Proof: There is no function $f in C^2(mathbb{R^3})$ with gradient $nabla f(x,y,z) = (yz, xz, xy^2)$












2














How can one show that there is no function, which is a continuously partially derivable function $f in C^2(mathbb{R^3})$ with this gradient



$$nabla f(x,y,z) = (yz, xz, xy^2)$$



I thought about using the Hessian matrix since one has to calculate all second partial derivatives of $f$ there.



Since only the gradient is given, can I calculate the antiderivatives first:



$yz = xyz$



$xz = xyz$



$xy^2 = xy^2z$



Now I want to calculate the antiderivatives of the antiderivatives:



$xyz = frac{yzx^2}{2}$



$xyz = frac{yzx^2}{2}$



$xy^2z = dfrac{y^2zx^2}{2}$



I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...






analysis functions derivatives continuity partial-derivative







share|cite|improve this question



























    2














    How can one show that there is no function, which is a continuously partially derivable function $f in C^2(mathbb{R^3})$ with this gradient



    $$nabla f(x,y,z) = (yz, xz, xy^2)$$



    I thought about using the Hessian matrix since one has to calculate all second partial derivatives of $f$ there.



    Since only the gradient is given, can I calculate the antiderivatives first:



    $yz = xyz$



    $xz = xyz$



    $xy^2 = xy^2z$



    Now I want to calculate the antiderivatives of the antiderivatives:



    $xyz = frac{yzx^2}{2}$



    $xyz = frac{yzx^2}{2}$



    $xy^2z = dfrac{y^2zx^2}{2}$



    I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...






    analysis functions derivatives continuity partial-derivative







    share|cite|improve this question

























      2












      2








      2


      1





      How can one show that there is no function, which is a continuously partially derivable function $f in C^2(mathbb{R^3})$ with this gradient



      $$nabla f(x,y,z) = (yz, xz, xy^2)$$



      I thought about using the Hessian matrix since one has to calculate all second partial derivatives of $f$ there.



      Since only the gradient is given, can I calculate the antiderivatives first:



      $yz = xyz$



      $xz = xyz$



      $xy^2 = xy^2z$



      Now I want to calculate the antiderivatives of the antiderivatives:



      $xyz = frac{yzx^2}{2}$



      $xyz = frac{yzx^2}{2}$



      $xy^2z = dfrac{y^2zx^2}{2}$



      I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...






      analysis functions derivatives continuity partial-derivative







      share|cite|improve this question













      How can one show that there is no function, which is a continuously partially derivable function $f in C^2(mathbb{R^3})$ with this gradient



      $$nabla f(x,y,z) = (yz, xz, xy^2)$$



      I thought about using the Hessian matrix since one has to calculate all second partial derivatives of $f$ there.



      Since only the gradient is given, can I calculate the antiderivatives first:



      $yz = xyz$



      $xz = xyz$



      $xy^2 = xy^2z$



      Now I want to calculate the antiderivatives of the antiderivatives:



      $xyz = frac{yzx^2}{2}$



      $xyz = frac{yzx^2}{2}$



      $xy^2z = dfrac{y^2zx^2}{2}$



      I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...






      analysis functions derivatives continuity partial-derivative




      analysis functions derivatives continuity partial-derivative






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 22 '18 at 22:13









      Bad At MathBad At Math

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          If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.






          share|cite|improve this answer





























            1














            In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.






            share|cite|improve this answer





















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              If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.






              share|cite|improve this answer


























                2














                If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.






                share|cite|improve this answer
























                  2












                  2








                  2






                  If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.






                  share|cite|improve this answer












                  If $$nabla f=(f_1(x,y,z),f_2(x,y,z),f_3(x,y,z))$$and $f_i$s are twice differentiable, we must have $${partial ^2 f_1over partial ypartial z}={partial ^2 f_2over partial xpartial z}={partial ^2 f_3over partial xpartial y}$$for all $x,y,z$ but this doesn't hold here since $$1=1ne 2yqquad forall x,y,zin Bbb R$$therefore such a function doesn't exist.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 '18 at 22:25









                  Mostafa AyazMostafa Ayaz

                  14.2k3937




                  14.2k3937























                      1














                      In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.






                      share|cite|improve this answer


























                        1














                        In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.






                        share|cite|improve this answer
























                          1












                          1








                          1






                          In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.






                          share|cite|improve this answer












                          In other words, you are asking to show that your vector field is not conservative. It is enough to show that the curl of your vector field is different to zero.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 22 '18 at 22:27









                          Daniel DuqueDaniel Duque

                          267




                          267






























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