Legendre's Condition

Theorem
Let $ y = y \left ( { x } \right )$ be a real function, such that:


 * $ y \left ( { a } \right )= A, \quad y \left ( { b } \right ) = B$

Let $ J \left [ { y } \right ]$ be a functional, such that:


 * $ \displaystyle J \left [ { y } \right ] = \int_a^b F \left ( { x, y, y' } \right ) \mathrm d x$

Then a necessary condition for $ J \left [ { y } \right ]$ to have a minimum for $ y = \hat { y }$ is


 * $ F_{ y' y' } \big \vert_{ y = \hat { y } } \ge 0 \quad \forall x \in \left [ { { a } \,. \,. \, { b } } \right ] $.

Lemma 1

 * $ \displaystyle \delta^2 J \left [ { y; h } \right ] = \int_a^b \left ( { P \left ( { x, y \left ( { x } \right ) } \right ) h'^2+Q \left ( { x, y \left ( { x } \right ) } \right ) h^2 } \right ) \mathrm d x$

where


 * $ \displaystyle P \left ( { x, y \left ( { x } \right ) } \right ) = \frac{ 1 }{ 2 } F_{ y' y' }, \quad Q \left ( { x, y \left ( { x } \right ) } \right ) = \frac{ 1 }{ 2 } \left ( { F_{ y y'} - \frac{ 1 }{ 2 } \frac{ \mathrm d }{ \mathrm d x } F_{ y y'} } \right )$

and


 * $ h \left ( { a } \right )= 0, \quad h \left ( { b } \right ) = 0$

Proof
where the overbar indicates derivatives being taken along some intermediate curves:


 * $ \overline F_{ y y } \left ( { x, y, y' } \right) = F_{ y y } \left ( { x, y + \theta h, y'+ \theta h' } \right)$
 * $ \overline F_{ y y' } \left ( { x, y, y' } \right) = F_{ y y' } \left ( { x, y + \theta h, y'+ \theta h' } \right)$
 * $ \overline F_{ y' y' } \left ( { x, y, y' } \right) = F_{ y' y' } \left ( { x, y + \theta h, y'+ \theta h' } \right)$

with $ 0 < \theta < 1$.

If $ \overline F_{ y y }$, $ \overline F_{ y y' }$, $ \overline F_{ y' y' } $ are to be replaced by $ F_{ y y }$, $ F_{ y y } $, $ F_{ y' y' }$ evaluated at the point $ \left ( { x, y \left ( { x } \right), y'\left ( { x } \right) } \right )$ then


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

where


 * $ \displaystyle \epsilon = \int_a^b \left [ { \epsilon_1 h^2 + \epsilon_2 h h' + \epsilon_3 h'^2 } \right ] $

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


 * $ \left \vert h \right \vert_1 \to 0 \implies \epsilon_1, \epsilon_2, \epsilon_3 \to 0$

Thus, $ \epsilon $ is an infinitesimal of the order higher than 2 $ \left \vert h \right \vert_1$.

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

Integrate the second term of $ \delta^2 J \left [ { y; h } \right ]$ by parts:

Therefore


 * $ \displaystyle \delta^2 J \left [ { y; h } \right ] = \int_a^b \left ( { \frac{ 1 }{ 2 } F_{ y' y' } h'^2 + \frac{ 1 }{ 2 } \left [ { F_{ y y' } - \frac{ \mathrm d }{ \mathrm d x} F_{ y y'}  } \right ]  } \right ) \mathrm d x$

Lemma 2
Let $ h$ be a real function such that:


 * $ h \in C^1 \left ( { a, b } \right ), \quad h \left ( { a } \right )= 0, \quad h \left ( { b } \right ) = 0 $

Let


 * $ \displaystyle \delta^2 J \left [ { y; h } \right ] = \int_a^b \left ( { P \left ( { x, y \left ( { x } \right ) } \right ) h'^2+Q \left ( { x, y \left ( { x } \right ) } \right ) h^2 } \right ) \mathrm d x $

Then a necessary condition for


 * $ \delta^2 J \left [ { y; h } \right ] \ge 0$

is


 * $ P \left ( { x, y \left ( { x } \right ) } \right ) \ge 0 \quad \forall x \in \left [ { a \,. \,. \, b } \right ]$

Proof
Assume that above is not true.

Then


 * $ \left( \exists x_0 \in \left [ { a \,. \,. \, b } \right ] \right ) \left( \exists \beta \in \R_{ < 0 } \right ) : F \left ( { x_0 } \right ) = - 2 \beta $

$ P $ is continuous.

Then


 * $ \exists \alpha \in \R_{ > 0 } : \left ( {a \ge x_0 - \alpha } \right ) \land \left ( {x_0 + \alpha \ge b } \right )$

Let


 * $ h = \begin{cases}

\sin^2 \left [ { \frac{ \pi \left ( { x-x_0 } \right)  }{ \alpha  } } \right ] > 0 & x_0 - \alpha \ge x \ge x_0 + \alpha \\ 0 & otherwise \end{cases}$

Then

where


 * $ \displaystyle M = \max_{ a \le x \le b } \left \vert Q \left ( { x } \right ) \right \vert $

For sufficiently small $ \alpha $ the is negative.

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