It is not clear to me what $\bar{b}$, $D$ and $\bar{U}$ are. And
is $U$ a single open set or a disjoint number of sets, $\bar{U}$ being its
closure? Assuming that $\bar{b}=\mathbf{b}\in \mathbb{R}^{n}$ and $%
D=\partial _{\mathbf{x}}$ then
$$
(-\partial _{\mathbf{x}}^{2}+\mathbf{b\cdot }\partial _{\mathbf{x}})u=0,\;%
\mathbf{x}\in U,\;u(\mathbf{x})=g(\mathbf{x}),\;\mathbf{x}\in \partial U.
$$
Writing
$$
v(\mathbf{x})=\exp [-\frac{1}{2}\mathbf{x\cdot b}]u(\mathbf{x}),
$$
we obtain
$$
(-\partial _{\mathbf{x}}^{2}+\frac{1}{4}b^{2})v(\mathbf{x})=0,\;\mathbf{x}%
\in U,\;v(\mathbf{x})=\exp [-\frac{1}{2}\mathbf{x\cdot b}]g(\mathbf{x})=h(%
\mathbf{x}),\;\mathbf{x}\in \partial U.
$$
In a few simple cases it is possible to obtain $v(\mathbf{x})$ and hence $u(%
\mathbf{x})$ explicitly and has the required properties.
In one dimension let $U=[-d,+d]$. Then, for $x\in U$ the general solution is
a linear combination of the eigenfunctions of $-\partial _{\mathbf{x}}^{2}$
at the eigenvalue $-\frac{1}{4}b^{2}$,
$$
v(x)=\alpha \exp [ikx]+\beta \exp [-ikx],\;x\in (-d,+d),\;k^{2}=-\frac{1}{4}%
b^{2},
$$
so
$$
v(x)=\alpha \exp [\frac{bx}{2}]+\beta \exp [-\frac{bx}{2}]
$$
Now
$$
v(-d)=\alpha \exp [-\frac{bd}{2}]+\beta \exp [+\frac{bd}{2}%
]=h(-d),\\v(+d)=\alpha \exp [+\frac{bd}{2}]+\beta \exp [-\frac{bd}{2}]=h(+d),
$$
which gives unique $\alpha $ and $\beta $, so there is only one solution,
which has the required properties.
In general $v(\mathbf{x})$ is again a linear combination of the
eigenfunctions of $-\partial _{\mathbf{x}}^{2}$ at the eigenvalue $-\frac{1}
{4}b^{2}$ for $\mathbf{x}$ in $U$.
In three dimensions they satisfy
$$
(-\partial _{\mathbf{x}}^{2}+\frac{1}{4}b^{2})\varphi =(-\frac{1}{x}\partial
_{x}^{2}x+\frac{\mathbf{l}^{2}}{x^{2}}+\frac{1}{4}b^{2})\varphi =0.
$$
Let ($\mathbf{x}=x\mathbf{e}_{\mathbf{x}}$)
$$\begin{eqnarray*}
\varphi _{l}^{m}(\mathbf{x}) &=&\rho _{l}(x)Y_{l}^{m}(\mathbf{e}_{\mathbf{x}%
}),\;l=0,1,2,\ldots \\
(-\frac{1}{x}\partial _{x}^{2}x+\frac{l(l+1)}{x^{2}}+\frac{1}{4}b^{2})\rho
_{l}(x) &=&0.
\end{eqnarray*}$$
The general expression for $v(\mathbf{x})$ is then (see, for instance, A.
Messiah, Quantum Mechanics I)
$$
v(\mathbf{x)=}\sum_{l,m}c_{lm}\varphi _{l}^{m}(\mathbf{x})=\sum_{l,m}c_{lm}%
\rho _{l}(x)Y_{l}^{m}(\mathbf{e}_{\mathbf{x}}),\;\mathbf{x}\in U.
$$
Now consider the special case that $U$ is a sphere with radius $r$ centered
in the origin and suppose further that $h(\mathbf{x})=h$, fixed, for all $%
\mathbf{x}\in \partial U$. Then
$$
v(r\mathbf{e}_{\mathbf{x}})=\sum_{l,m}c_{lm}\rho _{l}(r)Y_{l}^{m}(\mathbf{e}%
_{\mathbf{x}})=h,
$$
which requires $l=m=0$, leaving
$$
v(r\mathbf{e}_{\mathbf{x}})=c_{00}\rho _{0}(r)=h
$$
Here
$$
(-\frac{1}{x}\partial _{x}^{2}x+\frac{1}{4}b^{2})\rho _{0}(x)=0\;\Rightarrow
\;(-\partial _{x}^{2}+\frac{1}{4}b^{2})x\rho _{0}(x)=0,
$$
so
$$\begin{eqnarray*}
x\rho _{0}(x) &=&\alpha \exp [\frac{bx}{2}]+\beta \exp [-\frac{bx}{2}], \\
\rho _{0}(x) &=&\frac{1}{x}\{\alpha \exp [\frac{bx}{2}]+\beta \exp [-\frac{bx%
}{2}]\}
\end{eqnarray*}$$
Regularity in the origin requires $\beta =-\alpha $, so
$$\begin{eqnarray*}
\rho _{0}(x) &=&\alpha \frac{1}{x}\{\exp [\frac{bx}{2}]-\exp [-\frac{bx}{2}%
]\}, \\
v(\mathbf{x}) &=&c_{00}\alpha \frac{1}{x}\{\exp [\frac{bx}{2}]-\exp [-\frac{%
bx}{2}]\}, \\
v(r\mathbf{e}_{\mathbf{x}}) &=&c_{00}\alpha \frac{1}{r}\{\exp [\frac{br}{2}%
]-\exp [-\frac{br}{2}]\}=h,
\end{eqnarray*}$$
which fixes $c_{00}\alpha $.
For a more general $h=h(r\mathbf{e}_{\mathbf{x}
})$ it can be expanded in spherical harmonics, giving $c_{lm}.$ Now $\rho
_{l}(x)$ is associated with a spherical Bessel function.
In general, finding a solution for $v(\mathbf{x})$ in $U$ and then using
the boundary conditions on $\partial U$ should work but may be complicated
for more general $U$ and $h(\mathbf{x}).$
