I am reading a paper dealing with a general elliptic PDE that I need to transform from classical formulation to weak formulation: $$\left\{\begin{matrix} - \sum_{i=1}^n \sum_{j=1}^n (a_{ij} u_{x_i})_{x_j} + a_0u=f & in \ \Omega \quad \quad \quad \quad \quad \quad \boldsymbol{(^{*})}\\ \ \ \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ u =0 \quad \ \ \ \ & \ in \ \partial \Omega \quad \ \quad \quad \quad \quad \quad \quad \end{matrix}\right.$$

How is this equation transformed into the weak formulation?

$$ \quad \quad \quad \int_{\Omega} \sum_{i=1}^n \sum_{j=1}^n a_{ij} u_{x_i}v_{x_j} + \int_{\Omega} a_0uv=\int_{\Omega}fv \quad \forall v \in H_0^1(\Omega) \quad \quad \quad \boldsymbol{(^{**})}$$

I am aware of the fact that these are often deduced from Green's formulas, but I can't find how to do it.

I know I must to pass the derivative from the term $(a_{ij} u_{x_i}) $ on $ \boldsymbol{(^{*})}$ to $v$ by using the fact that $v=0 \ on \ \partial \Omega$, But What is the exact identity? I would also like some reference to the formulas.


The Gauss-Green theorem (the basis of the Green identities) states that if $u,v$ are sufficiently smooth on a nice domain $\Omega$ then $$\int_\Omega \frac{\partial u}{\partial x_j} v \, dx = - \int_\Omega u \frac{\partial v}{\partial x_j} \, dx + \int_{\partial \Omega} uv \nu_j dS$$ where $\nu_j$ is the $j$th component of the external normal unit vector and $dS$ is the surface area measure.

If either $u$ or $v$ happens to vanish on the boundary $\partial \Omega$ this becomes $$\int_\Omega \frac{\partial u}{\partial x_j} v \, dx = - \int_\Omega u \frac{\partial v}{\partial x_j} \, dx.$$ Using the standard approximation procedure this remains valid if $u,v \in H^1(\Omega)$ and one or the other belongs to $H^1_0(\Omega)$.

The idea is to take (*) and multiply both sides of the equation by a function $v \in H^1_0(\Omega)$. Then integrate over $\Omega$ and apply Green's identity to arrive at (**).

  • $\begingroup$ Thanks for the answer. One more question. Why is approximation ( I suppose you mean using density) used? According to the response in this question I made: math.stackexchange.com/questions/1785269/… If one can directly use these equalities for Sobolev functions, why us the density argument? $\endgroup$ – D1X Jun 14 '16 at 19:25
  • $\begingroup$ If you know the formulas for Sobolev functions you don't need density. If you start with a classical formula then simple density arguments work to make the extension. $\endgroup$ – Umberto P. Jun 14 '16 at 19:29
  • $\begingroup$ Okay, so both methods are correct. I just wanted to clarify that, thanks again. $\endgroup$ – D1X Jun 14 '16 at 19:30
  • $\begingroup$ I am a bit confused with the names and notation these theorems have. Is what you call Gauss-Green theorem what is called in this page en.wikipedia.org/wiki/Divergence_theorem Divergence theorem? Because there seems to be a slight difference. $\endgroup$ – D1X Jun 17 '16 at 18:56
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    $\begingroup$ The Gauss-Green theorem is pretty much equivalent to the divergence theorem. If $\mathbf F$ is a vectorfield with $uv$ in one component and $0$ in each of the other components, the divergence theorem $$\int_U \mathrm{div} \mathbf F \, dx = \oint_{\partial U} \mathbf F \cdot \mathbf \nu \, dS$$ takes the form $$\int_U u_{x_j}v + u v_{x_j} \, dx = \int_{\partial U} uv \nu_j \, dS.$$ $\endgroup$ – Umberto P. Jun 18 '16 at 3:28

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