Solution: Is $u$ a Constant Function on $S$?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: Let $S$ be a solid region in $\mathbb{R}^3$ with outward unit normal $\mathbf{n}$, and let $u$ be a function that satisfies Laplace's equation ($\nabla^2 u = 0$) and the boundary condition $\nabla u(x,y,z)\cdot\mathbf{n} = 0$ for $(x,y,z)\in\partial S$. Show that $u$ is a constant function.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

Delays this week brought to you by filling out paperwork for a new job and other annoying important things. >_>

No one answered this week's problem. You can find the solution below.

[sp]We will need to use Green's identity to answer this problem. Recall that if $f$ and $g$ are continuously differentiable scalar-valued functions on an open set $U$ in $\mathbb{R}^3$, then for any solid region $S$ contained in $U$,
\[\iiint\limits_S \nabla f\cdot\nabla g \,dV = \iint\limits_{\partial S} f\nabla g\cdot \mathbf{n}\,d\sigma -\iiint\limits_S f\nabla^2 g\,dV.\]
In our case, we take $f=g=u$ to see that
\[\iint\limits_S \nabla u \cdot \nabla u \,dV = \iint\limits_{\partial S} u\nabla u \cdot \mathbf{n}\,d\sigma - \iint\limits_S u\nabla^2 u \,dV.\]
Since $\nabla u\cdot\mathbf{n} = 0$ on $\partial S$, the first integral is zero. Likewise, since $u$ satisfies Laplace's equation $\nabla^2 u = 0$, it follows that the second integral is zero. Therefore, we now see that
\[\iiint\limits_S \nabla u \cdot \nabla u \,dV = \iiint\limits_S \|\nabla u\|^2\,dV = 0.\]
Since the integrand is continuous and non-negative, it must follow that $\nabla u =0$ which implies that $u$ is constant on $S$.$\hspace{.25in}\blacksquare$ [/sp]