]> https://gitweb.dealii.org/ - dealii-svn.git/commitdiff
Use a number of suggestions by Francesco Freddi.
authorwolf <wolf@0785d39b-7218-0410-832d-ea1e28bc413d>
Thu, 20 Oct 2005 22:04:32 +0000 (22:04 +0000)
committerwolf <wolf@0785d39b-7218-0410-832d-ea1e28bc413d>
Thu, 20 Oct 2005 22:04:32 +0000 (22:04 +0000)
git-svn-id: https://svn.dealii.org/trunk@11640 0785d39b-7218-0410-832d-ea1e28bc413d

75 files changed:
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.html
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro.pdf
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img100.png [new file with mode: 0644]
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img101.png [new file with mode: 0644]
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img28.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img29.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img30.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img31.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img32.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img34.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img35.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img36.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img37.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img38.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img39.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img40.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img41.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img42.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img43.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img44.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img45.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img46.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img47.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img48.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img49.png
deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img50.png
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deal.II/doc/tutorial/chapter-2.step-by-step/step-18.data/intro/img99.png

index 5b8eeb9c1a588b125fae160c2ae7e25176dc0d63..22a9f287bae5de3a91f8d025d27d6f5bf56b3ec4 100644 (file)
@@ -6,6 +6,7 @@
 file by <a href="step-18.data/intro.pdf">clicking here</a>]
 </p>
 
+
 <P>
 This tutorial program is another one in the series on the elasticity problem
 that we have already started with step-8 and step-17. It extends it into two
@@ -294,23 +295,31 @@ possibly time-varying force function <!-- MATH
  WIDTH="47" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
  SRC="step-18.data/intro/img27.png"
  ALT="$ \vec f(\vec x,t)$">
-.
+. The changes in
+configuration can therefore be considered as being stationary
+instantaneously. An alternative view of this is that <IMG
+ WIDTH="10" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
+ SRC="step-18.data/intro/img28.png"
+ ALT="$ t$">
+ is not really a time
+variable, but only a time-like parameter that governs the evolution of the
+problem.
 
 <P>
 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 <IMG
+introduce a tensorial stress variable <IMG
  WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img28.png"
+ SRC="step-18.data/intro/img29.png"
  ALT="$ \sigma$">
-, and write the differential equations in
-terms of the stress:
+, and write the differential
+equations in terms of the stress:
 <P></P>
 <DIV ALIGN="CENTER"><TABLE CELLPADDING="0" WIDTH="100%" ALIGN="CENTER">
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="RIGHT"><IMG
  WIDTH="54" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img29.png"
+ SRC="step-18.data/intro/img30.png"
  ALT="$\displaystyle - \div\sigma$"></TD>
 <TD NOWRAP ALIGN="LEFT"><IMG
  WIDTH="29" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
@@ -319,7 +328,7 @@ terms of the stress:
 <TD>&nbsp;</TD>
 <TD NOWRAP ALIGN="LEFT">in <IMG
  WIDTH="34" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img30.png"
+ SRC="step-18.data/intro/img31.png"
  ALT="$ \Omega(t)$">
 <IMG
  WIDTH="9" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
@@ -342,7 +351,7 @@ terms of the stress:
  -->
 <IMG
  WIDTH="86" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img31.png"
+ SRC="step-18.data/intro/img32.png"
  ALT="$ \Gamma_D\subset\partial\Omega(t)$">
 <IMG
  WIDTH="9" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
@@ -365,7 +374,7 @@ terms of the stress:
  -->
 <IMG
  WIDTH="116" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img32.png"
+ SRC="step-18.data/intro/img33.png"
  ALT="$ \Gamma_N=\partial\Omega(t)\backslash\Gamma_D$">
 <IMG
  WIDTH="9" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
@@ -377,7 +386,7 @@ terms of the stress:
 <BR CLEAR="ALL"><P></P>
 Note that these equations are posed on a domain <IMG
  WIDTH="34" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img30.png"
+ SRC="step-18.data/intro/img31.png"
  ALT="$ \Omega(t)$">
  that
 changes with time, with the boundary moving according to the
@@ -386,17 +395,17 @@ displacements <!-- MATH
  -->
 <IMG
  WIDTH="49" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img33.png"
+ SRC="step-18.data/intro/img34.png"
  ALT="$ \vec u(\vec x,t)$">
  of the points on the boundary. To
-complete this system, we have to specify the relationship between the
-stress and the strain, as follows:
+complete this system, we have to specify the incremental relationship between
+the stress and the strain, as follows:
 <P></P>
 <DIV ALIGN="CENTER"><TABLE CELLPADDING="0" WIDTH="100%" ALIGN="CENTER">
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="82" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img34.png"
+ SRC="step-18.data/intro/img35.png"
  ALT="$\displaystyle \dot\sigma = C \varepsilon (\dot{\vec u}),$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 <A NAME="eq:stress-strain">(11)</A></TD></TR>
@@ -404,7 +413,7 @@ stress and the strain, as follows:
 <BR CLEAR="ALL"><P></P>
 where a dot indicates a time derivative. Both the stress <IMG
  WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img28.png"
+ SRC="step-18.data/intro/img29.png"
  ALT="$ \sigma$">
  and the
 strain <!-- MATH
@@ -412,7 +421,7 @@ strain <!-- MATH
  -->
 <IMG
  WIDTH="34" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img35.png"
+ SRC="step-18.data/intro/img36.png"
  ALT="$ \varepsilon(\vec u)$">
  are symmetric tensors of rank 2.
 
@@ -427,7 +436,7 @@ 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 <IMG
  WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img36.png"
+ SRC="step-18.data/intro/img37.png"
  ALT="$ n$">
 :
 <P></P>
@@ -435,7 +444,7 @@ discrete equilibrium of force at time step <IMG
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="106" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img37.png"
+ SRC="step-18.data/intro/img38.png"
  ALT="$\displaystyle -\div\sigma^n = f^n,$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 (12)</TD></TR>
@@ -447,7 +456,7 @@ where
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="167" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img38.png"
+ SRC="step-18.data/intro/img39.png"
  ALT="$\displaystyle \sigma^n = \sigma^{n-1} + C \varepsilon (\Delta \vec u^n),$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 (13)</TD></TR>
@@ -458,25 +467,33 @@ and <!-- MATH
  -->
 <IMG
  WIDTH="37" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img39.png"
+ SRC="step-18.data/intro/img40.png"
  ALT="$ \Delta \vec u^n$">
  the incremental displacement for time step
 <IMG
  WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img36.png"
+ SRC="step-18.data/intro/img37.png"
  ALT="$ n$">
-. This way, if we want to solve for the displacement increment, we
+. In addition, we have to specify initial data <!-- MATH
+ $\vec u(\cdot,0)=\vec u_0$
+ -->
+<IMG
+ WIDTH="85" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
+ SRC="step-18.data/intro/img41.png"
+ ALT="$ \vec u(\cdot,0)=\vec u_0$">
+. 
+This way, if we want to solve for the displacement increment, we
 have to solve the following system:
 <P></P>
 <DIV ALIGN="CENTER"><TABLE CELLPADDING="0" WIDTH="100%" ALIGN="CENTER">
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="RIGHT"><IMG
  WIDTH="108" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img40.png"
+ SRC="step-18.data/intro/img42.png"
  ALT="$\displaystyle - \div C \varepsilon(\Delta\vec u^n)$"></TD>
 <TD NOWRAP ALIGN="LEFT"><IMG
  WIDTH="110" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img41.png"
+ SRC="step-18.data/intro/img43.png"
  ALT="$\displaystyle = \vec f + \div\sigma^{n-1}$"></TD>
 <TD>&nbsp;</TD>
 <TD NOWRAP ALIGN="LEFT">in <!-- MATH
@@ -484,7 +501,7 @@ have to solve the following system:
  -->
 <IMG
  WIDTH="59" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img42.png"
+ SRC="step-18.data/intro/img44.png"
  ALT="$ \Omega(t_{n-1})$">
 <IMG
  WIDTH="9" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
@@ -495,11 +512,11 @@ have to solve the following system:
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="RIGHT"><IMG
  WIDTH="71" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img43.png"
+ SRC="step-18.data/intro/img45.png"
  ALT="$\displaystyle \Delta \vec u^n(\vec x,t)$"></TD>
 <TD NOWRAP ALIGN="LEFT"><IMG
  WIDTH="196" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img44.png"
+ SRC="step-18.data/intro/img46.png"
  ALT="$\displaystyle = \vec d(\vec x,t_n) - \vec d(\vec x,t_{n-1}) \qquad$"></TD>
 <TD>&nbsp;</TD>
 <TD NOWRAP ALIGN="LEFT">on <!-- MATH
@@ -507,7 +524,7 @@ have to solve the following system:
  -->
 <IMG
  WIDTH="111" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img45.png"
+ SRC="step-18.data/intro/img47.png"
  ALT="$ \Gamma_D\subset\partial\Omega(t_{n-1})$">
 <IMG
  WIDTH="9" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
@@ -518,11 +535,11 @@ have to solve the following system:
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="RIGHT"><IMG
  WIDTH="119" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img46.png"
+ SRC="step-18.data/intro/img48.png"
  ALT="$\displaystyle \vec n \ C \varepsilon(\Delta \vec u^n(\vec x,t))$"></TD>
 <TD NOWRAP ALIGN="LEFT"><IMG
  WIDTH="196" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img47.png"
+ SRC="step-18.data/intro/img49.png"
  ALT="$\displaystyle = \vec b(\vec x,t_n)-\vec b(\vec x,t_{n-1}) \qquad$"></TD>
 <TD>&nbsp;</TD>
 <TD NOWRAP ALIGN="LEFT">on <!-- MATH
@@ -530,7 +547,7 @@ have to solve the following system:
  -->
 <IMG
  WIDTH="141" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img48.png"
+ SRC="step-18.data/intro/img50.png"
  ALT="$ \Gamma_N=\partial\Omega(t_{n-1})\backslash\Gamma_D$">
 <IMG
  WIDTH="9" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
@@ -547,7 +564,7 @@ finite element formulation, reads as follows: find <!-- MATH
  -->
 <IMG
  WIDTH="394" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img49.png"
+ SRC="step-18.data/intro/img51.png"
  ALT="$ \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})\}$">
 
@@ -557,7 +574,7 @@ such that
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="412" HEIGHT="78" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img50.png"
+ SRC="step-18.data/intro/img52.png"
  ALT="\begin{gather*}\begin{split}(C \varepsilon(\Delta\vec u^n), \varepsilon(\varphi)...
 ...in H^1(\Omega(t_{n-1}))^d: \vec v\vert _{\Gamma_D}=0\}. \end{split}\end{gather*}"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
@@ -570,7 +587,7 @@ are no boundary forces, i.e.&nbsp;<!-- MATH
  -->
 <IMG
  WIDTH="44" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img51.png"
+ SRC="step-18.data/intro/img53.png"
  ALT="$ \vec b = 0$">
 , and that the deformation of the
 body is driven by body forces <IMG
@@ -580,7 +597,7 @@ body is driven by body forces <IMG
  and prescribed boundary displacements
 <IMG
  WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img52.png"
+ SRC="step-18.data/intro/img54.png"
  ALT="$ \vec d$">
  alone. It is also worth noting that when integrating by parts, we
 would get terms of the form <!-- MATH
@@ -589,7 +606,7 @@ would get terms of the form <!-- MATH
  -->
 <IMG
  WIDTH="158" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img53.png"
+ SRC="step-18.data/intro/img55.png"
  ALT="$ (C \varepsilon(\Delta\vec u^n), \nabla \varphi
 )_{\Omega(t_{n-1})}$">
 , but that we replace it with the term involving the
@@ -598,14 +615,14 @@ symmetric gradient <!-- MATH
  -->
 <IMG
  WIDTH="35" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img54.png"
+ SRC="step-18.data/intro/img56.png"
  ALT="$ \varepsilon(\varphi)$">
  instead of <!-- MATH
  $\nabla\varphi$
  -->
 <IMG
  WIDTH="28" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img55.png"
+ SRC="step-18.data/intro/img57.png"
  ALT="$ \nabla\varphi$">
 . Due to
 the symmetry of <IMG
@@ -619,7 +636,7 @@ non-symmetric.
 <P>
 The system at time step <IMG
  WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img36.png"
+ SRC="step-18.data/intro/img37.png"
  ALT="$ n$">
 , to be solved on the old domain
 <!-- MATH
@@ -627,7 +644,7 @@ The system at time step <IMG
  -->
 <IMG
  WIDTH="59" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img42.png"
+ SRC="step-18.data/intro/img44.png"
  ALT="$ \Omega(t_{n-1})$">
 , has exactly the form of a stationary elastic
 problem, and is therefore similar to what we have already implemented
@@ -639,7 +656,7 @@ continuous finite elements.
 There are differences, however:
 
 <OL>
-<LI>We have to move the mesh after each time step, in order to be
+<LI>We have to move (update) the mesh after each time step, in order to be 
   able to solve the next time step on a new domain;
 
 <P>
@@ -649,7 +666,7 @@ There are differences, however:
  -->
 <IMG
  WIDTH="39" HEIGHT="16" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img56.png"
+ SRC="step-18.data/intro/img58.png"
  ALT="$ \sigma^{n-1}$">
  to compute the next incremental
   displacement, i.e.&nbsp;we need to compute it at the end of the time step
@@ -672,12 +689,12 @@ Updating the stress variable</A>
 <P>
 As indicated above, we need to have the stress variable <IMG
  WIDTH="23" HEIGHT="17" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img57.png"
+ SRC="step-18.data/intro/img59.png"
  ALT="$ \sigma^n$">
  available
 when computing time step <IMG
  WIDTH="41" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img58.png"
+ SRC="step-18.data/intro/img60.png"
  ALT="$ n+1$">
 , and we can compute it using
 <P></P>
@@ -685,7 +702,7 @@ when computing time step <IMG
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="167" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img59.png"
+ SRC="step-18.data/intro/img61.png"
  ALT="$\displaystyle \sigma^n = \sigma^{n-1} + C \varepsilon (\Delta \vec u^n).$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 <A NAME="eq:stress-update">(18)</A></TD></TR>
@@ -694,7 +711,7 @@ when computing time step <IMG
 There are, despite the apparent simplicity of this equation, two questions
 that we need to discuss. The first concerns the way we store <IMG
  WIDTH="23" HEIGHT="17" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img57.png"
+ SRC="step-18.data/intro/img59.png"
  ALT="$ \sigma^n$">
 : even
 if we compute the incremental updates <!-- MATH
@@ -702,7 +719,7 @@ if we compute the incremental updates <!-- MATH
  -->
 <IMG
  WIDTH="37" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img39.png"
+ SRC="step-18.data/intro/img40.png"
  ALT="$ \Delta \vec u^n$">
  using lowest-order
 finite elements, then its symmetric gradient <!-- MATH
@@ -710,7 +727,7 @@ finite elements, then its symmetric gradient <!-- MATH
  -->
 <IMG
  WIDTH="56" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img60.png"
+ SRC="step-18.data/intro/img62.png"
  ALT="$ \varepsilon(\Delta\vec u^n)$">
  is
 in general still a function that is not easy to describe. In particular, it is
@@ -723,7 +740,7 @@ stress-strain tensors <IMG
  it is not even a bi- or trilinear function. Thus, it
 is a priori not clear how to store <IMG
  WIDTH="23" HEIGHT="17" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img57.png"
+ SRC="step-18.data/intro/img59.png"
  ALT="$ \sigma^n$">
  in a computer program.
 
@@ -735,7 +752,7 @@ require the stress is in the term
  -->
 <IMG
  WIDTH="135" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img61.png"
+ SRC="step-18.data/intro/img63.png"
  ALT="$ (\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})}$">
 . In practice, we of
 course replace this term by numerical quadrature:
@@ -744,7 +761,7 @@ course replace this term by numerical quadrature:
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="518" HEIGHT="52" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img62.png"
+ SRC="step-18.data/intro/img64.png"
  ALT="$\displaystyle (\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})} = \sum_{K\s...
 ...athbb{T}}} \sum_q w_q \ \sigma^{n-1}(\vec x_q) : \varepsilon(\varphi(\vec x_q),$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
@@ -753,16 +770,16 @@ course replace this term by numerical quadrature:
 <BR CLEAR="ALL"><P></P>
 where <IMG
  WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img63.png"
+ SRC="step-18.data/intro/img65.png"
  ALT="$ w_q$">
  are the quadrature weights and <IMG
  WIDTH="21" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img64.png"
+ SRC="step-18.data/intro/img66.png"
  ALT="$ \vec x_q$">
  the quadrature points on
 cell <IMG
  WIDTH="19" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img65.png"
+ SRC="step-18.data/intro/img67.png"
  ALT="$ K$">
 . This should make clear that what we really need is not the stress
 <!-- MATH
@@ -770,7 +787,7 @@ cell <IMG
  -->
 <IMG
  WIDTH="39" HEIGHT="16" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img56.png"
+ SRC="step-18.data/intro/img58.png"
  ALT="$ \sigma^{n-1}$">
  in itself, but only the values of the stress in the quadrature
 points on all cells. This, however, is a simpler task: we only have to provide
@@ -782,7 +799,7 @@ end of each time step we then only have to evaluate <!-- MATH
  -->
 <IMG
  WIDTH="85" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img66.png"
+ SRC="step-18.data/intro/img68.png"
  ALT="$ \varepsilon(\Delta \vec u^n(\vec x_q))$">
 , multiply it by the stress-strain tensor <IMG
  WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
@@ -794,11 +811,11 @@ result to update the stress <!-- MATH
  -->
 <IMG
  WIDTH="52" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img67.png"
+ SRC="step-18.data/intro/img69.png"
  ALT="$ \sigma^n(\vec x_q)$">
  at quadrature point <IMG
  WIDTH="12" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img68.png"
+ SRC="step-18.data/intro/img70.png"
  ALT="$ q$">
 .
 
@@ -809,14 +826,14 @@ due to the fact that we compute <!-- MATH
  -->
 <IMG
  WIDTH="36" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img69.png"
+ SRC="step-18.data/intro/img71.png"
  ALT="$ \Delta u^n$">
  on the domain <!-- MATH
  $\Omega(t_{n-1})$
  -->
 <IMG
  WIDTH="59" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img42.png"
+ SRC="step-18.data/intro/img44.png"
  ALT="$ \Omega(t_{n-1})$">
 ,
 and then use this displacement increment to both update the stress as well as
@@ -825,7 +842,7 @@ move the mesh nodes around to get to <!-- MATH
  -->
 <IMG
  WIDTH="43" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img70.png"
+ SRC="step-18.data/intro/img72.png"
  ALT="$ \Omega(t_n)$">
  on which the next increment
 is computed. What we have to make sure, in this context, is that moving the
@@ -838,10 +855,11 @@ can be understood as follows: locally, the incremental deformation <!-- MATH
  -->
 <IMG
  WIDTH="28" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img71.png"
+ SRC="step-18.data/intro/img73.png"
  ALT="$ \Delta\vec u$">
  can be decomposed into three parts, a linear translation (the constant part
-of the displacement field in the neighborhood of a point), a dilational
+of the displacement increment field in the neighborhood of a point), a
+dilational 
 component (that part of the gradient of the displacement field that has a
 nonzero divergence), and a rotation. A linear translation of the material does
 not affect the stresses that are frozen into it - the stress values are
@@ -853,14 +871,14 @@ situation where <!-- MATH
  -->
 <IMG
  WIDTH="108" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img72.png"
+ SRC="step-18.data/intro/img74.png"
  ALT="$ \Delta\vec u=(y, -x)^T$">
 , with which <!-- MATH
  $\varepsilon(\Delta \vec u)=0$
  -->
 <IMG
  WIDTH="77" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img73.png"
+ SRC="step-18.data/intro/img75.png"
  ALT="$ \varepsilon(\Delta \vec u)=0$">
 ). Nevertheless, if the the material was pre-stressed in a certain
 direction, then this direction will be rotated along with the material.  To
@@ -869,20 +887,20 @@ this end, we have to define a rotation matrix <!-- MATH
  -->
 <IMG
  WIDTH="61" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img74.png"
+ SRC="step-18.data/intro/img76.png"
  ALT="$ R(\Delta \vec u^n)$">
  that
 describes, in each point the rotation due to the displacement increments. It
 is not hard to see that the actual dependence of <IMG
  WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img75.png"
+ SRC="step-18.data/intro/img77.png"
  ALT="$ R$">
  on <!-- MATH
  $\Delta \vec u^n$
  -->
 <IMG
  WIDTH="37" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img39.png"
+ SRC="step-18.data/intro/img40.png"
  ALT="$ \Delta \vec u^n$">
  can
 only be through the curl of the displacement, rather than the displacement
@@ -890,7 +908,7 @@ itself or its full gradient (as mentioned above, the constant components of
 the increment describe translations, its divergence the dilational modes, and
 the curl the rotational modes). Since the exact form of <IMG
  WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img75.png"
+ SRC="step-18.data/intro/img77.png"
  ALT="$ R$">
  is cumbersome, we
 only state it in the program code, and note that the correct updating formula
@@ -900,7 +918,7 @@ for the stress variable is then
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="299" HEIGHT="38" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img76.png"
+ SRC="step-18.data/intro/img78.png"
  ALT="$\displaystyle \sigma^n = R(\Delta \vec u^n)^T [\sigma^{n-1} + C \varepsilon (\Delta \vec u^n)] R(\Delta \vec u^n).$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 <A NAME="eq:stress-update+rot">(20)</A></TD></TR>
@@ -1008,18 +1026,19 @@ sequence of operations on the present mesh:
 <LI><TT>assemble_system ()</TT> [via <TT>solve_timestep ()</TT>]:
   This first function is also the most interesting one. It assembles the
   linear system corresponding to the discretized version of equation
-  (<A HREF="#eq:linear-system">17</A>). This leads to a system matrix <!-- MATH
+  (<A HREF="#eq:linear-system"><IMG  ALIGN="BOTTOM" BORDER="1" ALT="[*]"
+ SRC="step-18.data/intro/file:/home/bangerth/bin/share/lib/latex2html/icons/crossref.png"></A>). This leads to a system matrix <!-- MATH
  $A_{ij} = \sum_K
 A^K_{ij}$
  -->
 <IMG
  WIDTH="105" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img77.png"
+ SRC="step-18.data/intro/img79.png"
  ALT="$ A_{ij} = \sum_K
 A^K_{ij}$">
  built up of local contributions on each cell <IMG
  WIDTH="19" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img65.png"
+ SRC="step-18.data/intro/img67.png"
  ALT="$ K$">
  with entries
   <P></P>
@@ -1027,7 +1046,7 @@ A^K_{ij}$">
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="171" HEIGHT="38" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img78.png"
+ SRC="step-18.data/intro/img80.png"
  ALT="$\displaystyle A^K_{ij} = (C \varepsilon(\varphi_j), \varepsilon(\varphi_i))_K;$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 (21)</TD></TR>
@@ -1035,7 +1054,7 @@ A^K_{ij}$">
 <BR CLEAR="ALL"><P></P>
 In practice, <IMG
  WIDTH="29" HEIGHT="16" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img79.png"
+ SRC="step-18.data/intro/img81.png"
  ALT="$ A^K$">
  is computed using numerical quadrature according to the
   formula
@@ -1044,7 +1063,7 @@ In practice, <IMG
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="277" HEIGHT="52" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img80.png"
+ SRC="step-18.data/intro/img82.png"
  ALT="$\displaystyle A^K_{ij} = \sum_q w_q [\varepsilon(\varphi_i(\vec x_q)) : C : \varepsilon(\varphi_j(\vec x_q))],$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 (22)</TD></TR>
@@ -1052,11 +1071,11 @@ In practice, <IMG
 <BR CLEAR="ALL"><P></P>
 with quadrature points <IMG
  WIDTH="21" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img64.png"
+ SRC="step-18.data/intro/img66.png"
  ALT="$ \vec x_q$">
  and weights <IMG
  WIDTH="23" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img63.png"
+ SRC="step-18.data/intro/img65.png"
  ALT="$ w_q$">
 . We have built these
   contributions before, in step-8 and step-17, but in both of these cases we
@@ -1071,7 +1090,7 @@ with quadrature points <IMG
  -->
 <IMG
  WIDTH="84" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img81.png"
+ SRC="step-18.data/intro/img83.png"
  ALT="$ \varepsilon(\varphi_i),\varepsilon(\varphi_j)$">
 . This is not really
   convenient, in particular if we want to consider more complicated elasticity
@@ -1086,7 +1105,7 @@ with quadrature points <IMG
  -->
 <IMG
  WIDTH="241" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img82.png"
+ SRC="step-18.data/intro/img84.png"
  ALT="$ C_{ijkl} = \lambda \delta_{ij} \delta_{kl} + \mu (\delta_{ik} \delta_{jl}
 + \delta_{il} \delta_{jk})$">
 . While we in fact do not use a more complicated
@@ -1111,23 +1130,23 @@ What we then need is two things: a way to create the stress-strain rank-4
  as well as to create a symmetric tensor of rank 2 (the strain
   tensor) from the gradients of a shape function <IMG
  WIDTH="20" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img83.png"
+ SRC="step-18.data/intro/img85.png"
  ALT="$ \varphi_i$">
  at a quadrature
   point <IMG
  WIDTH="21" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img64.png"
+ SRC="step-18.data/intro/img66.png"
  ALT="$ \vec x_q$">
  on a given cell. At the top of the implementation of this
   example program, you will find such functions. The first one,
   <TT>get_stress_strain_tensor</TT>, takes two arguments corresponding to
   the Lam&#233; constants <IMG
  WIDTH="14" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img84.png"
+ SRC="step-18.data/intro/img86.png"
  ALT="$ \lambda$">
  and <IMG
  WIDTH="14" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img85.png"
+ SRC="step-18.data/intro/img87.png"
  ALT="$ \mu$">
  and returns the stress-strain tensor
   for the isotropic case corresponding to these constants (in the program, we
@@ -1137,11 +1156,11 @@ What we then need is two things: a way to create the stress-strain rank-4
   <TT>get_strain</TT> takes an object of type <TT>FEValues</TT> and indices
   <IMG
  WIDTH="10" HEIGHT="17" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img86.png"
+ SRC="step-18.data/intro/img88.png"
  ALT="$ i$">
  and <IMG
  WIDTH="12" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img68.png"
+ SRC="step-18.data/intro/img70.png"
  ALT="$ q$">
  and returns the symmetric gradient, i.e. the strain,
   corresponding to shape function <!-- MATH
@@ -1149,7 +1168,7 @@ What we then need is two things: a way to create the stress-strain rank-4
  -->
 <IMG
  WIDTH="49" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img87.png"
+ SRC="step-18.data/intro/img89.png"
  ALT="$ \varphi_i(\vec x_q)$">
 , evaluated on the cell
   on which the <TT>FEValues</TT> object was last reinitialized.
@@ -1193,17 +1212,18 @@ Assembling the local contributions
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="331" HEIGHT="71" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img88.png"
+ SRC="step-18.data/intro/img90.png"
  ALT="\begin{gather*}\begin{split}f^K_i &amp;= (\vec f, \varphi_i)_K -(\sigma^{n-1},\varep...
 ...gma^{n-1}_q : \varepsilon(\varphi_i(\vec x_q)) \right\} \end{split}\end{gather*}"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 (23)</TD></TR>
 </TABLE></DIV>
 <BR CLEAR="ALL"><P></P>
-to the right hand side of (<A HREF="#eq:linear-system">17</A>) is equally
+to the right hand side of (<A HREF="#eq:linear-system"><IMG  ALIGN="BOTTOM" BORDER="1" ALT="[*]"
+ SRC="step-18.data/intro/file:/home/bangerth/bin/share/lib/latex2html/icons/crossref.png"></A>) is equally
   straightforward (note that we do not consider any boundary tractions <IMG
  WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img89.png"
+ SRC="step-18.data/intro/img91.png"
  ALT="$ \vec b$">
  here). Remember that we only had to store the old stress in the
   quadrature points of cells. In the program, we will provide a variable
@@ -1213,7 +1233,7 @@ to the right hand side of (<A HREF="#eq:linear-system">17</A>) is equally
  -->
 <IMG
  WIDTH="39" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img90.png"
+ SRC="step-18.data/intro/img92.png"
  ALT="$ \sigma^{n-1}_q$">
  in each quadrature point. With this the code for the right
   hand side looks as this, again rather elegant:
@@ -1243,12 +1263,12 @@ for (unsigned int i=0; i&lt;dofs_per_cell; ++i)
  -->
 <IMG
  WIDTH="97" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img91.png"
+ SRC="step-18.data/intro/img93.png"
  ALT="$ \vec f(\vec x_q) \cdot \varphi_i(\vec x_q)$">
 , we have made use of the fact that for the chosen finite element, only
   one vector component (namely <TT>component_i</TT>) of <IMG
  WIDTH="20" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img83.png"
+ SRC="step-18.data/intro/img85.png"
  ALT="$ \varphi_i$">
  is
   nonzero, and that we therefore also have to consider only one component of
@@ -1257,7 +1277,7 @@ for (unsigned int i=0; i&lt;dofs_per_cell; ++i)
  -->
 <IMG
  WIDTH="41" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img92.png"
+ SRC="step-18.data/intro/img94.png"
  ALT="$ \vec f(\vec x_q)$">
 .
 
@@ -1285,10 +1305,12 @@ This essentially concludes the new material we present in this function. It
  -->
 <IMG
  WIDTH="37" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img39.png"
+ SRC="step-18.data/intro/img40.png"
  ALT="$ \Delta \vec u^n$">
  computed before, we update the stress values in all quadrature points
-  according to (<A HREF="#eq:stress-update">18</A>) and (<A HREF="#eq:stress-update+rot">20</A>),
+  according to (<A HREF="#eq:stress-update"><IMG  ALIGN="BOTTOM" BORDER="1" ALT="[*]"
+ SRC="step-18.data/intro/file:/home/bangerth/bin/share/lib/latex2html/icons/crossref.png"></A>) and (<A HREF="#eq:stress-update+rot"><IMG  ALIGN="BOTTOM" BORDER="1" ALT="[*]"
+ SRC="step-18.data/intro/file:/home/bangerth/bin/share/lib/latex2html/icons/crossref.png"></A>),
   including the rotation of the coordinate system.
 
 <P>
@@ -1346,7 +1368,8 @@ Plasticity models.</A>
 </H4> The most obvious extension is to use a more
 realistic material model for large-scale quasistatic deformation. The natural
 choice for this would be plasticity, in which a nonlinear relationship between
-stress and strain replaces equation (<A HREF="#eq:stress-strain">11</A>). Plasticity
+stress and strain replaces equation (<A HREF="#eq:stress-strain"><IMG  ALIGN="BOTTOM" BORDER="1" ALT="[*]"
+ SRC="step-18.data/intro/file:/home/bangerth/bin/share/lib/latex2html/icons/crossref.png"></A>). Plasticity
 models are usually rather complicated to program since the stress-strain
 dependence is generally non-smooth. The material can be thought of being able
 to withstand only a maximal stress (the yield stress) after which it starts to
@@ -1358,7 +1381,7 @@ elastic energy
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="311" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img93.png"
+ SRC="step-18.data/intro/img95.png"
  ALT="$\displaystyle E(\vec u) = (\varepsilon(\vec u), C\varepsilon(\vec u))_{\Omega} - (\vec f, \vec u)_{\Omega} - (\vec b, \vec u)_{\Gamma_N},$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 (24)</TD></TR>
@@ -1370,7 +1393,7 @@ subject to the constraint
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="88" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img94.png"
+ SRC="step-18.data/intro/img96.png"
  ALT="$\displaystyle f(\sigma(\vec u)) \le 0$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 (25)</TD></TR>
@@ -1413,7 +1436,7 @@ Incompressibility is characterized by Poisson's ratio
 <TR VALIGN="MIDDLE">
 <TD NOWRAP ALIGN="CENTER"><IMG
  WIDTH="101" HEIGHT="52" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img95.png"
+ SRC="step-18.data/intro/img97.png"
  ALT="$\displaystyle \nu = \frac{\lambda}{2(\lambda+\mu)},$"></TD>
 <TD NOWRAP WIDTH="10" ALIGN="RIGHT">
 &nbsp;&nbsp;&nbsp;</TD></TR>
@@ -1424,7 +1447,7 @@ where <!-- MATH
  -->
 <IMG
  WIDTH="30" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img96.png"
+ SRC="step-18.data/intro/img98.png"
  ALT="$ \lambda,\mu$">
  are the Lam&#233; constants of the material.
 Physical constraints indicate that <!-- MATH
@@ -1432,16 +1455,17 @@ Physical constraints indicate that <!-- MATH
  -->
 <IMG
  WIDTH="86" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img97.png"
+ SRC="step-18.data/intro/img99.png"
  ALT="$ -1\le \nu\le \tfrac 12$">
-. If <IMG
+ (the condition
+also follows from mathematical stability considerations). If <IMG
  WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
- SRC="step-18.data/intro/img98.png"
+ SRC="step-18.data/intro/img100.png"
  ALT="$ \nu$">
 
 approaches <IMG
  WIDTH="15" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
- SRC="step-18.data/intro/img99.png"
+ SRC="step-18.data/intro/img101.png"
  ALT="$ \tfrac 12$">
 , then the material becomes incompressible. In that
 case, pure displacement-based formulations are no longer appropriate for the
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