How calculate extremes of the functional?












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Is it also here to use the Euler-Lagrange equation? Could someone tell me how it will look like?



$${F}_{u} = int_{0}^{1} left( uu' + uu''^{2} + uu'' + u'u'' + 2u'' right) mbox{d}x$$



$$u(0) = u'(0) = u(1) = 0$$



$$u'(1) = 1$$










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$endgroup$

















    1












    $begingroup$


    Is it also here to use the Euler-Lagrange equation? Could someone tell me how it will look like?



    $${F}_{u} = int_{0}^{1} left( uu' + uu''^{2} + uu'' + u'u'' + 2u'' right) mbox{d}x$$



    $$u(0) = u'(0) = u(1) = 0$$



    $$u'(1) = 1$$










    share|cite|improve this question









    $endgroup$















      1












      1








      1


      0



      $begingroup$


      Is it also here to use the Euler-Lagrange equation? Could someone tell me how it will look like?



      $${F}_{u} = int_{0}^{1} left( uu' + uu''^{2} + uu'' + u'u'' + 2u'' right) mbox{d}x$$



      $$u(0) = u'(0) = u(1) = 0$$



      $$u'(1) = 1$$










      share|cite|improve this question









      $endgroup$




      Is it also here to use the Euler-Lagrange equation? Could someone tell me how it will look like?



      $${F}_{u} = int_{0}^{1} left( uu' + uu''^{2} + uu'' + u'u'' + 2u'' right) mbox{d}x$$



      $$u(0) = u'(0) = u(1) = 0$$



      $$u'(1) = 1$$







      functional-analysis euler-lagrange-equation






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      asked Jan 14 at 12:33









      SvsSvs

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      62






















          2 Answers
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          $begingroup$

          $${L} = fleft(x,u,u',u''right)$$



          $$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$



          But I don't know how calculate this diverates:
          $$frac{ partial {L} }{ mbox{d}u } =$$



          $$frac{ partial {L} }{ mbox{d}u' }= $$



          $$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$



          $$frac{ partial {L} }{ mbox{d}u'' } = $$



          $$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$






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            0












            $begingroup$

            Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
            $$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$






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              2 Answers
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              2 Answers
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              0












              $begingroup$

              $${L} = fleft(x,u,u',u''right)$$



              $$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$



              But I don't know how calculate this diverates:
              $$frac{ partial {L} }{ mbox{d}u } =$$



              $$frac{ partial {L} }{ mbox{d}u' }= $$



              $$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$



              $$frac{ partial {L} }{ mbox{d}u'' } = $$



              $$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $${L} = fleft(x,u,u',u''right)$$



                $$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$



                But I don't know how calculate this diverates:
                $$frac{ partial {L} }{ mbox{d}u } =$$



                $$frac{ partial {L} }{ mbox{d}u' }= $$



                $$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$



                $$frac{ partial {L} }{ mbox{d}u'' } = $$



                $$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $${L} = fleft(x,u,u',u''right)$$



                  $$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$



                  But I don't know how calculate this diverates:
                  $$frac{ partial {L} }{ mbox{d}u } =$$



                  $$frac{ partial {L} }{ mbox{d}u' }= $$



                  $$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$



                  $$frac{ partial {L} }{ mbox{d}u'' } = $$



                  $$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$






                  share|cite|improve this answer









                  $endgroup$



                  $${L} = fleft(x,u,u',u''right)$$



                  $$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$



                  But I don't know how calculate this diverates:
                  $$frac{ partial {L} }{ mbox{d}u } =$$



                  $$frac{ partial {L} }{ mbox{d}u' }= $$



                  $$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$



                  $$frac{ partial {L} }{ mbox{d}u'' } = $$



                  $$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 15 at 20:00









                  SvsSvs

                  62




                  62























                      0












                      $begingroup$

                      Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
                      $$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
                        $$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
                          $$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$






                          share|cite|improve this answer









                          $endgroup$



                          Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
                          $$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 17 at 20:58









                          QmechanicQmechanic

                          5,17711858




                          5,17711858






























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