Existence and Uniqueness of Poisson Equation with Robin Boundary Condition using First Variation Methods I'm currently stuck on the following exercise from Evans PDE Chapter 8 Exercise 11.

Let $\beta: \mathbb{R} \rightarrow \mathbb{R}$ be smooth with \begin{equation} 0 < a \leq \beta'(z) \leq b, \text{ } z \in \mathbb{R} \end{equation} for constants $a,b$. Let $f \in L^2(U)$ where $U$ is a bounded subset of $\mathbb{R}^n$ with smooth boundary. Formulate what it means for $u \in H^1(U)$ to be a weak solution of the non-linear boundary value problem \begin{equation*} \begin{cases} -\Delta u = f \text{ in } U\\ \frac{\partial u}{\partial \nu} + \beta(u) = 0 \text{ on } \partial U \end{cases} \end{equation*} Prove there exists a unique solution.($\nu$ is the outward normal vector)

Let $\mathrm{Tr}$ be the trace operator, then I was able to formulate what a weak solution meant e.g. for any $v \in H^1(U)$ \begin{equation*} \int_{\partial U} \beta\big(\mathrm{Tr}(u)\big) \mathrm{Tr}(v) + \int_{\Omega} Du \cdot Dv - fv = 0 \end{equation*} However, I have problems finding a corresponding energy for this PDE. From the condition that $\beta'(z)$ is strictly positive and that we want a unique solution, I deduced that our energy probably has an expression for the anti-derivative of $\beta$ to make the energy strictly convex. I believe the energy is \begin{equation*} E(u) := \int_{U} \frac{1}{2} |Du|^2 - fu \text{ } dx + \int_{\partial U}\int_{0}^{\mathrm{Tr}(u)} \beta'(t) \text{ } dt dx \end{equation*} and our admissible set $\mathcal{A} = H^1(U)$. Indeed, the Euler Lagrange Equation matches the weak formulation. And we know from joint convexity of the Lagrangian associated with the energy that any solution of the Euler-Lagrange is a minimizer, so there is at most one solution by Strict Convexity. However, I cannot prove there exists a solution e.g. I can't prove the minimizing sequence is bounded. Any hints or help would be appreciated.
 A: So I was able to prove existence and uniqueness of the PDE. I decided to prove uniqueness and existence from a first variational point of view (e.g. Chapter 8 of Evans). 
Given \begin{equation*} \tag{0.1} \begin{cases} -\Delta u = f \text{ on } U \\ \frac{\partial u}{\partial n} + \beta(u) = 0 \text{ in } \partial U \end{cases}  \end{equation*} with $0 < a \leq \beta'(z) \leq b$, we notice this implies the antiderivative of $\beta$ is strictly convex, so we hope to find an energy associated with $(0.1)$ such that the energy is strictly convex to get the uniqueness. 
To do this we observe that the energy \begin{equation*} E(u) := \int_{U} \frac{1}{2} |Du|^2 - fu \text{ }dx + \int_{\partial U} \int_{0}^{Tr(u)} \beta(t) \text{ }dt dH^{n-1} \end{equation*} (Tr is the trace operator, which can be defined since $\partial U$ is smooth and bounded) minimized over $H^1(U)$ has the following Euler-Lagrange Equation which can be obtained by taking the Frechet derivative with any smooth function $v \in C^{\infty}(\overline{U})$ \begin{equation*} \int_{U} Du \cdot Dv - fv \text{ } dx + \int_{\partial U} \beta(Tr(u)) Tr(v)  \end{equation*} where the last term is justified by using $\beta \in C^1$ to apply the fundamental theorem of calculus. Now we know from joint convexity on $(u,Du)$ of the Lagrangian associated to $E(u)$, $u$ solves $(0.1)$ then its a minimizer of $E(u)$ and and in fact $u$ solves $(0.1)$ if and only if it minimizes $E(u)$ over $H^1(U)$. So uniqueness follows since the minimizer of $E(u)$ is unique from strict convexity.
To show existence it suffices to show a minimizer exists. To do this we want to exploit the weak topology of $H^1(U)$ by showing the minimizing sequence of $E(u)$, $\{u_k\}$ is bounded in the $H^1(U)$ norm. 
This follows from the following lemma: Let $f \in H^1(U)$ then there exists a $C$ independent of $f$ such that \begin{equation*} ||f||_{H^1(U)} \leq C(||Tr(f)||_{L^2(\partial U)} + ||Df||_{L^2(U)})  \end{equation*} [which the proof for is very similar to the usual Poincare Inequality proof]. Then a routine inequality argument with Cauchy's Inequality $ab \leq \epsilon a^2 + \frac{b^2}{4 \epsilon}$ shows that the minimizing sequence is bounded. So we can extract a subsequence \begin{equation*} u_{n_k} \rightarrow u \text{ in } L^2(U) \end{equation*} \begin{equation*} Du_{n_k} \rightharpoonup Du \text{ in } L^2(U) \end{equation*}
Then as the Lagrangian of $E(u)$ is convex in $(Du)$ we see it is lower semicontinous with respect to weak convergence, so $u$ is in fact a minimizer, so there exists a min .
