= Solution
Use an exponential perturbation whose image under $L$ is strictly positive everywhere. The coefficient $r$ is bounded on the closed disk; put $R=\max_{\overline D}|r|$ and choose $k>R$, $k>0$. For $v(x,y)=e^{kx}$,
$$
Lv=(k^2+kr(x,y))e^{kx}\geq k(k-R)e^{kx}>0.
$$
Consequently $u_\epsilon=u+\epsilon v$ satisfies $Lu_\epsilon=f+\epsilon Lv>0$. Part (i) applies, so for every $z\in\overline D$,
$$
u(z)+\epsilon v(z)\leq\max_{\partial D}(u+\epsilon v)
\leq\max_{\partial D}u+\epsilon e^k.
$$
Letting $\epsilon\downarrow0$ gives
$$
\boxed{u(z)\leq\max_{\partial D}u.}
$$
Thus \b[the maximum is attained on the boundary also when $f\geq0$]. This is the <exponential perturbation proof of the weak maximum principle with drift>. It does not require differentiating the continuous coefficients $r,s$.
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