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Green's first identity at the boundary

  1. Nov 20, 2015 #1
    As required by the Green's identity, the integrated function has to be smooth and continuous in the integration region Ω.

    How about if the function is just discontinuous at the boundary?

    For example, I intend to make a volume integration of a product of electric fields, the field function is well-behaved in the volume but discontinue at the boundary (for the normal component of the field) due to a difference of permitivity occurs.

    In this case, can I still use the Green's first identity?
    If I can, what kind of surface divergence I should use at the boundary?

    Thanks for your help
     
  2. jcsd
  3. Nov 20, 2015 #2

    RUber

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    If your function is only discontinuous on the boundary, then I am pretty sure you can define your region Omega as the open region not including the boundary.
    To determine the divergence, I have seen it done using:
    ## \hat n \times ( \vec H_1 - \vec H_2 ) = \vec J ,## with ##\hat n ## the unit normal to the surface pointing into region 1 and ##\vec J## the current present on the boundary described by ##f##.
    Then, defining exterior and interior limit functions:
    ##
    \begin{align*} w_e (x) &= \lim_{p \to x} w(p) \quad p \text{ exterior to } \partial \Omega,\\

    w_i (x) &= \lim_{q \to x} w(q) \quad q \text{ interior to } \partial \Omega.

    \end{align*}
    ##
    Thus, the boundary condition is expressed as:

    ## \frac{\partial w_e}{\partial n} - \frac{\partial w_i}{\partial n}= f(x) \qquad \text{ for } x \in \partial \Omega.##
     
  4. Nov 26, 2015 #3
    Thanks to your reply

    Em...Exactly same to what I'm thinking about. Although the function is discontinuous at the boundary, that doesn't mean this differential result is not well behaved. Then we should be able to extend the function safely and then use the Green's identity.
     
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