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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\newcommand{\pl}{\partial}
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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 extremals for

$$I(y,z)=\ds\int_0^{\frac{\pi}{2}} \,dx({y'}^2+{z'}^2+2yz)$$

subject to $y(0)=z(0)=0,\
y\left(\ds\frac{\pi}{2}\right)=z\left(\ds\frac{\pi}{2}\right)=1$.

\medskip

{\bf Answer}

This is a (generalisation 2) type problem with

$$F=({y'}^2+{z'}^2+2yz)=F(y,y',z,z')$$

Thus $\ds\frac{\pl F}{\pl y'}=2y',\ \ds\frac{\pl F}{\pl y}=2z,\
\ds\frac{\pl F}{\pl z'}=2z',\ \ds\frac{\pl F}{\pl z}=2y$ etc.

and we have simultaneous E-l equations:

$\left\{\begin{array} {rcl} \ds\frac{\pl F}{\pl
y}-\ds\frac{d}{dx}\left(\ds\frac{\pl F}{\pl y'}\right) &amp; = &amp; 0\\
\ds\frac{\pl F}{\pl z}-\ds\frac{d}{dx}\left(\ds\frac{\pl F}{\pl
z'}\right) &amp; = &amp; 0 \end{array}\right\} \Rightarrow
\left\{\begin{array} {rcl} 2z-\ds\frac{d}{dx}(2y') &amp; = &amp; 0\\
2y-\ds\frac{d}{dx}(2z') &amp; = &amp; 0 \end{array}\right\}$

$\Rightarrow \left\{\begin{array} {rcl} y''-z &amp; = &amp; 0\\ z''-y &amp; =
&amp; 0 \end{array}\right\}
\stackrel{\Rightarrow}{\frac{d^2}{dx^2}}\left\{\begin{array} {rcl}
y^{(iv)}-z'' &amp; = &amp; 0\\ z^{(iv)}-y'' &amp; = &amp; 0 \end{array} \right\}
\Rightarrow \left\{\begin{array}{rcl} y^{(iv)}-y &amp; = &amp; 0\\
z^{(iv)}-z &amp; = &amp; = \end{array} \right\}$

\ \ \ \ \ \ (1)\ \ \ \ \ \ (2)\ \ \ \ \ \ \ \ (1) into (2)

Thus

$y=Ae^x+Be^{-x}+C\cos x+D\sin x$

$z=Ae^x+Be^{-x}-C\cos x-D\sin x$

Boundary conditions:

$y(0)=z(0)=0 \Rightarrow A+B=0,\ C=0$

$y\left(\ds\frac{\pi}{2}\right)=z\left(\ds\frac{\pi}{2}\right)=1
\Rightarrow A^{-1}=2\sinh \ds\frac{\pi}{2},\ D=0$

$\Rightarrow \un{y=\ds\frac{\sinh x}{\sinh \ds\frac{\pi}{2}},\
z=\ds\frac{\sinh x}{\sinh \ds\frac{\pi}{2}}}$

NB could probably guess similarity of solution from symmetry of
$f$ in $y,y'$ and $z,z'$.
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