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PDE harnack

  1. Jun 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Problem 2 http://math.mit.edu/~jspeck/18.152_Spring 2017/Exams/Practice Midterm Exam.pdf
    "Let ##u## such that ##Laplacian( u)=0##
    Show if ##u \le \sqrt{x}##, then ##u=0##

    2. Relevant equations
    At the solution http://math.mit.edu/~jspeck/18.152_Spring 2017/Exams/Practice Midterm Exam_Solutions.pdf
    define ##v=u + \sqrt{R}##

    3. The attempt at a solution

    equation 10 says, by Harnack
    ##\frac{R(R-|x|)}{(R+|x|)^2}v(0) \le v(x) \le \frac{R(R+|x|)}{(R-|x|)^2}v(0) ##

    but my question is, why in formula 10 change v(0) by sqrt{R} at the left and right side?
    ## ( \frac{R(R-|x|)}{(R+|x|)^2}-1) \sqrt{R} \le u(x) \le ( \frac{R(R+|x|)}{(R-|x|)^2}-1) \sqrt{R}##

    I understand that change ##v(x) \to u(x) + \sqrt{R}##, but, i dont understand why change v(0) by sqrt{R} at the left and right sides.
     
    Last edited: Jun 24, 2017
  2. jcsd
  3. Jun 24, 2017 #2

    haruspex

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    I agree with you, the algebra is flawed. I see no way to fix it.
     
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