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 ) \quad \left ( { 1 } \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} $

By Necessary and Sufficient Condition for Boundary Conditions to be Self-adjoint:


 * $ \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} \quad \left ( { 2 } \right ) $

$ F $ depends on $ x $ directly through its first variable and indirectly through its third vector variable together with boundary conditions $ \left ( { 1 } \right ) $:


 * $ \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 vector variable together with boundary conditions $ \left ( { 1 } \right ) $:


 * $ \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' } \frac{ \partial \psi_k }{ \partial \mathbf y } $

Substitution of the last three equations into $ \left ( { 2} \right ) $ leads to:


 * $ \displaystyle F_{ y_i' x } + F_{ y_i' \mathbf y' } \boldsymbol \psi_x = F_{ y_i } + F_{ \mathbf y' } \boldsymbol \psi_{ y_i } - \left ( { F_{ y_i' \mathbf y } + \sum_{ k = 1 }^N F_{ y_i' y_k' } \frac{ \partial \psi_k }{ \partial \mathbf y } } \right ) \boldsymbol \psi - F_{ \mathbf y' } \frac{ \partial \boldsymbol \psi }{ \partial y_i} $

which can be simplified to:


 * $ \displaystyle F_{ y_i } - F_{ y_i'x } - F_{ y_i' \mathbf y } \boldsymbol \psi - F_{ y_i' y_j' } \left ( { \frac{ \partial \psi_j }{ \partial x } + \frac{ \partial \psi_j }{ \partial y_j } \psi_j } \right ) = 0 $

By assumption:


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

the second total derivative $ x $ of which yields:

Hence:


 * $ \displaystyle F_{ y_i } - \left [ { F_{ y_i' x } + F_{ y_i' \mathbf y } \frac{ \mathrm d \mathbf y }{ \mathrm d x } + F_{ y_i' \mathbf y' } \frac{ \mathrm d \mathbf y' }{ \mathrm d x } } \right ] = 0 $

The second term is just a total derivative $ x $, thus:


 * $ \displaystyle F_{ y_i } - \frac{ \mathrm d F_{ y_i' } }{ \mathrm d x } = 0 \quad \left ( { 3 } \right ) $

Boundary conditions $ \left ( { 1 } \right ) $ are mutually consistent equation $ \left ( { 3 } \right ) $ because they hold $ \forall x \in \left [ { a \,. \,. \, b } \right] $.

By definition, they are consistent the functional $ J $.

Since the boundary conditions are consistent $ J $ and self-adjoint, by definition they constitute a field of $ J $.

Sufficient Condition
By assumption, $ \left ( { 1 } \right ) $ is a field of $ J $.

Hence, $ \left ( { 1 } \right ) $ is self-adjoint and mutually consistent $ J $.

Thus, by definition, they are consistent :


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

The can be rewritten as follows:

The vanishes.

Therefore:


 * $ \displaystyle \frac{ \partial \mathbf p }{ \partial x } = - \frac{ \partial H }{ \partial \mathbf y} $

By Necessary and Sufficient Condition for Boundary Conditions to be Self-adjoint:


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

By definition:


 * $ \displaystyle \mathbf p = \frac{ \partial F }{ \partial \mathbf y'} $

Hence:


 * $ \displaystyle \frac{ \partial p_k }{ \partial y_i } = \frac{ \partial p_i}{ \partial y_k } $