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PDE harnack

  • #1

Homework Statement


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##

Homework Equations


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

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:

Answers and Replies

  • #2
haruspex
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why change v(0) by sqrt{R} at the left and right sides.
I agree with you, the algebra is flawed. I see no way to fix it.
 

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