Conditions for Integral Functionals to have same Euler's Equations

From ProofWiki
Jump to navigation Jump to search

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

According to Necessary Condition for Integral Functional to have Extremum for given function/Dependent on N Functions:

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.

$\blacksquare$


Sources