Minimal surfaces of revolution

Transcription

1 5 April 013 Minimal surfaces of revolution Maggie Miller 1 Introduction In tis paper, we will prove tat all non-planar minimal surfaces of revolution can be generated by functions of te form f = 1 C cos(cx), x R (as illustrated in Figure 1). We will accomplis tis by Assuming a non-planar minimal surface of revolution exists Sowing tat te surface must be given by te above form Prove tat suc a surface is indeed minimal. By minimal surface, we mean a surface wose area (in te usual notion of surface area in R 3 ) will be increased by any sligt perturbation. Tis definition implies tat radius of our surface cannot approac zero as we move along te fixed axis toward infinity, so by translation and scaling, we can assume tat some segment of suc a surface as boundary of two circles of radius 1 in parallel planes wit centers along a common axis. Figure 1: A plot of te described surface wit boundary, wit = 1. In te derivation of te form of te surface connecting tese circles, we sow tat tere is a maximum allowable distance between te two boundaries, and find tis value in section 4. We will later give an alternate definition of a minimal surface via te notion of curvature. 1

2 Existence We aim to find te minimal surface wit a boundary of two circles of radius 1 parallel to te xy plane wit centers on te z-axis at z = and z =, wit te requirement tat te surface be connected. We assume tat te solution will be a surface of revolution, generated by revolution about te z-axis of te image of a C function. (We require te image to be C because a later alternate definition of minimal surfaces requires second partial derivatives). Hencefort, wen referring to surfaces of revolution, we mean specifically surfaces of revolution generated by C curves in tis fasion. Let B = { f C ([, ]) f(x) 0, f( ) = f( ) = 1} be an open subset of a affine space over vector space V = { f C ([ 1, ]) f( ) = f( ) = 0}. Ten if tere exists some element r(z) of B tat minimizes A : B R, A(r) = π r(z) (z) dx (1) ten r(z) must generate te minimal surface wit te given boundary. Suppose tat r B minimizes A. Let η V. We define g : R R by g(t) = A(r + tη). Note A is differentiable, so g is differentiable. Ten as r + tη B for all t and r minimizes A in B, we must ave g (0) = 0. () Noting tat (r + η) 1 + (r + tη ) is continuous wit continuous derivative, we can use te Leibnitz rule on (1) to find g (t) = d dt π = π = π g (0) = π So from (), we find tat for all η V, Via integration by parts, wit u = (r + η) 1 + (r + tη ) dz d dt (r + η) 1 + (r + tη ) dz ( η ) 1 + (r + tη ) + (r + tη)(r + tη )η dz 1 + (r + tη ) rr 1+r (η + rr η ) dz (η + rr η ) dz = 0. and dv = η dz,

3 η dz + rr η (η + rr η ) dz = 0 η dz + η dz + η dz + uv z= z= rr η dz = 0 z= z= z= z= udv = 0 vdu = 0 η (r rr )( ) rr r dz = 0 ( ) 3 Recall η V, so η = η = 0. Terefore, We claim r rr + 1 ( ) 3 η dz η (r rr )( ) rr r dz = 0 ( ) 3 0, introducing te following lemma: η r rr + 1 dz = 0 (3) ( ) 3 Fundamental Lemma of Calculus of Variations (du Bois-Reymond) Let f B suc tat fgdz = 0 for any g V. Ten f 0. Proof. Suppose tat f 0. Ten tere exists some t [, ] wit f(t) > 0. Since f is continuous, tis implies te existence of an open neigborood U [, ] of t suc tat f(u) > 0 for all u U. Let g be a nonzero, non-negative bump function on [, ] compactly supported on U. Ten g V and fgdz = U fgdz But U is not open and fg 0 on U, wit equality possible only on U, implying our supposition. Terefore, f 0. fgdz > 0, violating Tus from (3), we find r rr Tat is, r rr We solve te arising differential equation (1+r ) 3 troug separation of variables. Letting u = r, 3

4 r rr + 1 = u ru + 1 = u r du dr u + 1 u du 1 + u = dr r ln 1 + u = ln + c = Cr r = ± Cr dr 1 Cr 1 = ±dz 1 c 1 cos 1 (c 1 r) = ±z + c 0 r = 1 c 1 cos(c 1 z + c ) were c, C, c 0, c 1, c are arbitrary real constants. Our boundary conditions r = r = 1 yield c = 0, so relabeling C := c 1, r = 1 cos(cz). (4) C So in particular, C = cos. Suppose tere is a solution to tis equation. We note [ C ( ) (C + C)( + ) C + C cos ] C cos [ ] C cos = C sin = 1 sin ( [ C ] [ sin sin So on te curve C = cos, as grows large and approaces zero, [ C ] [ sin sin C C sin ( C sin ( ( C( + ) ) [ 1 + ) )] C 0 ) [ 1 sin ( C( + ) sin )] 1 ( C( + ) C But C = cos implies C 1. Terefore, tere exists some maximal 0 suc tat for > 0, tere exists no solution C. Figure suggests Or, more generally, if te radius of eac circular boundary is r, ten r. )] 1 )] C 4

5 Figure : vs. C. Note tat solutions for C exist only wen falls below some maximal value, approximately Altoug we ave only considered minimal surfaces of revolution, we ave already observed tat te family of surfaces studied did not yield a unique surface wit a desired boundary. Referring to Figure, we see tat for any < 0, tere exist two distinct values of C for wic r = 1 C cos(cz), z [, ] generates a minimal surface wit boundary two circles of radius 1 centered along a common axis at a distance. However, te coice of te smaller potential value of C will clearly yield a larger surface area tan will te coice of te larger potential C. We must now sow tat wen 0 < < 0, te surface generated by revolving r = 1 C cos(cz) around te z-axis is a minimal surface. We first prove tat tis surface is te minimal connected surface of revolution wit te given boundary. We will ten calculate te precise value of 0. 3 Minimality Let < 0, C be a solution to C = cos [, and r :, ] R 0 be defined by r(z) = 1 C cos(cz). To sow tat te surface M generated by r is a minimal surface wit te given boundary, we define te notion of mean curvature. Definition 1. Let S be an embedded orientable C -parametrizeable surface in R 3. Let s S and n be te positively oriented normal vector to S in R 3 at s. oose any two ortogonal planes A, B wo contain n and pass troug s, and let te intersections of A, B wit S be called α, β respectively. Ten α, β are C curves passing troug s in R 3, wit curvatures κ a, κ b at s. Ten te mean curvature of S at s is κa+κ b. Note tat tis does not depend on our coice of A and B: If we coose A so tat κ a as maximum possible value, ten for any plane C containing n, s separated from A by angle θ, we ave κ c = κ a cos θ + κ b sin θ. Ten if D is te plane containing n, s at an angle θ + π to A, 5

6 κ c + κ d = κ a (cos θ + sin θ) + κ b (cos θ + sin θ) = κ a + κ b. Tis value can be calculated troug Gaussian curvature, Definition. Let S be an embedded orientable C -parametrizeable surface of variables u, v in R 3. Let s S and n be te normal vector of S in R 3 at s. Let E = S u, S u, F = S u, S v, G = S v, S v be functions from S to R. Similarly define e = n, S u, f = n, S u v, g = n, S v. Ten te mean curvature of S at s is given by coosing our two planes A, B to contain tangent space at our cosen point in S.)[] e(s)g(s)+g(s)e(s) f(s)f (s) (E(s)G(s) F (s) ) u sqrte, Eu F v E(EG F ). (tis if obtained from te first definition by respectively, were u, v is some basis of te An alternate definition of minimal surfaces says tat a surface S is minimal if and only if its mean curvature is zero at any point in s S.[1] We parametrize an extension of M by M(u, v) = ( 1 C cos(cu) cos(v), 1 C cos(cu) sin(v), u), were u R and v [0, π]. Ten Terefore, n = = finally yielding 1 M u M ( 1 C cos4 (Cu) M = (sin(cu) cos(v), sin(cu) sin(v), 1) u M ( v = 1C cos(cu) sin(v), 1C ) cos(cu) cos(v), 0 M = (C cos(cu) cos(v), C cos(cu) sin(v), 0) u M = ( sin(cu) sin(v), sin(cu) cos(v), 0) u v M ( v = 1C cos(cu) cos(v), 1C ) cos(cu) sin(v), 0 M v u M v ) 1 ( 1C cos(cu) cos(v), 1C cos(cu) sin(v), 1C ) sin(cu) cos(cu) E = cos (Cu) F = 0 G = 1 C cos (Cu) C e = cos (Cu) f = 0 1 g = cos (Cu) 6

7 Tus, for any (u, v), we find Tus, M is a minimal surface. eg + ge ff (EG F ) = (1) = 0 4 Constraints on eigt Finally, we must find te value of 0. We let C vary and require = (C) to satisfy C = cos, as seen in Figure. As te maximum of te specified curve, d dc = 0. 0 sin Were C 0 is te value of C associated wit 0. Tat is, from (5), 1 C cos = 1 cos = C d dc cos = d dc C ( + C ) d = 1 dc 0 sin C0 0 = 1 (5) 1 C0 0 cos = 1 C 0 C 0 0 C0 0 tan = 1 Matematica yields wit precision up to 15 significant digits, C 0 0 ( ) 1 C0 0 0 = sin = Ten (5) yields 5 Conclusion In conclusion, all non-planar minimal surfaces of revolution must be generated by functions of te form f = 1 C cos(cx), x R. Moreover, we ave proven tat all suc surfaces are indeed minimal, and we ave found constraints on te lengt of line segments parallel to te distinguised axis wit endpoints on te surface, dependent on te distance of te segment from te distinguised axis. Finally, we conclude tat apart from one distinguised member (wen = 0 ) of tis family, tese surfaces come in pairs tat intersect in two circles of equal radius in parallel planes wit centers along te distinguised axis. 7

8 References [1] N. Korevaar, J. Ratzkin, N. Smale, and A. Treibergs. A survey of te classical teory of constant mean a survey of te classical teory of constant mean curvature surfaces in R 3., June 00. University of Uta. ttp:// ratzkin/papers/minicourse.pdf. [] C. Teleman. Riemann surfaces, 003. Berkeley University. ttp://mat.berkeley.edu/ teleman/mat/riemann.pdf. 8