# What is the weak formulation of this problem?

Find $u\in H_D^1(\Omega)$ such that

• $-\nabla\cdot(a\nabla u)=0$ in $\Omega$,
• $\dfrac{\partial u}{\partial n}=g$ on $\Gamma_N$,
• $u=0$ on $\Gamma_D$.

The function $a(x,y)$ is piecewise constant on each of the $9$ subdomains

$$\begin{array}{|c|c|c|} \hline \Omega_7&\Omega_8&\Omega_9\\ \hline \Omega_4&\Omega_5&\Omega_6\\ \hline \Omega_1&\Omega_2&\Omega_3\\ \hline \end{array}$$

(or see the picture here).

I tried multiplying by $v$ then integrating (the standard approach). I broke down the integral to have $9$ integrals, one for each subdomain, to use Green’s formula, but I’m stuck since now I have border integrals over all the subdomains...

Edit : $\Gamma_D$ is the top edge of the square, $\Gamma_N$ is the the union of the three other edges.

Edit 2 : I tried another approach, can someone JUST check if this is correct ? $$\int_{\Omega} a\nabla u \cdot \nabla v = \int_{\partial \Omega }\frac{\partial u}{\partial n}}$$

• Welcome to math.SE! This might help you learning the latex-based formatting program used in this site. – user228113 Jun 10 '15 at 1:35
• Thanks for the edit it is much readable now. – mdrlol Jun 10 '15 at 1:46

You have to keep the weighted Neumann jumps $[a \frac{\partial u}{\partial n}]$ on the internal edges. Usually, these jumps are explicitly specified in the strong formulation.
I don't understand the formulation that you provide in your second edit. If $a$ is a function, why is it outside the integral? And where is the test function $v$ in the boundary part?
• I don't think that you are allowed to expand $\nabla (a\nabla u)$ – Albe Jun 12 '15 at 9:45
• If $a$ is only piecewise constant, you have to rely on the notion of weak derivative and perform integration by parts. As for the link, you may assume that you have homogeneous weighted Neumann jumps. However, in general, these jump conditions are provided by the physical properties of the problem under consideration. – Albe Jun 12 '15 at 9:56