Hello again guys and gals! I'm stuck on a problem that I thought was going to be simple for me to prove, but I was wrong (yet again). I will try not to be so long-winded this time (see my previous questions, and you'll see what I mean). Here's the problem (exactly):

Problem: Let $D=\big\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\big\}$ and let $u$ be a non-constant, real-valued $C^2$ function on a neighborhood of $D$ which satisfies $u(x,y)=0$ for all $(x,y)\in\partial D$. Prove that $$\iint_D\!u\,\Delta u\,dA<0.$$

I've recently proven Green's first and second identities – as an attempt to sketch the entire proof, let $\Omega\subset\mathbb{R}^3$ (the proof easily generalizes to $n$ dimensions, but I'm stating everything relative to $3$-space for reference purposes coupled with the problem I stated above) be an open, bounded subset with smooth boundary, and let $u,v:\Omega\rightarrow\mathbb{R}$ be $C^2(\Omega)$ scalar functions. First we consider the vector field $\boldsymbol{F}=u\,\nabla v$ and we prove the identity div$(\boldsymbol{F}):=\nabla\cdot\boldsymbol{F}=u\Delta v+\nabla u\nabla v$, whereas $\Delta=\nabla^{2}$ is the Laplace operator. Then, by application of the Divergence Theorem, we have

$$\text{Green's First Identity: } \int_\Omega u \, \Delta v \, dV + \int_\Omega \nabla u\cdot\nabla v \, dV = \int_{\partial\Omega} u \dfrac{\partial v}{\partial\boldsymbol{\nu}}\,dA,$$ where $\dfrac{\partial v}{\partial\boldsymbol{\nu}} = \nabla v\cdot\boldsymbol{\nu}$ is the outward normal derivative (Thus $\dfrac{\partial v}{\partial\boldsymbol{\nu}}$ is the directional derivative of $v$ in the direction of the outward normal to $\partial\Omega$).

Thus, by reversing the roles of $u,v$, using the identity above, applying the Divergence Theorem in this case, and then subtracting the overall result with the expression we derived in Green's First Identity above yields Green's Second Identity.

As far as some preliminary work concerning the proof for the problem stated above, I keep thinking to use Green's first identity, and I started re-working through the proof of Green's First Identity with the vector field $\boldsymbol{F}=u\nabla u$, instead. My troubles are arising when trying to bring in the assumptions, while using Green's first identity, to prove this. What I was thinking is to prove this by contradiction, and assume ${\displaystyle{\iint_D u\Delta u\,dA\geq0}}$. Then use Green's first identity with $\Omega\supset D$ being an open, bounded set, and then since $u(x,y)=0$ whenever $x,y\in\partial D$ we can take $u(x,y)=0$ on $\partial\Omega$ which means ${\displaystyle{\int_{\partial\Omega} u \dfrac{\partial u}{\partial\boldsymbol{\nu}} \, dA}}=0$ [?] (I think I can derive a contradiction from here, but I'm not sure if I'm correct with the previous statement). Any help as to proving this is GREATLY appreciated, as one can see, I'm now stumped on this!!! I'll keep working in the meantime, but much thanks goes out for any hints, tips, suggestions, answers, recommendations, etc., that are provided!

As a final remark, this is an old prelim problem from the University of Pittsburgh Prelim Exam of 2010 -- click here for the problem (see the very last problem on the exam).

  • $\begingroup$ As a comment following the post of Umberto P.'s proof below - as well as after I wrote up own proof based on him setting me straight (and checking the statement w/ some specific cases for $u$) - please ignore the statement, "since $u(x,y)=0$ whenever $x,y\in\partial D$ we can take $u(x,y)=0$ for $x,y\in\partial\Omega$," above. The correction in lieu of Umberto's proof below is that since $u$ is non-constant on $\Omega\supset D$, then is it certainly non-constant on $D$ as well. This was my "foul-up" and my apologies goes out for any confusion brought upon from above. $\endgroup$ – Procore Apr 18 '16 at 1:05
  • $\begingroup$ Several times you wrote \int\!\!\int. I changed those to \iint, with two consecutive "i"s, with the results that you see. If I'm not mistaken, in a displayed as opposed to inline context, \iint (with two consecutive "i"s) has the same effect as \int \!\!\! \int. I also put in proper use of \text{} and a small space before dA and \Delta f and \nabla f where those are preceeded by something other than a binary operator. $\endgroup$ – Michael Hardy Jun 5 '18 at 18:11
  • $\begingroup$ @MichaelHardy The only thing I can say is thank you. Gotta love/hate the subtleties between LaTex and Mathjax. </iint> is what I'll use from now on, among other insightful suggestions. Looking back on this, I'm reminded of Royden's, $\textit{Real Analysis}$, reassuring the student's enjoyment out of proofs by contradiction. $\endgroup$ – Procore Jun 27 '18 at 2:07

It is rather straightforward. Since $u$ is nonconstant, $\displaystyle \int_D \nabla u \cdot \nabla u \, dx = \int_D |\nabla u|^2 \, dx \not= 0$ because otherwise $|\nabla u|$ would be identically zero. This integral is nonnegative, so you end up with $$\int_D u \Delta u \, dx < \int_D u \Delta u \, dx + \int_D |\nabla u|^2 \, dx = \int_{\partial D} u \frac{\partial u}{\partial \nu} \, dS = 0.$$

  • $\begingroup$ Sorry for my previous (now deleted) comment - overall, thank you for steering me away from the contradiction attempt to prove this, as I follow your proof completely. I knew I was making things too hard on myself (also yet again), but at least I was correct that using Green's First Identity was a step in the right direction. Thanks again! $\endgroup$ – Procore Apr 17 '16 at 23:21

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