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<div class="">On Dec 27, 2020, at 5:01 PM, Salazar De Troya, Miguel via petsc-users <<a href="mailto:petsc-users@mcs.anl.gov" class="">petsc-users@mcs.anl.gov</a>> wrote:</div>
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<span style="font-size: 11pt;" class="">Hello,<o:p class=""></o:p></span></div>
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<span style="font-size: 11pt;" class=""><o:p class=""> </o:p></span></div>
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<span style="font-size: 11pt;" class="">I am interested in calculating the gradients of an optimization problem with one goal and one constraint functions which need TSAdjoint for their adjoints. I’d like to call each of their adjoints in different calls, but
it does not seem to be possible without making compromises.</span></div>
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<div>If you are calculating the derivatives to the same set of parameters, the adjoints of all cost functionals can be done with a single backward run. </div>
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<span style="font-size: 11pt;" class="">For instance, one could set<span class="Apple-converted-space"> </span></span><span style="font-size: 11pt;" class="">TSCreateQuadratureTS() and TSSetCostGradients() with different quadratures (and their gradients) for
each adjoint call (one at a time). This would evaluate the cost functions in the backwards run though, whereas one typically computes the cost functions in a different routine than the adjoint call (like in line searches evaluations)</span></div>
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<div><span class="">The second argument of TSCreateQuadratureTS() allows you to choose if the quadrature is evaluated in the forward run or in the backward run. The choice typically depends on the optimization algorithms. Some optimization algorithms may expect
users to provide an objective function and its gradient as a bundle; in this case, the choice does not make a difference. Some other algorithms may occasionally evaluate the objective function without evaluating its gradient, then evaluating the quadrature
in the forward run is definitely a better choice.<br class="">
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<span style="font-size: 11pt;" class=""><o:p class=""></o:p></span></div>
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<span style="font-size: 11pt;" class=""><o:p class=""> </o:p></span></div>
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<span style="font-size: 11pt;" class="">One could also set TSCreateQuadratureTS() with the goal and the constraint functions to be evaluated at the forward run (as typically done when computing the cost function). The problem would be that the adjoint call
now requires two sets of gradients for TSSetCostGradients() and their adjoint are calculated together, costing twice if your routines for the cost and the constraint gradients are separated.</span></div>
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<div>You can put the two sets of gradients in vector arrays and pass them to TSSetCostGradients() together. Only one call to TSAdjointSolve() is needed. See the example src/ts/tutorials/ex20adj.c, where we have two independent cost functionals, and their adjoints
correspond to lambda[0]/mup[0] and lambda[1]/mup[1] respectively. After performing a TSAdjontSolve, you will get the gradients for both cost functionals.</div>
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<span style="font-size: 11pt;" class=""><o:p class=""></o:p></span></div>
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<span style="font-size: 11pt;" class="">The only solution I can think of is to set TSCreateQuadratureTS() with both the goal and constraint functions in the forward run. Then, in the adjoint calls, reset TSCreateQuadratureTS() with just the cost function I
am interested in (either the goal or the constraint) and set just a single TSSetCostGradients(). Will this work? Are there better alternatives?</span></div>
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<span class="">TSCreateQuadratureTS() is needed only when you have integral terms in the cost functionals. It has nothing to do with the procedure to compute the adjoints for multiple cost functionals simultaneously. Do you have integrals in both the goal and
the constraint? If so, you can create one quadrature TS and evaluate both integrals together. For example, you may have r[0] (the first element of the output vector in your cost integrand) for the goal and r[1] for the constraint. Just be careful that the
adjoint variables (array lambda[]/mup[]) should be organized in the same order.<br class="">
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<span style="font-size: 11pt;" class=""><o:p class=""></o:p></span></div>
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<span style="font-size: 11pt;" class="">Even if successful, there is the problem that the trajectory goes back to the beginning when we perform a TSAdjointSolve() call. Subsequent calls to TSAdjointSolve() (for instance for another cost function) are invalid
because the trajectory is not set at the end of the simulation. One needs to call the forward problem to bring it back to the end. Is there a quick way to set the trajectory state to the last time step without having to run the forward problem? I am attaching
an example to illustrate this issue. One can uncomment lines 120-122 to obtain the right value of the derivative.</span></div>
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<div>Most likely you need only one call to TSAdjointSolve(). Reusing the trajectory for multiple calls is also doable. But I doubt you would need it.</div>
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<div>Hong (Mr.)</div>
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<span style="font-size: 11pt;" class=""><o:p class=""></o:p></span></div>
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<span style="font-size: 11pt;" class="">Thanks<o:p class=""></o:p></span></div>
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<span style="font-size: 11pt;" class="">Miguel<o:p class=""></o:p></span></div>
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<span style="font-size: 11pt;" class=""><o:p class=""> </o:p></span></div>
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<span lang="ES" style="font-size: 9pt; font-family: Consolas;" class="">Miguel A. Salazar de Troya</span><span lang="ES" style="font-size: 10.5pt;" class=""><o:p class=""></o:p></span></div>
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<span style="font-size: 9pt; font-family: Consolas;" class="">Postdoctoral Researcher, Lawrence Livermore National Laboratory</span><span style="font-size: 10.5pt;" class=""><o:p class=""></o:p></span></div>
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<span style="font-size: 9pt; font-family: Consolas;" class="">B141</span><span style="font-size: 10.5pt;" class=""><o:p class=""></o:p></span></div>
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<span style="font-size: 9pt; font-family: Consolas;" class="">Rm: 1085-5</span><span style="font-size: 10.5pt;" class=""><o:p class=""></o:p></span></div>
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<span style="font-size: 9pt; font-family: Consolas;" class="">Ph: 1(925) 422-6411</span><o:p class=""></o:p></div>
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<span id="cid:B4CB317F30258C42B6EE9FC2778A2D8C@namprd09.prod.outlook.com"><simple-ode.py></span></div>
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