Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 5 / b

Write the elliptic differential operator as $L=\Delta +r{\partial }_x+s{\partial }_y$. At an interior maximum, $u_x=u_y=0$ and $u_{xx}+u_{yy}\le 0$. Therefore $Lu\le 0$ there, contradicting $Lu=f>0$.

The extreme value theorem ensures that a maximum exists on the compact closed disk. It cannot lie in the interior, so the maximum is attained on $\partial D$. The drift terms disappear at a stationary point, which is why the strict-sign proof is so short.

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 $v(x,y)=e^{kx}$,Consequently $u_{ϵ}=u+ϵv$ satisfies $L{u}_{ϵ}=f+ϵLv>0$. Part (i) applies, so for every $z\in \overline{D}$,Letting $ϵ\downarrow 0$ givesThus the maximum is attained on the boundary also when $f\ge 0$. This is the exponential perturbation proof of the weak maximum principle with drift. It does not require differentiating the continuous coefficients $r,s$.