Homogenization of Linear Elliptic Equations

1 Introduction to the Problem
- 1.1 Playing games with nonlinear PDEs: solving PDEs with unknown coefficients
2 Typical Basic and Advanced Problems
3 Discrete Mathematical Background: Markov Chains and Finite Differences
- 3.1 Markov Chains and the Invariant Measure
- 3.2 Examples of Markov Chains coming from Finite Differences
4 Continuous Mathematical Background: PDEs etc
5 Results and Conjectures
6 References

Introduction to the Problem

Playing games with nonlinear PDEs: solving PDEs with unknown coefficients

Suppose you are solving a linear elliptic PDE with unknown coefficients, in a square, with known boundary conditions. To be concrete, say at every point of the domain the equation is either

$L 1 u = f$ or $L 2 u = f,$

where, for example,

$L 1 u = 2 u x x + 1 u y y,$ and $L 2 u = 1 u x x + 2 u y y,$

What can be said about the solution? We will show, using elementary arguments (which can be backed up with rigorous analysis) how to make some progress on this problem.

In particular, we consider: - the antagonistic case, where the operators are chosen to be the worst possible, and - the random case, where the coefficients are completely random. these are the two most extreme situations, and give some idea as to the possibilities in the general case.

Typical Basic and Advanced Problems

Known Analytic solutions from the references

Pavilotis and Stuart, Layered materials for divergence case in two dimensions
Holmes, Introduction to Perturbation Methods Seperable case, again for divergence structure equations
Braides and Defranceschi Homogenization of Multiple Integrals, section 14.5 divergence structure elliptic, we need to check what they have done.

Most of these examples are for divergence structure. Question: do we get a different answer for non-divergence structure?

Basic: One Dimensional Homogenization

period case
random case

In both cases, the operator homogenizes to the harmonic mean, but the invariant measure, while it is p.w. constant, is non-trivial. See The 1D Problem for numerical results. See Justification for The 1D Problem for an explanation. ( we look at the transpose operator, and see that in regions where the invariant measure is constant, we get zero, but where it jumps, the equation gives a1 uL - 2a1 u0 + a2uR = -a1 w1 + a2 w2, here w1 = a1/(a1+a2), w2 = a2/(a1 + a2), so it works).

This is interesting because the invariant measure is entirely determined by what happens at a jump. So the structure of the coefficients (random or periodic) really doesn't matter. It's the same proof in one dimension.

We can understand the continuous result by approximation the equation by a markov process. So this can be made into a rigorous proof.

Basic: Two Dimensional Isotropic Homogenization

Now we consider the two dimensional problem, where we have a variable scalar coefficient in front of the Laplacian. I.e. $a (x)Δ u$ , where $a (x) = a 1 or a 2$ which are positive constants.

The most basic question is a product of one dimensional problems, i.e. if the coefficent a(x,y) = a1(x)a2(y). Then the problem splits.

In the divergence structure case, the solution is anisotropic with coefficients which a combination of the arithmetic and harmonic mean. [numerics for the separable case ] So typical example for this case, is [1 a] in the x-direction, [1 b] in the y-direction, and the pattern is [1 a; b ab]. Is the answer the same as in the divergence structure case?

In this case, the most basic question is the periodic checkerboard. So this pattern is [1 a; a 1]. What does it homogenize to? Is it the harmonic mean?

Our intuition suggests that we average the operators with coeffs 1, a, based on the weights which represent how much time is spent acting by each operator. So this suggests the harmonic mean again. Q: is the invariant measure p.w. constant? Is the result the same for the random case? Now we can have probabilities different from 1/2, and we can adjust the weights, to validate our guess. [periodic and random numerics for the checkerboard case ]

Advanced Examples: Anisotropic Homogenization

If we now consider a diffusion matrix, for example $u x x + 3 u y y$

for one of the operators, then we expect that the homogenized operator could also be anisotropic.

the basic example would be a symmetric situations, e.g. $L 1 = u x x + 3 u y y, L 2 = 3 u x x + u y y$ . Then symmetry considerations suggest an anisotropic operator. What is the coefficient? Must be the average of the operators, since equal time is spend in each domain (at least in the periodic case).

This is indeed what is obtained. [Two dimensional anisotropic homogenization results]

Discrete Mathematical Background: Markov Chains and Finite Differences

See Markov Chains and Generators

Markov Chains and the Invariant Measure

When we discretize the equation using a finite difference scheme, we get a markov chain. We want to find the invariant measure. This will tell us how much time the process spends in each of the states. For this we refer to Ch 5 of Stuart and Pavliotis, library link

Although this looks complicated, we can compute this stuff fairly easily. Some examples follow.

Examples of Markov Chains coming from Finite Differences

If you just discretize the heat equation, on say three points, see that you the invariant measure is a constant.

U n + 1 = U n + d t (U j - 1 - 2 U j + U j + 1)

The generator $L$ :

$\begin{pmatrix} \ddots & & \\ 1 & -2 & 1 \\ & & \ddots \\ \end{pmatrix}$

The transition matrix $P$ :

$\begin{pmatrix} \ddots & & \\ dt & 1-2 dt & dt \\ & & \ddots \\ \end{pmatrix}$

But if instead you discretize the head equation on 4 points and have different coefficients, then you get a non-trivial invariant measure.

The generator $L$ :

$\begin{pmatrix} \ddots & & & & & \\ 1 & -2 & 1 & & & \\ & 2 & -4 & 2 & & \\ & & 1 & -2 & 1 & \\ & & & 2 & -4 & 2 \\ & & & & & \ddots \\ \end{pmatrix}$

The transition matrix $P$ :

$\begin{pmatrix} \ddots & & & & & \\ dt & 1-2 dt& dt & & & \\ & 2 dt & 1-4 dt & 2 dt & & \\ & & dt & 1-2 dt & dt & \\ & & & 2 dt & 1-4 dt & 2 dt \\ & & & & & \ddots \\ \end{pmatrix}$

When $d t = 0.1$ , the above transition matrix $P$ become

$\begin{pmatrix} \ddots & & & & & \\ 0.1 & 0.8 & 0.1 & & & \\ & 0.2 & 0.6 & 0.2 & & \\ & & 0.1 & 0.8 & 0.1 & \\ & & & 0.2 & 0.6 & 0.2 \\ & & & & & \ddots \\ \end{pmatrix}$

Check out the following matlab example:

 % start with the finite difference matrix on three states
 D = [-2 1 1; 1 -2 1; 1 1 -2]
 % now multiply by the coefficients
 a = diag([1 2 5]);
 M = a*D;
 % now find the transition matrix
 dt = .1
 P = eye(3) + dt*M
 % since we advance time according to the largest intensity
 % we get the following transition matrix
 
 Pt = [ .8 .2 .5; .1 .6 .5; .1 .2 0]
 
 % now find the invariant measure as an eigenvalue
 
 [d, e] = eig(M);
 dd = d(:,1);
 dd = dd/sum(dd);

Continuous Mathematical Background: PDEs etc

Upper and Lower Solutions

To answer this question in the worst case, in other words, to get upper and lower bounds on the solution, we will use the comparison principle, to show that there is a nonlinear partial differential equation which gives the largest solutions. There is another equation which gives the smallest. These equations are

$max(L 1 u, L 2 u) = f,$ and $min(L 1 u, L 2 u) = f,$

Average Solution in the random or periodic case

To get an idea of the typical solution, we need some information. If we know that the material is periodic, say a checkerboard with either of the operators, then we can simply solve. On there other hand, if the material is random, say a checkerboard with either of the operators randomly with some probability p (=1/2, say), then we can might want to do a few trials and see if there is a reasonable average. We would also like to know the standard deviation.

This is typical, and has been done for more complicated problems. But in this case, we can say much more. To do so, we need to learn some homogenization.

The diffusion interpretation of the parabolic equation

Want to understand that generator for the diffusion. This is just generalizing that the heat equation is the solution of a random walk. We just weight the random walk when we have coefficients which are different.

See Stochastic Differential Equations

Theorems of the alternative

Special to the linear case. See Holmes or Stuart.

Example. If we take the operator to be $L (x) = a (x)Δ$ then there is a periodic solution of

$L (x) u = A$ if and only if $A = \bar a$ the harmonic mean. See Stuart Ch 12 for the divergence structure case.

The cell problem

Cite the review paper by Souganidis and Engquist, in the nonlinear case, which also applies here. The Fredholm alternative work for linear operators. A different existence of solutions theory is used in the nonlinear case, where now the homogenized operator is nonlinear. $F(x, D^2u + Q) = \bar F(Q)$

The Lie-Trotter Formula

Pretend instead we were solving the parabolic problem $u t = L u$ Where L is each of the operators for a very short time. What is the resulting operator, in the limit that we switch very often? The answer is given by the Lie-Trotter product.

Check out the wikipedia page http://en.wikipedia.org/wiki/Lie-Trotter

In fact, this is used in numerical methods where you split a sum of terms.

Ergodicity

Check out the wikipedia page http://en.wikipedia.org/wiki/Ergodic_theory We will use this to show that we can replace time averages with spatial averages.

Results and Conjectures

The following we have been able to show. In the case where $L (x) = a (x)Δ$ we know that

The homogenized operator is $\bar L = \bar a \Delta$ where $\bar a = \left (\text{average of }\frac 1 {a(x)}\right )^{-1}$

By approximating in the discrete setting we can show the following, the invariant measure is simply $\rho^\infty = \frac 1 {a(x)} \bar a$

We want to show this in the continuous case directly. To prove: Need to write down the adjoint equation directly and verify it.

We want to consider the anisotropic case, where we expect that the same formula holds, with suitable modifications, and that

$\bar L = \sum w_i L_i$ where the weight are the average of the invariant measure on the sets where each operator is active.

To prove: don't know. Possibilities: learn theorem of the alternative, and apply it with the right constant using the Cell Problem formulation. Use the idea of the generator of the process, and the ergodic theorem. Come up with exact correctors for the cell problem in special cases (less likely to work).

Idea for proof: go through the application of the theorem of the alternative in the divergence structure case, then do it for non-divergence structure case, and see if the invariant measure plays the role of the constants.

References

G. A. PAVLIOTIS, A. M. Stuart. Multiscale Methods: Averaging and Homogenization. Springer, 2008. [1] library link

Graeme Milton, The theory of composites, (online access at sfu) [2]

Homogenization of Linear Elliptic Equations - Adam Oberman Math

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