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If $u_1,u_2$ satisfy the same boundary value problem, their difference $w=u_1-u_2$ satisfies $Lw=0$ in $D$ and $w=0$ on the boundary, by linearity. Part (b)(ii) givesApply it again to $-w$, which also satisfies $L(-w)=0$ and zero Dirichlet boundary conditions. This gives $-w\le 0$. Hence $w=0$ throughout $\overline{D}$, provingThis proves uniqueness whenever a solution exists; existence is the separately supplied assumption. When the drift coefficients vanish, the same argument uses part (a).

The maximum principle for subharmonic functions here states that if $u\in C^2(D)\cap C(\overline{D})$ and $\Delta u=f\ge 0$ in $D$, thenIf $f\le 0$, apply this to $-u$ to obtain the corresponding minimum principle. In particular a harmonic function has both its maximum and minimum on the boundary. Without a sign condition on $f$, this conclusion need not hold.

First suppose $\Delta u>0$. By compactness and continuity, $u$ attains a maximum on $\overline{D}$. At an interior maximum, the first derivatives vanish and the Hessian matrix is negative semidefinite. Its trace is then nonpositive, contradicting $\Delta u>0$. Thus a maximum lies on the boundary.

For the non-strict case put $u_{ϵ}=u+ϵ(x^2+y^2)$, with $ϵ>0$. Then $\Delta u_{ϵ}=f+4ϵ>0$, so the strict case yields, for $z\in \overline{D}$,Let $ϵ\downarrow 0$. This gives $u(z)\le {\max }_{\partial D}u$, proving the theorem. The argument proves the weak maximum principle; it does not claim that every maximizer must be on the boundary when $u$ is constant.

The Picard-Lindelöf theorem gives local existence and uniqueness for an initial value problem when $f$ is continuous on a rectangle around $(a,b)$ and is Lipschitz continuous in its second variable there, with a constant uniform in the first variable. A sufficient hypothesis is that $f$ and $\partial f/\partial y$ are continuous on that rectangle.

More quantitatively, if the rectangle is $|x-a|\le h_0$, $|y-b|\le k$, and $|f|\le M$, a sufficiently small $h>0$ with $h\le h_0$ and $hM<k$ yields a unique differentiable solution on $[a-h,a+h]$ remaining inside the vertical strip. The proof uses the integral equationand Picard iteration, or the Banach fixed-point theorem after shrinking $h$ to make the integral operator a contraction. The continuity and Lipschitz hypotheses must hold near the actual initial point.