Δ ( x ⋅ ∇ u ) = Δ u + x ⋅ ( ∇ ( Δ u ) )Does this hold for x ∈ R n and u : R n → R ? If it does, why does it?I calculated the above formula by writing out all the variables from x 1 to x 2 and guessing that it must be correct. But I am not familiar with 'direct' calculations using ∇ and Δ, so I need some help.

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Δ  (  x  ⋅  ∇  u  )  =  Δ  u  +  x  ⋅  (  ∇  (  Δ  u  )  )Does this hold for   x  ∈            R        n   and   u  :            R        n    →      R   ? If it does, why does it?I calculated the above formula by writing out all the variables from       x    1   to       x    2    and guessing that it must be correct. But I am not familiar with &#039;direct&#039; calculations using   ∇ and   Δ, so I need some help.

barquegese2 · Accepted Answer

Since there is no standard writing of the derivative of a vector field I don&#039;t know of a way to do this in a coordinate free manner. I&#039;m coming to the following result which is different from your formula.The j in the calculation is fixed, and       δ          i      j       is the Kronecker delta. I write       x    i   for the       i          t      h       component of x and       u          .      i       for the partial derivative of the scalar function u with respect to       x    i  . Then                                          ∂                          ∂                              x                j                                                              (                      ∑            i                                x            i                                u                          .              i                                )                            =                  ∑          i                                      (                                    δ                      i            j                                    u                      .            i                          +                  x          i                          u                      .            i            j                                                )                                                    =                  u                      .            j                          +                  ∑          i                          x          i                          u                      .            i            j                                   .            Doing this a second time we obtain                                                        ∂              2                                      ∂                              x                j                2                                                              (                      ∑            i                                x            i                                u                          .              i                                )                            =                  u                      .            j            j                          +                              ∂                          ∂                              x                j                                                              ∑          i                          x          i                          u                      .            i            j                                                    =                  u                      .            j            j                          +                  ∑          i                                      (                                    δ                      i            j                                    u                      .            i            j                          +                  x          i                          u                      .            i            j            j                                                )                                                    =        2                  u                      .            j            j                          +                  ∑          i                          x          i                          u                      .            i            j            j                                   .            Summing over j we find that in fact                    (1)                    Δ        (        x        ⋅        ∇        u        )        =        2        Δ        u        +        x        ⋅        ∇        (        Δ        u        )                 .            Note that in the one-dimensional case we have            (        x      u    ′    (  x  )                    )              ″    =            (            u    ′    (  x  )  +  x      u    ″    (  x  )                    )              ′    =  2      u    ″    (  x  )  +  x            (            u    ′    (  x  )                    )              ″       ,as obtained in (1)

Delta(x * grad u)=Delta u+x *(grad(Delta u)) Does this hold for x in RR^n and u:RR^n to RR ? If it does, why does it?

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