Equivalence of Definitions of Conjugate Point

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Theorem

The following definitions of the concept of Conjugate Point in the context of Calculus of Variations are equivalent:

Definition 1

Let:

$-\map {\dfrac \d {\d x} } {P h'} + Q h = 0$

with boundary conditions:

$\map h a = 0, \quad \map h c = 0, \quad a < c \le b$

Suppose:

$\map h x = 0 \quad \neg \forall x \in \closedint a b$

Suppose:

$\map h a = 0, \quad \map h {\tilde a} = 0, \quad a \ne \tilde a$


Then the point $\tilde a$ is called conjugate to the point $a$ with respect to solution to the aforementioned differential equation.

Definition 2

Let $y = \map y x$ and $y^* = \map {y^*} x$ be extremal functions.

Let:

$M = \tuple {a, \map y a}$
$\tilde M = \tuple {\tilde a, \map y {\tilde a} }$

Let $y$ and $y^*$ both pass through the point $M$.

Let:

$\map {y^*} {x - \tilde a} - \map y {x - \tilde a} = \epsilon \size {\map {y^*} {x - \tilde a} - \map y {x - \tilde a} }_1$

where:

$\size {\map {y^*} {x - \tilde a} - \map y {x - \tilde a} }_1 \to 0 \implies \epsilon \to 0$


Then $\tilde M$ is conjugate to $M$.

Definition 3

Let $y = \map y x$ and $y = \map {\tilde y} x$ be extremal functions.

Let:

$M = \paren {a, \map y a}$
$\tilde M = \paren {\tilde a, \map y {\tilde a} }$

Let both $y = \map y x$ and $y = \map {\tilde y} x$ pass through the point $M$.

Let

$\ds \lim_{\norm {\map y x - \map {\tilde y} x}_{1, \infty} \to 0} \sqbrk {\paren {x, \map y x}: \map y x - \map {\tilde y} x = 0} = \tilde M$

In other words, let $\tilde M$ be the limit points of intersection of $y = \map y x$ and $y = \map {\tilde y} x$ as $\norm {\map y x - \map {\tilde y} x}_{1, \infty} \to 0$.


This article, or a section of it, needs explaining.
In particular: the notation $\norm {\map y x - \map {\tilde y} x}_{1, \infty}$ with particular regard to the subscript This is most likely the standard norm of Sobolev space, also know as np-norm. Original source never calls it by word; only by special symbols
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Then $\tilde M$ is conjugate to $M$.


Proof

Definition 2 implies Definition 3

Let the extremal $y = \map y x$ satisfy:

$\map y a = A$

Let the extremal $y^*_\alpha$ pass through $M = \tuple {a, A}$ and satisfy:

$\map { {y^*}'_\alpha} a - \map {y'} a = \alpha$

Then the following is a valid expression for $y^*_\alpha$:

$\map {y^*_\alpha} x = \map y x + \map h x \alpha + \epsilon$

where $\epsilon = k \alpha$ with:

$\alpha \to 0 \implies k \to 0$

and:

$\map h a = 0$
$\map {h'} a = 1$

where $\map h x$ is an appropriate solution to Jacobi's equation.

Let:

$\map h {\tilde a} = 0$
$\beta = \sqrt {\dfrac \epsilon \alpha}$


This needs considerable tedious hard slog to complete it.
In particular: Explain the rest of lemma
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Then:

$\map h x \ne 0 \quad \forall x \in \openint a b \implies \map {h'} a \ne 0$


This article, or a section of it, needs explaining.
In particular: rewrite the above unambiguously
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By Taylor's Theorem, the expression:

$\map {y_\alpha} x - \map y x = \map h x \alpha + \epsilon$

takes values with different signs at the points $\tilde a - \beta$ and $\tilde a + \beta$.

Since:

$\alpha \to 0 \implies \beta \to 0$

the limit of points of intersection of $y = \map {y_\alpha^*} x$ and $y = \map y x$, as $\alpha \to 0$ is $\map {\tilde M} {\tilde a, \map y {\tilde a} }$

$\Box$


Definition 3 implies Definition 2


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This needs considerable tedious hard slog to complete it.
In particular: Definition 1 to link in yet
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Sources

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