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:

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

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

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.


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.


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 $y$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.