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


 * $\map {F_y} {x, \hat y, \hat y'} - \dfrac \d {\d x} \map {F_{y'} } {x, \hat y, \hat y'} = 0$

which is derived from:


 * $\displaystyle \int_a^b \paren {\map {F_y} {x, \hat y, \hat y'} - \frac \d {\d x} \map {F_{y'} } {x, \hat y, \hat y'} } \rd x = 0$

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

By Taylor's Theorem:

where the omitted ordered tuple of variables is $\tuple {x, y, y'}$, and $\map {\hat y} x = \map y x$ has been used as a solution to $F_y - \dfrac \d {\d x} F_{y'} = 0$.

Therefore, Euler's equation is to be derived from


 * $\displaystyle \int_a^b \paren {\paren {F_{yy} - \frac \d {\d x} F_{y'y} } h - \map {\frac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2, h h', h'^2} } \rd x = 0$

By integration by parts,


 * $\displaystyle \int_a^b \map \OO {h^2, h h', h'^2} \rd x = \int_a^b \map \OO {h^2} \rd x$

Thus, the equivalent differential equation is:


 * $\paren {F_{yy} - \dfrac \d {\d x} F_{y'y} } h - \map {\dfrac \d {\d x} } {F_{y'y'} h'} + \map \OO {h^2} = 0$

Omission of $\map \OO {h^2}$ and multiplication of equation by $\frac 1 2$ yields Jacobi's equation.