Proving Complex Function Well Defined

[Milky]

What does it mean to prove a complex function is well defined?

 

[quasar987]

In general, when you hastily define a function by means other than specifying explicitly each value of the function associated with each value of the domain, it is a good idea to check that the function is well defined. That is to say, that the way you defined your function indeed defines a function, in the sense that to each value in the domain corresponds one and only one value in the codomain.

 

[Milky]

So essentially, if:

(x,y)=(x_0,y-0), \phi(x,y)=\phi(x_0,y_0)[\tex] <br /> <br /> ?

 

[quasar987]

No, this is a triviality. If a=b, then always f(a)=f(b).

you want to show that if (x,y) is in the domain of the function, then there is one and only one element b in the codomain sucht that f(x,y)=b.

 

[Milky]

I've done an example in complex variables earlier, which I've repasted here. If I show you my reasoning would you be able to tell me if i proved it right?

Homework Statement

Let D be a connected domain in R^2 and let u(x,y) be a continuous vector field defined on D. Suppose u has zero circulation and zero flux for any simple closed contour on D.

u(x,y) = (u_1(x,y),u_2(x,y))

\Gamma = \int_{c}(u\circ\gamma)tds = 0F=\int_{c}(u\circ\gamma)nds = 0[/tex]\phi(x,y)=\int_{c}(u\circ\gamma)tds\psi(x,y)=\int_{c}(u\circ\gamma)ndsProve that \phi, \psi are well defined.

The Attempt at a Solution

For \phi:

I think to prove its well defined means to prove that if (x,y)=(x_0,y_0), then \phi(x,y)=\phi(x_0,y_0)

Let C_1 and C_2 be two independent paths from (x,y) to (x_0,y_0)

Then, these two paths form a closed contour C_0, for which the integral is zero. Then,

\int_{c_0}(u\circ\gamma)tds = 0 = \int_{c_2}(u\circ\gamma)tds - \int_{c_1}(u\circ\gamma)tdsThen, \int_{c_2}(u\circ\gamma)tds = \int_{c_1}(u\circ\gamma)tdsSo, when (x,y)=(x_0,y_0) the integrals are equal as well.
Is this how to prove it is well-defined?

 

[quasar987]

Your notation for flux and circulation are too messy. And why do you invoke Cauchy when there is no mention of complex functions anywhere. The integrals over loops are simply all equal to 0 by hypothese. so, no, you did not prove anything.

 

[Milky]

This was the notation given to me by both my book and my professor. I have to work with what he wants me to work with.
Okay, I don't know where to start now. In my notes, I have that
\phi(x,y)=\int_{c}(u\circ\gamma)tds
would be well defined if the integral was independent of the choice of path.

Do you have any suggestions to start me off? Thanks.

 

[matt grime]

Pick two paths and show that they give the same answer.

 

[Milky]

Essentially, I thought that's what I was doing but now I'm confused.
I have revised post 5, and what I did was
1. Created two independent paths from (x,y) to (x_0,y_0)
2. Let the two paths form a closed contour C, in which C=C_2-C_1
3. Since the integral of a closed contour is zero, then C_1=C_2
Since they are equal, they are independent of the path. So, it is well defined.

Where have I gone wrong?

 

[Hurkyl]

No, this is a triviality.

Not always. For example, consider these attempts to define a function of rational numbers:

<br /> f\left(\frac{p}{q}\right) = \frac{p^2 + q^2}{2pq}<br />

<br /> f\left(\frac{p}{q}\right) = p^2 + q^2<br />

One of these is a function, and one is not. The one that isn't fails precisely because of an example where a=b, but f(a) \neq f(b).

 

[Milky]

Did I do it correctly?

 

[matt grime]

2. Let the two paths form a closed contour C, in which C=C_2-C_1

these are paths, subsets of R^2, what does it mean to subtract one from the other? Nothing in this case

3. Since the integral of a closed contour is zero, then C_1=C_2

No. The paths are not the same. But you weren't supposed to show that.

Since they are equal, they are independent of the path. So, it is well defined.
where have I gone wrong.

In confusing a path with the integral along that path - I think you have the right idea though. Pick those two paths, C_1 and C_2, jointly they form a closed path, C and the integral around the path is zero. Call the integals I(C), I(C_1) and I(C_2) in the natural way.

0=I(C)

and I(C) equals what?

 

[Milky]

I(C) would be the addition of I(C_1) and I(C_2), but since one of those integrals goes in the clockwise direction, it would be I(C_1) - I(C_2)
So, then the integrals are equal?