Let $d\in\left\{2,3\right\}$ and $\Omega_t$ be the bounded set occupied by a fluid at time $t\ge 0$. Let $c\in\Omega_0$ be a particle and $$[0,\infty)\to\mathbb R^d\;,\;\;\;t\mapsto X_t(c)\in\Omega_t\tag 1$$ be the movement of $c$ over time. Let $x:=X_t(c)$. Then, the speed of movement of $c$ along the path $(1)$ is described by $$v_t(x):=\frac{\partial X_t}{\partial t}(c)\;.\tag 2$$ [Note, that $(2)$ is a well-defined mapping $\Omega_t\to\mathbb R^d$]. Now, let $\eta_t:\Omega_t\to[0,\infty)$ be the concentration of imaginary things in the fluid at time $t\ge 0$. Then, $$N_t:=\int_{\Omega_t}\eta_t\;{\rm d}\lambda$$ is the amount of things in $\Omega$ at time $t\ge 0$.
It's easy to show, that \begin{equation} \begin{split} \frac{{\rm d}N_t}{{\rm d}t}&=\int_{\Omega_t}\frac{\partial\eta_t}{\partial t}+\nabla\cdot\eta_tv_t\;{\rm d}\lambda\\ &=\int_{\Omega_t}\left(\frac{\partial}{\partial t}+v_t\cdot\nabla\right)\eta_t+\eta_t(\nabla\cdot v_t)\;{\rm d}\lambda \end{split}\tag 3 \end{equation} [Note, that $(3)$ is Reynolds' transport theorem in its form for a material element].
Let's assume, that we're considering an incompressible Newtonian fluid with uniform density $\rho$ and viscosity $\nu$. By the instationary Navier-Stokes equations, $$\left\{\begin{matrix}\displaystyle\left(\frac\partial{\partial t}+v_t\cdot\nabla\right)v_t&=&\displaystyle\nu\Delta v_t-\frac 1\rho\nabla p_t+f_t&&\text{in }\Omega_t\\\nabla\cdot v_t&=&0&&\text{in }\Omega_t\end{matrix}\right.\;,\tag 4$$ for all $t>0$, where $p_t:\Omega_t\to\mathbb R$ is the pressure of the fluid and $f_t:\Omega_t\to\mathbb R^d$ is the sum of all external forces.
Substituting the second equation of $(4)$ into $(3)$ yields $$\frac{{\rm d}N_t}{{\rm d}t}=\int_{\Omega_t}\left(\frac{\partial}{\partial t}+v_t\cdot\nabla\right)\eta_t\;{\rm d}\lambda\tag 5$$
Now, I want to come up with an advection-diffusion equation of the form $$\left(\frac\partial{\partial t}+v_t\cdot\nabla\right)\eta_t=\kappa\Delta\eta_t+s_t\;\;\;\text{in }\Omega_t\tag 6$$ where $\kappa\in\mathbb R$ is somehow a diffusion rate and $s_t:\Omega_t\times[0,\infty)\to\mathbb R^d$ is the sum of all external sources.
However, $(5)$ is an "advection equation", which doesn't take diffusion into account. Taking a look at $(6)$, it seems like we only need to show, that $$\frac{{\rm d}N_t}{{\rm d}t}=\int_{\Omega_t}\kappa\Delta\eta_t+s_t\;{\rm d}\lambda\;.\tag 7$$
Let me try to do something stupid: Assuming conservation of mass, $$N_t=N_0\;\;\;\text{for all }t\ge 0$$ and thereby $$0=\frac{{\rm d}N_t}{{\rm d}t}\stackrel{(3)}=\int_{\Omega_t}\frac{\partial\eta_t}{\partial t}+\nabla\cdot\eta_tv_t\;{\rm d}\lambda\;.\tag 8$$ Using Fick's first law, we obtain $$\eta_tv_t=-d_t\nabla\eta_t\;,\tag 9$$ where $d_t:\Omega_t\to\mathbb R$ is the diffusion coefficient of the corresponding fluid. Combination of $(8)$ and $(9)$ yields $$\frac{\partial\eta_t}{\partial t}=\nabla\cdot(d_t\nabla\eta_t)\;.\tag{10}$$ However, we don't want to assume conservation of mass. If we would have $$\int_{\Omega_t}s_t-(v_t\cdot\nabla)\eta_t=\frac{{\rm d}N_t}{{\rm d}t}\tag{11}$$ instead of $(8)$, the same argumentation would yield $$\frac{{\rm d}N_t}{{\rm d}t}=\int_{\Omega_t}\nabla\cdot(d_t\nabla\eta_t)+s_t\;{\rm d}\lambda\tag{12}$$ which is exactly what I'm looking for.
Now, I need somebody to make sense of $(11)$.