I need a check on the following problem:
Consider the system
$$ \begin{cases} -(\mu u')' + b u' = 0 \\ u(0)=0 \\ u(1)=1 \end{cases} $$ with $b$ and $\mu$ functions in $L^2$. I want to find the weak formulation for this problem
I take $v \in H_0^1(0,1)$ and mutliply and integrate over $(0,1)$: $$\int_0^1 \mu u' v'dx + \int_0^1 b u' v dx=0$$
I want to have both the test function, and also the solution $u$, in the same functional space, so that I can apply Lax-Milgram lemma. To this aim, I consider $$\bar{u}(x) = u(x)-x$$ which satisfy $\bar{u}(0)=u(0)-0=0$ and $\bar{u}(1)=1-1=0$. So I apply the substitution in the weak formulation, i.e.:
$$\int_0^1 \mu (\bar{u} +x)' v' dx + \int_0^1 b (\bar{u}+x)'v dx = 0$$
which, after trivial computation becomes:
$$\int_0^1 \mu \bar{u}' v' dx + \int_0^1 b \bar{u}'v dx = - \int_0^1 bv dx - \int_0^1 \mu v' dx$$
So, the weak formulation is:
"Find $\bar{u} \in H_0^1$ s.t. $$\int_0^1 \mu \bar{u}' v' dx + \int_0^1 b \bar{u}'v dx = - \int_0^1 bv dx - \int_0^1 \mu v' dx$$ for all $v \in H_0^1$"
Once I find the solution $\bar{u}$, I recover $u(x)$ with $$u(x)=\bar{u}(x) + x$$
Is everything correct?