So I am confused about some details of obtaining a variational formulation specifically for Poisson's equation. I am in a Scientific Computing class and we just started discussing FEM for Poisson's equation in particular. However, we have a bunch of homework problems that are about finding variational formulations for slightly different PDE's or Poisson's with trickier boundary conditions.

  1. So as far as I can see the variational formulation always ends up in the form $a(u,v) = L(v)$ where $a(u,v)$ is a bilinear form on some Hilbert space V. From Analysis and Lax-Milgram I know that there exists a unique u such that this is satisfied $\forall v \in V$ if $a(u,v)$ is continuous and coercive and $L(v)$ is continuous. Now if I start with some PDE and multiply by test functions and integrate and get it into some form $a(u,v) = L(v)$ does this have any meaning if $a(u,v)$ and $L(v)$ don't meet those requirements for the Lax-Milgram theorem to apply? ie, is a variational formulation only valid/meaningful if those requirements are met?

  2. So when we move to a finite dimensional subspace $V_h \subset V$, I am confused with how much freedom we have specifying $V_h$. In some homework problems that are about Poisson equation with mixed boundary conditions I have a hard time maintaining those boundary conditions. In particular, lets say that $a(u,v)$ and $ L(v)$ don't meet all hypothesis of the Lax-Milgram, but if I constrain it to $v \in V_h$ and I force boundary conditions on all the elements of $V_h$ such that $a(u,v) = L(v) \forall v \in V_h$ and they then meet the requirements for Lax-Milgram then is this a valid Variational Formulation for FEM?

I realize that this might be a bit too general, so one of my homework questions is for Poisson's equation with Robin boundary conditions. I haven't been able to get it into a well behaved Variational formulation so I'm really wondering what requirements can be relaxed and still retain a valid Variational Formulation. Thanks for your time!


1 Answer 1


Question 1: In numerical PDE, Lax-Milgram lemma's requirement does not have to be met. Coercivity basically means the problem you got is invertible. There is a weaker version of coercivity which is called Babuska–Brezzi inf-sup condition, it is for mixed formulation of FEM. Another one is called Fredholm alternative, it is when both the coercivity and inf-sup condition fail. Though I doubt what you need are these right now. In your case, you wanna deliver a well-posed FEM problem for Robin boundary condition, the Lax-Milgram lemma has to be satisfied.

Question 2:

(1) For Dirichlet boundary value problem. In the continuous space, say if $-\Delta u= f$ in $\Omega$, $u|_{\partial \Omega}=g$. We could write $u= u_0 + u_g$, where $-\Delta u_0= f$ in $\Omega$, $u_0|_{\partial \Omega}=0$, and $-\Delta u_g= 0$ in $\Omega$, $u_g|_{\partial \Omega}=g$. In the variational problem, the test function is chosen to be 0 on the boundary, if we solve $$ (\nabla u,\nabla v) = (f,v) $$ for any $v\in H^1_0$, what we get is $u_0$. We don't know what the boundary is like solving the variational problem because the test function $v$ is always 0 there. However, we know what the exact $u$'value is on boundary, so we are good here.

In finite elements, we force the value of the FEM solution $u_h\in V_h$ on the boundary nodes to be equal to the exact value of the boundary data. The test function $v$ is chosen to be zero on the boundary, i.e.,$v\in V_h\cap H^1_0$ for Poisson equation. Lax-Milgram lemma's conditions are satisfied for both $H^1_0$ and $V_h \cap H^1_0$. The problem is well-posed.

To answer your question, we don't put any constraints in the interior nodes (or say Degrees of Freedom associated to interior nodes), we just solve the variational problem setting test function $v=0$ on the boundary. What we get is the finite elements approximation $u_{0,h}$ to $u_0$. Then we let $u_{g,h}(V) = g(V)$ for any boundary node $V$, and $u_{g,h}(V')=0$ for any interior nodes $V'$, this is the discrete version $u_g$. The equations they satisfy are: $$ (\nabla u_{0,h},\nabla v) = (f,v)\quad \forall v\in V_h\cap H^1_0, \text{ and } u_{0,h}|_{\partial \Omega} = 0 $$

$$ (\nabla u_{g,h},\nabla v) = 0\quad \forall v\in V_h\cap H^1_0, \text{ and } u_{g,h}|_{\partial \Omega} = \Pi_h g $$ where $\Pi_h g$ is the interpolation of $g$ on boundary which is defined by imposing the nodal value like we just did. And your final finite elements solution is $u_h = u_{0,h} + u_{g,h}$.

(2) For Pure Neumann boundary problem, the solution is unique up to be a constant, we have to consider the problem in a slightly different space: $\{v\in H^1: \displaystyle \int_{\Omega} v = 0\} = H^1/\mathbb{R}$. Thanks to Poincare inequality, Lax-Milgram lemma's conditions are again met. For FEM problem, we don't force any boundary condition for the test function, rather we just solve the variational equation (When solving the linear system, there are various ways to deal with the modulus of constant part). In this sense, we say that the Neumann boundary condition is satisfied "weakly".

Last Question:

Now for Robin boundary condition, say your equation is: $$ -\Delta u = f \quad \text{in } \Omega $$ with Robin boundary condition for all the boundary $\partial \Omega$: $$ \alpha u +\frac{\partial u}{\partial n} = g $$ where $\alpha$ is a positive constant ($\alpha$ has to be positive a.e. on the boundary), and $\partial u/\partial n = \nabla u\cdot n$ is the normal derivative on boundary. Multiply the equation by a test function $v\in H^1$, $$ -\int_{\Omega} \Delta u\,v = \int_{\Omega} fv $$ and perform integration by parts: $$\int_{\Omega} \nabla u\cdot \nabla v - \int_{\partial \Omega}(\nabla u\cdot n )v= \int_{\Omega} fv $$ plug in the Robin boundary data: $$\int_{\Omega} \nabla u\cdot \nabla v - \int_{\partial \Omega}(g-\alpha u)v= \int_{\Omega} fv $$ Rearrange: $$\int_{\Omega} \nabla u\cdot \nabla v + \int_{\partial \Omega}\alpha u v= \int_{\Omega} fv + \int_{\partial \Omega}gv $$ The left hand side $a(u,v) = \displaystyle \int_{\Omega} \nabla u\cdot \nabla v + \int_{\partial \Omega}\alpha u v$, right hand side is the $L(v)$. The proof of continuities of the right hand side and the first term in $a(u,v)$ is natural, just following the proof for Dirichlet and Neumann boundary problems. For second term on the left we would like to use the trace inequality: $$ \int_{\partial \Omega}\alpha u v \leq \alpha \|u\|_{L^2(\partial \Omega)} \|v\|_{L^2(\partial \Omega)} \leq C \alpha \|u\|_{H^1(\Omega)} \|v\|_{H^1(\Omega)} $$ thus $a(u,v) \leq C \|u\|_{H^1(\Omega)} \|v\|_{H^1(\Omega)}$.

The coercivity of $a(u,v)$ is a little bit more difficult to prove. The way I learned in FEM class is by what is called "compactness argument". The trick is using proof by contradiction: Suppose $a(u,v)$ is not coercive, then (here comes the compactness argument) there exists a sequence $\{v_n\}\subset H^1$, such that: $$ \|v_n\|_{H^1(\Omega)} = 1, \text{ and } a(v_n,v_n) \to 0.\tag{1} $$ Now since $\{v_n\}$ is a bounded sequence in $H^1$ (consider the distance being induced by the $H^1$-norm), then there exists a weakly convergent subsequence $v_{n_j}\to v$ in $H^1$ and $v_{n_j}\to v$ in $L^2$-norm (thanks user33869 for pointing out my previous error). Weak convergence is: $$ \int_{\Omega}(v-v_{n_j})w + \int_{\Omega}\nabla(v-v_{n_j})\cdot \nabla w \to 0, \; \text{ for all } w\in H^1(\Omega). $$ This implies that: $$ \int_{\Omega} |\nabla v|^2 = \lim_{j\to \infty}\int_{\Omega} \nabla v_{n_j}\cdot \nabla v \leq \lim_{j\to \infty}\left( \int_{\Omega} |\nabla v_{n_j}|^2 \right)^{1/2} \left(\int_{\Omega} |\nabla v |^2\right)^{1/2} $$ This is $$ |v|_{H^1(\Omega)} \leq \lim_{j\to \infty} |v_{n_j}|_{H^1(\Omega)}\tag{2} $$ Using another assumption now $$ a(v_n,v_n) = \int_{\Omega} |\nabla v_n|^2 + \int_{\partial \Omega}\alpha v_n^2 \to 0 \tag{3} $$ This implies that $\displaystyle\int_{\Omega} |\nabla v_n|^2 \to 0$. By (2), letting $n_j\to \infty$ yields $\displaystyle\int_{\Omega} |\nabla v|^2 = 0$. Now we have $ \|v\|_{H^1(\Omega)}=\|v\|_{L^2(\Omega)}$.

Now using $v_{n_j}\to v$ in $L^2$-norm, together with (3), $\|v\|_{L^2(\Omega)} = \lim\limits_{j\to \infty} \|v_{n_j}\|_{L^2(\Omega)}=1$

The $L^2$-norm of gradient of $v$ being zero implies $v$ is a constant, The $L^2$-norm of $v$ being 1 implies $v=c$ where the constant $c\neq 0$. However, again by (3): $$ \int_{\partial \Omega}\alpha v^2 = \lim_{n\to \infty}\int_{\partial \Omega}\alpha v_{n_j}^2 = 0 $$ together with the positivity assumption on $\alpha$, this says $v=0$ on $\partial \Omega$. Constradition.

Therefore there does not exist such sequence satisfying (1). The negation of the statement (1) is: For all sequence in $H^1(\Omega)$, if $\|v_n \|_{H^1(\Omega)}$ is bounded, $a(v_n,v_n)$ is always bounded away from 0. This can be translated to any function $v$ in $H^1$, for we could always find a bounded sequence that goes to $v$ by the completeness of a Hilbert space. Thus the coercivity is proved.

  • $\begingroup$ Thank you Shuhao! You obviously put time into that response. To clarify, in the bit about the Poisson's with Dirichlet Boundary Conditions, you talk about restricting $u_h \in V_h$ to have the required values on the boundary. So is there total freedom in making such restrictions to $u_h$. Also, doesn't $u_h$ only come up when we create the FEM from the continuous version, and then how is the continuous version satisfied without specifying these constrained $u_h$'s? $\endgroup$
    – Fractal20
    Apr 15, 2013 at 13:49
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    $\begingroup$ If $\{ u_n\}$ is bounded in $H^1$, why is that it has a convergent subsequence in $H^1$? I think by Rellich's compactness theorem, it only has a convergent subsequence in $L^2$? And it's well known that a space is finite dimension $\Leftrightarrow$ every bounded sequence has a convergent subsequence. (My attack is on your 5th line in the part that proves coercivity of $a(\cdot ,\cdot)$) $\endgroup$
    – user33869
    Oct 1, 2013 at 19:48
  • $\begingroup$ But weak convergence doesn't mean norm convergence. How do you get $\| v-v_{n_j}\|+\| \nabla (v-v_{n_j}\|\to 0$ if it's only weakly convergent? $\endgroup$
    – user33869
    Oct 2, 2013 at 0:26
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    $\begingroup$ @ellya In proof, yes. In implementation, you can just subtract the averages of the piecewise linear elements and use these as the test functions. $\endgroup$
    – Shuhao Cao
    Mar 3, 2015 at 15:54
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    $\begingroup$ @MathNewbie A non-zero constant, for example $1$, is in $H^1(\Omega)$ but never have zero trace in the sense that for any $\|v - 1\|_{H^1}\geq |\Omega|/2 $ (the proof can be done using Friedrichs' inequality) where $v \in C^{\infty}_c(\Omega)$, which is dense in $H^1_0(\Omega)$ in terms of $H^1$-norm. $\endgroup$
    – Shuhao Cao
    Dec 8, 2015 at 3:21

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