{"action":"created","comment":{"url":"https://api.github.com/repos/dealii/dealii/pulls/comments/80772560","pull_request_review_id":1803576,"id":80772560,"diff_hunk":"@@ -0,0 +1,426 @@\n+<h1>Results</h1>\n+\n+Now we use the method we discussed above to solve Navier Stokes equations with\n+viscosity 1/400 and 1/10000.\n+\n+<h3> Test case 1: Re=400 </h3>\n+\n+In the first test case the viscosity is set to be 1/400. As we disccussed in\n+introduction, the initial guess is the soluton to corresponding Stokes\n+problem. In the following table, the residuals at each Newton's iteration on\n+every mesh is shown. By the data in the table, it shows the convergence rate\n+of Newton's iteration is quadratic.\n+\n+<table align=\"center\" border=\"1\">\n+  \n+  <tr>\n+    <th>&nbsp;</th>\n+    <th colspan=\"2\">Mesh0</th>\n+    <th colspan=\"2\">Mesh1</th>\n+    <th colspan=\"2\">Mesh2</th>\n+    <th colspan=\"2\">Mesh3</th>\n+    <th colspan=\"2\">Mesh4</th>\n+  </tr>\n+\n+  <tr>\n+    <th>Newton's iter </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+  </tr>\n+\n+  <tr>\n+    <th>1         </th>\n+    <th>7.40396e-3</th>\n+    <th>3         </th>\n+    <th>1.05562e-3  </th>\n+    <th>3         </th>\n+    <th>4.94796e-4  </th>\n+    <th>3         </th>\n+    <th>2.5624e-4 </th>\n+    <th>2         </th>\n+    <th>1.26733e-4  </th>\n+    <th>2         </th>\n+  </tr>\n+\n+  <tr>\n+    <th>2         </th>\n+    <th>3.86766e-3  </th>\n+    <th>4         </th>\n+    <th>1.3549e-5  </th>\n+    <th>3         </th>\n+    <th>1.41981e-6  </th>\n+    <th>3        </th>\n+    <th>1.29108e-6  </th>\n+    <th>4         </th>\n+    <th>6.14794e-7  </th>\n+    <th>4         </th>\n+  </tr>\n+\n+  <tr>\n+    <th>3         </th>\n+    <th>1.60421e-3</th>\n+    <th>4         </th>\n+    <th>1.24836e-9  </th>\n+    <th>3         </th>\n+    <th>9.11557e-11  </th>\n+    <th>3         </th>\n+    <th>3.35933e-11 </th>\n+    <th>3         </th>\n+    <th>5.86734e-11 </th>\n+    <th>2         </th>\n+  </tr>\n+\n+  <tr>\n+    <th>4         </th>\n+    <th>9.26748e-4  </th>\n+    <th>4         </th>\n+    <th>2.75537e-14  </th>\n+    <th>4         </th>\n+    <th>1.39986e-14 </th>\n+    <th>5         </th>\n+    <th>2.18864e-14 </th>\n+    <th>5         </th>\n+    <th>3.38787e-14 </th>\n+    <th>5         </th>\n+  </tr>\n+\n+  <tr>\n+    <th>5          </th>\n+    <th>1.34601e-5</th>\n+    <th>4          </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+  </tr>\n+\n+  <tr>\n+    <th>6         </th>\n+    <th>2.5235e-8    </th>\n+    <th>5    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+  </tr>\n+\n+  <tr>\n+    <th>7         </th>\n+    <th>1.38899e-12    </th>\n+    <th>4    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+  </tr>\n+\n+  <tr>\n+    <th>8           </th>\n+    <th>4.68224e-15 </th>\n+    <th>4           </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;      </th>\n+    <th>&nbsp;      </th>\n+    <th>&nbsp;      </th>\n+    <th>&nbsp;      </th>\n+    <th>&nbsp;      </th>\n+    <th>&nbsp;      </th>\n+  </tr>\n+</table>\n+\n+The following is the generated grids during solving. For this case of Re=400,\n+the initial guess is obtianed by solving Stokes on $8 \\times 8$ mesh, and the\n+mesh is refined adaptively. Between meshs, the solution from coarse is\n+transfered to fine mesh as an initial guess.\n+\n+<TABLE WIDTH=\"60%\" ALIGN=\"center\">\n+  <tr>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re400_Mesh0.png\" alt=\"\">\n+    </td>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re400_Mesh1.png\" alt=\"\">\n+    </td>\n+  </tr>\n+\n+  <tr>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re400_Mesh2.png\" alt=\"\">\n+    </td>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re400_Mesh3.png\" alt=\"\">\n+    </td>\n+  </tr>\n+\n+  <tr>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re400_Mesh4.png\" alt=\"\">\n+    </td>\n+  </tr>\n+</table>\n+\n+This picture is the graphical result of lid-drive cavity with Re=400.\n+<img src=\"step-57_Re400_Streamline.png\" alt=\"\">\n+\n+Then the solution by Newton's iteration is compared with a reference solution\n+from High-Re Solutions for Incompressible Flow Using the Navier-Stokes\n+Equations and a Multigrid Method, U. Ghia, K.N.Ghia and C.T.Shin.\n+\n+<img src=\"step-57-compare-Re400.svg\" alt=\"\">\n+ \n+<h3> Test case 1: Re=10000 </h3>\n+\n+Newton's iteration requires a good initial guess. However, the nonlinear term\n+gets dominant when Reynold number is large, so that the solution to Stokes\n+equations may be far away from the exact solution. If the Stokes solution acts\n+as the initial guess, the convergence will be lost. The following picture\n+shows that the residual can not get lower as iteration going on.\n+\n+<img src=\"step-57_Re10000_loss_convergence.svg\" alt=\"\">\n+\n+The initial guess, therefore, have to be obtained via a staircase which has\n+been discuessed in introduction. Here the step size is 2000 and the initial\n+mesh is of size $32 \\times 32$. After obtaining an initial guess, the mesh is\n+refined as in previous test case. The following picture shows that at each\n+stair Newton's iteration has quadratic convergence and the convergence rate is\n+also quadratic on every refined mesh. 60 steps of Newton's iterations are\n+excuted for solving this test case.\n+\n+<img src=\"step-57_Re10000_get_convergence.svg\" alt=\"\">\n+\n+Also we show the residual from each step of Newton's iteration on every\n+mesh. The quadratic convergence is shown clearly in the table.\n+\n+<table align=\"center\" border=\"1\">\n+  \n+  <tr>\n+    <th>&nbsp;</th>\n+    <th colspan=\"2\">Mesh0</th>\n+    <th colspan=\"2\">Mesh1</th>\n+    <th colspan=\"2\">Mesh2</th>\n+    <th colspan=\"2\">Mesh3</th>\n+    <th colspan=\"2\">Mesh4</th>\n+  </tr>\n+\n+  <tr>\n+    <th>Newton's iter </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+    <th>Residual      </th>\n+    <th>FGMRES        </th>\n+  </tr>\n+\n+  <tr>\n+    <th>1          </th>\n+    <th>3.81015e-6  </th>\n+    <th>7          </th>\n+    <th>4.83666e-3  </th>\n+    <th>3          </th>\n+    <th>1.74361e-3  </th>\n+    <th>3          </th>\n+    <th>5.61146e-4  </th>\n+    <th>2          </th>\n+    <th>2.06674e-04  </th>\n+    <th>2         </th>\n+  </tr>\n+\n+  <tr>\n+    <th>2         </th>\n+    <th>1.39551e-8</th>\n+    <th>6         </th>\n+    <th>1.64745e-3 </th>\n+    <th>6         </th>\n+    <th>5.57647e-4 </th>\n+    <th>6         </th>\n+    <th>1.35722e-4 </th>\n+    <th>6         </th>\n+    <th>2.36666e-5 </th>\n+    <th>6         </th>\n+  </tr>\n+\n+  <tr>\n+    <th>3         </th>\n+    <th>2.00292e-13    </th>\n+    <th>6         </th>\n+    <th>4.34298e-4 </th>\n+    <th>6         </th>\n+    <th>4.3981e-5 </th>\n+    <th>6         </th>\n+    <th>1.78859e-6 </th>\n+    <th>6         </th>\n+    <th>3.55519e-08 </th>\n+    <th>6         </th>\n+  </tr>\n+\n+  <tr>\n+    <th>4         </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>1.5311e-4 </th>\n+    <th>6         </th>\n+    <th>7.54006e-8 </th>\n+    <th>6         </th>\n+    <th>2.45452e-10 </th>\n+    <th>3         </th>\n+    <th>1.64301e-12</th>\n+    <th>6        </th>\n+  </tr>\n+\n+  <tr>\n+    <th>5         </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>2.50725e-5 </th>\n+    <th>5         </th>\n+    <th>8.43023e-12 </th>\n+    <th>6         </th>\n+    <th>2.14919e-14 </th>\n+    <th>7         </th>\n+    <th>3.23876e-14 </th>\n+    <th>8         </th>\n+  </tr>\n+\n+  <tr>\n+    <th>6         </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>2.45113e-6</th>\n+    <th>6         </th>\n+    <th>1.24722e-14</th>\n+    <th>7         </th>\n+    <th>&nbsp; </th>\n+    <th>&nbsp;        </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+  </tr>\n+\n+  <tr>\n+    <th>7         </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>3.18648e-8 </th>\n+    <th>6         </th>\n+    <th>&nbsp;</th>\n+    <th>&nbsp;         </th>\n+    <th>&nbsp; </th>\n+    <th>&nbsp;         </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+  </tr>\n+\n+  <tr>\n+    <th>8         </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>3.35688e-12 </th>\n+    <th>6         </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;</th>\n+    <th>&nbsp;        </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+  </tr>\n+\n+  <tr>\n+    <th>9         </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>7.20278e-15</th>\n+    <th>6         </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;</th>\n+    <th>&nbsp;       </th>\n+    <th>&nbsp;    </th>\n+    <th>&nbsp;    </th>\n+  </tr>\n+</table>\n+The generated grids during solving is shown as below.\n+<TABLE WIDTH=\"60%\" ALIGN=\"center\">\n+  <tr>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re10000_Mesh0.png\" alt=\"\">\n+    </td>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re10000_Mesh1.png\" alt=\"\">\n+    </td>\n+  </tr>\n+\n+  <tr>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re10000_Mesh2.png\" alt=\"\">\n+    </td>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re10000_Mesh3.png\" alt=\"\">\n+    </td>\n+  </tr>\n+\n+  <tr>\n+    <td ALIGN=\"center\">\n+      <img src=\"step-57_Re10000_Mesh4.png\" alt=\"\">\n+    </td>\n+  </tr>\n+</table>\n+We compare the solution by Newtion's method with reference solution which can be found in Numerical Solutions of 2-D Steady incompressible driven cavity flow at high Reynolds numbers, E.Erturk, T.C.Corke and C.Gokcol.\n+<img src=\"step-57-compare-Re10000.svg\" alt=\"\">\n+The following picture presents the graphical result.\n+<img src=\"step-57_Re10000_Streamline.png\" alt=\"\">\n+\n+Furthermore, we should notice that the error consists of the nonlinear error,\n+which can be minished by Newton's iteration, and the discrete error. The\n+discrete error depends on the mesh size. That is why we have to refine the\n+mesh and repeat Newton's iteration on the new one. From above table, we can","path":"examples/step-57/doc/results.dox","position":400,"original_position":400,"commit_id":"526578e8e2e7c1391b73f04de7ab21e115bea970","original_commit_id":"526578e8e2e7c1391b73f04de7ab21e115bea970","user":{"login":"kronbichler","id":8276200,"avatar_url":"https://avatars.githubusercontent.com/u/8276200?v=3","gravatar_id":"","url":"https://api.github.com/users/kronbichler","html_url":"https://github.com/kronbichler","followers_url":"https://api.github.com/users/kronbichler/followers","following_url":"https://api.github.com/users/kronbichler/following{/other_user}","gists_url":"https://api.github.com/users/kronbichler/gists{/gist_id}","starred_url":"https://api.github.com/users/kronbichler/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/kronbichler/subscriptions","organizations_url":"https://api.github.com/users/kronbichler/orgs","repos_url":"https://api.github.com/users/kronbichler/repos","events_url":"https://api.github.com/users/kronbichler/events{/privacy}","received_events_url":"https://api.github.com/users/kronbichler/received_events","type":"User","site_admin":false},"body":"new one -> next finer mesh","created_at":"2016-09-27T19:16:30Z","updated_at":"2016-09-27T19:18:13Z","html_url":"https://github.com/dealii/dealii/pull/3116#discussion_r80772560","pull_request_url":"https://api.github.com/repos/dealii/dealii/pulls/3116","_links":{"self":{"href":"https://api.github.com/repos/dealii/dealii/pulls/comments/80772560"},"html":{"href":"https://github.com/dealii/dealii/pull/3116#discussion_r80772560"},"pull_request":{"href":"https://api.github.com/repos/dealii/dealii/pulls/3116"}}},"pull_request":{"url":"https://api.github.com/repos/dealii/dealii/pulls/3116","id":85303647,"html_url":"https://github.com/dealii/dealii/pull/3116","diff_url":"https://github.com/dealii/dealii/pull/3116.diff","patch_url":"https://github.com/dealii/dealii/pull/3116.patch","issue_url":"https://api.github.com/repos/dealii/dealii/issues/3116","number":3116,"state":"open","locked":false,"title":"Introduce step-57: Navier Stokes with nonlinear solver","user":{"login":"vastronzl","id":20875655,"avatar_url":"https://avatars.githubusercontent.com/u/20875655?v=3","gravatar_id":"","url":"https://api.github.com/users/vastronzl","html_url":"https://github.com/vastronzl","followers_url":"https://api.github.com/users/vastronzl/followers","following_url":"https://api.github.com/users/vastronzl/following{/other_user}","gists_url":"https://api.github.com/users/vastronzl/gists{/gist_id}","starred_url":"https://api.github.com/users/vastronzl/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/vastronzl/subscriptions","organizations_url":"https://api.github.com/users/vastronzl/orgs","repos_url":"https://api.github.com/users/vastronzl/repos","events_url":"https://api.github.com/users/vastronzl/events{/privacy}","received_events_url":"https://api.github.com/users/vastronzl/received_events","type":"User","site_admin":false},"body":"This tutorial is work in progress together with @tjhei . This work is not 100% completed. We are looking forward to your honest feedback. 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