Jacobi's Necessary Condition

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Theorem

Let $J$ be a functional, such that:

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

Let $\map y x$ correspond to the minimum of $J$.

Let:

$F_{y'y'}>0$

along $\map y x$.

Then the open interval $\openint a b$ contains no points conjugate to $a$.

Proof

By Necessary Condition for Twice Differentiable Functional to have Minimum, $J$ is minimised by $y = \map {\hat y} x$ if:

$\delta^2 J \sqbrk {\hat y; h} \ge 0$

for all admissable real functions $h$.

By lemma 1 of Legendre's Condition,

$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {P h'^2 + Q h^2} \rd x$

where:

$P = F_{y' y'}$

This article, or a section of it, needs explaining. In particular: and what is $Q$? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

By Nonnegative Quadratic Functional implies no Interior Conjugate Points, $\openint a b$ does not contain any conjugate points with respect to $J$.

$\blacksquare$

Source of Name

This entry was named for Carl Gustav Jacob Jacobi.

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