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Minimal Surfaces in R^3 Minimal Surfaces in R3 AJ Vargas May 10, 2018 1 Introduction The goal of this note is to give a brief introduction to the theory of minimal surfaces in R3, and to show how one would go about generalizing the theory to oriented k-dimensional hypersurfaces in Rn. We will also touch on some interesting problems, applications, and re- sults related to the field. We will touch on Plateau’s laws for soap films, Plateau’s problem, the Double Bubble theorem, and finally show some pictures. First let’s say what we mean by a minimal surface. Roughly, a minimal surface is one for which each point has neighborhood which, as itself a surface, has smallest possible area among surfaces sharing the same boundary. That is, a minimal surface is one which locally minimizes area. To have a concrete idea of how to come up with such things, let’s begin with the classical derivation of the minimal surface equation as the Euler-Lagrange equation for the area functional, which is a certain PDE condition due to Lagrange circa 1762 de- scribing precisely which functions can have graphs which are minimal surfaces. By viewing a function whose graph was a minimal surface as a minimizing function for a certain area functional (i.e. a smooth map from a function space into R which measures “area”), we are able to use techniques from the calculus of variations to show that for an oriented surface, locally minimizing area is equivalent with the vanishing of the divergence of the unit normal vector to that surface. It is from this perspective that we can generalize the situation to arbitrary k-dimensional submanifolds of Rn, and in fact to any k-dimensional submanifold of a Riemannian manifold. Recall that any surface is locally the graph of some smooth function. This is a relatively easy consequence of the inverse function theorem, based on what it means to be a regular parametrized surface, that is, where the two co-ordinate vectors for the parametrization span a two dimensional vector space. Then by the above discussion, to find more general minimal surfaces it will suffice to find minimal graphs of functions. Let U = ? be a bounded open 6 set in R2, and let φ : U φ(U) R3 be any di↵eomorphism. We should think of the the set φ(@U)asaclosed“wire”inspace.Inthesettingofaregularsurface,thisrolewill! ⇢ be filled by the boundary of the neighborhood in which the surface can be viewed as the graph of a function. In any case, let’s consider the set H of functions h C1(U)suchthat h(@U)=φ(@U)(i.e.whoseboundariesarethegivenwire).We’llsaythatthegraphof2 f H has minimal area with respect to @U if 2 1 dΓ dΓ f h ZU ZU for each h H (here Γ is just the graph h, with the usual parametrization, and dΓ is the 2 h h area element det(I) dx dy, I being the first fundamental form of Γh). Say that f has minimal area if its graph does.^ In other words, f has minimal area with respect to U if it is p a minimizer of the area functional a : H R defined by ! a(h)= dΓh. ZU Here is how the minimizing functions were characterized. Suppose f minimizes a. Then chose any other smooth ⌘ H so that ⌘’s value on @U is identically 0. We can get a 2 parametrized family of functions zt = f + t⌘ H. The term t⌘ is called a variation of f. Since f was chosen to be a minimizing function,2 t =0mustbeacriticalpointforthe function A(t)=a(zt)= d(Γzt ) ZU 2 2 = 1+(zt)x +(zt)y dx dy U ^ Z q = 1+ z 2 dxdy. |r t| ZZU That t =0iscriticalmeansthat p d d A(t) = 1+ z 2 dxdy dt dt |r t| t=0 ZZU t=0 d p (1) = 1+ z 2 dxdy dt |r t| ZZU t=0 p =0. It’s not hard to check that d 2 zt, ⌘ 1+ zt = hr r i dt |r | 1+ z 2 |r t| p z = p r t , ⌘ − 1+ z 2 r ⌧ |r t| so that after evaluating at t =0,equation(1)becomesp d d A(t) = 1+ z 2 dxdy dt dt |r t| t=0 ZZU t=0 p f = r , ⌘ dxdy 2 U − 1+ f r ZZ ⌧ |r | =0. p 2 Now, applying the formula div('F)= ', F +' div(F) (where ' : Rn R is smooth and F is a smooth vector field) we get hr i ! f f f r , ⌘ = ⌘ div r div ⌘ r − 1+ f 2 r 1+ f 2 − 1+ f 2 ⌧ |r | ✓ ✓ |r | ◆◆ ✓ ✓ |r | ◆◆ so that integratingp both sides and applyingp the divergence theorem,p we can rewrite equa- tion (1) again as d f A(t) = r , ⌘ dxdy 2 dt U − 1+ f r t=0 ZZ ⌧ |r | f f = ⌘ divp r dxdy div ⌘ r dxdy 2 2 U 1+ f − U 1+ f ZZ ✓ ✓ |r | ◆◆ ZZ ✓ ✓ |r | ◆◆ f f = ⌘ div p r dxdy ⌘ rp ds. 2 2 U 1+ f − @U 1+ f ZZ ✓ ✓ |r | ◆◆ Z ✓ |r | ◆◆ f = ⌘ div p r dxdy p 2 U 1+ f ZZ ✓ ✓ |r | ◆◆ =0 p f Notice that we could conclude ⌘ r ds =0because⌘ was chosen to vanish @U p1+ f 2 ✓ |r | ◆◆ on @U. We can now apply the fundamentalR lemma of the calculus of variations, which states that if ⌦ Rn is an open set, and f is a continuous function on ⌦, then if for all compactly 2 supported functions ' C1(⌦)onehas 2 '(x)f(x)dx, Z⌦ then f is identically 0 on ⌦. In our present situation, any function ' which is compactly supported on U will extend to a function ' C1(U)whichwillvanishon@U, so we can conclude, as desired, that 2 f @ fx @ fy div r = + = trace(D)=0 (2) 1+ f 2 @x 1+ f 2 @y 1+ f 2 ✓ |r | ◆ ✓ |r | ◆ ✓ |r | ◆ on all ofpU if and only if f is ap minimizer of the areap functional a. Equation (2) is called the minimal surface equation. Here, D is the Jacobian matrix of the function f . p1+ f 2 |r | If we compute all the partials and simplify, we get the following PDE f (1 + f 2) 2f f f + f (1 + f 2)=0. (3) xx y − x y xy yy x It can be shown that any solution to this equation must be real-analytic. This demon- strates that functions of minimal area are not so readily abundant. As a final note, it 3 turns out that the shape operator of any graph of minimal area parametrized via r(x, y)= (x, y, f(x, y)) has no trace anywhere, because the trace of this map turns out to be exactly the minimal surface equation (3) up to some non-zero multiple. Let’s see this. It’s not hard to compute the coefficients of the first and second fundamental forms of a function as 2 2 E =1+fx F = fxfy G =1+fy fxx fxy fyy e = 2 2 f = 2 2 g = 2 2 p1+fx +fy p1+fx +fy p1+fx +fy and therefore the mean curvature is 2 2 eG 2fF + gE fxx(1 + f ) 2fxfyfxy + fyy(1 + f ) y − x − 2 = 3 =0. 2(EG F ) 2(1 + f 2 + f 2) 2 − x y Since the trace of the shape operator of a surface is independent of choice of parametriza- tion, one may instead say a parametrized surface is minimal whenever its shape operator has no trace anywhere. This definition provides us with some more geometric perspective. Mean curvature tells us the average value of the normal curvatures of the surface over all possible directions. It is well known that the mean curvature is determined by the average of the eigenvalues of the shape operator, which correspond respectively to the maximum and min- imum of the principal curvatures of the surface at a point over all possible directions. This directions of maximal and minimal curvature are called principal directions. So namely, the trace of the shape operator being 0 everywhere tells us that the principal curvatures at each point cancel each other. This means that for all points on the surface, the curvature of acurveintheprincipaldirectionofthesurfacewillbethenegativeofthecurvatureinthe other principal direction. Let us now try to generalize the situation. Basically the same procedure we outlined above will make sense for an arbitrary k-submanifold of Rn, but first we have to say clearly what the “area” (volume more generally) element is for such a thing. For a regular surface, the most natural definition is that the area element would be the size of the area of the parallelogram spanned by the co-ordinate tangent vectors to the surface at a point times the usual area element dx dy. In general, any k-dimensional linear subspace of some n- dimensional inner product space^ is itself an inner product space via the induced inner product on the bigger space. We already know from linear algebra that the area of the parallelogram spanned by the linearly independent set v1,...,vk of vectors is gotten by taking the size of the determinant of the symmetric matrix{ which encodes} how the inner product operates on the subspace generated by the vectors. The matrix in other words encodes all of the information about lengths and angles in the span of the set. This matrix is called the Gram matrix of the linearly independent set. It is given by T G(v1,...,vk)=(v1 ... vk) (v1 ... vk), or perhaps more clearly, [G(v1,...,vk)]ij = vi,vj .
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