Given
$$ I[u(x,y)] = \frac12\int\int_G\left| \nabla u \right|^2$$
where G is defined by $x^2+y^2\lt R^2 $
Find u(x,y) that will provide the lowest possible value of the functional. Also given:
$$u(r=R,\theta) = ax^2 $$
So I used E-L relation, and found that I end up with Laplace's equation.
Using separation of variables, I get
$u(r,\theta) = R(r)\Theta (\theta): $
$$\frac{r}{R(r)}\frac{\partial}{\partial r}\left(r R(r)' \right) = -\frac{\Theta''}{\Theta} $$
I for $\Theta$ and got:
$$\Theta_n =\alpha_ncos(n\theta) + \beta sin(n\theta) $$
And for R i get two solutions depending on n:
If n = 0:
$$R_0 = Aln(r) + B$$
A has to be zero so the function doesn't diverge at r=0
When $n \ne 0$:
$$R_n = Cr^n + Dr^{-n} $$
Where d has to be zero, again so the function doesn't diverge at n=0.
I get the function:
$$u(r,\theta) = B_0 + \sum \left(\alpha_n cos(n\theta) + \beta_n sin(n\theta)\right)r^n$$
Using our BV, changing it to polar coordinates:
$$ u(r=R, \theta)= ar^2\left(\frac12 - \frac{cos2\theta}{2}\right) $$
$$u(r,\theta) = B_0 + \sum \left(\alpha_n cos(n\theta) + \beta_n sin(n\theta)\right)r^n = ar^2\left(\frac12 - \frac{cos2\theta}{2}\right) $$
Here we see that we only have terms when n=2:
$$ B_0 = \frac{aR^2}{2}$$
$$\beta_n = 0 $$
$$ \alpha_n = -a\frac{1}{2} $$
Therefore we get:
$$\frac{aR^2}{2} -a\frac{1}{2}r^2cos2\theta $$
Switching back to cartesian I get:
$$\frac{aR^2}{2} + \frac{a}{2}x^2 -\frac{a}{2}y^2$$
According to solutions, Im missing a term $by$.
I'd appreciate any type of help