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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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\begin{document}

{\bf Question}

\begin{description}
\item[(a)]
A curve $C$ joins two points $(a,\alpha),\ (b,\beta)$ and has
prescribed slopes at $x=a$, and $b$. Given that the functional

$$I=\ds\int_a^b F(y,y',y'',x) dx$$

must be stationary when evaluated along this curve, write down the
Euler-Lagrange equation which determines $C$.

\item[(b)]
If $F$ does not explicitly depend of $x$ or $y$, show that the
above equation for the extremal has a first integral

$$y''\ds\frac{\pl F}{\pl y''}-F=Ay'+B$$

where $A$,\ $B$ are constants. (Hints: After a simplification, try
a multiplication by $y''$ and then carry out a partial
integration, as in the \lq\lq special cases" section of the simple
E-L notes. The note that $F=F(y',y'')$ only and recognise the
first derivative of $F$ wrt $x$).

\item[(c)]
Derive a differential equation for the function $y(x0$ which makes

$$I=\ds\int_0^2 y'{y''}^2 dx$$

stationary. Solve this equation, given the boundary conditions
$y(0)=y'(0)=0,\ y(2)=1,\ y'(2)=1$. (Hint: use the answer of part b
above).
\end{description}

\medskip

\newpage
{\bf Answer}

\begin{description}
\item[(a)]
Bookwork: $\ds\frac{\pl F}{\pl y}-\ds\frac{\pl}{\pl
x}\left(\ds\frac{\pl F}{\pl y'}\right)-\ds\frac{\pl^2}{\pl
x^2}\left(\ds\frac{\pl F}{\pl y''}\right)=0$

\item[(b)]

If $F=F(y',y'')$ then

$-\ds\frac{d}{dx}\left(\ds\frac{\pl F}{\pl
y'}\right)+\ds\frac{d^2}{dx^2}\left(\ds\frac{\pl F}{\pl
y''}\right)=0$

$\Rightarrow \ds\frac{d}{dx}\left(\ds\frac{\pl F}{\pl
y''}\right)-\ds\frac{\pl F}{\pl y'}=const=A$ say

now multiply through by $y''$:

$\Rightarrow \underbrace{\ds\frac{d}{dx}\left(y''\ds\frac{\pl
F}{\pl y''}\right)-y'''\ds\frac{\pl F}{\pl y''}}-y''\ds\frac{\pl
F}{\pl y'}=Ay''$

\ \ \ spot this partial integration as in notes

$\Rightarrow \ds\frac{d}{dx}\left(y''\ds\frac{\pl F}{\pl
y''}\right)-\ds\frac{d}{dx}F(y',y'')=Ay''$

$\Rightarrow y''\ds\frac{\pl F}{\pl y''}-F(y',y'')=Ay'+B$ as
required.

\item[(c)]
$F=y'{y''}^2$ so (B) $\Rightarrow 2y'{y''}^2-y'{y''}^2=Ay'+B$
since $y'(0)=0 \Rightarrow B=0$

Therefore $y'=0$ or $y''=2\alpha$ say $\Rightarrow y=\alpha
x^2+\beta x+\gamma$ $y(0)=0,\ y'(0)=0 \Rightarrow \gamma=\beta=0;\
y(2)=1 \Rightarrow \alpha=\ds\frac{1}{4},\ y'(2)=1$ is satisfied

$\Rightarrow$ solution is \un{$y=\ds\frac{x^2}{4}$}
\end{description}
\end{document}
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