= Solution
{c}
On the two independent variables, the total <derivative> is $D_x=\partial_x+\epsilon^{-1}\partial_X$. Multiplication of $D_x(aD_xy)=f$ by $\epsilon^2$ gives exactly the stated multiple-scale equation.
At order one, $\partial_X(a\,\partial_Xy_0)=0$. Integrating twice gives
$$
y_0=B(x)+A(x)\int^X\frac{dS}{a(x,S)}.
$$
The periodic reciprocal coefficient has strictly positive mean. Unless $A=0$, the <integral> has a nonzero linear drift in $X$, violating the periodic cell ansatz and creating secular growth when $X=x/\epsilon$. Thus $A=0$ and $y_0=y_0(x)$.
At order $\epsilon$,
$$
\partial_X\{a(y_{1X}+y_0')\}=0,\qquad
y_{1X}=\frac{C(x)}{a(x,X)}-y_0'.
$$
Periodicity requires the <derivative> to have zero cell average. Consequently
$$
0=C(x)\int_0^1a^{-1}(x,X)dX-y_0',
\qquad C(x)=\hat a(x)y_0'(x).
$$
At order $\epsilon^2$, the equation is
$$
\partial_X\{a(y_{2X}+y_{1x})\}
+\partial_x\{a(y_{1X}+y_0')\}=f(x,X).
$$
The second flux inside braces is $C(x)$, independent of $X$. Integrating over one <period> annihilates the first <derivative>, by periodicity, and yields $C'(x)=\int_0^1f(x,X)dX$. Hence
$$
\boxed{\frac d{dx}\bigl(\hat a(x)y_0'(x)\bigr)=\hat f(x),\quad
\hat a(x)=\left(\int_0^1a(x,X)^{-1}dX\right)^{-1},\quad
\hat f(x)=\int_0^1f(x,X)dX.}
$$
The effective transport coefficient is the <harmonic mean> of $a$ over the fast <period>; the source is averaged arithmetically. Smoothness sufficient for these formal differentiations and uniform positivity on the cells are understood.
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