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## 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.

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