= Solution
{c}
Take the difference of two <weak solutions>, so $f=\psi_0=\psi_1=0$. At the stated $H^1(U_T)$ regularity, it is not justified simply to test by $u_t$, which need not belong to the spatial test space $H_0^1$. Instead use the antiderivative test underlying <uniqueness of Sobolev weak solutions of a wave equation>.
Fix $s\in(0,T]$ and define
$$
v(t)=\begin{cases}\displaystyle\int_t^su(r)dr,&0\leq t<s,\\0,&s\leq t\leq T,
\end{cases}
\qquad z(t)=\int_0^tu(r)dr.
$$
These are <Bochner integrals> in $H_0^1$. The function $v$ is an admissible space-time $H^1$ test, $v_t=-u$ before $s$, and $v(t)=z(s)-z(t)$. The time term in the <weak formulation> is
$$
-\int_0^s(u_t,v_t)dt=\frac12\|u(s)\|_2^2,
$$
because $u(0)=0$. Symmetry of $a^{ij}$ and <integration by parts> in time give
$$
\int_0^s\int_Ua^{ij}u_{x_i}v_{x_j}
=\frac12\int_Ua^{ij}(x,0)z_{x_i}(s)z_{x_j}(s)
+\frac12\int_0^s\int_Ua_t^{ij}v_{x_i}v_{x_j}.
$$
For the lower-order part, <integration by parts> in space gives
$$
\int_U(b^i u_{x_i}v+cuv)
=\int_Uu\bigl((c-\operatorname{div}b)v-b\cdot Dv\bigr).
$$
Bounded coefficients, the <Poincaré inequality> for $v$, and the <Young inequality> now imply
$$
\|u(s)\|_2^2+\theta\|Dz(s)\|_2^2
\leq C\int_0^s\bigl(\|u(t)\|_2^2+\|Dv(t)\|_2^2\bigr)dt.
$$
Since $Dv(t)=Dz(s)-Dz(t)$,
$$
\|u(s)\|_2^2+\theta\|Dz(s)\|_2^2
\leq C\int_0^s\bigl(\|u(t)\|_2^2+\|Dz(t)\|_2^2\bigr)dt
+Cs\|Dz(s)\|_2^2.
$$
For $s$ in a sufficiently short fixed interval, absorb the last term into the left-hand side. The <Gronwall inequality> applied to $\|u(s)\|_2^2+(\theta/2)\|Dz(s)\|_2^2$ proves $u=0$ on that interval. The equation gives continuity of $u_t$ in $H^{-1}$, so both initial traces at its endpoint are again zero. Repeating with the same uniform coefficient bounds covers $[0,T]$. Therefore
$$
\boxed{\text{the weak solution is unique whenever it exists.}}
$$
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