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

[yungman]

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

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

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 \Omega.

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

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


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

Thanks

Alan

 

[adriank]

(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).

 

[yungman]

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

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



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

To proof this:

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


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

That is all the book have, I typed word to word from the book. No more explanations of what are r, r_0 \hbox { and } R \;.

 

[yungman]

Anyone?

 

[adriank]

What's the issue?

 

[darkside00]

All I remember, if it equals zero, then your doing greens theorem wrong or supposed to get zero. closed curve...

 

[adriank]

No one mentioned Green's theorem...

 

[darkside00]

unrelated?

 

[adriank]

Yes, Green's functions are totally unrelated.

 

[yungman]

What's the issue?

The issue is by definition of harmonic function in a region\Omega \hbox { with closed boundary } \Gamma, 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 \Gamma 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!

 

[darkside00]

Yes, Green's functions are totally unrelated.

Doesn't all that derivation (Laplacian apparently) and divergence involve his theorem? how is it totally unrelated?

 

[yungman]

Doesn't all that derivation (Laplacian apparently) and divergence involve his theorem? how is it totally unrelated?

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:

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

This is like the first page of the whole nightmare I am going through!

 

[adriank]

The issue is by definition of harmonic function in a region\Omega \hbox { with closed boundary } \Gamma, 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 \Gamma 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!

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.

 

[yungman]

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.
What is Ω ∪ Γ and (Ω ∪ Γ) \ {(x₀, y₀)}?
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.
What is the meaning of compact?
Also, a harmonic function's max and min are not by definition on the boundary of its domain; this is a theorem.
But the textbook proofed it!

I am still trying to understand your answer.

Thanks

 

[adriank]

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.

 

[yungman]

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.

Let me clarify, since v \rightarrow -\infty \hbox { as } (x,y) \rightarrow\; (x_0,y_0) \;, G is not defined at (x0,y0). Therefore there is no max or min of G in the region?

 

[adriank]

There is a max, but no min. Otherwise, you are correct.

 

[yungman]

There is a max, but no min. Otherwise, you are correct.

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.

https://www.physicsforums.com/showthread.php?t=421564

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

Thanks

Alan