Conditions for Extremal Embedding in Field of Functional

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Theorem

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

Suppose:

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

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 $\closedint c b$:

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

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

By there exists a real point between two real points:

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

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 $\hointr{a-\epsilon} a$.



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

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.

$\blacksquare$


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