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 $x\in\left[{{ a}\,.\,.\,{c}} \right) \cap \left({{ c}\,.\,.\,{ b}} \right]$ 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:

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

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


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

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

Proof
Rewrite $J[ y]$ 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:

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

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

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.