= Solution
Differentiate the uniquely determined lower solution and the zero upper solution:
$$
u_y(x,y)=\begin{cases}-2\pi\cos(\pi x)\cos(\pi y),&y<0,\\0,&y>0.\end{cases}
$$
Their one-sided traces are
$$
\boxed{u_y(x,0^-)=-2\pi\cos(\pi x),\qquad u_y(x,0^+)=0.}
$$
They disagree, for example at $x=0$, so the <normal derivative> is not continuous across the interface. The jump, defined as upper trace minus lower trace, is $2\pi\cos(\pi x)$. It vanishes at the two interior points $x=\pm1/2$, but not identically on the interface.
This illustrates that <continuous transmission need not preserve normal derivatives>. In fact, at any point where these traces differ, the two-sided first derivative with respect to $y$ does not exist. The resulting function is continuous and smooth separately on each open side, but is not globally $C^1$. Consequently a globally classical $C^2$ solution satisfying all these data cannot exist. The calculation in part (c) evaluates the smooth lower formula on the interface; it does not supply the missing cross-interface regularity.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2013-mixed-type-interface.png]
{title=Continuous piecewise solution and unequal normal derivative traces}
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