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Exercise: is f(x,y) continuous and differentiable?

  1. Oct 23, 2013 #1
    1. The problem statement, all variables and given/known data

    could you please check if this exercise is correct?
    thank you very much :)

    ##f(x,y)=\frac{ |x|^θ y}{x^2+y^4}## if ##x \neq 0##
    ##f(x,y)=0## if ##x=0##
    where ##θ > 0## is a constant
    study continuity and differentiabilty of this function

    3. The attempt at a solution
    Being the quotient of continuous functions, it is always continuos except, maybe, in the point (0,0)
    I normally use the sequential continuity to tell is a function of this kind is continuous. However, I couldn't find a couple of suitable successions for this case, so I decided to proceed as follows:
    ## \left| \frac{|x|^θ y}{x^2+y^4} - f (0,0) \right| ≤ \left| |x|^θ y \right|##
    that goes to 0 if x,y go to 0. So the function is continuous in (0,0).
    Then it could be differentiable. I look for the partials:

    ##\frac{\partial f}{\partial x}\lim_{t \to 0} \frac{f(x_0+t)-f(0,0)}{t}=0##
    and
    ##\frac{\partial f}{\partial y}=0## as well. if they are continuous in 0,0 the function is differentiable for the theorem of total differential. but the partials ar every "complicated", so i prefer to apply the definition of differential:

    ##\lim_{(x,y) \to (0,0)} \frac{f(x,y)-f(0,0)-(∇f(0,0),(x,y)}{\sqrt{x^2+y^2}}##
    =## \left| \frac{|x|^θ y}{(x^2+y^4) \sqrt{x^2+y^2}}-0 \right| ## but i see it is my original function multiplied ##\frac{1}{\sqrt{x^2+y^2}}##. this last function goes to infinity for x,y going to 0,0, so the limit doesn't exist and the function is not differentiable.
     
  2. jcsd
  3. Oct 23, 2013 #2

    haruspex

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    I don't follow the logic of that. You're taking f(0,0) to be 0, I presume.
     
  4. Oct 23, 2013 #3

    arildno

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    Hint:
    Introduce polar coordinates, and see for which values of your constant "theta" the function is:
    a) Continuous
    b) And furthermore, differentiable.
     
  5. Oct 23, 2013 #4

    HallsofIvy

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    If a function is differentiable, then it is continuous. Since you are asked whether or not the function is differentiable and continuous, I think I would be inclined to look at "differentiable" first. If the answer is "yes, it is differentiable for all (x, y)", you get "continuous" automatically!

    Of course, "differentiable" for functions of two variables is more complicated than just saying the partial derivatives exist.
     
  6. Oct 23, 2013 #5

    arildno

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    Halls:
    Sure enough.
    But, in this case, you have parameter values for which the function is continuous, but not differentiable, and that range ought to be included when saying when the function is continuous.
     
  7. Oct 23, 2013 #6
    yes, exactly, it's part of the hypothesis given
     
  8. Oct 23, 2013 #7
    we haven't studied polar coordinates during the course (yet?). isn't there another way?
     
  9. Oct 23, 2013 #8

    haruspex

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    Sure, but how do you obtain the inequality I quoted? It seems to require the denominator, x2+y4, to be ≥ 1.
     
  10. Oct 23, 2013 #9
    yes, i just figured out later that it might be false.
    then sould i try to do this:
    ##(x-y^2)^2 \geq 0 \to x^2+y^4 \geq xy^2##
    so
    ## \left| \frac{|x|^θy}{y^4+x^2} \right| \leq \left| \frac{|x|^θ}{xy}\right|##
    if
    ##x=\frac{1}{n}## i get ##y=|\frac{1}{n}|^θ \cdot n## and for n to infinity the f(x,y) goes to 0 only if θ>2. so for θ<2 it is not continuous
     
  11. Oct 23, 2013 #10

    haruspex

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    Yes, that's better, but what about θ=2? Next, differentiability.
     
  12. Oct 24, 2013 #11
    it's not contnuous in (0,0) for θ=2, because it's homogeneus of degree 0.
    I know it can't be differentiable when it isn't continuous, so , using the same proceeding as that i posted at the beginning, i'll say that it is differentiable for every θ>2
     
  13. Oct 24, 2013 #12

    haruspex

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    You are right that it is not continuous for θ=2, but I don't understand your reasoning. Wouldn't f(x,y)=0 be homogeneous of degree 0?
    In your OP you concluded it was nowhere differentiable (which was wrong). If you now believe it is differentiable for every θ>2 pls post your proof.
     
  14. Oct 24, 2013 #13

    haruspex

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    Sorry, I didn't read this post properly the first time (jet lag). You claim to have shown f < some other function, g say, then because g tends to infinity at O for a certain theta you conclude f does too. That's illogical. You would need to show f >= g for that. And I should not have agreed that f is discontinuous at O for all theta < 2.
    Try considering lines of approach to the origin, like y = kx for a constant k.
     
  15. Oct 25, 2013 #14
    no worries :)
    unfortunately i have never heard about the method you mentioned, and our teacher wants us to solve it the way he taught us, which means by increasing/reducing inequalities and using a few theorems (lagrange, theorem of the total differential...)
    but i thought that I could show the greater inequality converges to 0, so that the first one also does:

    ## \left|\frac{|x|^θy}{x^2+y^4} \right| \leq \left|frac{|x|^θy}{xy^2} \right| = \left|frac{|x|^θ}{xy} \right|##
    now if ##x=\frac{1}{n}## i get ##y=\frac{1}{n^{θ-1}}##
    so the function becomes: ##\frac{\frac{1}{n}}{\frac{1}{n^{θ-1}}} \to ##n^{θ-2}## that goes to zero if ##θ-2<|1|## so ##θ-2>-1## and ##θ-2>1## that gives
    ##1<θ<3##. for these values it is continuous.
    I look for differentiabilty just like in the OP and i get the function is not differentiable in (0,0)
     
  16. Oct 25, 2013 #15

    haruspex

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    Fixing up your LaTex:
    I don't understand the logic of that. How does it give you a value for y? To deduce that you would need that the RHS = 1.
    The RHS is an upper bound for the function, so you cannot us it for showing discontinuity. At est, you can use it to show continuity by showing the RHS tends to 0 as the origin is approached. But that is not the case, since it is unbounded along y=0, which the original function was not.
    Can you describe in more detail the methods you have been taught for testing continuity of functions of two variables?
    The line-of-approach method is a quick and easy way to get an idea of what is going on. You consider approaching the point of interest along some curve - in this case a line like y = kx. That gets it down to a function of one variable. If this produces the right limit irrespective of k then you've a good chance it is continuous (but you at least also need to check the case of approaching along the y axis, which the values of k do not cover). If there is any value of k which produces the 'wrong' limit then it definitely is not continuous.
    Another useful way is converting to some sort of polar form. In this case, because you have y^4, I would suggest ##r^2 = x^2 + y^4##, so ## x = r \cos(\theta)##, ## y^2 = r \sin(\theta) ##. (There's clearly a problem with this in the 3rd and fourth quadrants - not sure how to handle that.) You'd then need to show two things: f tends to 0 as r tends to 0, regardless of theta; as |(x, y)| tends to 0 r tends to 0.
    The method in the OP is wrong. |(x,y)| tends to 0 does not imply f(x,y)/|(x,y)| tends to infinity. f might tend to zero even faster. I assure you there are values of theta for which f is differentiable at O.
     
  17. Oct 25, 2013 #16
    yes, i've been taught these theorems:
    lagrange (mean value)
    theorem of the total differential
    clauiraut/schwarz's theorem
    but they don't seem to be very helpful for the continuity problem, for which i should use just the manipulation of inequalities with the help of young's inequality, for instance, and using the sequential continuity, maybe by subdiving a limit in two limits one of those goes to zero and the other to a real number
    i'll give you an example: i want to study the continuity in (0,0) of
    f(x,y)=##\frac{x^3 y}{x^4+y^2}## if ##x,y\neq(0,0)##
    ##f(x,y)=0## if ##(x,y)=(0,0)##
    ##0 \leq \lim_{(x,y) \to (0,0)}\frac{|x|^3 |y|}{x^4+y^2}\leq \lim_{(x,y) \to (0,0)} |x|\frac{|x|^4+ |y|^2}{2(x^4+y^2)}## by cauchy's inequality
    and, finally:
    ##\lim_{(x,y) \to (0,0)}\frac{|x|}{2}=0## which means the function is continuous
     
  18. Oct 25, 2013 #17

    haruspex

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    Well I can't see how to use those either. Besides, fishing around for a useful inequality is a rather haphazard approach. In the present case, what tends to happen is that you show f is bounded above by some function g, but g is unbounded near the origin, so it proves nothing.
    My method, encapsulating the denominator in a single variable which, as it tends to zero, will force x and y to zero, is more deterministic.
    The 3rd and 4th quadrants aren't a problem after all. I'm defining ##r = \sqrt{x^2+y^4}## and ##\phi = \arctan(x/y^2)##. You can show f is bounded above by r to some power. As long as that power > 0 you have continuity. Next, you can show that, when that power = 0, f is not continuous, completing the proof.
    If you are determined to find a proof along the lines you have been taught, you could use my method to find what the critical value of theta is and see if that gives you any clues.
     
    Last edited: Oct 25, 2013
  19. Oct 26, 2013 #18
    yes, I really can't use polar coordinates. But using y=kx in ##\frac{|x|^{\Theta}}{xy}## i see that I have to have ##\Theta>2##, otherwise the denominator goes to 0 faster then the numerator. right?
     
  20. Oct 26, 2013 #19

    haruspex

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    ##\frac{|x|^{\Theta}}{xy}## is not the function you are given - it is only a bound which you have found for the function. Unfortunately it is of no use. If you come in on the x axis you will have y = 0, so your upper bound is infinite, and tells you nothing at all about f.
    Why can't you use polar coordinates? It's just a substitution into different variables.
     
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