MINIMAL GRAPHS PART I: EXISTENCE OF LIPSCHITZ WEAK SOLUTIONS TO THE DIRICHLET PROBLEM WITH C 2 BOUNDARY DATA SPENCER HUGHES In these notes we prove that for any given smooth function on the boundary of the unit ball in Euclidean space, there exists a Lipschitz weak solution of the Minimal Surface Equation in the unit ball that is equal to the given smooth function at the boundary. The approach we take here can be found in [Giu84, Chapter 12] and uses the direct method of the calculus of variations. In fact, this approach proves more. Firstly, the regularity of the boundary data need only be C 2. Secondly, the graph of the Lipschitz function that we construct is actually area minimizing in the class of Lipschitz graphs with the same boundary. And thirdly, we are able to remark that the method carries over easily to the case of a domain with mean convex, C 2 boundary. I would like to thank Larry Guth for giving me the opportunity to lecture the material here in his real analysis class at MIT. Balls whose centres are not labelled are centred at the origin in R n. For Lipschitz u : B := B1 n (0) R and Ω open B, define A(u; Ω) = Area (graph u Ω ) = 1 + Du 2, where, by area, we mean n-dimensional Hausdorff measure. The notation A(u) simply means Area (graph u). It can be established (firstly for C 1 functions using integration by parts and duality and then by approximating a Lipschitz function in the L 1 norm by a sequence of C 1 functions) that for any Lipschitz u : B R, we have (1) A(u) = sup g n+1 + u div(g 1,..., g n ). g C 1 c (B,Rn+1 ): B g 1 From here one can easily deduce that A is strictly convex on L and lower semicontinuous with respect to weak L 1 convergence. Given f C 2 (), if u C 2 (B) is such that u = f and A(u) A(ũ) for any ũ C 2 (B) with ũ = f, then u satisfies 0 = d (2) A(u + tη) dt t=0 for all η Cc (Ω), from which a short calculation gives that Du Dη (3) 0 = for every η C 1 + Du 2 c (Ω). B I have ambitiously called these notes Part I in the hope that I will eventually neaten up and upload various different notes that I have written over the last five years or so on minimal graphs. 1 Ω
2 SPENCER HUGHES After integrating by parts we deduce that u solves the Dirichlet problem for the Minimal Surface Equation, i.e. ( ) Du Mu := div = 0 in B 1 + Du 2 u = f on. Any function u W 1,2 (B) that satisfies (3) is called a weak solution of the minimal surface equation. Write L L for the set of all Lipschitz functions u : B R with u = f and Lip u L and write a L := infũ LL A(ũ). The symbols L and a are defined similarly, but without the condition that Lip u L. Lemma 1. If L L, then exists u L L L such that A(u L ) = a L and such that if Lip u L < L, then A(u L ) = a Proof. Since L L, we can take a sequence {u j } j=1 L L with A(u j ) a L as j. The fact that Lip u j L for every j means that {u j } j=1 is equicontinuous and, for sufficiently large j, this sequence must also be uniformly bounded (why?). Therefore the Arzela Ascoli theorem implies that there exists v L L and a subsequence {j } of {j} such that u j v uniformly as j. Setting u L := v, the first claim follows from the lower semicontinuity of A with respect to weak L 1 convergence. Now given ũ L observe that (tu L + (1 t)ũ) = f for all t (0, 1) and that if Lip u L < L, then Lip(tu L + (1 t)ũ) L for sufficiently small t > 0. Thus for sufficiently small t > 0 we have (tu L + (1 t)ũ) L L whence A(u L ) A(tu L + (1 t)ũ). The convexity of A then implies that A(u L ) ta(u L ) + (1 t)a(ũ), from which the second claim follows. For Ω open B, write L L (Ω) for the set of all Lipschitz functions u : Ω R with Lip u L. We will need more terminology for our next result: A function u L L (Ω) is called a supersolution (resp. subsolution) in L L (Ω) if for every ũ L L (Ω) with ũ Ω = u Ω and ũ u (ũ u) we have that A(ũ; Ω) A(u; Ω). In particular, a minimizer in L L (Ω) (i.e. a function the area of the graph of which is smallest among competitors with the same boundary values) is both a super- and a subsolution. Lemma 2. For Ω B, let v and u be super- and subsolutions in L L (Ω). If u v on Ω, then u v in Ω. Proof. Suppose for the sake of contradiction, that S = {x Ω : v(x) < u(x)} and write m = min{u, v}. Since m L L (Ω), m Ω = u Ω and m u in Ω, the fact that u is a subsolution in L L (Ω) means that A(u; Ω) A(m; Ω), which implies that A(u; S) A(v; S). Now, the strict convexity of the area functional tells us that (4) A( 1 2 u + 1 2 v; S) < 1 2 A(u; S) + 1 2A(v; S) A(v; S). But, since w = max{ 1 2 u + 1 2 v, v} satisfies w L L(Ω), w Ω = v Ω and w v in Ω, the fact that v is a supersolution means that A(v; Ω) A(w; Ω), which implies that A(v; S) A( 1 2 u + 1 2v; S), contradicting the strict inequality in (4).
MINIMAL GRAPHS PART I 3 Corollary 3. Let u and v be respectively a subsolution and supersolution in L L (Ω). Then sup[u(x) v(x)] = sup [u(x) v(x)]. x Ω x Ω Proof. For every α R, the function v(x) + α is also a supersolution and for any x Ω, we clearly have u(x) v(x) + sup y Ω [u(y) v(y)]. The result now follows from the previous lemma. We can now reduce our goal of controlling the Lipschitz constant to a boundary estimate: Lemma 4. With u L as in Lemma 1 we have u L (x) u L (y) (5) Lip u L = sup x B, y x y Proof. Let x 1, x 2 be distinct points in B = B 1 (0). Both u and the function x u L (x+x 2 x 1 ) minimize area in L L (B 1 (0) B 1 (x 1 x 2 )) and so both functions are supersolutions and subsolutions in L L (B 1 (0) B 1 (x 1 x 2 )). Applying corollary 3 in the domain B 1 (0) B 1 (x 1 x 2 ) and with u L (x+x 2 x 1 ) in place of v(x) implies that there exists z (B 1 (0) B 1 (x 1 x 2 )) for which u L (x 1 ) u L (x 2 ) u L (z) u L (z + x 2 x 1 ). At least one of z and z + x 2 x 1 belongs to, so on dividing by x 2 x 1, we get the result. Proposition 5 (Existence of Barriers). Given f C 2 (B), there exist constants c 1, c 2 and r > 0 (depending only on f and n) such that v + : B 1 (0) \ B 1 r (0) R given by v + (x) = f(x) + c 1 log(1 + c 2 dist(x, )) has the following properties. (1) v + = f. (2) v + 1 r(0) sup f. (3) v + is a supersolution in L(B 1 (0) \ B 1 r (0)). Proof. We will try a function of the form v + (x) = f(x) + ψ(d(x)), where d(x) := dist(x, ) and where ψ C 2 ([0, 1 r]) will be chosen to satisfy ψ(0) = 0, ψ 1, ψ < 0 and ψ(1 r) 2 sup Ω f. Such a v + will already automatically satisfy the first two conditions above. Now, by direct computation we have that (1 + Dv + 2 ) 3/2 Mv + is equal to: (6) (7) (8) (9) (10) (1 + Df 2 ) f D i fd j fd ij f + ψ [2D i fd i d f + (1 + Df 2 ) d 2D i dd j fd ij f D i fd j fd ij d] + ψ 2 [ f + 2D i fd i d d D ij fd i dd j d] + ψ 3 d + ψ [1 + Df 2 (D i fd i d) 2 ], where we have already used the fact that Dd 2 = 1 and therefore that D i dd ij d = 0. Since 1 ψ ψ 2 and f, d C 2 (B 1 (0) \ B 1 r (0)), we can bound the terms on lines (6), (7) and (8) by Cψ 2, for some C = C(f, n) > 0. We also see (using the fact that ψ < 0 together with (D i fd i d) 2 Df 2 ) that that the final term is at most ψ. Thus (11) (1 + Dv 2 ) 3/2 Mv + ψ + Cψ 2 + ψ 3 d.
4 SPENCER HUGHES Since d 0 in B 1 (0) \ B 1 r (0) (as we can check by hand), we have that (12) (1 + Dv 2 ) 3/2 Mv + ψ + Cψ 2. We will choose ψ so that this is less than or equal to 0. Set ψ(t) = c 1 log(1 + c 2 t) with c 1 = 1/C, where C is from the line above. This ensures immediately that Mv + 0, which, after multiplying by non-negative η Cc (B 1 (0) \ B 1 r (0)) and integrating by parts implies that the function g : t A(v + + tη; B 1 (0) \ B 1 r (0)) satisfies g (0) 0, whence, by convexity of A we deduce that v + is a supersolution in the sense we have defined. We now need to show that we can choose c 2 and r so that ψ 1 and ψ(1 r) 2 sup f. Since and ψ (t) = c 1c 2 1 + c 2 t c 1 c 2 1 + c 2 (1 r) ψ(1 r) = c 1 log(1 + c 2 (1 r)), we need c 2 /(1 + c 2 (1 r)) c 1 1 and c 2 (1 r) e 2c 1 1 sup f 1. Setting 1 r = c 1/2 2 means that we need c 2 /(1 + c 1/2 2 ) c 1 1 and c 1/2 2 e 2c 1 1 sup f 1. But these can both be achieved by now choosing c 2 to be large. This completes the construction. Remark 6. (i) The function v here is called an upper barrier. If one takes the negative of the function obtained by applying this Proposition with f in place of f, one gets a function v satisfying (1) v = f. (2) v 1 r(0) inf f. (3) v is a subsolution in L(B 1 (0) \ B 1 r (0)). Such a function is called a lower barrier. (ii). In more general domains, the geometry of the boundary of the domain plays an important role: For a C 2 domain Ω, the argument above requires that d C 2 ({x : d(x) < r} Ω) and that d 0 in {x : d(x) < r} Ω for sufficiently small r > 0. The first of these is always true for sufficiently small r > 0 and the second is true for sufficiently small r > 0 provided Ω is mean convex (has non-negative mean curvature). Proofs of both these facts can be found in [GT01, 14.6]. (iii). If the boundary of Ω is not mean convex, then there is smooth boundary data for which the Dirichlet problem cannot be solved (see [Giu84, Chapter 12]). Theorem 7. Given f C 2 (), there exists u L with A(u) A(ũ) for all ũ L Proof. By solving the Dirichlet problem for the Laplacian with boundary values f, we can ensure that L L is non-empty for some sufficiently large L and assume that f C 2 (B). Let us also pick L > Lip v +. As per Remark 6. (i), let v : B 1 (0) \ B 1 r (0) R be a lower barrier. Then (13) v (x) inf f u(x) sup f v + on 1 r (0). Since u L is minimizing in L L (B 1 (0) \ B 1 r (0) and since w is a subsolution in B 1 (0) \ B 1 r (0), Lemma 2 implies that v u v + on B 1 (0) \ B 1 r (0). Then, using the fact that u = v ± on 1 (0), we have (14) u(x) u(y) (Lip v) x y
MINIMAL GRAPHS PART I 5 for every x B 1 (0) \ B 1 r (0), y. On the other hand, if x B 1 r (0), we have that { } u(x) u(y) max sup f u(y), u(y) inf f Lip v, (using (2) in the defining properties of v ± ) which shows that (14) holds for all x B. Then, by Lemma 4 we have Lip u Lip v < L and so the conclusion follows from Lemma 1. References [Eva09] L.C. Evans, Partial differential equations (graduate studies in mathematics, vol. 19), Instructor (2009). [Giu84] E. Giusti, Minimal surfaces and functions of bounded variation, Birkhauser, 1984. [GT01] D. Gilbarg and N.S. Trudinger, Elliptic partial differential equations of second order, Springer Verlag, 2001. Massachusetts Institute of Technology, Department of Mathematics, 77 Massachusetts Avenue, Cambridge, MA 02139-4307, E-mail address: sthughes@mit.edu