2
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $$ (\nabla u, \nabla v)_\Omega. $$

$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$
    – bob_bill
    Jan 15 at 9:38
  • $\begingroup$ 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. $\endgroup$ Jan 15 at 14:41
  • $\begingroup$ 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 $\endgroup$
    – bob_bill
    Jan 15 at 21:02
  • $\begingroup$ 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 $\endgroup$
    – bob_bill
    Jan 15 at 21:11
  • $\begingroup$ 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. $\endgroup$ Jan 16 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.