Maximum principle for Delta u >0

[loesung]

Hello!

I would like to show the following: u\in C^2(U) \cap C(\bar{U}) satisfies \Delta u(x)>0 for any x\in U, then \max_U u cannot be achieved by any point in U. Here, u\in \mathbb{R}^n, i.e. it's not complex valued.

Apparently, one can use the Taylor expansion formula to show this. But how?

Thanks in advance!

Los

 

[Eynstone]

Apparently, one can use the Taylor expansion formula to show this. But how?
Los

Yes. Expand upto the first order , assume the contrary & apply the mean value theorem.

 

[loesung]

I was given the following explanation, though I really don't understand what's going on with eqs. (3) and (4):

Using Taylors theorem, write

u(x)=u(x_0)+\nabla u(x_0)\cdot(x-x_0)+\frac{1}{2}(x-x_0)^T\nabla^2 u(x^*)(x-x_0)

Because we suppose a max is obtained, we have from vector calculus that

Du(x_0)=0, \qquad (1)

\nabla^2 u (x_0)\leq 0 \mbox{ as a matrix}, \qquad (2)

However, somehow it's important to note, for reasons that to me are not clear, that u is positive definite...:


v^T\nabla^2u(x_0)v\leq 0 \mbox{ for all }v\in\mathbb{R}^n,\qquad (3)

and so u(x_0)= \text{tr}(D^2u(x_0))\leq 0, \qquad (4)

which contradicts the assumption that \Delta u\geq 0

In particular I don't understand where (1) and (2) come from, and supposedly (2)\Rightarrow (3)...? I would appreciate if this could be cleared up!

Thank you for your time,



Los