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<p>Hope this is not too technical. <span style="font-family:Wingdings">J</span><o:p></o:p></p>
<p><o:p> </o:p></p>
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<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">**Jed, A’=[A^-1 U B], not the transpose of A.</span><o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">I understand that it's not the transpose. I still don't have a clue what the notation [A^{-1} U B] means. I would think it means some block decomposed thing, but I don't think you mean just adding columns or taking a product of matrices.
Using some standard mathematical notation would help.<o:p></o:p></p>
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<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">>>
</span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">A’=[A^-1 U B] means matrix A’ consists all elements from A^-1, the inverted A, and all elements from B. One simple equation is
<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal" style="text-indent:11.55pt"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> K=[Kst] =A’ = ASSEMBLE_ij (A^-1_ij) + ASSEMBLE_i’j’(B_i’j’), here (s,t), (i,j) and (i’,j’) = different index sets<o:p></o:p></span></p>
<p class="MsoNormal" style="text-indent:11.55pt"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal" style="text-indent:11.55pt"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">i.e., entry K_st= A^-1_st+ B_st,
<o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in;text-indent:.5in"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> = 0 when A^-1_st=0 and B_st=0</span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in;text-indent:.5in"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">= A^-1_st when B_st =0</span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p></o:p></span></p>
<p class="MsoNormal" style="margin-left:.5in;text-indent:.5in"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">= B_st when A^-1_st =0</span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">
</span><o:p></o:p></p>
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<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> We need to solve for A’*x=b. A is dense matrix.</span><o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">Where does A come from? Why is it dense?<o:p></o:p></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoListParagraph" style="margin-left:0in"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">>> A comes from wave propagation. It is dense since it denotes a function of Green function.<o:p></o:p></span></p>
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<p class="MsoNormal"> <o:p></o:p></p>
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<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"> </span><o:p></o:p></p>
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<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">** As I mentioned A’=K is the assembled global matrix using the rule K=A’=SIGMA_ij (A^-1_ij)+SIGMA_i’j’(B_i’j’)
from FEM. Here, SIGMA_ij (A^-1_ij) denotes the assembling process for A^-1_ij into K according the DOFs of each node in the FEM model,</span><o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">Is A^{-1} an operator on some subdomain? Are you trying to implement a substructuring algorithm? What is B physically?<o:p></o:p></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">>> NO.
</span>A^{-1} denotes the inverted A. B is a sparse matrix of much larger order.<span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p></o:p></span></p>
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<p class="MsoNormal"> <o:p></o:p></p>
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<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto">Then there is no way A^{-1} should be stored as a dense matrix.
<o:p></o:p></p>
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<p><span style="font-size:11.0pt;font-family:"Courier New";color:#1F497D">o</span><span style="font-size:7.0pt;color:#1F497D">
</span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">It is stored as a dense matrix in scalapack. It stays in the same way in cores as scalapack finishes its inversion. We could define A^-1 as dense matrix in PETSc. </span><o:p></o:p></p>
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<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto">It should not be done this way. A^{-1} should not be stored explicitly. Store the sparse finite element matrix A. Then when you want to "apply A^{-1}", solve with the sparse matrix.<o:p></o:p></p>
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<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><o:p> </o:p></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"> <span style="color:#1F497D">** Jed, you are getting close to understand the problem related to QA. A has to be inverted explicitly. A^-1 has to be known entry by entry such that
each entry in B could be assembled with A^-1 to form A’=K. This is a requirement beyond technical issue. This is a QA issue.</span><o:p></o:p></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">What does QA stand for? Can you explain why B needs the entries of A^{-1}?<o:p></o:p></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">>> QA means quality assurance. It is a procedure to ensure product quality. In Eq.
<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal" style="text-indent:.5in"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">K=A’=ASSEMBLE_ij (A^-1_ij)+ ASSEMBLE_i’j’(B_i’j’)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">
</span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p></o:p></span></p>
<p class="MsoNormal" style="text-indent:.5in"><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Entry B_i’j’ and
</span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">A^-1_ij may or may not locate at the same row and col. That why we need explicitly each entry in
</span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">B_i’j’ and
</span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">A^-1_ij to assemble K. The big picture is that K is the final sparse matrix we need to solve K*x=A’*x=b. However, K indexed by (s,t) needs to be constructed in terms of
dense matrix A and sparse matrix B using index sets (i,j) and (i’,j’). </span><span style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D"><o:p></o:p></span></p>
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