\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Write down the Euler-Langrange equation appropriate to the
functional

$$I=\ds\int \!\!\! \ds\int_Z F(u,u_x,u_y,u_{xx},u_{xy},u_{yy},x,y)
dx dy$$

Find the Euler-Lagrange equations associated with the functionals

\begin{description}
\item[(i)]
$I=\ds\int \!\!\! \ds\int (\bigtriangledown ^2 u)^2 dxdy$

\item[(ii)]
$I=\ds\int \!\!\! \ds\int (u_{xx}u_{yy}-u_{xy}^2)dxdy$

\end{description}


\medskip

{\bf Answer}

The E-L equation is:

$0=\ds\frac{\pl F}{\pl u}-\ds\frac{\pl}{\pl x}\left(\ds\frac{\pl
F}{\pl u_x}\right)-\ds\frac{\pl}{\pl y}\left(\ds\frac{\pl F}{\pl
u_y}\right)$

$+\ds\frac{\pl^2}{\pl x^2}\left(\ds\frac{\pl F}{\pl
u_{xx}}\right)-\underbrace{\ds\frac{\pl^2}{\pl x\pl
y}\left(\ds\frac{\pl F}{\pl u_{xy}}\right)}+\ds\frac{\pl ^2}{\pl
y^2}\left(\ds\frac{\pl F}{\pl U_{yy}}\right)$

Assuming that both $F$ and its first derivatives are given on the
boundary. So for

\begin{description}
\item[(i)]
$F=(u_{xx}+u_{yy})^2=(\bigtriangledown ^2u)^2=F(u_{xx},u_{yy})$
only, so E-L is:

$$\ds\frac{\pl^2}{\pl x^2}[2(u_{xx}+u_{yy})]+\ds\frac{\pl^2}{\pl
y^2}[2(u_{xx}+u_{yy})]=0$$

$\Rightarrow u_{xxxx}+2u_{xxyy}+u_{yyyy}=0$

or $\un{\bigtriangledown}^2(\un{\bigtriangledown}^2u)=0$

or $\un{\un{\bigtriangledown}^4u=0}$

\item[(ii)]
$F=(u_{xx}u_{yy}-u_{xy}^2)=F(u_{xx},u_{yy},u_{xy})$ only so E-L
is:

$$\ds\frac{\pl^2}{\pl x^2}(u_{yy})+\ds\frac{\pl^2}{\pl x\pl
y}(-2u_{xy})+\ds\frac{\pl^2}{\pl y^2}(u_{xx})=0$$

$\Rightarrow un{0=0}!$

So $F$ is such that $I$ is \un{always} stationary!!!
\end{description}

\end{document}