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Evans-pde-laplace eqn

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data

    From the book Evans-PDE, p.24, equation (12),
    It is written that

    [tex]C ||D^2f||_{L_{\infty}(R^n)} \int_{B(0,\epsilon)}|\Phi(y)|dy
    \\ \leq \begin{cases} C \epsilon^2 |\log{}\epsilon| & (n=2) \\ C \epsilon^2 & (n \geq 3) \end{cases}[/tex]

    How is this?

    2. Relevant equations

    [tex]\Phi(y) = \begin{cases} -\frac{1}{2\pi}\log{}|y| & (n=2) \\ \frac{1}{n(n-2)\alpha(n)} \frac{1}{|y|^{n-2}} & (n \geq 3) \end{cases}[/tex]
    for [tex]y \in \mathbb{R}^n-0[/tex] (the fundamental solution of Laplace's equation).

    [tex]\alpha(n)[/tex] is the volume of the unit ball in [tex]\mathbb{R}^n[/tex].

    [tex]C[/tex] is a constant.

    3. The attempt at a solution

    Take n=3. Then,

    [tex]\int_{B(0,\epsilon)}|\Phi(y)|dy = C\int_{0}^{\epsilon}\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{1}{r} d\theta d\phi dr[/tex]

    [tex]= C \int_{0}^{\epsilon}\frac{1}{r} dr[/tex]

    [tex]= C (\log{}(\epsilon)-\log{}(0)) = \infty[/tex]

    What am I doing wrong?
    Last edited: Aug 10, 2010
  2. jcsd
  3. Aug 11, 2010 #2


    User Avatar
    Homework Helper

    I think you got your Jacobian transformation wrong. If memory serves:
    dxdydz=r^{2}\sin\theta drd\theta d\phi
  4. Aug 13, 2010 #3
    ya you're right. I didn't account for the change of coordinates. thanks.
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