# Conditions for Integral Functionals to have same Euler's Equations

## Theorem

Let $\mathbf y$ be a real $n$-dimensional vector-valued function.

Let $\map F {x, \mathbf y, \mathbf y'}$, $\map \Phi {x, \mathbf y}$ be real functions.

Let $\Phi$ be twice differentiable.

Let:

 $\displaystyle \Psi$ $=$ $\displaystyle \frac {\d \Phi} {\d x}$ $\displaystyle$ $=$ $\displaystyle \frac {\partial \Phi} {\partial x} + \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} y_i'$

Let $J_1$, $J_2$ be functionals such that:

$\displaystyle J_1 \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
$\displaystyle J_2 \sqbrk {\mathbf y} = \int_a^b \paren {\map F {x, \mathbf y, \mathbf y'} + \map \Psi {x, \mathbf y, \mathbf y'} } \rd x$

Then $J_1$ and $J_2$ have same Euler's Equations.

## Proof

Euler's Equations for functional $J_1$ are:

$\displaystyle F_{\mathbf y} - \frac \d {\d x} F_{\mathbf y'} = 0$

Equivalently, for $J_2$ we have

 $\displaystyle \paren {F_{\mathbf y} + \Psi_{\mathbf y} } - \map {\frac \d {\d x} } {F_{\mathbf y'} + \Psi_{\mathbf y'} }$ $=$ $\displaystyle \paren {F_{\mathbf y} - \frac \d {\d x} F_{\mathbf y'} } + \paren {\Psi_{\mathbf y} - \frac \d {\d x} \Psi_{\mathbf y'} }$ $\displaystyle$ $=$ $\displaystyle 0$ condition for the existence of extremum

Furthermore:

 $\displaystyle \Psi_{\mathbf y}$ $=$ $\displaystyle \frac {\partial^2 \Phi} {\partial \mathbf y \partial x} + \frac \partial {\partial \mathbf y} \sum_{j \mathop = 1}^n \frac {\partial \Phi} {\partial y_j} y_j'$ $\displaystyle$ $=$ $\displaystyle \frac {\partial^2 \Phi} {\partial \mathbf y \partial x} + \sum_{j \mathop = 1}^n \frac {\partial^2 \Phi} {\partial \mathbf y \partial y_j} y_j'$
 $\displaystyle \Psi_{\mathbf y'}$ $=$ $\displaystyle \frac {\partial^2 \Phi} {\partial \mathbf y' \partial x} + \frac \partial {\partial \mathbf y'} \sum_{j \mathop = 1}^n \frac {\partial \Phi} {\partial y_j} y_j'$ $\displaystyle$ $=$ $\displaystyle 0 + \sum_{j \mathop = 1}^n \paren {\frac {\partial^2 \Phi} {\partial \mathbf y' \partial y_j} y_j' + \frac {\partial \Phi} {\partial y_j} \frac {\partial y_j'} {\partial \mathbf y'} }$ $\displaystyle$ $=$ $\displaystyle \frac {\partial \Phi} {\partial \mathbf y}$ as $\frac {\partial y_i'} {\partial y_j'} = \delta_{i j}$, where $\delta_{i j}$ is the Kronecker Delta
 $\displaystyle \frac {\d \Psi_{\mathbf y'} } {\d x}$ $=$ $\displaystyle \frac {\partial \Psi_{\mathbf y'} } {\partial x} + \sum_{j \mathop = 1}^n \frac {\partial \Psi_{\mathbf y'} } {\partial y_j} \frac {\d y_j} {\d x} + \sum_{j \mathop = 1}^n \frac {\partial \Psi_{\mathbf y'} } {\partial y_j'} \frac {\d y_j'} {\d x}$ $\displaystyle$ $=$ $\displaystyle \frac {\partial^2 \Phi} {\partial x \partial \mathbf y} + \sum_{j \mathop = 1}^n \frac {\partial^2 \Phi} {\partial \mathbf y \partial y_j} y_j'$

Since $\Phi$ is twice differentiable, by Schwarz-Clairaut Theorem partial derivatives commute and:

$\Psi_{\mathbf y} - \dfrac \d {\d x} \Psi_{\mathbf y'} = 0$

Therefore, $J_1$ and $J_2$ have same Euler's Equations.

$\blacksquare$