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

1. Aug 9, 2010

### 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

Last edited: Aug 10, 2010
2. Aug 10, 2010

(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).

3. Aug 10, 2010

### 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 $0<r_0 < R$ such that $v (x_0,y_0,x,y)<-2M \hbox{ on } C_r$ for all $0<r_0 \leq r < R$. Hence $G (x_0,y_0,x,y)\leq -M \hbox{ on } \;C_r\;$, because G=h+v, $v (x_0,y_0,x,y)<-2M \hbox{ and }|h (x_0,y_0,x,y)|\leq M \hbox{ on } C_r$. 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 $0<r \leq r_0 \;$, 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 explainations of what are $r, r_0 \hbox { and } R \;$.

4. Aug 10, 2010

### yungman

Anyone?

5. Aug 11, 2010

What's the issue?

6. Aug 11, 2010

### darkside00

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

7. Aug 11, 2010

No one mentioned Green's theorem...

8. Aug 11, 2010

### darkside00

unrelated?

9. Aug 11, 2010

10. Aug 11, 2010

### yungman

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!!!!

11. Aug 11, 2010

### darkside00

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

12. Aug 11, 2010

### yungman

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!!!!

13. Aug 11, 2010

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.

14. Aug 11, 2010

### yungman

Thanks

15. Aug 11, 2010

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.

16. Aug 11, 2010

### yungman

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?

17. Aug 11, 2010

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

18. Aug 11, 2010

### yungman

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.