Pointwise bounds on the gradients in composite materials
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1 Pointwise bounds on the gradients in composite materials E. Bonnetier, Université de Grenoble-Alpes Outline: 1. Introduction/motivation 2. Pointwise bounds on the gradients in composite media with close-to-touching inhomogeneities : PDE approach for perfectly conducting inclusions 3. Layer potentials and the Neumann-Poincaré operator 4. Application to cloaking by anomalous localized resonance
2 1. Introduction : Why are we interested in large gradient in inhomogeneous media? 1.1. Preventing damage in composite materials Many composite materials are made of inclusions or fibers embedded in a (soft) matrix phase. The inclusions (usually expensive) are used as reinforcements or conducting materials, while the matrix is usually light and cheap Airbus A 350 XWB : 53 % of composites utilised in the fuselage and wing to reduces the need for fatigue-related inspections required on more traditional aluminium jetliners The composites and titanium reduce the aircrafts overall fatigue and corrosion maintenance tasks by 60 per cent
3 - The response of a composite material under some external sollicitation is usually described by an elliptic (or parabolic) PDE - In continuum mechanics, stress are related to the gradients of the solution to the PDE (the displacement field) - Most mechanical models for crack nucleation and propagation involve the local values of stress (and usually in a pointwise fashion) - Questions [Babuška]: Does the presence of narrow regions in between inclusions favor stress concentration that could lead to fracture? Are the gradients uniformly bounded as the inclusions get closer? How do the bounds depend on the material coefficient contrast?
4 1.2. Resonance phenomena in electro-magnetics and optics Plasmon resonances in nanoparticles [Gang Bi et al, Optics Comm., 285 (2012) 2472] Gold particles of diameter << than the wavelength of the exciting radiation Resonances (powerful, localized sources of light) may exist for only particular frequencies
5 Resonances = values of ω for which there exists non-trivial solutions to j E + ω 2 ε(ω, x)µ 0 E = 0 in (R 2 \ D) D [ nue] = [ε(ω, x)e] = 0 on D The dielectric permittivity has the form 8 >< ε 0! ε(ω, x) = ωp 2 >: ε 0 ε ω 2 + iγω in R 2 \ D in D Asymptotics as the size of the particles 0 : the electric potential satisfies at first order and integral equation of the form (λ(ω)i K D )ϕ = 0 where K D is the Neumann-Poincaré operator
6 Polaritons : localized surface or cavity modes - Ebbesen et al (Nature, 1998) have shown that, at certain frequencies, the rugosity of metallic surfaces can create fields with intensities that can be very large locally (up to a factor ) - The localization of the fields is due to resonance phenomena in the sub-wavelength cavities formed by the non-smooth surface, or to surface waves Surface Enhanced Raman Scaterring - Because it is sensitive to lengthscales smaller than the wavelength, this phenomenon has many potential applications: imaging of single molecules, optical filters,... - However, control of SERS requires great care in the design of the gratings (instability of the hot spots)
7 Experimental evidence A. Barbara, P. Quémerais, E. Bustarret, T. Lopez-Rios, PRB 66, (R) (2002) medskip G.A. Kriegsmann, SIAP, 65, (2001), E.B.-F. Triki
8 . The geometry of a model problem e w h c The metallic surface is assumed to be perfectly conducting. The width of the cavity is small compared to the wavelength. 3D configuration with invariance along x 3 direction, k = ω εµ Ω e = R 2,+, Ω c = ( w/2, w/2) ( h, 0) We are interested in the Green function G w defined in Ω := Ω e Ω i Γ w, solution to 8 ( + k 2 )G w(x, Y ) = δ Y (X) X Ω >< G w(x, Y ) = 0 X Ω n X >: radiation condition Resonances are (complex) frequencies for which the above PDE has no solution When w 0, the resonances come close to the real line, and the PDE has solutions with large gradients in in the vicinity of the cavity w
9 Representation of G w G e = Green function in the upper half-plane (Ỹ = (y 1, y 2 )) 8 G e(x, Y ) = i H (1) 0 (k X Y ) + H(1) 0 4 (k X Ỹ ), >< G e(x, Y ) = 0 X R 2+ n X >: X 1/2 ( r ik)g e(x, Y ) 0 X G c = Green function in the cavity 8 4 X X G c(x, Y ) = hw m=0 n=0 >< >: n X G c(x, Y ) = 0 X Ω i cos( mπ w (x 1 + w/2)) cos( mπ w (y 1 + w/2)) k 2 ( mπ w )2 ( nπ h )2 cos( nπ h (x 2 + w)) cos( nπ h (y 2 + w))
10 Representation for G w: for Z R 2+ 8 Z >< G w(y, Z) = G e(y, Z) + Z Γ w >: G w(y, Z) = Γ w n e G w(x 1, 0, Z)G e(x 1, 0, Y )dx 1, Y R 2+ n i G w(x 1, 0, Z)G c(x 1, 0, Y )dx 1, Y Ω i We obtain an integral equation for φ(y) = x 2 G w(wx, 0, wy, 0) S w(k)(y)φ(x) = 1 Ge(wy, 0, Z) w Z 1/2 with S w(k)φ(y) = φ(x)(g e + G c)(wx, 0, wy, 0)dx 1/2 Asymptotic study of the invertibility of w S w(k) [B-Triki, Babadjian-B-Triki] S w(k) = θ w(k) < 1,. > +S 1 + ws 2 + w 2 (ln(w)s 3 (k) + S 4,w (k))
11 1.3. Cloaking, meta-materials : anomalous localized resonance, superlensing Meta-materials are materials with negative diffractive indices. Their embedding in dielectrics or materials with positive indices, produces interesting phenomena (superlensing, cloaking) Model problem : Z Let f L 2 (R d ) with compact support, f = 0 R d j div(εs(x) vs) = f in R 2, v s(x) 0 as x ε s(x) = j 1 + is in the annulus A = {ri < r < r e} s > 0 small 1 elsewhere
12 Questions : - Identify the sources for which the energy Z E s = s v s 2 as s 0 A (this quantity is proportionnal to the electromagnetic power dissipated into heat) - When E s blows up, on what conditions do the associated fields v s vanish outside a certain region? (so that the source is cloaked)
13 [Costabel-Stephan 85, Bonnet-Chesnel-Ciarlet 12, Milton-Nicorovici 06, Ammari-Ciraolo-Kang-Lee-Milton 13, Kohn-Lu-Schweizer-Weinstein 14, H.M. Nguyen 14] q If the support of f is contained in a ball of critical radius r = re 3r 1 i, the field outside that region tends to 0 as s 0 v s is however very large in the annulus
14 Represent v s as Z v s = S i ϕ i + S eϕ e + G(x y)f(y)dy R 2 The layer potentials solve 0 is (λi K ) = B 2(2 is) I + K i νi S is νe S i 2(2 is) I + K e 1 C A ϕi ϕ e «= νi F νe F «However, the eigenvalues of K accumulate to 0 and λ (λi K ) 1 has an essential singularity there The cloaking is called anomalous because the resonance in this case is not due to the excitation by the source of a finite dimensional eigenmode of a linear operator
15 1.4. Make your own experiment Courtesy of Babak Sanii and Alice Muller-Egan Presented at the LBNL Nanofest, 2010
16 2. Pointwise bounds on the gradients in composite media with close-to-touching inhomogeneities
17 2.1. Corner singularities Ω polygonal domain in R 2, f L 2 (Ω) Seek a function u H 1 (Ω) solution to j u = f in Ω u = 0 on Ω Thm : about the structure of u u(x) = u reg(x) + X j c j S j (x) with u reg H 2 (Ω), S j singular functions associated to the corners of Ω Moreover, one has an estimate u reg H 2 (Ω) + X j c j f L 2 (Ω) [Kondratiev, Grisvard, Maz ya-plamenevski, Costabel-Dauge-Nicaise,... ]
18 Case of a sector (with Dirichlet BC s) Ω = {(r cos(θ), r sin(θ)), 0 < r < R, 0 < θ < ω} Separation of variables : seek u(x, y) = v(r, θ) = ψ(r)ϕ(θ) u = ` 2 rr + 1/r r + 1/r 2 θθ v = (ψ (r) + 1/rψ (r))ϕ(θ) + ψ(r)ϕ (θ) = 0 so that for some constant λ It follows that ψ (r) + r 1 ψ (r) r 2 ψ(r) ϕ(θ) = α cos(λθ) + β sin(λθ) = ϕ (θ) ϕ(θ) = λ 2 ψ(r) = α r λ + β r λ
19 The condition that u(x, y) = v(r, θ) lies in H 1 (Ω) implies that Z R Z ω v(r, θ) 2 rdrdθ 0 0 and Z R Z ω [ rv(r, θ)] 2 + ˆr 1 θ v(r, θ) 2 rdrdθ 0 0 are bounded, which eliminates the modes r λ with negative exponents It follows that u(r, θ) = r λ (a cos(λθ) + b sin(λθ) Applying the boundary conditions on the edges θ = 0 and θ = ω gives a = 0 and one obtains the dispersion relation so that λ = λ(ω) = kπ, k = 1, 2,... ω sin(λω) = 0 Singular modes only when Ω is non-convex (ω > π), for which 0 < λ < 1 In this case, u blows up like r λ 1
20 Case of a transmission problem : Assume Ω is a 2D disk, filled with 2 phases with conductivities a 1, a 2 Seek a solution to 8>< div(a(x) u(x)) = 0 in Ω u(x) = g(x) on Ω >: a(x) = a 1 0 < θ < ω a(x) = a 2 ω < θ < 2π Look for u in the form 8 < u 1 (x) = r λ (α 1 cos(λθ) + β 1 sin(λθ)) 0 < θ < ω u(x) = : u 2 (x) = r λ (α 2 cos(λθ) + β 2 sin(λθ)) ω < θ < 2π
21 The transmission conditions write u 1 (r, 0 + ) = u 2 (r, 2π ) u 1 (r, ω ) = u 2 (r, ω + ) a 1 nu 1 (r, 0 + ) = a 2 nu 2 (r, 2π ) a 1 nu 1 (r, 0 + ) = a 1 nu 2 (r, 2π ) from which one deduces cos(2λπ) sin(2λπ) 6 cos(λω) sin(λω) cos(λω) sin(λω) 4 0 a 1 a 2 sin(2λπ) a 2 cos(2λπ) a 1 sin(λω) a 1 cos(λω) a 2 sin(λω) a 2 cos(λω) B α 1 β 1 α 2 β 2 1 C A = 0
22 Nontrivial solutions exist provided the following dispersion relation is satisfied (k 2 + 1) sin(ωλ) sin((2π ω)λ) + 2k (1 cos((2π ω)λ) cos(ωλ)) = 0 where k = a 1 /a 2 is the coefficient contrast One can check that the smallest positive root of the dispersion relation satisfies λ = λ(k, ω) 1/2
23 Transmission problem : two touching corners Again 2 phases with conductivities a 1, a 2 Look for u in the form u(x) = r λ (α j cos(λθ) + β j sin(λθ)) in the j-th sector The smallest positive root of the corresponding dispersion relation satisfies λ(k, π/2) 0 as k 0 or k Things get worse when inhomogeneities interact!
24 2.2. Regularity in piecewise Hölder media 2 disks in 2D, for a conduction problem [B.-Vogelius, 2001] j div(a(x) u(x)) = 0 in Ω u(x) = ϕ on Ω j a(x) = k 1 in the disks a(x) = 1 outside When 0 < k <, then u W 1, (Ω) C for some C independent of δ
25 Piecewise Hölder conductivities, scalar equations and strongly elliptic systems [Li-Vogelius 01, Li-Nirenberg 03, Nakamura 10, Ammari-B-Triki-Vogelius 15] Ω = D 0 ` N i=1 D i, each Di has C 1,α boundary 0 < λ γ Λ and γ Di has regularity C 0,µ NX u C 1,α C ( boundary data ) (D i Ω ε) i=0 where α α = inf(µ, ) and C is independent of δ 2(1 + α)
26 Perfectly conducting case : k = + (also k = 0 in 2D) gradients may blow up as δ 0 [Yun 07, Bao-Li-Yin 09] u L (Ω\D 1 D 2 ) C/ δ ϕ C 2 ( Ω) n = 2 u L (Ω\D 1 D 2 ) C δ ln(δ) ϕ C 2 ( Ω) n = 3 u L (Ω\D 1 D 2 ) C/δ ϕ C 2 ( Ω) n 4 (see also [Ammari-Dassios-Kang-Lim 07, Lim-Yun 09, Yun 09])
27 Upper and lower bounds for special symmetries in 2D [Ammari-Kang et al. 05, 07, 10] 2 disks at a distance δ, of same radius r and conductivity k, meeting at X = 0 tangentially to the direction T Assume u(x) = H(x) + O(x 1 ) as x Then, u(x) = u r(x) + u s(x) with u r C, independent of δ and k if k < 1 if k > 1 8 >< >: 8 >< >: u s C 1 H(0) T 2k + p δ/r u s(0) C 2 H(0) T 2k + p δ/r u s C 1 H(0) N 2k 1 + p δ/r u s(0) C 2 H(0) N 2k 1 + p δ/r
28 Asymptotic expansion of u as δ 0 when k = [Kang-Lim-Yun 2013] u(x) = 4πr 1r 2 r 1 + r 2 (n H)(0)w 0 (x) + b(x) for 2 circular inclusions touching at 0 when δ = 0 where b is uniformly bounded wrt δ, and where w 0 solves 8 >< >: w 0 = 0 in R 2 \ D 1 D 2 w Z 0 Bj = constant j j = 1, 2 νw 0 dσ B j = ( 1) j j = 1, 2 w 0 = O( x 1 ) as x
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