Jacobi's Equation is Variational Equation of Euler's Equation

Theorem
The Variational equation of Euler's equation is Jacobi's equation.

Proof
Let Euler's equation be


 * $ \displaystyle F_y \left ( { x, \hat y, \hat y' } \right ) - \frac{ \mathrm d }{ \mathrm d x } F_{ y' } \left ( { x, \hat y, \hat y' } \right ) = 0$

which is derived from:


 * $ \displaystyle \int_a^b \left ( { \displaystyle F_y \left ( { x, \hat y, \hat y' } \right ) - \frac{ \mathrm d }{ \mathrm d x } F_{ y' } \left ( { x, \hat y, \hat y' } \right ) } \right ) \mathrm d x = 0 $

Let $ \hat y \left ( { x } \right ) = y \left ( { x } \right ) $ and $ \hat y \left ( { x } \right ) = y \left ( { x } \right ) + h \left ( { x } \right ) $ be solutions of Euler's equation.

By Taylor's theorem:

where the ommited ordered set of variables is $ \left ( { x, y, y' } \right )$, and $ \hat y \left ( { x } \right ) = y \left ( { x } \right ) $ has been used as a solution to $ \displaystyle F_y - \frac{ \mathrm d }{ \mathrm d x } F_{ y' } = 0 $.

Therefore, Euler's equation is to be derived from


 * $ \displaystyle \int_a^b \left ( { \left ( { F_{ y y } - \frac{ \mathrm d }{ \mathrm d x }F_{ y' y } } \right ) h - \frac{ \mathrm d }{ \mathrm d x} \left ( { F_{ y' y' } h'  } \right ) + \mathcal O \left ( { h^2, hh', h'^2 } \right )  } \right ) \mathrm d x= 0$

By integration by parts,


 * $ \displaystyle \int_a^b \mathcal O \left ( { h^2, hh', h'^2 } \right ) \mathrm d x = \int_a^b \mathcal O \left ( { h^2 } \right ) \mathrm d x $

Thus, the equivalent differential equation is


 * $ \displaystyle \left ( { F_{ y y } - \frac{ \mathrm d }{ \mathrm d x }F_{ y' y } } \right ) h - \frac{ \mathrm d }{ \mathrm d x} \left ( { F_{ y' y' } h' } \right ) + \mathcal O \left ( { h^2 } \right ) = 0 $

Omission of $ \mathcal O \left ( { h^2 } \right ) $ and multiplication of equation by $ \frac{ 1 }{ 2 } $ yields Jacobi's equation.