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Differentiability of a function -- question on bounding

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data
    I need to see if the function defined as


    ##f(x,y) = \left\{
    \begin{array}{lr}
    \frac{xy^2}{x^2 + y^2} & (x,y)\neq{}(0,0)\\
    0 & (x,y)=(0,0)
    \end{array}
    \right.##

    is differentiable at (0,0)

    2. Relevant equations

    A function is differentiable at a point, if It can be approximated at that point by a linear transformation,

    ##\lim_{(u, v) \rightarrow (0, 0)} \frac {f(u, v) - df_{(0, 0)}(u, v) - f(0, 0)} {||(u, v)||} =0##

    Useful bounds:

    ##|u|\leq ||(u, v)||;
    |v|\leq ||(u, v)||##
    ##|u|^2=u^2\leq ||(u, v)||^2= (\sqrt{u^2+v^2})^2 = (u^2+v^2)##
    3. The attempt at a solution

    Both partial derivatives are zero at the point, so all I have left is

    ##\lim_{(u, v) \rightarrow (0, 0)} \frac {\frac{uv^2}{u^2 + v^2}} {||(u, v)||} =\lim_{(u, v) \rightarrow (0, 0)} \frac {uv^2}{(u^2 + v^2)||(u, v)||} ##

    Now the bounds:

    ##\frac {uv^2}{(u^2 + v^2)||(u, v)||}\leq\frac {u||(u, v)||^2}{(u^2 + v^2)||(u, v)||} = \frac {u||(u, v)||}{(u^2 + v^2)} \leq \frac {||(u, v)||^2}{(u^2 + v^2)}= \frac{||(u, v)||^2}{||(u, v)||^2}=1##


    Does this means that the function is not differentiable at (0,0), or did I make a mistake along the way ? Thanks in advance for the help.
     
  2. jcsd
  3. Mar 9, 2016 #2

    stevendaryl

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    In polar coordinates, the function looks a lot better-behaved:

    Letting [itex]x=r cos(\theta)[/itex], [itex]y = r sin(\theta)[/itex], then your function in polar coordinates becomes [itex]r sin(\theta) cos(\theta)[/itex], which seems very innocuous.
     
  4. Mar 9, 2016 #3
    Thanks for the help!
     
  5. Mar 9, 2016 #4

    Ray Vickson

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    That should be ##r \cos(\theta) \sin^2(\theta)##.
     
  6. Mar 9, 2016 #5

    Samy_A

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    How does showing that ##\frac {uv^2}{(u^2 + v^2)||(u, v)||} \leq 1## prove that the function is not differentiable in (0,0)?
    Use @stevendaryl 's suggestion (as corrected by @Ray Vickson ) to show that the limit isn't 0 or doesn't exist.
     
  7. Mar 9, 2016 #6
    You're right, Samy_A, it doesn't, and with the polar coordinates I have a function which is the product of two functions ##H(r,\theta)=F(r)G(\theta)## and while ##\lim_{(r) \rightarrow (0)} F(r) = 0##, ##G(\theta)## is bounded in [0,2pi], therefore the original function is differentiable in (0,0). Thank you, everybody.
     
  8. Mar 9, 2016 #7

    Ray Vickson

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    Isn't the definition of differentiability at (0,0) that there exist constants ##a## and ##b## such that
    [tex] f(x,y) = f(0,0) + a x + b y + o\left(\sqrt{x^2+y^2} \right)?[/tex]
    That is, we have ##f(x,y) \approx f(0,0) + a x + by## "to first order in small ##|x|,|y|##".

    What would that say in polar coordinates? Does your function satisfy that property?
     
  9. Mar 10, 2016 #8

    Samy_A

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    You had to prove that ##\displaystyle \lim_{(u, v) \rightarrow (0, 0)} \frac {uv^2}{(u^2 + v^2)||(u, v)||}=0##.
    As has been suggested, switching to polar coordinates makes live easier here.
    If the limit is 0, your function is differentiable in (0,0). (That is assuming the partial derivatives in (0,0) are both 0. You stated this without proof, but I think it is indeed correct.)

    But is the limit 0? Does the limit even exist?
     
  10. Mar 11, 2016 #9
    I'm sorry, that happens for hurrying. I forgot to divide by ##||(u, v)||## in the definition, which gives me an extra ##r## in the denominator:

    ##\frac {r^3 cos(\theta) sin(\theta)^2} {r^3} = cos(\theta) sin(\theta)^2##

    which means it doesn't exist, therefore is not differentiable in (0,0). Thanks again for your patience.
     
  11. Mar 11, 2016 #10

    Ray Vickson

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    No: the limit DOES exist along any direction, but is not linear in ##\sin(\theta)## and ##\cos(\theta)##. That is why the function is not differentiable: its directional derivative is not a linear function of the direction.
     
  12. Mar 18, 2016 #11
    Hi Ray Vickson, I was trying to use the theorem that states that if I can decompose the function like this ##F(r,\theta)=H(r)G(\theta)## and ##\lim_{(r) \rightarrow (0)} H(r) = 0## and ##G(\theta)## is bounded in [0,2pi] then the original function is differentiable, but I see now that since the r part is cancelled, all I have left is
    ##F(\theta)=cos(\theta)sin(\theta)^2## and therefore, I cannot use this theorem.

    Do you mean that is not linear in the variable ##\theta##? Or, for example, if what I get is ##F(\theta)=sin(\theta)## I could say it is differentiable, because it is linear in the variable ##sin(\theta)## (Would this variable come from the definition in the transformation to polar coordinates?)

    And another question, is there some theorem or text you can direct me to that states that if the directional derivative is not a linear function of the direction then the function is not differentiable? I didn't know that, and my notes don't mention a case where the ##r## part is gone. Thank you for your patience.
     
  13. Mar 18, 2016 #12

    Ray Vickson

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    NO, as in #10, I said it is not linear in ##\cos(\theta)## and ##\sin(\theta)##. Look again at #7, where it says that we need ##f(x,y) \approx f(0,0) + a x + by## to first order in small ##x,y##; now put ##x = r \cos(\theta)## and ##y = r \sin(\theta)##. You do not have ##f \approx a r \cos(\theta) + b r \sin(\theta)## for constants ##a,b##, so you do not satisfy that differentiability criterion.

    I do not know how your textbook/notes/instructor defines the concept of "differentiability" in the multivariate case (which is trickier and more stringent than in the univariate case), so I don't know how to answer your general question in that regard.
     
  14. Mar 19, 2016 #13
    Thanks Ray, I understand now what you meant.
     
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