C^1 or C^2? Investigating Vector Identity

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SUMMARY

The vector identity \nabla \cdot (\nabla f \times \nabla g) = 0 holds true when the functions f and g are at least C^{2} functions. This conclusion is based on the requirement for the equality of mixed partial derivatives, which necessitates that both functions be twice differentiable. Therefore, for the identity to be valid, it is essential that f and g meet the C^{2} criteria.

PREREQUISITES
  • Understanding of vector calculus
  • Familiarity with C^{1} and C^{2} function classes
  • Knowledge of partial derivatives
  • Concept of mixed partial derivatives
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  • Study the properties of C^{1} and C^{2} functions in depth
  • Learn about the equality of mixed partial derivatives
  • Explore applications of vector identities in physics
  • Investigate the implications of differentiability in vector calculus
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Mathematicians, physics students, and anyone studying advanced calculus or vector analysis will benefit from this discussion.

madachi
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One of the basic vector identities is

[itex]\nabla \cdot (\nabla f \times \nabla g) = 0[/itex]

Is this true if [itex]f[/itex] and [itex]g[/itex] are [itex]C^{1}[/itex] ? (Or they must be [itex]C^{2}[/itex] functions?

Thanks!
 
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I think this follows from the equality of mixed partial derivatives. So I think that f and g must be al least twice differentiable for this to hold.
 

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