Legendre's Condition/Lemma 1

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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

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

with respect to 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:

\(\displaystyle \Delta J\sqbrk{y;h}\) \(=\) \(\displaystyle J\sqbrk{y+h}-J\sqbrk y\) Definition of increment of functional
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \paren { \map F{x,y+h,y'+h'} - \map F{x,y,y'} }\rd x\) form of $J$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\paren{F+\sqbrk{F_y h+F_{y'}h'}+\frac 1 2\sqbrk {\overline F_{yy} h^2+\overline F_{yy'}hh'+\overline F_{y'y'}h'^2}-F}\rd x\) By Taylor's theorem
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\sqbrk{F_y h+F_{y'}h'}\rd x+\frac 1 2 \int_a^b \sqbrk{\overline F_{yy}h^2+\overline F_{yy'}hh'+\overline F_{y'y'} h'^2}\rd x\) cancel $F$

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

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

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 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:

\(\displaystyle \int_a^b 2 F_{yy'}hh'\rd x\) \(=\) \(\displaystyle \int_a^b 2 F_{yy'}h\rd h\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b F_{yy'}\rd h^2\)
\(\displaystyle \) \(=\) \(\displaystyle F_{yy'}h^2\big\vert_{x=a}^{x=b}-\int_a^b \frac \d {\d x} \paren {F_{yy'} } h^2 \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle -\int_a^b\frac \d {\d x} \paren { F_{yy'} } h^2\rd x\)

Therefore

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


$\Box$


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