1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Mean Value property (harmonic functions) with a source?

  1. Feb 1, 2012 #1
    Mean Value property (harmonic functions) with a source??

    1. The problem statement, all variables and given/known data

    I understand that the heat equation may yield ∂u/∂t=0 on the LHS and on the RHS we may still have Uxx+Q where Uxx is partial with respect to x twice and Q is a heat source. U in our case may be the temperature function.

    Now my teacher introduces the mean value property for harmonic functions and I ask, when there is a heat source the mean value property breaks down, is this because our RHS is not a harmonic function or am I wrong in assuming the property doesn't still hold. My reasoning is that if we take a cross section of let's say a cylinder with a wire (sourse) going through the middle and we take a point close to the source, create a ball around it, then due to the conservation of energy we will find that the point has a temperature hotter than the average temperature along the boundary of the ball we created.

    Much input is needed/appreciated as my teacher wasn't able to address my question.

    2. Relevant equations

    1.4 Mean-Value Property: If u is harmonic on B(a,r), then u equals the average of u over ∂B(a,r).

    3. The attempt at a solution

    Hours of thinking
  2. jcsd
  3. Feb 1, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    Re: Mean Value property (harmonic functions) with a source??

    Nothing wrong with your thinking. If a function is harmonic then Uxx=0. If your equation is Uxx+Q=0 and Q is nonzero then you have a heat source and U is no longer harmonic. So the mean value property no longer applies. U is harmonic only where Q=0.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook