Asymptotic Expansion of a Multiscale Partial Differential Equation

Question

I'm trying to understand how to solve $$-\nabla\cdot(K(\frac{x}{\epsilon})\nabla u(x,\frac{x}{\epsilon})=f \text{ in } \Omega$$ $$u(x,\frac{x}{\epsilon})=u_D \text{ in } \partial \Omega$$
where $K(\frac{x}{\epsilon})$ is a rapidly oscillating Y-peridic in $R^n$ with periodicity cell $$Y\equiv\{y=(y_1,...,y_n) \text{ } | \text { } 0<y_i<1 \text{ for } i=1,...,n\}$$ and $\epsilon$ is a scale parameter. As $\epsilon\rightarrow 0$, $K(x,\frac{x}{\epsilon})$ oscillates faster and faster. So, I'm seeking the behavior of the solution $u(x,\frac{x}{\epsilon})$ in the limit as $\epsilon\rightarrow 0$ through a process called homogenization.

To understand my question, I need to show some of the steps in homogenization:

Ansatz:
Let $y=\frac{x}{\epsilon}$. I can write the power series expansion of u in terms of epsilon:
$$u(x,y)=u_0(x,y)+\epsilon u_1(x,y) + \epsilon^2 u_2(x,y)+...$$ where $u_i$ are all Y-periodic functions with respect to the variable y. Since $y=\frac{x}{\epsilon}$, we can use the chain rule to define the gradient operator as $\nabla = \nabla_x + \frac{1}{\epsilon} \nabla_y$. Applying this operator to $u(x,y)$ and after substituting this into the PDE and collecting like terms together, we obtain the equation

$\epsilon^{-2}\left[\nabla_y\cdot(a(y)\nabla_yu_0\right]$
$+\epsilon^{-1}\left[ \nabla_y\cdot(a(y)\nabla_yu_1)+\nabla_y\cdot(a(y)\nabla_xu_0) + a(y)\nabla_x\cdot(\nabla_yu_0)\right]$ $+\epsilon^0\left[ \nabla_y\cdot(a(y)\nabla_yu_2 + a(y)\nabla_xu_1) +a(y)\nabla_x\cdot(\nabla_yu_1) + a(y)\nabla_x\cdot(\nabla_xu_0) \right]$ $+\epsilon^1(...)+... = -f(x)$

Equating Coefficients

Now comes the fun part. To solve, we need equate like coefficients of $\epsilon$ on both sides of this equation. Let's start with the $\epsilon^2$ terms. Since there are no epsilon terms on the right hand side of the equation above, we obtain:

$$\nabla_y\cdot(a(y)\nabla_yu_0)=0 \text{ for } y\in Y$$

I read several texts on this subject, and all make the claim that since $u^0$ is Y-peridic, this equation implies that $u_0(x,y)$ is independent of y, ergo we can redefine $u_0(x,y)\equiv u_0(x)$. Each text i've read seems to treat this as a trivial deduction, but I can't see the connection between Y-periodicity and $u_0$ being independent of the variable $y$. I'm sure it's just staring me in the face, but I just can't see it. Any help making the connection between the two would be greatly appreciated!

Note: See this (pp. 8) for an example of one person's take on the triviality of this claim.

Answer 1

Being periodic, the function $u_0(x,\cdot)$ must attain its global maximum at some point. By the maximum principle for elliptic equations (e.g., Chapter 3 of the book by Gilbarg and Trudinger) a solution that attains its maximum must be constant.

Answer 2

I think I found the answer to my question by looking at this presentation.

Assuming Y-periodicity in $u_i$ is not enough to show that $u_0(x,y)$ is independent of y. We need to impose an additional condition. To understand this, we need to look at the asymptotic expansion once again:

$$u(x,y)=u_0(x,y)+\epsilon u(x,y)+\epsilon^2 u(x,y)+...$$

As noted in the question, we can write the gradient of u(x,y) as

$$\nabla u(x,y)=\nabla_x u_0(x,y) + \frac{1}{\epsilon}\nabla_y u_0(x,y)+\nabla_y u_1(x,y) + O(\epsilon)$$.

We must impose the condition that $\nabla u(x,y)$ is bounded as $\epsilon\rightarrow 0$ (otherwise, solutions may blow up to infinity). Observing the three terms in $\nabla u(x,y)$, the only troublesome term as $\epsilon\rightarrow 0$ is $\frac{1}{\epsilon}\nabla_y u_0(x,y)$, so we impose that $\nabla_y u_0(x,y)=0$. But $\nabla_y u_0(x,y)=0$ must imply that $u_0(x,y)$ is independent of $y$.