Geometric PDE and The Magic of Maximum Principles Alexandrov s Theorem John McCuan December 12, 2013
Outline 1. Young s PDE 2. Geometric Interpretation 3. Maximum and Comparison Principles 4. Alexandrov s Theorem 1
Variation of Area G[u] = Ω 1 + Du 2 (Nonparametric) Volume Constrained Plateau Problem: (P): Minimize G over A = {u C 2 (Ω) C 0 ( Ω) : u Ω = u 0, Ω u = V 0}. 2
Lagrange Multiplier Problem (P) λ : Minimize over G[u] = Ω 1 + Du 2 λ u Ω Ã = {u C 2 (Ω) C 0 ( Ω) : u Ω = u 0 }. 3
Lagrange Lemma If u solves (P) λ, and it happens that Ω u = V 0, then u solves Problem (P). Proof: 4
Euler-Lagrange PDE δ G[η] = Ω Du Dη λη 1 + Du 2 Theorem: If u solves (P) λ, then div Du 1 + Du 2 = λ. 5
Constrained Plateau BVP: div ( ) Du 1+ Du 2 = λ on Ω u Ω = u 0 Young Operator: Du Mu = div 1 + Du 2 1 [ = (1 + u 2 (1 + Du 2 ) 3/2 y )u xx 2u x u y u xy + (1 + u 2 x )u yy] 6
Quasilinear Elliptic Operator(s) Linear Partial Differential Operator: Lu = a ij (x)d ij u + b j (x)d j u + c(x)u Quasilinear: Qu = A ij (x, y, Du)D ij u + C(x, u, Du) Examples: Laplace a ij = δ ij Young (A ij ) = 1 (1 + Du 2 ) 3/2 ( 1 + u 2 y u x u y u x u y 1 + u 2 x ) 7
Ellipticity λ 0 > 0 such that aij (x)ξ i ξ j λ 0 ξ 2 (This is called uniform ellipticity.) Young s Operator on R 2 : 1 [ (1 + u 2 (1 + Du 2 ) 3/2 y )ξ1 2 2u xu y ξ 1 ξ 2 + (1 + u 2 x)ξ2] 2 1 = (1 + Du 2 ) 3/2 [ ξ Theorem: M is uniformly elliptic if there is some M > 0 such that Du < M on Ω. (Bounded Gradient Condition) 8
Geometric Interpretations a. Solutions minimize area with respect to a volume constraint. They are pieces of soap bubbles or surfaces bounding liquids in equilibrium in zero gravity. b. Mean Curvature: The Young operator is also called the mean curvature operator 9
Surface Geometry Calculation v 0 T x0 R 2 w 0 T p S α(t) = (x 0 + v 0 t, u(x 0 + v 0 t)) w 0 = α (0) = (a, b, au x + bu y ) 10
Arclength s = t 0 α (τ) dτ dt ds = 1 α = 1 a 2 + b 2 + (au x + bu y ) 2 (re)parameterization by arclength: γ(s) = α(t(s)) γ = α α 11
Curvature Vector k = d 2 γ ds 2 = [ α α α α 3(α α ) ] 1 α Applicable for any curve in R 3 12
Normal Curvature For a curve on a surface, the normal curvature is k = k N = N α α 2 Musnier s Theorem: All curves with the same direction in T p S have the same normal curvature. 13
Calculation α = (a, b, au x (x 0 + v 0 t) + bu y (x 0 + v 0 t)) α = (0, 0, a 2 u xx + 2abu xy + b 2 u yy ) N = ( u x, u y, 1) 1 + Du 2 Normal Curvature: k = 1 a2 u xx + 2abu xy + b2 u yy 1 + Du 2 a 2 + b 2 + (au x + bu y ) 2 14
Next Task Find the normal curvature associated with a direction w orthogonal to w 0 in T p S Basis for T p S: {(1, 0, u x ), (0, 1, u y )} ONB: u 1 = (1, 0, u x )/ 1 + u 2 x u 2 = ( u xu y, 1 + u 2 x, u y ) (1 + u 2 x)(1 + u 2 x + u 2 y) 15
Components of w 0 = (a, b, au x + bu y ): w 0 u 1 = a + au2 x + bu x u y 1 + u 2 x w 0 u 2 = b 1 + u 2 x + u 2 y 1 + u 2 x
w = w 0 = (w 0 u 2 )u 1 + (w 0 u 1 )u 2 w 0 u 1 = a + au2 x + bu x u y, w 0 u 2 = b 1 + u 2 x + u 2 y 1 + u 2 x 1 + u 2 x u 1 = (1, 0, u x) 1 + u 2 x, u 2 = ( u xu y, 1 + u 2 x, u y ) (1 + u 2 x )(1 + u2 x + u2 y ) 16
ã = b 1 + u 2 x + u 2 y 1 + u 2 x 1 1 + u 2 x u xu y (a + au 2 x + bu x u y ) (1 + u 2 x) 1 + u 2 x + u 2 y = b + bu2 y + au x u y 1 + u 2 x + u 2 y b = a + au2 x + bu x u y 1 + u 2 x + u 2 y
Notation W = 1 + u 2 x + u2 y, R = bu x au y, p = u x, q = u y bw pr ã = W b = aw + qr W 17
Orthogonal Normal Curvature k = 1 ã2 u xx + 2ã bu xy + b2 u yy 1 + Du 2 ã 2 + b 2 + (ãu x + bu y ) 2 We know w = w 0 = w 0. So ã 2 + b 2 + (ãu x + bu y ) 2 = a 2 + b 2 + (au x + bu y ) 2 (Caution: This is tricky to check directly!) 18
Sum of Orthogonal Normal Curvatures: k + k = 1 (a2 + ã2 )u xx + 2(ab + ã b)u xy + (b2 + b2 )u yy 1 + Du 2 a 2 + b 2 + (au x + bu y ) 2 19
Check/Simplify: ã 2 = b2 W 2 2bpW R + p 2 R 2 W ã b = abw 2 + (bq ap)w R pqr 2 W b 2 = a2 W 2 2aqW R + q 2 R 2 W k + k = 1 (a2 + ã2 )u xx + 2(ab + ã b)u xy + (b2 + b2 )u yy 1 + Du 2 a 2 + b 2 + (au x + bu y ) 2 20
Check/Simplify: a 2 + ã 2 = 1 W (a2 W + b 2 W 2 2bpW R + p 2 R 2 ) = 1 W [a2 W + b 2 W 2 2pW (b 2 p abq) + p 2 (b 2 p 2 2abpq + a 2 q 2 )] = 1 W [a2 (W + p 2 q 2 ) + b 2 (W 2 2p 2 W + p 4 ) + 2abpq(1 + q 2 )] = 1 W [a2 (1 + p 2 )(1 + q 2 ) + b 2 (1 + q 2 ) 2 + 2abpq(1 + q 2 )] = 1 + q2 W [a2 (1 + p 2 ) + b 2 (1 + q 2 ) + 2abpq] = 1 + q2 W [a2 + b 2 + (ap + bq) 2 ] 21
Conclusion (Geometric Interpretation) k + k = 1 (1 + Du 2 ) 3/2 [ (1 + u 2 y )u xx + ] Check: ab+ã b = pq W [a2 +b 2 +(ap+bq) 2 ] Then: b 2 = 1 + p2 W [a2 +b 2 +(ap+bq) 2 ] k + k = Mu H = k + k 2 Mean Curvature 22
3. Maximum and Comparison Principles 0. Weak Maximum Principle Assumptions: u C 2 (Ω) C 0 ( Ω) Ω open, bounded Lu 0 with Lu = a ij D ij u + b j D j u (Note: c = 0) uniformly elliptic Conclusion: max Ω u = max Ω u 23
Proof in case of strict inequality: Assume Lu > 0 on Ω. Then u(x) < max Ω IDEA: At an interior max: u for all x Ω Du(x 0 ) = 0 D 2 u(x 0 ) 0 = aij (x 0 ) D ij u(x 0 ) > 0 24
Details: Second condition: D 2 u(x 0 ) 0 Multivariable Taylor s Formula: Lu(x 0 ) = trace[(aij)d 2 u] u(x) = u(x 0 ) + Du(x 0 )(x x 0 ) + 1 2 D2 u(x )(x x 0 ), x x 0 Take v = (x x 0 )/ x x 0 fixed. Then at max D 2 u(x ) x x 0 x x 0, x x 0 x x 0 = D 2 u(x )v, v 0 (Such a matrix is negative semi-definite.) = 2[u(x) u(x 0)] x x 0 2 25
Getting the Contradiction (more details) A = (a ij ) (positive definite) H = (b ij ) (negative semi definite) (both symmetric) Diagonalize H: UHU 1 = Λ = U orthogonal and λ 1,..., λ n 0 λ 1... λ n 26
Getting the contradiction (continued) trace(ah) = trace(ahu 1 U) = trace(uahu 1 ) = trace(uau 1 Λ) Check that M = UAU 1 has positive diagonal entries. trace(mλ) = m jj λ j 0. 27
1. E. Hopf Boundary Point Lemma Assumptions: u C 2 (Ω 0 ) C 0 ( Ω 0 ), Ω 0 = B r (ξ 0 ), Lu 0, L uniformly elliptic, c = 0 Know: max Ω 0 u = max Ω 0 u We can take x 0 B r (ξ 0 ) with u(x 0 ) = max u Conclusion: If u(x) < u(x 0 ) for x Ω 0, then lim h 0 u(x 0 ) u(x 0 h(x 0 ξ 0 )/r) h > 0 28
2. E. Hopf Strong Maximum Principle Assumptions: u C 2 (Ω) C 0 ( Ω), Ω open, bounded Lu 0, L uniformly elliptic, c = 0 Conclusion: Either u max u or then u(x) < max Ω u x Ω 29
3. Comparison Principle Assumptions: u, v C 2 (Ω) C 0 ( Ω), (alt. u, v C 2 ( Ω)) Ω open, bounded Mu Mv, Mu = A ij (Du)D ij u uniformly elliptic Conclusion: If u Ω v Ω, then u v or u(x) < v(x) x Ω 30
Proof: L(u v) = Mu Mv = A ij (Du)D ij u A ij (Dv)D ij v = A ij ((1 t)dv + tdu)[(1 t)d ij v + td ij u] 1 d = 1 0 = ( 1 + i,j,k dt t=0 { Aij ((1 t)dv + tdu)[(1 t)d ij v + td ij u] } dt 0 A ij((1 t)dv + tdu) dt ( 1 0 ) D ij (u v) A ij p k ((1 t)dv + tdu)[(1 t)d ij v + td ij u] dt = a ij (x)d ij (u v) + k b k (x)d k (u v) ) D k (u 31
where a ij = 1 0 A ij((1 t)dv + tdu) dt b k = i,j 1 0 A ij p k ((1 t)dv + tdu)[(1 t)d ij v + td ij u] dt We need a ij, b k continuous (and bounded) in Ω. (alt. u, v C 2 ( Ω))
Rest of the proof: w = u v Lw 0 w Ω 0 Strong Maximum Principle = w max w or w(x) < max w 0 for x Ω 32
4. Boundary Point Comparison Principle Assumptions: u, v C 2 ( Ω 0 ), Ω 0 = B r (ξ 0 ) Mu Mv, Mu = A ij (Du)D ij u uniformly elliptic Conclusion: If u Ω v Ω with equality at x 0 Ω 0, then u v or lim h 0 v(x 0 h(x 0 ξ 0 )/r) u(x 0 h(x 0 ξ 0 )/r) h > 0 33
Alexandrov s Theorem Assumptions: S is a parametric surface compact, embedded, no boundary CMC (constant mean curvature) Conclusion: S is a (round) sphere. 34
Reflection Lemma Given any direction v R 3 we can show that S has a plane Π v of reflective symmetry orthogonal to v. Moreover, Π v divides S into two graphs of locally bounded gradient over Π v. 35
Proof: S encloses a volume V R 3. S = V Move Π up and reflect. Initially (C): Ŝ is a graph of bounded gradient, and Ŝ V. t = sup{t : (C) holds} 36
Proof of Alexandrov s Theorem (continued) There are two ways condition (C) can cease to hold Interior Touching: Mu = Mv u v Comparison Principle = u v Conclusion: {p Ŝ S} is open (and closed) in Ŝ. 37
Proof of Alexandrov s Theorem (continued) Second way (C) can cease to hold Boundary Touching: Mu = Mv on a half neighborhood. u v Du(x 0 ) = Dv(x 0 ) Boundary Comparison Principle = u v 38
Comments on other details 39
Have a good break! See you in the Spring. 40