Lecture 7 Minimal Surface equations non-solvability strongly convex functional further regularity Consider minimal surface equation div + u = ϕ on ) = 0 in The solution is a critical point or the minimizer of inf +. u =ϕ But the integrand F p) = + p is not strongly convex, that is D F I, only D F > 0. The loss of strong convexityor convexity causes non-solvability, or non minimizer for general domains, unlike u = 0 with case. Eg. Let the boundary data be u = t on B and u = 0 on B with = B \B. catenoid on annulus figure. The minimizer should be radial by symmetry), or just we consider radial solutions. Necessarily we have a constraint 0 = div = r da r da. B r\b + B r + B + We infer u r + u r r n = C 0 u r = Cr n ) with C. Then u r) = r dρ Cρ n ) eg n= = ch ). dρ < ρ n ) 0 May, 00
Now u = t, say 00000000 on B contradicts the above inequality. RMK. Mean curvature of is necessary and sufficient use distance to the boundary as barrier) in solving minimal surface equation with arbitrary boundary condition. In our lecture, we only consider strongly convex and ϕ C, ). Eg. Consider the non-convex functional inf u0)=0 u)= 0 { F u x ) dx with F p) = double well F figure p p ) for p 0 quadratic extension for p > 0, The Euler-Lagrangian equation is D x F p u x )) = F u x ) u xx = 0. u = x is a critical point, not minimizer, 0 F x ) dx =. v = with v = 0 or, minimizers, not smooth, not unique, 0 F v ) dx = 0. various critical pts figure Next we solve the minimal surface equation via strongly convex functionals. Step. Let F p) = + p + p, then I D F ) + ) I. + x ) = x x +x = sin θ, + x ) = cos θ θ xx x = +x ) Parallel to the minimizing process to, we minimize +x J [u] = F ). Let the minimizing sequence u k H with u k = ϕ on J [ u k] inf J [u] = m. We claim: { u k} is a Cauchy sequence in H. convexity figure For any small) positive ε, we have for all large k and l m J [ u k] m + ε [ m J u k + u l)] = F k + l)) [ F k) + F l)] min D F k l m + ε k l. 4
Hence k l ε. 4 So we know u k u in H the minimizer is unique by setting ε = 0) and satisfies for all φ H0 ) Dxi φ F ) p i = 0. Step 0. Now inf + = inf. u W, u=ϕ on To make sense the functional, W, is the right space. Note W, functions also have L trace on. Going with the W, minimizing sequence woud not lead to a W, Cauchy sequence. Even if one has a W, minimizer, the minimizer is only in W, space, not in W, space.) For any ε > 0, there exists v W, with v = ϕ on such that + Dv inf +ε. To move from W, v to W,, let V η C be the approximation for v, then + DV η inf +ε + ε. Also there is = ε, DV η L ) such that + DV η + DV η inf +ε + ε + ε. So the minimizer u H with u = ϕ on for + + satisfies Thus + + inf +3ε. + 0 provided L) C independent of to be justified). *) Observe a dependent bound of L is already there. Further there is u H ) such that u u in H weakly by weak compactness of H space. Let us show that + lim inf + = inf. 0 3 inf
This is because the functional + is convex and a convex combination of u u in H strongly. Then + + D convex combination of u Jessen convex combination of inf +C. + At this point, we have already obtained a minimizer u and its H regularity then its uniqueness ε = 0 argument pushed further), if we furnish *). Indeed we show a stronger claim by taking advantage of the boundary data C, at this point, further relaxed C 0 condition depending on Bombieri-De Giorgi-Miranda s a priori gradient estimate.) and C, boundary: L ) C ϕ C,, ), independent of. RMK. First note from De Giorgi-Nash, we already know u C,α inside, though we do not know yet a uniform C,α norm. Further by a similar, but simpler argument than De Giorgi-Nash, one can get C ε regularity of u up to the Lipschitz boundary with C β boundary data ϕ. In the following we are just drawing uniform estimates independent of parameter. Boundary. For any linear function L, D i F pi DL) ) = 0. We compare L to u satisfying D )) i F pi = 0. We have Di Fp i p j ) D )) j L u = 0 ) I F pipj + ) I. The strong) maximum principle implies that the inf and sup of L u achieves on the boundary. Recall/Exercise: For C, boundary strongly κ 0 -convex, that is the principle curvatures κ,, κ n ) κ 0 componentwise, and C, boundary data ϕ, we have x n = x boundary figure L {}}{ ϕ 0) + D x ϕ 0) x Mx n ϕ x) L u L + L + {}}{ ϕ 0) + D x ϕ 0) x +Mx n on. Hint: x n κ 0 x. Apply the maximum principle either after Moser, or a simpler argument to be found in the end of this lecture), we get L u L + in. 4
It implies D xnu 0) M. Thus = D u, D n u ) M ϕ C,, κ 0 ) on, -free. Bdry Lip) Interior to Boundary. For any e R n, for any x, really boundary of { εe} with ε small. By the boundary Lip Bdry Lip), we have for any fixed boundary point x = x 0 and for all ε ε 0 x 0 ) u x + εe) u x) + Mε. By the compactness of, we have the above inequality at all boundary points of { εe} for all ε ε. Observe that both u x + εe) and u x) are W, weak solutions to Dxi F pi Dv) ) = 0 in { εe}. By the strong) maximum principle u x + εe) u x) + Mε in { εe}, from which we infer for all x { εe} Similarly we obtain By letting ε 0, we get u x + εe) u x) ε M. M u x + εe) u x). ε L ) M, -free. Let go to zero, Summary:) we have got the minimizer for u = ϕ C, ) on the strongly convex boundary, such that L ) C ϕ C,, κ 0 )). Step C,α. Regularity for the critical point u. First we have )) u x + εe) u x) Dxi F pi p j ) D xj = 0. ε De Giorgi-Nash implies u x + εe) u x) ε C Ca B /) u x + εe) u x) ε C L B ) C ϕ C,, κ 0 )). + Dv with L B ) 5
Thus u C,α and u is a weak solution to D i F pi )) = 0. Next we show that u C,α, then F pi p j ) D ij u = 0. The proof is through the C,α solution to a Dirichlet problem by Schauder theory. Let a ij x) = F pi p j x)) like the regularity for viscosity/perron solution to u = 0), we know how to solve { aij x) D ij w = 0 in B η Schauder) w = u on B η by weighted norm method. Then we have C,α B η ) solution w. Proposition Let u C,α be a weak solution to D i F pi )) = 0 in B η and w C,α solution to Schauder). Then u w in B η. The idea of the proof is to show D i F pi Dw)) = 0 by a viscosity way. The technical execution is to modify w to v C,α such that Di F pi Dv)) = F pi p j Dv) D ij v 0 v u and v x ) > u x ) v = u on boundary. Then contradicts D i F pi )) = 0. Proof. Suppose u w in B η, say max Bη w u) = w u) x 0 ) = t > 0 and x 0 B η. w over u at x 0 figure Frist step toward a sub soluution v : Let w t = w + t x η ), then w t = u on B η and w t x 0 ) = w x 0 ) + t x0 η ) = u x 0 ) + t + t x0 η ) > u x 0 ), where we assumed that we started with η. Then there exists C t such that w t C t u in B η w t C t = u at x B η x may not be x 0 ) Dw t x ) = x ). v over u over w t figure Second step toward a sub solution v : Let v = w t C t t 4 x x + γ, then v u in a neighborhood N γ of x. And N γ shrinks to the point x as γ goes to zero. 6
Since w C,α, a ij x) C α, and Dv x ) = x ), we can choose γ small so that N γ small, then Dv is close to in N γ and eventually so that Di F pi Dv)) = F pi p j Dv) D ij v = Fpi p j Dv) F pi p j ) ) D ij v }{{}}{{} o) bounded = o ) + aij x) D ij w 0 + t a ij x) }{{} 4 ij µ + a ij x) D ij v µ t 8 for small γ. Now Di F pi )) = 0 in N Di F pi Dv)) 0 γ, u = v on N γ or Di Fpipj ) D j v u) ) 0. Take a test function v u) + H0 N γ ), we get 0 v u) + D i Fpi p j ) D j v u) ) N γ Sard = Di v u) + F pi p j ) D j v u) N γ µ Di v u) +. N γ It follows that Di N γ v u) + = 0, then v u) + 0 or v u in N γ. But v u = γ > 0 at x in N γ. This contradiction shows that u w C,α in B η. Exercise: Let u be a C,α solution to F pi p j ) D ij u = 0 and µi ) F pi p j µ I. Show that u C 3,α. 7