\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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\begin{document}
In what follows you may assume that the following notation applies

$$y=y(x),\ y'=\ds\frac{dy}{dx}.$$

You may also assume that, unless otherwise stated, $y$ is a
sufficiently continuously differentiable function.

{\bf Question}

Find the curves $y=y(x)$ which make the following functional
integrals stationary

\begin{description}
\item[(i)]
$\ds\int_0^1 y' \,dx$

\item[(ii)]
$\ds\int_0^1 yy' \,dx$

\item[(iii)]
$\ds\int_0^1 xyy' \,dx$

\end{description}

where is each case $y$ satisfies the boundary conditions $y(0)=0,
y(1)=1$.


\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$F(y,y',x)=y'$:

EL equation is $\ds\frac{\pl F}{\pl
y}-\ds\frac{d}{dx}\left(\ds\frac{\pl F}{\pl y'}\right)=0$

Thus stationary integral occurs when

$\ds\frac{\pl F}{\pl y}=0;\ \ds\frac{\pl F}{\pl y'}=1 \Rightarrow
0-\ds\frac{d}{dx}(1)=0$

i.e., $0=0$

\un{i.e., for any $y(x)$ you like!}

[Why?

$\ds\int_0^1 y' \,dx=\ds\int_0^1 \ds\frac{dy}{dx}
\,dx=\ds\int_{y(0)}^{y(1)} \,dy=|ds\int_0^1 \,dy=1$

for \un{all} $y=y(x)$. Same answer so always stationary.]

\newpage
\item[(ii)]
$F(y,y',x)=y'$

$\ds\frac{\pl F}{\pl y}=y';\ \ds\frac{\pl F}{\pl y'}=y$

so E-L equation becomes

\begin{eqnarray*} y'-\ds\frac{d}{dx}(y) & = & 0\\ \Rightarrow
y'-y' & = & 0\\ 0 & = & 0 \end{eqnarray*}

$\Rightarrow$ stationary integral for any $y(x)$ you like (as
above)

[Why?

\begin{eqnarray*} \ds\int_0^1 y'y \,dx & = & \ds\int_0^1
\ds\frac{d}{dx}\left(\ds\frac{y^2}{2}\right) \,dx\\ & = &
\left[\ds\frac{y^2}{2}\right]_{y(0)}^{y(1)}\\ & = &
\left[\ds\frac{y^2}{2}\right]_0^1\\ & = & \ds\frac{1}{2}
\end{eqnarray*}

Same value for whatever $y$ you pick, so stationary!]

\item[(iii)]
$F(y,y',x)=xyy'$

$\ds\frac{\pl F}{\pl y}=xy';\ \ds\frac{\pl F}{\pl y'}=xy$ so E-L
equation becomes

$\begin{array} {crcl} & xy'-\ds\frac{d}{dx}(xy) & = & 0\\
\Rightarrow & xy'-y-xy' & = & 0\\ \Rightarrow & y & = & 0
\end{array}$

So apparently $y=0$ for all $x$ is the curve which minimises the
integral. Indeed it satisfies the E-L equation, but \un{NOT} the
boundary condition that $y(1)=1$. Consequently there is \un{NO}
continuous curve which makes $\ds\int_0^1 xyy' \,dx$ stationary
\un{and} satisfies the boundary conditions.
\end{description}

\end{document}
