VERTICAL BLOW UPS OF CAPILLARY SURFACES IN R 3, PART 1: CONVEX CORNERS

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1 Electronic Journal of Differential Equations, Vol. 2007(2007), No. 152, pp ISSN: URL: or ftp ejde.math.txstate.edu (login: ftp) VERTIL LW UPS F PILLRY SURFES IN R 3, PRT 1: NVEX RNERS THLI JEFFRES, KIRK LNSTER bstract. ne technique which is useful in the calculus of variations is that of blowing up. This technique can contribute to the understanding of the boundary behavior of solutions of boundary value problems, especially when they involve mean curvature and a contact angle boundary condition. ur goal in this note is to investigate the structure of blown up sets of the form P R and N R when P, N R 2 and P (or N ) minimizes an appropriate functional; sets like P R can be the limits of the blow ups of subgraphs of solutions of mean curvature problems, for example. In Part ne, we investigate blown up sets when the domain has a convex corner. s an application, we illustrate the second author s proof of the oncus-finn onjecture by providing a simplified proof when the mean curvature is zero. 1. Introduction onsider the nonparametric prescribed mean curvature problem with contact angle boundary data in the cylinder Ω R where n 2, Ω R n is bounded and open, T f = Nf = H(x, f) for x Ω (1.1) T f ν = cos γ on Ω, (1.2) f, Nf = T f, ν is the 1+ f 2 exterior unit normal on Ω, γ : Ω [0, π] and f 2 (Ω). Examples show that even if Ω is smooth, a finitely valued solution f of (1.1)-(1.2) need not exist (e.g. [4]). If Ω is locally Lipschitz, then one might consider formulating (1.1)-(1.2) as a variational problem. If Ω is not smooth at x 0 Ω and a generalized solution f of (1.1)-(1.2) exists, the behavior of f near x 0 is often of great interest; when f is a variational solution of (1.1)-(1.2) and H(, f( )) L (Ω), then the blow ups of f at x 0 will (usually) be minimal hypersurfaces and specific information about the behavior of these blow ups can contribute to an understanding of the behavior of f near x 0. Suppose f V (Ω) minimizes the functional F : V (Ω) R given by g F(g) = 1 + Dg 2 dx + H(x, t) dt dx cos(γ(x))g dx; (1.3) Ω Ω 2000 Mathematics Subject lassification. 49Q20, 5310, Key words and phrases. low-up sets; capillary surface; oncus-finn conjecture. c 2007 Texas State University - San Marcos. Submitted January 9, Published November 13, Ω

2 2 T. JEFFRES, K. LNSTER EJDE-2007/152 that is, f is a variational solution of (1.1)-(1.2). ssume Ω is locally Lipschitz, = (0,..., 0) Ω R n and H(, f( )) L (Ω). Let {ɛ j } and {z j } be real sequences with ɛ j 0 as j. For each integer j 1, set Ω j = {x R n : ɛ j x Ω} and define f j V (Ω j ) by notice that f j minimizes F j (g) = 1 + Dg 2 dx + Ω j f j (x) = f(ɛ jx) z j ɛ j ; (1.4) Ω j g 0 H j (x, t) dt dx cos(γ j (x))g dx Ω j and is a variational solution of Ng = H j (x, g) for x Ω j and T g ν j = cos γ j on Ω, where H j (x, z) = ɛ j H(ɛ j x, ɛ j z), γ j (x) = γ(ɛ j x) and ν j = ν(ɛ j x). Notice that H j (, f j ( )) 0 as j. Suppose Ω = lim j Ω j, γ = lim j γ j and ν = lim j ν j exist. s in [9] (also [17, Lemma 1.2]), we can find a subsequence of {f j }, which we continue to denote {f j }, which converges locally to a function f : Ω [, ] in the sense that φ Uj φ U in L 1 loc (Ω R) as j, where U j = {(x, t) Ω j R : t < f j (x)} denotes the subgraph of f j, U = {(x, t) Ω R : t < f (x)} denotes the subgraph of f and φ V denotes the characteristic function of a set V (e.g. [5], [6], [12]). Furthermore, f is a generalized solution of the functional F (g) = 1 + Dg 2 dx cos(γ )g dh n (1.5) Ω Ω in the sense that for each compact subset K of R n+1 with finite perimeter, U minimizes the functional F K defined on subsets of Ω R by F K (V ) = Dφ V cos(γ )φ V dh n. The sets K (Ω R) K ( Ω R) P = {x Ω : f (x) = }, (1.6) N = {x Ω : f (x) = }, (1.7) have a special structure which is of principal interest to us. The set P minimizes the functional Φ() = Dφ cos(γ )φ dh n. (1.8) Ω Ω and the set N minimizes the functional Ψ() = Dφ + cos(γ )φ dh n (1.9) Ω Ω in the appropriate sense (e.g. [5], [16]). fter modification on a set of measure zero, we may assume that U, P and N coincide with the essential boundaries of U, P and N respectively (e.g. [5, Theorem 1.1]). When the limiting contact angle γ satisfies certain conditions (depending on Ω ), the continuity at and the behavior near of f are unknown. It would be valuable to understand the geometry of sets P and N in Ω which minimizes Φ and Ψ respectively when such conditions hold. ur goal is to determine the possible geometries of P and N when n = 2 and γ satisfies appropriate conditions and to

3 EJDE-2007/152 VERTIL LW UPS 3 illustrate the application of this knowledge by proving the oncus-finn onjecture in the special case that the prescribed mean curvature H is zero. 2. Statement of Results Set n = 2; we will denote elements of R 2 by (x, y) rather than by x = (x 1, x 2 ) as in the previous section. Let Ω be an open subset of R 2 with a corner at = (0, 0) Ω such that, for some δ 0 > 0, Ω is piecewise smooth in δ0 () and Ω δ0 () consists of two 1 arcs + Ω and Ω whose tangent lines approach the lines L + = {θ = α} and L = {θ = α}, respectively, as the point is approached. Let ν + and ν denote the exterior unit normals on + Ω and Ω respectively. Here α [0, π], polar coordinates relative to are denoted by r and θ and δ () is the ball in R 2 of radius δ about. Let (x + (s), y + (s)) be an arclength parametrization of + Ω and (x (s), y (s)) be an arclength parametrization of Ω, where s = 0 corresponds to the point for both parametrizations. We will assume γ 1 = lim s 0 γ(x + (s), y + (s)), γ 2 = lim s 0 γ(x (s), y (s)) both exist and γ 1, γ 2 (0, π). In this case, ( Ω ) \ {} = Σ 1 Σ 2 with Ω = {(r cos θ, r sin θ) : r > 0, α < θ < α}, (2.1) Σ j = {(r cos θ, r sin θ) : r > 0, θ = ( 1) j+1 α}, j = 1, 2, and the limiting contact angle γ equals γ 1 on Σ 1 and γ 2 on Σ 2. set P Ω minimizes Φ if and only if for each T > 0, Φ T (P) Φ T (P S), Φ T (P) Φ T (P \ S) for every S Ω T, where Ω T = T () Ω, Σ T j = T () Σ j, j = 1, 2, and Φ T () = Dφ cos(γ 1 ) φ dh 1 cos(γ 2 ) Ω T Σ T 1 Σ T 2 φ dh 1 = H 1 (Ω T ) cos(γ 1 )H 1 (Σ T 1 ) cos(γ 2 )H 1 (Σ T 2 ). set N Ω minimizes Ψ if and only if for each T > 0, Ψ T (N ) Ψ T (N S), Ψ T (N ) Ψ T (N \ S) for everys Ω T where Ψ T () = Dφ + cos(γ 1 ) φ dh 1 + cos(γ 2 ) φ dh 1 Ω T Σ T 1 Σ T 2 = H 1 (Ω T ) + cos(γ 1 )H 1 (Σ T 1 ) + cos(γ 2 )H 1 (Σ T 2 ). If P minimizes Φ, then after modification on a set of measure zero, we may assume P coincides with the essential boundary of P (e.g. [5, Theorem 1.1]) and Ω P consists of a union of rays. If N minimizes Ψ, then the same holds for N and Ω N. We may also assume P and N are open. It is convenient to introduce some notation. If and are points in R 2, then denotes the open line segment with endpoints and and denotes the closed line segment with endpoints and. When the context is clear as in an arithmetic equality or inequality, will denote the length of the line segments

4 4 T. JEFFRES, K. LNSTER EJDE-2007/152 and. If,, Ω, then denotes the open triangular region in R 2 with vertices, and. Theorem 2.1. Suppose α π/2 and P Ω minimizes Φ. Let (r, θ) be polar coordinates about. Then one of the following holds: (i) P = or P = Ω ; (ii) α < π 2, γ 1 + γ 2 π = 2α and P =, where Σ 1, Σ 2 and angles and have measures γ 1 and γ 2 respectively; (iii) α < π 2, γ 1 + γ 2 π = 2α and P = Ω \, where Σ 1, Σ 2 and angles and have measures π γ 1 and π γ 2 respectively; (iv) γ 1 + π γ 2 2α and there exists Σ 1 such that Ω P = Σ 1 \, Ω P is the ray L in Ω starting at and making an angle of measure γ 1 with Σ 1 \ and P is the open sector between Σ 1 \ and L; (v) γ 2 + π γ 1 2α and there exists Σ 1 such that Ω P = Σ 2, Ω P is the ray L in Ω starting at and making an angle of measure γ 1 with and P is the open connected region with boundary Σ 2 L; (vi) γ 1 + π γ 2 2α and there exists Σ 2 such that Ω P = Σ 1, Ω P is the ray L in Ω starting at and making an angle of measure γ 2 with and P is the open connected region with boundary Σ 1 L; (vii) γ 2 + π γ 1 2α and there exists Σ 2 such that Ω P = Σ 2 \, Ω P is the ray L in Ω starting at and making an angle of measure γ 2 with Σ 2 \ and P is the open sector between Σ 2 \ and L; (viii) γ 1 + π γ 2 2α, Ω P = Σ 1 {}, Ω P is a ray L = {θ = β} in Ω starting at which makes an angle of measure greater than or equal to γ 1 with Σ 1 and an angle of measure greater than or equal to π γ 2 with Σ 2 (i.e. π α γ 2 β α γ 1 ) and P = {β < θ < α}; or (ix) γ 2 + π γ 1 2α, Ω P = Σ 2 {}, Ω P is a ray L = {θ = β} in Ω starting at which makes an angle of measure greater than or equal to π γ 1 with Σ 1 and an angle of measure greater than or equal to γ 2 with Σ 2 (i.e. γ 2 α β α + γ 1 π) and P = { α < θ < β}. Theorem 2.2. Suppose α π 2 and N Ω minimizes Ψ. Let (r, θ) be polar coordinates about. Then one of the following holds: (i) N = or N = Ω ; (ii) α < π 2, γ 1 + γ 2 π = 2α and N = Ω \, where Σ 1, Σ 2 and angles and have measures γ 1 and γ 2 respectively; (iii) α < π 2, γ 1 + γ 2 π = 2α and N =, where Σ 1, Σ 2 and angles and have measures π γ 1 and π γ 2 respectively; (iv) γ 1 + π γ 2 2α and there exists Σ 1 such that Ω N = Σ 2, Ω N is the ray L in Ω starting at and making an angle of measure π γ 1 with and N is the open connected region with boundary Σ 2 L; (v) γ 2 + π γ 1 2α and there exists Σ 1 such that Ω N = Σ 1 \, Ω N is the ray L in Ω starting at and making an angle of measure π γ 1 with Σ 1 \ and N is the open sector between Σ 1 \ and L; (vi) γ 1 + π γ 2 2α and there exists Σ 2 such that Ω N = Σ 2 \, Ω N is the ray L in Ω starting at and making an angle of measure π γ 2 with Σ 2 \ and N is the open sector between Σ 2 \ and L;

5 EJDE-2007/152 VERTIL LW UPS 5 (vii) γ 2 + π γ 1 2α and there exists Σ 2 such that Ω N = Σ 1, Ω N is the ray L in Ω starting at and making an angle of measure π γ 2 with and N is the open connected region with boundary Σ 1 L; (viii) γ 1 + π γ 2 2α, Ω N = Σ 2 {}, Ω N is a ray L = {θ = β} in Ω starting at which makes an angle of measure greater than or equal to γ 1 with Σ 1 and an angle of measure greater than or equal to π γ 2 with Σ 2 (i.e. π α γ 2 β α γ 1 ) and P = { α < θ < β}; or (ix) γ 2 + π γ 1 2α, Ω N = Σ 1 {}, Ω N is a ray L = {θ = β} in Ω starting at which makes an angle of measure greater than or equal to π γ 1 with Σ 1 and an angle of measure greater than or equal to γ 2 with Σ 2 (i.e. γ 2 α β α + γ 1 π) and P = {β < θ < α}. Figure 1 illustrates the geometry of P in Theorem 2.1 and N in Theorem 2.2. In each case, the shaded region illustrates the geometry of P and the unshaded region illustrates the geometry of N. 3. pplication to capillarity onsider the stationary liquid-gas interface formed by an incompressible fluid in a vertical cylindrical tube with cross-section Ω. For simplicity, we assume that near (0, 0), Ω has straight sides (as in [14]) and so we may assume Ω = {(r cos(θ), r sin(θ)) : 0 < r < 1, α < θ < α}. (3.1) In a microgravity environment or in a downward-oriented gravitational field, this interface will be a nonparametric surface z = f(x, y) which is a solution of the boundary value problem (1.1)-(1.2) with H(z) = κz + λ; that is, Nf = κf + λ in Ω (3.2) T f ν = cos γ a.e. on Ω (3.3) where T f = f/ 1 + f 2, Nf = T f, ν is the exterior unit normal on Ω, κ and λ are constants with κ 0, γ = γ(x, y) [0, π] is the angle at which the liquid-gas interface meets the vertical cylinder ([3]) and γ 1, γ 1 (0, π) are as in 2. Lemma 3.1. Suppose α π 2 and γ 1 + π γ 2 < 2α. Let f 2 (Ω) 1 (Ω \ {}) satisfy (3.2) and (3.3) and define n(x, y) = ( T f(x, y), 1 ) 1 + f(x, y) 2 (3.4) to be the (downward) unit normal to the graph of f at (x, y, f(x, y)). Let β ( α, α) and let {(x j, y j )} be a sequence in Ω satisfying lim j (x j, y j ) = (0, 0) and lim j (x j, y j ) x 2 j + y2 j = (cos(β), sin(β)). (3.5) (i) If β [ α + π γ 2, α γ 1 ], then lim j n(x j, y j ) = ( sin(β), cos(β), 0). (ii) If β ( α, α + π γ 2 ], then lim j n(x j, y j ) = ( sin( α + π γ 2 ), cos( α + π γ 2 ), 0). (iii) If β [α γ 1, α), lim j n(x j, y j ) = ( sin(α γ 1 ), cos(α γ 1 ), 0).

6 6 T. JEFFRES, K. LNSTER EJDE-2007/152 case (ii) case (iii) case (iv) case (v) case (vi) case (vii) case (viii) case (ix) Figure 1. Illustration for Theorems 2.1 and 2.2 Proof. Suppose {(x j, y j ) : j N} is a sequence in Ω converging to (0, 0) as j and satisfying (3.5); we may assume x j > 0 for all j N. For each j N, define f j 2 (Ω j ) 1 (Ω j \ {}) by (1.4) with z j = f(x j, y j ) and ɛ j = x 2 j + y2 j, so

7 EJDE-2007/152 VERTIL LW UPS 7 that f j (x, y) = f(ɛ jx, ɛ j y) f(x j, y j ) ɛ j. Let U j be the subgraph of f j as in 1 and n j be the downward unit normal to the graph of f j ; that is, n j (x, y) = ( T f j (x, y), fj (x, y) 2 ), (x, y) Ωj. s in 1, there exists a subsequence of {(x j, y j )}, still denoted {(x j, y j )}, and a generalized solution f : Ω [, ] of (1.5) such that f j converges to f in the sense that φ Uj φ U in L 1 loc (Ω R) as j, where Ω is given in (2.1). Let P and N be given by (1.6) and (1.7) respectively. Notice that f j (x j /ɛ j, y j /ɛ j ) = 0 for all j N and so f (cos(β), sin(β)) = 0. Using density lower bounds (e.g. [18, Lemma 3.1]), we see that (cos(β), sin(β), 0) U and (cos(β), sin(β)) Ω \ (P N ); hence P = Ω and N Ω. Since γ 1 + π γ 2 < 2α, either P = or one of cases (iv), (vi) or (viii) of Theorem 2.1 holds and either N = or one of cases (iv), (vi) or (viii) of Theorem 2.2 holds. We claim that (Ω R) U is the portion of a plane Π in Ω R, with (cos(β), sin(β), 0) Π. There are two possibilities to consider. The first is that there exists ɛ > 0 such that ɛ (cos(β), sin(β)) (P N ) =. The second is that (cos(β), sin(β)) P N. Suppose the first case holds and let G = Ω \ (P N ). Notice that f is a classical (i.e. 2 (G)) solution of the minimal surface equation in G (e.g. [5], [11]). Let h(x, y) = f(x, y) Rf(β), where Rf(β) = lim r 0 f(r cos(β), r sin(β)) ([9]), and define h j : Ω j R by h j (x, y) = f(ɛ jx, ɛ j y) Rf(β) ɛ j = h(ɛ jx, ɛ j y) ɛ j for j N. Set c j = f(xj,yj) Rf(β) ɛ j and observe that h j (x, y) = f j (x, y) + c j. Hence From [10, Theorem 3], we see that h j (x, y) = f j (x, y), (x, y) G. lim f j(x, y) = f (x, y) for (x, y) G (3.6) j and so lim j h j (x, y) = f (x, y) for (x, y) G. Set E = {(x, y, z) Ω R : z < h(x, y)} and V j = {(x, y, z) Ω j R : z < h j (x, y)}, j N. Notice that (r cos(β), r sin(β), h(r cos(β), r sin(β))) E for r > 0 and so (0, 0, 0) E since (r cos(β), r sin(β), h(r cos(β), r sin(β))) (0, 0, 0) as r 0. y [14, Theorem 4.5], E t = {(x, y, z) : t(x, y, z) E} converges to a (solid) minimal cone (with vertex at (0, 0, 0)) in R 3 in the sense that φ Et φ in L 1 loc (R3 ) as r 0. Now (Ω R) is a minimal surface which is a cone with vertex at (0, 0, 0) and hence is a portion of a plane. Notice that V j = E ɛj and so h j converges to h, where h denote the generalized solution with subgraph φ, in the sense that φ Vj φ in L 1 loc (R3 ) as j. y [10, Theorem 3] and (3.6), we see that lim h j(x, y) = h (x, y) for (x, y) G, j

8 8 T. JEFFRES, K. LNSTER EJDE-2007/152 is a nonvertical plane, f = h is a constant vector and (Ω R) U is the portion of a nonvertical plane Π in Ω R. Since n j (x j /ɛ j, y j /ɛ j ) = n(x j, y j ), we see that lim j n(x j, y j ) = ( h, 1)/ h Suppose the second case holds. We may assume (cos(β), sin(β)) P; then (cos(β), sin(β), 0) is a point on P R and, in a neighborhood of (cos(β), sin(β), 0), U is a real analytic surface. Since P R U, this real analytic surface contains a portion Λ of P R with (cos(β), sin(β), 0) Λ. This implies that the component of U (Ω R) which contains (cos(β), sin(β), 0) is ( P Ω ) R. Since n j (x j /ɛ j, y j /ɛ j ) = n(x j, y j ), we see from [10, Theorem 3] that lim n(x j, y j ) = ξ = (ξ 1, ξ 2, 0), (3.7) j where ξ is the normal to P R at (cos(β), sin(β), 0) which points into P R. If we let ξ = (ξ 1, ξ 2, ξ 3 ) denote the unit normal to Π at (cos(β), sin(β), 0) pointing into U, then it is easy to see (as oncus and Finn observed in [2]) that ξ 3 cannot be less than 0 (i.e. no plane can meet Σ 1 R in angle γ 1 and Σ 2 R in angle γ 2 ) and so ξ 3 = 0. Hence U = P R, where P is given in one of (iv), (vi) or (viii) of Theorem 2.1 and N = Ω \ P. Suppose β [ α + π γ 2, α γ 1 ] holds. From Theorem 2.1, we see that case (viii) must hold. Since P is a line going through (0, 0) and (cos(β), sin(β)), we have ξ = ( sin(β), cos(β), 0) and (i.) of Lemma 3.1 follows from (3.7). Suppose β ( α, α + π γ 2 ] holds. Then case (vi) of Theorem 2.1 holds, ξ = ( sin( α + π γ 2 ), cos( α + π γ 2 ), 0) and (ii.) of Lemma 3.1 is established. Finally, suppose β [α γ 1, α). Then case (iv) of Theorem 2.1 holds, ξ = ( sin(α γ 1 ), cos(α γ 1 ), 0) and (iii.) of Lemma 3.1 is established. Remark 3.2. In [13] and [14], Shi assumes Ω is a wedge domain (i.e. (3.1), although with π 2 < α < π); for simplicity, we consider (convex) wedge domains. However, as noted in [13] and [14], the assumption that Ω has straight sides near can be relaxed. She also assumes γ = γ 1 on {θ = α} and γ = γ 2 on {θ = α}. Neither of these assumptions has a significant impact on the critical monotonicity estimates (i.e. (8) and page 339, [14]) which imply that the limit of the {V j } is a (solid, minimal) cone with vertex at (0, 0, 0). lso, the hypothesis of [14, Theorem 4.5] only requires f to satisfy (1.1) (with n = 2 and H bounded near ) rather than (3.2). Thus, we may assume in this section that Ω and γ are as described in 2 and f satisfies (1.1) and (3.3). Using (v), (vii) and (ix) of Theorems 2.1 and 2.2 and the techniques used in the proof of Lemma 3.1, we see that the following holds. Lemma 3.3. Suppose α π 2 and γ 2 + π γ 1 < 2α. Let f 2 (Ω) 1 (Ω \ {}) satisfy (3.2) and (3.3) and define n(x, y) by (3.4). Let β ( α, α) and let {(x j, y j )} be a sequence in Ω satisfying lim j (x j, y j ) = (0, 0) and (3.5). (i) If β [ α + γ 2, α + γ 1 π], then lim j n(x j, y j ) = (sin(β), cos(β), 0). (ii) If β ( α, α + γ 2 ], then lim j n(x j, y j ) = (sin( α + γ 2 ), cos( α + γ 2 ), 0). (iii) If β [α + γ 1 π, α), then lim n(x j, y j ) = (sin(α + γ 1 π), cos(α + γ 1 π), 0). j

9 EJDE-2007/152 VERTIL LW UPS 9 The oncus-finn onjecture is the conjecture that if κ 0, 0 < α < π 2 and π γ 1 γ 2 < 2α, then every solution of (3.2)-(3.3) must be discontinuous at = (0, 0) (see, for example, [8, 13] and [14]). The second author proved the oncus- Finn onjecture in [8]. Here we wish to illustrate the applicability of Theorems 2.1 and 2.2 to calculus of variations problems in R 3 by providing a proof of the oncus-finn onjecture in the special case that κ = 0 and λ = 0. (We note that while searching for a proof of the conjecture, this case was the first which the second author considered and its proof provided the roadmap for the proof in the general case.) Theorem 3.4 (oncus-finn onjecture with κ = λ = 0). Suppose 0 < α < π 2 and κ = λ = 0 in (3.2). Suppose further that π γ 1 γ 2 < 2α. Then every solution of (3.2)-(3.3) must be discontinuous at = (0, 0). Proof. We may suppose γ 1 + π γ 2 < 2α, κ = λ = 0 in (3.2), f 2 (Ω) 1 (Ω \ {}) satisfies (3.2) and (3.3) and f is continuous at. The graph of f, G = {(x, y, f(x, y)) : (x, y) Ω}, may be represented in isothermal coordinates. It follows from the arguments in [7] and [9] that there exists X : R 3, X(u, v) = (x(u, v), y(u, v), z(u, v)), (with = {(u, v) : u 2 + v 2 < 1}) such that X is a homeomorphism of and G and a diffeomorphism of and G, X u X v = 0 and X u = X v on, X(u, v) = (0, 0, 0) for (u, v). Since f is continuous at, we know, in addition, K(u, v) = (x(u, v), y(u, v)) is a homeomorphism of and Ω; if f were discontinuous at, then there would be a closed, nontrivial interval I such that K(u, v) = (0, 0) for each (u, v) I [7, 9]. We will assume X(1, 0) = (0, 0, f(0, 0)). onsider the Gauss map n K of G; in particular, recall that since G has zero mean curvature, the stereographic projection (from the north pole) of the Gauss map, which we denote g : R 2, is an analytic function of u + iv when we consider and R 2 as being in the complex plane. Notice that g :. Let m θ : [0, δ] be defined by K(m θ (t)) = t(cos(θ), sin(θ)) for each θ ( α, α). Let θ 1 and θ 2 satisfy α + π γ 2 < θ 1 < θ 2 < α γ 1. From Lemma 3.1, we see that lim g(m θ (t)) = sin(θ) + i cos(θ) for each θ [θ 1, θ 2 ] t 0 and, in particular, lim t 0 g(m θ (t)) = 1 for θ (θ 1, θ 2 ). n the other hand, the Phragmen-Lindelof Theorem for (bounded) analytic functions applied to g (when restricted to a suitable subdomain of and composed with a conformal map) implies that lim t 0 g(m θ (t)) lies on the line segment joining sin(θ 1 ) + i cos(θ 1 ) and sin(θ 2 ) + i cos(θ 2 ) and hence lim t 0 g(m θ (t)) < 1 for θ (θ 1, θ 2 ) (e.g. [1]). This contradiction implies that our assumption that f was continuous at was incorrect. 4. uxiliary Lemmas In this section, let P denote a minimizer of Φ and N denote a minimizer of Ψ. We do not assume α π 2 since the results here will also be used in Part Two, where π 2 < α < π. fter modification on a set of measure zero, we may assume (i) each component of Ω P (and of Ω N ) is a connected component of the intersection of Ω and a line;

10 10 T. JEFFRES, K. LNSTER EJDE-2007/152 (ii) Ω P consists of the union of rays (and lines) in Ω which do not intersect in Ω and (iii) Ω N consists of the union of rays (and lines) in Ω which do not intersect in Ω. These items follow since the sets P R and N R are ruled minimal surfaces in R 3 which are area minimizing with respect to compact perturbations (e.g. [16]). Lemma 4.1. For each r, T R with 0 < r < T, there are only finitely many components of Ω P (or of Ω N ) which intersect Ω T \ Ω r. Proof. Suppose {M n : n N} is a countably infinite collection of distinct components of Ω P, 0 < r < T and M n Ω T \ Ω r for each n N. Let D Ω T \ Ω r be an accumulation point of {M n }. If D Ω, pick ɛ > 0 such that 2ɛ (D) Ω. If D Σ 1, pick ɛ > 0 such that 2ɛ (D) Σ 2 =. If D Σ 2, pick ɛ > 0 such that 2ɛ (D) Σ 1 =. Since D is an accumulation point of {M n }, there is a subsequence {M nk } of {M n } such that M nk ɛ (D) for each k N. Hence H 1 (Ω M nk ) ɛ for each k. Since H 1 (Σ T 1 Σ T 2 ) <. we see that Φ T (P) =, a contradiction. The argument for Ω N is similar. Remark 4.2. We will later show in Lemma 4.18 that for each T > 0, Ω P and Ω N each have only finitely many components which intersect Ω T. We note that if α π/2 or if an infinite number of components of Ω P (or of Ω N ) contain as an endpoint, then the proof of Lemma 4.1 already yields this result. However, if α < π/2, the fact that Φ T (P) < (or Ψ T (N ) < ) does not suffice to exclude an infinite number of components of Ω P (or of Ω N ) from intersecting Ω T and this is illustrated in the following example. Example 4.3. Let 0 < α < π/2 and set S = n=1(2 2n 1, 2 2n ) R and U = Ω S. Notice that Ω U = n=2m n, where M n = {2 n } ( 2 n tan(α), 2 n tan(α)), and is an accumulation point of the sequence {M n : n N}. If T > 1, then H 1 (Ω T U) = tan(α) and Φ T (U) < and Ψ T (U) < for any choice of γ 1 and γ 2. Lemma 4.4. If M 1 and M 2 are two distinct components of Ω P (or of Ω N ) and M 1 M 2 = {}, then = and the angle between M 1 and M 2 has measure greater than or equal to π. Proof. Suppose two components of Ω P intersect at a point Ω and, if =, then they meet in an angle of measure less than π. Then either one component P 1 of P is a convex wedge with vertex or one component N 1 of Ω \ P is a convex wedge with vertex. Suppose the first case holds and let P Figure 2. onvex wedge 1

11 EJDE-2007/152 VERTIL LW UPS 11 I 1 and I 2 be the components of Ω P 1 ; then I 1 I 2 = {}. Let I 1, I 2 and T > max{,, } (see Figure 2). Since P minimizes Φ, φ T (P) φ T (P \ ) and so +, which is a contradiction. Thus the first case cannot hold. P P Figure 3. onvex wedge 2 Suppose the second case holds and let I 1 and I 2 be the components of Ω N 1 and I 1, I 2 (see Figure 3). Let T be as above. Since P minimizes Φ, φ T (P) φ T (P ) and so +, which is a contradiction. Thus the second case cannot hold. The argument for components of Ω N is essentially the same. Remark 4.5. We may state Lemma 4.4 informally as follows: Two components of Ω P (or of Ω N ) cannot meet on Σ 1 Σ 2. Two components of Ω P (or of Ω N ) which meet at meet in a angle of measure greater than or equal to π. Lemma 4.6. Suppose Σ 1 P (Ω \ P) and Σ 1 \ {} with P. Let P 1 denote the connected component of P whose closure contains. Then the measure of the angle P 1 makes at is greater than or equal to γ 1. θ γ 1 P Figure 4. θ < γ 1 Proof. From Lemma 4.4 and (i)-(iii), we see that only one component of P contains in its closure; let us denote this component by P 1. Denote by θ the measure of the angle at formed by P 1 and assume θ < γ 1. Let be the point on Ω P 1 for which the angle has measure π γ 1 (see Figure 4). Since P minimizes Φ, φ T (P) φ T (P \ ) for T large. Hence cos(γ 1 ). Now = (sin(γ 1 )/ sin(γ 1 θ)) and = (sin(θ)/ sin(γ 1 θ)) and so sin(γ 1 ) cos(γ 1 ) sin(γ 1 θ) sin(θ). short calculation shows that this implies 1 cos(γ 1 θ), which contradicts the assumption that θ < γ 1. Lemma 4.7. Suppose Σ 2 P (Ω \ P) and Σ 2 \ {} with P. Let P 1 denote the connected component of P whose closure contains. Then the measure of the angle P 1 makes at is greater than or equal to γ 2.

12 12 T. JEFFRES, K. LNSTER EJDE-2007/152 Lemma 4.8. Suppose Σ 1 P (Ω \ P). Then lies in the closure of exactly one connected component P 1 of P and P 1 makes an angle of measure exactly γ 1 at. γ 1 θ P Figure 5. θ > γ 1 Proof. Since, Lemma 4.4 implies that only one component P 1 of P contains in its closure. Let θ be the measure of the angle P 1 makes at and assume θ > γ 1. Let Σ 1 and Ω P 1 such that / P 1 and angle has measure γ 1 (see Figure 5). Now φ T (P) φ T (P ) for large T and so cos(γ 1 ). Now = (sin(γ 1 )/ sin(θ γ 1 )) and = (sin(θ)/ sin(θ γ 1 )) and so sin(γ 1 ) sin(θ) cos(γ 1 ) sin(θ γ 1 ). short calculation shows that this implies 1 cos(θ γ 1 ), which contradicts the assumption that θ > γ 1. Lemma 4.9. Suppose P (Ω \ P), Ω P, Σ 1, Ω \ P and θ 1 is the measure of the angle. Then θ 1 π γ 1. N 1 θ 1 P Figure 6. P (Ω \ P) Proof. ssume θ 1 < π γ 1. Let N 1 be the component of Ω \ P which contains. Pick Ω N 1 such that angle has measure γ 1 (see Figure 6). Now φ T (P) φ T (P ) for large T and so cos(γ 1 ). Since = (sin(γ 1 )/ sin(θ 1 + γ 1 )) and = (sin(θ 1 )/ sin(θ 1 + γ 1 )), we see that sin(γ 1 ) sin(θ 1 ) cos(γ 1 ) sin(θ 1 + γ 1 ). short calculation shows that this implies 1 cos(π θ 1 γ 1 ), which contradicts the assumption that θ 1 < π γ 1.

13 EJDE-2007/152 VERTIL LW UPS 13 Lemma Suppose Σ 2 P (Ω \ P). Then lies in the closure of exactly one connected component P 1 of P and P 1 makes an angle of measure exactly γ 2 at. Lemma Suppose P (Ω \ P), Ω P, Σ 2, Ω \ P and θ 2 is the measure of the angle. Then θ 2 π γ 2. Lemma Suppose Σ 1 N (Ω \N ) and Σ 1 \{} with N. Let N 1 denote the connected component of N whose closure contains. Then the measure of the angle N 1 makes at is greater than or equal to π γ 1. θ N 1 γ 1 Figure 7. π θ < π γ 1 Proof. From Lemma 4.4 and (i)-(iii), we see that only one component of N contains in its closure; let us denote this component by N 1. Denote by π θ the measure of the angle at formed by N 1 and assume θ > γ 1. Let be the point on Ω N 1 for which the angle has measure γ 1 (see Figure 7). Since N minimizes Ψ, ψ T (N ) ψ T (N \ ) for T large. Hence +cos(γ 1 ). Now = (sin(γ 1 )/ sin(θ γ 1 )) and = (sin(θ)/ sin(θ γ 1 ) and so sin(γ 1 ) + cos(γ 1 ) sin(θ γ 1 ) sin(θ). short calculation shows that this implies 1 cos(θ γ 1 ), which contradicts the assumption that θ > γ 1. Thus θ γ 1 and so the measure π θ of the angle which N 1 makes at is greater than or equal to π γ 1. Lemma Suppose Σ 2 N (Ω \N ) and Σ 2 \{} with N. Let N 1 denote the connected component of N whose closure contains. Then the measure of the angle N 1 makes at is greater than or equal to π γ 2. Lemma Suppose Σ 1 N (Ω \ N ). Then lies in the closure of exactly one connected component N 1 of N and N 1 makes an angle of measure exactly π γ 1 at. Lemma Suppose Σ 2 N (Ω \ N ). Then lies in the closure of exactly one connected component N 1 of N and N 1 makes an angle of measure exactly π γ 2 at. Lemma Suppose N (Ω \ N ), Ω N, Σ 1, Ω \ N and θ 1 is the measure of the angle. Then θ 1 γ 1. Lemma Suppose N (Ω \ N ), Ω N, Σ 2, Ω \ N and θ 2 is the measure of the angle. Then θ 2 γ 2. The proofs of Lemmas 4.7, 4.10, 4.11 and 4.13 follow those of Lemmas 4.6, 4.8, 4.9 and 4.12 respectively. The proofs of Lemmas 4.14 and 4.15 are similar to that of Lemma 4.8 while the proofs of Lemmas 4.16 and 4.17 are similar to that of Lemma 4.9.

14 14 T. JEFFRES, K. LNSTER EJDE-2007/152 Lemma For each T > 0, only finitely many components of Ω P (or of Ω N ) intersect Ω T. Proof. Example 4.3 illustrates the only case with which we need to deal. Suppose 0 < α < π/2, T > 0 and Ω T P = n=1m n, where M n = n n, n Σ 1, n Σ 2, n+1 < n and n+1 < n for each n N and n, n as n. Let P 1 be a component of P whose boundary contains M 3. Then either P 1 = M 2 M or P 1 = M 3 M Let us assume the first case holds. Notice then by Lemmas 4.8 and 4.9 that P makes angles of measure γ 1 at 2 and 3 and angles of measure γ 2 at 2 and 3. Hence 2γ 1 + 2γ 2 = 2π or γ 1 + γ 2 = π; this implies Σ 1 and Σ 2 are parallel, a contradiction. similar argument applied to Ω N shows that 2(π γ 1 ) + 2(π γ 2 ) = 2π or γ 1 + γ 2 = π again. Lemma Suppose α π 2. Then: (a) Two components of Ω P for which the closure of one (or both) of the components is disjoint from Ω cannot be parallel. (b) Two components of Ω N for which the closure of one (or both) of the components is disjoint from Ω cannot be parallel. Proof. Suppose M 1 and M 2 are distinct components of Ω P such that M 1 is a line or ray in Ω, M 2 is a line in Ω and M 1 and M 2 are parallel. We may assume that either (i) M 1 and M 2 lie on the boundary of a component P 1 of P or (ii) M 1 and M 2 lie on the boundary of a component N 1 of Ω \ P. Let M 1 and M 2 be fixed with orthogonal to M 1 and M 2. Now pick M 1 and D M 2 such that and D are parallel and = D < = D. Fix T > max{,,, D}. ssuming (i.), the minimality of P implies Φ T (P) Φ T (P \D) or +D +D. However +D > +D and we have a contradiction. ssuming (ii.), the minimality of P implies Φ T (P) Φ T (P D) or + D + D. However + D > + D and we have a contradiction. This proves the Lemma when M 1 and M 2 are distinct components of P. The proof when M 1 and M 2 are distinct components of N is similar. Lemma Let Ω 1 = {0 < θ < α} and Ω 2 = { α < θ < 0}. Then no component of Ω 1 P nor of Ω 1 N can be parallel to Σ 1 and no component of Ω 2 P nor of Ω 2 N can be parallel to Σ 2. Proof. Suppose L is a component of Ω P such that L 1 = L Ω and L 1 is parallel to Σ 1. From (i)-(iii) of 4, we know that L is either a ray starting at a point on Σ 2 (if α < π 2 ) or a line in Ω (if α π 2 ) and, in either case, L 1 is a ray in Ω 1 starting at a point = (a, 0). Let Σ 1 and set (= ()) = + (a, 0) L 1. Select T > max{,, }. Let denote the open region with boundary. Now the open region U Ω 1 with boundary Σ 1 L 1 is a component of Ω 1 P or a component of Ω 1 N. Suppose first that U is a component of Ω 1 P. Since we obtain Φ T (P) Φ T (P \ ), cos(γ 1 ) +.

15 EJDE-2007/152 VERTIL LW UPS 15 Σ 1 U L 1 Figure 8. U P Since = is fixed (independent of ), = and cos(γ 1 ) < 1, we obtain 2 1 cos(γ 1 ) for all Σ 1, which is impossible for sufficiently large. Suppose second that U is a component of Ω 1 \ P. Since Φ T (P) Φ T (P ), we obtain + cos(γ 1 ). Since = is fixed (independent of ), = and cos(γ 1 ) < 1, we obtain cos(γ 1 ) for all Σ 1, which is impossible for sufficiently large. The case in which L is a component of Ω N and L 1 = L Ω is parallel to Σ 1 follows by a similar argument. The cases in which L is a component of Ω P or of Ω N such that L 2 = L Ω is parallel to Σ 2 follow using similar comparison arguments. 5. Proofs for convex corners: Theorems 2.1, 2.2 In this section, we assume α π 2 and let P and N denote minimizers of Φ and Ψ respectively. Notice that the only situations in which differences between the geometry when α = π 2 and α < π 2 occur first in cases (ii) and (iii), where we assume α < π 2, second when α = π 2 and components of P or N might be distinct parallel lines in Ω, a possibility eliminated by Lemma 4.19, and third when α = π 2 and P (or N ) contains a single line parallel to Ω. This last possibility cannot occur as a simple comparison argument shows. Thus we may assume α < π 2 subsequently. The arguments used here are similar to those of Tam ([15]). The conclusions of Theorems 2.1 and 2.2 follow from the eight claims proven here. laim (1). Every component of Ω P is unbounded and every component of Ω N is unbounded unless γ 1 + γ 2 π = 2α. Suppose that γ 1 + γ 2 π 2α and M = is a bounded component of Ω P, where Σ 1 and Σ 2. Notice that no component of Ω P can have in its closure since otherwise Ω P would contain a ray in Ω with as one endpoint and this ray would intersect M, in contradiction to (i)-(iii). Hence P \ Ω \ P or Ω \ P \ P.

16 16 T. JEFFRES, K. LNSTER EJDE-2007/152 In the first case, there exist Σ 1 and D Σ 2 such that D is a component of P and D. From Lemmas 4.8 and 4.9, we see that D makes angles of measure γ 1 at, γ 2 at D and 2α at. Therefore γ 1 + γ 2 + 2α = π or γ 1 + γ 2 π = 2α, a contradiction. In the second case, there exist Σ 1 and D Σ 2 such that D is a component of Ω \P. From Lemmas 4.8 and 4.9, we see that D makes angles of measure π γ 1 at, π γ 2 at D and 2α at. Therefore π γ 1 +π γ 2 +2α = π or γ 1 + γ 2 π = 2α, a contradiction. The argument for Ω \ N is similar. laim (2). Ω P and Ω N have at most one component. Suppose M 1 and M 2 are distinct components of Ω P such that M 1 and M 2 lie on the boundary of a component P 1 of P or a component N 1 of Ω \ P. Now Lemma 4.4 implies M 1 and M 2 are either parallel or the lines L 1 and L 2 on which they respectively lie intersect at a point E outside of Ω. We will consider the various cases individually and show that each leads to a contradiction. D Figure 9. ase (a) (a) M 1 and M 2 are parallel, M 1 M 2 P 1, L 1 Σ 1 and L 2 Σ 2. Let L 1 Σ 1 and L 2 Σ 2. We will pick M 1 and D M 2 such that and D are parallel (see Figure 9); we let T > max{,,, D}. Now Φ T (P) Φ T (P \ D), where D denotes the open polygon with vertices,,, D and. Then + D cos(γ 1 ) + cos(γ 2 ) + D. Since the lengths, and D are fixed, this inequality is false when and D are sufficiently large. D Figure 10. ase (b) (b) M 1 and M 2 are parallel, M 1 M 2 N 1, L 1 Σ 1 and L 2 Σ 2. Let L 1 Σ 1 and L 2 Σ 2. We will pick M 1 and D M 2 such that and D are parallel (see Figure 10); we let T > max{,,, D}. Now Φ T (P) Φ T (P D), where D denotes the open polygon with vertices,,, D and. Then + D D cos(γ 1 ) cos(γ 2 ).

17 EJDE-2007/152 VERTIL LW UPS 17 D Figure 11. ase (c) Since the lengths, and D are fixed, this inequality is false when and D are sufficiently large. (c) M 1 and M 2 are parallel, M 1 M 2 P 1, L 1 Σ 1 and L 2 Σ 1. Let L 1 Σ 1 and L 2 Σ 1. We will pick M 1 and D M 2 such that and D are parallel (see Figure 11); we let T > max{,,, D}. Now Φ T (P) Φ T (P \ D), where D denotes the open polygon with vertices,, D and. Then + D cos(γ 1 ) + D. Since the lengths and D are fixed, this inequality is false when and D are sufficiently large. D Figure 12. ase (d) (d) M 1 and M 2 are parallel, M 1 M 2 N 1, L 1 Σ 1 and L 2 Σ 1. Let L 1 Σ 1 and L 2 Σ 1. We will pick M 1 and D M 2 such that and D are parallel (see Figure 12); we let T > max{,,, D}. Now Φ T (P) Φ T (P D), where D denotes the open polygon with vertices,, D and. Then + D D cos(γ 1 ). Since the lengths and D are fixed, this inequality is false when and D are sufficiently large. (e) M 1 and M 2 are parallel, M 1 M 2 P 1, L 1 Σ 2 and L 2 Σ 2. Let L 1 Σ 2 and L 2 Σ 2. We will pick M 1 and D M 2 such that and D are parallel (see Figure 13); we let T > max{,,, D}. Now Φ T (P) Φ T (P \ D), where D denotes the open polygon with vertices,, D and. Then + D cos(γ 2 ) + D.

18 18 T. JEFFRES, K. LNSTER EJDE-2007/152 D Figure 13. ase (e) Since the lengths and D are fixed, this inequality is false when and D are sufficiently large. D Figure 14. ase (f) (f) M 1 and M 2 are parallel, M 1 M 2 N 1, L 1 Σ 2 and L 2 Σ 2. Let L 1 Σ 2 and L 2 Σ 2. We will pick M 1 and D M 2 such that and D are parallel (see Figure 14); we let T > max{,,, D}. Now Φ T (P) Φ T (P D), where D denotes the open polygon with vertices,, D and. Then + D D cos(γ 2 ). This inequality is false when and D are sufficiently large. (g) M 1 and M 2 are not parallel, M 1 M 2 P 1, L 1 Σ 1 and L 2 Σ 1. Let L 1 Σ 1 and L 2 Σ 1. We will pick M 1 and D M 2 such that and D are parallel (see Figure 15); we let T > max{,,, D}. Now Φ T (P) Φ T (P \ D), where D denotes the open polygon with vertices,, D and. Then + D cos(γ 1 ) D. E D Figure 15. ase (g)

19 EJDE-2007/152 VERTIL LW UPS 19 Now = E E and D = ED E and we have E + ED D + E + E + cos(γ 1 ). Since the lengths E, E and are fixed, this inequality is false when E and ED are sufficiently large. E D Figure 16. ase (h) (h) M 1 and M 2 are not parallel, M 1 M 2 N 1, L 1 Σ 1 and L 2 Σ 1. Let L 1 Σ 1 and L 2 Σ 1. We will pick M 1 and D M 2 such that and D are parallel (see Figure 16); we let T > max{,,, D}. Now Φ T (P) Φ T (P D), where D denotes the open polygon with vertices,, D and. Then + D D cos(γ 1 ). Now = E E and D = ED E and we have E + ED D + E + E cos(γ 1 ). This inequality is false when E and ED are sufficiently large. E D Figure 17. ase (i) (i) M 1 and M 2 are not parallel, M 1 M 2 P 1, L 1 Σ 1 and L 2 Σ 2 and Let L 1 Σ 1 and L 2 Σ 2. We will pick M 1 and D M 2 such that and D are parallel (see Figure 17); we let T > max{,,, D}. Now Φ T (P) Φ T (P \D), where D denotes the open polygon with vertices,,, D and. + D cos(γ 1 ) cos(γ 2 ) D, = E E and D = ED E implies E + ED D + E + E + cos(γ 1 ) + cos(γ 2 ). Since the lengths E, E, and are fixed, this inequality is false when E and ED are sufficiently large. (j) M 1 and M 2 are not parallel, M 1 M 2 N 1, L 1 Σ 1 and L 2 Σ 2. Let L 1 Σ 1 and L 2 Σ 2. We will pick M 1 and D M 2 such that

20 20 T. JEFFRES, K. LNSTER EJDE-2007/152 E D Figure 18. ase (j) and D are parallel (see Figure 18); we let T > max{,,, D}. Now Φ T (P) Φ T (P D), where D denotes the open polygon with vertices,,, D and. + D D cos(γ 1 ) cos(γ 2 ), = E E and D = ED E implies E + ED D + E + E cos(γ 1 ) cos(γ 2 ). This inequality is false when E and ED are sufficiently large. E D Figure 19. ase (k) (k) M 1 and M 2 are not parallel, M 1 M 2 P 1, L 1 Σ 2 and L 2 Σ 2. Let L 1 Σ 2 and L 2 Σ 2. We will pick M 1 and D M 2 such that and D are parallel (see Figure 19); we let T > max{,,, D}. Now Φ T (P) Φ T (P \ D), where D denotes the open polygon with vertices,, D and. Then + D cos(γ 2 ) D. Now = E E and D = ED E and we have E + ED D + E + E + cos(γ 2 ). Since the lengths E, E and are fixed, this inequality is false when E and ED are sufficiently large. (l) M 1 and M 2 are not parallel, M 1 M 2 N 1, L 1 Σ 2 and L 2 Σ 2. Let L 1 Σ 2 and L 2 Σ 2. We will pick M 1 and D M 2 such that and D are parallel (see Figure 20); we let T > max{,,, D}. Now Φ T (P) Φ T (P D), where D denotes the open polygon with vertices,, D and. Then + D D cos(γ 2 ).

21 EJDE-2007/152 VERTIL LW UPS 21 E D Figure 20. ase (l) Now = E E and D = ED E and we have E + ED D + E + E cos(γ 2 ). This inequality is false when E and ED are sufficiently large. This proves laim (2.) for Ω P. The proof for Ω N is essentially the same. laim (3). Suppose Ω P has Σ 1 as an endpoint. Then either (i) γ 1 + γ 2 π = 2α and P =, where Σ 2 and angles and have measures γ 1 and γ 2 respectively, (ii) γ 1 + γ 2 π = 2α and P = Ω \, where Σ 2 and angles and have measures π γ 1 and π γ 2 respectively, or (iii) Ω P is a ray with endpoint which makes angles of measure γ 1 and π γ 1 with Σ 1 and P is the component of Ω \ P which forms an angle of measure γ 1 at. If γ 1 + γ 2 π 2α, laim (1) implies Ω P is an infinite ray with endpoint. Even when γ 1 + γ 2 π = 2α, Ω P might be an infinite ray with endpoint. We shall suppose this is the case. Then (iii) follows from Lemma 4.8. laim (4). Suppose Ω P has Σ 2 as an endpoint. Then either (i) γ 1 + γ 2 π = 2α and P =, where Σ 1 and angles and have measures γ 1 and γ 2 respectively, (ii) γ 1 + γ 2 π = 2α and P = Ω \, where Σ 1 and angles and have measures π γ 1 and π γ 2 respectively, or (iii) Ω P is a ray with endpoint which makes angles of measure γ 2 and π γ 2 with Σ 2 and P is the component of Ω \ P which forms an angle of measure γ 2 at. Let us suppose Ω P is an infinite ray with endpoint. Then our claim follows from Lemma laim (5). Suppose Ω P has = (0, 0) as an endpoint. Then Ω P is a ray with endpoint which makes angles of measure θ 1 and θ 2 with Σ 1 and Σ 2 respectively (and θ 1 + θ 2 = 2α) such that (i) θ 1 γ 1 and θ 2 π γ 2 if P is the open (infinite) sector whose boundary contains Σ 1 and (ii) θ 1 π γ 1 and θ 2 γ 2 if P is the open (infinite) sector whose boundary contains Σ 2.

22 22 T. JEFFRES, K. LNSTER EJDE-2007/152 The proof of our claim follows from Lemmas 4.6 and 4.11 when P is the open (infinite) sector whose boundary contains Σ 1 and from Lemmas 4.9 and 4.7 when P is the open (infinite) sector whose boundary contains Σ 2. laim (6). Suppose Ω N has Σ 1 as an endpoint. Then either (i) γ 1 + γ 2 π = 2α and N =, where Σ 2 and angles and have measures π γ 1 and π γ 2 respectively, (ii) γ 1 + γ 2 π = 2α and N = Ω \, where Σ 2 and angles and have measures γ 1 and γ 2 respectively, or (iii) Ω N is a ray with endpoint which makes angles of measure γ 1 and π γ 1 with Σ 1 and N is the component of Ω \ N which forms an angle of measure π γ 1 at. Let us suppose Ω N is an infinite ray with endpoint. Then our claim follows from Lemma laim (7). Suppose Ω N has Σ 2 as an endpoint. Then either (i) γ 1 + γ 2 π = 2α and N =, where Σ 1 and angles and have measures π γ 1 and π γ 2 respectively, (ii) γ 1 + γ 2 π = 2α and N = Ω \, where Σ 1 and angles and have measures γ 1 and γ 2 respectively, or (iii) Ω N is a ray with endpoint which makes angles of measure γ 2 and π γ 2 with Σ 2 and N is the component of Ω \ N which forms an angle of measure π γ 2 at. Let us suppose Ω N is an infinite ray with endpoint. Then our claim follows from Lemma laim (8). Suppose Ω N has = (0, 0) as an endpoint. Then Ω N is a ray with endpoint which makes angles of measure θ 1 and θ 2 with Σ 1 and Σ 2 respectively (and θ 1 + θ 2 = 2α) such that (i) θ 1 π γ 1 and θ 2 γ 2 if N is the open (infinite) sector whose boundary contains Σ 1 and (ii) θ 1 γ 1 and θ 2 π γ 2 if N is the open (infinite) sector whose boundary contains Σ 2. The proof of our claim follows from Lemmas 4.12 and 4.17 when N is the open (infinite) sector whose boundary contains Σ 1 and from Lemmas 4.16 and 4.13 when N is the open (infinite) sector whose boundary contains Σ 2. The conclusions of Theorem 2.1 follow from the results in 4 and the laims proven above except for the restrictions on γ 1 and γ 2 in (iv)-(vii). For example, one consequence of (v) is that if γ 2 + π γ 1 > 2α, then the conclusion of (v), illustrated in Figure 1 (v), cannot hold. To see this, assume the inequality above and the conclusion of (v) both hold. Set θ = 2α+γ 1 π and β = γ 1 +π γ 2 2α > 0. Let D be the point of intersection of the lines on which rays L and Σ 2 lie; notice that D / Ω. Fix L and let be the point on Σ 2 determined by the condition that angle has measure β. Then angle has measure π γ 2. Let be the open subset of Ω whose boundary is the quadrilateral. For T > max{,, }, Φ T (P) Φ T (P \ ) or + cos(γ 1 ) + cos(γ 2 ).

23 EJDE-2007/152 VERTIL LW UPS 23 Notice that = D D, = D D, D = ( sin(β)/ sin(γ 2 ) ) D and = ( sin(θ)/ sin(γ 2 ) ) D. Set m = D + cos(γ 1 ) cos(γ 2 )D. Then or D sin(θ) sin(γ 2 ) D + cos(γ 2) sin(β) sin(γ 2 ) D + m sin(γ 2 ) sin(θ) + cos(γ 2 ) sin(β) + m sin(γ 2) D. Since β = γ 2 θ, we obtain sin(γ 2 ) sin(θ) cos 2 (γ 2 ) sin(θ) + cos(γ 2 ) sin(γ 2 ) cos(θ) + m sin(γ 2) D or, after dividing by sin(γ 2 ) and simplifying, 1 cos(γ 2 θ) + m D. (5.1) Now 0 < γ 2 θ < 2π and so cos(γ 2 θ) < 1. Therefore, for D sufficiently large, cos(γ 2 θ) + m D < 1, in contradiction to (5.1). The other restrictions on γ 1 and γ 2 in (iv) - (vii) of Theorem 2.1 follow using similar arguments. The conclusions of Theorem 2.2 follow from the results in 4 and the laims proven previously in this section except that the restrictions on γ 1 and γ 2 in (iv) - (vii) follow as above. cknowledgements. The first author thanks the Mathematical Sciences Research Institute for its stimulating environment and generous support. The second author is indebted to the Max-Planck-Institut für Mathematik in den Naturwissenschaften, in Leipzig, and the entre for Mathematics and its pplications, ustralian National University, in anberra, for their hospitality during some of the course of this study. References [1] H. ear and G. Hile, ehavior of Solutions of Elliptic Differential Inequalities near a point of Discontinuous oundary Data, omm. PDE 8 (1983), [2] P. oncus and R. Finn, apillary wedges revisited, SIM J. Math. nal. 27 (1996), no. 1, [3] R. Finn, Equilibrium apillary Surfaces, Springer-Verlag, New York, [4] R. Finn and R. Neel, -singular solutions of the capillary problem, J. Reine ngew. Math. 512 (1999), [5] E. Giusti, Generalized solutions of prescribed mean curvature equations, Pacific J. Math. 88 (1980), [6] E. Guisti, Minimal Surfaces and Functions of ounded Variation, irkhäuser, oston, [7] K. E. Lancaster, oundary behavior of a nonparametric minimal surface in R 3 at a nonconvex point, nalysis 5 (1985), [8] K. E. Lancaster, Proof of the oncus-finn onjecture, preprint. [9] K. E. Lancaster and D. Siegel, Existence and ehavior of the Radial Limits of a ounded apillary Surface at a orner, Pacific J. Math. Vol. 176, No. 1 (1996), orrection to figures, Pacific J. Math. Vol. 179, No. 2 (1997), [10] U. Massari and L. Pepe, Successioni convergenti di ipersuperfici di curvatura media assegnata, Rend. Sem. Univ. Padova 53, (1975), [11] M. Miranda, Un principio di massimo forte per le frontiere minimali e una sua applicazione alla risoluzione del problema al contorno per l equazione delle superfici di area minima, Rend. Sem. Mat. Univ. Padova 45 (1971), [12] M. Miranda, Superfici minime illimitate, nn. Scuola Norm. Sup. Pisa 4 (1977), [13] Danzhu Shi, ehavior of apillary Surfaces at a Reentrant orner, Ph.D. Dissertation, Stanford University, February 2005.

24 24 T. JEFFRES, K. LNSTER EJDE-2007/152 [14] Danzhu Shi, apillary Surfaces at a Reentrant orner, Pacific J. Math. Vol. 224, No. 2 (2006), [15] L. F. Tam, The ehavior of apillary Surfaces as Gravity Tends to Zero, PhD Dissertation, Stanford University [16] L. F. Tam, The behavior of a capillary surface as gravity tends to zero, omm. Partial Differential Equations 11 (1986), [17] L. F. Tam, Regularity of capillary surfaces over domains with corners: borderline case, Pacific J. Math. 124 (1986), [18] L. F. Tam, n Existence riteria for apillary Free Surfaces without gravity, Pacific J. Math. 125, No. 2 (1986), Thalia Jeffres Department of Mathematics and Statistics, Wichita State University, Wichita, Kansas, , US address: jeffres@math.wichita.edu Kirk Lancaster Department of Mathematics and Statistics, Wichita State University, Wichita, Kansas, , US address: lancaster@math.wichita.edu