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

[yungman]

I want to verify my understand of this.

Let $u$ defined in region $\Omega$ with boundary $\Gamma$.

If $u = 0 \hbox { on the boundary } \Gamma$, then $u = 0 \hbox { in the region } \Omega$.


The way to look at this, suppose $u$ 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 $\Gamma$. 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 $\Gamma$ and equal to x+y anywhere else, then the assertion cannot not be true! Please help.

 

[adriank]

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.

 

[yungman]

Thanks for your reply, I am getting desperate!

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

The example of u=XY or u=X+Y is only an example that I gave where u is a function of both $X(x) \; \hbox{ and }\; Y(y)$. There u can be XY or X+Y to cover all cases.

 

[adriank]

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.

 

[yungman]

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.

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 $\nabla u =0$.

BTW, how do you get symboms like Ω and Γ direct without going into latex?

 

[adriank]

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.)

 

[yungman]

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.)

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.

Thanks

Alan