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Let $u\in C^2(D)$, $D$ is the closed unit disk in $\mathbf{R}^2$. Assume that $\Delta u>0$. Show that $u$ cannot have a maximum point in $D\setminus\partial D$.

This statement is in a calculus book, after the discussion of extremal values of multivariable functions. So my guess is that I should use the Hessian of $u$ somehow. I started to proof indirectly. Assume that $(x_0,y_0)\in D\setminus\partial D$ is a maximum point. Then $\frac{\partial u}{\partial x}(x_0,y_0),\,\frac{\partial u}{\partial y}(x_0,y_0)=0$. Now I want to investigate the positive/negative definiteness of Hessian and deduce contradiction, but I got stuck.

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If this is a maximum, what can you say about definiteness of the Hessean matrix? And starting at the other hand, how is $\Delta u$ computed from the Hesse matrix? –  Hans Engler Nov 28 '12 at 18:06
    
@HansEngler $\Delta u=tr(Hesse)=\lambda_1+\lambda_2$ (sum of eigvalues). If it is a maximum, then Hesse is negative (semi?)definite, it implies $\lambda_1<0,\,\lambda_2<0$ (or $\leq 0$ ?). So $\Delta u\leq 0$, a contradiction, if I'm correct. Thanks. –  vesszabo Nov 28 '12 at 19:35

1 Answer 1

up vote 2 down vote accepted

To not leave this look unanswered: suppose $a\in \Omega$ is a point of local maximum. (I change the notation to reserve $D$ for derivatives.) Since $u$ is $C^2$ smooth, we have second-order Taylor expansion $$u(x)=u(a)+Du(a) (x-a)+ \frac12 (x-a)^T D^2u(a) (x-a)+o(|x-a|^2)$$ The maximality implies that $Du(a)=0$ and $D^2u(a)$ is negative semidefinite. Therefore, the trace of $D^2u(a)$ is non-positive. This trace is $\Delta u(a)$.

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