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Maximum principle for Delta u >0

  1. Sep 21, 2010 #1

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

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

    Thanks in advance!

    Last edited: Sep 21, 2010
  2. jcsd
  3. Sep 23, 2010 #2
    Yes. Expand upto the first order , assume the contrary & apply the mean value theorem.
  4. Sep 25, 2010 #3
    I was given the following explaination, though I really don't understand what's going on with eqs. (3) and (4):

    Using Taylors theorem, write

    [tex] 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) [/tex]

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

    [tex]Du(x_0)=0, \qquad (1) [/tex]

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

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

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

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

    which contradicts the assumption that [tex]\Delta u\geq 0[/tex]

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

    Thank you for your time,

    Last edited: Sep 26, 2010
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