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

[Happiness]

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:

 

[DrClaude]

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

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.

 

[RUber]

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.

 

[Samy_A]

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.

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.

 

[RUber]

Thanks, Samy...I spoke too soon.