$\phi: \mathbb{R} \to \mathbb{R}$ is a given, smooth gradient field and I come from the equation $\frac{\partial}{\partial t} u(x,t) = \text{div}(\phi(x,t))$ for $x\in \mathbb{R}$, $t\geq 0$ and some initial conditions for $u$.
My problem is different from that in so far as I would like $u$ to locally increase due to incoming flux (as stated by the equation) BUT not to decrease with outgoing flux:
On a small cell $[x,x+\epsilon]$ there is $\text{div}(\phi(x,t))=\frac{\phi(x+\epsilon,y,t)-\phi_1(x,y,t)}{\epsilon}$ which in the limit gives $\frac{\partial}{\partial x} \phi$.
Assume that at some $x$ there is $\phi(x,t)>0$ at time $t$, i.e. the field points to the right. Numerically, in a small time step $\Delta t$ the quantity $\frac{\phi(x+\epsilon,t)}{\epsilon} \cdot \Delta t$ would move to the cell on the right, $[x+\epsilon,x+2\epsilon]$, and $\frac{\phi(x,t)}{\epsilon} \cdot \Delta t$ would come from the cell on the left, $[x-\epsilon,x]$.
Now what I would like is that $u$ in the cell on the right, $[x+\epsilon,x+2\epsilon]$, still increases by $\frac{\phi_1(x+\epsilon,t)}{\epsilon} \cdot \Delta t$ due to the flux from the cell $[x,x+\epsilon]$. And that $u$ in $[x,x+\epsilon]$ cell still increases by $\frac{\phi(x,t)}{\epsilon} \cdot \Delta t$, which comes from its left neighbor. (So far everything is like in the original equation) BUT, that $u$ in $[x,x+\epsilon]$ cell is NOT decreased by $\frac{\phi(x+\epsilon,t)}{\epsilon} \cdot \Delta t$. That is, in this time step, the quantity $\frac{\phi(x+\epsilon,t)}{\epsilon} \cdot \Delta t$ is effectivly created in $x$ out of nowhere.
I hope this gets my aim across. What I described however is clearly wrong because as $\epsilon \to 0$, the generated quantity goes to $\infty$, which is not what I intend. Does anyone see what I mean and if (or if not) it is feasible to formulate this well?
Thanks in advance!
I think the equation $$\frac{\partial}{\partial t} u(x,t) = k |\phi(x,t)|$$ captures your informal description: the rate of mass creation in a cell is proportional to the rate of flow through the cell. There is no derivative of $\phi$ involved; since we are not taking differences of $\phi$-values, we should not be dividing by $dx$ either.