Higher Order Macro Coefficients in Periodic Homogenization
M.VanninathanTIFR - Centre for Applicable Mathematics, Bangalore
Collaborators: Conca, San Martin (Chile)Smaranda (Romania)
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Heat Conduction Problem
Conductivity matrix: a(y) = (akl(y)), y ∈ IRN
Symmetric matrixBoundedness |akl(y)| ≤ MPositive definiteness a(y) ≥ a0I ∀y with a0 > 0Periodicity: akl(y) is Y− periodicExample ( Two phase conductor with phases (α0, α1))
akl(y) = a(y)δkl , a(y) = α0χT c (y) + α1χT (y)
Volume Proportion is (1− γ, γ) where γ = |T ||Y |
χT constitutes microstructure
Hear Conduction Problem
Basic operator A ≡ −div(a(y)∇) ≡ − ∂∂yk
{akl(y) ∂
∂yl
}Aε ≡ − ∂
∂xk
{akl
(xε
) ∂
∂xl
}We consider the equation Aεuε = f in IRN
uε is temperature
f is Heat source
Classical Homogenization Problem
Restrictions on oscillations
Assume uεH1
⇀ u
Aεuε ≡ − ∂
∂xk
{akl
(xε
) ∂uε∂xl
}= f in IRN
Question: What is the limit of the constitutive relation?
akl
(xε
) ∂uε∂xl
L2⇀??
How to take the average 〈akl( xε )∂uε
∂xl〉?
What is the equation satisfied by u?
Homogenization Theorem
There are two steps:
(a) Existence of q:Fourier conduction law is preserved.
(b) Evaluation of q using a cell test problem.
(a) There exist homogenized coefficients/matrix q = (qkl)
akl(x
ε)∂uε
∂xl
L2⇀ qkl
∂u
∂xl∀k = 1 . . .N
Qu ≡ − ∂
∂xk
{qkl
∂u
∂xl
}= f in IRN
(b)
qkl =1
|Y |
∫Y
amn(y)∂
∂yn(χk + yk)
∂
∂ym(χl + yl)
Cell Test functions :
{Aχk = ∂akl
∂ylχk is Y − periodic
Note the dependence on the microstructure
Above result establishes the existence of mixtures of materials withconductivities α0, α1 . . . in volume proportions γ0, γ1 . . . by varyingmicrostructures in a periodic manner.
Conductivities of the mixture are given by the eigenvalues of thematrix qExample (N = 1, 1 d Homogenization)Mixture of 2 phases (α0, α1) in proportion (1− γ, γ)
Homogenized Coefficient: q =
(1− γα0
+γ
α1
)−1(Harmonic Mean)
No dependence on the microstructure (represented by χT )
Bloch Wave Approach
How to go beyond the usual homogenized coefficients?Homogeneous medium (IRN− translation invariance) :Plane waves e iη.y
Periodic medium ( ZN - translation invariance) : Bloch WavesNew condition is (η − Y ) periodicity with η ∈ Y ′ =]− 1
2 ,12 [N
Periodic cell Y =]0, 2π[N
Dual cell Y ′ =]− 1
2,
1
2[N
We consider the eigenvalue problem for (λ(η), ψ(y , η))
Aψ ≡ − ∂
∂yk
{akl(y)
∂ψ
∂yl
}= λ(η)ψ in IRN
ψ is (η − Y ) periodic
i.e,ψ(y + 2πp) = ψ(y)e2πip.η ∀p ∈ ZN
change of variables ψ(y) = φ(y)e iη.y
Parametrized eigenvalue problems
A(η)φ = λ(η)φ, φ is Y − periodic
Shifted operator A(η) ≡ −(
∂∂yk
+ iηk
){akl(y)
(∂∂yl
+ iηl
)}Eigenvalues: 0 ≤ λ1(η) ≤ λ2(η) . . .→∞Eigenvalues: φ1(y , η)φ2(y , η) . . . (0.n basis)
Connection with Homogenization
At η = 0, λ1(0) = 0 is a simple eigenvalue of the operatorA(0) = A with eigenvector = constant.Regular perturbation theory then implies that
η 7−→ (λ1(η), φ1(y , η))
is analytic when |η| is small
λ1(η) = λ1(0) + λ′1(0)η + 12!λ
21(0)η2 + 1
3!λ(3)1 (0)η3 + 1
4!λ(4)1 (0)η4 + . . .
φ1(η) = φ1(0) + ∂φ1
∂ηk(0)ηk + . . .
Results λ1(0) = 0,All odd order derivatives of λ1(η) at η = 0 vanish
12!λ
(2)1 (0) = q, homogenized matrix
∂φ1∂ηk
(y , 0) = iχk(y), cell test functions
Interpretation of Homogenization: Lowest energy level (m = 1)Largest scale (η ≈ 0)
There are two possibilities to go further.
(i) High energy homogenization ( at level m > 1)
(ii) Keep m = 1, decrease scale i.e., |η| is not very small)
In case (ii), we can see the importance of higher order macro
coefficients d = 14!λ
(4)1 (0) . . . ( apart from q = 1
2!λ(2)1 (0))
Various derivatives of φ1(y , η) at η = 0 yield new cell testfunctions.
Aim Study properties of d , compare with q. In particular, seekproperties independent of microstructures. They are useful inOptimal Design Problems
High Energy Homogenization (level m)
Structural Conditions
(a) minη∈Y ′
λm(η) is attained at η = 0
It is the unique minimizer
(b) λm(0) is a simple eigenvalue of A(0) = A
(c) λ(2)m (0) is a positive definite matrix.
The corresponding homogenized matrix is 12!λ
(2)m (0).
Acoustic Wave Propagation
v εtt + Aεv ε = 0
v ε(x , 0) = φ1
(xε, εξ)v0(x)
v εt (x , 0) = φ1
(xε, εξ)v1(x).
λε1(ξ) =1
2!λ(2)1 (0)ξ2 +
1
4!ε2λ
(4)1 (0)ξ4 + O(ε4|ξ|6)
If |ξ| = o(ε−12 ) then ε2|ξ|4 = o(1).
In this case, q is relevant.Homogenized Medium is good enough:
Aε ≈ −q d2
dx2
For shorter waves |ξ| = o(ε−23 ) then ε2|ξ|4 = o(ε−
23 ) and
ε4|ξ|6 = o(1)In this case, both (q, d) are relevant: Homogenized Medium alone
is not good enough. Aε ≈ −q d2
dx2+ ε2d d4
dx4
With the decrease of scales, short waves undergo dispersion.
Physical Space Representation
We already have (energy - momentum) space representation
q =1
2!λ(2)1 (0), d =
1
4!λ(4)1 (0)
Basic operator A ≡ − ∂∂yk
{akl(y) ∂
∂yl
}Bilinear Form a(φ, ψ) =
∫Y
akl(y) ∂φ∂yl∂ψ∂yk
Cell test functions (χ(1), χ(2)) Y− periodic
Aχ(1) = ηk∂akl∂yl
Note χ(1) = ηkχk
Aχ(2) = (akl − qkl)ηkηl − ηkCkχ(1),
Ckφ ≡ −akl ∂φ∂yl −∂∂yl
(aklφ)
Physical space Representation
qη.η = 1|Y |a(χ(1) + η.y , χ(1) + η.y)
dη4 = − 1|Y |a(χ(2) − 1
2(χ(1))2, χ(2) − 12(χ(1))2)
Consequences
q is positive qη.η ≥ c|η|2
d is negative dη4 ≤ 0
These hold universally, in particular independent of microstructureMicrostructure contributes to q through χ(1)
Microstructure contributes to d through (χ(1), χ(2))In d , we see interaction of microstructures, including a self-interaction represented by (χ(1))2.
Question We would like to exploit negativity of d .How negative d can be?What is the minimum value of d when we vary microstructure?What is the microstructure at which this min value is attained ?Is it classical ( i.e., given by characteristic function ) ?Is it generalized / relaxed?
Results in one dimension
Two phases with conductivities (α0, α1) in proportion (1− γ, γ)
1. Value of q = (1−γα0+ γ
α1)−1
q is insensitive to microstructure
2. The values of d over classical microstructures range over theentire interval
− 1
12|Y |2γ2(1− γ)2q3(
1
α1− 1
α0)2 ≤ d < 0
3. Max value for d is attained at the followinggeneralized/relaxed microstructure ( not a characteristicfunction)
θ∗max(y) ≡ γ (local density of α1 − phase)
4. Min value for d is attained at the following classicalmicrostructure
θ∗min(y) = χ[0,γ|Y |](y)
5. As we introduce more and more interfaces between phases,the value of d increases.At min value, we have just one interface.At max value , there is a continuum of interfaces.
Remark Taken together the values of (q, d) provide Harmonicmean and a correction to it at the next scale.What about Arithmetic Mean ?
Murat-Tartar Theorem (α0 > α1)
µ−(γ) ≤ µj ≤ µ+(γ),
N∑j=1
1
µj − α1≤ 1
µ−(γ)− α1+
N − 1
µ+(γ)− α1,
N∑j=1
1
α0 − µj≤ 1
α0 − µ−(γ)+
N − 1
α0 − µ+(γ).
Here, we have used the following notations for the Arithmetic andHarmonic Means:
µ+(γ) = (1− γ)α0 + γα1,
1
µ−(γ)=
1− γα0
+γ
α1.
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Case of Laminates
As microstructure varies among classical laminates, we have
1. qη.η = µ−(γ)η21 + µ+(γ)|η̃|2 ( fixed value)
2. dη4 fills up the following interval
− 1
12|Y |2γ2(1−γ)2µ−(γ)−1
{µ−(γ))2η21(
1
α1− 1
α0) + |η̃|2(α1 − α0)
}2
≤ dη4 < 0
New feature: With the decrease of scale, we are able to seeinteraction between longitudinal and transverse modes.
Legendre- Hadamard Condition
Even though dη4 ≤ 0, d does not satisfy L-H condition.There are vectors η̃ and η such that d(η ⊗ η̃) · (η ⊗ η̃) > 0. Moreprecisely, we obtain the spectral decomposition of the map
d : SymN → SymN , (Mmn) 7→ (dklmnMmn)
Case : N = 2. Above map has two negative eigenvalues λ1 and λ2
and a third eigenvalue λ3 which is positive.They are given by
λ1,2 = −A2 + B2
2±
√(A2 − B2
2
)2
+1
9A2B2.
λ3 =2
3AB.
A =1
m2(1/α)
(1
α1− 1
α0
)√m(1/α)(m(b2T )−m2(bT )) > 0,
B = (α0 − α1)√m(1/α)(m(b2T )−m2(bT )) > 0.
bT :]0, 2π[→ IR is defined by
bT (y1) =
y1∫0
(χT1(s)− γ)ds.
T1 ⊂ [0, 2π]
T = T1 × [0, 2π]N−1
γ =|T1|2π
=|T ||Y |
.
Blossoming Principle
For any multivariate polynomial pn : IRm → IR of total degree n,there exists a unique symmetric n-affine functionPn : IRm × . . . IRm → IR satisfying
Pn(x , . . . , x) = pn(x) for all x ∈ IRm.
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Homogenization of Hashin Structures
Hashin - Shtrikman method/Formula
qH − α1
qH + (N − 1)α1= (1− γ)
(α0 − α1)
α0 + (N − 1)α1
q = qH is the ( scalar) homogenized coefficient
Questions What is minimum value of d over all Hashin structures?What is the minimizer?
We applied the principle from one - dimension:
d minimizes interface area(By duality, we maximize volume)
Our Guess Hashin structure with minimum number of Hashininclusions: APOLLONIAN - HASHIN structureNumerics confirm that our guess is correct.
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