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

Let $\mathbf y$ be an N-dimensional vector.
Let $J$ be a (real) functional such that:
:$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
Let the corresponding momenta and Hamiltonian be:
{{begin-eqn}}
{{eqn | l = \map {\mathbf p} {x, \mathbf y, \mathbf y'}
      | r = \dfrac {\partial \map F {x, \mathbf y, \mathbf y'} } {\partial \mathbf y'}
}}
{{eqn | l = \map H {x, \mathbf y, \mathbf y'}
      | r = -\map F {x, \mathbf y, \mathbf y'} + \mathbf p \mathbf y'
}}
{{end-eqn}}
Let the following be a family of boundary conditions:
:$(1): \quad \map {\mathbf y'} x = \map \bspsi {x, \mathbf y}$
Then a family of boundary conditions is a field for the functional $J$ {{iff}} $\forall x \in \closedint a b$ the following self-adjointness and consistency relations hold:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial \map {p_i} {x, \mathbf y, \map \bspsi {x, \mathbf y} } } {\partial y_k}
      | r = \dfrac {\partial \map {p_k} {x, \mathbf y, \map \bspsi {x, \mathbf y} } } {\partial y_i}
}}
{{eqn | l = \dfrac {\partial \map {\mathbf p} {x, \mathbf y, \map \bspsi {x, \mathbf y} } } {\partial x}
      | r = -\dfrac {\partial \map H {x, \mathbf y, \map \bspsi {x, \mathbf y} } } {\partial \mathbf y}
}}
{{end-eqn}}
{{explain|Why "consistency"?}}

Proof:
=== Necessary Condition ===
Set $\mathbf y = \map \bspsi {x, \mathbf y}$ in the definition of momenta and Hamiltonian.
Substitute corresponding definitions into the consistency relation.
On the {{LHS}} we have:
:$\dfrac {\partial \map {p_i} {x, \mathbf y, \map \bspsi {x, \mathbf y} } } {\partial x} = \dfrac {\partial^2 \map F {x, \mathbf y, \map \bspsi {x,\mathbf y} } } {\partial x \partial y_i'}$
On the {{RHS}} we have:
{{begin-eqn}}
{{eqn | l = -\dfrac {\partial \map H {x, \mathbf y, \map \bspsi {x, \mathbf y} } } {\partial y_i}
      | r = -\dfrac {\partial \paren {-\map F {x, \mathbf y, \map \bspsi {x, \mathbf y} } + \map {\mathbf p} {x, \mathbf y, \map \bspsi {x, \mathbf y} } \map {\mathbf y'} x} } {\partial y_i} 
}}
{{eqn | r = \frac {\partial F} {\partial y_i} - \frac {\partial^2 F} {\partial y_i \partial \mathbf y'} \bspsi - \frac {\partial F} {\partial \mathbf y'} \frac {\partial \bspsi} {\partial y_i}
}}
{{end-eqn}}
Together they imply:
:$\dfrac {\partial^2 F} {\partial x \partial y_i'} = \dfrac {\partial F} {\partial y_i} - \dfrac {\partial^2 F} {\partial y_i \partial \mathbf y'} \bspsi - \dfrac {\partial F} {\partial \mathbf y'} \dfrac {\partial \bspsi} {\partial y_i}$
By Necessary and Sufficient Condition for Boundary Conditions to be Self-adjoint:
:$\dfrac {\partial^2 F} {\partial y_i \partial y_k'} = \dfrac {\partial^2 F} {\partial y_k \partial y_i'}$
Then:
:$\dfrac {\partial^2 F} {\partial x \partial y_i'} = \dfrac {\partial F} {\partial y_i} - \dfrac {\partial^2 F} {\partial \mathbf y \partial y_i'} \bspsi - \dfrac {\partial F} {\partial \mathbf y'} \dfrac {\partial \bspsi} {\partial y_i}$
$F$ depends on $\mathbf y'$ only through its third vector variable, thus $\dfrac {\partial F} {\partial y_k'} = F_{y_k'}$:
:$(2): \quad \dfrac {\partial F_{y_i'} } {\partial x} = \dfrac {\partial F} {\partial y_i} - \dfrac {\partial F_{y_i'} } {\partial \mathbf y} \bspsi - F_{\mathbf y'} \dfrac {\partial \bspsi} {\partial y_i}$
$F$ depends on $x$ directly through its first variable and indirectly through its third vector variable together with boundary conditions $(1)$:
:$\dfrac {\partial F_{y_i'} } {\partial x} = F_{y_i' x} + F_{y_i' \mathbf y'} \bspsi_x$
$F$ depends on $\mathbf y$ directly through its second vector variable and indirectly through its third vector variable together with boundary conditions $(1)$:
:$\dfrac {\partial F} {\partial y_i} = F_{y_i} + F_{\mathbf y'} \bspsi_{y_i}$
:$\ds \dfrac {\partial F_{y_i'} } {\partial \mathbf y} = F_{y_i' \mathbf y} + \sum_{k \mathop = 1}^N F_{y_i'y_k'} \dfrac {\partial \psi_k} {\partial \mathbf y}$
Substitution of the last three equations into $\paren 2$ leads to:
:$\ds F_{y_i' x} + F_{y_i' \mathbf y'} \bspsi_x = F_{y_i} + F_{\mathbf y'} \bspsi_{y_i} - \paren {F_{y_i' \mathbf y} + \sum_{k \mathop = 1}^N F_{y_i'y_k'} \dfrac {\partial \psi_k} {\partial \mathbf y} } \bspsi - F_{\mathbf y'} \dfrac {\partial \bspsi} {\partial y_i}$
which can be simplified to:
:$F_{y_i} - F_{y_i' x} - F_{y_i' \mathbf y} \bspsi - F_{y_i' y_j'} \paren {\dfrac {\partial \psi_j} {\partial x} + \dfrac {\partial \psi_j} {\partial y_j} \psi_j} = 0$
By assumption:
:$\dfrac {\d y_k} {\d x} = \psi_k$
the second total derivative {{WRT|Differentiation}} $x$ of which yields:
{{begin-eqn}}
{{eqn | l = \frac {\d^2 y_k} {\d x^2}
      | r = \frac {\d y_k'} {\d x}
}}
{{eqn | r = \frac {\d \psi_k} {\d x}
}}
{{eqn | r = \frac {\partial \psi_k} {\partial x} + \frac {\partial \psi_k} {\partial \mathbf y} \frac {\d \mathbf y} {\d x}
      | c = Total Derivative of $\psi_k$ {{WRT|Differentiation}} $x$ 
}}
{{eqn | r = \frac {\partial \psi_k} {\partial x} + \frac {\partial \psi_k} {\partial \mathbf y} \bspsi
}}
{{end-eqn}}
Hence:
:$F_{y_i} - \paren {F_{y_i' x} + F_{y_i' \mathbf y} \dfrac {\d \mathbf y} {\d x} + F_{y_i' \mathbf y'} \dfrac {\d \mathbf y'} {\d x} } = 0$
The second term is just a total derivative {{WRT|Differentiation}} $x$, thus:
:$(3): \quad F_{y_i} - \dfrac {\d F_{y_i'} } {\d x} = 0$
Boundary conditions $(1)$ are mutually consistent {{WRT}} equation $(3)$ because they hold $\forall x \in \closedint a b$.
By definition, they are consistent {{WRT}} the functional $J$.
Since the boundary conditions are consistent {{WRT}} $J$ and self-adjoint, by definition they constitute a field of $J$.
{{qed|lemma}}