Let $B\subset \mathbb R^2$ be a unit ball. let $v\in W^{1,2}(B)$ be given. We know that $0\leq v\leq 1$ and it is possible that $v=0$ on some positive $\mathcal L^2$ measurable set in $B$.
Let $w\in W^{1,2}(B)$ be given as well.
Define $$ \bar u:=\operatorname{argmin}\left\{\int_B|\nabla u|^2v^2,\,u\in W^{1,2}(B),\,\, T[u]=T[w]\right\} $$ where $T$ denote the standard trace operator.
My question: do we have $\bar u\in W^{1,2}(B)$ exist? (I do not care about uniqueness)
Here is an idea, which I came up after @Svetoslav comment to my last answer. I think it can be improved to your case. Let $I=(0,1)$ and $v(x)=x$ for $x\in I$. Let $$W=\{u\in W^{1,2}(I):\ u(0)=1,\ u(1)=0\}.$$
Define $J$ by $$J(u)=\int x^2 |u'|^2.\ \forall\ u\in W.$$
Because $u$ is absolutely continuous and $u(0)=1$, $u(1)=0$, we must conclude that $u'$ is not equal to zero a.e., which implies that $J(u)>0$.
I am gonna show that there is a sequence $u_k\in W$ such that $J(u_k)\to 0$. Indeed, let
$$ u_k(x) = \begin{cases} -kx+1 & \quad \text{if } x\in [0,1/k], \\ 0 & \quad \text{if } x\in [1/k,1].\\ \end{cases} $$
Note that $$J(u_k)=\int_0^{1/k} x^2k^2=\frac{1}{3k},$$
therefore $J(u_k)\to0$ if $k\to \infty$. We conclude that $J$ does not have a minimum in $W$.