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:


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


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

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:


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


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

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:


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

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


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

Then:


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

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


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

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


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

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


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


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

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


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

which can be simplified to:


 * $\displaystyle F_{y_i}-F_{y_i'x}-F_{y_i'\mathbf y}\boldsymbol\psi-F_{y_i'y_j'}\paren{\frac{\partial\psi_j}{\partial x}+\frac{\partial\psi_j}{\partial y_j}\psi_j}=0$

By assumption:


 * $\displaystyle\frac{\d y_k}{\d x}=\psi_k$

the second total derivative $x$ of which yields:

Hence:


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

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


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

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 :


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

The can be rewritten as follows:

The vanishes.

Therefore:


 * $\displaystyle\frac{\partial\mathbf p}{\partial x}=-\frac{\partial H}{\partial\mathbf y}$

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


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

By definition:


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

Hence:


 * $\displaystyle\frac{\partial p_k}{\partial y_i}=\frac{\partial p_i}{\partial y_k}$