From: wolf Date: Fri, 2 Sep 2005 15:23:01 +0000 (+0000) Subject: Regenerate formulas. X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=76fc0ebaf1476c24a430d1f174b0e9b86d3c1f49;p=dealii-svn.git Regenerate formulas. git-svn-id: https://svn.dealii.org/trunk@11359 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.html b/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.html index 57ad2e5ffe..5b8eeb9c1a 100644 --- a/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.html +++ b/deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.html @@ -46,13 +46,14 @@ elastic wave equation
$\displaystyle \rho \frac{\partial^2 \vec u}{\partial t^2} + c \frac{\partial \vec u}{\partial t} - \div ( C \varepsilon(\vec u)) = \vec f$   in $ \Omega$$ \Omega$ +$\displaystyle ,$ @@ -63,31 +64,36 @@ where $ \vec u=\vec u (\vec x,t)$ is the deformation of the body, $ \vec u=\vec u (\vec x,t)$ + is the deformation of the body, $ \rho$ + and $ c$ the density and attenuation coefficient, and $ c$ + the density and attenuation coefficient, and $ \vec f$ external forces. + ALT="$ \vec f$"> + external forces. In addition, initial conditions

$\displaystyle \vec u(\cdot, 0) = \vec u_0(\cdot)$   on $ \Omega$$ \Omega$ +$\displaystyle ,$ @@ -100,11 +106,11 @@ to be specified for a unique solution:
@@ -112,21 +118,22 @@ to be specified for a unique solution: $\Gamma_D\subset\partial\Omega$ --> $ \Gamma_D\subset\partial\Omega$$ \Gamma_D\subset\partial\Omega$ +$\displaystyle ,$ + ALT="$\displaystyle \vec n \ C \varepsilon(\vec u(\vec x,t))$"> @@ -134,10 +141,11 @@ to be specified for a unique solution: $\Gamma_N=\partial\Omega\backslash\Gamma_D$ --> $ \Gamma_N=\partial\Omega\backslash\Gamma_D$$ \Gamma_N=\partial\Omega\backslash\Gamma_D$ +$\displaystyle .$
$\displaystyle \vec u(\vec x,t)$ $\displaystyle = \vec d(\vec x,t) \qquad$   (3)
$\displaystyle \vec n C \varepsilon(\vec u(\vec x,t))$ $\displaystyle = \vec b(\vec x,t) \qquad$   @@ -149,43 +157,50 @@ In above formulation, $ \varepsilon(\vec u)= \tfrac 12 (\nabla \vec u + \nabla -\vec u^T)$ is the symmetric gradient of the displacement, also called the +\vec u^T)$"> + is the symmetric gradient of the displacement, also called the strain. $ C$ is a tensor of rank 4, called the stress-strain + ALT="$ C$"> + is a tensor of rank 4, called the stress-strain tensor that contains knowledge of the elastic strength of the material; its symmetry properties make sure that it maps symmetric tensors of rank 2 (``matrices'' of dimension $ d$, where $ d$ +, where $ d$ is the spatial dimensionality) onto + ALT="$ d$"> + is the spatial dimensionality) onto symmetric tensors of the same rank. We will comment on the roles of the strain and stress tensors more below. For the moment it suffices to say that we interpret the term $ \div ( C \varepsilon(\vec u))$ as the vector with + ALT="$ \div ( C \varepsilon(\vec u))$"> + as the vector with components $ \tfrac \partial{\partial x_j} C_{ijkl} \varepsilon(\vec u)_{kl}$, + ALT="$ \tfrac \partial{\partial x_j} C_{ijkl} \varepsilon(\vec u)_{kl}$"> +, where summation over indices $ j,k,l$ is implied. + ALT="$ j,k,l$"> + is implied.

The quasistatic limit of this equation is motivated as follows: each small @@ -194,71 +209,76 @@ forcing function, will result in a corresponding change in the configuration of the body. In general, this will be in the form of waves radiating away from the location of the disturbance. Due to the presence of the damping term, these waves will be attenuated on a time scale of, say, $ \tau$. Now, assume + ALT="$ \tau$"> +. Now, assume that all changes in external forcing happen on times scales that are much larger than $ \tau$. In that case, the dynamic nature of the change is + ALT="$ \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

+ ALT="$\displaystyle \vec n \ C \varepsilon(\vec u(\vec x,t))$">
$\displaystyle - \div ( C \varepsilon(\vec u))$ $\displaystyle = \vec f$   in $ \Omega$$ \Omega$ +$\displaystyle ,$ (5)
$\displaystyle \vec u(\vec x,t)$ $\displaystyle = \vec d(\vec x,t) \qquad$   on $ \Gamma_D$$ \Gamma_D$ +$\displaystyle ,$ (6)
$\displaystyle \vec n C \varepsilon(\vec u(\vec x,t))$ $\displaystyle = \vec b(\vec x,t) \qquad$   on $ \Gamma_N$$ \Gamma_N$ +$\displaystyle .$ @@ -271,46 +291,49 @@ possibly time-varying force function $ \vec f(\vec x,t)$. + ALT="$ \vec f(\vec x,t)$"> +.

While these equations are sufficient to describe small deformations, computing large deformations is a little more complicated. To do so, let us first introduce a stress variable $ \sigma$, and write the differential equations in + ALT="$ \sigma$"> +, and write the differential equations in terms of the stress:

@@ -318,21 +341,22 @@ terms of the stress: $\Gamma_D\subset\partial\Omega(t)$ --> $ \Gamma_D\subset\partial\Omega(t)$$ \Gamma_D\subset\partial\Omega(t)$ +$\displaystyle ,$ + ALT="$\displaystyle \vec n \ C \varepsilon(\vec u(\vec x,t))$"> @@ -340,10 +364,11 @@ terms of the stress: $\Gamma_N=\partial\Omega(t)\backslash\Gamma_D$ --> $ \Gamma_N=\partial\Omega(t)\backslash\Gamma_D$$ \Gamma_N=\partial\Omega(t)\backslash\Gamma_D$ +$\displaystyle .$
$\displaystyle - \div\sigma$ $\displaystyle = \vec f$   in $ \Omega(t)$$ \Omega(t)$ +$\displaystyle ,$ (8)
$\displaystyle \vec u(\vec x,t)$ $\displaystyle = \vec d(\vec x,t) \qquad$   (9)
$\displaystyle \vec n C \varepsilon(\vec u(\vec x,t))$ $\displaystyle = \vec b(\vec x,t) \qquad$   @@ -351,24 +376,26 @@ terms of the stress:

Note that these equations are posed on a domain $ \Omega(t)$ that + ALT="$ \Omega(t)$"> + that changes with time, with the boundary moving according to the displacements $ \vec u(\vec x,t)$ of the points on the boundary. To + ALT="$ \vec u(\vec x,t)$"> + of the points on the boundary. To complete this system, we have to specify the relationship between the stress and the strain, as follows:

$\displaystyle \dot\sigma = C \varepsilon (\dot{\vec u}),$ @@ -376,16 +403,18 @@ stress and the strain, as follows:

where a dot indicates a time derivative. Both the stress $ \sigma$ and the + ALT="$ \sigma$"> + and the strain $ \varepsilon(\vec u)$ are symmetric tensors of rank 2. + ALT="$ \varepsilon(\vec u)$"> + are symmetric tensors of rank 2.

@@ -397,14 +426,15 @@ Time discretization Numerically, this system is solved as follows: first, we discretize the time component using a backward Euler scheme. This leads to a discrete equilibrium of force at time step $ n$: + ALT="$ n$"> +:

$\displaystyle -\div\sigma^n = f^n,$ @@ -416,7 +446,7 @@ where
$\displaystyle \sigma^n = \sigma^{n-1} + C \varepsilon (\Delta \vec u^n),$ @@ -427,23 +457,25 @@ and $ \Delta \vec u^n$ the incremental displacement for time step + ALT="$ \Delta \vec u^n$"> + the incremental displacement for time step $ n$. This way, if we want to solve for the displacement increment, we + ALT="$ n$"> +. This way, if we want to solve for the displacement increment, we have to solve the following system:

@@ -451,21 +483,22 @@ have to solve the following system: $\Omega(t_{n-1})$ --> $ \Omega(t_{n-1})$$ \Omega(t_{n-1})$ +$\displaystyle ,$ @@ -473,21 +506,22 @@ have to solve the following system: $\Gamma_D\subset\partial\Omega(t_{n-1})$ --> $ \Gamma_D\subset\partial\Omega(t_{n-1})$$ \Gamma_D\subset\partial\Omega(t_{n-1})$ +$\displaystyle ,$ + ALT="$\displaystyle \vec n \ C \varepsilon(\Delta \vec u^n(\vec x,t))$"> @@ -495,10 +529,11 @@ have to solve the following system: $\Gamma_N=\partial\Omega(t_{n-1})\backslash\Gamma_D$ --> $ \Gamma_N=\partial\Omega(t_{n-1})\backslash\Gamma_D$$ \Gamma_N=\partial\Omega(t_{n-1})\backslash\Gamma_D$ +$\displaystyle .$
$\displaystyle - \div C \varepsilon(\Delta\vec u^n)$ $\displaystyle = \vec f + \div\sigma^{n-1}$   (14)
$\displaystyle \Delta \vec u^n(\vec x,t)$ $\displaystyle = \vec d(\vec x,t_n) - \vec d(\vec x,t_{n-1}) \qquad$   (15)
$\displaystyle \vec n C \varepsilon(\Delta \vec u^n(\vec x,t))$ $\displaystyle = \vec b(\vec x,t_n)-\vec b(\vec x,t_{n-1}) \qquad$   @@ -511,10 +546,11 @@ finite element formulation, reads as follows: find $ \Delta \vec u^n \in \{v\in H^1(\Omega(t_{n-1}))^d: v\vert _{\Gamma_D}=\vec d(\cdot,t_n) - \vec d(\cdot,t_{n-1})\}$ + such that

@@ -533,58 +569,67 @@ are no boundary forces, i.e. $ \vec b = 0$, and that the deformation of the + ALT="$ \vec b = 0$"> +, and that the deformation of the body is driven by body forces $ \vec f$ and prescribed boundary displacements + ALT="$ \vec f$"> + and prescribed boundary displacements $ \vec d$ alone. It is also worth noting that when integrating by parts, we + ALT="$ \vec d$"> + alone. It is also worth noting that when integrating by parts, we would get terms of the form $ (C \varepsilon(\Delta\vec u^n), \nabla \varphi -)_{\Omega(t_{n-1})}$, but that we replace it with the term involving the +)_{\Omega(t_{n-1})}$"> +, but that we replace it with the term involving the symmetric gradient $ \varepsilon(\varphi)$ instead of $ \nabla\varphi$. Due to + ALT="$ \nabla\varphi$"> +. Due to the symmetry of $ C$, the two terms are equivalent, but the symmetric version + ALT="$ C$"> +, the two terms are equivalent, but the symmetric version avoids a potential for round-off to render the resulting matrix slightly non-symmetric.

The system at time step $ n$, to be solved on the old domain + ALT="$ n$"> +, to be solved on the old domain $ \Omega(t_{n-1})$, has exactly the form of a stationary elastic + ALT="$ \Omega(t_{n-1})$"> +, has exactly the form of a stationary elastic problem, and is therefore similar to what we have already implemented in previous example programs. We will therefore not comment on the space discretization beyond saying that we again use lowest order @@ -603,9 +648,10 @@ There are differences, however: $\sigma^{n-1}$ --> $ \sigma^{n-1}$ to compute the next incremental + ALT="$ \sigma^{n-1}$"> + to compute the next incremental displacement, i.e. we need to compute it at the end of the time step to make sure it is available for the next time step. Essentially, the stress variable is our window to the history of deformation of @@ -625,18 +671,20 @@ Updating the stress variable

As indicated above, we need to have the stress variable $ \sigma^n$ available + ALT="$ \sigma^n$"> + available when computing time step $ n+1$, and we can compute it using + ALT="$ n+1$"> +, and we can compute it using

$\displaystyle \sigma^n = \sigma^{n-1} + C \varepsilon (\Delta \vec u^n).$ @@ -645,34 +693,39 @@ when computing time step

There are, despite the apparent simplicity of this equation, two questions that we need to discuss. The first concerns the way we store $ \sigma^n$: even + ALT="$ \sigma^n$"> +: even if we compute the incremental updates $ \Delta \vec u^n$ using lowest-order + ALT="$ \Delta \vec u^n$"> + using lowest-order finite elements, then its symmetric gradient $ \varepsilon(\Delta\vec u^n)$ is + ALT="$ \varepsilon(\Delta\vec u^n)$"> + is in general still a function that is not easy to describe. In particular, it is not a piecewise constant function, and on general meshes (with cells that are not rectangles parallel to the coordinate axes) or with non-constant stress-strain tensors $ C$ it is not even a bi- or trilinear function. Thus, it + ALT="$ C$"> + it is not even a bi- or trilinear function. Thus, it is a priori not clear how to store $ \sigma^n$ in a computer program. + ALT="$ \sigma^n$"> + in a computer program.

To decide this, we have to see where it is used. The only place where we @@ -681,40 +734,45 @@ require the stress is in the term $(\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})}$ --> $ (\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})}$. In practice, we of + ALT="$ (\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})}$"> +. In practice, we of course replace this term by numerical quadrature:

+...athbb{T}}} \sum_q w_q \ \sigma^{n-1}(\vec x_q) : \varepsilon(\varphi(\vec x_q),$">
$\displaystyle (\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})} = \sum_{K\s... -...thbb{T}}} \sum_q w_q \sigma^{n-1}(\vec x_q) : \varepsilon(\varphi(\vec x_q)),$ (19)

where $ w_q$ are the quadrature weights and $ w_q$ + are the quadrature weights and $ \vec x_q$ the quadrature points on + ALT="$ \vec x_q$"> + the quadrature points on cell $ K$. This should make clear that what we really need is not the stress + ALT="$ K$"> +. This should make clear that what we really need is not the stress $ \sigma^{n-1}$ in itself, but only the values of the stress in the quadrature + ALT="$ \sigma^{n-1}$"> + in itself, but only the values of the stress in the quadrature points on all cells. This, however, is a simpler task: we only have to provide a data structure that is able to hold one symmetric tensor of rank 2 for each quadrature point on all cells (or, since we compute in parallel, all @@ -723,22 +781,26 @@ end of each time step we then only have to evaluate $ \varepsilon(\Delta \vec u^n(\vec x_q))$, multiply it by the stress-strain tensor $ \varepsilon(\Delta \vec u^n(\vec x_q))$ +, multiply it by the stress-strain tensor $ C$, and use the + ALT="$ C$"> +, and use the result to update the stress $ \sigma^n(\vec x_q)$ at quadrature point $ \sigma^n(\vec x_q)$ + at quadrature point $ q$. + ALT="$ q$"> +.

The second complication is not visible in our notation as chosen above. It is @@ -746,23 +808,26 @@ due to the fact that we compute $ \Delta u^n$ on the domain $ \Omega(t_{n-1})$, + ALT="$ \Omega(t_{n-1})$"> +, and then use this displacement increment to both update the stress as well as move the mesh nodes around to get to $ \Omega(t_n)$ on which the next increment + ALT="$ \Omega(t_n)$"> + on which the next increment is computed. What we have to make sure, in this context, is that moving the mesh does not only involve moving around the nodes, but also making corresponding changes to the stress variable: the updated stress is a variable @@ -772,9 +837,10 @@ can be understood as follows: locally, the incremental deformation $ \Delta\vec u$ can be decomposed into three parts, a linear translation (the constant part + ALT="$ \Delta\vec u$"> + can be decomposed into three parts, a linear translation (the constant part of the displacement field in the neighborhood of a point), a dilational component (that part of the gradient of the displacement field that has a nonzero divergence), and a rotation. A linear translation of the material does @@ -786,48 +852,54 @@ situation where $ \Delta\vec u=(y, -x)^T$, with which $ \varepsilon(\Delta \vec u)=0$). Nevertheless, if the the material was pre-stressed in a certain + ALT="$ \varepsilon(\Delta \vec u)=0$"> +). Nevertheless, if the the material was pre-stressed in a certain direction, then this direction will be rotated along with the material. To this end, we have to define a rotation matrix $ R(\Delta \vec u^n)$ that + ALT="$ R(\Delta \vec u^n)$"> + that describes, in each point the rotation due to the displacement increments. It is not hard to see that the actual dependence of $ R$ on $ \Delta \vec u^n$ can + ALT="$ \Delta \vec u^n$"> + can only be through the curl of the displacement, rather than the displacement itself or its full gradient (as mentioned above, the constant components of the increment describe translations, its divergence the dilational modes, and the curl the rotational modes). Since the exact form of $ R$ is cumbersome, we + ALT="$ R$"> + is cumbersome, we only state it in the program code, and note that the correct updating formula for the stress variable is then

$\displaystyle \sigma^n = R(\Delta \vec u^n)^T [\sigma^{n-1} + C \varepsilon (\Delta \vec u^n)] R(\Delta \vec u^n).$ @@ -941,34 +1013,37 @@ sequence of operations on the present mesh: A^K_{ij}$ --> $ A_{ij} = \sum_K -A^K_{ij}$ built up of local contributions on each cell + built up of local contributions on each cell $ K$ with entries + ALT="$ K$"> + with entries

+ ALT="$\displaystyle A^K_{ij} = (C \varepsilon(\varphi_j), \varepsilon(\varphi_i))_K;$">
$\displaystyle A^K_{ij} = (C \varepsilon(\varphi_i), \varepsilon(\varphi_j))_K;$ (21)

In practice, $ A^K$ is computed using numerical quadrature according to the + ALT="$ A^K$"> + is computed using numerical quadrature according to the formula

$\displaystyle A^K_{ij} = \sum_q w_q [\varepsilon(\varphi_i(\vec x_q)) : C : \varepsilon(\varphi_j(\vec x_q))],$ @@ -976,46 +1051,53 @@ In practice,

with quadrature points $ \vec x_q$ and weights $ \vec x_q$ + and weights $ w_q$. We have built these + ALT="$ w_q$"> +. We have built these contributions before, in step-8 and step-17, but in both of these cases we have done so rather clumsily by using knowledge of how the rank-4 tensor $ C$ + is composed, and considering individual elements of the strain tensors $ \varepsilon(\varphi_i),\varepsilon(\varphi_j)$. This is not really + ALT="$ \varepsilon(\varphi_i),\varepsilon(\varphi_j)$"> +. This is not really convenient, in particular if we want to consider more complicated elasticity models than the isotropic case for which $ C$ had the convenient form + ALT="$ C$"> + had the convenient form $ c_{ijkl} = \lambda \delta_{ij} \delta_{kl} + \mu (\delta_{ik} \delta_{jl} -+ \delta_{il} \delta_{jk})$. While we in fact do not use a more complicated + ALT="$ C_{ijkl} = \lambda \delta_{ij} \delta_{kl} + \mu (\delta_{ik} \delta_{jl} ++ \delta_{il} \delta_{jk})$"> +. While we in fact do not use a more complicated form than this in the present program, we nevertheless want to write it in a way that would easily allow for this. It is then natural to introduce classes that represent symmetric tensors of rank 2 (for the strains and stresses) and 4 (for the stress-strain tensor $ C$). Fortunately, deal.II + ALT="$ C$"> +). Fortunately, deal.II provides these: the SymmetricTensor<rank,dim> class template provides a full-fledged implementation of such tensors of rank rank (which needs to be an even number) and dimension dim. @@ -1023,54 +1105,63 @@ with quadrature points What we then need is two things: a way to create the stress-strain rank-4 tensor $ C$ as well as to create a symmetric tensor of rank 2 (the strain + ALT="$ C$"> + as well as to create a symmetric tensor of rank 2 (the strain tensor) from the gradients of a shape function $ \varphi_i$ at a quadrature + ALT="$ \varphi_i$"> + at a quadrature point $ \vec x_q$ on a given cell. At the top of the implementation of this + ALT="$ \vec x_q$"> + on a given cell. At the top of the implementation of this example program, you will find such functions. The first one, get_stress_strain_tensor, takes two arguments corresponding to the Lamé constants $ \lambda$ and $ \lambda$ + and $ \mu$ and returns the stress-strain tensor + ALT="$ \mu$"> + and returns the stress-strain tensor for the isotropic case corresponding to these constants (in the program, we will choose constants corresponding to steel); it would be simple to replace this function by one that computes this tensor for the anisotropic case, or taking into account crystal symmetries, for example. The second one, get_strain takes an object of type FEValues and indices $ i$ and $ i$ + and $ q$ and returns the symmetric gradient, i.e. the strain, + ALT="$ q$"> + and returns the symmetric gradient, i.e. the strain, corresponding to shape function $ \varphi_i(\vec x_q)$, evaluated on the cell + ALT="$ \varphi_i(\vec x_q)$"> +, evaluated on the cell on which the FEValues object was last reinitialized.

Given this, the innermost loop of assemble_system computes the local contributions to the matrix in the following elegant way (the variable stress_strain_tensor, corresponding to the tensor $ C$, has + ALT="$ C$"> +, has previously been initialized with the result of the first function above): 

 for (unsigned int i=0; i<dofs_per_cell; ++i)
@@ -1101,7 +1192,7 @@ Assembling the local contributions
 

@@ -1111,18 +1202,20 @@ Assembling the local contributions
 

 to the right hand side of (17) is equally
   straightforward (note that we do not consider any boundary tractions $ \vec b$ here). Remember that we only had to store the old stress in the
+ ALT="$ \vec b$">
+ here). Remember that we only had to store the old stress in the
   quadrature points of cells. In the program, we will provide a variable
   local_quadrature_points_data that allows to access the stress
   $ \sigma^{n-1}_q$ in each quadrature point. With this the code for the right
+ ALT="$ \sigma^{n-1}_q$">
+ in each quadrature point. With this the code for the right
   hand side looks as this, again rather elegant:
   
 for (unsigned int i=0; i<dofs_per_cell; ++i)
@@ -1149,21 +1242,24 @@ for (unsigned int i=0; i<dofs_per_cell; ++i)
  $\vec f(\vec x_q) \cdot \varphi_i(\vec x_q)$
  -->
 $ \vec f(\vec x_q) \cdot \varphi_i(\vec x_q)$, we have made use of the fact that for the chosen finite element, only
+ ALT="$ \vec f(\vec x_q) \cdot \varphi_i(\vec x_q)$">
+, we have made use of the fact that for the chosen finite element, only
   one vector component (namely component_i) of $ \varphi_i$ is
+ ALT="$ \varphi_i$">
+ is
   nonzero, and that we therefore also have to consider only one component of
   
 $ \vec f(\vec x_q)$.
+ ALT="$ \vec f(\vec x_q)$">
+.
 
 

 This essentially concludes the new material we present in this function. It
@@ -1188,9 +1284,10 @@ This essentially concludes the new material we present in this function. It
  $\Delta \vec u^n$
  -->
 $ \Delta \vec u^n$ computed before, we update the stress values in all quadrature points
+ ALT="$ \Delta \vec u^n$">
+ computed before, we update the stress values in all quadrature points
   according to (18) and (20),
   including the rotation of the coordinate system.
 
@@ -1260,7 +1357,7 @@ elastic energy
 

 

 \begin{gather*}\begin{split}f^K_i &= (\vec f, \varphi_i)_K -(\sigma^{n-1},\varep...
 ...gma^{n-1}_q : \varepsilon(\varphi_i(\vec x_q)) \right\} \end{split}\end{gather*}
 

 

 $\displaystyle E(\vec u) = (\varepsilon(\vec u), C\varepsilon(\vec u))_{\Omega} - (\vec f, \vec u)_{\Omega} - (\vec b, \vec u)_{\Gamma_N},$
 
@@ -1272,7 +1369,7 @@ subject to the constraint
 

 

 $\displaystyle f(\sigma(\vec u)) \le 0$
 
@@ -1315,7 +1412,7 @@ Incompressibility is characterized by Poisson's ratio
 

 

 $\displaystyle \nu = \frac{\lambda}{2(\lambda+\mu)},$
 
@@ -1326,23 +1423,27 @@ where 
 $ \lambda,\mu$ are the Lamé constants of the material.
+ ALT="$ \lambda,\mu$">
+ are the Lamé constants of the material.
 Physical constraints indicate that 
 $ -1\le \nu\le \tfrac 12$. If $ -1\le \nu\le \tfrac 12$
+. If $ \nu$
+
 approaches $ \tfrac 12$, then the material becomes incompressible. In that
+ ALT="$ \tfrac 12$">
+, then the material becomes incompressible. In that
 case, pure displacement-based formulations are no longer appropriate for the
 solution of such problems, and stabilization techniques have to be employed
 for a stable and accurate solution. The book and paper cited above give
@@ -1366,16 +1467,16 @@ you wish. The main complication one has to overcome is that one has to
 transfer the data that is stored in the quadrature points of the cells of the
 old mesh to the new mesh, preferably by some sort of projection scheme. This
 is only slightly messy in the sequential case; in fact, the functions
-FETools::get_projection_from_quadrature_points_matrix will do
-the projection, and the FiniteElement::get_restriction_matrix and
-FiniteElement::get_prolongation_matrix functions will do the
-transfer between mother and child cells. However, it becomes complicated once
-we run the program in parallel, since then each process only stores this data
-for the cells it owned on the old mesh, and it may need to know the values of
-the quadrature point data on other cells if the corresponding cells on the new
-mesh are assigned to this process after subdividing the new mesh. A global
-communication of these data elements is therefore necessary, making the entire
-process a little more unpleasant.
+FETools :: get_projection_from_quadrature_points_matrix will do
+the projection, and the FiniteElement :: get_restriction_matrix and
+FiniteElement :: get_prolongation_matrix functions will do the
+transfer between mother and child cells. However, it becomes complicated
+once we run the program in parallel, since then each process only stores this
+data for the cells it owned on the old mesh, and it may need to know the
+values of the quadrature point data on other cells if the corresponding cells
+on the new mesh are assigned to this process after subdividing the new mesh. A
+global communication of these data elements is therefore necessary, making the
+entire process a little more unpleasant.
 
 

 
@@ -1411,4 +1512,3 @@ although this is already done in the distributed version of the Makefile.
 Conversely, if you want to run the program in 2d (after making the necessary
 changes to accommodate for a 2d geometry), you have to change the Makefile
 back to allow for 2d.
-
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