For an open set $\Omega \subset\mathbb{R}^n$, consider the following problem:

$$\left\{\begin{matrix} - \Delta u + u=f & \mbox{ in }\ \Omega \quad \quad \quad \quad \quad \\ \ \ \ \quad \quad \ u=0 & \ \mbox{ in }\ \partial \Omega \quad \quad \quad \quad \quad \end{matrix}\right.$$

and its weak formulation $$\int_{\Omega} \nabla u \nabla v +\int_{\Omega}uv= \int_{\Omega} fv.$$

The variational approach to solve this PDE involves solving the weak formulation by means of Lax-Milgram theorem. That is, we want some $u \in H_0^1(\Omega)$ which verifies the last equation for every $v \in H_0^1(\Omega)$. Then, Lax-Milgram theorem applied to the bilinear form $$a(u,v)= \int_{\Omega} \nabla u \nabla v+ \int_{\Omega} uv$$ and the linear functional $\varphi(v)= \int_{\Omega} fv$ yields a unique solution to the weak equation.

Then it is needed to show the equivalence of weak and classical formulation to show existence and uniqueness of solution.

Now, the question is:

Why is that precise weak formulation chosen? Why not use, for example, $$\int_{\Omega} \Delta u v +\int_{\Omega}uv= \int_{\Omega} fv \ ?$$

As far as I know, it is for the sole purpose of lowering the requeriments of $u$. For the first one we need $u$ to have (at least) one derivative. For the second one we need (at least) two derivatives. Is there any other reason I am missing?

  • 2
    $\begingroup$ The idea is to require less regularity for $u$. If you want to use $\Delta u$ (under an integral) you have to require that $u$ is twice (weakly) differentiable $($perhaps you'll look for a solution $u \in H^2(\Omega))$. However, if you only use $\nabla u$, you can search for solutions $u$ in the larger space $H^1(\Omega)$. So to answer your question: no, I don't think there is any other reason. $\endgroup$ – User8128 Jul 3 '16 at 19:17
  • $\begingroup$ Another perspective, albeit not really an answer: ideally the test function space is "about as big" as the solution space. In this case you have "about as many equations as unknowns". In infinite dimensions this doesn't make strict sense, but in the finite element setting this is exactly what is going on. $\endgroup$ – Ian Jul 3 '16 at 19:21

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