Integration by parts (Green's identities) Im kinda lost with IBP using green identities involving vector field and scalar field
u - solution of PDE
$-\Delta u + \nabla\cdot (\vec b u) = f$ in $U$
$u = g$ in $\partial U$
trying to convert to weak form, to make use of the given condition a) $||\vec b||_{L^p}\leq C$ to use Lax-Milgram theorem to prove uniqueness of smooth solution
multiplying by arbitrary $v \in W_0^{1,q}(U) $
$\int_{U} [-\Delta u  + \nabla \cdot (\vec b u)]vdx=<f,v>$
first term is $\int \nabla u \cdot \nabla v dx=\int Du \cdot Dv dx$ since
$\int_{U} [-\Delta u v]dx=-\int_{\partial U} \frac{\partial u}{\partial n}\cdot v dS +\int_U \nabla u \cdot \nabla v dx$ boundary integral cancels since v = 0 on $\partial U$
now dealing with low order term
$\int_U\nabla \cdot (\vec b u)vdx=\int_U[u(\nabla \cdot \vec b)v+\vec b \cdot (\nabla u)v]dx$
$\int_{U} \bigg( Dv\cdot Du + u(\nabla \cdot \vec b)v+\vec b \cdot (\nabla u)v \bigg) dx=B[u,v]$
Will be proving uniqueness using Lax-Milgram theorem that states if B is bounded and coercive then solution is unique:
$B[u,v]\leq\int_{U} \bigg( |Dv\cdot Du + u(\nabla \cdot \vec b)v+\vec b \cdot (\nabla u)v |\bigg) dx$
using Holder inequality :
$\leq ||Du||^2_{L2}||Dv||^2_{L2}+||u||^2_{L2}||v||^2_{L2}||(\nabla \cdot \vec{b})||^2_{L2} + ||\vec b||^2_{L2} ||\nabla u||^2_{L2} ||v||^2_{L2}$
now
$\leq ||u||_{W12}||v||_{W12}+||u||^2_{L2}||v||^2_{L2}||(\nabla \cdot \vec{b})||^2_{L2} + ||\vec b||^2_{L2} ||u||_{W12} ||v||^2_{L2}$
how to proceed then? We only given that Ln norm of b  is bounded, but norm of $\nabla \cdot \vec b $ is unknown?
Trying to apply Green's identity to make up weak formulation to prove uniqueness using Lax-Milgram but messing up with vector and scalar fields. :(
another condition b) part is if given $\nabla \cdot \vec b\geq 0$ prove uniqueness maybe applying Lax Milgram is wrong because there is no upper bound and boundedness of B[u,v] hence can't be proven hence can't guarantee uniqns existence
Any help would be appreciated.
UPD: I figured out that applying divergence formula $div([bu]v) = div(bu)v + bu \cdot \nabla v$, integrating and using divergence theorem that
$\int_U [\nabla \cdot (bu)]vdx = \int_{dU} \vec \nu \cdot [(bu) v]dS - \int_U (bu) \cdot \nabla vdx$
UPD2:
since from UPD 1 we have integrating initial problem
$\int_U \nabla u \cdot \nabla v dx + \int_{\partial U} (\vec b u)v\cdot \vec n dS - \int_U \vec b u \cdot \nabla v dx - \int_U fv dx=0$
v = 0 at bdry
$\int_U (\nabla u -\vec b u) \cdot \nabla v dx - \int_U fv dx=0$
hence u in $H^1_0$ is weak solution if $\forall v\in H^1_0$ satisfies:
$\int_U (\nabla u -\vec b u) \cdot \nabla v dx - \int_U fv dx=0$
So trying to set up the Dirichlet minimisation problem from  is $DL_p = \nabla u -\vec b u$ integrating this wrt p=Du and this wrt z=u $DL_z=fv\implies I[w]=\int_U L dx = \int_U 0.5 (\nabla w)^2 -\vec b u \nabla u - fu dx$
 A: Under the assumption that the condition $\Vert \boldsymbol{b}\Vert_{L^p} \leq C < \infty$ holds also for $p \to \infty$  you can show continuity for the bilinearform
\begin{align} B[u,v] := \int_U \nabla u \cdot \nabla v + v \boldsymbol{b} \cdot \nabla u \mathrm d x\end{align}
through
\begin{align} \big \vert  B[u,v] \big \vert =& \Bigg \vert \int_U \nabla u \cdot \nabla v + v \boldsymbol{b} \cdot \nabla u \mathrm d x \Bigg \vert = \Bigg \vert \int_U \nabla u \cdot \nabla v \mathrm d x + \int_U  v \boldsymbol{b} \cdot \nabla u \mathrm d x \Bigg \vert \\
= & \Big \vert \big\langle \nabla u, \nabla v \big\rangle_{L^2(U)} + \big \langle v,  \boldsymbol{b} \cdot \nabla u \big \rangle_{L^2(U)}\Big\vert \\
\overset{\text{Triangle-Ineq.}}{\leq} & \Big \vert \big\langle \nabla u, \nabla v \big\rangle_{L^2(U)} \Big \vert + \Big \vert \big \langle v, \boldsymbol{b} \cdot \nabla u \big \rangle_{L^2(U)}\Big\vert \\
\overset{\text{Cauchy-Schwarz}}{\leq} & \Vert \nabla u \Vert_{L^2(U)} \Vert \nabla v \Vert_{L^2(U)} + \Vert v \Vert_{L^2(U)} \Vert \boldsymbol{b} \cdot \nabla u \Vert_{L^2(U)} \\
\leq & \Vert \nabla u \Vert_{L^2(U)} \Vert \nabla v \Vert_{L^2(U)} + \Vert v \Vert_{L^2(U)} \big \Vert \Vert \boldsymbol{b} \Vert_{L^\infty(U)} \nabla u \big \Vert_{L^2(U)} \\
\leq & \Vert  u \Vert_{W^{1,2}(U)} \Vert v \Vert_{W^{1,2}(U)} + \Vert v \Vert_{W^{1,2}(U)} \Vert \boldsymbol{b} \Vert_{L^\infty(U)} \Vert u  \Vert_{W^{1,2}(U)} \\
\leq & \max \big \{1, \Vert \boldsymbol{b} \Vert_{L^\infty(U)} \big\} \Vert_{L^\infty(U)} \Vert u  \Vert_{W^{1,2}(U)}
\end{align}
Concerning your second assumption, $\nabla \cdot \boldsymbol{b} \geq 0$, I am not sure if you can show coercivity for this case. I am more familiar with the case $c -0.5 \nabla \cdot \boldsymbol{b} \geq 0 $ for the elliptic PDE $\Delta u + \boldsymbol{b} \cdot \nabla u + c u = f$, which would translate here ($c = 0$) to $\nabla \cdot \boldsymbol{b} \leq 0$.
In that case, you can show first
$$\int_U u \boldsymbol{b} \cdot \nabla u \mathrm d x = \int_U \nabla \cdot (\boldsymbol{b}u^2) - u \boldsymbol{b}\cdot \nabla u - u^2 \nabla \cdot \boldsymbol{b} \mathrm d x $$
Thus, with $u \in W^{1, 2}_0(U)$
$$\int_U u \boldsymbol{b} \cdot \nabla u \mathrm d x = 0.5 \int_U - u^2 \nabla \cdot \boldsymbol{b} \mathrm d x \geq 0 $$
which would imply
\begin{align} \big\vert B[u,u] \big\vert =& \Bigg \vert \int_U \nabla u \cdot \nabla u \mathrm d x + \int_U  u \boldsymbol{b} \cdot \nabla u \mathrm d x \Bigg \vert \\
\geq& \Bigg \vert \int_U \nabla u \cdot \nabla u \mathrm d x \Bigg \vert = 0.5  \Bigg\vert \bigg[  \int_U \nabla u \cdot \nabla u \mathrm d x  +\int_U \nabla u \cdot \nabla u \mathrm d x \bigg] \Bigg\vert \\
\overset{\text{Poincare-Friedrichs}}{\geq}& 0.5 \Big[ \Vert \nabla u \Vert^2_{L^2(U)} + \frac{1}{C} \Vert  u \Vert^2_{L^2(U)} \Big] \\
\geq & 0.5 \min \{1, 1/C \} \Vert u \Vert^2_{W^{1, 2}(U)}\end{align}
