]> https://gitweb.dealii.org/ - dealii.git/commitdiff
Added description of C and a solution example
authorGiovanni Alzetta <giovannialzetta@hotmail.it>
Wed, 2 May 2018 15:59:45 +0000 (17:59 +0200)
committerMatthias Maier <tamiko@43-1.org>
Tue, 8 May 2018 04:44:55 +0000 (23:44 -0500)
examples/step-60/doc/intro.dox
examples/step-60/doc/results.dox

index dd02773084d32af52621e2b4e769ac6bf9624b01..221f2b5c5a6e277129cd17864b357b709179cde3 100644 (file)
@@ -201,6 +201,19 @@ C_{\alpha j} := (v_j, q_\alpha)_\Gamma  = \sum_{K\in \Gamma} \int_{\hat K}
 \sum_{K\in \Gamma} \sum_{i=1}^{n_q}  \big(\hat q_\alpha(\hat x_i)  (v_j \circ F_{K}) (\hat x_i) J_K (\hat x_i) w_i \big)
 \f]
 
+Computing this sum is difficult because we have to evaluate $(v_j \circ F_{K}) (\hat x_i)$; this
+complicated process is illustrated in the following picture:
+
+<p align="center">
+  <img src="https://www.dealii.org/images/steps/developer/step-60_C_interpolation.png" alt="">
+</p>
+
+This translates in the following algorithm:
+- Fix the cell $K$ in $\Gamma$ and compute the real point $y_i \coloneqq F_{K} (\hat x_i)$
+- Find the cell of $\Omega$ in which $y_i$ lies, say it's $T$.
+- To evaluate the basis function use invert shape function $G_{T}$ on $T$:
+ $v_j(y_i) = \hat v_j \circ G^{-1}_{T} (y_i)$.
+
 We solve the above saddle point problem by iterating over the Schur complement
 (which is described, for example, in step-20),
 and we construct such complement using LinearOperator classes.
index 85f416278d32d9c4e8c98765536f320021666e5a..c48f2a2d22e72bfd64f883f635617e8a2d20387a 100644 (file)
@@ -2,8 +2,20 @@
 
 <h3> Test case 1: </h3>
 
-For the default problem the value of u on Gamma is 1 and on $\partial\Omega$
-is 0. In fact this is the solution:
+For the default problem the value of u on Gamma is 1: this is like imposing a constant
+Dirichlet boundary on the portion of $\Omega$ inside gamma.
+<p align="center">
+  <img src="https://www.dealii.org/images/steps/developer/step-60_1_no_grid.png" alt=""
+   style="width: 50%; height: 50%">
+</p>
+In this second image we can appreciate how, using GridTools::compute_point_locations ,
+we have been able to adaptively refine $\Omega$ exactly where the solution is varying
+the most:
+
+<p align="center">
+  <img src="https://www.dealii.org/images/steps/developer/step-60_1_grid.png" alt=""
+   style="width: 50%; height: 50%">
+</p>
 
 <a name="extensions"></a>
 <h3>Possibilities for extensions</h3>

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