How can we calculate the Euler-lagrange equations?

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 ?

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

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.