From a34ae1bfcd7f121e31d7fe380efd98c4d7d42d47 Mon Sep 17 00:00:00 2001 From: Marc Fehling Date: Tue, 4 Jun 2024 16:16:09 +0200 Subject: [PATCH] step-72: fix doxygen formatting. See also: https://www.doxygen.nl/manual/markdown.html#mddox_emph_spans --- examples/step-72/doc/intro.dox | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/examples/step-72/doc/intro.dox b/examples/step-72/doc/intro.dox index 354946d227..6011369bab 100644 --- a/examples/step-72/doc/intro.dox +++ b/examples/step-72/doc/intro.dox @@ -115,8 +115,8 @@ using automatic differentiation to compute the linearization of the residual vector. To this end, let us change notation for a moment and denote by $F(U)$ not the residual of the differential equation, but in fact the *residual vector* -- -i.e., the *discrete residual*. We do so because that is what we -*actually* do when we discretize the problem on a given mesh: We solve +i.e., the *discrete residual*. We do so because that is what we *actually* +do when we discretize the problem on a given mesh: We solve the problem $F(U)=0$ where $U$ is the vector of unknowns. More precisely, the $i$th component of the residual is given by @@ -153,8 +153,8 @@ clear that they are distinct from each other and from $j$ above. Because in this formula, $F(U)$ only depends on the coefficients $U_j$, we can compute the derivative $J(U)_{ij}^K$ as a matrix via automatic differentiation of $F(U)_i^K$. By the same argument as we -always use, it is clear that $F(U)^K$ does not actually depend on -*all* unknowns $U_j$, but only on those unknowns for which $j$ is a +always use, it is clear that $F(U)^K$ does not actually depend on *all* +unknowns $U_j$, but only on those unknowns for which $j$ is a shape function that lives on cell $K$, and so in practice, we restrict $F(U)^K$ and $J(U)^K$ to that part of the vector and matrix that corresponds to the *local* DoF indices, and then distribute from the -- 2.39.5