- #1
PeteSampras
- 44
- 2
Homework Statement
Problem 2 http://math.mit.edu/~jspeck/18.152_Spring%202017/Exams/Practice%20Midterm%20Exam.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%202017/Exams/Practice%20Midterm%20Exam_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 don't understand why change v(0) by sqrt{R} at the left and right sides.
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