# Distributed Lagrange multiplier approach to impose constraint in Poisson equation

I'm trying to understand how Lagrange multipliers are applied in order to impose constraints in PDEs. Consider $$B \subset \Omega$$. For instance, a square inside another square domain $$\Omega$$. Let's say we want the solution to the Poisson problem $$-\Delta u = f \text{ in } \Omega$$ subject to the constraint $$u = g \text{ on } \partial B$$

with homogeneous Dirichlet BC on the boundary of the exterior square. The weak formulation corresponding to this is obtained by using a Lagrange multiplier $$\lambda$$. In particular, the weak form is the saddle point of the following Lagrangian $$L(u ,\lambda)= \frac{1}{2} |\nabla u|^2 -(f,u)- \langle\lambda, u-g \rangle_{\partial B}$$ and its weak form is to find $$(u ,\lambda)$$ s.t. $$(\nabla u, \nabla v) + \langle \lambda,v \rangle = (f,v) \qquad \forall v \in V \\ \langle u,q \rangle = \langle g,q \rangle \ \qquad \forall q \in Q$$

and I've found that deal.ii step-60 is implementing exactly this, with $$V=H^1(\Omega)$$ and $$Q = L^2(\partial B)$$.

I was trying to derive the above weak form without using directly that Lagrangian. My idea was to decompose the domain as $$\Omega = \Omega_1 \cup \Omega_2$$ where $$\Omega_1 = \Omega \setminus B$$ and $$\Omega_2=B$$. Then an equivalent reformulation of the problem is the union of the following two subproblems:

$$\begin{cases} - \Delta u_1 = f \text{ in } \Omega_1 \\ u_1 = g \text{ on } \partial B \\ u_1 = 0 \text{ on } \Gamma^D \end{cases}$$ (where $$\Gamma^D$$ is the boundary of the exterior square)

$$\begin{cases} - \Delta u_2 = f \text{ in } \Omega_2 \\ u_2 = u_1 \text{ on } \partial B \end{cases}$$

Now I write the weak forms on both domains.

## In $$\Omega_1$$:

$$(\nabla u_1, \nabla v_1)_{\Omega_1} + \langle \lambda, v_1 \rangle = (f,v_1)_{\Omega_1} \qquad \forall v_1 \in H^1(\Omega_1) \\ \langle u_1, q \rangle = \langle g,q \rangle$$

## In $$\Omega_2$$:

$$(\nabla u_2, \nabla v_2)_{\Omega_2} = (f,v_2)_{\Omega_2} \qquad \forall v_2 \in H^1(\Omega_2)$$

Now I'd like to sum those two formulations and recover the original one, but I have a crucial problem in the sum $$(\nabla u_1, \nabla v_1)_{\Omega_1} + (\nabla u_2, \nabla v_2)_{\Omega_2}$$ which I don't know how to handle. What am I missing?

You define a function $$u$$ on $$\Omega=\Omega_1 \cup \Omega_2$$ so that on $$\Omega_1$$ you have $$u=u_1$$ and similarly on the other part of the domain. You'd do the same with a function $$v$$. Then the term you have trouble with is simply $$(\nabla u, \nabla v)_\Omega.$$
• Thanks @WolfgangBangerth. However, what is striking me is that there's that smooth interface $\partial B$ between $\Omega_1$ and $\Omega_2$. And along that interface the gradient of $u$ is not continuous, as seen in the plots in step-60. What is the mathematical reason of the fact that I can basically "ignore" this and sum those two terms? Jan 15 at 9:38
• You just get a function with a kink. When we write $\nabla u$, we mean the weak gradient, which is not defined on $B$ but that doesn't matter because we then integrate over $\Omega$ within which $B$ is of measure zero. Jan 15 at 14:41
• Therefore in the end, the weak form will have test functions $v \in H^1(\Omega_1 \cup \Omega_2)$. Isn't this different from $H^1(\Omega)$? I mean, it seems to me I'm not taking into account $\partial B$, since I'd say that $\Omega = \Omega_1 \cup \partial B \cup \Omega_2$ . I noticed that indeed you wrote $\Omega=\Omega_1 \cup \Omega_2$, but again it seems that $\partial B$ is left out. After this point everything is clear to me, thanks @WolfgangBangerth Jan 15 at 21:02
• Maybe I got it: you assume that the two subdomains cover the whole domain and they share an interface, i.e. $\Omega_1 \cap \Omega_2 = \partial B$, right ? @WolfgangBangerth Jan 15 at 21:11
• It was sloppy notation. I should have written $\text{int}(\bar\Omega_1 \cup \bar\Omega_2)$, taking the closure of the two subdomains and then taking the interior of the union to ensure that the resulting domain is open again. This is the same as $\Omega_1\cup\partial B\cup\Omega_2$, assuming that $\partial B$ is an open set. Jan 16 at 18:54