# 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 $J$ if and only if $\forall x\in\closedint a b$ the following self-adjointness and consistency relations hold:

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

## Proof

### 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 left hand side we have:

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

On the right hand side we have:

 $\displaystyle -\frac{\partial H\sqbrk{x,\mathbf y,\map{\boldsymbol\psi}{x,\mathbf y} } }{\partial y_i}$ $=$ $\displaystyle -\frac{\partial\sqbrk{-F\sqbrk{x,\mathbf y,\map{\boldsymbol\psi}{x,\mathbf y} }+\mathbf p\sqbrk{x,\mathbf y, \map{\boldsymbol\psi}{x,\mathbf y} }\map{\mathbf y'} x} }{\partial y_i}$ $\displaystyle$ $=$ $\displaystyle \frac{\partial F}{\partial y_i}-\frac{\partial^2 F}{\partial y_i\partial\mathbf y'}\boldsymbol\psi-\frac{\partial F}{\partial\mathbf y'}\frac{\partial\boldsymbol\psi}{\partial y_i}$

Together they imply:

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

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 with respect to $x$ of which yields:

 $\displaystyle \frac{\d^2 y_k}{\d x^2}$ $=$ $\displaystyle \frac{\d y_k'}{\d x}$ $\displaystyle$ $=$ $\displaystyle \frac{\d \psi_k}{\d x}$ $\displaystyle$ $=$ $\displaystyle \frac{\partial\psi_k}{\partial x}+\frac{\partial\psi_k}{\partial\mathbf y}\frac{\d \mathbf y}{\d x}$ Total Derivative of $\psi_k$ with respect to $x$ $\displaystyle$ $=$ $\displaystyle \frac{\partial\psi_k}{\partial x}+\frac{\partial\psi_k}{\partial\mathbf y}\boldsymbol\psi$

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 with respect to $x$, thus:

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

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

By definition, they are consistent with respect to the functional $J$.

Since the boundary conditions are consistent with respect to $J$ and self-adjoint, by definition they constitute a field of $J$.

$\Box$

### Sufficient Condition

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

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

Thus, by definition, they are consistent with respect to:

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

The left hand side can be rewritten as follows:

 $\displaystyle F_{y_i}-\frac{\d F_{y_i'} }{\d x }$ $=$ $\displaystyle F_{y_i}-\sqbrk{F_{y_i'x}+F_{y_i'\mathbf y}\frac{\d \mathbf y}{\d x}+F_{y_i'\mathbf y'}\frac{\d \mathbf y'}{\d x} }$ $\displaystyle$ $=$ $\displaystyle F_{y_i}-\sqbrk{F_{y_i'x}+F_{y_i'\mathbf y}\boldsymbol\psi+F_{y_i'\mathbf y'}\paren{\frac{\partial\boldsymbol\psi}{\partial x}+\sum_{k=1}^N\frac{\partial\boldsymbol\psi}{\partial y_k}\psi_k} }$ $\displaystyle\frac{\d \mathbf y'}{\d x}=\frac{\partial\boldsymbol\psi}{\partial x}+\sum_{k=1}^N\frac{\partial\boldsymbol \psi}{\partial y_k}\psi_k$ $\displaystyle$ $=$ $\displaystyle \frac{\partial F}{\partial y_i}-F_{\mathbf y'}\boldsymbol\psi_{y_i}-\sqbrk{\frac{\partial F_{y_i'} }{\partial x}+F_{y_i'\mathbf y}\boldsymbol\psi+F_{y_i'\mathbf y'}\sum_{k=1}^N\frac{\partial\boldsymbol\psi}{\partial y_k} \psi_k}$ $\displaystyle\frac{\partial F}{\partial y_i}=F_{y_i}+F_{\mathbf y'}\boldsymbol\psi_{y_i}$ $\displaystyle$ $=$ $\displaystyle \frac{\partial F}{\partial y_i}-F_{\mathbf y'}\boldsymbol\psi_{y_i}-\sqbrk{\frac{\partial F_{y_i'} }{\partial x}+\paren{ {\frac{\partial F_{y_i'} }{\partial\mathbf y}-\sum_{k=1}^N F_{y_i'y_k'}\frac{\partial\psi_k}{\partial \mathbf y} } }\boldsymbol\psi+F_{y_i'\mathbf y'}\sum_{k=1}^N\frac{\partial\boldsymbol\psi}{\partial y_k}\psi_k}$ $\displaystyle\frac{\partial F_{y_i'} }{\partial\mathbf y}=F_{y_i'\mathbf y}+\sum_{k=1}^N F_{y_i'y_k'}\frac{\partial\psi_k}{\partial\mathbf y}$ $\displaystyle$ $=$ $\displaystyle \frac{\partial F}{\partial y_i}-F_{\mathbf y'}\boldsymbol\psi_{y_i}-\frac{\partial F_{y_i'} }{\partial x}-\frac{\partial F_{y_i'} }{\partial\mathbf y}\boldsymbol\psi$ $\displaystyle$ $=$ $\displaystyle \frac{\partial F}{\partial y_i}-\frac{\partial F}{\partial\mathbf y'}\frac{\partial\boldsymbol\psi}{\partial y_i}-\frac{\partial^2 F}{\partial x\partial y_i'}-\frac{\partial^2 F}{\partial\mathbf y\partial y_i'}\boldsymbol\psi$ $F$ depends on $\mathbf y'$ only through its third vector variable $\displaystyle$ $=$ $\displaystyle \frac{\partial F}{\partial y_i}-\frac{\partial F}{\partial\mathbf y'}\frac{\partial\boldsymbol\psi}{\partial y_i}-\frac{\partial^2 F}{\partial x\partial y_i'}-\frac{\partial^2 F}{\partial y_i\partial\mathbf y'}\boldsymbol\psi$ By assumption, boundary conditions are self-adjoint $\displaystyle$ $=$ $\displaystyle \frac{\partial F}{\partial y_i}-\mathbf p\frac{\partial\mathbf y'}{\partial y_i}-\frac{\partial p_i}{\partial x}-\frac{\partial\mathbf p}{\partial y_i}\mathbf y'$ $\displaystyle$ $=$ $\displaystyle \frac{\partial F}{\partial y_i}-\frac{\partial p_i}{\partial x}-\frac{\partial\paren{\mathbf p\mathbf y'} }{\partial y_i}$ $\displaystyle$ $=$ $\displaystyle -\frac{\partial p_i}{\partial x}-\frac{\partial\paren{-F+\mathbf p\mathbf y'} }{\partial y_i}$ $\displaystyle$ $=$ $\displaystyle -\frac{\partial p_i}{\partial x}-\frac{\partial H}{\partial y_i}$

The right hand side vanishes.

Therefore:

$\displaystyle\frac{\partial\mathbf p}{\partial x}=-\frac{\partial H}{\partial\mathbf y}$
$\displaystyle\frac{\partial^2 F}{\partial y_i\partial y_k'}=\frac{\partial^2 F}{\partial y_k\partial y_i'}$

By definition:

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

Hence:

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

$\blacksquare$