Question #126909

Prove that

(v.del)v = 1/2 del v^2 - v×(del×v)

v is a vector

Expert's answer

SInce, (https://en.wikipedia.org/wiki/Del)

∇(u⃗⋅v⃗)=(u⃗⋅∇)v⃗+(v⃗⋅∇)u⃗+u⃗×(∇×v⃗)+v⃗×(∇×u⃗){\displaystyle \nabla ({\vec {u}}\cdot {\vec {v}})=({\vec {u}}\cdot \nabla ){\vec {v}}+({\vec {v}}\cdot \nabla ){\vec {u}}+{\vec {u}}\times (\nabla \times {\vec {v}})+{\vec {v}}\times (\nabla \times {\vec {u}})}

Now, put u→=v→\overrightarrow{u}=\overrightarrow{v} ,we get

∇(v⃗⋅v⃗)=(v⃗⋅∇)v⃗+(v⃗⋅∇)v⃗+v⃗×(∇×v⃗)+v⃗×(∇×v⃗){\displaystyle \nabla ({\vec {v}}\cdot {\vec {v}})=({\vec {v}}\cdot \nabla ){\vec {v}}+({\vec {v}}\cdot \nabla ){\vec {v}}+{\vec {v}}\times (\nabla \times {\vec {v}})+{\vec {v}}\times (\nabla \times {\vec {v}})}

  ⟹  ∇(v2)=2(v⃗⋅∇)v⃗+2v⃗×(∇×v⃗)  ⟹  (v⃗⋅∇)v⃗=12∇(v2)−v⃗×(∇×v⃗)\implies {\displaystyle \nabla ( {v^2})=2({\vec {v}}\cdot \nabla ){\vec {v}}+2{\vec {v}}\times (\nabla \times {\vec {v}})}\\ \implies{ ({\vec {v}}\cdot \nabla ){\vec {v}}= \frac{1}{2}{\displaystyle \nabla ( {v^2})}-{\vec {v}}\times (\nabla \times {\vec {v}})}


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