[proofplan]
We apply the [Lax-Milgram Theorem](/theorems/91) to the shifted bilinear form $B_\mu$ associated with $L + \mu I$ on the [Hilbert space](/pages/1083) $H^1_0(U)$. Boundedness of $B_\mu$ follows from the $L^\infty$ bounds on the coefficients and the Cauchy–Schwarz inequality. Coercivity follows from the [Garding Inequality](/theorems/92): for $\mu \ge \gamma$, the shift $\mu$ absorbs the lower-order penalty, and the Poincare inequality converts the gradient bound into a full $H^1_0$ bound. With both hypotheses verified, Lax-Milgram produces the unique weak solution.
[/proofplan]
[step:Define the shifted bilinear form $B_\mu$ and reformulate the PDE]
Define the bilinear form $B_\mu: H^1_0(U) \times H^1_0(U) \to \mathbb{R}$ associated with $L + \mu I$:
\begin{align*}
B_\mu[u, v] := \int_U \sum_{i,j=1}^n a_{ij}\, \partial_{x_i} u\, \partial_{x_j} v \, d\mathcal{L}^n + \int_U \sum_{i=1}^n b_i\, (\partial_{x_i} u)\, v \, d\mathcal{L}^n + \int_U (c + \mu)\, u\, v \, d\mathcal{L}^n.
\end{align*}
A function $u \in H^1_0(U)$ is a weak solution of $Lu + \mu u = f$ with $u = 0$ on $\partial U$ if and only if:
\begin{align*}
B_\mu[u, v] = (f, v)_{L^2(U)} \qquad \text{for all } v \in H^1_0(U).
\end{align*}
[/step]
[step:Verify boundedness of $B_\mu$ on $H^1_0(U)$]
Set $\Lambda := \max_{i,j}\bigl(\|a_{ij}\|_{L^\infty(U)},\, \|b_i\|_{L^\infty(U)},\, \|c\|_{L^\infty(U)}\bigr)$. By the Cauchy–Schwarz inequality applied to each integral:
\begin{align*}
|B_\mu[u, v]| &\le \Lambda \sum_{i,j=1}^n \|\partial_{x_i} u\|_{L^2(U)} \|\partial_{x_j} v\|_{L^2(U)} + \Lambda \sum_{i=1}^n \|\partial_{x_i} u\|_{L^2(U)} \|v\|_{L^2(U)} + (\Lambda + \mu) \|u\|_{L^2(U)} \|v\|_{L^2(U)}.
\end{align*}
Since $\|\partial_{x_i} u\|_{L^2(U)} \le \|u\|_{H^1_0(U)}$ and $\|u\|_{L^2(U)} \le \|u\|_{H^1_0(U)}$, there exists a constant $K = K(n, \Lambda, \mu) > 0$ such that:
\begin{align*}
|B_\mu[u, v]| \le K \|u\|_{H^1_0(U)} \|v\|_{H^1_0(U)}.
\end{align*}
[/step]
[step:Verify coercivity of $B_\mu$ for $\mu \ge \gamma$ using the Garding inequality]
Write $B_\mu[u, u] = B[u, u] + \mu \|u\|_{L^2(U)}^2$, where $B$ is the bilinear form of $L$ (without the shift). By the [Garding Inequality](/theorems/92), there exist constants $\beta_0 > 0$ and $\gamma \ge 0$ (depending on $U$, $\theta$, and the coefficients) such that:
\begin{align*}
B[u, u] \ge \beta_0 \|u\|_{H^1_0(U)}^2 - \gamma \|u\|_{L^2(U)}^2.
\end{align*}
Adding $\mu \|u\|_{L^2(U)}^2$:
\begin{align*}
B_\mu[u, u] = B[u, u] + \mu \|u\|_{L^2(U)}^2 \ge \beta_0 \|u\|_{H^1_0(U)}^2 + (\mu - \gamma) \|u\|_{L^2(U)}^2.
\end{align*}
For $\mu \ge \gamma$, the second term is non-negative, so:
\begin{align*}
B_\mu[u, u] \ge \beta_0 \|u\|_{H^1_0(U)}^2.
\end{align*}
[guided]
The [Garding Inequality](/theorems/92) tells us that $B[u, u]$ is almost coercive — it controls $\|u\|_{H^1_0}^2$ up to a correction $-\gamma \|u\|_{L^2}^2$. The shift $\mu u$ adds exactly $\mu \|u\|_{L^2}^2$ to $B_\mu[u, u]$, which cancels the penalty when $\mu \ge \gamma$.
Concretely, $B_\mu[u, u] = B[u, u] + \mu \|u\|_{L^2(U)}^2$. By the [Garding Inequality](/theorems/92):
\begin{align*}
B_\mu[u, u] \ge \beta_0 \|u\|_{H^1_0(U)}^2 - \gamma \|u\|_{L^2(U)}^2 + \mu \|u\|_{L^2(U)}^2 = \beta_0 \|u\|_{H^1_0(U)}^2 + (\mu - \gamma)\|u\|_{L^2(U)}^2.
\end{align*}
When $\mu \ge \gamma$, the $L^2$ term is non-negative and can be dropped:
\begin{align*}
B_\mu[u, u] \ge \beta_0 \|u\|_{H^1_0(U)}^2.
\end{align*}
This is the coercivity condition required by [Lax-Milgram](/theorems/91).
[/guided]
[/step]
[step:Apply Lax-Milgram to obtain the unique weak solution]
The linear functional $\ell_f: H^1_0(U) \to \mathbb{R}$ defined by $\ell_f(v) := (f, v)_{L^2(U)} = \int_U f\, v \, d\mathcal{L}^n$ is bounded on $H^1_0(U)$: by Cauchy–Schwarz, $|\ell_f(v)| \le \|f\|_{L^2(U)} \|v\|_{L^2(U)} \le \|f\|_{L^2(U)} \|v\|_{H^1_0(U)}$, so $\ell_f \in (H^1_0(U))^*$.
We have verified that $B_\mu$ is bounded and coercive on the [Hilbert space](/pages/1083) $H^1_0(U)$. By the [Lax-Milgram Theorem](/theorems/91), there exists a unique $u \in H^1_0(U)$ satisfying:
\begin{align*}
B_\mu[u, v] = \ell_f(v) = (f, v)_{L^2(U)} \qquad \text{for all } v \in H^1_0(U).
\end{align*}
This $u$ is the unique weak solution of $Lu + \mu u = f$ with $u = 0$ on $\partial U$.
[/step]
