From ed7ccb5fd9bca554ded603ec9002232c06d56262 Mon Sep 17 00:00:00 2001 From: kronbichler Date: Fri, 14 Mar 2008 13:14:17 +0000 Subject: [PATCH] Corrected some right hand sides in step-20 and step-21 as remarked in the mailing list. git-svn-id: https://svn.dealii.org/trunk@15890 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-20/doc/intro.dox | 4 ++-- deal.II/examples/step-21/doc/intro.dox | 23 ++++++++++++----------- 2 files changed, 14 insertions(+), 13 deletions(-) diff --git a/deal.II/examples/step-20/doc/intro.dox b/deal.II/examples/step-20/doc/intro.dox index 97bcd8cda1..7fa21d7a1f 100644 --- a/deal.II/examples/step-20/doc/intro.dox +++ b/deal.II/examples/step-20/doc/intro.dox @@ -61,8 +61,8 @@ areas of high pressure to areas of low pressure. With this second variable, one then finds an alternative version of the Laplace equation, called the mixed formulation: @f{eqnarray*} - K^{-1} {\mathbf u} - \nabla p &=& 0 \qquad {\textrm{in}\ } \Omega, \\ - -{\textrm{div}}\ {\mathbf u} &=& 0 \qquad {\textrm{in}\ }\Omega, \\ + K^{-1} {\mathbf u} + \nabla p &=& 0 \qquad {\textrm{in}\ } \Omega, \\ + -{\textrm{div}}\ {\mathbf u} &=& -f \qquad {\textrm{in}\ }\Omega, \\ p &=& g \qquad {\textrm{on}\ } \partial\Omega. @f} diff --git a/deal.II/examples/step-21/doc/intro.dox b/deal.II/examples/step-21/doc/intro.dox index d08c107a41..ea09ee7345 100644 --- a/deal.II/examples/step-21/doc/intro.dox +++ b/deal.II/examples/step-21/doc/intro.dox @@ -64,7 +64,7 @@ so-called pressure equation: @f{eqnarray*} - \nabla \cdot (\mathbf{K}\lambda(S) \nabla p)= q. @f} -Here, $q$ is th sum source term, and +Here, $q$ is the sum source term, and @f[ \lambda(S) = \frac{k_{rw}(S)}{\mu_{w}}+\frac{k_{ro}(S)}{\mu_{o}} @f] @@ -145,8 +145,8 @@ derived above by going back to the first order, mixed formulation. To this end, we re-introduce the total velocity $\mathbf u$ and write the equations in the following form: @f{eqnarray*} - \mathbf{u}+\mathbf{K}\lambda(S) \nabla p&=&0 \\ - \nabla \cdot\mathbf{u} &=& 0 \\ + \mathbf{u}-\mathbf{K}\lambda(S) \nabla p&=&0 \\ + \nabla \cdot\mathbf{u} &=& q \\ S_{t} + \mathbf{u} \cdot \nabla F(S) &=& 0. @f} This formulation has the additional benefit that we do not have to express the @@ -155,7 +155,7 @@ of the pressure, but can rather take the primary variable for it. Given the saddle point structure of the first two equations and their similarity to the mixed Laplace formulation we have introduced in @ref step_20 "step-20", it will come as no surprise that we will use a mixed discretization again. - + But let's postpone this for a moment. The first business we have with these equations is to think about the time discretization. In reservoir simulation, there is a rather standard algorithm that we will use here. It first solves @@ -165,7 +165,7 @@ Explicit Saturation. In a slightly modified form, this algorithm can be written as follows: for each time step, solve @f{eqnarray*} \mathbf{u}^{n+1}+\mathbf{K}\lambda(S^n) \nabla p^{n+1}&=&0 \\ - \nabla \cdot\mathbf{u}^{n+1} &=& 0 \\ + \nabla \cdot\mathbf{u}^{n+1} &=& q^{n+1} \\ \frac {S^{n+1}-S^n}{\triangle t} + \mathbf{u}^{n+1} \cdot \nabla F(S^n) &=& 0, @f} where $\triangle t$ is the length of a time step. Note how we solve the @@ -174,14 +174,14 @@ saturation $S^n$, and then do an explicit time step for $S^{n+1}$ that only depends on the previously known $S^n$ and the just computed $\mathbf{u}^{n+1}$. We can then state the problem in weak form as follows, by multiplying each -equation with test functions $\mathbf v$, $q$, and $\sigma$ and integrating +equation with test functions $\mathbf v$, $\phi$, and $\sigma$ and integrating terms by parts: @f{eqnarray*} \left((\mathbf{K}\lambda(S^n))^{-1} \mathbf{u}^{n+1},\mathbf v\right)_\Omega - (p^{n+1}, \nabla\cdot\mathbf v)_\Omega &=& - (p^{n+1}, \mathbf v)_{\partial\Omega} \\ - (\nabla \cdot\mathbf{u}^{n+1}, q)_\Omega &=& 0 + (\nabla \cdot\mathbf{u}^{n+1}, \phi)_\Omega &=& (q^{n+1},\phi)_\Omega @f} Note that in the first term, we have to prescribe the pressure $p^{n+1}$ on the boundary $\partial\Omega$ as boundary values for our problem. $\mathbf n$ @@ -201,20 +201,21 @@ For the saturation equation, we obtain after integrating by parts &=& (S^n,\sigma)_\Omega. @f} -Using the fact that $\nabla \cdot \mathbf{u}^{n+1}=0$, we can rewrite the cell -term to get an equation as follows: +Using the fact that $\nabla \cdot \mathbf{u}^{n+1}=q^{n+1}$, we can rewrite the +cell term to get an equation as follows: @f{eqnarray*} (S^{n+1}, \sigma)_\Omega - \triangle t \sum_K \left\{ - \left(F(S^n) \mathbf{u}^{n+1}, \nabla \sigma)\right)_K + \left(F(S^n) \mathbf{u}^{n+1}, \nabla \sigma\right)_K - \left(F(S^n) (\mathbf n \cdot \mathbf{u}^{n+1}), \sigma\right)_{\partial K} \right\} &=& - (S^n,\sigma)_\Omega. + (S^n,\sigma)_\Omega + + \triangle t \sum_K \left(F(S^n), q^{n+1} \sigma\right). @f} -- 2.39.5