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Question on why the book claimed Green's function =< 0.

  1. Aug 9, 2010 #1
    Green's function [itex]G(x_0,y_0,x,y) =v(x_0,y_0,x,y) + h(x_0,y_0,x,y)[/itex] in a region [itex]\Omega \hbox { with boundary } \Gamma[/itex]. Also [itex]v(x_0,y_0,x,y) = -h(x_0,y_0,x,y)[/itex] on boundary [itex]\Gamma[/itex] and both [itex]v(x_0,y_0,x,y) \hbox { and }h(x_0,y_0,x,y)[/itex] are harmonic function in [itex]\Omega[/itex]

    [tex]v=\frac{1}{2}ln[(x-x_0)^2 + (y-y_0)^2] [/tex]

    From the Max and Min proterty of Harmonic function in a region. The max and min value of the function is on the boundary of the region it is in.

    In this case, G is defined as G=v+h and h=-v on the boundary. G=0 on the boundary so both max and min equal to zero. Why is the book claimed G is negative or zero inside the region [itex]\Omega[/itex].

    The book stated G is harmonic function in [itex] \Omega \;[/itex] and G=0 on [itex]\Gamma[/itex]. That pretty much lock in G=0 in [itex] \Omega \;[/itex].

    See my post below what the book said word to word.

    If G=0 in [itex] \Omega \;[/itex], then it is pretty useless!!! I am confused!!! Please help.


    Last edited: Aug 10, 2010
  2. jcsd
  3. Aug 10, 2010 #2
    (x0, y0) is not in the domain of v, so there is no mystery. The function never attains a minimum, which is permissible because its domain is not compact (it excludes an isolated point).
  4. Aug 10, 2010 #3
    Thanks for the reply. Yes, I understand (x0,y0) is not in the domain. Let me just type exactly what the book written:

    Green's function [itex]G(x_0,y_0,x,y) =v(x_0,y_0,x,y) + h(x_0,y_0,x,y)[/itex] in a region [itex]\Omega \hbox { with boundary } \Gamma[/itex]. Also [itex]v(x_0,y_0,x,y) = -h(x_0,y_0,x,y)[/itex] on boundary [itex]\Gamma[/itex] and both [itex]v(x_0,y_0,x,y) \hbox { and }h(x_0,y_0,x,y)[/itex] are harmonic function in [itex]\Omega[/itex]

    [tex]v=\frac{1}{2}ln[(x-x_0)^2 + (y-y_0)^2] [/tex]

    Properties in question: [itex]G (x_0,y_0,x,y)\leq 0\; [/itex] for all (x,y) except [itex](x_0,y_0)[/itex]

    To proof this:

    G=v+h. Fix a closed disk [itex]D_R\;[/itex] centered at [itex](x_0,y_0)\;[/itex] and contained in [itex]\Omega\;[/itex]. Since h is harmonic in [itex]\Omega\;[/itex], it is continuous and hence bounded in [itex]D_R\;[/itex], say [itex] |h (x_0,y_0,x,y)|\leq M \hbox{ on } \Omega [/itex]. Now, v tends to [itex]-\infty\;[/itex] as (x,y) approaches [itex](x_0,y_0)[/itex]. So we can find [itex]0<r_0 < R[/itex] such that [itex]v (x_0,y_0,x,y)<-2M \hbox{ on } C_r[/itex] for all [itex]0<r_0 \leq r < R[/itex]. Hence [itex]G (x_0,y_0,x,y)\leq -M \hbox{ on } \;C_r\; [/itex], because G=h+v, [itex]v (x_0,y_0,x,y)<-2M \hbox{ and }|h (x_0,y_0,x,y)|\leq M \hbox{ on } C_r [/itex]. Denote the region [itex] \Omega [/itex] minus the disk of radius [itex] r\;[/itex] centered at [itex](x_0,y_0)\;\hbox { by } \Omega_r[/itex]. The boundary of [itex] \Omega_r \; [/itex] consists of [itex]\Gamma\; [/itex] and [itex]C_r\; [/itex]. The function [itex]G\; \hbox { is harmonic in }\; \Omega_r \;[/itex] and we just proved that it is [itex]\leq \;0\;[/itex] on it's boundary. By the maximum principle for harmonic functions, it follows that [itex] G \leq 0 \;\hbox { on } \Omega_r [/itex]. Since this is true for all [itex] 0<r \leq r_0 \;[/itex], letting [itex] r \rightarrow 0[/itex], we see that [itex] G \leq 0 \;\hbox { on } \Omega \; [/itex] minus [itex](x_0,y_0)\;[/itex]. This prove [itex]G (x_0,y_0,x,y)\leq 0\; [/itex] for all (x,y) except [itex](x_0,y_0)[/itex]

    the diagram is just a region [itex]\Omega[/itex] with boundary [itex]\Gamma[/itex]. Inside have a disk [itex]D_R\;[/itex] centered at [itex](x_0,y_0)[/itex]. then a circle [itex] C_r \;[/itex] centered at [itex](x_0,y_0)[/itex] inside [itex]D_R\;[/itex].

    That is all the book have, I typed word to word from the book. No more explainations of what are [itex] r, r_0 \hbox { and } R \;[/itex].
  5. Aug 10, 2010 #4
  6. Aug 11, 2010 #5
    What's the issue?
  7. Aug 11, 2010 #6
    All I remember, if it equals zero, then your doing greens theorem wrong or supposed to get zero. closed curve...
  8. Aug 11, 2010 #7
    No one mentioned Green's theorem...
  9. Aug 11, 2010 #8
  10. Aug 11, 2010 #9
  11. Aug 11, 2010 #10
    The issue is by definition of harmonic function in a region[itex]\Omega \hbox { with closed boundary } \Gamma [/itex], The max and min of the function are on the boundary. By definition of Green's function G is harmonic function and G=v+h where h=-v on boundary [itex]\Gamma[/itex] G=v+(-v)=0 on boundary. This mean the max and min of G are both equal to zero. By definition of the harmonic function, all the value of G inside the region lies between the max and min which mean the only value G in the region is zero!!!

    The book stated the property of G =< 0. So it is not zero. And the most important of it all, G is useless if it is zero in the region and on it's boundary!!!!
  12. Aug 11, 2010 #11
    Doesn't all that derivation (Laplacian apparently) and divergence involve his theorem? how is it totally unrelated?
  13. Aug 11, 2010 #12
    Green's Theorem used in multi-variables calculus is only the introduction of the Green's function. The whole thing is a way to find the value of the function inside a region by the values on the boundary. It get much much deeper than just:

    [tex] \int_{\Gamma} Mdx + Ndy= \int_{\Omega} \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} dx dy [/tex]

    This is like the first page of the whole nightmare I am going through!!!!
  14. Aug 11, 2010 #13
    The max and min of the function, if they exist, are on the boundary. And G is not defined on all of Ω; it's not defined at (x0, y0). And even though Ω ∪ Γ (where Γ is the boundary of Ω) may be compact, the domain is actually (Ω ∪ Γ) \ {(x0, y0)}, which isn't compact because of the missing isolated point. Thus the function G may not attain a minimum.

    Here's another example of a nonzero harmonic function that's zero on the boundary of its domain: Let f be the function f(x, y) = x, defined on the half-plane x ≥ 0. Then the boundary of the domain of f is the line x = 0, and f is certainly harmonic, but not zero. Notice that this function has no maximum.

    Also, a harmonic function's max and min are not by definition on the boundary of its domain; this is a theorem.
  15. Aug 11, 2010 #14
    I am still trying to understand your answer.

  16. Aug 11, 2010 #15
    Ω ∪ Γ is the union of Ω and Γ. (Ω ∪ Γ) \ {(x0, y0)} is the set Ω ∪ Γ excluding the point (x0, y0). These are basic notations for sets.

    In this case, compact is equivalent to closed and bounded. The set (Ω ∪ Γ) \ {(x0, y0)} is not closed. In my other example, the half-plane x ≥ 0 is not bounded. It's a theorem that a continuous function with compact domain has a maximum and minimum (because it has a compact image).

    The point is that in both examples, the function does not have both a maximum and minimum. If a harmonic function had a maximum and minimum, then those would be attained on the boundary.
  17. Aug 11, 2010 #16
    Let me clarify, since [itex] v \rightarrow -\infty \hbox { as } (x,y) \rightarrow\; (x_0,y_0) \;[/itex], G is not defined at (x0,y0). Therefore there is no max or min of G in the region?
  18. Aug 11, 2010 #17
    There is a max, but no min. Otherwise, you are correct.
  19. Aug 11, 2010 #18
    I really appreciate your help. I am studying on my own and all I have is a few PDE books, apparently this subject is beyond the normal under grad PDE book. The book I have just added this section in and look like it has typos and not very complete and I have a really hard time with it. The book never mention anything like what you said and this make a whole world of sense. Thanks for you help.

    I have another question that I posted in the homework section if you have time to take a look, I would really appreciated. It is about Green's function also.


    Do you have any suggestion of a book on Green's function, something that is easy to understand?


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