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:


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


 * $\map H {x,\mathbf y,\mathbf y'}=-\map F {x,\mathbf y,\mathbf y'}+\mathbf p\mathbf y'$

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:


 * $\frac {\partial p_i \sqbrk{x,\mathbf y,\map{\boldsymbol\psi} {x,\mathbf y} } } {\partial y_k}=\frac{\partial p_k \sqbrk{x,\mathbf y,\map{\boldsymbol\psi} {x,\mathbf y} } } {\partial y_i}$


 * $\frac {\partial\mathbf p\sqbrk{x,\mathbf y,\map{\boldsymbol\psi} {x,\mathbf y} } } {\partial x}=-\frac {\partial H\sqbrk{x,\mathbf y,\map{\boldsymbol\psi} {x,\mathbf y} } } {\partial\mathbf y}$

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:

On the we have:

Together they imply:


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

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


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

Then:


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

$F$ depends on $\mathbf y'$ only through its third vector variable, thus $\frac{\partial F} {\partial y_k'}=F_{y_k'}$:


 * $(2):\quad\frac{\partial F_{y_i'} } {\partial x}=\frac{\partial F} {\partial y_i}-\frac{\partial F_{y_i'} } {\partial\mathbf y} \boldsymbol\psi-F_{\mathbf y'}\dfrac{\partial\boldsymbol\psi} {\partial y_i}$

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


 * $\frac{\partial F_{y_i'} } {\partial x}=F_{y_i'x}+F_{y_i'\mathbf y'}\boldsymbol\psi_x$

$F$ depends on $\mathbf y$ directly through its second vector variable and indirectly through its third vector variable together with boundary conditions $(1)$:


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


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

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:


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