Equivalence of Definitions of Conjugate Point

Theorem
Definitions of Conjugate Point are equivalent

Defintion 2 implies Definition 3
Let the extremal $ y = y \left ( { x } \right ) $ satisfy


 * $ y \left ( { a } \right ) = A $

Let the extremal $ y^*_\alpha $ pass through $ M = \left ( { a, A } \right ) $ and satisfy


 * $ {y^*}'_\alpha \left ( { a } \right ) - y' \left ( { a } \right )  = \alpha $

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


 * $ y^*_\alpha \left ( { x } \right ) = y \left ( { x } \right ) + h \left ( { x } \right ) \alpha + \epsilon $

where $ \epsilon = k \alpha $ with


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

and


 * $h \left ( { a } \right ) = 0, \quad h' \left ( { a } \right ) = 1 $

where $ h \left ( { x } \right ) $ is an appropriate solution to Jacobi's equation.

Let


 * $h \left ( { \tilde a } \right ) = 0, \quad \beta = \sqrt {\dfrac{ \epsilon }{ \alpha } }$

Then


 * $ h \left ( { x } \right ) \ne 0 \quad \forall x \in \left ( {a \,. \,. \, b } \right ) \implies h' \left ( { a } \right ) \ne 0$

By Taylor's theorem, the expression


 * $ y_\alpha \left ( { x } \right ) - y \left ( { x } \right ) = h \left ( { x } \right ) \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 = y_\alpha^* \left ( { x } \right ) $ and $ y = y \left ( { x } \right )$, as $ \alpha \to 0 $ is $ \tilde M \left ( { \tilde a, y \left ( { \tilde a } \right ) } \right ) $