Equivalence of Definitions of Conjugate Point
Theorem
The following definitions of the concept of Conjugate Point 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
- $\displaystyle \lim_{\norm {\map y x - \map {\tilde y} x}_{1, \infty} \to 0} \sqbrk {\paren {x, \map y x}: \map y x - \map {\tilde y} x = 0} = \tilde M$
In other words, let $\tilde M$ be the limit points of intersection of $y = \map y x$ and $y = \map {\tilde y} x$ as $\norm {\map y x - \map {\tilde y} x}_{1, \infty} \to 0$.
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:
- $\map h {\tilde a} = 0$
- $\beta = \sqrt {\dfrac \epsilon \alpha}$
Then:
- $\map h x \ne 0 \quad \forall x \in \openint a b \implies \map {h'} a \ne 0$
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
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 5.27$: Jacobi's Necessary Condition. More on Conjugate Points