=== Necessary Condition === Set $\mathbf y = \map {\boldsymbol \psi} {x, \mathbf y}$ in the definition of momenta and Hamiltonian. Substitute corresponding definitions into the consistency relation. On the {{LHS}} we have: :$\dfrac {\partial \map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial x} = \dfrac {\partial^2 \map F {x, \mathbf y, \map {\boldsymbol \psi} {x,\mathbf y} } } {\partial x \partial y_i'}$ On the {{RHS}} we have: {{begin-eqn}} {{eqn | l = -\dfrac {\partial \map H {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial y_i}       | r = -\dfrac {\partial \paren {-\map F {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } + \map {\mathbf p} {x, \mathbf y, \map {\boldsymbol \psi} {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'} \boldsymbol \psi - \frac {\partial F} {\partial \mathbf y'} \frac {\partial \boldsymbol \psi} {\partial y_i} }} {{end-eqn}} Together they imply: :$\dfrac {\partial^2 F} {\partial x \partial y_i'} = \dfrac {\partial F} {\partial y_i} - \dfrac {\partial^2 F} {\partial y_i \partial \mathbf y'} \boldsymbol \psi - \dfrac {\partial F} {\partial \mathbf y'} \dfrac {\partial \boldsymbol \psi} {\partial y_i}$ 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: :$\dfrac {\partial^2 F} {\partial x \partial y_i'} = \dfrac {\partial F} {\partial y_i} - \dfrac {\partial^2 F} {\partial \mathbf y \partial y_i'} \boldsymbol \psi - \dfrac {\partial F} {\partial \mathbf y'} \dfrac {\partial \boldsymbol \psi} {\partial y_i}$ $F$ depends on $\mathbf y'$ only through its third vector variable, thus $\dfrac {\partial F} {\partial y_k'} = F_{y_k'}$: :$(2): \quad \dfrac {\partial F_{y_i'} } {\partial x} = \dfrac {\partial F} {\partial y_i} - \dfrac {\partial F_{y_i'} } {\partial \mathbf y} \boldsymbol \psi - F_{\mathbf y'} \dfrac {\partial \boldsymbol \psi} {\partial y_i}$ $F$ depends on $x$ directly through its first variable and indirectly through its third vector variable together with boundary conditions $(1)$: :$\dfrac {\partial F_{y_i'} } {\partial x} = F_{y_i' x} + F_{y_i' \mathbf y'} \boldsymbol \psi_x$ $F$ depends on $\mathbf y$ directly through its second vector variable and indirectly through its third vector variable together with boundary conditions $(1)$: :$\dfrac {\partial F} {\partial y_i} = F_{y_i} + F_{\mathbf y'} \boldsymbol \psi_{y_i}$ :$\ds \dfrac {\partial F_{y_i'} } {\partial \mathbf y} = F_{y_i' \mathbf y} + \sum_{k \mathop = 1}^N F_{y_i'y_k'} \dfrac {\partial \psi_k} {\partial \mathbf y}$ Substitution of the last three equations into $\paren 2$ leads to: :$\ds F_{y_i' x} + F_{y_i' \mathbf y'} \boldsymbol \psi_x = F_{y_i} + F_{\mathbf y'} \boldsymbol \psi_{y_i} - \paren {F_{y_i' \mathbf y} + \sum_{k \mathop = 1}^N F_{y_i'y_k'} \dfrac {\partial \psi_k} {\partial \mathbf y} } \boldsymbol \psi - F_{\mathbf y'} \dfrac {\partial \boldsymbol \psi} {\partial y_i}$ which can be simplified to: :$F_{y_i} - F_{y_i' x} - F_{y_i' \mathbf y} \boldsymbol \psi - F_{y_i' y_j'} \paren {\dfrac {\partial \psi_j} {\partial x} + \dfrac {\partial \psi_j} {\partial y_j} \psi_j} = 0$ By assumption: :$\dfrac {\d y_k} {\d x} = \psi_k$ the second total derivative {{WRT}} $x$ 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}} $x$  }} {{eqn | r = \frac {\partial \psi_k} {\partial x} + \frac {\partial \psi_k} {\partial \mathbf y} \boldsymbol \psi }} {{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}} $x$, 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}} === Sufficient Condition === By assumption, $(1)$ is a field of $J$. Hence, $(1)$ is self-adjoint and mutually consistent {{WRT}} $J$. Thus, by definition, they are consistent {{WRT}}: :$F_{y_i} - \dfrac {\d F_{y_i'} } {\d x} = 0$ The {{LHS}} can be rewritten as follows: {{begin-eqn}} {{eqn | l = F_{y_i} - \frac {\d F_{y_i'} } {\d x}       | r = F_{y_i} - \paren {F_{y_i' x} + F_{y_i' \mathbf y} \frac {\d \mathbf y} {\d x} + F_{y_i' \mathbf y'} \frac {\d \mathbf y'} {\d x} } }} {{eqn | r = F_{y_i} - \paren {F_{y_i' x} + F_{y_i' \mathbf y} \boldsymbol \psi + F_{y_i' \mathbf y'} \paren {\frac {\partial \boldsymbol \psi} {\partial x} + \sum_{k \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_k} \psi_k} }       | c = as $\ds \frac {\d \mathbf y'} {\d x} = \frac {\partial \boldsymbol \psi} {\partial x} + \sum_{k \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_k} \psi_k$ }} {{eqn | r = \frac {\partial F} {\partial y_i} - F_{\mathbf y'} \boldsymbol \psi_{y_i} - \paren {\frac {\partial F_{y_i'} } {\partial x} + F_{y_i' \mathbf y} \boldsymbol \psi + F_{y_i' \mathbf y'} \sum_{k \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_k} \psi_k}       | c = as $\dfrac {\partial F} {\partial y_i} = F_{y_i} + F_{\mathbf y'} \boldsymbol \psi_{y_i}$ }} {{eqn | r = \frac {\partial F} {\partial y_i} - F_{\mathbf y'} \boldsymbol \psi_{y_i} - \paren {\frac {\partial F_{y_i'} } {\partial x} + \paren { {\frac {\partial F_{y_i'} } {\partial \mathbf y} - \sum_{k \mathop = 1}^N F_{y_i' y_k'} \frac {\partial \psi_k} {\partial \mathbf y} } } \boldsymbol \psi + F_{y_i' \mathbf y'} \sum_{k \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_k} \psi_k}       | c = as $\dfrac {\partial F_{y_i'} } {\partial \mathbf y} = F_{y_i' \mathbf y} + \sum_{k \mathop = 1}^N F_{y_i' y_k'} \frac {\partial \psi_k} {\partial \mathbf y}$ }} {{eqn | r = \frac {\partial F} {\partial y_i} - F_{\mathbf y'} \boldsymbol \psi_{y_i} - \frac {\partial F_{y_i'} } {\partial x} - \frac {\partial F_{y_i'} } {\partial \mathbf y} \boldsymbol \psi }} {{eqn | r = \frac {\partial F} {\partial y_i} - \frac {\partial F} {\partial \mathbf y'} \frac {\partial \boldsymbol \psi} {\partial y_i} - \frac {\partial^2 F} {\partial x \partial y_i'} - \frac {\partial^2 F} {\partial \mathbf y \partial y_i'} \boldsymbol \psi       | c = as $F$ depends on $\mathbf y'$ only through its third vector variable }} {{eqn | r = \frac {\partial F} {\partial y_i} - \frac {\partial F} {\partial \mathbf y'} \frac {\partial \boldsymbol \psi} {\partial y_i} - \frac {\partial^2 F} {\partial x \partial y_i'} - \frac {\partial^2 F} {\partial y_i \partial \mathbf y'} \boldsymbol \psi       | c = By assumption, boundary conditions are self-adjoint }} {{eqn | r = \frac {\partial F} {\partial y_i} - \mathbf p \frac {\partial \mathbf y'} {\partial y_i} - \frac {\partial p_i} {\partial x} - \frac {\partial \mathbf p} {\partial y_i} \mathbf y' }} {{eqn | r = \frac {\partial F} {\partial y_i} - \frac {\partial p_i} {\partial x} - \frac {\partial \paren {\mathbf p \mathbf y'} } {\partial y_i} }} {{eqn | r = -\frac {\partial p_i} {\partial x} - \frac {\partial \paren {-F + \mathbf p \mathbf y'} } {\partial y_i}  }} {{eqn | r = -\frac {\partial p_i} {\partial x} - \frac {\partial H} {\partial y_i}  }} {{end-eqn}} The {{RHS}} vanishes. Therefore: :$\dfrac {\partial \mathbf p} {\partial x} = -\dfrac {\partial H} {\partial \mathbf y}$ By Necessary and Sufficient Condition for Boundary Conditions to be Self-adjoint: :$\dfrac {\partial^2 F} {\partial y_i \partial y_k'} = \dfrac {\partial^2 F} {\partial y_k \partial y_i'}$ By definition of canonical variable: :$\mathbf p = \dfrac {\partial F} {\partial \mathbf y'}$ Hence: :$\dfrac {\partial p_k} {\partial y_i} = \dfrac {\partial p_i} {\partial y_k}$ {{qed}}