# Legendre's Condition/Lemma 2

## Lemma

Let $h$ be a real function such that:

$h \in C^1 \openint a b$
$\map h a = 0$
$\map h b = 0$

Let:

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

where $P \in C^0 \closedint a b$.

Then a necessary condition for:

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

is:

$\forall x\in\closedint a b: \map P {x, \map y x} \ge 0$

## Proof

Assume that above is not true.

Then:

$\paren {\exists x_0 \in \closedint a b} \land \paren {\exists \beta \in \R_{<0} }: \map P {x_0} = -2 \beta$

$P$ is continuous.

Thus:

$\exists \alpha \in \R_{>0}: \paren {a \ge x_0 - \alpha} \land \paren {x_0 + \alpha \ge b}$

and:

$\forall x \in \openint {x_0 - \alpha} {x_0 + \alpha}: \map P x < -\beta$

In other words:

$\map P x \begin {cases} = 0 & : x \in \closedint a {x_0 - \alpha} \lor \closedint {x_0 + \alpha} b \\ < 0 & : x \in \closedint {x_0 - \alpha} {x_0 + \alpha} \end {cases}$

Let

$h = \begin {cases} \sin^2 \paren {\dfrac {\map \pi {x - x_0} } \alpha} & : x_0 - \alpha \ge x \ge x_0 + \alpha \\ 0 & : \text {otherwise} \end{cases}$

It belongs to $C^1 \openint a b$ because:

 $\displaystyle \lim_{x \to x_0 - \alpha + 0^+} h$ $=$ $\displaystyle \lim_{x \to x_0 - \alpha + 0^+} \map {\sin^2} {\frac {\map \pi {x - x_0} } \alpha}$ $\displaystyle$ $=$ $\displaystyle \map {\sin^2} {\map \pi {\frac {0^+} \alpha - 1} }$ $\displaystyle$ $=$ $\displaystyle 0$

 $\displaystyle \lim_{x \to x_0 - \alpha + 0^+}h'$ $=$ $\displaystyle \lim_{x \to x_0 - \alpha + 0^+} \map \sin {\frac {2 \map \pi {x - x_0} } \alpha} \frac \pi \alpha$ $\displaystyle$ $=$ $\displaystyle \map \sin {2 \map \pi {\frac {0^+} \alpha - 1} } \frac \pi \alpha$ $\displaystyle$ $=$ $\displaystyle 0$

 $\displaystyle \lim_{x \to x_0 - \alpha + 0^+}h''$ $=$ $\displaystyle \lim_{x \to x_0 - \alpha + 0^+} \map \cos {\frac {2 \map \pi {x - x_0} } \alpha} \frac {2 \pi^2} {\alpha^2}$ $\displaystyle$ $=$ $\displaystyle \map \cos {2 \map \pi {\frac {0^+} \alpha - 1} } \frac {2 \pi^2} {\alpha^2}$ $\displaystyle$ $=$ $\displaystyle \frac {2 \pi^2} {\alpha^2}$
 $\displaystyle \lim_{x \to x_0 + \alpha + 0^-} h$ $=$ $\displaystyle \lim_{x \to x_0 + \alpha + 0^-} \map {\sin^2} {\frac {\map \pi {x - x_0} } \alpha}$ $\displaystyle$ $=$ $\displaystyle \map {\sin^2} {\map \pi {\frac {0^-} \alpha + 1} }$ $\displaystyle$ $=$ $\displaystyle 0$

 $\displaystyle \lim_{x \to x_0 + \alpha + 0^-}h'$ $=$ $\displaystyle \lim_{x \to x_0 - \alpha + 0^-} \map \sin {\frac {2 \map \pi {x - x_0} } \alpha} \frac \pi \alpha$ $\displaystyle$ $=$ $\displaystyle \map \sin {2 \map \pi {\frac {0^-} \alpha + 1} } \frac \pi \alpha$ $\displaystyle$ $=$ $\displaystyle 0$

 $\displaystyle \lim_{x \to x_0 + \alpha + 0^-}h''$ $=$ $\displaystyle \lim_{x \to x_0 - \alpha + 0^-} \map \cos {\frac {2 \map \pi {x - x_0} } \alpha} \frac {2 \pi^2} {\alpha^2}$ $\displaystyle$ $=$ $\displaystyle \map \cos {2 \map \pi {\frac {0^-} \alpha + 1} } \frac {2 \pi^2} {\alpha^2}$ $\displaystyle$ $=$ $\displaystyle \frac {2\pi^2} {\alpha^2}$

In other words, only $h$ and $h'$ are continuous in $\closedint a b$

Then:

 $\displaystyle \int_a^b \paren {P h'^2 + Q h^2} \rd x$ $=$ $\displaystyle \int_{x_0 - \alpha}^{x_0 + \alpha} P \frac {\pi^2} {\alpha^2} \map {\sin^2} {\frac {2 \map \pi {x - x_0} } \alpha} \rd x + \int_{x_0 - \alpha}^{x_0 + \alpha} Q \map {\sin^4} {\frac {\map \pi {x - x_0} } \alpha} \rd x$ $\displaystyle$ $<$ $\displaystyle -\beta \frac {\pi^2} {\alpha^2} \int_{x_0 - \alpha}^{x_0 + \alpha} \map {\sin^2} {\frac {2 \map \pi {x - x_0} } \alpha} \rd x + \max_{a \le x \le b} \size {\map Q x} \int_{x_0 - \alpha}^{x_0 + \alpha} \map {\sin^4} {\frac {\map \pi {x - x_0} } \alpha}\rd x$ as $\displaystyle \map P x < -\beta, \quad \map Q x < \max_{a \mathop \le x \mathop \le b} \size {\map Q x}$ $\displaystyle$ $=$ $\displaystyle -\beta \frac {\pi^2} \alpha + \max_{a \mathop \le x \mathop \le b} \size {\map Q x} \int_{x_0 - \alpha}^{x_0 + \alpha} \map {\sin^4} {\frac {\map \pi {x - x_0} } \alpha} \rd x$ $\displaystyle$ $=$ $\displaystyle -\beta \frac {\pi^2} \alpha + \frac 3 4 \alpha M$

where:

$\displaystyle M = \max_{a \mathop \le x \mathop \le b} \size {\map Q x}$

For sufficiently small $\alpha$ the right hand side is negative.

Hence, $\delta^2 J$ is negative for the corresponding $h$.

To conclude, it has been shown that

$P \ge 0 \quad \neg \forall x \in \closedint a b \implies \delta^2 J<0$

Then, by contrapositive statement this is equivalent to:

$\forall x \in \closedint a b: \delta^2 J \ge 0 \implies P \ge 0$

$\blacksquare$

## 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