In the context of electrical impedance tomography, one encounters the following partial differential equation.
Let $ \Omega $ be a bounded Lipschitz domain in $ \mathbb{R}^2 $ with $C^1$ boundary $ \partial \Omega $ with outward normal $ n $. Assuming $ \Omega $ contains material with electrical conductivity $ q(x) $ satisfying $ q(x) \geq q_0 > 0 $. Then the electrical potential $ u(x) $ inside $ \Omega $ satisfies \begin{gather} \label{eqn:eitpde} - \nabla \cdot (q \nabla u) = 0 \text{ in } \Omega \tag{1} \\ q \nabla u \cdot n = f \text{ on }\partial \Omega \tag{2} \label{eqn:eitbc} \end{gather} where $ f(x) $ is the applied current density on $ \partial \Omega $ satisfying $ \int_{\partial \Omega} f(x) \operatorname{d}\! x = 0 $.
In solving the inverse problem, one may define the following mismatching functional \begin{equation} \tag{3} \label{eqn:mf} F(q) = \frac{1}{2} \int_{\partial \Omega} (u - m)^2 \end{equation} (The simplified version is stated here for easier discussion, for the complete mismatching functional, please refer to the above reference.) Then it remains to compute the gradient of $F$, now according to my understanding, the gradient of $F$ is the functional derivative, i.e. Gateaux derivative of $F$.
From the above reference, the gradient of $F$ is computed as \begin{equation} \tag{4} \label{eqn:gradf} \frac{\operatorname{d}\!F}{\operatorname{d}\!q} = - \nabla u \cdot \nabla z \end{equation} where $ z $ satisfies \begin{gather} \label{eq:avpde} - \nabla \cdot (q \nabla z) = 0 \text{ in }\Omega \tag{5} \\ q \nabla z \cdot n = u - m \tag{6} \label{eq:avbc} \end{gather} with the constraint \begin{equation} \tag{7} \label{eqn:zcons} \int_{\partial \Omega} z = 0. \end{equation} My question is
How do one obtain \eqref{eqn:gradf}?
In attempt to compute the functional derivative, I have first assumed a perturbation in $ q $ by $ \epsilon \tilde{q} $ which induces a corresponding perturbation in $ u $ by $ \epsilon \tilde{u} $. The change in $F$, denoted by $ \delta F $ is then \begin{equation} \tag{8} \label{eqn:Fturb} \delta F(q) = \epsilon \int_{\partial \Omega} \tilde{u}(u - m) + O(\epsilon^2). \end{equation} But then, the perturbations $ \tilde{u} $ and $\tilde{q}$ are related by \begin{equation} \tag{9} \label{eqn:uqturb} - \nabla \cdot [(q + \epsilon \tilde{q}) \nabla (u + \epsilon \tilde{u})] = 0 \end{equation} which transforms to (with ignoring higher order terms) \begin{equation} \tag{10} \label{eqn:uqturbfinal} - \epsilon [ \nabla \cdot (\tilde{q} \nabla u) + \nabla \cdot (q \nabla \tilde{u})] = 0 \end{equation} I tried to multiply \eqref{eqn:uqturbfinal} by $ z $ and integrate over $ \Omega$, with applying Green's first identity to yield \begin{equation} \tag{11} \label{eqn:asq} \int_{\partial \Omega} q z \nabla \tilde{u} \cdot n - \int_\Omega q \nabla z \cdot \nabla \tilde{u} - \int_\Omega \tilde{q} \nabla z \cdot \nabla u = 0. \end{equation} But this is the place where I got stuck, it seems like there is no evident choice of $ z $ such that when \eqref{eqn:asq} is added to \eqref{eqn:Fturb}, the perturbations are decoupled. I would appreciate very much if anyone can give a guide on how should I proceed or point out what did I miss or what went wrong.
Update (2020-02-10)
So basically I reworked and apply appropriate Green's identities instead, Instead of \eqref{eqn:asq}, I've got \begin{equation} \tag{12} \label{eqn:asqnew} \int_{\partial \Omega} z \tilde{q} \nabla u \cdot n - \int_\Omega \tilde q \nabla z \cdot \nabla u + \int_{\partial \Omega} q z \nabla \tilde u \cdot n - \int_{\partial \Omega} q \tilde u \nabla z \cdot n - \int_\Omega \tilde{u} \nabla \cdot ( q \nabla z ) = 0. \end{equation} So by adding \eqref{eqn:asqnew} to \eqref{eqn:Fturb}, one obtain \begin{multline} \label{Fturbnew} \tag{13} \delta F(q)/\epsilon = \int_{\partial \Omega} \tilde{u}(u - m) + \int_{\partial \Omega} z \tilde{q} \nabla u \cdot n - \int_\Omega \tilde q \nabla z \cdot \nabla u \\ + \int_{\partial \Omega} q z \nabla \tilde u \cdot n - \int_{\partial \Omega} q \tilde u \nabla z \cdot n - \int_\Omega \tilde{u} \nabla \cdot ( q \nabla z ) + O(\epsilon^2). \end{multline} Now it is evident that if one assumes that $ \nabla u \cdot n $ is given, then $ \tilde q = 0 $ on $ \partial \Omega$ by \eqref{eqn:eitbc}. From the given gradient one also knows that the term $ \int_{\partial \Omega} q z \nabla \tilde u \cdot n $ should be zero, So that, if $ z $ indeed satisfies \eqref{eq:avpde} and \eqref{eq:avbc}, then \eqref{Fturbnew} becomes \begin{equation} \delta F(q)/\epsilon = - \int_\Omega \tilde q \nabla z \cdot \nabla u + O(\epsilon^2). \end{equation} So, one picks $ \tilde q = \nabla z \cdot \nabla u $ to ensure the functional decreases along the gradient direction. My new question would be:
How do you know the term $ \int_{\partial \Omega} q z \nabla \tilde u \cdot n $ is zero?