From 131afbb05007fec94acc25547beb0704531f2039 Mon Sep 17 00:00:00 2001 From: Wolfgang Bangerth Date: Fri, 21 Jan 2022 08:45:08 -0700 Subject: [PATCH] Fix a sign error in step-22's introduction. While there also do minimal maintenance in a couple of other place. --- examples/step-22/doc/intro.dox | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/examples/step-22/doc/intro.dox b/examples/step-22/doc/intro.dox index 002bc0dd5c..0f9a9da7a8 100644 --- a/examples/step-22/doc/intro.dox +++ b/examples/step-22/doc/intro.dox @@ -43,7 +43,7 @@ we will focus on the simpler Stokes system. Note that when deriving the more general compressible Navier-Stokes equations, the diffusion is modeled as the divergence of the stress tensor @f{eqnarray*} - \tau = - \mu (2\varepsilon(\textbf{u}) - \frac{2}{3}\nabla \cdot \textbf{u} I), + \tau = - \mu \left(2\varepsilon(\textbf{u}) - \frac{2}{3}\nabla \cdot \textbf{u} I\right), @f} where $\mu$ is the viscosity of the fluid. With the assumption of $\mu=1$ (assume constant viscosity and non-dimensionalize the equation by dividing out @@ -60,7 +60,7 @@ are continuous), the formulations are equivalent: @f{eqnarray*} \textrm{div}\; \tau = -2\textrm{div}\;\varepsilon(\textbf{u}) - = -\triangle \textbf{u} + \nabla \cdot (\nabla\textbf{u})^T + = -\triangle \textbf{u} - \nabla \cdot (\nabla\textbf{u})^T = -\triangle \textbf{u}. @f} This is because the $i$th entry of $\nabla \cdot (\nabla\textbf{u})^T$ is given by: @@ -70,7 +70,8 @@ This is because the $i$th entry of $\nabla \cdot (\nabla\textbf{u})^T$ is given = \sum_j \frac{\partial}{\partial x_j} [(\nabla\textbf{u})]_{j,i} = \sum_j \frac{\partial}{\partial x_j} \frac{\partial}{\partial x_i} \textbf{u}_j = \sum_j \frac{\partial}{\partial x_i} \frac{\partial}{\partial x_j} \textbf{u}_j -= \frac{\partial}{\partial x_i} \textrm{div}\; \textbf{u} += \frac{\partial}{\partial x_i} + \underbrace{\textrm{div}\; \textbf{u}}_{=0} = 0. @f} If you can not assume the above mentioned regularity, or if your viscosity is -- 2.39.5