# Weak form of an elliptic boundary value problem.

I am studying finite elements and I am not understanding the higher-dimensional version of the weak form.

Say we have a BVP \begin{align} -\nabla \cdot (a(x,y) \nabla u(x,y)) + c(x,y) u(x,y) &= f(x,y), x \in \Omega \subset \mathbb{R}^2\\ (a(x,y) \nabla u(x,y)) \cdot \textbf{n}& = 0, \text{ for }(x,y) \in \partial \Omega. \end{align}

The steps to get to the weak form are to

1. Formally multiply the differential equation by a test function (x,y)
2. Integrate both sides
3. Apply a higher dimensional analogue of "integration by parts"
4. Apply boundary conditions ("if necessary")

The notes I am using has this as their BVP (very similiar to mine) \begin{align} - \nabla \cdot (a(x,y)\nabla u(x,y)) + c(x,y) u(x,y) = f(x,y), x \in \Omega \subset \mathbb{R}^2, \\ u|_{\partial \Omega D} = g_D, \quad a\nabla u \cdot \textbf{n}|_{\partial \Omega N} = a \frac{\partial u}{\partial \textbf{n}} = g_N. \end{align} They skip to the weak form \begin{align} \text{Find $u \in S$ such that } \\ \int_{\Omega} (av_x u_x + av_y u_y + cvu) dxdy - \int_{\partial \Omega_N} vg_N ds = \int_{\Omega} vfdxdy, \\ \forall v \in T \text{ with } T=\{v \in H^1 (\Omega) : v|_{\partial \Omega_D} = 0 \}, \text{ and } S = \{v \in H^1 (\Omega) : v|_{\partial \Omega_D} = g_D \}. \end{align} Can anyone write out all the steps they used to get here, or explain how I could get to the weak form from my problem?

• Hint: Use integration by parts/Divergence theorem. Commented May 6, 2017 at 21:44
• What happens to the $a(x,y)$ ? Commented May 6, 2017 at 21:52

Multiplying the given PDE by a test function $v\in T$ and integrating over the domain gives: $$\int_\Omega \Big[-v\nabla\cdot(a\nabla u) + cuv\Big]\, dxdy = \int_\Omega fv\, dxdy. \tag{1}$$ We do not need to do anything with $\displaystyle\int_\Omega cuv\, dxdy$ and $\displaystyle\int_\Omega fv\, dxdy$ since they are already in the form of what we want, so we only need to investigate the first integral. Using integration by parts, we obtain \begin{align*} \int_\Omega v\nabla\cdot(a\nabla u)\, dxdy & = \int_{\partial\Omega} v(a\nabla u)\cdot\mathbf{n}\, ds- \int_\Omega \nabla v\cdot (a\nabla u)\, dxdy \\ & = \int_{\partial\Omega_D} v(a\nabla u)\cdot\mathbf{n}\, ds + \int_{\partial\Omega_N} v(a\nabla u)\cdot\mathbf{n}\, ds- \int_\Omega \nabla v\cdot (a\nabla u)\, dxdy. \end{align*} Now, since $v\in T$, $v|_{\partial\Omega_D} = 0$ and the first integral on the right vanishes. Since $u$ satisfies the given PDE, we have the boundary condition $a\nabla u\cdot\mathbf{n} = g_N$ on $\partial\Omega_N$. Thus, the above reduces to: \begin{align*} \int_\Omega v\nabla\cdot(a\nabla u)\, dxdy & = \int_{\partial\Omega_N} vg_N\, ds - \int_\Omega \nabla v\cdot a\nabla u\, dxdy \\ & = \int_{\partial\Omega_N} vg_N\, ds - \int_\Omega (au_xv_x + au_yv_y)\, dxdy, \end{align*} and $$\int_\Omega -v\nabla\cdot(a\nabla u)\, dxdy = \int_\Omega (au_xv_x + au_yv_y)\, dxdy -\int_{\partial\Omega_N} vg_N\, ds.$$ Hence, we obtain the required weak form by substituting the above equation into (1).
• I have one question. In the example I am doing, the entire boundary term goes away, and I have no Dirichlet boundaries. Is my $S$ set (containing the solution $u$) just $H^1(\Omega)$? Also, is my set $T$ (containing $v$) also just $H^1 (\Omega)$, or do I need to impose extra conditions? My guess is that I don't, since the both boundary integrals terms go away, but I thought the set $T$ was suppose to be a space (so we need $v(0) = 0$). Does $v(0) = 0$ need to be a requirement to be in the set $T$? What exactly is $S$ and $T$? Commented May 6, 2017 at 22:48
• It depends on how the boundary term vanishes. If it's due to the boundary condition on $u$, then you are right; if it's due to some condition on $v$, then you need to alter $T$ so that it contains that information. You could view $T$ as the space of "test functions" and $S$ to be the space of admissible functions (your solution space). We say that $u\in S$ solves the given PDE in the weak sense if $u$ satisfies the weak form for all $v\in T$. Commented May 6, 2017 at 23:20
• In my case, the boundary terms vanish due to the boundary condition on $u$, so $T = H^1(\Omega)$. If the boundary term vanished due to a condition on $v$, then $T$ has to include that condition (but it didn't in my case). Also, my solution space $S$ can be properly chosen to just be $H^1(\Omega)$ as well, with no addition restrictions. Thanks for all your help. Commented May 6, 2017 at 23:47