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

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Theorem

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

Let $J \sqbrk {\mathbf y}$ be a functional of the form:

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

Let:

$\mathbf y \in C^1 \closedint a b$

where $C^1 \closedint a b$ denotes that $\mathbf y$ is continuously differentiable in $\closedint a b$

Let $\mathbf y$ satisfy boundary conditions:

$\map {\mathbf y} a = \mathbf A$
$\map {\mathbf y} b = \mathbf B$

where $\mathbf A$, $\mathbf B$ are real vectors.


Then a necessary condition for $J \sqbrk {\mathbf y}$ to have an extremum (strong or weak) for a given $\mathbf y$ is that they satisfy Euler's equations:

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


Proof

From Condition for Differentiable Functional of N Functions to have Extremum:

$\displaystyle \bigvalueat {\delta J \sqbrk {\mathbf y; \mathbf h} } {\mathbf y \mathop = \hat{\mathbf y} } = 0$

For the variation to exist it has to satisfy the requirement for a differentiable functional.

Note that the endpoints of $\map {\mathbf y} x$ are fixed. $\map {\mathbf h} x$ is not allowed to change values of $\map {\mathbf y} x$ at those points.

Hence $\map {\mathbf h} a = 0$ and $\map {\mathbf h} b = 0$.

We will start from the increment of a functional:

\(\displaystyle \Delta J \sqbrk {\mathbf y; \mathbf h}\) \(=\) \(\displaystyle J \sqbrk {\mathbf y + \mathbf h} - J \sqbrk {\mathbf y}\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'} \rd x - \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \paren {\map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'} - \map F {x, \mathbf y, \mathbf y'} } \rd x\)

Using multivariate Taylor's theorem, one can expand $\map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'}$ with respect to functions $\map {\mathbf h} x$ and $\map {\mathbf h'} x$:

\(\displaystyle \map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'}\) \(=\) \(\displaystyle \bigvalueat {\map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'} } {\mathbf h \mathop = \mathbf 0, \mathbf h' \mathop = \mathbf 0}\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \valueat {\sum_{i \mathop = 1}^n \frac {\partial {\map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'} } } {\partial {y_i} } } {h_i \mathop = 0, h_i' \mathop = 0} h_i\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \valueat {\sum_{i \mathop = 1}^n \frac {\partial {\map F {x, \mathbf y + \mathbf h, \mathbf y + \mathbf h} } } {\partial {y_i}'} } {h_i \mathop = 0, h_i' \mathop = 0} {h_i}'\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \map \OO {h_i h_j, h_i h_j', h_i' h_j'}\) for $i, j \in \openint 1 n$


We can substitute this back into the integral.

Note that the first term in the expansion and the negative one in the integral will cancel out.

Hence:

$\displaystyle \Delta J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \sum_{i \mathop = 1}^n \paren {F_{y_i} h_i + F_{y_i'} h_i' + \map \OO {h_i h_j,h_i h_j',h_i' h_j'} } \rd x$ for $i, j \in \openint 1 n$



By definition, the integral not counting in $\map \OO {h_i h_j,h_i h_j',h_i' h_j'}$ for $i, j \in \openint 1 n$ is a variation of functional:



$\displaystyle \delta J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \paren {F_{\mathbf y} \mathbf h + F_{\mathbf y'} \mathbf h'} \rd x$

The variation vanishes if for all functions $h_i$ every term containing $h_i$ vanishes independently.

Therefore, we discover a set of Euler's Equations being satisfied simultaneously:

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

$\blacksquare$



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