Conditions for Extremal Embedding in Field of Functional

Theorem

Let $J$ be a functional such that:

$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$

Let $\gamma$ be an extremal of $J$, defined by $\mathbf y = \map {\mathbf y} x$ for $x \in \closedint a b$.

Suppose:

$\forall x \in \closedint a b: \det \paren {F_{\mathbf y' \mathbf y'} } \ne 0$

Suppose no points conjugate to $\paren {a, \map {\mathbf y} a}$ lie on $\gamma$.

Then $\gamma$ can be embedded in a field.

Proof

Let $c \in \R$ be conjugate to $a$, such that $c < a$.

By assumption:

$c \notin \closedint a b$

Hence, there exists a set $\closedint c b$ such that:

$\closedint c b = \closedint c a \cup \closedint a b$

where $\size {c - a} > 0$.

$\exists \epsilon: \size {c - a} > \epsilon > 0$

Therefore, there exists $\epsilon > 0$ such that:

$\gamma$ can be extended onto the whole interval $\closedint {a - \epsilon} b$, where extension means definition of some mapping in $\hointr {a - \epsilon} a$

the interval $\closedint {a - \epsilon} b$ contains no points conjugate to $a$.

Consider a family of extremals leaving the point $\tuple {a - \epsilon, \map {\mathbf y} {a - \epsilon} }$.

There are no points conjugate to $a - \epsilon$ in $\closedint {a - \epsilon} b$.

Hence, for $x \in \closedint a b$ no two extremals in this family which are sufficiently close to the original extremal $\gamma$ can intersect.

Since all the functions are extremals, they satisfy same differential equations.

Lack of intersection implies different boundary conditions.

Denote these conditions collectively as $\map {\boldsymbol \psi} {x, \mathbf y} = \map {\mathbf y'} x$.

Thus, in some region $R$ containing $\gamma$ extremals sufficiently close to $\gamma$ define a central field in which $\gamma$ is embedded.

By Central Field is Field of Functional, $\gamma$ can be embedded in the field of functional.

$\blacksquare$