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Scalar field 2 dimensional discontinuous but differentiable

  1. Nov 4, 2014 #1
    Hi,

    [tex]f(X)=\frac{xy^2}{x^2+y^4}[/tex] is the function in question, this is the value of the function at ##X=(x,y)## when ##x\neq0##, and ##f(X)=0## when ##X=(0,y)## for any ##y## even ##y=0##.

    Now, along any vector or line from the origin the directional derivative ##f'(Y,0)## (where ##Y=(a,b)## is the vector or line along witch the derivative is being taken and ##0## is the point at witch it is taken, in this case.) exists for any vector ##Y=(a,b)## (this is easily verifiable).

    But the function is not continuous at ##X=(0,0)##, since the value of the function is ##0## at ##X=(0,0)## but along the parabola ##x=y^2## the value of the function is ##\frac{1}{2}## and the parabola comes arbitrary close to the value ##(0,0)##.

    So how is this possible im trying to get my head around this but cant figure it out??
     
  2. jcsd
  3. Nov 5, 2014 #2
    Attached is a plot of what your function z = f(X) looks like for points near (0,0). Points very close to (0,0) have been avoided by the computer, hence the hole. Notice the parabolic plateau near (0,0).
    Also notice that lines drawn through the origin never pass through more than one point of the parabola, and the parabola does not pass through (0,0), hence all directional derivatives exist, as the directional derivatives only care about values arbitrarily close to the point (0,0) along a particular line (not the total set of all values of f(X) around (0,0)). Ie., it is like there is a pinhole at (0,0) that you are shining a laser pointed in a single direction through. The light beam will travel for some non-zero distance until it hits the parabola, no matter what angle you choose. It is along that distance that we take the directional derivative. The second illustration illustrates one of these lines.
     

    Attached Files:

    Last edited: Nov 5, 2014
  4. Nov 5, 2014 #3
    That's a great plot, thanks a lot.

    but I think I figured it out myself. the mistake I did was to assume that the vector tangent to the parabola would have one more point, besides the origin in common with the parabola.

    the expression depends on how one approaches the point ##(0,0)## and that's the only right way to look at this one should not make analogies with discreet points.
     
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