Discontinuous Galerkin: confusion about the weak formulation for linear advection equation

Question

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.

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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

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EDIT:

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

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} $

Edit:

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

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

Then:

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

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

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

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

Regards