Necessary and Sufficient Condition for First Order System to be Field for Functional

Theorem
Let $ \mathbf y $ be an N-dimensional vector.

Let $ J $ be a functional such that:


 * $ \displaystyle J \left [ { \mathbf y } \right ] = \int_a^b F \left ( { x, \mathbf y, \mathbf y' } \right ) \mathrm d x $

Let the corresponding momenta and Hamiltonian be:


 * $ \displaystyle \mathbf p \left ( { x, \mathbf y, \mathbf y' } \right ) = F_{ \mathbf y' } \left ( { x, \mathbf y, \mathbf y' } \right) $


 * $ \displaystyle H \left ( { x, \mathbf y, \mathbf y' } \right ) = - F \left ( { x, \mathbf y, \mathbf y' } \right) + \mathbf p \mathbf y' $

Let the following be a family of boundary conditions:


 * $ \displaystyle \mathbf y' \left ( { x } \right ) = \boldsymbol \psi \left ( { x, \mathbf y } \right ) $

Then a family of boundary conditions is a field for the functional $ J $ iff $ \forall x \in \left [ { a \,. \,. \, b } \right ] $ the following self-adjointness and consistency relations hold:


 * $ \displaystyle \frac{ \partial p_i \left [ { x, \mathbf y, \boldsymbol \psi \left ( { x, \mathbf y } \right ) } \right ] }{ \partial y_k } = \frac{ \partial p_k \left [ { x, \mathbf y, \boldsymbol \psi \left ( { x, \mathbf y } \right ) } \right ] }{ \partial y_i } $


 * $ \displaystyle \frac{ \partial \mathbf p \left [ { x, \mathbf y, \boldsymbol \psi \left ( { x, \mathbf y } \right ) } \right ] }{ \partial x } = - H_{ \mathbf y } \left [ { x, \mathbf y, \boldsymbol \psi \left ( { x, \mathbf y } \right ) } \right ] $

Necessary Condition
Set $ \mathbf y' = \boldsymbol \psi \left ( { x, \mathbf y } \right ) $ in the definition of momenta and Hamiltonian.

Substitute corresponding definitions into the consistency relation.

On the we have:

On the we have:

Together they imply:


 * $ \displaystyle F_{ y_i' x } \vert_{ \boldsymbol \psi = const } + F_{ y_i' \boldsymbol \psi } \frac{ \partial \boldsymbol \psi }{ \partial x } = F_{ y_i } \vert_{ \boldsymbol \psi = const } + F_{ \boldsymbol \psi } \frac{ \partial \boldsymbol \psi }{ \partial y_i } -  F_{ \mathbf y' y_i } \vert_{ \boldsymbol \psi = const } \boldsymbol \psi  - F_{ \mathbf y' } \boldsymbol \psi_{ y_i } $

Use self-adjointness conditions:


 * $ \displaystyle \frac{ \partial F_{ y_k' } }{ \partial y_i } = \frac{ \partial F_{ y_i' } }{ \partial y_k }$

Then:


 * $ \displaystyle F_{ y_i } = F_{ y_i' x } + \sum_{ k = 1 }^N F_{ y_i' y_k' } \frac{ \partial \psi_k }{ \partial x } + \sum_{ k = 1 }^N \psi_k \frac{ \partial F_{ y_i' } }{ \partial y_k } $

Also, since:


 * $ \displaystyle \frac{ \partial F_{ y_i' } }{ \partial y_k } = F_{ y_i' y_k } + \sum_{ j = 1 }^N F_{ y_i' y_j' } \frac{ \partial \psi_j }{ \partial y_k }$

the previous equation becomes


 * $ F_{ y_i } = F_{ y_i' x } + \sum_{ k = 1 }^N F_{ y_i' y_k } \psi_k + \sum_{ k = 1 }^N F_{ y_i' y_k' } \left ( { \frac{ \partial \psi_k }{ \partial x } + \sum_{ j = 1 }^N \frac{ \partial \psi_k }{ \partial y_j } \psi_j } \right ) $

Along trajectories of the field we have:


 * $ \displaystyle \frac{ \mathrm d y_k }{ \mathrm d x } = \psi_k $

Hence:


 * $ \displaystyle \frac{ \mathrm d^2 y_k }{ \mathrm d x^2 } = \frac{ \partial \psi_k }{ \partial x } + \sum_{ j = 1 }^N \frac{ \partial \psi_k }{ \partial y_j } \psi_j$

Thus, the main equation reduces to


 * $ \displaystyle F_{ y_i } = F_{ y_i' x } + \sum_{ k = 1 }^N F_{ y_i' y_k } \frac{ \mathrm d y_k }{ \mathrm d x } + \sum_{ k = 1 }^N F_{ y_i' y_k' } \frac{ \mathrm d^2 y_k }{ \mathrm d x^2 }$

or


 * $ \displaystyle F_{ y_i } - \frac{ \mathrm d }{ \mathrm d x } F_{ y_i' } = 0 $

These are the extremals of $ J $.