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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##?

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

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