How to write summation of squared divergence terms in index summation notation?












0












$begingroup$


Sorry if this has already been asked before, but it's really difficult to try and explain the problem in words.



Anyways, I want to express the following:



$$phi = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2$$



My first instinct would be to square the divergence of $u_i$, but this would result in:



$$ left(frac{partial u_k}{partial x_k}right)^2 = left(frac{partial u_1}{partial x_1} + frac{partial u_2}{partial x_2}+ frac{partial u_3}{partial x_3}right)^2$$



I've narrowed it down to either of the following, but they both seem more like "hacks" than correct. I've also included the general line of thinking behind the ideas as well.



Idea 1:



$$ text{Idea 1: }quad frac{partial (u_k)^2}{partial (x_k)^2}=frac{partial (u_1)^2}{partial (x_1)^2} + frac{partial (u_2)^2}{partial (x_2)^2} + frac{partial (u_3)^2}{partial (x_3)^2} = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 $$



This being analogous to:



$$ text{Inspiration for Idea 1:} quad frac{(A)^2}{(B)^2} = left(frac{A}{B}right)^2$$



Idea 2:



$$ text{Idea 2: } quad frac{partial u_k}{partial x_k}frac{partial u_k}{partial x_k} = frac{partial u_1}{partial x_1}frac{partial u_1}{partial x_1} + frac{partial u_2}{partial x_2}frac{partial u_2}{partial x_2} + frac{partial u_3}{partial x_3}frac{partial u_3}{partial x_3} = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 $$



This idea was inspired from the incompressible form of the shear stress equations for fluids:



$$ text{Inspiration for Idea 2:} quad frac{partial^2 tau_{ij}}{partial x_j partial x_j} = frac{partial^2 tau_{i1}}{partial x_1^2} + frac{partial^2 tau_{i2}}{partial x_2^2} + frac{partial^2 tau_{i3}}{partial x_3^2}$$




On a slightly related note, if anyone knows of a good resource that goes over common index summation notation forms of common math expressions or the "rules" of index summation notation, let me know!






index-notation







share|cite|improve this question









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    0












    $begingroup$


    Sorry if this has already been asked before, but it's really difficult to try and explain the problem in words.



    Anyways, I want to express the following:



    $$phi = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2$$



    My first instinct would be to square the divergence of $u_i$, but this would result in:



    $$ left(frac{partial u_k}{partial x_k}right)^2 = left(frac{partial u_1}{partial x_1} + frac{partial u_2}{partial x_2}+ frac{partial u_3}{partial x_3}right)^2$$



    I've narrowed it down to either of the following, but they both seem more like "hacks" than correct. I've also included the general line of thinking behind the ideas as well.



    Idea 1:



    $$ text{Idea 1: }quad frac{partial (u_k)^2}{partial (x_k)^2}=frac{partial (u_1)^2}{partial (x_1)^2} + frac{partial (u_2)^2}{partial (x_2)^2} + frac{partial (u_3)^2}{partial (x_3)^2} = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 $$



    This being analogous to:



    $$ text{Inspiration for Idea 1:} quad frac{(A)^2}{(B)^2} = left(frac{A}{B}right)^2$$



    Idea 2:



    $$ text{Idea 2: } quad frac{partial u_k}{partial x_k}frac{partial u_k}{partial x_k} = frac{partial u_1}{partial x_1}frac{partial u_1}{partial x_1} + frac{partial u_2}{partial x_2}frac{partial u_2}{partial x_2} + frac{partial u_3}{partial x_3}frac{partial u_3}{partial x_3} = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 $$



    This idea was inspired from the incompressible form of the shear stress equations for fluids:



    $$ text{Inspiration for Idea 2:} quad frac{partial^2 tau_{ij}}{partial x_j partial x_j} = frac{partial^2 tau_{i1}}{partial x_1^2} + frac{partial^2 tau_{i2}}{partial x_2^2} + frac{partial^2 tau_{i3}}{partial x_3^2}$$




    On a slightly related note, if anyone knows of a good resource that goes over common index summation notation forms of common math expressions or the "rules" of index summation notation, let me know!






    index-notation







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Sorry if this has already been asked before, but it's really difficult to try and explain the problem in words.



      Anyways, I want to express the following:



      $$phi = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2$$



      My first instinct would be to square the divergence of $u_i$, but this would result in:



      $$ left(frac{partial u_k}{partial x_k}right)^2 = left(frac{partial u_1}{partial x_1} + frac{partial u_2}{partial x_2}+ frac{partial u_3}{partial x_3}right)^2$$



      I've narrowed it down to either of the following, but they both seem more like "hacks" than correct. I've also included the general line of thinking behind the ideas as well.



      Idea 1:



      $$ text{Idea 1: }quad frac{partial (u_k)^2}{partial (x_k)^2}=frac{partial (u_1)^2}{partial (x_1)^2} + frac{partial (u_2)^2}{partial (x_2)^2} + frac{partial (u_3)^2}{partial (x_3)^2} = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 $$



      This being analogous to:



      $$ text{Inspiration for Idea 1:} quad frac{(A)^2}{(B)^2} = left(frac{A}{B}right)^2$$



      Idea 2:



      $$ text{Idea 2: } quad frac{partial u_k}{partial x_k}frac{partial u_k}{partial x_k} = frac{partial u_1}{partial x_1}frac{partial u_1}{partial x_1} + frac{partial u_2}{partial x_2}frac{partial u_2}{partial x_2} + frac{partial u_3}{partial x_3}frac{partial u_3}{partial x_3} = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 $$



      This idea was inspired from the incompressible form of the shear stress equations for fluids:



      $$ text{Inspiration for Idea 2:} quad frac{partial^2 tau_{ij}}{partial x_j partial x_j} = frac{partial^2 tau_{i1}}{partial x_1^2} + frac{partial^2 tau_{i2}}{partial x_2^2} + frac{partial^2 tau_{i3}}{partial x_3^2}$$




      On a slightly related note, if anyone knows of a good resource that goes over common index summation notation forms of common math expressions or the "rules" of index summation notation, let me know!






      index-notation







      share|cite|improve this question









      $endgroup$




      Sorry if this has already been asked before, but it's really difficult to try and explain the problem in words.



      Anyways, I want to express the following:



      $$phi = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2$$



      My first instinct would be to square the divergence of $u_i$, but this would result in:



      $$ left(frac{partial u_k}{partial x_k}right)^2 = left(frac{partial u_1}{partial x_1} + frac{partial u_2}{partial x_2}+ frac{partial u_3}{partial x_3}right)^2$$



      I've narrowed it down to either of the following, but they both seem more like "hacks" than correct. I've also included the general line of thinking behind the ideas as well.



      Idea 1:



      $$ text{Idea 1: }quad frac{partial (u_k)^2}{partial (x_k)^2}=frac{partial (u_1)^2}{partial (x_1)^2} + frac{partial (u_2)^2}{partial (x_2)^2} + frac{partial (u_3)^2}{partial (x_3)^2} = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 $$



      This being analogous to:



      $$ text{Inspiration for Idea 1:} quad frac{(A)^2}{(B)^2} = left(frac{A}{B}right)^2$$



      Idea 2:



      $$ text{Idea 2: } quad frac{partial u_k}{partial x_k}frac{partial u_k}{partial x_k} = frac{partial u_1}{partial x_1}frac{partial u_1}{partial x_1} + frac{partial u_2}{partial x_2}frac{partial u_2}{partial x_2} + frac{partial u_3}{partial x_3}frac{partial u_3}{partial x_3} = left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 $$



      This idea was inspired from the incompressible form of the shear stress equations for fluids:



      $$ text{Inspiration for Idea 2:} quad frac{partial^2 tau_{ij}}{partial x_j partial x_j} = frac{partial^2 tau_{i1}}{partial x_1^2} + frac{partial^2 tau_{i2}}{partial x_2^2} + frac{partial^2 tau_{i3}}{partial x_3^2}$$




      On a slightly related note, if anyone knows of a good resource that goes over common index summation notation forms of common math expressions or the "rules" of index summation notation, let me know!






      index-notation




      index-notation






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      asked Jan 28 at 2:02









      u2berggeistu2berggeist

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          $begingroup$

          I think I've found a "correct" way of doing it.



          Using the Kronecker delta, $delta_{ij}$:



          $$ delta_{ij} =left{
          begin{array}{ll}
          0 & inot=j \
          1 & i=j \
          end{array}
          right.
          $$



          Using that, we can get:



          $$ left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 = frac{partial u_j}{partial x_k} frac{partial u_j}{partial x_k} delta_{jk} $$



          This way, we use don't repeat indices more than twice and ensure that the two pair of indices are the same (and so square themselves).






          share|cite|improve this answer









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            1 Answer
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            active

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            0












            $begingroup$

            I think I've found a "correct" way of doing it.



            Using the Kronecker delta, $delta_{ij}$:



            $$ delta_{ij} =left{
            begin{array}{ll}
            0 & inot=j \
            1 & i=j \
            end{array}
            right.
            $$



            Using that, we can get:



            $$ left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 = frac{partial u_j}{partial x_k} frac{partial u_j}{partial x_k} delta_{jk} $$



            This way, we use don't repeat indices more than twice and ensure that the two pair of indices are the same (and so square themselves).






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I think I've found a "correct" way of doing it.



              Using the Kronecker delta, $delta_{ij}$:



              $$ delta_{ij} =left{
              begin{array}{ll}
              0 & inot=j \
              1 & i=j \
              end{array}
              right.
              $$



              Using that, we can get:



              $$ left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 = frac{partial u_j}{partial x_k} frac{partial u_j}{partial x_k} delta_{jk} $$



              This way, we use don't repeat indices more than twice and ensure that the two pair of indices are the same (and so square themselves).






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I think I've found a "correct" way of doing it.



                Using the Kronecker delta, $delta_{ij}$:



                $$ delta_{ij} =left{
                begin{array}{ll}
                0 & inot=j \
                1 & i=j \
                end{array}
                right.
                $$



                Using that, we can get:



                $$ left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 = frac{partial u_j}{partial x_k} frac{partial u_j}{partial x_k} delta_{jk} $$



                This way, we use don't repeat indices more than twice and ensure that the two pair of indices are the same (and so square themselves).






                share|cite|improve this answer









                $endgroup$



                I think I've found a "correct" way of doing it.



                Using the Kronecker delta, $delta_{ij}$:



                $$ delta_{ij} =left{
                begin{array}{ll}
                0 & inot=j \
                1 & i=j \
                end{array}
                right.
                $$



                Using that, we can get:



                $$ left(frac{partial u_1}{partial x_1}right)^2 + left(frac{partial u_2}{partial x_2}right)^2 + left(frac{partial u_3}{partial x_3}right)^2 = frac{partial u_j}{partial x_k} frac{partial u_j}{partial x_k} delta_{jk} $$



                This way, we use don't repeat indices more than twice and ensure that the two pair of indices are the same (and so square themselves).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 22:41









                u2berggeistu2berggeist

                83




                83






























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