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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 $r,s$.

The weak maximum principle for elliptic operators gives the clipped boundary boundsIf $c=0$, constants solve the equation, and applying these bounds after subtracting the boundary maximum or minimum gives the sharper conclusion that both extrema occur on the boundary. For $c<0$, the clipping by zero is necessary in the general statement.

To prove the upper bound, write $B=\Vert b_1{\Vert }_{\infty }$ and $C_0=\Vert c{\Vert }_{\infty }$, and choose $\alpha >0$ large enough that $\lambda {\alpha }^2-B\alpha -C_0>0$. For $\varphi (x)=e^{\alpha x_1}$, uniform ellipticity givesPut $u_{\epsilon }=u+\epsilon \varphi$. Then $L{u}_{\epsilon }>0$. If $u_{\epsilon }$ had a positive maximum at an interior point, its gradient would vanish there and its Hessian matrix would be negative semidefinite. Only the symmetric part of $a_{ij}$ contracts with this Hessian, and its positive definiteness implies $a_{ij}{D}_{ij}{u}_{\epsilon }\le 0$. Also $c{u}_{\epsilon }\le 0$. These facts would give $L{u}_{\epsilon }\le 0$, a contradiction.

Since $\overline{\Omega }$ is compact and $u_{\epsilon }$ is continuous on it, its maximum exists. It is therefore either nonpositive or attained on the boundary, proving ${\sup }_{\Omega }{u}_{\epsilon }\le \max \{0,{\sup }_{\partial \Omega }{u}_{\epsilon }\}$. The exponential is bounded on the bounded domain; letting $\epsilon \downarrow 0$ proves the upper estimate. Apply the same argument to $-u$ for the lower estimate. This is the exponential perturbation proof of the weak maximum principle with drift and requires no continuity of the coefficient functions.

Assume real bounded coefficients, with a principal part defining a uniformly elliptic operator, say $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ for some $\lambda >0$, and assume $c\le 0$. No smoothness of the boundary is needed for this classical weak maximum principle for elliptic operators. Its conclusion isIf $c=0$, the zero may be omitted. Only the symmetric part of the principal coefficients contributes to the Hessian matrix contraction.

For a strict inequality $Lu>0$, a positive interior maximum is impossible: there $Du=0$, $D^{2}u$ is negative semidefinite, and $cu\le 0$, so $Lu\le 0$. To reduce the non-strict case to this one, let $q(x)=e^{\kappa x_1}$. Boundedness of $b^1,c$ permits choosing $\kappa$ so large thatThus $L(u+\epsilon q)>0$. The preceding maximum argument and continuity on the compact closure giveLet $\epsilon \downarrow 0$; the exponential is bounded on the bounded domain. This is the exponential perturbation proof of the weak maximum principle with drift, with the zeroth-order term included. For $c=0$, any interior maximum, of either sign, is impossible for the perturbed function, giving the sharper boundary bound. The sign restriction on $c$ is essential: a positive first Dirichlet Laplacian eigenfunction violates the conclusion for $L=\Delta +{\lambda }_1$.

Assume uniform ellipticity $a^{ij}{\xi }_i{\xi }_j\ge \lambda |\xi {|}^2$ with $\lambda >0$, bounded coefficients, and $c\le 0$. The weak maximum principle for elliptic operators isWhen $c=0$, zero can be omitted. The sign restriction is necessary, as shown by the failure of the weak maximum principle with a positive zeroth-order coefficient.

Let $M=\max (0,{\sup }_{\partial \Omega }u)$ and $w=u-M$, so $Lw=Lu-cM\ge 0$ and $w\le 0$ on the boundary. For sufficiently large $k$, $q=e^{k{x}_1}$ satisfiesIf $w$ is positive somewhere, $w+\epsilon q$ retains a positive interior maximum above all its boundary values for small $\epsilon >0$. At that point its gradient is zero and its Hessian matrix is negative semidefinite; ellipticity and $c\le 0$ give $L(w+\epsilon q)\le 0$, contradicting its strict positivity. This exponential perturbation proof of the weak maximum principle with drift also works with $M={\sup }_{\partial \Omega }u$ of either sign when $c=0$.