# Legendre's Condition/Lemma 1

## Lemma

Let $y = \map y x$ be a real function, such that:

$\map y a = A$
$\map y b = B$

Let $J \sqbrk y$ be a functional, such that:

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

where:

$F \in C^2 \closedint a b$

with respect to all its variables.

Then:

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

where:

 $\ds \map P {x, \map y x}$ $=$ $\ds \frac 1 2 F_{y'y'}$ $\ds \map Q {x, \map y x}$ $=$ $\ds \frac 1 2 \paren {F_{yy} - \frac \d {\d x} F_{yy} }$

## Proof

The minimising function $y$ has fixed end-points.

Therefore, consider an increment of a functional with $h$ such that:

$h \in C^1 \closedint a b: \paren {\map h a = 0} \land \paren {\map h b = 0}$

Then:

 $\ds \Delta J \sqbrk {y; h}$ $=$ $\ds J \sqbrk {y + h} - J \sqbrk y$ Definition of Increment of Functional $\ds$ $=$ $\ds \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x$ form of $J$ $\ds$ $=$ $\ds \int_a^b \paren {F + \paren {F_y h + F_{y'} h'} + \frac 1 2 \paren {\overline F_{yy} h^2 + \overline F_{yy'} h h' + \overline F_{y'y'} h'^2} - F} \rd x$ Taylor's Theorem $\ds$ $=$ $\ds \int_a^b \paren {F_y h + F_{y'} h'} \rd x + \frac 1 2 \int_a^b \paren {\overline F_{yy} h^2 + \overline F_{yy'} h h' + \overline F_{y'y'} h'^2} \rd x$ cancel $F$

where omitted variables are $\paren {x, y, y'}$, and the overbar indicates derivatives being taken along some intermediate curves:

 $\ds \overline {\map {F_{yy} } {x,y,y'} }$ $=$ $\ds \map {F_{yy} } {x, y + \theta h, y' + \theta h'}$ $\ds \overline {\map {F_{yy'} } {x,y,y'} }$ $=$ $\ds \map {F_{yy'} } {x, y + \theta h, y' + \theta h'}$ $\ds \overline {\map {F_{y'y'} } {x,y,y'} }$ $=$ $\ds \map {F_{y'y'} } {x, y + \theta h, y' + \theta h'}$

with $0 < \theta < 1$.

If $\overline F_{yy}$, $\overline F_{yy'}$, $\overline F_{y'y'}$ are to be replaced by $F_{yy}$, $F_{yy}$, $F_{y'y'}$ evaluated at the point $\tuple {x, \map y x, \map {y'} x}$, then:

$\ds \Delta J \sqbrk {y; h} = \int_a^b \paren {\map {F_y} {x, y, y'} h + \map {F_{y'} } {x, y, y'} h'} \rd x + \frac 1 2 \int_a^b \paren {\map {F_{yy} } {x, y, y'} h^2 + 2 \map {F_{yy'} } {x, y, y'} h h'+ \map {F_{y'y'} } {x, y, y'} h'^2} \rd x + \epsilon$

where:

$\ds \epsilon = \int_a^b \paren {\epsilon_1 h^2 + \epsilon_2 h h' + \epsilon_3 h'^2}$

By continuity of $F_{yy}$, $F_{yy}$, $F_{y'y'}$:

$\size h_1 \to 0 \implies \epsilon_1, \epsilon_2, \epsilon_3 \to 0$

Thus, $\epsilon$ is an infinitesimal of the order higher than 2 with respect to $\size h$.

The first and second term on the right hand side of $\Delta J \sqbrk {y; h}$ are $\delta J \sqbrk {y; h}$ and $\delta^2 J \sqbrk {y; h}$ respectively.

Integrate the second term of $\delta^2 J \sqbrk {y; h}$ by parts:

 $\ds \int_a^b 2 F_{yy'} h h' \rd x$ $=$ $\ds \int_a^b 2 F_{yy'} h \rd h$ $\ds$ $=$ $\ds \int_a^b F_{yy'} \rd h^2$ $\ds$ $=$ $\ds \bigintlimits {F_{yy'} h^2} {x \mathop = a} {x \mathop = b} - \int_a^b \map {\frac \d {\d x} } {F_{yy'} } h^2 \rd x$ $\ds$ $=$ $\ds -\int_a^b \map {\frac \d {\d x} } {F_{yy'} } h^2 \rd x$

Therefore:

$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\frac 1 2 F_{y'y'} h'^2 + \frac 1 2 \paren {F_{yy} - \frac \d {\d x} F_{yy'} } h^2} \rd x$

$\Box$

## Mistake

1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations: $\S 5.25$: The Formula for the Second Variation. Legendre's Condition p. 102

states that

$P = \dfrac 1 2 F_{y'y'} \quad Q = \dfrac 1 2 \paren {F_{yy'} - \dfrac \d {\d x} F_{yy'} }$

This is a mistake, since the second variation should contain both pure and mixed partial derivatives of the order 2.

However, $F_{yy}$ is missing and could not have been lost during derivation of the proof.

It should be:

$Q = \dfrac 1 2 \paren {F_{yy} - \dfrac \d {\d x} F_{yy'} }$

## Sources

1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 5.25$: The Formula for the Second Variation. Legendre's Condition