The proof has six steps: (1) choose a basis and write down the finite-dimensional approximation, (2) show the approximation exists by reducing to an ODE, (3) derive energy estimates that are uniform in the approximation parameter, (4) extract a convergent subsequence by weak compactness, (5) upgrade to strong convergence via the [Aubin–Lions lemma](/theorems/615), and (6) pass to the limit in the weak formulation.
**Step 1: Galerkin ansatz.** Let $\{w_k\}_{k=1}^\infty$ be an orthonormal basis of $L^2(U)$ with each $w_k \in H^1_0(U)$. (For instance, the $L^2$-normalised eigenfunctions of $-\Delta$ with Dirichlet [boundary](/page/Boundary) conditions on $U$; these exist and are smooth by the spectral theory and regularity theory for the Laplacian.) For each integer $m \ge 1$, we seek an approximate solution of the form
\begin{align*}
u_m(x, t) := \sum_{k=1}^m d_m^k(t)\, w_k(x),
\end{align*}
where the unknown coefficient [functions](/page/Function) $d_m^k: [0, T] \to \mathbb{R}$ are determined by requiring the PDE to hold when tested against each of the first $m$ basis functions:
\begin{align*}
\bigl((u_m)_t(\cdot, t),\; w_k\bigr)_{L^2(U)} + B\bigl[u_m(\cdot, t),\; w_k;\; t\bigr] = f(t)(w_k), \quad k = 1, \ldots, m, \quad \text{for a.e. } t \in (0, T),
\end{align*}
with the initial condition $d_m^k(0) = (g, w_k)_{L^2(U)}$ for each $k$. Here $f(t)(w_k)$ denotes the duality pairing between $f(t) \in H^{-1}(U)$ and $w_k \in H^1_0(U)$, and $B[\cdot, \cdot\,; t]$ is the bilinear form associated to $L$ at time $t$.
The initial condition ensures $u_m(\cdot, 0) = \sum_{k=1}^m (g, w_k)_{L^2} w_k = P_m g$, where $P_m: L^2(U) \to \operatorname{span}\{w_1, \ldots, w_m\}$ is the orthogonal projection. Since $\{w_k\}$ is a complete orthonormal system, $P_m g \to g$ in $L^2(U)$ as $m \to \infty$.
**Step 2: The Galerkin system is a linear ODE.** Since $\{w_k\}$ is $L^2$-orthonormal, $((u_m)_t, w_k)_{L^2} = \dot{d}_m^k(t)$ (the ordinary [derivative](/page/Derivative) of the coefficient). The bilinear form is linear in $u_m$:
\begin{align*}
B[u_m, w_k; t] = B\Bigl[\sum_{j=1}^m d_m^j(t) w_j,\; w_k;\; t\Bigr] = \sum_{j=1}^m B[w_j, w_k; t]\, d_m^j(t).
\end{align*}
Define the $m \times m$ matrix $A(t)$ by $A_{kj}(t) := B[w_j, w_k; t]$ and the vector $F(t)$ by $F_k(t) := f(t)(w_k)$. Since $a_{ij}, b_i, c \in L^\infty(U_T)$, each entry $A_{kj}$ is a bounded measurable function of $t$, so $A \in L^\infty(0, T; \mathbb{R}^{m \times m})$. Since $f \in L^2(0, T; H^{-1}(U))$ and $w_k \in H^1_0(U)$, each $F_k \in L^2(0, T)$. The Galerkin system becomes:
\begin{align*}
\dot{d}_m^k(t) + \sum_{j=1}^m A_{kj}(t)\, d_m^j(t) = F_k(t), \quad k = 1, \ldots, m, \quad d_m^k(0) = (g, w_k)_{L^2}.
\end{align*}
This is a system of $m$ linear first-order ODEs with bounded measurable coefficients and square-[integrable](/page/Integral) forcing. By Carathéodory's existence and uniqueness theorem, there exists a unique absolutely continuous solution $d_m = (d_m^1, \ldots, d_m^m): [0, T] \to \mathbb{R}^m$.
**Step 3: Uniform energy estimates.**
[claim:Energy Bound Uniform In $m$]
There exists $C > 0$, depending only on $U$, $T$, $\theta$, and $\|a_{ij}\|_\infty, \|b_i\|_\infty, \|c\|_\infty$, such that for every $m$:
\begin{align*}
\sup_{0 \le t \le T} \|u_m(t)\|_{L^2(U)}^2 + \int_0^T \|u_m(t)\|_{H^1_0(U)}^2\, dt &\le C\left(\|g\|_{L^2(U)}^2 + \int_0^T \|f(t)\|_{H^{-1}(U)}^2\, dt\right), \tag{E1} \\
\int_0^T \|(u_m)_t(t)\|_{H^{-1}(U)}^2\, dt &\le C\left(\|g\|_{L^2(U)}^2 + \int_0^T \|f(t)\|_{H^{-1}(U)}^2\, dt\right). \tag{E2}
\end{align*}
[/claim]
[proof]
*Proof of (E1).* Multiply the Galerkin equation for index $k$ by $d_m^k(t)$ and sum over $k = 1, \ldots, m$. The left side becomes
\begin{align*}
\sum_{k=1}^m \dot{d}_m^k(t)\, d_m^k(t) + \sum_{k=1}^m B[u_m, w_k; t]\, d_m^k(t).
\end{align*}
The first sum equals $\frac{1}{2}\frac{d}{dt}\sum_{k=1}^m |d_m^k(t)|^2 = \frac{1}{2}\frac{d}{dt}\|u_m(t)\|_{L^2(U)}^2$, where we used $\|u_m(t)\|_{L^2}^2 = \sum_k |d_m^k(t)|^2$ (orthonormality). This identity holds classically because each $d_m^k$ is absolutely continuous — there is no infinite-dimensional subtlety here. The second sum equals $B[u_m, \sum_k d_m^k w_k; t] = B[u_m, u_m; t]$ by bilinearity. The right side is $\sum_k F_k(t) d_m^k(t) = f(t)(u_m(t))$. So:
\begin{align*}
\frac{1}{2}\frac{d}{dt}\|u_m\|_{L^2}^2 + B[u_m, u_m; t] = f(t)(u_m(t)). \tag{$\star$}
\end{align*}
By uniform ellipticity, the bilinear form satisfies (Gårding's inequality): $B[v, v; t] \ge \beta\|v\|_{H^1_0}^2 - \gamma\|v\|_{L^2}^2$ for constants $\beta > 0$ and $\gamma \ge 0$ depending on $\theta$ and the $L^\infty$ norms of the coefficients. Therefore:
\begin{align*}
\frac{1}{2}\frac{d}{dt}\|u_m\|_{L^2}^2 + \beta\|u_m\|_{H^1_0}^2 \le \gamma\|u_m\|_{L^2}^2 + f(t)(u_m(t)).
\end{align*}
For the forcing term, the duality pairing satisfies $|f(t)(u_m)| \le \|f(t)\|_{H^{-1}} \|u_m\|_{H^1_0}$. Apply Young's inequality $ab \le \frac{\beta}{2}a^2 + \frac{1}{2\beta}b^2$ with $a = \|u_m\|_{H^1_0}$ and $b = \|f\|_{H^{-1}}$:
\begin{align*}
|f(t)(u_m)| \le \frac{\beta}{2}\|u_m\|_{H^1_0}^2 + \frac{1}{2\beta}\|f(t)\|_{H^{-1}}^2.
\end{align*}
Substituting and absorbing $\frac{\beta}{2}\|u_m\|_{H^1_0}^2$ into the left side:
\begin{align*}
\frac{d}{dt}\|u_m\|_{L^2}^2 + \beta\|u_m\|_{H^1_0}^2 \le 2\gamma\|u_m\|_{L^2}^2 + \frac{1}{\beta}\|f(t)\|_{H^{-1}}^2.
\end{align*}
Define $\varphi(t) := \|u_m(t)\|_{L^2}^2$. Dropping the non-negative $\beta\|u_m\|_{H^1_0}^2$ term on the left gives $\varphi'(t) \le 2\gamma\, \varphi(t) + \frac{1}{\beta}\|f(t)\|_{H^{-1}}^2$. By Gronwall's inequality:
\begin{align*}
\varphi(t) \le e^{2\gamma T}\left(\varphi(0) + \frac{1}{\beta}\int_0^T \|f(s)\|_{H^{-1}}^2\, ds\right) \quad \text{for all } t \in [0, T].
\end{align*}
Since $\varphi(0) = \|u_m(0)\|_{L^2}^2 = \|P_m g\|_{L^2}^2 \le \|g\|_{L^2}^2$, this gives $\sup_t \|u_m(t)\|_{L^2}^2 \le C(\|g\|_{L^2}^2 + \|f\|_{L^2(0,T;H^{-1})}^2)$.
For the gradient bound: integrate ($\star$) over $[0, T]$ instead of applying Gronwall:
\begin{align*}
\frac{1}{2}\|u_m(T)\|_{L^2}^2 + \beta\int_0^T \|u_m\|_{H^1_0}^2\, dt \le \frac{1}{2}\|u_m(0)\|_{L^2}^2 + \gamma\int_0^T \|u_m\|_{L^2}^2\, dt + \frac{1}{2\beta}\int_0^T \|f\|_{H^{-1}}^2\, dt + \frac{\beta}{2}\int_0^T \|u_m\|_{H^1_0}^2\, dt.
\end{align*}
Absorbing and using the $L^\infty_t L^2_x$ bound on $u_m$ just derived gives $\int_0^T \|u_m\|_{H^1_0}^2\, dt \le C(\|g\|_{L^2}^2 + \|f\|_{L^2(0,T;H^{-1})}^2)$.
*Proof of (E2).* We must bound $(u_m)_t$ in $L^2(0,T; H^{-1}(U))$. For any $v \in H^1_0(U)$ with $\|v\|_{H^1_0} \le 1$, write $v = v_m + v_m^\perp$ where $v_m := P_m v \in \operatorname{span}\{w_1, \ldots, w_m\}$ and $v_m^\perp \perp \operatorname{span}\{w_1, \ldots, w_m\}$ in $L^2$. Since $P_m$ is an orthogonal projection in $L^2$ and the $w_k$ belong to $H^1_0$, we have $\|v_m\|_{L^2} \le \|v\|_{L^2}$ and (crucially) $\|v_m\|_{H^1_0} \le \|v\|_{H^1_0} \le 1$.
Now $(u_m)_t$ takes values in $\operatorname{span}\{w_1, \ldots, w_m\}$, so $((u_m)_t, v_m^\perp)_{L^2} = 0$, and thus $(u_m)_t(v) = ((u_m)_t, v)_{L^2} = ((u_m)_t, v_m)_{L^2}$. From the Galerkin equation:
\begin{align*}
((u_m)_t, v_m)_{L^2} = f(t)(v_m) - B[u_m, v_m; t].
\end{align*}
Taking absolute values: $|((u_m)_t, v_m)_{L^2}| \le \|f(t)\|_{H^{-1}}\|v_m\|_{H^1_0} + C_B\|u_m(t)\|_{H^1_0}\|v_m\|_{H^1_0} \le \|f(t)\|_{H^{-1}} + C_B\|u_m(t)\|_{H^1_0}$, where $C_B$ is the [continuity](/page/Continuity) constant of $B$. Taking the supremum over $\|v\|_{H^1_0} \le 1$:
\begin{align*}
\|(u_m)_t(t)\|_{H^{-1}} \le \|f(t)\|_{H^{-1}} + C_B\|u_m(t)\|_{H^1_0}.
\end{align*}
Squaring, integrating over $[0, T]$, and using $(a + b)^2 \le 2a^2 + 2b^2$:
\begin{align*}
\int_0^T \|(u_m)_t\|_{H^{-1}}^2\, dt \le 2\int_0^T \|f\|_{H^{-1}}^2\, dt + 2C_B^2 \int_0^T \|u_m\|_{H^1_0}^2\, dt,
\end{align*}
which is bounded by the (E1) estimate.
[/proof]
**Step 4: Weak compactness.** The bounds (E1) and (E2) show that $\{u_m\}$ is bounded in $L^2(0, T; H^1_0(U))$ and $\{(u_m)_t\}$ is bounded in $L^2(0, T; H^{-1}(U))$, both with bounds independent of $m$. Since $L^2(0, T; H^1_0(U))$ is a [separable](/page/Separable) [Hilbert space](/page/Hilbert%20Space), the [Sequential Banach–Alaoglu theorem](/theorems/496) gives a subsequence (which we relabel as $u_m$) and an element $u$ such that:
\begin{align*}
u_m \rightharpoonup u \quad &\text{weakly in } L^2(0, T; H^1_0(U)), \\
(u_m)_t \rightharpoonup u_t \quad &\text{weakly in } L^2(0, T; H^{-1}(U)).
\end{align*}
In particular, $u \in \mathcal{W} := \{v \in L^2(0,T; H^1_0(U)) : v_t \in L^2(0,T; H^{-1}(U))\}$.
**Step 5: Strong convergence via Aubin–Lions.** [Weak convergence](/page/Weak%20Convergence) alone is not enough to pass to the limit in the weak formulation — the bilinear form $B$ contains products of $u_m$ and [test functions](/page/Test%20Function), and weak [limits](/page/Limit) do not commute with products in general. We need strong convergence in a weaker norm.
Apply the [Aubin–Lions Compactness Lemma](/theorems/615) with $X_0 = H^1_0(U)$, $X = L^2(U)$, $X_1 = H^{-1}(U)$, and $p = q = 2$. The hypotheses are:
- $\{u_m\}$ bounded in $L^2(0, T; H^1_0(U))$: this is (E1). ✓
- $\{(u_m)_t\}$ bounded in $L^2(0, T; H^{-1}(U))$: this is (E2). ✓
- $H^1_0(U) \subset\subset L^2(U)$: this is the [Rellich–Kondrachov theorem](/theorems/64) (valid since $U$ is bounded with $C^1$ boundary). ✓
The conclusion is that (passing to a further subsequence):
\begin{align*}
u_m \to u \quad \text{strongly in } L^2(0, T; L^2(U)).
\end{align*}
This strong convergence is the key tool for Step 6.
**Step 6: Passage to the limit in the weak formulation.** We must show that $u$ satisfies the weak formulation. Fix an index $\ell \in \mathbb{N}$ and a smooth time function $\varphi \in C^1([0, T])$ with $\varphi(T) = 0$. For every $m \ge \ell$, the Galerkin equation holds with test function $w_\ell$:
\begin{align*}
((u_m)_t, w_\ell)_{L^2}\, + B[u_m, w_\ell; t] = f(t)(w_\ell) \quad \text{for a.e. } t.
\end{align*}
Multiply both sides by $\varphi(t)$ and integrate over $[0, T]$. Integrate the time derivative term by parts: since $((u_m)_t, w_\ell)_{L^2} = \frac{d}{dt}(u_m, w_\ell)_{L^2}$,
\begin{align*}
\int_0^T \frac{d}{dt}(u_m, w_\ell)_{L^2}\, \varphi(t)\, dt = \bigl[(u_m, w_\ell)_{L^2}\, \varphi\bigr]_0^T - \int_0^T (u_m, w_\ell)_{L^2}\, \varphi'(t)\, dt.
\end{align*}
The boundary term at $t = T$ vanishes because $\varphi(T) = 0$. The boundary term at $t = 0$ is $-(u_m(0), w_\ell)_{L^2}\, \varphi(0) = -(P_m g, w_\ell)_{L^2}\, \varphi(0)$. For $m \ge \ell$, $(P_m g, w_\ell) = (g, w_\ell)$ since $w_\ell \in \operatorname{span}\{w_1, \ldots, w_m\}$. So the integrated Galerkin equation is:
\begin{align*}
-\int_0^T (u_m, w_\ell)_{L^2}\, \varphi'\, dt + \int_0^T B[u_m, w_\ell; t]\, \varphi\, dt = \int_0^T f(t)(w_\ell)\, \varphi\, dt + (g, w_\ell)_{L^2}\, \varphi(0). \tag{G$_m$}
\end{align*}
We now pass $m \to \infty$ in each term of (G$_m$):
[claim:Convergence Of Each Term]
As $m \to \infty$ (along the subsequence from Steps 4–5):
(i) $\int_0^T (u_m, w_\ell)_{L^2}\, \varphi'\, dt \to \int_0^T (u, w_\ell)_{L^2}\, \varphi'\, dt$.
(ii) $\int_0^T B[u_m, w_\ell; t]\, \varphi\, dt \to \int_0^T B[u, w_\ell; t]\, \varphi\, dt$.
[/claim]
[proof]
*(i)* By the strong convergence $u_m \to u$ in $L^2(0,T; L^2(U))$ from Step 5:
\begin{align*}
\left|\int_0^T \bigl((u_m, w_\ell)_{L^2} - (u, w_\ell)_{L^2}\bigr)\, \varphi'\, dt\right| &\le \|\varphi'\|_{L^\infty} \int_0^T |(u_m - u, w_\ell)_{L^2}|\, dt \\
&\le \|\varphi'\|_{L^\infty}\, \|w_\ell\|_{L^2} \int_0^T \|u_m - u\|_{L^2}\, dt \\
&\le \|\varphi'\|_{L^\infty}\, \|w_\ell\|_{L^2}\, \sqrt{T}\, \|u_m - u\|_{L^2(0,T; L^2)} \to 0.
\end{align*}
Here the last step uses $\int_0^T \|u_m - u\|_{L^2}\, dt \le \sqrt{T}(\int_0^T \|u_m - u\|_{L^2}^2\, dt)^{1/2}$ by Cauchy–Schwarz.
*(ii)* The bilinear form $B$ has the structure $B[v, w_\ell; t] = \int_U \sum_{i,j} a_{ij} \partial_{x_j} v\, \partial_{x_i} w_\ell + \sum_i b_i \partial_{x_i} v\, w_\ell + c\, v\, w_\ell\, d\mathcal{L}^n$. Define the linear functional $\Lambda: L^2(0, T; H^1_0(U)) \to \mathbb{R}$ by
\begin{align*}
\Lambda(v) := \int_0^T B[v, w_\ell; t]\, \varphi(t)\, dt.
\end{align*}
This is bounded: $|\Lambda(v)| \le C_B \|w_\ell\|_{H^1_0} \|\varphi\|_{L^\infty} \|v\|_{L^2(0,T; H^1_0)}$ (where $C_B$ depends on the $L^\infty$ norms of $a_{ij}, b_i, c$). Since $u_m \rightharpoonup u$ weakly in $L^2(0, T; H^1_0(U))$ and $\Lambda$ is a bounded linear functional, $\Lambda(u_m) \to \Lambda(u)$ by the definition of weak convergence.
[/proof]
The right side of (G$_m$) is independent of $m$ (for $m \ge \ell$). Therefore, passing to the limit:
\begin{align*}
-\int_0^T (u, w_\ell)_{L^2}\, \varphi'\, dt + \int_0^T B[u, w_\ell; t]\, \varphi\, dt = \int_0^T f(t)(w_\ell)\, \varphi\, dt + (g, w_\ell)_{L^2}\, \varphi(0). \tag{G$_\infty$}
\end{align*}
This holds for every $\ell \in \mathbb{N}$ and every $\varphi \in C^1([0, T])$ with $\varphi(T) = 0$.
Now we extend to all test functions. An arbitrary test function in $L^2(0, T; H^1_0(U))$ can be approximated by finite linear combinations $v(x, t) = \sum_{\ell=1}^N \varphi_\ell(t)\, w_\ell(x)$ with $\varphi_\ell \in C^1([0,T])$, $\varphi_\ell(T) = 0$. (This is because $\{w_\ell\}$ is a basis of $H^1_0(U)$ and smooth functions are dense in $L^2(0,T)$.) By linearity of (G$_\infty$) in $\ell$ and $\varphi$, (G$_\infty$) holds for all such finite combinations. By the continuity of each term in the $L^2(0,T; H^1_0)$ norm (using the bounds on $u$ and $f$), the identity extends by density to all $v \in L^2(0, T; H^1_0(U))$.
Integrating the first term of (G$_\infty$) back by parts in $t$ (which is justified since $u_t \in L^2(0,T; H^{-1})$) recovers the original weak formulation:
\begin{align*}
\int_0^T \bigl[u_t(t)(v(t)) + B[u(t), v(t); t]\bigr]\, dt = \int_0^T f(t)(v(t))\, dt
\end{align*}
for all $v \in L^2(0, T; H^1_0(U))$, and $u(\cdot, 0) = g$ in $L^2(U)$ (extracted from the boundary term at $t = 0$).
**Step 7: Uniqueness.** Suppose $u_1, u_2 \in \mathcal{W}$ are both weak solutions with the same data $(f, g)$. Their difference $w := u_1 - u_2$ satisfies $w_t + Lw = 0$ with $w(\cdot, 0) = 0$. The [Parabolic Energy Estimate](/theorems/603) with $f = 0$ and $g = 0$ gives
\begin{align*}
\sup_{0 \le t \le T} \|w(t)\|_{L^2}^2 + \int_0^T \|w(t)\|_{H^1_0}^2\, dt \le C\bigl(\|0\|_{L^2}^2 + \|0\|_{L^2(0,T;H^{-1})}^2\bigr) = 0.
\end{align*}
Therefore $w = 0$ in $\mathcal{W}$, so $u_1 = u_2$.
