Proof - Vector Calculus - Curl

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SUMMARY

The discussion centers on proving the vector calculus identity: \( u \times (\nabla \times u) = \frac{1}{2}\nabla(u^2) - (u \cdot \nabla)u \). Participants clarify the derivation of the term \( \frac{1}{2} \) in the equation. The expression \( \partial_i(u_j u_j) = 2u_j \partial_i(u_j) \) is highlighted as a key step in the proof, confirming the relationship between the curl and the gradient of the squared magnitude of the vector field.

PREREQUISITES
  • Understanding of vector calculus concepts, specifically curl and gradient.
  • Familiarity with the notation and operations involving partial derivatives.
  • Knowledge of vector fields and their properties in three-dimensional space.
  • Proficiency in manipulating mathematical expressions involving vectors and scalars.
NEXT STEPS
  • Study the derivation of vector calculus identities, focusing on the curl and gradient operations.
  • Learn about the physical interpretations of curl and gradient in fluid dynamics.
  • Explore advanced topics in vector calculus, such as Stokes' theorem and its applications.
  • Practice solving problems involving vector identities and their proofs in various contexts.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are looking to deepen their understanding of vector calculus and its applications in theoretical and applied contexts.

cristina89
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I need to prove this: u x ([itex]\nabla[/itex] x u) = [itex]\frac{1}{2}[/itex][itex]\nabla[/itex](u²) - (u [itex]\cdot[/itex] [itex]\nabla[/itex])u.

I've came to this: uj∂iuj - uj∂jui (i think it's correct)
But how this 1/2 appears?
 
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welcome to pf!

hi cristina89! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
cristina89 said:
I need to prove this: u x ([itex]\nabla[/itex] x u) = [itex]\frac{1}{2}[/itex][itex]\nabla[/itex](u²) - (u [itex]\cdot[/itex] [itex]\nabla[/itex])u.

I've came to this: uj∂iuj - uj∂jui (i think it's correct)
But how this 1/2 appears?

'cos ∂i(ujuj)= 2uji(uj) :wink:
 

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