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Question about function defined in a region using Green's identities.

  1. Jul 31, 2010 #1
    I want to verify my understand of this.

    Let [itex] u[/itex] defined in region [itex] \Omega[/itex] with boundary [itex] \Gamma[/itex].

    If [itex] u = 0 \hbox { on the boundary } \Gamma[/itex], then [itex] u = 0 \hbox { in the region } \Omega[/itex].

    The way to look at this, suppose [itex] u[/itex] is function of x component called Xand y component called Y. So either u=XY or u=X+Y.

    1) If u=XY, u=0 mean either X or Y equal zero on [itex] \Gamma[/itex]. That can only happen if X or Y is identically equal to zero within the range of x or y.

    2) If u=X+Y, u=0 means both X and Y identically equal to zero within the range of x and y.

    Am I correct?

    I am confuse, if I declare u=0 only on [itex] \Gamma[/itex] and equal to x+y anywhere else, then the assertion cannot not be true!!!! Please help.
    Last edited: Jul 31, 2010
  2. jcsd
  3. Aug 1, 2010 #2
    Please give more information. What are u and Ω? It looks like Ω is some nice subset of R2, but what conditions are there on u? Why is u = XY or u = X + Y?

    You are certainly correct if there is no restriction on u. However, the statement is true if, say, u is harmonic.
  4. Aug 1, 2010 #3
    Thanks for your reply, I am getting desperate!!!

    [itex]\Omega[/itex] is a simple or multiply connect open region and [itex]\Gamma[/itex] is the boundary of [itex]\Omega[/itex] on xy plane. [itex]u[/itex] is a harmonic function on [itex]\Gamma[/itex]. Why if u is harmonic function, then u=0 in [itex]\Omega[/itex]

    The example of u=XY or u=X+Y is only an example that I gave where u is a function of both [itex]X(x) \; \hbox{ and }\; Y(y)[/itex]. There u can be XY or X+Y to cover all cases.
  5. Aug 1, 2010 #4
    That statement is true if Ω is also bounded. The point is that a harmonic function cannot have any local maxima or minima, so if u has a maximum or minimum in the closure of Ω (which is the case if Ω is bounded, because then its closure is compact), then that must be on the boundary Γ. But if u is 0 on Γ, then for all x in Ω, 0 ≤ u(x) ≤ 0.
  6. Aug 1, 2010 #5
    Ω is totally surounded by Γ if that is what you mean. Think of Γ is a circle surounding Ω in the middle where Ω is a simple region.

    Can you explain the harmonic function a little bit more or give me a link on this? My book and a few links I looked up only said harmonic function has to have continuous first and second derivative and [itex]\nabla u =0[/itex].

    BTW, how do you get symboms like Ω and Γ direct without going into latex?
  7. Aug 1, 2010 #6
    By bounded I mean the usual definition that it doesn't go off to infinity. For example, the statement isn't true if Ω is the half-plane x > 0. (Then the function u(x, y) = x is harmonic, takes the value 0 on the boundary of Ω, but is not zero on Ω.) But if Ω is bounded, then its closure (which is the union of Ω and Γ) is compact, and then it works.

    Here's a Wikipedia link.

    (As for the symbols, probably the easiest way is to look up the name on Wikipedia and copy+paste. There is also Unicode Character Search.)
  8. Aug 1, 2010 #7
    After I wrote the last post, I decided to move on to the next section in the book. They start talking Gauss mean value and maximun minimum principle, I guess the question is ahead of the chapter, need to study the next section to understand the question.


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