[proofplan] We convert the differential eigenvalue problem $Lw = \lambda w$ into an equivalent problem for a compact self-adjoint operator on $L^2(U)$. We shift $L$ to the coercive operator $L_\gamma$, invert it to obtain the compact Green's operator $S: L^2(U) \to L^2(U)$ (compactness via [Rellich-Kondrachov](/theorems/64)), verify that symmetry of $L$ forces $S$ to be self-adjoint and positive, apply the [Spectral Theorem for Compact Self-Adjoint Operators](/theorems/538) to $S$, and translate the eigenvalues back via $\lambda_k = 1/\sigma_k - \gamma$. [/proofplan] [step:Construct the compact Green's operator $S$ via the shifted operator $L_\gamma$] By the [Garding Inequality](/theorems/92), there exist $\beta > 0$ and $\gamma \ge 0$ such that $B_\gamma[u,v] := B[u,v] + \gamma(u,v)_{L^2}$ is bounded and coercive on $H^1_0(U)$. By the [Lax-Milgram Theorem](/theorems/91), $L_\gamma = L + \gamma I: H^1_0(U) \to H^{-1}(U)$ is an isomorphism. Define the Green's operator as the composition: \begin{align*} S: L^2(U) \overset{\iota}{\hookrightarrow} H^{-1}(U) \xrightarrow{L_\gamma^{-1}} H^1_0(U) \overset{j}{\hookrightarrow} L^2(U). \end{align*} The operator $S$ is compact because $j: H^1_0(U) \hookrightarrow L^2(U)$ is compact by the [Rellich-Kondrachov Theorem](/theorems/64) and the remaining maps are bounded. [/step] [step:Verify $S$ is self-adjoint on $L^2(U)$] [claim:Self-Adjointness of $S$] $(Sf, g)_{L^2} = (f, Sg)_{L^2}$ for all $f, g \in L^2(U)$. [/claim] [proof] Let $u := Sf$ and $v := Sg$, so $B_\gamma[u, \phi] = (f, \phi)_{L^2}$ and $B_\gamma[v, \phi] = (g, \phi)_{L^2}$ for all $\phi \in H^1_0(U)$. Then: \begin{align*} (Sf, g)_{L^2} = (u, g)_{L^2} = B_\gamma[v, u] = B_\gamma[u, v] = (f, v)_{L^2} = (f, Sg)_{L^2}, \end{align*} where the third equality uses the symmetry of $B_\gamma$ (which holds because $a_{ij} = a_{ji}$ and $b_i = 0$). [/proof] [/step] [step:Verify $S$ is positive] [claim:Positivity of $S$] $(Sf, f)_{L^2} > 0$ for all $f \in L^2(U) \setminus \{0\}$. [/claim] [proof] Set $u = Sf \in H^1_0(U)$. Then $(Sf, f)_{L^2} = (u, f)_{L^2} = B_\gamma[u, u] \ge \beta\|u\|_{H^1_0}^2 \ge 0$. If $(Sf, f)_{L^2} = 0$, then $u = 0$, so $(f, \phi)_{L^2} = B_\gamma[0, \phi] = 0$ for all $\phi \in H^1_0(U)$. Since $H^1_0(U)$ is dense in $L^2(U)$, this forces $f = 0$, a contradiction. [/proof] [/step] [step:Apply the Spectral Theorem to obtain the eigensequence of $S$] Since $S$ is compact, self-adjoint, and positive on the separable Hilbert space $L^2(U)$, the [Spectral Theorem for Compact Self-Adjoint Operators](/theorems/538) yields: eigenvalues $\sigma_1 \ge \sigma_2 \ge \dots > 0$ with $\sigma_k \to 0$, and orthonormal eigenfunctions $\{w_k\}$ forming a basis of $L^2(U)$. [/step] [step:Translate eigenvalues of $S$ to eigenvalues of $L$] [claim:Eigenvalue Correspondence] $Sw = \sigma w$ with $\sigma \neq 0$ if and only if $w \in H^1_0(U)$ and $Lw = \lambda w$ with $\lambda = 1/\sigma - \gamma$. [/claim] [proof] If $Sw = \sigma w$ with $\sigma \neq 0$, then $w \in \operatorname{Range}(S) \subset H^1_0(U)$ and $L_\gamma w = (1/\sigma) w$, giving $Lw = (1/\sigma - \gamma) w$. Conversely, if $Lw = \lambda w$ with $w \neq 0$, then $\lambda + \gamma \neq 0$ (since $L_\gamma$ is an isomorphism) and $Sw = w/(\lambda + \gamma)$. [/proof] Setting $\lambda_k = 1/\sigma_k - \gamma$: since $\sigma_k \to 0^+$, we have $\lambda_k \to +\infty$; since $\sigma_1 \ge \sigma_2 \ge \dots$, we have $\lambda_1 \le \lambda_2 \le \dots$. Positivity of $\lambda_1$ follows from the Rayleigh quotient and the Poincare inequality. [/step] [step:Establish smoothness of eigenfunctions by elliptic regularity] Each $w_k$ satisfies $Lw_k = \lambda_k w_k$ weakly with $\lambda_k w_k \in L^2(U)$. If $\partial U$ is $C^\infty$ and coefficients are $C^\infty(\bar{U})$, boundary regularity gives $w_k \in H^2(U)$. Iterating: $\lambda_k w_k \in H^2$ gives $w_k \in H^3$, and by induction $w_k \in H^m$ for every $m$. By the [Sobolev embedding theorem](/theorems/903), $w_k \in C^\infty(\bar{U})$. [/step]