Corrected some right hand sides in step-20 and step-21 as remarked in the mailing...

git-svn-id: https://svn.dealii.org/trunk@15890 0785d39b-7218-0410-832d-ea1e28bc413d

diff --git a/deal.II/examples/step-20/doc/intro.dox b/deal.II/examples/step-20/doc/intro.dox
index 97bcd8cda161ffe914418e7636ae9dbe2990081e..7fa21d7a1f949b0415314f881d70de08b04ba0e4 100644 (file)
--- 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 d08c107a41168cf5c89376453a355ac13918c1ea..ea09ee734574bd3568484ac0c90a181aee80b549 100644 (file)
--- 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}