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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.

Story123
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    I had not the time/ability to find an answer last week, therefore I am giving here only two suggestions. The first one is to have a look at the monograph of Pao I have cited in this answer: despite dealing with boundary problems with nonlinear boundary conditions (in particular check §4.4 pp. 154-161) it deals only with Ḧolder classes. However perhaps you may adapt the method used to your $H^1$ framework. – Daniele Tampieri Mar 11 '19 at 21:33
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    The second one is on your formulation of the problem. Your weak formulation is correct: however I'd not use it to prove existence, uniqueness and constructibility of the solution. I'd rather use the know results on the classical (linear) Robin problem and apply it to solve a sequence of problems constructed as below: use the scale of boundary conditions $$\begin{split}\frac{\partial u_0}{\partial \nu} + bu_0 &= 0\ \frac{\partial u_1}{\partial \nu} + bu_1 &= \beta(u_0)\ &\vdots\ \frac{\partial u_n}{\partial \nu} + bu_n &= \beta(u_n)\&\vdots \end{split}$$ and solve the equation. – Daniele Tampieri Mar 11 '19 at 21:47
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    The scale of linear problems is the approach I am trying: as it is stated above, it is only a sketch. I have to prove the convergence of the entire process, but I was not able to prove it last week: also note that the second member should not be the simple $\beta(u_n)$. I used it in this comment just as an example. – Daniele Tampieri Mar 11 '19 at 21:55
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    Hmm thats an interesting perspective on how to approach it. I'll definitely try to come up with an alternative proof using this to show existence. – Story123 Mar 12 '19 at 05:26

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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 .

Story123
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