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Let G a bounded open periodic set in $\mathbb{R}^2$, it's period $\epsilon$ (a positive number which is assumed to be very small in comparaison with the size of the domain, and the rescaled unit periodic cell $Y=]0,1[ \times ]0,1[$. Let the matrix $A_{kj}$ symetric and positive-definite at each point of $\xi$: $$ A_{kj}= A_{jk} $$ $$ A_{kj} \eta_k \eta_j \geq K_1 \eta_k \eta_j \ \forall \eta=(\eta_1,\eta_2) \in \mathbb{R}^2, $$ where the constant $K_1 > 0$ is independent of $\epsilon$ and $\xi$. We have the following Lemma:

Lemma 1: The necessary and sufficient condition for $1$-periodic solution of the equation $$ \sum_{k,j=1}^2 \dfrac{\partial}{\partial \xi_k} (A_{kj} (\xi) \dfrac{\partial}{\partial \xi_j} N(\xi))= F(\xi)..... (3) $$ to exists is: $$ \langle F \rangle= \displaystyle\int_0^1 \displaystyle\int_0^1 F(\xi) d\xi_1 d \xi_2 =0, $$ and the general $1$ periodic solution of equation (3) is written as $N(\xi)= \overline{N(\xi)}+C$, where $\overline{N(\xi)}$ is a solution of (3) with zero mean over period $\langle \overline{N} \rangle=0$, and $C$ is an arbitrary cnstant.

My question concerns the follwing equation: $$ \sum_{k,j=1}^2 \dfrac{\partial}{\partial \xi_k} (A_{kj}(\xi) \dfrac{\partial}{\partial \xi_j} u_0(x,\xi))=0 ...(1) $$

How we deduce that Lemma 1 implies that the $1$-periodic solution of equation (1)is independent of $\xi$ that is $u_0(x,\xi)= v_0(x)$?

Thank you in advance for the help

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I don't think that you can deduce from Lemma 1 that the 1-periodic solution of $(1)$ is independent of $\xi$. What you need is a uniqueness result for equation $(3)$ (uniqueness up to an additive constant). This can, e.g., be obtained via the Lax-Milgram lemma.

Let us now assume that solutions to $(3)$ are unique. Then also solutions to $(1)$ are unique (up to an additive constant) and $u_0(x, \xi) = v(x)$, where $v$ is some function of $x$, is a solution to $(1)$. This proves your claim.

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