1. Not finding help here? Sign up for a free 30min 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!

Solution to Laplace Equation across Boundaries

  1. Aug 21, 2008 #1
    Hi everyone,

    I read somewhere that solutions to Laplace's Equation must agree at a shared boundary.
    So for example if [tex]\Phi_{1}[/tex] and [tex]\Phi_{2}[/tex] are two solutions to the Laplace equation in two different regions which share a boundary, then on the boundary [tex]\Phi_{1}[/tex] = [tex]\Phi_{2}[/tex]
    Is this true?
    Can you help me see why?

    Thanks in advance,
     
  2. jcsd
  3. Aug 21, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In order that a function satisfy Laplace's equation, it must be differentiable. Otherewise the derivatives don't make sense! In particular, that means it must be continuous.
     
  4. Aug 21, 2008 #3
    Though I dont disagree with your statement HallsofIvy, it wouldn't necessarily apply in this situation. One function [tex]\Phi_{1}[tex] is continuous in region 1 and the other [tex]\Phi_{2}[tex] is continuous in region 2, where region 1 and 2 only share a boundary.

    Would it be true that these two different functions have the same value on this boundary, given that they both satisfy the Laplace equation?
     
  5. Aug 22, 2008 #4

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

    If this true? Let f : S1->R = 37. Let g: S2->R = 92. Both f and g satisfy Laplace's equation on S1 and S2 respectively. Suppose S1 = {(x,y) | x^2 + y^2 <= 2}, and S2={(x,y) | 1< x^2 + y^2 <= 2}.
    S1 and S2 share a boundary, namely x^2 + y^2 = 2, but f and g are different functions on this boundary.
     
  6. Aug 22, 2008 #5

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    No, there is NO requirement that solutions of Laplace's equations in different regions (with shared boundary) coincide on the boundary. Rather, Laplace's equation itself makes no requirement. Within a region, a solution must be differentiable, but at the boundaries, it is usually not, and can even be discontinuous.

    The question as to whether two adjoining solutions must be "stitched together" in a continuous way depends entirely on the physical problem at hand. In electromagnetism, for example, the electric potential must be continuous in most cases; therefore, we must choose Laplace solutions that match up on the boundaries. Even so, the potential will usually NOT be differentiable at the boundaries (consider the potential of a conducting shell, which is constant inside, but falls as 1/r outside).

    There are some cases, however, in which the potential is also discontinuous: e.g., a dipole layer. A dipole layer is a surface which is considered to have a certain dipole moment per unit area (perpendicular to the surface). Across a dipole layer, the potential makes a discontinuous jump in proportion to the local dipole moment.
     
  7. Aug 26, 2008 #6
    Thank you kindly for your replies.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?