# 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

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

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

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

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

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

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

Furthermore:

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

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

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

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

$\blacksquare$