The original functional was J [ x ] = ∫ 0 1 x ⋅ f ( x ) ⋅ f ′ ( x ) d x with h ( x ) = x. Solved for the Gateau derivative: ∂ J [ f , h ] = ∫ 0 1 x ⋅ f ( x ) ⋅ h ′ ( x ) d x + ∫ 0 1 x ⋅ f ′ ( x ) ⋅ h ( x ) d x. To use the Euler method, L = x ⋅ f ⋅ x + x ⋅ f ′ ⋅ x such that L f = x 2 , L f ′ = x 2 , d d x L f ′ = 2 x. However, I must have made a mistake that I can't see with that because it yields x 2 − 2 x = 0 which doesn't tell me anything about what f must be.

Question

The original functional was      J  [  x  ]  =      ∫          0              1        x  ⋅  f  (  x  )  ⋅      f    ′    (  x  )  d  x with      h  (  x  )  =  x. Solved for the Gateau derivative:       ∂  J  [  f  ,  h  ]  =      ∫          0              1        x  ⋅  f  (  x  )  ⋅      h    ′    (  x  )  d  x  +      ∫          0              1        x  ⋅      f    ′    (  x  )  ⋅  h  (  x  )  d  x. To use the Euler method,      L  =  x  ⋅  f  ⋅  x  +  x  ⋅      f    ′    ⋅  x such that           L          f        =      x    2    ,      L                  f        ′              =      x    2    ,      d          d      x            L                  f        ′              =  2  x. However, I must have made a mistake that I can&#039;t see with that because it yields          x    2    −  2  x  =  0 which doesn&#039;t tell me anything about what      f must be.

Abigayle Lynn · Accepted Answer

The functional is  J  [  f  ]  =      ∫    0    1    L  (  x  ,  f  (  x  )  ,      f    ′    (  x  )  )    d  x  =      ∫    0    1    x  f  (  x  )      f    ′    (  x  )    d  xThe Euler-Lagrange equation for this functional, with   L  =  x  f      f    ′   is given by                              ∂                      ∂            f                          L        (        x        ,        f        ,                  f          ′                )        −                  d                      d            x                                    ∂                      ∂                          f              ′                                      L        (        x        ,        f        ,                  f          ′                )                            =                  ∂                      ∂            f                          (        x        f                  f          ′                )        −                  d                      d            x                                    ∂                      ∂                          f              ′                                      (        x        f                  f          ′                )                                                        =        x                  f          ′                −                  d                      d            x                          (        x        f        )                                                        =        −        f            Therefore, the functional has a stationary point for   f  (  x  )  ≡  0.At   f  =  0,   J  [  f  ]  =  0, which is not necessarily an extreme value. That is to say, there exist       f    1    (  x  ) and       f    2    (  x  ) for which   L  [      f    1    ]  &amp;gt;  0 and   L  [      f    2    ]  &amp;lt;  0 (e.g.,       f    1    (  x  )  =  x and       f    2    (  x  )  =  1  −  x).

The original functional was J[x]=int_0^1 (x)f(x)f′(x)dx with h(x)=x.

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