# PDE harnack

## 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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