In an introduction to Discontinuous galerkin methods, I have some problems in checking the weak formulation. I'm looking at page 16 here

The context is the advection reaction equation: $$\operatorname{div}(bu) + cu = f$$ $$u = g_D \in \Gamma_{-}$$ where $\Gamma_{-}$ is the inflow boundary: $\{ x \in \partial \Omega: b(x) \cdot n(x) \ <0\}$

Here $\partial_{-K} = \{ x \in \partial K: b(x)\cdot n(x) <0\}$ (see pag. 15)

They write the weak formulation in this way (with whom I agree so far), which you can find at (2.8): $$ \underbrace{\sum_{K \in T_h} \int_{K}( \operatorname{div}(bu_h) + c u_h )v_h - \int_{\partial_{-K} \setminus \partial \Omega} (b \cdot n)[u_h]v_h^+ds - \int_{\partial_{-K} \cap \partial \Omega} (b \cdot n)u_h^+ v_h^+ds}_{B(u_h,v_h)} = \sum_{K \in T_h} \int_K f v_h - \int_{\partial_{-K}\cap \partial \Omega} (b \cdot n)g_D v_h^+$$

Then they integrate by parts the first term in the bilinear form and they write (2.9)

$$ \sum_{K \in T_h} \int_{K}( -u_h (b \cdot \nabla v_h) + c u_h )v_h + \int_{\partial_{-K} \setminus \partial \Omega} (b \cdot n)u_h^{-}v_h^+ds + \int_{\partial_{+K}} (b \cdot n)u_h^+ v_h^+ds = \sum_{K \in T_h} \int_K f v_h - \int_{\partial_{-K}\cap \partial \Omega} (b \cdot n)g_D v_h^+$$

What I cannot understand is how to obtain the new bilinear form. More precisely, I don't know why they have the $\partial_{+K}$ on the last term of the bilinear form.

My attempt:

The integration by parts of the first term yields the boundary term $$\sum_{K} \int_{\partial K} (b \cdot n) u_h v_h$$ which I would split as $$\int_{\partial_{-K} \cap \partial \Omega} (b \cdot n) u_h^+ v_h^+ + \int_{\partial_{-K} \setminus \partial \Omega} (b \cdot n) u_h^+ v_h^+ + \int_{\partial_{+K} \cap \partial \Omega} (b \cdot n) u_h^+ v_h^+ + \int_{\partial_{+K} \setminus \partial \Omega} (b \cdot n) u_h^+ v_h^+$$

Now, inserting this in the bilinear form of (2.8) I obtain almost what they wrote in (2.9), except that their integral is on $\partial_{+K}$ only, and I cannot see why they wrote so.

Any help is highly appreciated


Following the kind suggestions by @ConvexHull:

After integration by parts:

$$\sum_{K} \int_{\partial_{-K}} (b \cdot n) u_h v_h = \sum_{K} \int_{\partial_{-K} \setminus \partial \Omega}(b \cdot n) u_h^+v_h^+ + \int_{\partial_{-K}\cap \partial \Omega} (b \cdot n) u_h^+ v_h^+$$

which plugged into (2.8) (simplifying the boundary terms and one part of the jump term) gives only $\sum_K \int_{\partial_{-K} \setminus \partial \Omega}(b \cdot n) u_h^{-} v_h^+$


1 Answer 1


Some definitions:

$\int_{\partial_{-K} \setminus \partial \Omega} $ = all element outer faces which do not coincide with the BC faces

$\int_{\partial_{-K}\cap \partial \Omega} $ = all element outer faces which do coincide with the BC faces

What they do:

They only divide the jump term $\left[u_h \right]$ in two integral expressions and rearrange the equations.

$- \int_{\partial_{+K}} = \int_{\partial_{-K}} = \int_{\partial_{-K} \setminus \partial \Omega} + \int_{\partial_{-K}\cap \partial \Omega} $


I think the surface integral cancels to zero. They motivate their definitions from a FEM point of view.

$$\sum_{K \in T_h} \int_{\partial K} (b \cdot n) u_h v_h ds = 0 $$


$$ \sum_{K \in T_h} \cdots - \int_{\partial_{-K} \setminus \partial \Omega} (b \cdot n)(u_h^+ - u_h^-)v_h^+ds - \int_{\partial_{-K} \cap \partial \Omega} (b \cdot n)u_h^+ v_h^+ds = \text{RHS} $$

$$ \sum_{K \in T_h} \cdots + \int_{\partial_{-K} \setminus \partial \Omega} (b \cdot n) u_h^-v_h^+ds \underbrace{ - \int_{\partial_{-K} \setminus \partial \Omega} (b \cdot n)u_h^+v_h^+ds - \int_{\partial_{-K} \cap \partial \Omega} (b \cdot n)u_h^+ v_h^+ds} = \text{RHS} $$

$$ \sum_{K \in T_h} \cdots + \int_{\partial_{-K} \setminus \partial \Omega} (b \cdot n) u_h^-v_h^+ds \underbrace{- \int_{\partial_{-K} } (b \cdot n)u_h^+v_h^+ds} = \text{RHS} $$

$$ \sum_{K \in T_h} \cdots + \int_{\partial_{-K} \setminus \partial \Omega} (b \cdot n) u_h^-v_h^+ds + \int_{\partial_{+K} } (b \cdot n)u_h^+v_h^+ds = \text{RHS} $$


  • $\begingroup$ Thanks for your answer. However, it's not clear to me how I should split starting from (2.8). Could you please be more explicit? $\endgroup$
    – FEGirl
    Aug 8, 2021 at 7:28
  • $\begingroup$ @bobinthebox You only have to understand that you have to choose either $\partial_{-K}$ or $\partial_{+K}$ when using the divergence theorem. $\endgroup$
    – ConvexHull
    Aug 8, 2021 at 7:40
  • $\begingroup$ Oh, I see. I'm trying to follow your advice, as you can see in my edit. However, I don't see how to get that $\int_{\partial_{+K}}$ term. What am I missing? @ConvexHull $\endgroup$
    – FEGirl
    Aug 8, 2021 at 7:53
  • $\begingroup$ @bobinthebox I think i also have missunderstood some definitions. Your splitting seems to be correct. One intuitive answer may be that the surface integral cancels to zero, as already mentioned. With this, you can rearrange the equation as mentioned in the book. They motivate their definitions from FEM point of view. $\endgroup$
    – ConvexHull
    Aug 8, 2021 at 8:33
  • $\begingroup$ You are referring to the split in my last edit, right? I'm so sorry again, but I can't really understand how to get the integral of $\partial_{+K}$. @ConvexHull $\endgroup$
    – FEGirl
    Aug 8, 2021 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.