DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

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102 Chapter 3. Surfaces: Further TopicsProof. Let ξ :[0, 1] → R 2 be a C 1 curve with ξ(0) = ξ(1) = 0. Then, using the fact that wecan pull the derivative under the integral sign (see Exercise 1) and then the chain rule, we haveddε∣ F (u ∗ + εξ) = d ∫ 1ε=0dε∣ f(t, u ∗ (t)+εξ(t), u ∗′ (t)+εξ ′ (t))dtε=0 0∫ 1∂=0 ∂ε∣ f(t, u ∗ (t)+εξ(t), u ∗′ (t)+εξ ′ (t))dtε=0∫ 1 ( ∂f=∂u (t, u∗ (t), u ∗′ (t)) · ξ(t)+ ∂f)∂u ′ (t, u∗ (t), u ∗′ (t)) · ξ ′ (t) dtand so, integrating by parts, we have==0∫ 10∫ 10( ∂f∂u · ξ(t) − d dt( ∂f∂u − d dt( ) ∂f)∂u ′ · ξ(t) dt + ∂f ] 1∂u ′ · ξ(t) 0( ∂f∂u ′ ) )· ξ(t)dt.Now, applying Exercise 2, since this holds for all C 1 ξ with ξ(0) = ξ(1) = 0, weinfer that∂f∂u − d ( ) ∂fdt ∂u ′ = 0,as desired. □Of course, the Euler-Lagrange equations really give a system of differential equations:∂f∂u = d ( ) ∂fdt ∂u(♣)′ ∂f∂v = d ( ) ∂fdt ∂v ′ .Example 1. Recall that for the unit sphere in the usual parametrization we have E =1,F =0,and G = sin 2 u. To find the shortest path from (u 0 ,v 0 )=(u 0 ,v 0 )tothepoint (u 1 ,v 1 )=(u 1 ,v 0 ),we want to minimize the functionalF (u, v) =∫ 10√(u ′ (t)) 2 + sin 2 u(t)(v ′ (t)) 2 dt.Assuming our critical path u ∗ is parametrized at constant speed, the equations (♣) give us v ′ (t) =const and u ′′ (t) =sin u(t) cos u(t)v ′ (t) 2 . (Cf. Example 6(b) in Section 4 of Chapter 2.) ▽We now come to two problems that interest us here: What is the surface of least area with agiven boundary curve? And what is the surface of least area containing a given volume? For thiswe must consider parametrized surfaces and hence functionals defined on functions of two variables.In particular, for functions x: D → R 3 defined on a given domain D ⊂ R 2 ,weconsider∫∫F (x) = ‖x u × x v ‖dudv.We seek a function x ∗ so that, for all variations ξ : D → R 3 with ξ = 0 on ∂D,D ξ F (u ∗ F (u ∗ + εξ) − F (u ∗ ))=lim= d ε→0 εdε∣ F (u ∗ + εξ) =0.ε=0D
§4. Calculus of Variations and Surfaces of Constant Mean Curvature 103Now we compute: Recalling that d dt ‖f(t)‖ = f(t) · f ′ (t)and setting x = x ∗ + εξ, wehave‖f(t)‖d1 (dε∣ ‖x u × x v ‖ =(ξuε=0‖x ∗ u × x ∗ × x ∗ v + x ∗ u × ξ v ) · (x ∗ u × x ∗ v) )v‖Next we observe that=(ξ u × x ∗ v + x ∗ u × ξ v ) · n.(ξ u × x ∗ v) · n = ( (ξ × x ∗ v) · n ) u − (ξ × x∗ uv) · n − (ξ × x ∗ v) · n u(x ∗ u × ξ v ) · n = ( (x ∗ u × ξ) · n ) v − (x∗ uv × ξ) · n − (x ∗ u × ξ) · n v ,and so, adding these equations, we obtain(ξ u × x ∗ v + x ∗ u × ξ v ) · n = ( (ξ × x ∗ v) · n ) u + ( (x ∗ u × ξ) · n ) v − ( (ξ × x ∗ v) · n u +(x ∗ u × ξ) · n v)= ( (ξ × x ∗ v) · n ) u − ( (ξ × x ∗ u) · n ) v − ( (ξ × x ∗ v) · n u +(x ∗ u × ξ) · n v)= ( (ξ × x ∗ v) · n ) u − ( (ξ × x ∗ u) · n ) v − ξ · (x ∗ v × n u + n v × x ∗ u).At the last step, we’ve used the identity (U × V) · W =(W × U) · V =(V × W) · U. Theappropriate way to integrate by parts in the two-dimensional setting is to apply Green’s Theorem,Theorem 2.6 of the Appendix, and so we let P =(ξ × x ∗ u) · n and Q =(ξ × x ∗ v) · n and obtain∫∫(ξ u × x ∗ v + x ∗ u × ξ v ) · ndudvD∫∫ ( ((ξ= × x∗v ) · n ) − ( (ξ × x ∗ uu) · n ) ) ∫∫dudv − ξ · (x ∗vv × n u + n v × x ∗ u)dudvD } {{ } } {{ }DQ uP v∫=∂D(ξ × x ∗ u) · n du +(ξ × x ∗} {{ }v) · n} {{ }PQ∫∫dv − ξ · (x ∗ v × n u + n v × x ∗ u)dudv.DSince ξ = 0 on ∂D, the line integral vanishes. Using the equations (††) onp.56, we find thatx ∗ v × n u = a(x ∗ u × x ∗ v) and n v × x ∗ u = d(x ∗ u × x ∗ v), so, at long last, we obtain∫∫∫∫ddε∣ ‖x u × x v ‖dudv = (ξ u × x ∗ v + x ∗ u × ξ v ) · ndudvε=0 DD∫∫= − ξ · (x ∗ v × n u + n v × x ∗ u)dudvD∫∫∫∫= − (a + d)ξ · (x ∗ u × x ∗ v)dudv = − 2Hξ · ndA,DDsince H = 1 2 trS P .We conclude from this, using a two-dimensional analogue of Exercise 2, the followingTheorem 4.2. Among all (parametrized) surfaces with a given boundary curve, the one ofleast area is minimal, i.e., has H =0.This result, indeed, is the origin of the terminology.Next, suppose we wish to characterize those closed surfaces of least area containing a givenvolume V .Tomake a parametrized surface closed, we require that x(u, v) =x 0 for all (u, v) ∈ ∂D.But how do we express the volume constraint in terms of x? The answer comes from the Divergence
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DIFFERENTIALGEOMETRY:A First Course
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4 Chapter 1. Curves2πb2πaFigure 1
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6 Chapter 1. CurvesAlternatively, s
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8 Chapter 1. CurvesAn important obs
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10 Chapter 1. Curvesof the road, st
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12 Chapter 1. CurvesSummarizing,κ(
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14 Chapter 1. Curvescurvature κ. I
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16 Chapter 1. CurvesFigure 2.2Conve
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18 Chapter 1. Curvesalso Exercise 2
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20 Chapter 1. Curvesto the curve β
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22 Chapter 1. Curvesthe osculating
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24 Chapter 1. CurvesWe turn now to
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26 Chapter 1. Curvesintersect C eit
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28 Chapter 1. Curvesh(s,t)α(t)α(s
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30 Chapter 1. Curvesy( x(s) ,y(s))
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32 Chapter 1. Curvesɛ2ɛ 1ɛ 3CFig
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CHAPTER 2Surfaces: Local Theory1. P
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36 Chapter 2. Surfaces: Local Theor
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