1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Well defined

  1. Oct 21, 2007 #1
    What does it mean to prove a complex function is well defined?
     
  2. jcsd
  3. Oct 21, 2007 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 fucntion 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.
     
  4. Oct 21, 2007 #3
    So essentially, if:

    [tex](x,y)=(x_0,y-0), \phi(x,y)=\phi(x_0,y_0)[\tex]

    ?
     
  5. Oct 21, 2007 #4

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
    Last edited: Oct 21, 2007
  6. Oct 21, 2007 #5
    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?
    1. The problem statement, all variables and given/known data

    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.

    [tex] u(x,y) = (u_1(x,y),u_2(x,y))[/tex]

    [tex]\Gamma = \int_{c}(u\circ\gamma)tds = 0[/tex]


    [tex]F=\int_{c}(u\circ\gamma)nds[/tex] = 0[/tex]


    [tex] \phi(x,y)=\int_{c}(u\circ\gamma)tds [/tex]


    [tex] \psi(x,y)=\int_{c}(u\circ\gamma)nds [/tex]


    Prove that [tex]\phi, \psi[/tex] are well defined.

    3. The attempt at a solution
    For [tex]\phi[/tex]:

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

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

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

    [tex]\int_{c_0}(u\circ\gamma)tds = 0 = \int_{c_2}(u\circ\gamma)tds - \int_{c_1}(u\circ\gamma)tds[/tex]


    Then, [tex]\int_{c_2}(u\circ\gamma)tds[/tex] = [tex]\int_{c_1}(u\circ\gamma)tds[/tex]


    So, when [tex](x,y)=(x_0,y_0)[/tex] the integrals are equal as well.
    Is this how to prove it is well-defined?
     
    Last edited: Oct 21, 2007
  7. Oct 21, 2007 #6

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  8. Oct 21, 2007 #7
    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
    [tex] \phi(x,y)=\int_{c}(u\circ\gamma)tds [/tex]
    would be well defined if the integral was independent of the choice of path.

    Do you have any suggestions to start me off? Thanks.
     
  9. Oct 21, 2007 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Pick two paths and show that they give the same answer.
     
  10. Oct 21, 2007 #9
    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 [tex](x,y) to (x_0,y_0)[/tex]
    2. Let the two paths form a closed contour C, in which [tex]C=C_2-C_1[/tex]
    3. Since the integral of a closed contour is zero, then [tex]C_1=C_2[/tex]
    Since they are equal, they are independent of the path. So, it is well defined.

    Where have I gone wrong?
     
  11. Oct 21, 2007 #10

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    [tex]
    f\left(\frac{p}{q}\right) = \frac{p^2 + q^2}{2pq}
    [/tex]

    [tex]
    f\left(\frac{p}{q}\right) = p^2 + q^2
    [/tex]

    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 [itex]f(a) \neq f(b)[/itex].
     
  12. Oct 21, 2007 #11
    Did I do it correctly?
     
  13. Oct 22, 2007 #12

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

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

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

    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?
     
  14. Oct 22, 2007 #13
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Well defined
Loading...