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Directional Derivative of a discontinuous function

  1. Nov 3, 2011 #1
    I was reviewing basic calculus of of functions of several variables. It struck me that in one of the exercises, it was required to show that a given function has directional derivative for all directions at the origin (0,0) even though it is not continuous at (0,0).
    Without getting into the details , I thought that this does not make sense as the function has to be differentiable at (0,0) for the directional derivatives to exist. And differentiability implies the continuity of the function at (0,0).
    Am I missing something here?
     
  2. jcsd
  3. Nov 3, 2011 #2

    mathman

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    If you define derivative the usual way, you need a value for the function at the origin, which you don't have. However near the origin it is possible to have the same derivative in all directions.
     
  4. Nov 4, 2011 #3
    Indeed the function given in that excercise is defined at the origin but it is not continuous at it. Does that imply that my conclusion is wrong?
     
  5. Nov 4, 2011 #4

    mathman

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    If you have derivatives at all points near the origin, and they all have the same limit at the origin, this limit may be called the derivative at the origin, even if the function is discontinuous there. At this point, it is more a matter of convention.
     
  6. Nov 4, 2011 #5

    mathwonk

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    "the function has to be differentiable at (0,0) for the directional derivatives to exist."

    this statement is false. to be differentiable, the directional derivatives not only have to exist, they have to depend linearly on the derivatives in the coordinate axis directions.

    e.g. the derivative in the direction (1,1) has to be the sum of the derivatives in the directions (1,0) and (0,1).

    the existence of all directional derivatives at (0,0) only implies continuity along all lines through the origin. one can cook up examples that are continuous along all lines, but not along some curve, say a parabola, through the origin.

    e.g.: f (x, y) = [xy^2]/[x^2+y^4], if x ≠ 0, and f (0, y) = 0.
     
    Last edited: Nov 4, 2011
  7. Nov 6, 2011 #6
    Thank you mathman and mathwonk for your help!
    Indeed the exercise is exactly the one that mathwonk has cooked!
    I kept thinking in terms of the theorems that helps in evaluating the directional derivative using the gradient. These theorems requires the differentiability of the function. However, the definition of the directional derivative does not require the differentiability, and using the definition I managed to show that the directional derivative is (0) for f(x,y) given in the above post.
    It is always useful to think in definitions first before jumping to theorems!
     
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