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In this paper https://arxiv.org/pdf/1907.09605.pdf \ let $\Omega \subset \mathbb{R}^n$ with $n \geq 1$ be a bounded Lipschitz domain with boundary $\partial \Omega$, $f: \Omega \rightarrow \mathbb{R}$ be an $L^2(\Omega)$ function (given datum), $K: L^2(\Omega) \rightarrow L^2(\Omega)$ be a bounded linear operator, and $X$ be a Banach space. Then a standard regularized variational model is given by $$ \min _{u \in X_{a d} \subseteq X} J(u):=\frac{1}{2}\|K u-f\|_{L^2(\Omega)}^2+\lambda\int_{\Omega}|\nabla u|, $$ where $X_{a d}$ is a closed, convex, nonempty admissible set which is contained in the solution space $X$, and $u$ is the solution that we want to reconstruct or recover.\ the corresponding Euler-Lagrange equations for (2.2) are: Find $u \in X_{a d} \subset X$ such that $$ \left\langle-\lambda \operatorname{div}\left(\frac{\nabla u}{|\nabla u|}\right)+K^*(K u-f), \hat{u}-u\right\rangle_{X^{\prime}, X} \geq 0, \quad \forall \hat{u} \in X_{a d} $$ How this variational derivative is calculated ?

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    $\begingroup$ The second part with $K^*$ is the rigorous derivative of the first part of $J$. The first part in the EL equations is, I would say, rather formal. You get it by taking the derivative of the regularization term on a formal level, and the negative divergence is meant distributionally. However, in order to justify this properly, much more work is necessary, after all, $\nabla u$ is only a measure, so at this stage I would just take the statement as "morally, it should look like that". $\endgroup$
    – Hannes
    Commented Mar 28 at 12:23

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