Sharp Korn inequalities in thin domains: The rst and a half Korn inequality

Transcription

1 Sarp Korn inequalities in tin domains: Te rst and a alf Korn inequality Davit Harutyunyan (University of Uta) joint wit Yury Grabovsky (Temple University) SIAM, Analysis of Partial Dierential Equations, Decamber 7-10, Scottsdale, Arizona

2 Korn's Inequalities Assume Ω R n is open, bounded, connected and Lipscitz and u H 1 (Ω, R n ), were u = (u 1, u 2,..., u n ) and u = ( u i ) n x. j i,j=1 Set e(u) = 1 2 ( u + ( u)t ), e ij (u) = u i + u j. x j x i Denote skew(r n ) = {L = Ax + b : A M n n, A T = A, b R n }. Assume V is a closed subspace of H 1 (Ω, R n ) suc tat V skew(r n ) = {0}.

3 Korn's Inequalities Korn's First and Second Inequalities 1. Tere exists a constant K 1 depending only on Ω suc tat ( K 1 (Ω) u 2 u 2 + e(u) ), 2 for any u H 1 (Ω, R n ) Ω Ω Ω 2. Tere exists a constant K 2 depending only on Ω and V suc tat K 2 (Ω, V ) u 2 e(u) 2, for any u V Ω Ω. 3. Tere exists a constant K > 0 depending only on Ω suc tat for any u H 1 (Ω, R n ), tere exists a skew-symmetric matrix A u suc tat K(Ω) u A u 2 e(u) 2 Ω Ω

4 Inequalities of Our Interest We are interested in sarp Korn inequalities. Question: How do K 1 (Ω) and K 2 (V, Ω) depend on Ω and V, wen Ω is tin? Ω is a tin domain (strips, rods, sells,...) wit tickness, ten K 1 (Ω) α and K 2 (V, Ω) β as 0. Find α and β. If for instance β is known and K 2 (V, Ω) C(V, Ω) β, wen is suciently small, ten wat is C(V, Ω)? Goal: Find te optimal constants in Korn's inequalities.

5 Examples Example 1 (Zero boundary conditions). If V = {u H 1 (Ω, R n ) : u(x) = 0 on Ω}, ten K 2 (V, Ω) = 1 2. Example 2 (Tin rectangle). If Ω = [0, ] [0, l], V = {u H 1 (Ω, R 2 ) : u(x, 0) = u(x, l) = 0}, ten K 2 (V, Ω) C 2.

6 Motivation Wy optimal constants? Te problem we were interested in: Buckling of cylindrical sells under axial compression, (2011). Critical buckling load, deformation modes? Koiter's formula (1945). λ() = C, were is te tickness of te sell, and C depends on te material. Te buckling modes are given by "Koiter's circle". It was known, tat te buckling load is igly sensitive to imperfections (sape, load). We aim to derive Koiter's formula and understand te sensitivity to imperfections applying te teory of buckling of slender structures, Grabovsky, Truskinovsky (2007).

7 Motivation If C = {(r, θ, z) : r [R, R + ], θ [0, 2π], z [0, L]}, and u = u r ē r + u θ ē θ + u z ē z, in cylindrical coordinates, te we impose te B.C.: Fixed bottom boundary conditions: u r (r, θ, 0) = u θ (r, θ, 0) = u z (r, θ, 0) = u r (r, θ, L) = u θ (r, θ, L) = 0, V 1, Breating cylinder u θ (r, θ, 0) = u z (r, θ, 0) = u θ (r, θ, L) = 0, u z (r, θ, L) = c, V 2.

8 Motivation, Problem Te teory of Grabovsky and Truskinovsky implies λ() ck(c ), were K(C ) is te optimal Korn's constant in te second Korn inequality for V 1 or V 2. Weter one as K(V i, C )? Answer: NO! K(V i, C ). Teorem (Grabovsky, H., 2012) If K(C ) lim 0 λ cl () = 0, ten te constitutively linearized quotient captures bot, te critical load and te buckling modes.

9 Korn's inequalities for perfect cylindrical sells Teorem (Grabovsky, H., 2012) Tere exist absolute constants C i > 0, i = 1, 2 suc tat for any u V i, tere olds u 2 C i C e(u) 2. C Tese estimates are sarp, in te sense tat te power of is optimal. If u = u r ē r + u θ ē θ + u z ē z, ten u = u r,r u θ,r u z,r u r,θ u θ r u θ,θ +u r u r z,θ r u r,z u θ,z u z,z.

10 Korn's inequalities for perfect cylindrical sells Ansatz. We assume R = 1, ten ( φ r (r, θ, z) = W,ηη φ θ (r, θ, z) = r 4 ( W,η θ 4, z θ 4, z ( φ z(r, θ, z) = (r 1)W,ηηz ) ) + r 1 θ 4, z ( ) θ 4 W,ηηη 4, z, ) ( θ W,z 4, z were te function W (η, z) is a smoot compactly supported function on ( 1, 1) (0, L), wile te function φ (θ, z) is extended as a 2π-periodic function in θ R. ),

11 Remarks on te Korn inequality, strategy If u = u r ē r + u θ ē θ + u z ē z, ten u = u r,r u θ,r u z,r u r,θ u θ r u θ,θ +u r u r z,θ r u r,z u θ,z u z,z. Prove te inequality block by block, wic means xing r, θ and z and proving 2D inequalities. For r, θ, z = const, we ave te blocks u θ,θ+u r r u θ,z, respectively. u z,θ r u z,z u r,r u r,z, u z,r u z,z u r,θ u θ r u θ,θ +u r r u r,r u θ,r,

12 Available Tools We needed Korn's inequalities wit constants decaying like or slower! For instance te cross section θ = const gives a Korn's second inequality on a tin rectangle: u r,r u r,z. u z,r u z,z Wat is available? θ = const gives a tin rectangle, Korn's second inequality on rectangles, K 2 Ryzak 2001?: not applicable. z = const gives a tin annulus, again optimal constant scales like 2, Dafermos 1968, for normalization conditions: not applicable. Uniform Korn-Poincaré inequality in tin domains, Lewicka, Müller 2011, tangential boundary conditions: not applicable. New inequalities are needed.

13 A Korn type inequality Standard approac: It is sucient to prove a second Korn inequality subject to Diriclet type boundary conditions for armonic displacements. Teorem (Grabovsky, H., 2012) Suppose w(x, y) is armonic in [0, ] [0, L], and satises w(x, 0) = w(x, L). Ten Te equality is attained at w y w w x + w x 2. w(x) = cos ( ( π x )) ( πy ) sin, L 2 L

14 Te rst and a alf Korn inequality for rectangles Teorem (Grabobsky, H., 2012) Suppose tat te vector eld U = (u, v) H 1 (Ω, R 2 ), were Ω = [0, ] [0, L], satises u(x, 0) = u(x, L). Ten for any (0, 1) and any L > 0 tere olds: ( ) u e(u) U e(u) 2. Tere are no boundary conditions imposed on v(x, y). Tis implies bot te rst (via Scwartz inequality) and te second (via Friedrics inequality) Korn inequalities. Te scaling of te constant is as needed.

15 Te rst and a alf Korn inequality for cylindrical sells Teorem (Grabobsky, H., 2012) Suppose U V 1 or U V 2. Ten tere exists a universal constant C > 0 suc, tat for any (0, 1) and any L > 0 tere olds: ( ) U 2 ur e(u) C + e(u) 2. Tis implies te second Korn inequality, but wit 2. Combine wit u r 3 C U 2 e(u).

16 Extensions An extension to R n for tin domains wit nonconstant tickness. Assume te operator n L(u) = i,j=1 a ij 2 u x i x j wit constant coecients satises n a ij x i x j λ x 2 for all x R n, (1) i,j=1 were λ > 0, and, n a ij Λ for all 1 j n. (2) i=1

17 Extensions For x = (x 1, x 2,..., x n ), let x = (x 2,..., x n ). Teorem (H., 2014) Let ω R n 1 be a bounded and simply-connected Lipscitz domain, let x 1 = ϕ(x ): ω R be a positive Lipscitz function wit H = sup x ω ϕ(x ) and = inf x ω ϕ(x ) > 0. Denote Ω = {x R n : x ω, 0 < x 1 < ϕ(x )} and assume tat te operator L(u) = n i,j=1 a 2 u ij x i x wit constant coecients satises conditions (1) j and (2). Ten tere exists a constant C depending on n, Λ, λ, L = Lip(ϕ) and te ratio m = H/ suc tat any u C 3 ( Ω) solution of L(u) = 0 satisfying te boundary conditions u(x) = 0 on te portion Γ = {x Ω : x ω} of te boundary of Ω fullls te inequality ( ) u u 2 ux1 C + u x1 2. C = C(n, λ, Λ, L, m).

18 Extensions Teorem (H., 2014) Let l > 0, let ϕ 1 C 1 [0, l] and let ϕ 2 and ϕ 1 be Lipscitz functions dened on [0, l]. Assume furtermore tat 0 < = min y [0,l] (ϕ 2 (y) ϕ 1 (y)) and H = min y [0,l] (ϕ 2 (y) ϕ 1 (y)). Denote Ω = {(x, y) R 2 : y (0, l), ϕ 1 (y) < x < ϕ 2 (y)}. Ten tere exists a constant C depending on m = H/, ρ 1 = ϕ 1 L (Ω), ρ 2 = ϕ 2 L (Ω) and ρ 1 = ϕ 1 L (Ω) suc tat if te rst component of te displacement U = (u, v) W 1,2 (Ω) satises te boundary conditions u(x) = 0 on te boundary portion Γ = {(x, y) Ω : y = 0 or y = l} in te trace sense, ten te strong second Korn inequality olds: ( ) u e(u) U 2 + e(u) 2. Te estimate is sarp. C = C(m, ρ 1, ρ 2, ρ 1).

19 Extensions Teorem (H., 2014) Let L > 0, ϕ 1, ϕ 2, Ω,, H, m, ρ 1, ρ 2 and ρ 1 be as in te previous teorem. Ten tere exists a constant C depending on m, ρ 1, ρ 2 and ρ 1 suc tat if te rst component u of te displacement U = (u, v) W 1,2 (Ω) is L-periodic, ten te second Korn inequality olds: ( ) u e(u) U 2 + e(u) 2. C = C(m, ρ 1, ρ 2, ρ 1). L periodicity is te periodicity of bot te function and te gradient.

20 Recent progress Consider a sell in te (r, θ, z) variables (θ and z are te principal directions): [ C = 2, ] [0, s] [0, L], 2 wit κ z = 0. (tis yields a zero Gaussian curvature). If U = (u r, u θ, u z ), ten 1 1 u r,r A θ u r,θ κ θ u θ A z u r,z 1 U = u θ,r A θ u θ,θ + A θ,z 1 A θ A z u z + κ θ u r A z u θ,z. 1 u z,z A θ u z,θ A θ,z 1 A θ A z u θ A z u z,z Assume K = sup κ θ <, K 1 = sup κ θ,θ <, 0 < a θ A θ b θ, 0 < a z A z b z, A z, A θ A. were a θ, a z, b θ, b z, A are constants. Tis includes cut cones and straigt cylinders wit arbitrary cross sections.

21 Recent progress Te spaces V 1 and V 2 are te same as before. C will be a constant depending only on te constants K, k, K 1, a z, a θ, b z, b θ and A. Teorem (Grabovsky, H., 2015) For any (0, 1) and any U V i, tere olds: ( ) U 2 ur e(u) C + e(u) 2 + u r 2.

22 Recent progress Teorem (Grabovsky, H., 2015) If κ θ > 0, ten for any (0, 1) and any U V i, tere olds: U 2 If κ θ = 0 in a box [θ 1, θ 2 ] [z 1, z 2 ], ten C e(u) 2. U 2 C 2 e(u) 2.

23 Work in progress Assume Ω is a sell of revolution given by C = [ r(z) 2, r(z) + ] [0, 2π] [0, L]. 2 Ten (conjecture) If κ z < 0, ten for any (0, 1) and any U V i, tere olds: If κ z > 0, ten U 2 C 4/3 e(u) 2. U 2 C e(u) 2. Teorem (Grabovsky, H., 2015) For any (0, 1) and any U V i, tere olds: ( ) U e(u) U 2 C + e(u) 2 + U 2.