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Must the 1st derivative of phi be undetermined at V=infinity

  1. Dec 7, 2015 #1
    Screen Shot 2015-12-07 at 10.26.48 pm.png

    I think it is not true that a discontinuous ##\nabla^2\psi## implies a discontinuous ##\nabla\psi##, because a continuous function can have a discontinuous derivative, eg. ##y=|x|##.

    Is it true that ##\nabla\psi## must always be undetermined at the boundary where ##V=\infty##?

    Attached below is the whole paragraph that the sentence appears in:
    Screen Shot 2015-12-07 at 10.27.02 pm.png
     
    Last edited: Dec 7, 2015
  2. jcsd
  3. Dec 7, 2015 #2

    DrClaude

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    That's not what it says. It says that the discontinuity of ##\nabla^2\psi## is the same as that of the potential, hence is a discontinuity of the second kind (infinite jump). Therefore the first derivative will be discontinuous.
     
  4. Dec 7, 2015 #3

    RUber

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    While it is true that a discontinuous second derivative does not imply a discontinuous first derivative, it is true that a continuous function can not have an infinite derivative.
     
  5. Dec 7, 2015 #4

    Samy_A

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    Is there an implicit condition for this to be true?
    Take ##f(x)=x^{1/3}##. This function is continuous, but the first derivative in x=0 is infinite.
     
    Last edited: Dec 7, 2015
  6. Dec 7, 2015 #5

    RUber

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    Thanks, Samy...I spoke too soon.
     
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