Legendre's Condition/Lemma 1

Lemma 1
Let $y=\map y x$ be a real function, such that:


 * $\map y a= A,\quad \map y b=B$

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


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

where


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

all its variables.

Then


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

where


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

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:

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


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

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 $\paren{x,\map y x,\map {y'} x}$ then


 * $\displaystyle\Delta J\sqbrk{y;h}=\int_a^b\left [ { F_y \left ( { x, y, y' } \right) h + \map {F_{y'} } {x,y,y'}h'} \right ] \rd x + \frac 1 2 \int_a^b \sqbrk {\map {F_{yy} } {x,y,y'}h^2+2 \map {F_{yy'} } {x,y,y'}hh'+\map {F_{y'y'} } {x,y,y'} h'^2}\rd x+\epsilon$

where


 * $\displaystyle\epsilon=\int_a^b \sqbrk {\epsilon_1 h^2+\epsilon_2 hh'+\epsilon_3 h'^2}$

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 $\size h$.

The first and second term on the 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:

Therefore


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

Mistake

 * $\S 5.25$: 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:


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