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Continuity for Two Variable Function

  1. Jul 30, 2011 #1
    In whatever little I have learnt about calculus of two variable functions I have been having some serious problems in the way continuity of a function is defined.
    We say that a function is not continuous if we can find two paths of approach along which the value of the independent variable is different.Now if I have to find such paths leading to a say a point,say the origin(0,0,0) the simplest way I can think of is to put the function equal to some constant c and obtain a contour curve for that value.Now in the neighborhood of the point under consideration(origin here),if I can limit the z-value of the contour lines crossing the neighborhood :
    f(0,0)-ε< cε <f(0,0)+ε
    where ε>0 is an arbitrary number,I can say that the function is continuous at that point.This follows from the formal definition of continuity and is easily understood.

    But I will put forward two examples:

    1.f(x,y)=x3+y3/x-y

    This function is not continuous at (0,0).A few contour lines for z={1,5,3} are here:


    http://rechneronline.de/function-graphs/" [Broken]

    By this I can conclude that the function is non-continuous at (0,0).

    2.f(x,y)=xy/√ (x2+y2)

    This function comes out to be continuous at (0,0) by the formal δ-ε method.However the corresponding contour lines look like this:
    http://rechneronline.de/function-graphs/" [Broken]

    Now I have paths to (0,0) that have different z-values even though the function is continuous at
    (0,0).Can somebody help me?

    Edit:The graphs linked are not working.Please use these functions:
    Function 1:
    1.x^3+y^3-x+y
    2.x^3+y^3-3*x+3*y
    3.x^3+y^3-5*+5*y
    Function 2:
    4.(x*y)-sqr(x^2+y^2)
    5.(x*y)-3*sqr(x^2+y^2)
    6.(x*y)-3*sqr(x^2+y^2)
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 30, 2011 #2
    Hello, I think it might be the problem of the online plotter... I plot the figure of this function with Matlab together with its contour. They look fine... The contour line for z = 0 are two axes... Your can show that the value of the function converges to 0 around (0,0).
     
  4. Jul 30, 2011 #3
    I didn't read the rest of this thread, but could you missing a set of parenthesis?
     
    Last edited: Jul 30, 2011
  5. Jul 30, 2011 #4

    I like Serena

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    Hi aim1732! :smile:

    You have a couple of problems here.

    First off, could you add proper parentheses?
    Your expressions are now very ambiguous (actually they are not ambiguous, but I suspect that what you wrote it not what you meant).

    Then, the functions you suggest to put as expression on the web site, do not correspond with the function you mentioned (not in any of the various ways you might have intended it).
    For instance, the web site draws "x^3+y^3-x+y" as "x^3-x".
    Apparently it implicitly sets y to zero without saying so.
    I recommend using WolframAlpha, where you can also see how your expressions are interpreted:
    http://www.wolframalpha.com/input/?i=x3+y3/x-y"
    http://www.wolframalpha.com/input/?i=z=xy/√+(x2+y2)"

    Finally, formally, none of these functions can be continuous in (0,0), since they are not defined there, which is a condition for being continuous.
    I suspect you intended to extend the function definitions, setting it to zero in (0,0).
     
    Last edited by a moderator: Apr 26, 2017
  6. Aug 21, 2011 #5
    Admittedly,I meant z= (x^3+y^3)/(x-y).And I didn't know that the website was setting y to zero so sorry about that.
    What I basically meant was that if I can find in principle at least two paths that pass through (0,0) and correspond to different values of z I can say that the function is not continuous.So I set the functions mentioned equal to a constant c and see that the curves in x-y plane generated all pass through (0,0) regardless of the value of c.Hence I conclude that the functions are not continuous--even though one is and the other is not.
     
  7. Aug 21, 2011 #6

    I like Serena

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    Umm, do you have a question?

    Note that according to the definition of continuity, neither function is continuous at (0,0) because neither function is defined at (0,0).
    This can be said without calculating limits, or looking at the contour graphs.
     
  8. Aug 23, 2011 #7

    HallsofIvy

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    One obvious point is that on the line y= x, the function is not defined so giving some special definition at (0, 0) would not help. Approaching (0, 0) along the line y= -x, the limit becomes
    [tex]\lim_{x\to 0}\frac{x^3- x^3}{2x}= 0[/tex]
    so that we might try defining y= 0 if y= x.

    Of course, finding two paths that give different results will tell us that there is not limit, but NOT finding two such paths does NOT tell us there is a limit! No matter how many paths we check we cannot check them all.

    What you could try is converting to polar coordinates. That way, the distance to (0, 0) is given by the single variable, r. If you can show that the limit, as r goes to 0, is 0 for any [itex]\theta[/itex], then the limit is 0.
     
  9. Aug 23, 2011 #8
    You probably meant dependent, so I'll assume that. This is not quite true, you've left out part of the definition.

    • The function must be morphism to the sequence-limit relation, which is another way of putting what you said.
    • We now know that all of those "paths" converge to the same point in space. For the function to be continuous, that point has to be the actual value of the function.

    If only the first requirement holds, then the function can be extended to a continuous function (it has a removable discontinuity, we say), but it is not actually continuous unless that limit is equal to the function at that point. (Which also requires that the function is defined there.)
     
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