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


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

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

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


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

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

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

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:

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

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

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:

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