# Understanding orthogonality in two-scale asymptotic expansion (cf. G. Allaire)

This question is about equation (2.16) of Lecture 2 on Homogenization in Porous Media_ by Allaire page 28.

There are two spacial scales: $x$ being macroscopic and $y=\dfrac{x}{\varepsilon}$ being microscopic.

As well there are two time scales: $t$ the short time scale and $\tau=\varepsilon^2t$ the long time scale.

Let $(\lambda, w(y))$ be the first eigenvalue/-function pair of the cell spectral problem (2.9, page 25). The first function of the asymptotic expansion (2.13, page 27), $u_0(t,\tau,x,y)$, can then be written in terms of the long-time behaviour of the solution $u(\tau,x)e^{-\lambda t}w(y)$.

Deriving the cascade of equations with different powers of $\varepsilon$ from plugging series (2.13) into the initial problem equation (2.8) is easy. Thus equation (2.15) is clear: $$c(y)\frac{\partial u_1}{\partial t} - \mathrm{div}_y(A(y)\nabla_y u_1)=\sigma(y)u_1+g_1$$ with $$g_1=\mathrm{div}_x(A(y)\nabla_y u_0)+\mathrm{div}_y(A(y)\nabla_x u_0)$$

Though the argumentation, that $g_1$ must be orthogonal to the first eigenfunction $w(y)$ is reasonable, I'm struggling with solving (2.16, page 28) $$\int_Y g_1(y)w(y)dy=0$$ into $$\int_Y g_1(y)w(y)dy \\=e^{-\lambda t}\int_Y A(y)\nabla_y w(y)\cdot\nabla_x u(x)w(y)-A(y)w(y)\nabla_x u(x)\cdot\nabla_y w(y)dy=0$$ withverwarnung the symmetric diffusion tensor $A(y)$.

Especially I'm confused by the change of sign. As well, I would assume, that the first term can be written as $$A(y)\nabla_y w(y)u(x)\cdot\nabla_xw(y),$$ which would equal to zero as $\nabla_xw(y)=0$. And following it would read $$\int_Y A(y)w(y)\nabla_xu(x)\cdot\nabla_yw(y)dy=0,$$
which I don't see to be true in general.

Thank you for any help and hints.