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

Theorem

Let $y, F$ be real functions.

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

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

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:

 $\displaystyle F_y - \dfrac \d {\d x} F_{y'}$ $=$ $\displaystyle 0$ $\displaystyle \bigvalueat {F_{y'} } {x \mathop = c \mathop - 0}$ $=$ $\displaystyle \bigvalueat {F_{y'} } {x \mathop = c \mathop + 0}$ $\displaystyle \bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop - 0}$ $=$ $\displaystyle \bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop + 0}$

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

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

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:

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

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

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:

 $\displaystyle \delta J$ $=$ $\displaystyle \delta J_1 + \delta J_2$ $\displaystyle$ $=$ $\displaystyle \bigvalueat {F_{y'} } {x \mathop = c \mathop - 0} \delta y_1 + \bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop - 0} \delta x_1 - \bigvalueat {F_{y'} } {x \mathop = c \mathop + 0} \delta y_1 - \bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop + 0} \delta x_1$ $\displaystyle$ $=$ $\displaystyle \bigvalueat {\paren{F_{y'} } {x \mathop = c \mathop - 0} - \bigvalueat {F_{y'} } {x \mathop = c \mathop + 0} } \delta y_1 + \paren {\bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop - 0} - \bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop + 0} }$ $\displaystyle$ $=$ $\displaystyle 0$

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

$\blacksquare$