@f[
(\beta \cdot \nabla u_h, v_h + \delta \beta\cdot\nabla v_h)_\Omega
-
- (\beta\cdot {\mathbf n} u_h, v_h)_{\partial\Omega_-}
+ (u_h, \beta\cdot {\mathbf n} v_h)_{\partial\Omega_-}
=
(f, v_h + \delta \beta\cdot\nabla v_h)_\Omega
-
- (\beta\cdot {\mathbf n} g, v_h)_{\partial\Omega_-}.
+ (g, \beta\cdot {\mathbf n} v_h)_{\partial\Omega_-}.
@f]
(\beta \cdot \nabla \varphi_i,
\varphi_j + \delta \beta\cdot\nabla \varphi_j)_\Omega
-
- (\beta\cdot {\mathbf n} \varphi_i, \varphi_j)_{\partial\Omega_-},
+ (\varphi_i, \beta\cdot {\mathbf n} \varphi_j)_{\partial\Omega_-},
@f]
with basis functions $\varphi_i,\varphi_j$. However, this is a
pitfall that happens to every numerical analyst at least once
(including the author): we have here expanded the solution
-$u_h = \sum_i u_i \varphi_i$, but if we do so, we will have to solve the
+$u_h = \sum_i U_i \varphi_i$, but if we do so, we will have to solve the
problem
@f[
- {\mathbf u}^T A = {\mathbf f}^T,
+ U^T A = F^T,
@f]
-where ${\mathbf u}=(u_i)$, i.e., we have to solve the transpose problem of
-what we might have expected naively.
+where $U$ is the vector of expansion coefficients, i.e., we have to
+solve the transpose problem of what we might have expected naively.
This is a point we made in the introduction of step-3. There, we argued that
to avoid this very kind of problem, one should get in the habit of always