750 * sparsity_pattern.copy_from(dsp);
752 * system_matrix.reinit(sparsity_pattern);
759 * <a name=
"ElasticProblemassemble_system"></a>
760 * <h4>ElasticProblem::assemble_system</h4>
764 * The big changes in
this program are in the creation of
matrix and right
765 * hand side, since they are problem-dependent. We will go through that
766 * process step-by-step, since it is a bit more complicated than in previous
771 * The
first parts of
this function are the same as before, however: setting
772 * up a suitable quadrature formula, initializing an
FEValues object for the
773 * (vector-valued) finite element we use as well as the quadrature
object,
774 * and declaring a number of auxiliary arrays. In addition, we declare the
775 * ever same two abbreviations: <code>n_q_points</code> and
776 * <code>dofs_per_cell</code>. The number of degrees of freedom per cell we
777 * now obviously ask from the composed finite element rather than from the
778 * underlying scalar Q1 element. Here, it is <code>dim</code> times the
779 * number of degrees of freedom per cell of the Q1 element, though
this is
780 * not
explicit knowledge we need to care about:
784 *
void ElasticProblem<dim>::assemble_system()
789 * quadrature_formula,
793 *
const unsigned int dofs_per_cell = fe.dofs_per_cell;
794 *
const unsigned int n_q_points = quadrature_formula.size();
799 * std::vector<types::global_dof_index> local_dof_indices(dofs_per_cell);
803 * As was shown in previous examples as well, we need a place where to
804 * store the values of the coefficients at all the quadrature points on a
805 * cell. In the present situation, we have two coefficients,
lambda and
809 * std::vector<double> lambda_values(n_q_points);
810 * std::vector<double> mu_values(n_q_points);
814 * Well, we could as well have omitted the above two arrays since we will
815 * use constant coefficients
for both
lambda and
mu, which can be declared
816 * like this. They both represent
functions always returning the constant
817 *
value 1.0. Although we could omit the respective factors in the
818 * assemblage of the
matrix, we use them here
for purpose of
826 * Like the two constant
functions above, we will
call the
function
827 * right_hand_side just once per cell to make things simpler.
830 * std::vector<Tensor<1, dim>> rhs_values(n_q_points);
834 * Now we can
begin with the
loop over all cells:
837 *
for (
const auto &cell : dof_handler.active_cell_iterators())
842 * fe_values.reinit(cell);
846 * Next we get the values of the coefficients at the quadrature
847 * points. Likewise
for the right hand side:
850 *
lambda.value_list(fe_values.get_quadrature_points(), lambda_values);
851 *
mu.value_list(fe_values.get_quadrature_points(), mu_values);
852 * right_hand_side(fe_values.get_quadrature_points(), rhs_values);
856 * Then
assemble the entries of the local stiffness
matrix and right
857 * hand side vector. This follows almost
one-to-
one the pattern
858 * described in the introduction of
this example. One of the few
859 * comments in place is that we can compute the number
860 * <code>comp(i)</code>, i.e. the index of the only
nonzero vector
861 * component of shape function <code>i</code>
using the
862 * <code>fe.system_to_component_index(i).first</code>
function call
867 * (By accessing the <code>
first</code> variable of the
return value
868 * of the <code>system_to_component_index</code>
function, you might
869 * already have guessed that there is more in it. In fact, the
870 *
function returns a <code>std::pair@<
unsigned int,
unsigned
871 *
int@></code>, of which the
first element is <code>comp(i)</code>
872 * and the
second is the value <code>base(i)</code> also noted in the
873 * introduction, i.e. the index of
this shape
function within all the
875 * i.e. <code>base(i)</code> in the diction of the introduction. This
876 * is not a number that we are usually interested in, however.)
884 *
for (
const unsigned int i : fe_values.dof_indices())
886 *
const unsigned int component_i =
887 * fe.system_to_component_index(i).first;
889 *
for (
const unsigned int j : fe_values.dof_indices())
891 *
const unsigned int component_j =
892 * fe.system_to_component_index(j).first;
894 *
for (
const unsigned int q_point :
895 * fe_values.quadrature_point_indices())
900 * The
first term is @f$\lambda \partial_i u_i, \partial_j
901 * v_j) + (\
mu \partial_i u_j, \partial_j v_i)@f$. Note
902 * that <code>shape_grad(i,q_point)</code> returns the
903 * gradient of the only
nonzero component of the i-th
904 * shape
function at quadrature
point q_point. The
905 * component <code>comp(i)</code> of the gradient, which
906 * is the derivative of
this only
nonzero vector
907 * component of the i-th shape
function with respect to
908 * the comp(i)th coordinate is accessed by the appended
913 * (fe_values.shape_grad(i, q_point)[component_i] *
914 * fe_values.shape_grad(j, q_point)[component_j] *
915 * lambda_values[q_point])
917 * (fe_values.shape_grad(i, q_point)[component_j] *
918 * fe_values.shape_grad(j, q_point)[component_i] *
919 * mu_values[q_point])
923 * The
second term is @f$(\
mu \nabla u_i, \nabla
924 * v_j)@f$. We need not access a specific component of
925 * the gradient, since we only have to compute the
926 * scalar product of the two gradients, of which an
927 * overloaded version of <tt>
operator*</tt> takes
928 * care, as in previous examples.
932 * Note that by
using the <tt>?:</tt>
operator, we only
933 *
do this if <tt>component_i</tt> equals
934 * <tt>component_j</tt>, otherwise a
zero is added
935 * (which will be optimized away by the compiler).
938 * ((component_i == component_j) ?
939 * (fe_values.shape_grad(i, q_point) *
940 * fe_values.shape_grad(j, q_point) *
941 * mu_values[q_point]) :
944 * fe_values.JxW(q_point);
951 * Assembling the right hand side is also just as discussed in the
955 *
for (
const unsigned int i : fe_values.dof_indices())
957 *
const unsigned int component_i =
958 * fe.system_to_component_index(i).first;
960 *
for (
const unsigned int q_point :
961 * fe_values.quadrature_point_indices())
962 * cell_rhs(i) += fe_values.shape_value(i, q_point) *
963 * rhs_values[q_point][component_i] *
964 * fe_values.JxW(q_point);
969 * The transfer from local degrees of freedom into the global
matrix
970 * and right hand side vector does not depend on the equation under
971 * consideration, and is thus the same as in all previous
975 * cell->get_dof_indices(local_dof_indices);
976 * constraints.distribute_local_to_global(
977 *
cell_matrix, cell_rhs, local_dof_indices, system_matrix, system_rhs);
986 * <a name=
"ElasticProblemsolve"></a>
987 * <h4>ElasticProblem::solve</h4>
991 * The solver does not care about where the system of equations comes, as
992 *
long as it stays positive definite and
symmetric (which are the
993 * requirements
for the use of the CG solver), which the system indeed
994 * is. Therefore, we need not change anything.
998 *
void ElasticProblem<dim>::solve()
1004 * preconditioner.
initialize(system_matrix, 1.2);
1006 * cg.solve(system_matrix, solution, system_rhs, preconditioner);
1008 * constraints.distribute(solution);
1015 * <a name=
"ElasticProblemrefine_grid"></a>
1016 * <h4>ElasticProblem::refine_grid</h4>
1020 * The
function that does the refinement of the grid is the same as in the
1021 * @ref step_6
"step-6" example. The quadrature formula is adapted to the linear elements
1022 * again. Note that the error estimator by
default adds up the estimated
1023 * obtained from all components of the finite element solution, i.e., it
1024 * uses the displacement in all directions with the same weight. If we would
1025 * like the grid to be adapted to the x-displacement only, we could pass the
1026 *
function an additional parameter which tells it to
do so and
do not
1027 * consider the displacements in all other directions
for the error
1028 * indicators. However,
for the current problem, it seems appropriate to
1029 * consider all displacement components with
equal weight.
1032 *
template <
int dim>
1033 *
void ElasticProblem<dim>::refine_grid()
1041 * estimated_error_per_cell);
1044 * estimated_error_per_cell,
1055 * <a name=
"ElasticProblemoutput_results"></a>
1056 * <h4>ElasticProblem::output_results</h4>
1060 * The output happens mostly as has been shown in previous examples
1061 * already. The only difference is that the solution
function is vector
1062 * valued. The
DataOut class takes care of this automatically, but we have
1063 * to give each component of the solution vector a different name.
1067 * To
do this, the DataOut::add_vector() function wants a vector of
1068 * strings. Since the number of components is the same as the number
1069 * of dimensions we are working in, we use the <code>switch</code>
1074 * We note that some graphics programs have restriction on what
1075 * characters are allowed in the names of variables. deal.II therefore
1076 * supports only the minimal subset of these characters that is supported
1077 * by all programs. Basically, these are letters,
numbers, underscores,
1078 * and some other characters, but in particular no whitespace and
1079 * minus/hyphen. The library will throw an exception otherwise, at least
1084 * After listing the 1
d, 2
d, and 3
d case, it is good style to let the
1085 * program die if we
run upon a case which we did not consider. Remember
1086 * that the
Assert macro generates an exception if the condition in the
1087 *
first parameter is not satisfied. Of course, the condition
1088 * <code>false</code> can never be satisfied, so the program will
always
1089 *
abort whenever it gets to the default statement:
1092 * template <
int dim>
1093 *
void ElasticProblem<dim>::output_results(const
unsigned int cycle) const
1098 * std::vector<std::string> solution_names;
1102 * solution_names.emplace_back(
"displacement");
1105 * solution_names.emplace_back(
"x_displacement");
1106 * solution_names.emplace_back(
"y_displacement");
1109 * solution_names.emplace_back(
"x_displacement");
1110 * solution_names.emplace_back(
"y_displacement");
1111 * solution_names.emplace_back(
"z_displacement");
1119 * After setting up the names
for the different components of the
1120 * solution vector, we can add the solution vector to the list of
1121 * data vectors scheduled
for output. Note that the following
1122 *
function takes a vector of strings as
second argument, whereas
1123 * the
one which we have used in all previous examples accepted a
1124 *
string there. (In fact, the
function we had used before would
1125 * convert the single
string into a vector with only
one element
1126 * and forwards that to the other
function.)
1132 * std::ofstream output(
"solution-" +
std::to_string(cycle) +
".vtk");
1141 * <a name=
"ElasticProblemrun"></a>
1146 * The <code>
run</code>
function does the same things as in @ref step_6
"step-6",
for
1147 * example. This time, we use the square [-1,1]^
d as domain, and we
refine
1148 * it globally four times before starting the
first iteration.
1152 * The reason
for refining is a bit accidental: we use the
QGauss
1153 * quadrature formula with two points in each direction
for integration of the
1154 * right hand side; that means that there are four quadrature points on each
1155 * cell (in 2D). If we only
refine the
initial grid once globally, then there
1156 * will be only four quadrature points in each direction on the
1157 * domain. However, the right hand side
function was chosen to be rather
1158 * localized and in that
case, by pure chance, it happens that all quadrature
1159 * points lie at points where the right hand side
function is
zero (in
1160 * mathematical terms, the quadrature points happen to be at points outside
1161 * the <i>support</i> of the right hand side
function). The right hand side
1162 * vector computed with quadrature will then contain only zeroes (even though
1163 * it would of course be
nonzero if we had computed the right hand side vector
1164 * exactly
using the integral) and the solution of the system of
1165 * equations is the
zero vector, i.e., a finite element
function that is
zero
1166 * everywhere. In a sense, we
1167 * should not be surprised that
this is happening since we have chosen
1168 * an
initial grid that is totally unsuitable
for the problem at hand.
1172 * The unfortunate thing is that
if the discrete solution is constant, then
1174 *
for each cell as well, and the
call to
1176 * for refinement (why should it if the indicated error is
zero for each
1177 * cell?). The grid in the next iteration will therefore consist of four
1178 * cells only as well, and the same problem occurs again.
1182 * The conclusion needs to be: while of course we will not choose the
1183 *
initial grid to be well-suited for the accurate solution of the problem,
1184 * we must at least choose it such that it has the chance to capture the
1185 * important features of the solution. In this case, it needs to be able to
1186 * see the right hand side. Thus, we
refine globally four times. (Any larger
1187 * number of global refinement steps would of course also work.)
1190 * template <
int dim>
1191 *
void ElasticProblem<dim>::
run()
1193 *
for (
unsigned int cycle = 0; cycle < 8; ++cycle)
1195 * std::cout <<
"Cycle " << cycle <<
':' << std::endl;
1205 * std::cout <<
" Number of active cells: "
1210 * std::cout <<
" Number of degrees of freedom: " << dof_handler.n_dofs()
1213 * assemble_system();
1215 * output_results(cycle);
1223 * <a name=
"Thecodemaincodefunction"></a>
1224 * <h3>The <code>main</code>
function</h3>
1228 * After closing the <code>Step8</code>
namespace in the last line above, the
1229 * following is the main
function of the program and is again exactly like in
1230 * @ref step_6
"step-6" (apart from the changed
class names, of course).
1237 * Step8::ElasticProblem<2> elastic_problem_2d;
1238 * elastic_problem_2d.run();
1240 *
catch (std::exception &exc)
1242 * std::cerr << std::endl
1244 * <<
"----------------------------------------------------"
1246 * std::cerr <<
"Exception on processing: " << std::endl
1247 * << exc.what() << std::endl
1248 * <<
"Aborting!" << std::endl
1249 * <<
"----------------------------------------------------"
1256 * std::cerr << std::endl
1258 * <<
"----------------------------------------------------"
1260 * std::cerr <<
"Unknown exception!" << std::endl
1261 * <<
"Aborting!" << std::endl
1262 * <<
"----------------------------------------------------"
1270 <a name=
"Results"></a><h1>Results</h1>
1274 There is not much to be said about the results of
this program, other than
1275 that they look nice. All images were made
using VisIt from the
1276 output files that the program wrote to disk. The
first two pictures show
1277 the @f$x@f$- and @f$y@f$-displacements as scalar components:
1279 <table width=
"100%">
1282 <img src=
"https://www.dealii.org/images/steps/developer/step-8.x.png" alt=
"">
1285 <img src=
"https://www.dealii.org/images/steps/developer/step-8.y.png" alt=
"">
1291 You can clearly see the sources of @f$x@f$-displacement around @f$x=0.5@f$ and
1292 @f$x=-0.5@f$, and of @f$y@f$-displacement at the origin.
1294 What
one frequently would like to
do is to show the displacement as a vector
1295 field, i.e., vectors that
for each point illustrate the direction and magnitude
1296 of displacement. Unfortunately, that
's a bit more involved. To understand why
1297 this is so, remember that we have just defined our finite element as a
1298 collection of two components (in <code>dim=2</code> dimensions). Nowhere have
1299 we said that this is not just a pressure and a concentration (two scalar
1300 quantities) but that the two components actually are the parts of a
1301 vector-valued quantity, namely the displacement. Absent this knowledge, the
1302 DataOut class assumes that all individual variables we print are separate
1303 scalars, and VisIt and Paraview then faithfully assume that this is indeed what it is. In
1304 other words, once we have written the data as scalars, there is nothing in
1305 these programs that allows us to paste these two scalar fields back together as a
1306 vector field. Where we would have to attack this problem is at the root,
1307 namely in <code>ElasticProblem::output_results()</code>. We won't
do so here but
1308 instead refer the reader to the @ref step_22
"step-22" program where we show how to
do this
1309 for a more
general situation. That said, we couldn
't help generating the data
1310 anyway that would show how this would look if implemented as discussed in
1311 @ref step_22 "step-22". The vector field then looks like this (VisIt and Paraview
1312 randomly select a few
1313 hundred vertices from which to draw the vectors; drawing them from each
1314 individual vertex would make the picture unreadable):
1316 <img src="https://www.dealii.org/images/steps/developer/step-8.vectors.png" alt="">
1319 We note that one may have intuitively expected the
1320 solution to be symmetric about the @f$x@f$- and @f$y@f$-axes since the @f$x@f$- and
1321 @f$y@f$-forces are symmetric with respect to these axes. However, the force
1322 considered as a vector is not symmetric and consequently neither is
1328 <a name="PlainProg"></a>
1329 <h1> The plain program</h1>
1330 @include "step-8.cc"