). So we can let v τ hβi v be a test function, which gives that R n + b v (v τ hβi v) dy + R n + v(v τ hβi v) dy = R n + ( g + v)(v τ hβi v) dy + Ñ f N (v τ hβi v) dy. Let s look at just the first term on the left-hand side to see, R n + b v (v τ hβi v) dy = 2 { R n + b (v τ hβi v) (v τ hβi v) dy + R n + b τ hβi v (v τ hβi v) dy + R n + b v (v τ hβi v) dy} by straight forward calculation and utilizing that b is symmetric. Now using ellipticity on the first term of the right-hand side gives 27

36 b v (v τ hβi v) dy R n + 2 {c (v τ hβi v) 2 dy + R b τ hβi v v dy n + R n + b τ hβi v τ hβi v dy + R b v v dy n + R n + b v τ hβi v dy} R n + = 2 {c (v τ hβi v) 2 dy R b τ hβi v τ hβi v dy n + R n + + b v v dy}. R n + Now we will focus on the last two terms on the right-hand side. By changing variables on the last term we have that R n + b v v dy R n + b τ hβi v τ hβi v dy = R n + (b τ hβi b) v v dy. Altogether, we now have that R n + (v τ hβi v) 2 dy c { R n + (τ hβi b b) v v dy + 2 R n + ( g + v)(v τ hβi v) dy Now, notice that +2 Ñ f N (v τ hβi v) dy 2 v(v τ hβi v) dy}. R n + since b is Lipschitz. Hence, R n + (τ hβi b b) v v dy ch R n + v 2 dy R n + (v τ hβi v) 2 dy c {ch R n + v 2 dy + 2 R n + ( g + v)(v τ hβi v) dy +2 Ñ f N (v τ hβi v) dy 2 v(v τ hβi v) dy} R n + ch R n + v 2 dy + c R n + g v τ hβi v dy +c R n + fn (v τ hβi v) dy where we notice some terms cancel. For the second term, note that R n + g v τ hβi v dy g L 2 (R n +) v τ hβi v L 2 (R n +) g L 2 (R n +)(ch v L 2 (R n +)) 28

37 where we used Lemma 2.9 and the fact that v is supported in a unit ball. Next, using that ab /2(a 2 + b 2 ), we have R n + g v τ hβi v dy ch g 2 L 2 (R n +) + ch v 2 L 2 (R n +). Hence, putting it all back together we now have R n + (v τ hβi v) 2 dy ch R n + v 2 dy + ch R n + g 2 dy + c R n + fn (v τ hβi v) dy. (2.9) Now we consider the last term, R n + fn (v τ hβi v) dσ(y) = R n + (u Φ)b η ν(v τ hβi ) dσ+ R n + κ + ϕ 2 (v τ hβi v) dσ using the definition of fn. For the second term, changing variables to push the difference onto + ϕ 2, using Cauchy-Schwarz, and recalling that ϕ is Lipschitz, gives κ + ϕ 2 (v τ hβi v) dσ R N + ch κ v L 2 (R n +) ch κ 2 + ch v 2 L 2 (R n +) where we again use that ab 2 (a2 + b 2 ). By the divergence theorem we have (u Φ)b η ν(v τ hβi v) dσ = R div((u Φ)(v τ hβi v)b η) dy. n + R n + Straight forward calculation gives (u Φ)b η ν(v τ hβi v) dσ R n + = b η ((u Φ)(v τ hβi v)) dy R n + + (u Φ)(v τ hβi v)div(b η) dy R n + ((u Φ) (v τ hβi v) R n + +(v τ hβi v) (u Φ))b η dy +c u Φ L 2 (Υ /M (x,0)) v τ hβi v L 2 (R n +) where we use the product rule in the first integral and for the second term we use Cauchy-Shwarz and the fact that η is bounded. Again, using that ab 2 (a2 + b 2 ) on the last term as well as Lemma 2.9, gives 29

38 (u Φ)b η ν(v τ hβi v) dσ R n + ((u Φ) (v τ hβi v) + (v τ hβi v) (u Φ))b η dy R n + + ch u Φ 2 L 2 (Υ /M (x,0)) + ch v 2 L 2 (R n +). We look at the first term on the right-hand side, to see ((u Φ) (v τ hβi v) + (v τ hβi v) (u Φ))b η dy R n + = (u Φ) (v τ hβi v) b η dy + R (v τ hβi v) (u Φ) b η dy n + R n + R n + (u Φ) (v τ hβi v) b η dy + ch (u Φ) 2 L 2 (Υ /M (x,0)) + ch v 2 L 2 (R n +). Combining these bounds with (2.9), we have R n + (v τ hβi v) 2 dy ch R n + v 2 dy + ch R n + g 2 dy + ch u Φ 2 W,2 (Υ /M (x,0)) + R n + (u Φ) (v τ hβi v) b η dy + ch κ 2. Now we turn our attention to the second to last term. splitting up terms gives Changing variables after R n + (u Φ) (v τ hβi v) b η dy = R n + ((u Φ)b η τ hβi ((u Φ)b η)) v dy ch (u Φ)b η L 2 (Υ /M (x,0)) v L 2 (R n +). Using the same ideas as before, we have R n + (u Φ) (v τ hβi v) b η dy ch u Φ 2 L 2 (Υ /M (x,0)) + ch v 2 L 2 (R n +). Together we now have R n + (v τ hβi v) 2 dy ch ( R n + v 2 dy + R n + g 2 dy + u Φ 2 W,2 (Υ /M (x,0)) + κ 2 ). Now, we need to deal with g term. Recalling the definition of g, we have 30

39 g L 2 (R n +) (u Φ)div(b η) L 2 (Υ /M (x,0)) + 2b (u Φ) η L 2 (Υ /M (x,0)) + det Φ g φ L 2 (Υ /M (x,0)) c u Φ L 2 (Υ /M (x,0)) + c (u Φ) L 2 (Υ /M (x,0)) +c g Φ L 2 (Υ /M (x,0)). Altogether now we have, R n + (v τ hβi v) 2 dy ch ( v 2 L 2 (R n +) + g Φ 2 L 2 (Υ /M (x,0)) + u Φ 2 W,2 (Υ /M (x,0)) + κ 2 ). Recalling the definition of v and changing variables back to Ω gives (v τ hβi v) 2 dy ch R { u 2 dy + n + Υ u 2 dy + g 2 L (x) Υ (x) 2 (Υ (x)) + κ 2 }, which gives us a bound for the D norm. Using the definition of the norm of the 3/2 Besov space given in (2.7), we have the result by bounding the Bv term its L 2 norm and then by v W,2 (R n and changing variables back to Ω. +) To prove our Reverse Hölder Inequality, we need to prove a Poincaré Inequality. The following domains are useful in that they allow us to find a family of domains for which the Poincaré inequality holds. We say that Ω is star-shaped Lipschitz with constant M and scale r if there is function, ρ (B (x)) [, M + ], that is Lipschitz with constant M such that Ω = {y x x y < rρ ( y x )} {x} y x for some x Ω, which is the domain s center. For the following lemmas, we will assume that our domains are centered at x = 0 to simplify the proofs. Lemma 2.9. If Ω is star-shaped Lipschitz with constant M and scale r, then there is a bi-lipschitz function f that maps B r (0) onto Ω. Proof. (of Lemma 2.9) Define f B r (0) R n by If y = 0, easily f(y) Ω. If y 0, then yρ ( y y f(y) = ) if y 0 0 if y = 0. 3

40 f(y) y ρ ( y y ) rρ ( y y ). Hence f(y) Ω. Let z Ω. If z = 0, then easily z is in the range of f. If z 0, define y = z ρ ( z z ). Since y < r and f(y) = z easily, we have that f is onto Ω. Since it is clear that y = 0 is the only point that maps to 0, let y, w 0 by two points in B r (0) with f(w) = f(y). Now we know that yρ ( y y ) = wρ ( w w ). (2.20) Now, since ρ maps into [, M + ], it must be that y and w are in the same direction, meaning y y = w w. This gives that ρ ( y y ) = ρ ( w w ). Dividing these out of (2.20), gives that y = w. Thus we know that f has an inverse and it is given by z if z 0 f ρ( (z) = z z ) 0 if z = 0. Now we want to show that f and f are Lipschitz. Let x and y be in B r (0). Note, f(x) f(y) = xρ ( x x ) yρ ( y y ) = xρ ( x x ) xρ ( y y ) + xρ ( y y ) yρ ( y y ) x ρ ( x x ) ρ ( y y ) + x y ρ ( y y ) M x x x y + (M + ) x y y where we use that ρ is Lipschitz and bounded by M +. Now, 32

41 x y y x f(x) f(y) M x + (M + ) x y x y x y y y + y y y x = M + (M + ) x y y y x y + y x y M + (M + ) x y y 2M x y + (M + ) x y = (3M + ) x y. We have assumed that x and y are not zero, but the result is even easier if one or both of them equal zero. Hence we now have that f is Lipscitz. Now to show that f is Lipschitz, note that ρ maps into [, M + ], so ρ maps into [ +M, ]. Now if y = 0 and x 0, then f (x) f (y) = f x (x) = x 0 ρ ( x x ) so the Lipschitz condition is satisfied. If both x, y 0, then we have f (x) f (y) = = x ρ ( x x ) y ρ ( y y ) xρ ( y y ) yρ ( x x ) ρ ( x x ) ρ ( y y ) xρ ( y y ) yρ ( x x ) using that ρ maps into [ M+, ]. Next f (x) f (y) xρ ( y y ) xρ ( x x ) + xρ ( x x ) yρ ( x x ) x ρ ( y y ) ρ ( x x ) + ρ ( x ) x y x (3M + ) x y by the work we already did. Hence f is Lipschitz. This gives the result. 33

42 Lemma Suppose that u C ( Ω), where Ω is star-shaped Lipschitz with constant M and scale r and u = 0 on S Ω, where S has positive measure, then ( u np p n n p dy) Ω c ( u p p dy) Ω where c depends on σ(s), if and only if u f where f is from Lemma 2.9, satisfies the same inequality on B r (0). This result follows from a change of variables. Note that the previous two lemmas enable us to get the desired inequality on our star-shaped domains by proving that we have the result on a ball. Given a set S of positive measure, we using the following notation to denote the average value of a function over S ū S = S u dy. Now, we introduce the following well-known result. Lemma 2.2. Let u C ( Ω), where Ω is star-shaped Lipschitz with constant M and scalar r, then for < p < n, we have u ū S L q (Ω) where p q = n, d = diam(ω), and S = B ɛr(x). cd n n m(s) u L p (Ω) Proof. (of Lemma 2.2) Let x Ω and y S. Calculus, we have By the Fundamental Theorem of u(x)m(s) S u(y) dy = S 0 y x u (x + y x y x t) y x dt dy. y x Now, letting ξ = y x and ˆω = y x y x, we can switch to polar coordinates by using that dy = ξ n dξ dˆω. Dividing by m(s) gives u(x) S u(y) dy = r ξ m(s) u(x + tˆω) ˆωχ Ω (x + ˆωt) dt ξ n dˆω dξ. 0 B (0) 0 Taking the absolute value gives u(x) ū S r ξ m(s) u(x + tˆω) ˆωχ Ω (x + ˆωt) dt ξ n dˆω dξ 0 B (0) 0 d ξ m(s) v(x + ˆωt) dt ξ n dˆω dξ 0 B (0) 0 where we use that ˆω =, set v(z) = χ Ω (z) u(z), and use that d = diam(ω), r < d. Switching the order of integration gives, 34