# Maximum principle question in partial differential equation

Problem is

Let $U$ be a bounded domain in $\mathbb{R}^{n}$ and $\vec{b} : \mathbb{R}^{n} \to \mathbb{R}^{n}$ and $g: \mathbb{R}^{n} \to \mathbb{R}$ be continuous. Show that there can be at most one solution $u \in C^{2}(U) \cap C(\bar{U})$ of the equation \ $-\bigtriangleup u + \bar{b}\cdot Du = 0$ in $U$ and $u = g$ in $\partial U$.

Suppose U is bounded and open. Suppose $u$ has a maximum or minimum value $z_0$ at $x_{0} \in U$. Then, $\bigtriangleup u(x_{0}) \leq 0$, and $Du = 0$, i.e. zero $n-$tuple on $\mathbb{R}$. Hence $-\bigtriangleup u (x_{0}) + \overrightarrow{b}(x) \cdot Du(x_{0}) = -\bigtriangleup u(x_{0}) = 0$. Since $Du(x_0) = \bigtriangleup u(x_0) =0$ and $u$ is continuous, there exists $\epsilon > 0$ such that $\forall x \in B(x_0,\epsilon)$, $u(x) = z_0$. Hence If U is connected, then for every point $x \in U, u(x) = z_0$. In this case $u$ has a constant solution if $g =z_{0}$ on $\partial U$. (If not, there is no solution.)
If U is not connected, but collection of connected subset $\{U_{i}\}$, then each of $U_{i}$ has a constant solution of u. Hence $u$ is step function. Also it is unique since if there is another step function solution $v$ exists, then $v(x)=u(x)$ for all $x \in U$ and $Dv = Du = \bigtriangleup v = \bigtriangleup u = 0$. Hence the equation has at most one solution exists.
• It looks wrong, why does there exist an open ball where $u(x)=z_0$? – JLA Jan 24 '14 at 6:48
• That's what I want to know, too, why the open ball where $u(x) = z_0$? – Robert Lewis Jan 24 '14 at 6:56
• @JLA Hmm, Since Derivative and second derivative is 0 at $x_0$, so it can't be changed near $x_0$. – user122655 Jan 24 '14 at 7:19
• @user122655 $x^3$ is a counterexample to that statement. – JLA Jan 24 '14 at 15:00