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:


 * $\displaystyle J \left[{\mathbf y}\right] = \int_a^b F \left({x, \mathbf y, \mathbf y'}\right) \rd x$

Let the corresponding momenta and Hamiltonian be:


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


 * $H \left({x, \mathbf y, \mathbf y'}\right) = -F \left({x, \mathbf y, \mathbf y'}\right) + \mathbf p \mathbf y'$

Let the following be a family of boundary conditions:


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

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


 * $\dfrac {\partial p_i \left[{x, \mathbf y, \boldsymbol \psi \left({x, \mathbf y}\right)}\right]} {\partial y_k} = \dfrac {\partial p_k \left[{x, \mathbf y, \boldsymbol \psi \left({x, \mathbf y}\right)}\right]} {\partial y_i}$


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

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:


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

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 $(1)$:


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


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

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:


 * $\dfrac {\d y_k} {\d x} = \psi_k$

the second total derivative $ x $ of which yields:

Hence:


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

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 $\forall x \in \left[{a \,.\,.\, b}\right]$.

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, $(1)$ 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:


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

By definition:


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

Hence:


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