We utilize the Lax-Milgram Theorem. To do so, we must associate the PDE with a bilinear form on the Hilbert space $H^1_0(U)$ and verify boundedness and coercivity.

**Step 1: Formulation of the Bilinear Form** We define the bilinear form $B_\mu: H^1_0(U) \times H^1_0(U) \to \mathbb{R}$ associated with the operator $L + \mu I$: \begin{align*} B_\mu[u, v] := \int_U \left( \sum_{i=1}^n \sum_{j=1}^n a_{ij} \partial_{x_i} u \partial_{x_j} v + \sum_{i=1}^n b_i (\partial_{x_i} u) v + (c + \mu) uv \right) \, dx. \end{align*} Solving the PDE weakly corresponds to finding $u \in H^1_0(U)$ such that $B_\mu[u, v] = \langle f, v \rangle_{L^2}$ for all $v \in H^1_0(U)$.

**Step 2: Boundedness of $B_\mu$** We verify that $|B_\mu[u, v]| \le \alpha \|u\|_{H^1_0} \|v\|_{H^1_0}$ for some $\alpha > 0$. Since $a_{ij}, b_i, c \in L^\infty(U)$, let $\Lambda = \max(\|a_{ij}\|_\infty, \|b_i\|_\infty, \|c\|_\infty)$. Using the Hölder inequality: \begin{align*} |B_\mu[u, v]| &\le \sum_{i,j} \int_U |a_{ij}| |\partial_{x_i} u| |\partial_{x_j} v| \, dx + \sum_{i} \int_U |b_i| |\partial_{x_i} u| |v| \, dx + \int_U |c + \mu| |u| |v| \, dx \\ &\le \Lambda \sum_{i,j} \|\partial_{x_i} u\|_{L^2} \|\partial_{x_j} v\|_{L^2} + \Lambda \sum_{i} \|\partial_{x_i} u\|_{L^2} \|v\|_{L^2} + (\Lambda + \mu) \|u\|_{L^2} \|v\|_{L^2}. \end{align*} Since $\|u\|_{L^2} \le C \|u\|_{H^1_0}$ (Poincaré inequality) and $\|\partial_{x_i} u\|_{L^2} \le \|u\|_{H^1_0}$, there exists a constant $K$ depending on $U, \Lambda, \mu$ such that: \begin{align*} |B_\mu[u, v]| \le K \|u\|_{H^1_0} \|v\|_{H^1_0}. \end{align*}

**Step 3: Coercivity (Gårding's Inequality)** We seek $\beta > 0$ such that $B_\mu[u, u] \ge \beta \|u\|_{H^1_0}^2$. Using the uniform ellipticity condition: \begin{align*} \int_U \sum_{i,j} a_{ij} \partial_{x_i} u \partial_{x_j} u \, dx \ge \theta \int_U |\nabla u|^2 \, dx = \theta \|\nabla u\|_{L^2}^2. \end{align*} For the lower-order terms, we use Cauchy's inequality with $\epsilon$. For any $\epsilon > 0$: \begin{align*} \left| \int_U \sum_{i} b_i (\partial_{x_i} u) u \, dx \right| &\le \Lambda \int_U |\nabla u| |u| \, dx \\ &\le \Lambda \left( \epsilon \|\nabla u\|_{L^2}^2 + \frac{1}{4\epsilon} \|u\|_{L^2}^2 \right). \end{align*} We choose $\epsilon = \frac{\theta}{2\Lambda}$. Then: \begin{align*} \left| \int_U \sum_{i} b_i (\partial_{x_i} u) u \, dx \right| \le \frac{\theta}{2} \|\nabla u\|_{L^2}^2 + C_1 \|u\|_{L^2}^2. \end{align*} Also, trivially: \begin{align*} \left| \int_U c u^2 \, dx \right| \le \Lambda \|u\|_{L^2}^2. \end{align*} Combining these into the expansion of $B_\mu[u, u]$: \begin{align*} B_\mu[u, u] &= \int_U \sum_{i,j} a_{ij} \partial_{x_i} u \partial_{x_j} u \, dx + \int_U \sum_{i} b_i (\partial_{x_i} u) u \, dx + \int_U c u^2 \, dx + \mu \int_U u^2 \, dx \\ &\ge \theta \|\nabla u\|_{L^2}^2 - \left( \frac{\theta}{2} \|\nabla u\|_{L^2}^2 + C_1 \|u\|_{L^2}^2 \right) - \Lambda \|u\|_{L^2}^2 + \mu \|u\|_{L^2}^2 \\ &= \frac{\theta}{2} \|\nabla u\|_{L^2}^2 + (\mu - C_1 - \Lambda) \|u\|_{L^2}^2. \end{align*} We define $\gamma := C_1 + \Lambda$. If we choose $\mu \ge \gamma$, the term $(\mu - \gamma)$ is non-negative. Thus: \begin{align*} B_\mu[u, u] \ge \frac{\theta}{2} \|\nabla u\|_{L^2}^2. \end{align*} By the Poincaré inequality, there exists $C_p > 0$ such that $\|u\|_{L^2}^2 \le C_p \|\nabla u\|_{L^2}^2$. Therefore, $\|\nabla u\|_{L^2}^2 \ge \frac{1}{1+C_p} \|u\|_{H^1_0}^2$. It follows that: \begin{align*} B_\mu[u, u] \ge \frac{\theta}{2(1+C_p)} \|u\|_{H^1_0}^2. \end{align*} This confirms coercivity with $\beta = \frac{\theta}{2(1+C_p)}$.

**Step 4: Conclusion** The bilinear form $B_\mu$ is bounded and coercive on $H^1_0(U)$. The linear functional $L_f(v) = \int_U fv \, dx$ is bounded on $H^1_0(U)$. By the Lax-Milgram Theorem, there exists a unique $u \in H^1_0(U)$ such that $B_\mu[u, v] = L_f(v)$ for all $v \in H^1_0(U)$.