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:


 * $\displaystyle J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$

Let the corresponding momenta and Hamiltonian be:

Let the following be a family of boundary conditions:


 * $(1): \quad \map {\mathbf y'} x = \map {\boldsymbol \psi} {x, \mathbf y}$

Then a family of boundary conditions is a field for the functional $J$ $\forall x \in \closedint a b$ the following self-adjointness and consistency relations hold:

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 we have:


 * $\dfrac {\partial p_i \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial x} = \dfrac {\partial^2 F \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x,\mathbf y} } } {\partial x \partial y_i'}$

On the we have:

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 $(1)$:


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


 * $\displaystyle \dfrac {\partial F_{y_i'} } {\partial \mathbf y} = F_{y_i' \mathbf y} + \sum_{k \mathop = 1}^N F_{y_i'y_k'} \dfrac {\partial \psi_k} {\partial \mathbf y}$

Substitution of the last three equations into $\paren 2$ leads to:


 * $\displaystyle F_{y_i' x} + F_{y_i' \mathbf y'} \boldsymbol \psi_x = F_{y_i} + F_{\mathbf y'} \boldsymbol \psi_{y_i} - \paren {F_{y_i' \mathbf y} + \sum_{k \mathop = 1}^N F_{y_i'y_k'} \dfrac {\partial \psi_k} {\partial \mathbf y} } \boldsymbol \psi - F_{\mathbf y'} \dfrac {\partial \boldsymbol \psi} {\partial y_i}$

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 $x$ of which yields:

Hence:


 * $F_{y_i} - \sqbrk {F_{y_i' x} + F_{y_i' \mathbf y} \dfrac {\d \mathbf y} {\d x} + F_{y_i' \mathbf y'} \dfrac {\d \mathbf y'} {\d x} } = 0$

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


 * $(3): \quad F_{y_i} - \dfrac {\d F_{y_i'} } {\d x} = 0$

Boundary conditions $(1)$ are mutually consistent equation $(3)$ because they hold $\forall x \in \closedint a b$.

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, $(1)$ is a field of $J$.

Hence, $(1)$ is self-adjoint and mutually consistent $J$.

Thus, by definition, they are consistent :


 * $F_{y_i} - \dfrac {\d F_{y_i'} } {\d x} = 0$

The can be rewritten as follows:

The 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:


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

By definition of canonical variable:


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

Hence:


 * $\dfrac {\partial p_k} {\partial y_i} = \dfrac {\partial p_i} {\partial y_k}$