Let $X=x/\epsilon$, $a(X)=1/d(X)$, and seek
$$
u=u_0(x,t)+\epsilon u_1(x,X,t)+\cdots
$$
with periodic correctors. The leading cell equation makes $u_0$ independent of $X$. At the next order, continuity of microscopic heat flux gives
$$
a(X)(u_{0x}+u_{1X})=J(x,t).
$$
Averaging over one period and using $\langle u_{1X}\rangle=0$ yields
$$
u_{0x}=J\int_0^1d(X)dX=J\overline d.
$$
Thus the <periodic homogenization of a diffusion equation> has effective diffusivity $1/\overline d$:
$$
\boxed{u_{0t}=\frac1{\overline d}u_{0xx}.}
$$

For the sawtooth profile, the two triangular areas give
$$
\overline d=\int_0^p\frac{2X}{p}dX
+\int_p^1\frac{2(1-X)}{1-p}dX=p+(1-p)=1.
$$
The homogenized problem is therefore the unit-diffusivity <heat equation>. Its half-line step solution is
$$
\boxed{u(x,t)=T_0\operatorname{erfc}\left(\frac{x}{2\sqrt t}\right)
=T_0\left[1-\operatorname{erf}\left(\frac{x}{2\sqrt t}\right)\right].}
$$
It has the required initial and boundary limits, and direct differentiation verifies $u_t=u_{xx}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 336 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-336/2/b/solution