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 ?

  • 2
    $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Mar 27 at 12:20
  • $\begingroup$ Do you understand how the variational derivative is calculated or is the problem just in how the optimality conditions are written? $\endgroup$
    – whpowell96
    Mar 27 at 15:49
  • $\begingroup$ I want to know how the variational derivative is calculated $\endgroup$
    – Mohamed
    Mar 27 at 16:42

1 Answer 1


We start with $$J[u] = \frac{1}{2}\int_\Omega (Ku-f)^2 + \lambda \sqrt{\nabla u\cdot\nabla u}~dx.$$ We now perturb this functional to the value $J[u+\epsilon v]$ and collect all the terms linear in $\epsilon$. This is the directional derivative which we will then manipulate into the functional derivative. I will skip some algebra and basic calculus steps to illustrate the main point.

$$ J[u+\epsilon v] = J[u] + \epsilon \int_\Omega Kuv + uKv + \lambda \frac{\nabla u\cdot \nabla v}{\|\nabla u\|}~dx + \mathcal{O}(\epsilon^2). $$ The last term in the integrand is obtained by differentiating $\sqrt{\nabla u\cdot\nabla u}$ with respect to $u$. The directional derivative is then $$ \delta J[u]v = \int_\Omega (Ku-f)Kv + \lambda \frac{\nabla u\cdot \nabla v}{\|\nabla u\|}~dx. $$ If we apply Green's identity (or divergence theorem) to the second term, we have $$ \delta J[u]v = \int_\Omega (Ku-f)Kv - \lambda \mathrm{div}\left(\frac{\nabla u}{\|\nabla u\|}\right)v~dx + \oint_{\partial\Omega} v \frac{\nabla u\cdot \hat{n}}{\|\nabla u\|}~ds. $$ For optimality, we can restrict $u$ to having homogeneous Neumann boundary conditions so that this boundary term disappears. Our directional derivative can now be written using a duality pairing as $$ \begin{aligned} \delta J[u]v &= \langle Ku-f, Kv \rangle -\lambda \left\langle \mathrm{div}\left(\frac{\nabla u}{\|\nabla u\|}\right), v\right\rangle \\ &= \left\langle K^* (Ku-f) - \lambda\mathrm{div}\left(\frac{\nabla u}{\|\nabla u\|}\right), v \right\rangle. \end{aligned} $$ If we write $v = \hat{u}-u$ and require that this directional derivative is positive ($J$ is increasing) in any feasible direction, then we obtain the desired result.

  • $\begingroup$ Thank you very much for the clarification $\endgroup$
    – Mohamed
    Mar 27 at 18:45

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