Necessary Condition for Integral Functional to have Extremum for given Function/Non-differentiable at Intermediate Point

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Theorem

Let $y, F$ be real functions.

Let $y$ be continuously differentiable for $x \in \hointr a c \cap \hointl c b$ and satisfy

$\map y a = A, \quad \map y b = B$

Let $J\sqbrk y$ be a functional of the form

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

Then the functional $J$ has a weak extremum if $y$ satisfies the following system of equations:

$\begin{cases} & \displaystyle F_y - \dfrac \d {\d x}F_{y'} = 0 \\ & \displaystyle F_{y'} \big \rvert_{x \mathop = c \mathop - 0} = F_{y'} \big \rvert_{x \mathop = c \mathop + 0} \\ & \displaystyle \paren{F - y' F_{y'} } \big \rvert_{x \mathop = c \mathop - 0} = \paren{F - y' F_{y'} } \big \rvert_{x \mathop = c \mathop + 0} \end{cases}$

where, by the use of limit from the left and from the right, the following abbreviations are denoted as follows:

$\displaystyle \map y x \rvert_{x \mathop = c \mathop + 0} = \lim_{x \to c^+} \map y x,\quad \map y x \rvert_{x \to x \mathop = c \mathop - 0} = \lim_{x \to c^-} \map y x$

The last two equations are known as the Weierstrass-Erdmann corner conditions.


Proof

Rewrite $J\sqbrk y$ as a sum of two functionals:

\(\displaystyle J \sqbrk y\) \(=\) \(\displaystyle \int_a^b \map F {x, y, y'} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^c \map F {x, y, y'} \rd x + \int_c^b \map F {x, y, y'} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle J_1 \sqbrk y + J_2 \sqbrk y\)

Recall that end points $x = a,x = b$ are fixed.

The function $\map y x$ has to be $C^0$ at $x = c$, but otherwise this point can move freely.

From general variation of functional, and noting that $y = \map y x$ is an extremal, write down variations for $J_1 \sqbrk y$ and $J_2 \sqbrk y$ separately:

$\displaystyle \delta J_1 = F_{y'} \rvert_{x \to c \mathop - 0} \delta y_1 + \sqbrk{F - y' F_{y'} } \big \rvert_{x \to c \mathop - 0} \delta x_1$

$\displaystyle \delta J_2 = - F_{y'} \rvert_{x \to c \mathop + 0} \delta y_1 - \sqbrk{F - y'F_{y'} } \big \rvert_{x \to c \mathop + 0} \delta x_1$

Note that $\delta J_1$ and $\delta J_2$ involve the same increments $\delta x_1$ and $\delta y_1$.



Since $y= \map y x$ is an extremum of $J$, we have:

\(\displaystyle \delta J\) \(=\) \(\displaystyle \delta J_1 + \delta J_2\)
\(\displaystyle \) \(=\) \(\displaystyle F_{y'} \rvert_{x \mathop = c \mathop - 0} \delta y_1 + \sqbrk{F - y' F_{y'} } \big \rvert_{x \mathop = c \mathop - 0} \delta x_1-F_{y'} \rvert_{x \mathop = c \mathop + 0}\delta y_1 - \sqbrk{F - y' F_{y'} } \big \rvert_{x \mathop = c \mathop + 0} \delta x_1\)
\(\displaystyle \) \(=\) \(\displaystyle \paren{F_{y'} \rvert_{x \mathop = c \mathop - 0} - F_{y'} \rvert_{x \mathop = c \mathop + 0} } \delta y_1 +\sqbrk{\paren{F - y' F_{y'} } \big \rvert_{x \mathop = c \mathop - 0} - \paren{F - y' F_{y'} } \big \rvert_{x \mathop = c \mathop + 0} }\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)

Since $ \delta x_1$ and $ \delta y_1$ are arbitrary, both collections of terms have to vanish independently.

$\blacksquare$


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