computed for boundary indicator 0. The result will be that we will get
discontinuous boundary values, zero on three sides of the square, and one on
the fourth.
+
+ <li>
+ Observe convergence: We will only discuss computing errors in norms in
+ @ref step_7 "step-7", but it is easy to check that computations converge
+ already here. For example, we could evaluate the value of the solution in a
+ single point and compare the value for different %numbers of global
+ refinement (the number of global refinement steps is set in
+ <code>LaplaceProblem::make_grid_and_dofs</code> above). To evaluate the
+ solution at a point, say at $(\frac 13, \frac 13)$, we could add the
+ following code to the <code>LaplaceProblem::output_results</code> function:
+ @code
+ std::cout << "Solution at (1/3,1/3): "
+ << VectorTools::point_value (dof_handler, solution,
+ Point<2>(1./3, 1./3))
+ << std::endl;
+ @endcode
+ For 1 through 8 global refinement steps, we then get the following sequence
+ of point values:
+ <table align="center">
+ <tr> <td># of refinements</td> <td>$u_h(\frac 13,\frac13)$</td> </tr>
+ <tr> <td>1</td> <td>0.166667</td> </tr>
+ <tr> <td>2</td> <td>0.227381</td> </tr>
+ <tr> <td>3</td> <td>0.237375</td> </tr>
+ <tr> <td>4</td> <td>0.240435</td> </tr>
+ <tr> <td>5</td> <td>0.241140</td> </tr>
+ <tr> <td>6</td> <td>0.241324</td> </tr>
+ <tr> <td>7</td> <td>0.241369</td> </tr>
+ <tr> <td>8</td> <td>0.241380</td> </tr>
+ <tr> <td>9</td> <td>0.241383</td> </tr>
+ </table>
+ By noticing that the difference between each two consecutive values reduces
+ by about a factor of 4, we can conjecture that the "correct" value may be
+ $u(\frac 13, \frac 13)\approx 0.241384$. In fact, if we assumed this to be
+ the correct value, we could show that the sequence above indeed shows ${\cal
+ O}(h^2)$ convergence — theoretically, the convergence order should be
+ ${\cal O}(h^2 |\log h|)$ but the symmetry of the domain and the mesh may lead
+ to the better convergence order observed.
+
+ A slight variant of this would be to repeat the test with quadratic
+ elements. All you need to do is to set the polynomial degree of the finite
+ element to two in the constructor
+ <code>LaplaceProblem::LaplaceProblem</code>.
</ul>