Timeline for Adjoint sensitivity analysis for a cost functional under an ODE constraint
Current License: CC BY-SA 4.0
9 events
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| Mar 23, 2023 at 21:10 | comment | added | fedja | @Abhi.A That is a common thing. Try some book in variational calculus. Technically it is a combination of two things: 1) The derivative with respect to a function argument is formally a linear functional on the corresponding linear space of admissible perturbations and 2) Most of the time (but not always!) that linear functional can be written as an integral against another function, which, in that case, is identified with the notation $\frac{dX}{d\rho}$ or something like that. | |
| Mar 23, 2023 at 21:00 | comment | added | Abhi. A | Thanks Fedja! It makes much more sense now. That was super helpful -- just one last thing, is there a reference I could cite for the definition that you gave? | |
| Mar 23, 2023 at 20:59 | comment | added | fedja | @Abhi.A Exactly like that :-) Only I would add parentheses: $(dX_{\rho}(x,1)/d\rho)(t)$ not to confuse the reader about which function takes $t$ as an argument on the LHS. | |
| Mar 23, 2023 at 20:59 | vote | accept | Abhi. A | ||
| Mar 23, 2023 at 20:53 | comment | added | Abhi. A | Thanks! that is very useful! | |
| Mar 23, 2023 at 20:51 | comment | added | fedja | @Abhi.A To simplify the matters and to relate it to something you know, let's say we have a real valued function $G(z)$ of an $n$-dimensional vector $z=(z_1,\dots,z_n)$. Then one can write $\frac{dG}{dz}=w=(w_1,\dots,w_n)$ (which is nothing but the gradient vector in a fancy notation) or one can write $G(z+dz)-G(z)\approx \sum_j w_j\, dz_j$. The first formula has no sum symbol in it, the second does, but it is the same statement! | |
| Mar 23, 2023 at 20:43 | comment | added | fedja | @Abhi.A You don't need to get rid of the integral at all, that's the whole point! On the contrary, you should get the linearized (in $\Delta\rho$) expression for the increment of the final output as $\int_0^1 D(t)\Delta\rho(t)\,dt$ and then you'll be able to claim (see the definition I provided) that the derivative of a certain functional with respect to the function argument is $D(t)$. That claim includes integration, if you expand the corresponding definition, which I provided. It is just not written explicitly in that shorthand notation. | |
| Mar 23, 2023 at 20:23 | comment | added | Abhi. A | Thanks, Fedja! Could you elaborate a little however? Should I be calculating $X_{\rho+d\rho}-X_{\rho}=\int_{0}^1 F(X_{\rho+d\rho}(x,t);\rho+d\rho)-F(X_{\rho};\rho) dt$ I am unsure how to do that, I thought the adjoint method would be useful, but I am unable to get rid of the integral. Do they mean the derivative to be taken with respect to $\rho_t=\rho(\cdot,t)$ for some given $t$? | |
| Mar 23, 2023 at 19:59 | history | answered | fedja | CC BY-SA 4.0 |