Necessary and Sufficient Condition for First Order System to be Field for Functional

From ProofWiki
Jump to navigation Jump to search

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:

\(\displaystyle \map {\mathbf p} {x, \mathbf y, \mathbf y'}\) \(=\) \(\displaystyle \dfrac {\partial \map F {x, \mathbf y, \mathbf y'} } {\partial \mathbf y'}\)
\(\displaystyle \map H {x, \mathbf y, \mathbf y'}\) \(=\) \(\displaystyle -\map F {x, \mathbf y, \mathbf y'} + \mathbf p \mathbf y'\)


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:

\(\displaystyle \dfrac {\partial \map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial y_k}\) \(=\) \(\displaystyle \dfrac {\partial \map {p_k} {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial y_i}\)
\(\displaystyle \dfrac {\partial \map {\mathbf p} {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial x}\) \(=\) \(\displaystyle -\dfrac {\partial \map H {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial \mathbf y}\)




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:

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

On the right hand side we have:

\(\displaystyle -\dfrac {\partial \map H {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial y_i}\) \(=\) \(\displaystyle -\dfrac {\partial \paren {-\map F {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } + \map {\mathbf p} {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } \map {\mathbf y'} x} } {\partial y_i}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\partial F} {\partial y_i} - \frac {\partial^2 F} {\partial y_i \partial \mathbf y'} \boldsymbol \psi - \frac {\partial F} {\partial \mathbf y'} \frac {\partial \boldsymbol \psi} {\partial y_i}\)


Together they imply:

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

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 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:

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

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

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

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:

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

The left hand side can be rewritten as follows:

\(\displaystyle F_{y_i} - \frac {\d F_{y_i'} } {\d x}\) \(=\) \(\displaystyle F_{y_i} - \paren {F_{y_i' x} + F_{y_i' \mathbf y} \frac {\d \mathbf y} {\d x} + F_{y_i' \mathbf y'} \frac {\d \mathbf y'} {\d x} }\)
\(\displaystyle \) \(=\) \(\displaystyle F_{y_i} - \paren {F_{y_i' x} + F_{y_i' \mathbf y} \boldsymbol \psi + F_{y_i' \mathbf y'} \paren {\frac {\partial \boldsymbol \psi} {\partial x} + \sum_{k \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_k} \psi_k} }\) as $\displaystyle \frac {\d \mathbf y'} {\d x} = \frac {\partial \boldsymbol \psi} {\partial x} + \sum_{k \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_k} \psi_k$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\partial F} {\partial y_i} - F_{\mathbf y'} \boldsymbol \psi_{y_i} - \paren {\frac {\partial F_{y_i'} } {\partial x} + F_{y_i' \mathbf y} \boldsymbol \psi + F_{y_i' \mathbf y'} \sum_{k \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_k} \psi_k}\) as $\dfrac {\partial F} {\partial y_i} = F_{y_i} + F_{\mathbf y'} \boldsymbol \psi_{y_i}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\partial F} {\partial y_i} - F_{\mathbf y'} \boldsymbol \psi_{y_i} - \paren {\frac {\partial F_{y_i'} } {\partial x} + \paren { {\frac {\partial F_{y_i'} } {\partial \mathbf y} - \sum_{k \mathop = 1}^N F_{y_i' y_k'} \frac {\partial \psi_k} {\partial \mathbf y} } } \boldsymbol \psi + F_{y_i' \mathbf y'} \sum_{k \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_k} \psi_k}\) as $\dfrac {\partial F_{y_i'} } {\partial \mathbf y} = F_{y_i' \mathbf y} + \sum_{k \mathop = 1}^N F_{y_i' y_k'} \frac {\partial \psi_k} {\partial \mathbf y}$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\partial F} {\partial y_i} - F_{\mathbf y'} \boldsymbol \psi_{y_i} - \frac {\partial F_{y_i'} } {\partial x} - \frac {\partial F_{y_i'} } {\partial \mathbf y} \boldsymbol \psi\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\partial F} {\partial y_i} - \frac {\partial F} {\partial \mathbf y'} \frac {\partial \boldsymbol \psi} {\partial y_i} - \frac {\partial^2 F} {\partial x \partial y_i'} - \frac {\partial^2 F} {\partial \mathbf y \partial y_i'} \boldsymbol \psi\) as $F$ 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:

$\dfrac {\partial \mathbf p} {\partial x} = -\dfrac {\partial H} {\partial \mathbf y}$

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 of canonical variable:

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

Hence:

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

$\blacksquare$


Sources