I have a question about the weak formulation of a PDE in finite element analysis. Suppose we have the following two-dimensional PDE:

$$ \Delta \cdot u(x,y) = q(x,y) $$ where $q$ is given, $u$ is unknown, and $\Delta$ is the Laplacian operator $\Delta = \left(\dfrac{\partial ^2}{\partial x^2}, \dfrac{\partial ^2}{\partial y^2}\right)$. The domain is an open set $\Omega \subset \mathbb{R}^2$. The boundary condition is that $u = 0$ on $\partial \Omega$.

The weak formulation is something like

\begin{equation} \int_\Omega \Delta u(x,y) \cdot v(x,y) d\mathbf{x} = \int_\Omega q(x,y) \cdot v(x,y) d\mathbf{x}\end{equation} for every $v$ in $H^1_0(0,1)$ (or perhaps a difference space?) However, we're going to reduce $H^1_0(0,1)$ to considering only a finite-dimensional subspace (in this case, suppose all the basis functions are affine.)

Here comes my question. It seems we always turn the left-hand side into

$$-\int_\Omega \nabla u(x,y) \cdot \nabla v(x,y) d\mathbf{x}$$

using Green's Identity, basically integration by parts. But why do we have to change it? Why can't we solve the system as originally stated?

I can see there's something fishy going on, since in the original system $\Delta u \equiv 0$ if we assume $u$ is piecewise-linear. But shouldn't both forms be the same? ... What's going on here?


You almost certainly can formally believe the equivalence of the two formulations if you have seen the proof of the Green's identity, right? However, when you "balance the derivatives" of the solution $u$ and the test function $v$ as you have done (in the beginning $u$ has two derivatives and $v$ zero, and in the weak formulation both have one derivatives and are, thus, balanced out), you actually end up requiring less smoothness from the solution $u$.

This implies that solutions to the weak problem are not, in general, equivalent to the solutions of the strong problem. The weak solution is equivalent to the strong solution only if $q \in L^2(\Omega)$ and the boundary of the domain satisfies some smoothness criteria. See, e.g., the wonderful book of Lawrence Evans, Partial Differential Equations, for more information on this matter.

But why do we have to change it? Why can't we solve the system as originally stated?

This is something which you will realize when implementing the finite element solver. It is much more convenient to have only single derivative in the weak formulation than two derivatives since then the finite element is only required to be $H^1$-conforming. (Intuitively, second derivative of piecewise-linear function is zero just as you mentioned and therefore the stiffness matrix would be zero.) You could solve the problem directly with $H^2$-conforming elements but those are much more harder to construct.

Furthermore, if the weak formulation is symmetric then you end up having a symmetric stiffness matrix which is faster to solve.

  • 1
    $\begingroup$ Everything you said is absolutaly right. Furthermore, when we use the Green's identity, we put the problem in a variational (weak) formulation, which is related to a problem of minimization of a functional energy. Then, we can use a powerful mathematical theory assernal to solve the problem or make proofs of convergence of finite element methods. $\endgroup$ – Jardel Vieira Mar 30 '15 at 3:11
  • $\begingroup$ @knl l I have a very similar question to what is posted here and I would appreciate it if you could comment on my post math.stackexchange.com/q/3186878/374907 I am a bit confused how how symmetry of the weak form impacts the solution found. $\endgroup$ – AzJ Apr 14 at 22:10

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