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