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 ) = \frac{ \partial F \left ( { x, \mathbf y, \mathbf y' } \right) }{ \partial \mathbf y' } $


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

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 \frac{ \partial^2 F }{ \partial x \partial y_i' } = \frac{ \partial F }{ \partial y_i } - \frac{ \partial^2 F }{ \partial y_i \partial \mathbf y' } \boldsymbol \psi - \frac{ \partial F }{ \partial \mathbf y' } \frac{ \partial \boldsymbol \psi }{ \partial y_i} $

Use self-adjointness conditions:


 * $ \displaystyle \frac{ \partial^2 F }{ \partial y_i \partial y_k' } = \frac{ \partial^2 F }{ \partial y_k \partial y_i' } $

Then:


 * $ \displaystyle \frac{ \partial^2 F }{ \partial x \partial y_i' } = \frac{ \partial F }{ \partial y_i } - \frac{ \partial^2 F }{ \partial \mathbf y \partial y_i' } \boldsymbol \psi - \frac{ \partial F }{ \partial \mathbf y' } \frac{ \partial \boldsymbol \psi }{ \partial y_i} $

$ F $ depends on $ \mathbf y' $ only through its third vector variable, thus $ \displaystyle \frac{ \partial F }{ \partial y_k } = F_{ y_k } $:


 * $ \displaystyle \frac{ \partial F_{ y_i' } }{ \partial x } = \frac{ \partial F }{ \partial y_i } - \frac{ \partial F_{ y_i' } }{ \partial \mathbf y } \boldsymbol \psi - F_{ \mathbf y' } \frac{ \partial \boldsymbol \psi }{ \partial y_i} $

$ F $ depends on $ x $ directly through its first variable and indirectly through its third variable together with boundary conditions:


 * $ \displaystyle \frac{ \partial F_{ y_i' } }{ \partial x } = F_{ y_i' x } + F_{ y_i' \mathbf y' } \boldsymbol \psi_x $

$ F $ depends on $ \mathbf y $ directly through its second vector variable and indirectly through its third variable together with boundary conditions:


 * $\displaystyle \frac{ \partial F }{ \partial y_i } = F_{ y_i } + F_{ \mathbf y' } \boldsymbol \psi_{ y_i } $


 * $\displaystyle \frac{ \partial F_{ y_i' } }{ \partial \mathbf y } = F_{ y_i' \mathbf y } + \sum_{ k = 1 }^N F_{ y_i' y_k' } \boldsymbol \psi_{ \mathbf y } $