Rayleigh-Ritz method for an extremum problem I am trying to use the Rayleigh-Ritz method to calculate an approximate solution to the extremum problem with the following functional:
$$ L[y]=\int\int_D (u_x^2+u_y^2+u^2-2xyu)\,dx\,dy, $$
$D$ is the unit square i.e. $0 \leq x \leq 1, 0 \leq y \leq 1.$
Also $u=0$ on the boundary of $D$.
I have chosen to use the trial function:
$$ \phi(x,y)=cxy(1-x)(1-y) $$
Where $c$ is a constant that I need to find.
I am familiar with using the Rayleigh-Ritz method most of the time, however this question I am not sure of. Is it possible to convert the problem to a Sturm-Liouville ration type?
Thanks for your help.

Your integral is in the form of
$$L(x,y,u)=\int\int_D (u_x^2+u_y^2+u^2-2xyu)\,dx\,dy$$
$$0 \leq x \leq 1, 0 \leq y \leq 1$$
Due to homogenous boundary conditions it is possible to use your approximation function
$$u(x,y)=cxy(1-x)(1-y)$$
When substituted into integral equation
$$L(x,y,u)=\int_0^1\int_0^1 (u_x^2+u_y^2+u^2-2xyu)\,dx\,dy=\frac{7}{300}c^2-\frac{1}{72}c$$
and taking first derivative condition and solving for c
$$\frac{d\,L}{d\,c}=\frac{7}{150}c-\frac{1}{72}=0\Rightarrow c=\frac{25}{84}$$
and
$$u(x,y)=\frac{25}{84}xy(1-x)(1-y)$$
Since the second derivative is positive it is a minimum.
