From: Timo Heister Date: Tue, 23 Feb 2016 18:04:17 +0000 (-0500) Subject: step-22: document symmetric gradient X-Git-Tag: v8.5.0-rc1~1283^2 X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=refs%2Fpull%2F2238%2Fhead;p=dealii.git step-22: document symmetric gradient --- diff --git a/examples/step-22/doc/intro.dox b/examples/step-22/doc/intro.dox index 4b00e31674..5c0ce904dd 100644 --- a/examples/step-22/doc/intro.dox +++ b/examples/step-22/doc/intro.dox @@ -40,6 +40,45 @@ valid; taking into account inertia effects then leads to the nonlinear Navier-Stokes equations. However, in this tutorial program, 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), +@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 +$\mu$) and assuming incompressibility ($\textrm{div}\; \textbf{u}=0$), we +arrive at the formulation from above: +@f{eqnarray*} + \textrm{div}\; \tau = -2\textrm{div}\;\varepsilon(\textbf{u}). +@f} +A different formulation uses the Laplace operator ($-\triangle \textbf{u}$) +instead of the symmetrized gradient. A big difference here is that the +different components of the velocity do not couple. If you assume additional +regularity of the solution $\textbf{u}$ (second partial derivatives exist and +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}. +@f} +This is because the $i$th entry of $\nabla \cdot (\nabla\textbf{u})^T$ is given by: +@f{eqnarray*} +[\nabla \cdot (\nabla\textbf{u})^T]_i += \sum_j \frac{\partial}{\partial x_j} [(\nabla\textbf{u})^T]_{i,j} += \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} += 0. +@f} +If you can not assume the above mentioned regularity, or if your viscosity is +not a constant, the equivalence no longer holds. Therefore, we decided to +stick with the more physically accurate symmetric tensor formulation in this +tutorial. + + To be well-posed, we will have to add boundary conditions to the equations. What boundary conditions are readily possible here will become clear once we discuss the weak form of the equations. @@ -66,7 +105,7 @@ form as \end{pmatrix}, @f} forming the dot product from the left with a vector-valued test -function $\phi = \begin{pmatrix}\textbf v \\ q\end{pmatrix}$ and integrating +function $\phi = \begin{pmatrix}\textbf{v} \\ q\end{pmatrix}$ and integrating over the domain $\Omega$, yielding the following set of equations: @f{eqnarray*} (\mathrm v, @@ -76,13 +115,13 @@ over the domain $\Omega$, yielding the following set of equations: = (\textbf{v}, \textbf{f})_\Omega, @f} -which has to hold for all test functions $\phi = \begin{pmatrix}\textbf v +which has to hold for all test functions $\phi = \begin{pmatrix}\textbf{v} \\ q\end{pmatrix}$. In practice, one wants to impose as little regularity on the pressure variable as possible; consequently, we integrate by parts the second term: @f{eqnarray*} - (\mathrm v, -2\; \textrm{div}\; \varepsilon(\textbf{u}))_{\Omega} + (\textbf{v}, -2\; \textrm{div}\; \varepsilon(\textbf{u}))_{\Omega} - (\textrm{div}\; \textbf{v}, p)_{\Omega} + (\textbf{n}\cdot\textbf{v}, p)_{\partial\Omega} - @@ -92,9 +131,9 @@ variable as possible; consequently, we integrate by parts the second term: @f} Likewise, we integrate by parts the first term to obtain @f{eqnarray*} - (\nabla \mathrm v, 2\; \varepsilon(\textbf{u}))_{\Omega} + (\nabla \textbf{v}, 2\; \varepsilon(\textbf{u}))_{\Omega} - - (\textbf{n} \otimes \mathrm v, 2\; \varepsilon(\textbf{u}))_{\partial\Omega} + (\textbf{n} \otimes \textbf{v}, 2\; \varepsilon(\textbf{u}))_{\partial\Omega} - (\textrm{div}\; \textbf{v}, p)_{\Omega} + (\textbf{n}\cdot\textbf{v}, p)_{\partial\Omega} - @@ -105,20 +144,20 @@ Likewise, we integrate by parts the first term to obtain where the scalar product between two tensor-valued quantities is here defined as @f{eqnarray*} - (\nabla \mathrm v, 2\; \varepsilon(\textbf{u}))_{\Omega} + (\nabla \textbf{v}, 2\; \varepsilon(\textbf{u}))_{\Omega} = 2 \int_\Omega \sum_{i,j=1}^d \frac{\partial v_j}{\partial x_i} \varepsilon(\textbf{u})_{ij} \ dx. @f} Because the scalar product between a general tensor like -$\nabla\mathrm v$ and a symmetric tensor like +$\nabla\textbf{v}$ and a symmetric tensor like $\varepsilon(\textbf{u})$ equals the scalar product between the symmetrized forms of the two, we can also write the bilinear form above as follows: @f{eqnarray*} - (\varepsilon(\mathrm v), 2\; \varepsilon(\textbf{u}))_{\Omega} + (\varepsilon(\textbf{v}), 2\; \varepsilon(\textbf{u}))_{\Omega} - - (\textbf{n} \otimes \mathrm v, 2\; \varepsilon(\textbf{u}))_{\partial\Omega} + (\textbf{n} \otimes \textbf{v}, 2\; \varepsilon(\textbf{u}))_{\partial\Omega} - (\textrm{div}\; \textbf{v}, p)_{\Omega} + (\textbf{n}\cdot\textbf{v}, p)_{\partial\Omega} - @@ -129,7 +168,7 @@ above as follows: We will deal with the boundary terms in the next section, but it is already clear from the domain terms @f{eqnarray*} - (\varepsilon(\mathrm v), 2\; \varepsilon(\textbf{u}))_{\Omega} + (\varepsilon(\textbf{v}), 2\; \varepsilon(\textbf{u}))_{\Omega} - (\textrm{div}\; \textbf{v}, p)_{\Omega} - (q,\textrm{div}\; \textbf{u})_{\Omega} @@ -153,8 +192,8 @@ possibilities for imposing boundary conditions: @f{eqnarray*} \textbf u = \textbf g_D \qquad\qquad \textrm{on}\ \Gamma_D. @f} - Because test functions $\textbf v$ come from the tangent space of - the solution variable, we have that $\textbf v=0$ on $\Gamma_D$ + Because test functions $\textbf{v}$ come from the tangent space of + the solution variable, we have that $\textbf{v}=0$ on $\Gamma_D$ and consequently that @f{eqnarray*} -(\textbf{n} \otimes \mathrm @@ -198,11 +237,11 @@ possibilities for imposing boundary conditions: (n_i v_j,p \delta_{ij} - 2\; \varepsilon(\textbf{u})_{ij})_{\Gamma_N} \\ &=& - (\textbf{n} \otimes \mathrm v, + (\textbf{n} \otimes \textbf{v}, p \textbf{1} - 2\; \varepsilon(\textbf{u}))_{\Gamma_N}. \\ &=& - (\mathrm v, + (\textbf{v}, \textbf{n}\cdot [p \textbf{1} - 2\; \varepsilon(\textbf{u})])_{\Gamma_N}. @f} In other words, on the Neumann part of the boundary we can @@ -215,7 +254,7 @@ possibilities for imposing boundary conditions: If the boundary is subdivided into Dirichlet and Neumann parts $\Gamma_D,\Gamma_N$, this then leads to the following weak form: @f{eqnarray*} - (\varepsilon(\mathrm v), 2\; \varepsilon(\textbf{u}))_{\Omega} + (\varepsilon(\textbf{v}), 2\; \varepsilon(\textbf{u}))_{\Omega} - (\textrm{div}\; \textbf{v}, p)_{\Omega} - (q,\textrm{div}\; \textbf{u})_{\Omega} @@ -235,7 +274,7 @@ possibilities for imposing boundary conditions: @f} with a rank-2 tensor (matrix) $\textbf S$. The associated weak form is @f{eqnarray*} - (\varepsilon(\mathrm v), 2\; \varepsilon(\textbf{u}))_{\Omega} + (\varepsilon(\textbf{v}), 2\; \varepsilon(\textbf{u}))_{\Omega} - (\textrm{div}\; \textbf{v}, p)_{\Omega} - (q,\textrm{div}\; \textbf{u})_{\Omega} @@ -299,7 +338,7 @@ boundary conditions on $\Gamma_D$ and $\Gamma_N$ reads like this: find $\textbf u\in \textbf V_g = \{\varphi \in H^1(\Omega)^d: \varphi_{\Gamma_D}=\textbf g_D\}, p\in Q=L^2(\Omega)$ so that @f{eqnarray*} - (\varepsilon(\mathrm v), 2\; \varepsilon(\textbf{u}))_{\Omega} + (\varepsilon(\textbf{v}), 2\; \varepsilon(\textbf{u}))_{\Omega} - (\textrm{div}\; \textbf{v}, p)_{\Omega} - (q,\textrm{div}\; \textbf{u})_{\Omega} @@ -309,7 +348,7 @@ g_D\}, p\in Q=L^2(\Omega)$ so that (\textbf{v}, \textbf g_N)_{\Gamma_N} @f} for all test functions -$\textbf v\in \textbf V_0 = \{\varphi \in H^1(\Omega)^d: \varphi_{\Gamma_D}=0\},q\in +$\textbf{v}\in \textbf V_0 = \{\varphi \in H^1(\Omega)^d: \varphi_{\Gamma_D}=0\},q\in Q$. These equations represent a symmetric saddle point problem. It is well known @@ -331,7 +370,7 @@ pressures. This then leads to the following discrete problem: find $\textbf u_h,p_h$ so that @f{eqnarray*} - (\varepsilon(\mathrm v_h), 2\; \varepsilon(\textbf u_h))_{\Omega} + (\varepsilon(\textbf{v}_h), 2\; \varepsilon(\textbf u_h))_{\Omega} - (\textrm{div}\; \textbf{v}_h, p_h)_{\Omega} - (q_h,\textrm{div}\; \textbf{u}_h)_{\Omega} @@ -340,7 +379,7 @@ that - (\textbf{v}_h, \textbf g_N)_{\Gamma_N} @f} -for all test functions $\textbf v_h, q_h$. Assembling the linear system +for all test functions $\textbf{v}_h, q_h$. Assembling the linear system associated with this problem follows the same lines used in @ref step_20 "step-20", step-21, and explained in detail in the @ref vector_valued module.