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


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

which is derived from:


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

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 ommited ordered set of variables is $\paren{x,y,y'}$, and $\map {\hat y} x=\map y x$ has been used as a solution to $\displaystyle F_y-\frac \d {\d x}F_{y'}=0 $.

Therefore, Euler's equation is to be derived from


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

By integration by parts,


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

Thus, the equivalent differential equation is


 * $\displaystyle\paren {F_{yy}-\frac \d {\d x}F_{y'y} }h-\frac \d {\d x} \paren { F_{y'y'}h'}+\map {\mathcal O} {h^2}=0$

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