TOPICS IN HOMOGENIZATION OF DIFFERENTIAL EQUATIONS 7

2. Two-scale asymptotic expansions method

We will illustrate this method by applying it to an elliptic boundary value prob-lem for the unknown u

Á : � æ R given by

(1)

Y]

[≠ div

1A

1x

Á

2Òu

Á

2= f in �,

u

Á = 0 on ˆ�.

The above model can be an e�cient model for various phenomena in continuummechanics. In the below table, we list some of the scientific domains in which theabove elliptic problem can be used and the meaning of the coe�cient A(x) and theunknown u

Á(x) in those respective fields.

Domain Unknown u

Á(x) Material coe�cient A(x)Electrostatics Electric potential Dielectric coe�cientMagnetostatics Magnetic potential Magnetic permeabilitySteady heat transfer Temperature Thermal conductivity

As an helpful example, let us consider a one-dimensional problem which I bor-rowed from a lecture by Daniel Peterseim2 (Bonn).

Example 2.1. For the unknown u

Á(x), consider the boundary value problem

(2)

Y_]

_[

≠ ddx

3a

1x

Á

2 duÁ

dx

4= 1 in (0, 1),

u

Á(0) = u

Á(1) = 0,

with the conductivity coe�cient given as

a

1x

Á

2= 1

2 + cos! 2fix

Á

"

which is plotted below

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 3. Plot of!2 + cos

! 2fix

Á

""≠1 with Á = 10≠1 (left), 10≠2 (right).

2He gave this example during a scientific workshop at the Hausdor� Research Institute for Math-ematics, Bonn.

8 HARSHA HUTRIDURGA

Observe that the coe�cient a

!x

Á

"given above has more and more oscillations in

the Á π 1 regime. It is indeed of interest to learn the e�ect of these high frequencyoscillations on the solution u

Á(x). A lucky accident in this example is that we canexplicitly integrate the di�erential equation (2) yielding

u

Á(x) = x ≠ x

2 + Á

31

4fi

sin3

2fix

Á

4≠ x

2fi

sin3

2fix

Á

4≠ Á

4fi

2 cos3

2fix

Á

4+ Á

4fi

2

4

which is plotted below

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.05

0.1

0.15

0.2

0.25

0.3

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

Figure 4. Plot of u

Á(x) on [0, 1] with Á = 10≠1 (left), 10≠2 (right).

Remark that the solution u

Á(x) has a parabolic profile x≠x

2 with small amplitudeoscillations of O(Á) along the profile. Indeed we have the uniform convergence

u

Á(x) æ x ≠ x

2 as Á æ 0.

So, the above parabolic profile is the homogenized approximation of the solutionprofile u

Á(x).

In the above example, instead of characterizing the homogenized operator, wehave explicitly computed the homogenized profile. This is possible, thanks to theexplicit integration of the di�erential equation (2). There is no such luck in obtainingexplicit analytical expressions for solutions to di�erential equations in general, evenin one dimension.

Let us briefly describe the setting of periodic homogenization. Denote the unitcube Y := (0, 1)d µ Rd. Consider a Y -periodic function A : � æ Rd◊d, i.e.,

A(x + kei

) = A(x) for each x œ Rd

, ’k œ Z ’i œ {1, . . . , d}where {e

i

} denotes the canonical basis in Rd. Taking Á > 0 as a parameter, considerthe dilated sequence defined as

A

Á(x) := A

1x

Á

2.

Remark that the dilated function A

Á(x) is ÁY -periodic because

A

Á(x + kÁei

) = A

3x + kÁe

i

Á

4= A

1x

Á

+ kei

2= A

1x

Á

2= A

Á(x)

for all k œ Z.In all the theory that is being presented in this course, a fundamental assumptionis that of separated scales. To be precise, we consider x to be a slow variable

TOPICS IN HOMOGENIZATION OF DIFFERENTIAL EQUATIONS 9

and introduce a new auxiliary variable y := x

Á

which we call a fast variable. Theassumption of separated scales essentially implies that the variables x and y are tobe treated as independent variables.With the above understanding of the slow and fast variables, in the periodic setting,we let the variable y œ Y , the unit cube in Rd introduced earlier. Then, the periodicconductivity coe�cient A can be treated as a function of the fast spatial variabley. We further assume that the conductivity coe�cient A(y) is uniformly boundedand uniformly positive definite, i.e., there exist constants –, — > 0 (uniform in they variable) such that

– |›|2 Æ A(y)› · › Æ — |›|2 ’y œ Y and ’› œ Rd \ {0}.(3)Employing the Lax-Milgram theorem, we can prove that the boundary value prob-lem (1) is well-posed.

Theorem 2.2. Suppose the source term f œ L2(�). Suppose that the periodicconductivity matrix A(y) is positive definite and that it is uniformly bounded, i.e., itsatisfies (3). Then for each fixed Á > 0, there exists a unique solution u

Á œ H10(�) to

the elliptic boundary value problem (1). Furthermore, we have the uniform estimateÎu

ÁÎH10(�) Æ C ÎfÎL2(�)(4)

with constant C > 0 being independent of the parameter Á.

2.1. A postulate and matched asymptotic. The model conductivity problem(1) has two spatial scales and it has a small parameter Á > 0. Hence the essentialidea of the method of asymptotic expansions is to consider a power series expansionfor the unknown u

Á in the parameter Á and with coe�cients that depend on both thespatial scales, i.e., on both the slow and fast variables. More precisely, we postulatethat the solution to (1) can be written as

u

Á(x) =ÿ

iØ0Á

i

u

i

1x,

x

Á

2(5)

under the assumption that the coe�cient functions u

i

(x, y) are Y -periodic in they variable. The coe�cient functions u

i

!x,

x

Á

"are referred to as locally periodic

functions. At least formally, for Á π 1, we have from (5) the approximation

u

Á(x) ¥ u0(x, y)---y= x

Á

Then, in the Á π 1 regime, one can argue that the y dependance of u0 gets “aver-aged” out over the periodic cell Y . One could then find the homogenized approxi-mation for u

Á asuhom(x) :=

⁄

Y

u0(x, y) dy.

The above heuristics essentially motivates the postulated asymptotic expansion (5)for the solution u

Á(x). These hand-wavy heuristics are not at all rigorous. Moreanalysis oriented approach to derive homogenized limits will be the objective ofthe chapters to come. As far as this chapter is concerned, we admit the two-scaleasymptotic expansion for the solution and carry on with the computations. Observe

10 HARSHA HUTRIDURGA

that the explicit solution obtained in Example 2.1 has the structure postulated abovein (5). That is a mere coincidence. The principle behind the asymptotic expansionsmethod is to plug the expansion (5) in the model problem (1) and match thecoe�cients of various powers of Á. Before we proceed, we need to keep the followingin mind while applying the asymptotic expansion method.

• For any locally periodic function Â

!x,

x

Á

", its total derivative becomes

ÒËÂ

1x,

x

Á

2È= Ò

x

Â(x, y)---y= x

Á

+ 1Á

Òy

Â(x, y)---y= x

Á

Note that the above expression is a consequence of the chain rule for di�er-entiation.

• Suppose �(x, y) is a function of both the slow and fast variables. Then thefollowing are equivalent

(i) �1

x,

x

Á

2= 0 ’x œ �, ’Á > 0,

(ii) �(x, y) = 0 ’x œ �, ’y œ Y.

Let us get on with the task of plugging the asymptotic expansion (5) in the modelproblem (1) yielding

(6)

≠div1

A

1x

Á

2Òu

Á

2= ≠ 1

Á

2 divy

1A (y) Ò

y

u0(x, y)2---

y= x

Á

≠ 1Á

divy

1A (y)

!Òx

u0(x, y) + Òy

u1(x, y)"2---

y= x

Á

≠ 1Á

divx

1A (y) Ò

y

u0(x, y)2---

y= x

Á

≠ divy

1A (y)

!Òx

u1(x, y) + Òy

u2(x, y)"2---

y= x

Á

≠ divx

1A (y)

!Òx

u0(x, y) + Òy

u1(x, y)"2---

y= x

Á

≠ Á divy

1A (y)

!Òx

u2(x, y) + Òy

u3(x, y)"2---

y= x

Á

≠ Á divx

1A (y)

!Òx

u1(x, y) + Òy

u2(x, y)"2---

y= x

Á

≠ · · ·where we will not explicitly write down the O(Á2) terms and further on (but thereaders can naturally extrapolate). In the computations to follow, only the O(1)terms and the terms of higher order play significant role. In the boundary valueproblem (1), f(x) is a given source term of O(1). Hence equating the coe�cients ofvarious powers of Á, we obtain a cascade of equations. As it will become apparentin the computations to follow, the orders in Á that are essential for arriving at the

TOPICS IN HOMOGENIZATION OF DIFFERENTIAL EQUATIONS 11

homogenized equation are that of O(Á≠2), O(Á≠1) and O(1).At O(Á≠2), we obtain

(7)

Y]

[div

y

1A (y) Ò

y

u0(x, y)2

= 0 for y œ Y,

y ‘æ u0(x, y) is Y -periodic.

Note that the periodic boundary conditions are because of our periodicity assump-tion on the coe�cient functions in the postulated asymptotic expansion. The vari-able x is treated as a parameter in (7) which is treated as a periodic boundary valueproblem in the y variable.Before proceeding further, let us remark that u0 is a function independent of the y

variable. To see that, let us multiply the equation (7) by u0 and integrate over Y

yielding⁄

Y

A(y)Òy

u0(x, y) · Òy

u0(x, y) dy ≠⁄

ˆY

A(y)Òy

u0(x, y) · n(y) d‡(y) = 0.

By assumption A(·) and u0(x, ·) are periodic. As the normal n(y) takes oppositesigns on the opposite ends of the boundary ˆY , we have

⁄

ˆY

A(y)Òy

u0(x, y) · n(y) d‡(y) = 0.

Furthermore, the coercivity assumption on A(y) implies

– ÎÒy

u0(x, ·)Î2L2(Y ) Æ

⁄

Y

A(y)Òy

u0(x, y) · Òy

u0(x, y) dy

with the right hand side of the above inequality vanishing by the above energycalculations. Hence we indeed have that

u0(x, y) © u0(x).

Continuing our method of matched asymptotic, at O(Á≠1), we have

(8)

Y]

[≠div

y

1A (y) Ò

y

u1(x, y)2

= divy

1A (y) Ò

x

u0(x)2

for y œ Y,

y ‘æ u1(x, y) is Y -periodic.

Note that we have dropped the term involving the Òy

u0(x, y) from (6) even thoughit was apparently of order O(Á≠1). This is because of our earlier observation thatu0 is independent of the y variable. Equation (8) should be treated as an equationfor the unknown u1(x, y). Because of linearity, observe that we can separate theslow and fast variables in u1 as

u1(x, y) =dÿ

i=1Ê

i

(y)ˆu0ˆx

i

(x)(9)

12 HARSHA HUTRIDURGA

where the unknown functions w

i

(y) for each i œ {1, . . . , d} solve the so-called cellproblems

(10)

Y]

[≠div

y

1A (y) Ò

y

Ê

i

(y)2

= divy

1A (y) e

i

2for y œ Y,

y ‘æ Ê

i

(y) is Y -periodic.

Here {ei

}d

i=1 denotes the canonical basis of Rd. The d equations in (10) are calledcell problems as they are posed on the periodicity cell Y . The solutions to (10) arehenceforth called cell solutions.Finally, at O(1), we obtain(11)Y

_____]

_____[

≠divy

1A (y) Ò

y

u2(x, y)2

= divy

1A (y) Ò

x

u1(x, y)2

+ divx

1A (y)

1Ò

x

u0(x) + Òy

u1(x, y)22

+ f(x),

y ‘æ u1(x, y) is Y -periodic.

So far, we have not spoken about the solvability of the periodic boundary valueproblems (7)-(8)-(10)-(11) that we derived by plugging the asymptotic expansionin the model problem. The following lemma will address this issue.

Lemma 2.3. Suppose the periodic conductivity coe�cient A(y) is positive definiteand that it is uniformly bounded, i.e., it satisfies (3). Let g œ L2(Y ) be a given func-tion. Then there exists a unique solution v œ H1

per(Y )/R to the following periodicboundary value problem

(12)

Y]

[≠div

y

1A (y) Ò

y

v(y)2

= g(y) for y œ Y,

y ‘æ v(y) is Y -periodic

if and only if the source term g(y) in (12) satisfies the compatibility condition

ÈgÍ :=⁄

Y

g(y) dy = 0.(13)

Proof. We know that the the quotient space H1per(Y ) has the equivalent norm

ÎvÎH1per(Y ) = ÎÒ

y

vÎL2(Y ) .

Solving the periodic boundary value problem (12) is equivalent to solving the vari-ational problem:

Find a unique v œ H1per(Y ) such that ’w œ H1

per(Y ), we have

a(v, w) = l(w)

with a(v, w) :=⁄

Y

A(y)Òy

v(y) · Òy

w(y) dy and l(w) :=⁄

Y

g(y)w(y) dy.

TOPICS IN HOMOGENIZATION OF DIFFERENTIAL EQUATIONS 13

To employ the Lax-Milgram theorem, we need to show that the bilinear form a(·, ·)is coercive on H1

per(Y ). This is straightforward. Next, we need to show that thelinear form l(·) is bounded. Compute the modulus of the linear form

|l(w)| =----⁄

Y

g(y)w(y) dy

---- =----⁄

Y

1g(y)w(y) ≠ ÈwÍg(y) + ÈwÍg(y)

2dy

----

=⁄

Y

---1

g(y)w(y) ≠ ÈwÍg(y)2--- dy

Æ Îw ≠ ÈwÍÎL2(Y ) ÎgÎL2(Y )

Æ C ÎÒy

wÎL2(Y ) ÎgÎL2(Y )

where we have used the Holder inequality and the Poincare inequality. Note thatthe above computation only works under the assumption that g(y) is of zero mean.The existence and uniqueness of v œ H1

per(Y )/R then follows from Lax-Milgram. ⇤

Remark 2.4. This remark is about the solvability of the boundary value problems(7)-(8)-(10)-(11). As Lemma 2.3 asserts, only detail that one needs to check is thatthe source terms in all these periodic boundary value problems are of zero averageon the periodicity cell.

• The source term in the O(Á≠2) equation (7) isg(y) © 0.

Hence the compatibility condition (13) is trivially satisfied. The result ofLemma 2.3 then implies that the unique solution to (7) is

0 œ H1per(Y )/R i.e., u0(x, y) © u0(x)

which is what we deduced earlier via energy computations.• The source term in the O(Á≠1) equation (8) is

g(y) = divy

1A (y) Ò

x

u0(x)2

.

As the variable x plays the role of a parameter, we will not make explicitthe dependence of the source g on the x variable. Let us now compute theaverage of g(y) over the periodic cell⁄

Y

divy

1A (y) Ò

x

u0(x)2

dy =⁄

ˆY

A (y) Òx

u0(x) · n(y) d‡(y) = 0.

Thus the compatibility condition (13) is satisfied. Hence the existence of aunique solution u1(x, ·) œ H1

per(Y )/R to the O(Á≠1) equation (8).• A similar arguments work for the cell problems (10) whose source terms are

g(y) = divy

1A (y) e

i

2.

• Finally, to solve the periodic boundary value problem for u2(x, ·) œ H1per(Y )/R,

let us make sure that the source term in (11) satisfies the compatibility con-dition. In fact, writing down the compatibility condition for (11) yields the

14 HARSHA HUTRIDURGA

homogenized equation for u0(x).The source term in the O(1) equation (11) is

g(y) = divy

1A (y) Ò

x

u1(x, y)2

+ divx

1A (y)

!Òx

u0(x) + Òy

u1(x, y)"2

+ f(x)

Integrating the above given g(y) over the periodicity cell yields⁄

Y

g(y) dy = I1 + I2 + I3 :=⁄

Y

divy

1A (y) Ò

x

u1(x, y)2

dy

+⁄

Y

divx

1A (y)

!Òx

u0(x) + Òy

u1(x, y)"2

dy

+⁄

Y

f(x) dy.

Remark that, because A(·) and u1(x, ·) are Y -periodic, we haveI1 = 0.

Furthermore, as f is independent of the y variable,I3 = f(x).

From Lemma 2.3, we can solve for u2(x, ·) uniquely, provided we satisfy thecompatibility condition which translates here as

≠I2 = I3

which is nothing but

≠ divx

⁄

Y

1A (y)

1Ò

x

u0(x) + Òy

u1(x, y)22

dy = f(x) for x œ �.(14)

Observe that substituting for u1(x, y) in the above equation in terms of thecell solutions Ê

i

(y) and the gradient of u0(x), i.e., using (9) results in asecond order di�erential equation for u0(x).

Let us pick up our calculations from (14). Substituting for u1(x, y) using (9)results in

≠ divx

⁄

Y

1A (y)

1Ò

x

u0(x, y) + Òy

u1(x, y)22

dy

= ≠ divx

⁄

Y

AA (y)

AÒ

x

u0(x, y) +dÿ

k=1Ò

y

Ê

k

(y) ˆu0ˆx

k

(x)BB

dy

= ≠ divx

1AhomÒ

x

u0(x)2

with the constant matrix Ahom – called the homogenized matrix – which is given as

Ahom› =⁄

Y

A (y)A

› +dÿ

k=1›

k

Òy

Ê

k

(y)B

dy

TOPICS IN HOMOGENIZATION OF DIFFERENTIAL EQUATIONS 15

for any › œ Rd. Note in particular that the elements of the matrix Ahom are

[Ahom]ij

=⁄

Y

A (y)1

ej

+ Òy

Ê

j

(y)2

· ei

dy(15)

for i, j œ {1, . . . , d}. The calculations so far are summarised below.Proposition 2.5. The solution u

Á(x) to the boundary value problem (1) is approx-imated as

u

Á(x) ¥ u0(x) + Á

dÿ

j=1Ê

j

1x

Á

2ˆu0ˆx

j

(x)

where u0(x) is called the homogenized approximation which solves the homogenizedboundary value problem

(16)

Y]

[≠div

1AhomÒu0(x)

2= f in �,

u0 = 0 on ˆ�,

with Ahom being the constant homogenized conductivity whose elements are

[Ahom]ij

=⁄

Y

A (y)1

ej

+ Òy

Ê

j

(y)2

· ei

dy

for i, j œ {1, . . . , d}. The functions Ê

j

(y) are called the cell solutions which solvethe cell problems

Y]

[≠div

y

1A (y) Ò

y

Ê

i

(y)2

= divy

1A (y) e

i

2for y œ Y,

y ‘æ Ê

i

(y) is Y -periodicfor each i œ {1, . . . , d}. Furthermore, the homogenized conductivity is uniformlypositive definite, i.e., there exists a positive – > 0 such that

Ahom› · › Ø – |›|2(17)

for all › œ Rd \ {0}.Proof. A lion share of the proof of Proposition 2.5 is already present in the calcula-tions preceding the statement of the proposition. We are left, however, to show thatthe homogenized approximation u0(x) satisfies zero Dirichlet data on the boundaryand to show that Ahom is uniformly positive definite.As we are postulating an asymptotic expansion (5), let us simply equate the expan-sion to zero on the boundary

0 = u0(x) + Á u1

1x,

x

Á

2+ Á

2u2

1x,

x

Á

2+ . . . for x œ ˆ�.

As the above equality should hold in the Á æ 0 limit, at least formally, we obtainu0(x) = 0 for x œ ˆ�.

In order to prove the positive definiteness of Ahom, let us multiply by Ê

i

(y) the cellproblem for Ê

j

(y) and integrate over the periodicity cell Y yielding⁄

Y

A(y)1

Òy

Ê

j

(y) + ej

2· Ò

y

Ê

i

(y) dy = 0.

16 HARSHA HUTRIDURGA

Add the above zero to the expression (15) yielding an alternate expression for theelements of the homogenized conductivity

[Ahom]ij

=⁄

Y

A (y)1

ej

+ Òy

Ê

j

(y)2

·1

ei

+ Òy

Ê

i

(y)2

dy(18)

for i, j œ {1, . . . , d}. Using the above alternate expression for the homogenizedconductivity, let us compute the dot product

Ahom› · › =⁄

Y

A (y)1

› + Òy

Ê

›

(y)2

·1

› + Òy

Ê

›

(y)2

dy

where the function Ê

›

(y) is defined as

Ê

›

(y) =dÿ

k=1Ê

k

(y)›k

.

The coercivity assumption (3) on the conductivity coe�cient A(y) implies

Ahom› · › Ø –

⁄

Y

|› + Òy

Ê

›

(y)|2 dy

= –

⁄

Y

|Òy

Ê

›

(y)|2 dy

¸ ˚˙ ˝Ø0

+–

⁄

Y

|›|2 dy + 2–

⁄

Y

› · Òy

Ê

›

(y) dy

¸ ˚˙ ˝=0 as Ê

›

periodicHence

Ahom› · › Ø – |›|2 .

Suppose now thatAhom› · › = 0

which implies that› + Ò

y

Ê

›

(y) = 0 ’y œ Y

i.e.Ê

›

(y) = C ≠ › · y

for some constant C. This says that Ê

›

(y) is an a�ne function in the y variable.This is a contradiction to the periodicity of Ê

›

(y) except if › = 0. ⇤

Remark 2.6. If the conductivity coe�cient A(y) is assumed to be symmetric, thenthe expression (18) for the homogenized conductivity implies that Ahom is symmetricas well.

Exercise 2.7. Using the method of two-scale asymptotic expansions, prove a resultsimilar in flavour to Proposition 2.5 for the following periodic parabolic problem:

(19)

Y____]

____[

fl

1x

Á

2ˆ

t

u

Á(t, x) ≠ div1

A

1x

Á

2Òu

Á(t, x)2

= f(t, x) in (0, T ) ◊ �,

u

Á(t, x) = 0 on (0, T ) ◊ ˆ�,

u

Á(0, x) = u

in(x) in �,

TOPICS IN HOMOGENIZATION OF DIFFERENTIAL EQUATIONS 17

where the coe�cients fl(·) and A(·) are Y -periodic. Assume that the coe�cient fl isbounded from above and away from zero, i.e.,

0 < c1 Æ fl(·) Æ c2 < Œ.

Assume further that the coe�cient A(·) satisfies the assumption (3). The bulk sourcef(t, x) and the initial datum u

in(x) are nice functions and are as regular as thecomputations demand.Hint: Treat the time variable t as a parameter while performing the asymptoticexpansions, i.e., postulate that the solution u

Á(t, x) to the above initial boundaryvalue problem can be written as

u

Á(t, x) = u0

1t, x,

x

Á

2+ Á u1

1t, x,

x

Á

2+ Á

2u2

1t, x,

x

Á

2+ . . .

where the coe�cient functions u

i

(t, x, y) are assumed to be Y -periodic in the y

variable.

Exercise 2.8. Assume that the coe�cients fl(·), A(·) satisfy the same assumptionsas in Exercise 2.7. Assume further that the data f(t, x), u

in(x) and u

in1 (x) are

nice functions and that they are as regular as the computations demand. Usingthe method of two-scale asymptotic expansions, prove a result similar in flavour toProposition 2.5 for the following periodic hyperbolic problem:

(20)

Y________]

________[

fl

1x

Á

2ˆ

2tt

u

Á(t, x) ≠ div1

A

1x

Á

2Òu

Á(t, x)2

= f(t, x) in (0, T ) ◊ �,

u

Á(t, x) = 0 on (0, T ) ◊ ˆ�,

u

Á(0, x) = u

in(x) in �,

ˆ

t

u

Á(0, x) = u

in1 (x) in �.

2.2. One dimensional problem. In this section, we will present some explicitcomputations in a one dimensional setting. Let the strictly positive and boundedconductivity coe�cient be denoted by a(y) which is 1-periodic. Then one dimen-sional analogue of the boundary value problem (1) is

(21)

Y_]

_[

≠ ddx

3a

1x

Á

2 duÁ

dx

4= f(x) in (¸1, ¸2),

u

Á(¸1) = u

Á(¸2) = 0,

for an interval (¸1, ¸2) µ R.The one-dimensional analogue of Proposition 2.5 is stated here.

Corollary 2.9. The cell solution in the one-dimensional setting is given by

Ê(y) = C ≠ y + ahom

⁄y

0

d·

a(·)(22)

18 HARSHA HUTRIDURGA

for a constant C. Furthermore, the constant ahom is the homogenized conductivitygiven by

ahom =3⁄ 1

0

dy

a(y)

4≠1

(23)

Proof. Let us write down the one-dimensional analogue of the cell problem (10).Y_]

_[

≠ ddy

3a (y)

3dÊ

dy

(y) + 144

= 0 in (0, 1),

Ê(0) = Ê(1).Explicitly integrating the above di�erential equation yields for some constant C1,

a (y)3

dÊ

dy

(y) + 14

= C1 =∆ dÊ

dy

(y) = ≠1 + C1a(y)

Integrating again would yield

Ê(y) = C ≠ y +⁄

y

0

C1a(·) d·.

The constant of integration C is a dummy variable (for now). It can be chosenby normalization as we look for solutions in the quotient space H1

per(0, 1)/R. Theconstant C1, however, is determined by the periodic boundary condition in the cellproblem.

C1 =3⁄ 1

0

dy

a(y)

4≠1

.

Now, let us write down the one dimensional analogue of the homogenized conduc-tivity (15).

ahom =⁄ 1

0a (y)

3dÊ

dy

(y) + 14

dy

which is nothing but

C1

⁄ 1

0dy.

Hence we have deduced that the one dimensional homogenized conductivity is noth-ing but the harmonic average over the periodicity cell, i.e.,

ahom =3⁄ 1

0

dy

a(y)

4≠1

This proves the corollary. ⇤Remark 2.10. Note that the following strict inequality holds

3⁄ 1

0

dy

a(y)

4≠1

<

⁄ 1

0a(y) dy

for any non-trivial function a(y). By, trivial, we mean functions that are identicallyconstants in which case the harmonic average and the simple average coincide. Thisobservation implies that the homogenized conductivity of a periodic composite is

TOPICS IN HOMOGENIZATION OF DIFFERENTIAL EQUATIONS 19

always less than the associated simple average (arithmetic average), at least in onedimension.

Remark 2.11. Let us reconsider the one dimensional illustrative Example 2.1. Wechose the periodic conductivity to be

a(y) = 12 + cos(2fiy) .

Computing the homogenized conductivity in this case yields

ahom =3⁄ 1

0

12 + cos(2fiy)

2dy

4≠1

= 12 .

Hence the homoegnized equation for the Example 2.1 becomesY_]

_[

≠d2u0

dx

2 = 2 in (0, 1),

u0(0) = u0(1) = 0which integrates explicitly as

u0(x) = x ≠ x

2

This is the parabolic profile which we found as a macroscopic approximation to thesolution u

Á(x) in Example 2.1.

Exercise 2.12. Compute the homogenized conductivity ahom for the following pe-riodic conductivity which is piecewise constant:

a(y) =

Y_____]

_____[

1 for y œ !0,

13"

2 for y œ # 13 ,

12"

3 for y œ # 12 ,

34"

4 for y œ # 34 , 1

"

Compute also its arithmetic average.

2.3. Laminated geometries in higher dimensions. The one dimensional set-ting of section 2.2 is pretty special in the sense that given a periodic conductivitya(y), we can straightaway compute the homogenized conductivity ahom withouthaving to go though the cell problem. In this section, we consider the examples ofperiodic laminated structures in higher dimensions, i.e., scenarios where the conduc-tivity coe�cient depends on a single direction (i.e. varying only in one direction).We take

A(y) = A(y1).In this section, we further make the assumption that the above conductivity coe�-cient is isotropic, i.e., there exists a scalar function a(y1) such that

A(y) = a(y1) Id.(24)We will prove the following interesting result on the above chosen laminated struc-tures.

20 HARSHA HUTRIDURGA

Proposition 2.13. The homogenized conductivity associated with the isotropic con-ductivity coe�cient (24) is the following diagonal matrix

[Ahom]ii

=

Y____]

____[

3⁄ 1

0

dy1a(y1)

4≠1

for i = 1,

⁄ 1

0a(y1) dy1 for i ”= 1.

Remark 2.14. It is to be noted that even though we chose an isotropic periodicconductivity coe�cient to begin with, the homogenization procedure destroys theisotropy as can be noticed in Proposition 2.13.

Proof of Proposition 2.13. Let us rewrite the cell problem (10) with the chosenisotropic conductivity (24) as

Y]

[≠div

y

1a (y1) Ò

y

Ê

i

(y)2

= divy

1a (y1) e

i

2for y œ Y,

y ‘æ Ê

i

(y) is Y -periodicfor each i œ {1, . . . , d}. In particular, for i = 1, we have

Y_]

_[

≠divy

1a (y1) Ò

y

Ê1(y)2

= ˆ

ˆy1a (y1) for y œ Y,

y ‘æ Ê1(y) is Y -periodicObserve that the solution Ê1(y) depends only on the y1 direction, i.e.,

Ê1(y) = ‰(y1)where ‰(y1) solves the one dimensional cell problem

Y_]

_[

≠ ddy1

3a (y1)

3d‰

dy1(y1) + 1

44= 0 in (0, 1),

‰(0) = ‰(1).Next, for i ”= 1, we have

Y_]

_[

≠divy

1a (y1) Ò

y

Ê

i

(y)2

= ˆ

ˆy

i

a (y1) = 0 for y œ Y,

y ‘æ Ê

i

(y) is Y -periodicObserve that the solutions to the above (d ≠ 1) cell problems are all in the zeroequivalence class (see Lemma 2.3). Hence we have

Ê

i

(y) © 0 for i ”= 1.

The above information on the cell solutions help us get some insight into the homog-enized coe�cient which in this case is a diagonal matrix. Let us compute [Ahom]11using the formula (15).

[Ahom]11 =⁄

Y

a(y1)3

d‰

dy1(y1) + 1

4dy1 =

3⁄ 1

0

dy1a(y1)

4≠1

TOPICS IN HOMOGENIZATION OF DIFFERENTIAL EQUATIONS 21

where the last equality follows from Corollary 2.9. Next, let us compute [Ahom]ii

for i ”= 1 using the formula (15).

[Ahom]ii

=⁄

Y

a(y1)1

Òy

Ê

i

(y) + ei

2· e

i

dy =⁄ 1

0a(y1) dy1

because the cell solutions Ê

i

(y) © 0 for i ”= 1. ⇤Exercise 2.15. Using Proposition 2.13, compute the homogenized conductivityAhom for the following laminar periodic conductivity:

A(y) =I

d1 for y1 œ !0,

12"

d2 for y1 œ # 12 , 1

"

with strictly positive constants d1, d2.

Exercise 2.16. Mimicking the proof of Proposition 2.13, obtain analytic expres-sions for the homogenized conductivity Ahom in two dimensions when the periodicconductivity coe�cient is laminar, i.e., A(y) = A(y1), but not necessarily isotropicas was the case in (24).