Conditions for Integral Functionals to have same Euler's Equations

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Theorem

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

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

According to the necessary conditions for an integral functional dependent on N functions to have an extremum for given function,

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.


$\blacksquare$

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