<h1>Results</h1>


There is not much to be said about the results of this program, other than
that they look nice. All images were made using VisIt from the
output files that the program wrote to disk. The first two pictures show
the $x$- and $y$-displacements as scalar components:

<table width="100%">
<tr>
<td>
<img src="https://www.dealii.org/images/steps/developer/step-8.x.png" alt="">
</td>
<td>
<img src="https://www.dealii.org/images/steps/developer/step-8.y.png" alt="">
</td>
</tr>
</table>


You can clearly see the sources of $x$-displacement around $x=0.5$ and
$x=-0.5$, and of $y$-displacement at the origin.

What one frequently would like to do is to show the displacement as a vector
field, i.e., vectors that for each point illustrate the direction and magnitude
of displacement. Unfortunately, that's a bit more involved. To understand why
this is so, remember that we have just defined our finite element as a
collection of two  components (in <code>dim=2</code> dimensions). Nowhere have
we said that this is not just a pressure and a concentration (two scalar
quantities) but that the two components actually are the parts of a
vector-valued quantity, namely the displacement. Absent this knowledge, the
DataOut class assumes that all individual variables we print are separate
scalars, and VisIt and Paraview then faithfully assume that this is indeed what it is. In
other words, once we have written the data as scalars, there is nothing in
these programs that allows us to paste these two scalar fields back together as a
vector field. Where we would have to attack this problem is at the root,
namely in <code>ElasticProblem::output_results()</code>. We won't do so here but
instead refer the reader to the step-22 program where we show how to do this
for a more general situation. That said, we couldn't help generating the data
anyway that would show how this would look if implemented as discussed in
step-22. The vector field then looks like this (VisIt and Paraview
randomly select a few
hundred vertices from which to draw the vectors; drawing them from each
individual vertex would make the picture unreadable):

<img src="https://www.dealii.org/images/steps/developer/step-8.vectors.png" alt="">


We note that one may have intuitively expected the
solution to be symmetric about the $x$- and $y$-axes since the $x$- and
$y$-forces are symmetric with respect to these axes. However, the force
considered as a vector is not symmetric and consequently neither is
the solution.
