Use balanced braces for doxygen formulas.

examples/step-11/doc/intro.dox

@@ -3,7 +3,7 @@
 
 The problem we will be considering is the solution of Laplace's problem with
 Neumann boundary conditions only:
-@f{eqnarray*}
+@f{eqnarray*}{
   -\Delta u &=& f \qquad \mathrm{in}\ \Omega,
   \\
   \partial_n u &=& g \qquad \mathrm{on}\ \partial\Omega.

examples/step-14/doc/intro.dox

@@ -79,7 +79,7 @@ sum over cells where each cell's contribution can then be used as an
 error indicator for this cell.
 Thus, we split the scalar products into terms for each cell, and
 integrate by parts on each of them:
-@f{eqnarray*}
+@f{eqnarray*}{
   J(e)
   &=&
   \sum_K (\nabla (u-u_h), \nabla (z-\varphi_h))_K
@@ -98,7 +98,7 @@ with which this term could cancel, the weight $z-\varphi_h$ can be chosen as
 zero, and the whole term disappears.
 
 Thus, we have
-@f{eqnarray*}
+@f{eqnarray*}{
   J(e)
   &=&
   \sum_K (f+\Delta u_h, z-\varphi_h)_K
@@ -109,7 +109,7 @@ the value of this quantity as taken from this side of the cell (for the usual
 Lagrange elements, derivatives are not continuous across edges). We then
 rewrite the above formula by exchanging half of the edge integral of cell $K$
 with the neighbor cell $K'$, to obtain
-@f{eqnarray*}
+@f{eqnarray*}{
   J(e)
   &=&
   \sum_K (f+\Delta u_h, z-\varphi_h)_K
@@ -126,7 +126,7 @@ the jump of the normal derivative by
 and get the final form after setting the discrete function $\varphi_h$, which
 is by now still arbitrary, to the point interpolation of the dual solution,
 $\varphi_h=I_h z$:
-@f{eqnarray*}
+@f{eqnarray*}{
   J(e)
   &=&
   \sum_K (f+\Delta u_h, z-I_h z)_K

examples/step-15/doc/intro.dox

@@ -40,7 +40,7 @@ have to use Newton's method to compute the solution iteratively.
 In a classical sense, the problem is given in the following form:
 
 
-  @f{align*}
+  @f{align*}{
     -\nabla \cdot \left( \frac{1}{\sqrt{1+|\nabla u|^{2}}}\nabla u \right) &= 0 \qquad
     \qquad &&\textrm{in} ~ \Omega
     \\
@@ -53,7 +53,7 @@ space. In this example, we choose $\Omega$ as the unit disk.
 As described above, we solve this equation using Newton's method in which we
 compute the $n$th approximate solution from the $(n-1)$th one, and use
 a damping parameter $\alpha^n$ to get better global convergence behavior:
-  @f{align*}
+  @f{align*}{
     F'(u^{n},\delta u^{n})&=- F(u^{n})
     \\
     u^{n+1}&=u^{n}+\alpha^n \delta u^{n}

examples/step-15/doc/results.dox

@@ -221,11 +221,11 @@ the form
   @f]
 we use a Newton iteration that requires us to repeatedly solve the
 linear partial differential equation
-  @f{align*}
+  @f{align*}{
     F'(u^{n},\delta u^{n}) &=- F(u^{n})
   @f}
 so that we can compute the update
-  @f{align*}
+  @f{align*}{
     u^{n+1}&=u^{n}+\alpha^n \delta u^{n}
   @f}
 with the solution $\delta u^{n}$ of the Newton step. For the problem

examples/step-18/doc/intro.dox

@@ -48,7 +48,7 @@ In addition, initial conditions
 @f]
 and Dirichlet (displacement) or Neumann (traction) boundary conditions need
 to be specified for a unique solution:
-@f{eqnarray*}
+@f{eqnarray*}{
   \mathbf{u}(\mathbf{x},t) &=& \mathbf{d}(\mathbf{x},t)
   \qquad
   \textrm{on}\ \Gamma_D\subset\partial\Omega,
@@ -82,7 +82,7 @@ that all changes in external forcing happen on times scales that are
 much larger than $\tau$. In that case, the dynamic nature of the change is
 unimportant: we can consider the body to always be in static equilibrium,
 i.e. we can assume that at all times the body satisfies
-@f{eqnarray*}
+@f{eqnarray*}{
   - \textrm{div}\  ( C \varepsilon(\mathbf{u})) &=& \mathbf{f}(\mathbf{x},t)
   \qquad
   \textrm{in}\ \Omega,
@@ -127,7 +127,7 @@ implementing a <i>real</i> model, as we will see in step-44.
 To come back to defining our "artificial" model, let us first
 introduce a tensorial stress variable $\sigma$, and write the differential
 equations in terms of the stress:
-@f{eqnarray*}
+@f{eqnarray*}{
   - \textrm{div}\  \sigma &=& \mathbf{f}(\mathbf{x},t)
   \qquad
   \textrm{in}\ \Omega(t),
@@ -172,7 +172,7 @@ and $\Delta \mathbf{u}^n$ the incremental displacement for time step
 $n$. In addition, we have to specify initial data $\mathbf{u}(\cdot,0)=\mathbf{u}_0$.
 This way, if we want to solve for the displacement increment, we
 have to solve the following system:
-@f{align*}
+@f{align*}{
   - \textrm{div}\   C \varepsilon(\Delta\mathbf{u}^n) &= \mathbf{f} + \textrm{div}\  \sigma^{n-1}
   \qquad
   &&\textrm{in}\ \Omega(t_{n-1}),
@@ -190,7 +190,7 @@ finite element formulation, reads as follows: find $\Delta \mathbf{u}^n \in
 \{v\in H^1(\Omega(t_{n-1}))^d: v|_{\Gamma_D}=\mathbf{d}(\cdot,t_n) - \mathbf{d}(\cdot,t_{n-1})\}$
 such that
 <a name="step_18.linear-system"></a>
-@f{align*}
+@f{align*}{
   (C \varepsilon(\Delta\mathbf{u}^n), \varepsilon(\varphi) )_{\Omega(t_{n-1})}
   &=
   (\mathbf{f}, \varphi)_{\Omega(t_{n-1})}
@@ -207,7 +207,7 @@ Using that $\sigma^{n-1} \mathbf{n}
             = [C \varepsilon(\mathbf{u}^{n-1})] \mathbf{n}
             = \mathbf{b}(\mathbf x, t_{n-1})$,
 these equations can be simplified to
-@f{align*}
+@f{align*}{
   (C \varepsilon(\Delta\mathbf{u}^n), \varepsilon(\varphi) )_{\Omega(t_{n-1})}
   &=
   (\mathbf{f}, \varphi)_{\Omega(t_{n-1})}
@@ -529,7 +529,7 @@ for (unsigned int i=0; i<dofs_per_cell; ++i)
   between symmetric tensors of even rank here.
 
   Assembling the local contributions
-  @f{eqnarray*}
+  @f{eqnarray*}{
       f^K_i &=&
       (\mathbf{f}, \varphi_i)_K -(\sigma^{n-1},\varepsilon(\varphi_i))_K
       \\

examples/step-20/doc/intro.dox

@@ -8,7 +8,7 @@ particular Raviart-Thomas elements -- and using block matrices to define
 solvers, preconditioners, and nested versions of those that use the
 substructure of the system matrix. The equation we are going to solve is again
 the Poisson equation, though with a matrix-valued coefficient:
-@f{eqnarray*}
+@f{eqnarray*}{
   -\nabla \cdot K({\mathbf x}) \nabla p &=& f \qquad {\textrm{in}\ } \Omega, \\
   p &=& g \qquad {\textrm{on}\ }\partial\Omega.
 @f}
@@ -89,7 +89,7 @@ areas of high pressure to areas of low pressure (thus the negative sign).
 
 With this second variable, one then finds an alternative version of the
 Laplace equation, called the <i>mixed formulation</i>:
-@f{eqnarray*}
+@f{eqnarray*}{
   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.
@@ -103,11 +103,11 @@ under the common assumption that $K$ is a symmetric tensor.
 
 The weak formulation of this problem is found by multiplying the two
 equations with test functions and integrating some terms by parts:
-@f{eqnarray*}
+@f{eqnarray*}{
   A(\{{\mathbf u},p\},\{{\mathbf v},q\}) = F(\{{\mathbf v},q\}),
 @f}
 where
-@f{eqnarray*}
+@f{eqnarray*}{
   A(\{{\mathbf u},p\},\{{\mathbf v},q\})
   &=&
   ({\mathbf v}, K^{-1}{\mathbf u})_\Omega - ({\textrm{div}}\ {\mathbf v}, p)_\Omega
@@ -141,7 +141,7 @@ elements, for example the classic book by Brenner and Scott, also state the
 relevant results. In any case, with appropriate choices of function
 spaces, the discrete formulation reads as follows: Find ${\mathbf
 u}_h,p_h$ so that
-@f{eqnarray*}
+@f{eqnarray*}{
   A(\{{\mathbf u}_h,p_h\},\{{\mathbf v}_h,q_h\}) = F(\{{\mathbf v}_h,q_h\})
   \qquad\qquad \forall {\mathbf v}_h,q_h.
 @f}
@@ -156,20 +156,20 @@ zero everywhere else. If we also choose ${\mathbf v}=0$ everywhere
 (remember that the weak form above has to hold for <i>all</i> discrete
 test functions $q,v$), then putting these choices of test functions
 into the weak formulation above implies in particular that
-@f{eqnarray*}
+@f{eqnarray*}{
   - (1,{\textrm{div}}\ {\mathbf u}_h)_K
   =
   -(1,f)_K,
 @f}
 which we can of course write in more explicit form as
-@f{eqnarray*}
+@f{eqnarray*}{
   \int_K {\textrm{div}}\ {\mathbf u}_h
  =
   \int_K f.
 @f}
 Applying the divergence theorem results in the fact that ${\mathbf
 u}_h$ has to satisfy, for every choice of cell $K$, the relationship
-@f{eqnarray*}
+@f{eqnarray*}{
   \int_{\partial K} {\mathbf u}_h\cdot{\mathbf n}
   =
   \int_K f.
@@ -202,7 +202,7 @@ The deal.II library (of course) implements Raviart-Thomas elements $RT(k)$ of
 arbitrary order $k$, as well as discontinuous elements $DG(k)$. If we forget
 about their particular properties for a second, we then have to solve a
 discrete problem
-@f{eqnarray*}
+@f{eqnarray*}{
   A(x_h,w_h) = F(w_h),
 @f}
 with the bilinear form and right hand side as stated above, and $x_h=\{{\mathbf u}_h,p_h\}$, $w_h=\{{\mathbf v}_h,q_h\}$. Both $x_h$ and $w_h$ are from the space
@@ -263,7 +263,7 @@ return a non-zero value for more than just one component.
 We could now attempt to rewrite the bilinear form above in terms of vector
 components. For example, in 2d, the first term could be rewritten like this
 (note that $u_0=x_0, u_1=x_1, p=x_2$):
-@f{eqnarray*}
+@f{eqnarray*}{
   ({\mathbf u}_h^i, K^{-1}{\mathbf u}_h^j)
   =
   &\left((x_h^i)_0, K^{-1}_{00} (x_h^j)_0\right) +
@@ -463,7 +463,7 @@ In view of the difficulties using standard solvers and preconditioners
 mentioned above, let us take another look at the matrix. If we sort our
 degrees of freedom so that all velocity come before all pressure variables,
 then we can subdivide the linear system $Ax=b$ into the following blocks:
-@f{eqnarray*}
+@f{eqnarray*}{
   \left(\begin{array}{cc}
     M & B \\ B^T & 0
   \end{array}\right)
@@ -483,7 +483,7 @@ to the gradient.
 By block elimination, we can then re-order this system in the following way
 (multiply the first row of the system by $B^TM^{-1}$ and then subtract the
 second row from it):
-@f{eqnarray*}
+@f{eqnarray*}{
   B^TM^{-1}B P &=& B^TM^{-1} F - G, \\
   MU &=& F - BP.
 @f}
@@ -713,7 +713,7 @@ We will try something along the second approach, as much to improve the
 performance of the program as to demonstrate some techniques. To this end, let
 us recall that the ideal preconditioner is, of course, $S^{-1}$, but that is
 unattainable. However, how about
-@f{eqnarray*}
+@f{eqnarray*}{
   \tilde S^{-1} = [B^T ({\textrm{diag}\ }M)^{-1}B]^{-1}
 @f}
 as a preconditioner? That would mean that every time we have to do one
@@ -782,12 +782,12 @@ formulation as stated above. Since we want to monitor convergence of the
 solution inside the program, we choose right hand side, boundary conditions,
 and the coefficient so that we recover a solution function known to us. In
 particular, we choose the pressure solution
-@f{eqnarray*}
+@f{eqnarray*}{
   p = -\left(\frac \alpha 2 xy^2 + \beta x - \frac \alpha 6 x^3\right),
 @f}
 and for the coefficient we choose the unit matrix $K_{ij}=\delta_{ij}$ for
 simplicity. Consequently, the exact velocity satisfies
-@f{eqnarray*}
+@f{eqnarray*}{
   {\mathbf u} =
   \left(\begin{array}{cc}
     \frac \alpha 2 y^2 + \beta - \frac \alpha 2 x^2 \\

examples/step-21/doc/intro.dox

@@ -38,7 +38,7 @@ however.
 The velocity with which molecules of each of the two phases move is
 determined by Darcy's law that states that the velocity is
 proportional to the pressure gradient:
-@f{eqnarray*}
+@f{eqnarray*}{
   \mathbf{u}_{j}
   =
   -\frac{k_{rj}(S)}{\mu_{j}} \mathbf{K} \cdot \nabla p
@@ -61,7 +61,7 @@ each phase,
 with a source term for each phase. By summing over the two phases,
 we can express the governing equations in terms of the
 so-called pressure equation:
-@f{eqnarray*}
+@f{eqnarray*}{
 - \nabla \cdot (\mathbf{K}\lambda(S) \nabla p)= q.
 @f}
 Here, $q$ is the sum source term, and
@@ -80,12 +80,12 @@ The second part of the equations is the description of the
 dynamics of the saturation, i.e., how the relative concentration of the
 two fluids changes with time. The saturation equation for the displacing
 fluid (water) is given by the following conservation law:
-@f{eqnarray*}
+@f{eqnarray*}{
   S_{t} + \nabla \cdot (F(S) \mathbf{u}) = q_{w},
 @f}
 which can be rewritten by using the product rule of the divergence operator
 in the previous equation:
-@f{eqnarray*}
+@f{eqnarray*}{
   S_{t} + F(S) \left[\nabla \cdot \mathbf{u}\right]
         + \mathbf{u} \cdot \left[ \nabla F(S)\right]
   = S_{t} + F(S) q + \mathbf{u} \cdot \nabla F(S) = q_{w}.
@@ -104,7 +104,7 @@ where the fractional flow is often parameterized via the (heuristic) expression
 @f]
 Putting it all together yields the saturation equation in the following,
 advected form:
-@f{eqnarray*}
+@f{eqnarray*}{
   S_{t} + \mathbf{u} \cdot \nabla F(S) = 0,
 @f}
 where $\mathbf u$ is the total velocity
@@ -124,7 +124,7 @@ $\mathbf u (1-F'(S))$. $F(S)$ is consequently often referred to as the
 <i>fractional flow</i>.
 
 In summary, what we get are the following two equations:
-@f{eqnarray*}
+@f{eqnarray*}{
   - \nabla \cdot (\mathbf{K}\lambda(S) \nabla p) &=& q
   \qquad \textrm{in}\ \Omega\times[0,T],
   \\
@@ -169,7 +169,7 @@ In the reservoir simulation community, it is common to solve the equations
 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*}
+@f{eqnarray*}{
   \mathbf{u}+\mathbf{K}\lambda(S) \nabla p&=&0 \\
   \nabla \cdot\mathbf{u} &=& q \\
   S_{t} + \mathbf{u} \cdot \nabla F(S) &=& 0.
@@ -194,7 +194,7 @@ L. Stone and A. O. Gardner Jr: <i>Analysis of gas-cap or dissolved-gas
 reservoirs</i>, Trans. SPE AIME, 222 (1961), pp. 92-104).
 In a slightly modified form, this algorithm can be
 written as follows: for each time step, solve
-@f{eqnarray*}
+@f{eqnarray*}{
   \mathbf{u}^{n+1}+\mathbf{K}\lambda(S^n) \nabla p^{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,
@@ -211,7 +211,7 @@ splitting" method. step-58 has a long description of the idea behind this.)
 We can then state the problem in weak form as follows, by multiplying each
 equation with test functions $\mathbf v$, $\phi$, and $\sigma$ and integrating
 terms by parts:
-@f{eqnarray*}
+@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}
@@ -223,7 +223,7 @@ the boundary $\partial\Omega$ as boundary values for our problem. $\mathbf n$
 denotes the unit outward normal vector to $\partial K$, as usual.
 
 For the saturation equation, we obtain after integrating by parts
-@f{eqnarray*}
+@f{eqnarray*}{
   (S^{n+1}, \sigma)_\Omega
   -
   \triangle t
@@ -238,7 +238,7 @@ For the saturation equation, we obtain after integrating by parts
 @f}
 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*}
+@f{eqnarray*}{
   (S^{n+1}, \sigma)_\Omega
   -
   \triangle t
@@ -272,7 +272,7 @@ terms on the interfaces between cells, since discontinuous functions are not
 really defined there. In particular, we have to give a meaning to the last
 term on the left hand side of the saturation equation. To this end, let us
 define that we want to evaluate it in the following sense:
-@f{eqnarray*}
+@f{eqnarray*}{
   &&\left(F(S^n) (\mathbf n \cdot \mathbf{u}^{n+1}), \sigma\right)_{\partial K}
   \\
   &&\qquad =
@@ -321,7 +321,7 @@ B &    0 & 0\\
 where the individual matrices and vectors are defined as follows using
 shape functions $\mathbf v_i$ (of type Raviart Thomas $RT_k$) for
 velocities and $\phi_i$ (of type $DGQ_k$) for both pressures and saturations:
-@f{eqnarray*}
+@f{eqnarray*}{
 M^u(S^n)_{ij} &=&
 \left((\mathbf{K}\lambda(S^n))^{-1} \mathbf{v}_i,\mathbf
 v_j\right)_\Omega,
@@ -471,7 +471,7 @@ boundary conditions for the saturation on the inflow part of the boundary,
                      \mathbf{n} \cdot \mathbf{u}(\mathbf{x},t) < 0\}.
 @f]
 On this inflow boundary, we impose the following saturation values:
-@f{eqnarray}
+@f{eqnarray}{
   S(\mathbf{x},t) = 1 & \textrm{on}\ \Gamma_{in}\cap\{x_1=0\},
   \\
   S(\mathbf{x},t) = 0 & \textrm{on}\ \Gamma_{in}\backslash \{x_1=0\}.
@@ -540,7 +540,7 @@ functions introduced at the end of the results section of @ref step_20
   linear solvers will no longer converge properly.
 
   <li>A function that models a somewhat random medium. Here, we choose
-  @f{eqnarray*}
+  @f{eqnarray*}{
     k(\mathbf x)
     &=&
     \min \left\{ \max \left\{ \sum_{i=1}^N \sigma_i(\mathbf{x}), 0.01 \right\}, 4\right\},