Soft enforcement of boundary constraints in linear PDE

Question

Suppose I have a region $\Omega\subset\mathbb R^2$ with smooth boundary curve $\partial\Omega$. Consider the PDE \begin{cases} \Delta u(x) = 0\ \forall x\in\mathrm{int}\ \Omega \\ au(x)+b\frac{\partial u}{\partial n}(x)=g(x)\ \forall x\in\partial\Omega. \end{cases} In general, we can think of the Dirichlet equation $\Delta u=0$ as minimizing the energy $\int_\Omega \|\nabla u(x)\|_2^2\ dA(x)$, the "weak form" version of the PDE, in a particular Sobolev space.

I am wondering about a particular means of building in the Robin boundary conditions above. In particular, suppose we enforce the boundary conditions using a "soft constraint," by optimizing an integral such as the following: $$ \min_{u_\alpha\in H^1(B)} \left[ \int_\Omega \|\nabla u_\alpha(x)\|_2^2\ dA(x) +\alpha \oint_{\partial\Omega} \left( a u_\alpha (x)+b\frac{\partial u_\alpha}{\partial n}(x)-g(x) \right)^2 \ d\ell(x) \right] $$ Here, $B\supset \Omega$ is some larger ball in $\mathbb R^2$ containing $\Omega$.

Is there any theory on whether such a variational problem is well-posed? It seems this energy should be zero if $u_\alpha$ is harmonic in $\Omega$ and satisfies the Robin conditions, for any $\alpha>0$.

Pointers to any relevant theory---or similar if not identical setups---is much appreciated. The reality is we are solving a nonlinear version of the same thing. I'm mostly concerned with an objective that integrates over a region and a curve summed together, and how this affects variational calculus.

Answer 1

The soft enforcement aspect is interesting, and I can't say whether or not it would give acceptable results, but there's also seemingly no guarantee that the solution to the Robin problem should minimize this energy. In fact, you said it should be zero, but notice that even if $\Delta u=0$, the gradient might not be zero (the whole solution would actually be a constant in that case).

The main reason that the energy $E[u] := \frac{1}{2} \int_\Omega \Vert \triangledown u(x) \Vert^2_2 dA(x)$ works for Dirichlet problems is that it's directly related to its weak formulation. If you consider a minimizer $u\in H^1_0(\Omega)$ and any function $v\in H^1_0(\Omega)$, then since $u$ minimizes $E$: $$ 0 = \frac{d}{d\varepsilon} E[u+\varepsilon v] \vert_{\varepsilon=0} = \int_\Omega \triangledown u \cdot \triangledown v\ dA, $$

which is exactly the weak formulation of the Dirichlet problem. For the Robin problem however, the weak formulation would be: $$ B[u,v] := \int_\Omega \triangledown u \cdot \triangledown v\ dA - \frac{1}{b} \int_{\partial\Omega} (g-au)v\ d\ell = 0, \ \ \ \ \forall v \in H^1(\Omega). $$ A quick search online only got me this similar setup with Neumann boundary conditions, but it can easily be adapted for Robin conditions. Defining the corresponding energy as such: $$ E[u] := \frac{1}{2} \int_\Omega \Vert \triangledown u \Vert^2_2 dA - \frac{1}{b} \int_{\partial\Omega} (g-\frac{a}{2}u)u\ d\ell, $$ you can check that $\frac{d}{d\varepsilon} E[u+\varepsilon v]\vert_{\varepsilon=0} = B[u,v]$, and so any minimizer $u$ of $E$ would solve the Robin problem.

Conversely, to see that the solution $u$ to the Robin problem minimizes $E$, we can see that for any $w\in H^1(\Omega)$: $$ E[w] = E[u] + B[u,v] + \frac{1}{2} \int_{\Omega} \Vert \triangledown v \Vert^2_2\ dA + \frac{a}{2b} \int_{\partial\Omega} v^2\ d\ell, $$ with $v:=w-u$. Since $B[u,v]=0$, then $E[w]\geq E[u]$ (assuming $a$ and $b$ are non-zero and of the same sign).