Let
- $f \in L^2(\mathbb R^d)^d$
- $u_\epsilon : \mathbb R^d \rightarrow \mathbb R^d$
- $a: \mathbb R^d \rightarrow \mathbb R^{d \times d}$ such that there exists a $\lambda >0$, $\forall \xi \in \mathbb R^d$ $\xi a \xi \ge \lambda \vert \xi \vert^2$ and $\vert a \xi \vert \le \vert\xi \vert$
We look at the equation:
$$\operatorname{div}(a(\frac{.}{\epsilon}) \nabla(u_\epsilon)) = -\operatorname{div}(f)$$
First we would like to show the existence and uniqueness of the (weak) solution $u_\epsilon$ at $\epsilon >0$ fixed.
I think we should use Lax-Milgram theorem, but other propositions could maybe work too (like minimization on $a$ ? At least if we suppose $a$ symmetric, no ? This solution would be welcome)
Lax-Milgram: Let $H$ be a Hilbert space and $V$ a normed space. Let $B : H × V → \mathbb R$ be a continuous, bilinear function. Then the following are equivalent:
(coercivity) for some constant $c > 0$, $$\inf\limits_{‖ v ‖_V = 1} \sup\limits_{‖ h ‖_H ≤ 1} | B ( h , v ) | ≥ c $$
(existence of a "weak inverse") for each continuous linear functional $f ∈ V^∗$, there is an element $h ∈ H$ such that $B ( h , v ) = ⟨ f , v ⟩$ for all $v ∈ V $.
So we need to fix a Hilbert space $H$, a normed space $V$ and find a continuous coercive bilinear function $B : H × V → \mathbb R$
Intuition: take $H=V=\{ v \in H^1_{loc}(\mathbb R^d) \mid \nabla v \in L^2(\mathbb R^d)^d\}/\mathbb R$
$H$ is a Hilbert space (didn't check, but it looks like it. I only know that the sobolev space $H^1$ is a Hilbert space). The norm equipped on $H$ is
$$‖ v ‖_H=‖ \nabla v ‖_{L^2(\mathbb R^d)}$$
I don't really know what to choose for $B$. Maybe something like
$$B(v,w)=\int_{\mathbb R^d} \nabla v \, a(\frac{.}{\epsilon}) \nabla w$$
Then $u_\epsilon$ is a weak solution of the equation iff $\forall v \in H$,
$$B(u_\epsilon,v)=-\int_{\mathbb R^d}\nabla v \, f$$
right ? I am going into this completely blind, never looked at PDEs before.
Checking continuity of $B$:
$$\vert B(v,w) \vert \le \int_{\mathbb R^d} \vert \nabla v \vert \, \vert a(\frac{.}{\epsilon}) \nabla w \vert \le ‖ v ‖_H ‖ w ‖_H $$
Checking coercivity of $B$:
$$B(v,v) = \int_{\mathbb R^d} \nabla v \, a(\frac{.}{\epsilon}) \nabla v \ge \lambda \int_{\mathbb R^d} \vert \nabla v \vert^2 = \lambda ‖ v ‖_H$$
Are we done ? Is this a classical way of showing existence of solutions ? Are my arguments viable ? Even if my proof is good, I would like to see other arguments, so if you have ideas, don't hesitate.
Let's begin the second part: solving this equation.
Simplify the problem by assuming that $a$ is $[0,1]^d$-periodic.
Suppose god exists and that he tells us the solution looks like this:
$$u_\epsilon (x) = \sum\limits_{k=0}^\infty \epsilon^k u_k(x,\frac{x}{\epsilon})$$
where $u_k(x,.)$ is also $[0,1]^d$-periodic for all $k$.
Question: how can we find the functions $u_k$ ?
Thank you for reading this long post.