I'm having difficulty proving the formula: u × ω = ∇ ( u ⋅ u 2 ) − u ⋅ ∇ uI should be using tensor notation. Given is that: ω = ∇ × uand ∇ ⋅ u = 0I've done this so far: ( u × ω ) i = ( u × ( ∇ × u ) ) i = ϵ i j k u j ( ϵ k l m ∂ ∂ x l u m ) = ϵ i j k ϵ k l m u j ∂ ∂ x l u m = ϵ k i j ϵ k l m u j ∂ ∂ x l u m = ( δ i l δ j m − δ i m δ j l ) u j ∂ ∂ x l u m = u j ∂ ∂ x i u j − u j ∂ ∂ x j u i But that is as far as I come.

Question

I&#039;m having difficulty proving the formula:  u  ×  ω  =  ∇     (            u      ⋅             u        2    )  −  u  ⋅  ∇     uI should be using tensor notation. Given is that:  ω     =  ∇  ×     uand  ∇  ⋅     u     =  0I&#039;ve done this so far:  (  u  ×  ω      )    i    =  (  u  ×     (  ∇  ×     u  )      )    i    =      ϵ          i      j      k            u    j    (      ϵ          k      l      m            ∂          ∂                     x        l                  u    m    )  =      ϵ          i      j      k            ϵ          k      l      m               u    j        ∂          ∂                     x        l                  u    m    =      ϵ          k      i      j            ϵ          k      l      m               u    j        ∂          ∂                     x        l                  u    m    =  (      δ          i      l            δ          j      m        −      δ          i      m            δ          j      l        )      u    j        ∂          ∂                     x        l                  u    m    =      u    j        ∂          ∂                     x        i                  u    j    −      u    j        ∂          ∂                     x        j                  u    i  But that is as far as I come.

Sheldon Castillo · Accepted Answer

If             w      →        =  ∇  ×            u      →       , then using implied summation notation reveals                                          u            →                          ×                              w            →                                              =                              u            →                          ×        ∇        ×                              u            →                                                                          =                  u          i                                                    x              ^                                i                ×                  ∂          j                (                                            x              ^                                j                ×                                            x              ^                                k                          u          k                )                                                        =        (                  δ                      i            k                                                              x              ^                                j                −                  δ                      i            j                                                              x              ^                                k                )                  u          i                          ∂          j                (                  u          k                )                                                        =                                            x              ^                                j                          u          i                          ∂          j                (                  u          i                )        −                                            x              ^                                k                          u          i                          ∂          i                (                  u          k                )                                                        =                  1          2                ∇        (                  |                                      u            →                                                |                    2                )        −        (                              u            →                          ⋅        ∇        )                              u            →                              as was to be shown!Alternativley, using the Levi-Civita notation, we can write                    (                              u            →                          ×                              w            →                                    )          i                                    =        (                              u            →                          ×        ∇        ×                              u            →                                    )          i                                                                =                  ϵ                      i            j            k                                    u          j                (        ∇        ×                              u            →                                    )          k                                                                =                  ϵ                      i            j            k                                    u          j                          ϵ                      k            ℓ            m                                    ∂          ℓ                (                  u          m                )                                                        =        (                  δ                      i            ℓ                                    δ                      j            m                          −                  δ                      i            m                                    δ                      j            ℓ                          )                  u          j                          ∂          ℓ                (                  u          m                )                                                        =                  u          j                          ∂          i                (                  u          j                )        −                  u          j                          ∂          j                (                  u          i                )                                                        =                  1          2                          ∂          i                (                  u          j                          u          j                )        −        (                  u          j                          ∂          j                )        (                  u          i                )                                                        =                              (                          1              2                        ∇            (                                          u                →                                      ⋅                                          u                →                                      )            −            (                                          u                →                                      ⋅            ∇            )                                          u                →                                      )                    i                    Hence, we conclude that            u      →        ×            w      →        =      1    2    ∇  (            u      →        ⋅            u      →        )  −  (            u      →        ⋅  ∇  )            u      →      as expected!

Proof of u xx omega=grad ((uu)/2)−u grad u

Answered question

Answer & Explanation

New Questions in Precalculus