One Sided Approximation for Mixed Derivatives












3












$begingroup$


Consider the function u(x,y,z)



I am trying to approximate the partial derivative at point (i,j,k) by one sided finite difference method.



Now using one sided 2nd order finite difference approxmation for the first derivative, we have



$u_y(x,y,z) = frac{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}{Delta{y}}$



Now applying this for second derivative, we have



$u_{xy} = frac{3u_y(x,y,z) - 4u_y(x-Delta{x},y,z) + u_y(x-2Delta{x},y,z)}{Delta{x}}$



$ = frac{3[frac{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}{Delta{y}}] - 4[frac{3u(x-Delta{x},y,z) - 4u(x-Delta{x},y-Delta{y},z) + u(x-Delta{x},y-2Delta{y},z)}{Delta{y}} ] + [frac{3u(x-2Delta{x},y,z) - 4u(x-2Delta{x},y-Delta{y},z) + u(x-2Delta{x},y-2Delta{y},z)}{Delta{y}}]}{Delta{x}}$



$ = frac{3[{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}] - 4[{3u(x-Delta{x},y,z) - 4u(x-Delta{x},y-Delta{y},z) + u(x-Delta{x},y-2Delta{y},z)} ] + [{3u(x-2Delta{x},y,z) - 4u(x-2Delta{x},y-Delta{y},z) + u(x-2Delta{x},y-2Delta{y},z)}]}{Delta{x}{Delta{y}}} $



$ = $ resulting exp



My query is that for one sided approximation, is this the simplest form of mixed derivative that is obtainable or have I missed something that simplifies the resulting equation?






numerical-methods partial-derivative finite-differences







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Consider the function u(x,y,z)



    I am trying to approximate the partial derivative at point (i,j,k) by one sided finite difference method.



    Now using one sided 2nd order finite difference approxmation for the first derivative, we have



    $u_y(x,y,z) = frac{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}{Delta{y}}$



    Now applying this for second derivative, we have



    $u_{xy} = frac{3u_y(x,y,z) - 4u_y(x-Delta{x},y,z) + u_y(x-2Delta{x},y,z)}{Delta{x}}$



    $ = frac{3[frac{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}{Delta{y}}] - 4[frac{3u(x-Delta{x},y,z) - 4u(x-Delta{x},y-Delta{y},z) + u(x-Delta{x},y-2Delta{y},z)}{Delta{y}} ] + [frac{3u(x-2Delta{x},y,z) - 4u(x-2Delta{x},y-Delta{y},z) + u(x-2Delta{x},y-2Delta{y},z)}{Delta{y}}]}{Delta{x}}$



    $ = frac{3[{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}] - 4[{3u(x-Delta{x},y,z) - 4u(x-Delta{x},y-Delta{y},z) + u(x-Delta{x},y-2Delta{y},z)} ] + [{3u(x-2Delta{x},y,z) - 4u(x-2Delta{x},y-Delta{y},z) + u(x-2Delta{x},y-2Delta{y},z)}]}{Delta{x}{Delta{y}}} $



    $ = $ resulting exp



    My query is that for one sided approximation, is this the simplest form of mixed derivative that is obtainable or have I missed something that simplifies the resulting equation?






    numerical-methods partial-derivative finite-differences







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Consider the function u(x,y,z)



      I am trying to approximate the partial derivative at point (i,j,k) by one sided finite difference method.



      Now using one sided 2nd order finite difference approxmation for the first derivative, we have



      $u_y(x,y,z) = frac{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}{Delta{y}}$



      Now applying this for second derivative, we have



      $u_{xy} = frac{3u_y(x,y,z) - 4u_y(x-Delta{x},y,z) + u_y(x-2Delta{x},y,z)}{Delta{x}}$



      $ = frac{3[frac{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}{Delta{y}}] - 4[frac{3u(x-Delta{x},y,z) - 4u(x-Delta{x},y-Delta{y},z) + u(x-Delta{x},y-2Delta{y},z)}{Delta{y}} ] + [frac{3u(x-2Delta{x},y,z) - 4u(x-2Delta{x},y-Delta{y},z) + u(x-2Delta{x},y-2Delta{y},z)}{Delta{y}}]}{Delta{x}}$



      $ = frac{3[{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}] - 4[{3u(x-Delta{x},y,z) - 4u(x-Delta{x},y-Delta{y},z) + u(x-Delta{x},y-2Delta{y},z)} ] + [{3u(x-2Delta{x},y,z) - 4u(x-2Delta{x},y-Delta{y},z) + u(x-2Delta{x},y-2Delta{y},z)}]}{Delta{x}{Delta{y}}} $



      $ = $ resulting exp



      My query is that for one sided approximation, is this the simplest form of mixed derivative that is obtainable or have I missed something that simplifies the resulting equation?






      numerical-methods partial-derivative finite-differences







      share|cite|improve this question









      $endgroup$




      Consider the function u(x,y,z)



      I am trying to approximate the partial derivative at point (i,j,k) by one sided finite difference method.



      Now using one sided 2nd order finite difference approxmation for the first derivative, we have



      $u_y(x,y,z) = frac{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}{Delta{y}}$



      Now applying this for second derivative, we have



      $u_{xy} = frac{3u_y(x,y,z) - 4u_y(x-Delta{x},y,z) + u_y(x-2Delta{x},y,z)}{Delta{x}}$



      $ = frac{3[frac{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}{Delta{y}}] - 4[frac{3u(x-Delta{x},y,z) - 4u(x-Delta{x},y-Delta{y},z) + u(x-Delta{x},y-2Delta{y},z)}{Delta{y}} ] + [frac{3u(x-2Delta{x},y,z) - 4u(x-2Delta{x},y-Delta{y},z) + u(x-2Delta{x},y-2Delta{y},z)}{Delta{y}}]}{Delta{x}}$



      $ = frac{3[{3u(x,y,z) - 4u(x,y-Delta{y},z) + u(x,y-2Delta{y},z)}] - 4[{3u(x-Delta{x},y,z) - 4u(x-Delta{x},y-Delta{y},z) + u(x-Delta{x},y-2Delta{y},z)} ] + [{3u(x-2Delta{x},y,z) - 4u(x-2Delta{x},y-Delta{y},z) + u(x-2Delta{x},y-2Delta{y},z)}]}{Delta{x}{Delta{y}}} $



      $ = $ resulting exp



      My query is that for one sided approximation, is this the simplest form of mixed derivative that is obtainable or have I missed something that simplifies the resulting equation?






      numerical-methods partial-derivative finite-differences




      numerical-methods partial-derivative finite-differences






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Jun 12 '13 at 23:44









      rajaditya_mrajaditya_m

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          $begingroup$

          It's never too late for an answer.



          The answer is YES. This is the simplest (=shortest) form of a second order finite difference for the mixed derivative in $x$ and $y$ direction, if you want both directions to be one-sided. (I assume this is what you want, although you are talking about a second derivitive.)



          Since you have



          $u(x,y,z), (x,y-Delta{y},z), u(x,y-2Delta{y},z), u(x-Delta{x},y,z), u(x-Delta{x},y-Delta{y},z), u(x-Delta{x},y-2Delta{y},z), u(x-2Delta{x},y,z), u(x-2Delta{x},y-Delta{y},z), u(x-2Delta{x},y-2Delta{y},z)$



          only once in your representation, there cannot be a shorter (=simpler) one. The only thing you could do is to multiply out the brackets.






          share|cite|improve this answer











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            1 Answer
            1






            active

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            active

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            active

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            $begingroup$

            It's never too late for an answer.



            The answer is YES. This is the simplest (=shortest) form of a second order finite difference for the mixed derivative in $x$ and $y$ direction, if you want both directions to be one-sided. (I assume this is what you want, although you are talking about a second derivitive.)



            Since you have



            $u(x,y,z), (x,y-Delta{y},z), u(x,y-2Delta{y},z), u(x-Delta{x},y,z), u(x-Delta{x},y-Delta{y},z), u(x-Delta{x},y-2Delta{y},z), u(x-2Delta{x},y,z), u(x-2Delta{x},y-Delta{y},z), u(x-2Delta{x},y-2Delta{y},z)$



            only once in your representation, there cannot be a shorter (=simpler) one. The only thing you could do is to multiply out the brackets.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              It's never too late for an answer.



              The answer is YES. This is the simplest (=shortest) form of a second order finite difference for the mixed derivative in $x$ and $y$ direction, if you want both directions to be one-sided. (I assume this is what you want, although you are talking about a second derivitive.)



              Since you have



              $u(x,y,z), (x,y-Delta{y},z), u(x,y-2Delta{y},z), u(x-Delta{x},y,z), u(x-Delta{x},y-Delta{y},z), u(x-Delta{x},y-2Delta{y},z), u(x-2Delta{x},y,z), u(x-2Delta{x},y-Delta{y},z), u(x-2Delta{x},y-2Delta{y},z)$



              only once in your representation, there cannot be a shorter (=simpler) one. The only thing you could do is to multiply out the brackets.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                It's never too late for an answer.



                The answer is YES. This is the simplest (=shortest) form of a second order finite difference for the mixed derivative in $x$ and $y$ direction, if you want both directions to be one-sided. (I assume this is what you want, although you are talking about a second derivitive.)



                Since you have



                $u(x,y,z), (x,y-Delta{y},z), u(x,y-2Delta{y},z), u(x-Delta{x},y,z), u(x-Delta{x},y-Delta{y},z), u(x-Delta{x},y-2Delta{y},z), u(x-2Delta{x},y,z), u(x-2Delta{x},y-Delta{y},z), u(x-2Delta{x},y-2Delta{y},z)$



                only once in your representation, there cannot be a shorter (=simpler) one. The only thing you could do is to multiply out the brackets.






                share|cite|improve this answer











                $endgroup$



                It's never too late for an answer.



                The answer is YES. This is the simplest (=shortest) form of a second order finite difference for the mixed derivative in $x$ and $y$ direction, if you want both directions to be one-sided. (I assume this is what you want, although you are talking about a second derivitive.)



                Since you have



                $u(x,y,z), (x,y-Delta{y},z), u(x,y-2Delta{y},z), u(x-Delta{x},y,z), u(x-Delta{x},y-Delta{y},z), u(x-Delta{x},y-2Delta{y},z), u(x-2Delta{x},y,z), u(x-2Delta{x},y-Delta{y},z), u(x-2Delta{x},y-2Delta{y},z)$



                only once in your representation, there cannot be a shorter (=simpler) one. The only thing you could do is to multiply out the brackets.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 13 at 11:13









                Hirak

                2,57411437




                2,57411437










                answered Sep 22 '14 at 9:52









                RobRob

                257110




                257110






























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