I'm reading an article in imaging processing and i've been stuck in some notation details that i think that are not allowing me to procede my mathematical adventure...
So we know that $$u(t)=-\frac{d}{dt}\Phi(t)$$
Where $\Phi:(0,T)\to\mathbb{R}^L$. After some rather interestating calculations we get $$\Phi(t)=\Psi_P(p(t))+\Psi_E(p(t))$$ Where $p$ the a trajectory of a particle. These $\Psi$ functions are given by $$\Psi_P(p(t))=\mu_0\int_{\mathbb{R}^d}\mathbf{p}_R(x)^Tc(x)\tilde{m}(H_S(x-p(t)))\:dx$$ and $$\Psi_E(p(t))=\mu_0\int_{\mathbb{R}^d}\mathbf{p}_R(x)^TH_S(x-p(t))\:dx$$ Where $\mathbf{p}_R:\mathbb{R}^d\to\mathbb{R}^{d\times L}$, $c:\mathbb{R}^d\to\mathbb{R}$, $\tilde{m}:\mathbb{R}^d\to\mathbb{R}^d$ and $H_S:\mathbb{R}^d\to\mathbb{R}^d$.
We assume that $\frac{d}{dt}\Psi_E(p(t))=0$. Then we can say (This is were the problem arises)
$$u(t)=-\nabla_Z\Psi_P(p(t))\frac{d}{dt}p(t)$$
(i can see that these is some sort of chain rule but i don't really get the $-\nabla_Z$ part) where (these is the other issue)
$$\nabla_Z\Psi_P(p(t))=\mu_0R^T\int_{\mathbb{R}^d} c(x)D_Z\tilde{m}(H_S(x-p(t)))\:dx$$ (Why $D_Z$?) Oh, here we made another assumption, it was that $\mathbf{p}_R(x)=R$.
Thank you so much for your time, if there is any function that you think its necessary for me to describe better, let me know.