I will be adding my solutions to the questions I can answer given the content from my advanced PDEs module.
Prove there is at most one smooth solution of this initial/boundary value problem for the heat equation with Neumann boundary conditions: \begin{align*} \begin{cases} u_t - \Delta u = f, \quad& \text{in } \, U_T \\ \frac{\partial u}{\partial \nu} = 0, \quad& \text{on } \, \partial U \times [0, T] \\ u = g, \quad& \text{on } \, U \times \{t = 0\} \end{cases} \end{align*}
Let $u^{(1)}$ and $u^{(2)}$ be smooth solutions to the Neumann problem and write $w = u^{(1)} - u^{(2)}$, we observe that \begin{equation*} w_t - \Delta w = u^{(1)}_t - u^{(2)}_t - \Delta u^{(1)} + \Delta u^{(2)} = f -f = 0. \end{equation*} So $w$ is a solution to the homogeneous Neumann problem \begin{align*} \begin{cases} w_t - \Delta w = 0, \quad& \text{in } \, U_T \\ \frac{\partial w}{\partial \nu} = 0, \quad& \text{on } \, \partial U \times [0, T] \\ w = 0, \quad& \text{on } \, U \times \{t = 0\}, \end{cases} \end{align*} it suffices to prove $w\equiv0$, uniqueness follows as a direct consequence. As is typical with a problem like this we appeal to the so called energy identity \begin{align*} \frac{d}{dt} \frac12 \int_U w^2 \, dx &= \int_U w w_t \, dx \\ &= \int_U w \Delta w \, dx \\ &= \int_{\partial U} w \frac{\partial w}{\partial \nu} \, dS - \int_U |\nabla w|^2 \, dx \\ &= - \int_U |\nabla w|^2 \, dx \\ &\leq 0. \end{align*} Since \[ \frac{d}{dt} \frac12 \int_U w^2 \, dx \leq 0 \] the integral as a function of $t$ is non-increasing. We compute the initial value \[ \frac12 \int_U w(x, 0)^2 \, dx = 0, \] since the integrand is non-negative, this implies $w(\cdot,t)=0$ almost everywhere in $U$ for all $t\in[0,T]$. By continuity, $w\equiv 0$ in $U_T$.
Assume $u$ is a smooth solution of \begin{align*} \begin{cases} u_t - \Delta u = 0, \quad& \text{in } \, U_T \\ u = 0, \quad& \text{on } \, \partial U \times [0, T] \\ u = g, \quad& \text{on } \, U \times \{t = 0\}. \end{cases} \end{align*} Prove the exponential decay estimate: \[ \|u(\cdot, t)\|_{L^2(U)} \leq e^{-\lambda_1 t} \|g\|_{L^2(U)} \quad (t \geq 0), \] where $\lambda_1 > 0$ is the pricipal eigenvalue of $-\Delta u$ (with zero boundary conditions) on $U$.
Let $\{\phi_n\}_{n\in \mathbb{N}}$ be the orthonormal basis we can construct from the Dirichlet Laplacian with eigenvalues satisfying $0 < \lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$. We write $g$ in terms of an expansion of this basis \begin{equation*} g = \sum_{n \in \mathbb{N}} a_n \phi_n , \quad a_n = \int_U g \phi_n \, dx. \end{equation*} By separation of variables we have the solution \begin{equation*} u(x, t) = \sum_{n \in \mathbb{N}} a_n e^{-\lambda_n t} \phi_n(x), \end{equation*} we apply Parceval's theorem and the result follows after some manipulation and applying Parceval's theorem again \begin{equation*} \|u(\cdot, t)\|^2_{L^2(U)} = \sum_{n \in \mathbb{N}}a_n^2 e^{-2\lambda_n t}. \end{equation*} By the ordering of eigenvalues, $\lambda_1 \leq \lambda_n$ for all $n \in \mathbb{N}$ so \begin{equation*} e^{-2\lambda_n t} \leq e^{-2 \lambda_1 t}. \end{equation*} Hence, \begin{align*} \|u(\cdot, t)\|^2_{L^2(U)} &\leq e^{-2\lambda_1 t}\sum_{n \in \mathbb{N}}a_n^2 \\ &= e^{-2\lambda_1 t} \|g\|^2_{L^2(U)}. \end{align*} Taking square roots finishes the proof \begin{equation*} \|u(\cdot, t)\|_{L^2(U)} \leq e^{-\lambda_1 t} \|g\|_{L^2(U)}. \end{equation*}
Suppose $f \in L^2(U)$ and assume that $u_m = \sum_{k = 1}^m d_m^k w_k$ solves \begin{equation*} \int_U \nabla u_m \cdot \nabla w_k \, dx = \int_U f w_k \, dx \end{equation*} for $k=1, \dots, m$. Show that a subsequence of $\{u_m\}_{m \in \mathbb{N}}$ converges weakly in $H^1_0(U)$ to the weak solution $u$ of \begin{align*} \begin{cases} -\Delta u = f, \quad& \text{in } U \\ u = 0, \quad& \text{on } \partial U. \end{cases} \end{align*}
Let $R: H^1_0(U) \to (H^1_0(U))^*$ be the Riesz map. Identifying the usual biliner form $B: H^1_0(U) \times H^1_0(U) \to \mathbb{R}$. We claim that $(u_m)_{m \in \mathbb{N}}$ is bounded in $H^1_0(U)$. Recall that $\|\cdot\|_{H^1_0(U)} \sim \|\nabla \cdot\|_{L^2(U)}$ and note $\|\nabla u_m\|^2_{L^2(U)} = B[u_m, u_m]$. \begin{align*} |(f, u_m)_{L^2(U)}| &\leq \|f\|_{L^2(U)} \|u_m\|_{L^2(U)} \\ &\leq \|f\|_{L^2(U)} \|u_m\|_{H^1_0(U)} \\ &\leq C_1 \|f\|_{L^2(U)} \|\nabla u_m\|_{L^2(U)}, \end{align*} where we have used Cauchy-Schwarz, the definition of $\|\cdot\|_{H^1_0(U)}$ and then equivalence of the norms. substituting this into the weak formulation of the problem gives \begin{align*} \|\nabla u_m\|^2_{L^2(U)} &\leq C_1 \|f\|_{L^2(U)} \|\nabla u_m\|_{L^2(U)} \\ \|\nabla u_m\|_{L^2(U)} &\leq C_1 \|f\|_{L^2(U)}. \end{align*} Now we use the equivalence of the norms again to obtain \begin{equation*} \|u_m\|_{H^1_0(U)} \leq C_2 \|f\|_{L^2(U)}, \end{equation*} so $(u_m)_{m \in \mathbb{N}}$ is bounded in $H^1_0(U)$. Writing $\tilde u_m = R(u_m)$ we have $(\tilde u_m)_{m \in \mathbb{N}}$ is bounded in $(H^1_0(U))^*$, we apply Banach-Alaoglu to obtain a weak-* convergent subsequence, $(\tilde u_{m_n})_{n \in \mathbb{N}}$, to some limit $\tilde u \in (H^1_0(U))^*$. Equivalently written \begin{equation*} (u_{m_n}, v)_{H^1_0(U)} \longrightarrow (u, v)_{H^1_0(U)} , \quad \forall v \in H^1_0(U), \end{equation*} where $u = R^{-1}(\tilde u)$. This is the definition of weak convergence in $H^1_0(U)$, so $(u_{m_n})_{n\in \mathbb{N}}$ converges weakly to $u$ in $H^1_0(U)$, this also implies weak convergence to $u$ in $L^2(U)$. Since $H^1_0(U)$ is compactly embedded in $L^2(U)$ there exists a further subsequence $(u_{m_{n_k}})_{k \in \mathbb{N}}$ that converges strongly to some $v \in L^2(U)$, as strong convergence implies weak convergence it is the case that $v = u$. By continuity of $B$ \begin{equation*} B[u_{m_{n_k}}, v] \longrightarrow B[u, v] , \quad \text{in } L^2(U). \end{equation*} We also have \begin{equation*} B[u_{m_{n_k}}, v] = \int_U f v \, dx , \quad \forall v \in \text{span}\{w_1, \dots, w_{m_{n_k}}\}, \end{equation*} since this span is dense in $H^1_0(U)$, passing to the limit gives \begin{equation*} \int_U \nabla u \cdot \nabla v \, dx = \int_U f v \, dx. \end{equation*}
Suppose $H$ is a Hilbert space and $u_k \rightharpoonup u$ in $L^2(0, T, H)$. Assume further we have the uniform bounds \begin{equation*} \underset{t \in [0, T]}{\text{ess sup}} \|u_k(t)\| \leq C \quad (k=1,\dots) \end{equation*} for some constant $C$. Prove \begin{equation*} \underset{t \in [0, T]}{\text{ess sup}} \|u(t)\| \leq C. \end{equation*}
Since $\{u_k\}_{k \in \mathbb{N}}$ is bounded in $L^\infty(0, T; H)$ we can apply Banach-Alaoglu and get a subsequence $\{u_{k_j}\}_{j \in \mathbb{N}}$ and some $\tilde u \in L^\infty(0, T; H)$ such that \begin{equation*} u_{k_j} \overset*\rightharpoonup \tilde u \quad \text{in } L^\infty(0, T; H), \end{equation*} equivalently written \begin{equation*} \int_0^T ( u_{k_j}(t), \varphi(t))_H \, dt \longrightarrow \int_0^T (\tilde u(t), \varphi(t))_H \, dt \quad \forall \varphi \in L^1(0, T; H). \end{equation*} Recall that $L^2(0, T; H) \subset L^1(0, T; H)$ with the embedding $\|\cdot\|_{L^1(0,T;H)} \leq\sqrt T \|\cdot\|_{L^2(0, T;H)}$. So every $L^2(0, T;H)$ function is an admissible test function for weak-* convergence which gives the definition of weak convergence in $L^2(0, T;H)$. Since $u_k \rightharpoonup u$ in $L^2(0, T;H)$ by uniqueness of weak limits $u = \tilde u$ and hence $u \in L^\infty(0, T;H)$ and by weak-* lower semi-continuity of the norm we have the result \begin{equation*} \underset{t \in[0, T]}{\text{ess sup}} \|u(t)\| \leq \liminf_{k \to \infty} \underset{t \in[0, T]}{\text{ess sup}}\|u_k(t)\| \leq C. \end{equation*}