Proper advection equation for non-conservative values in polar cylindrical coordinates. I'm numerically solving a system of PDE's consiting of some conservation laws (Euler equations) along with advection of non-additive values (like molar mass or unit heat capacity, for example). The question is concerned with advection equations. In 2D cartesian coords, advection of value $\phi(x,y,t)$ is governed by this equation:
$$
\frac{\partial \phi}{\partial t} + u \frac{\partial \phi}{\partial x} + v \frac{\partial \phi}{\partial y} = 0
$$
where $u$ and $v$ are velocity components along $x$ and $y$ axes. The velocity vector is provided by solving Euler equations.
This equation may be written in quasi-conservative form with source term:
$$
\frac{\partial \phi}{\partial t} + \frac{\partial \phi u}{\partial x} + \frac{\partial \phi v}{\partial y} = \phi\frac{\partial u}{\partial x} + \phi\frac{\partial v}{\partial y}
$$
In this form it is being successfully solved by some numerical method. By "succesfully" I mean that the solution looks physical enough: values of $\phi$ are advected with the gas flow and are not "accumulated" in the areas where gas flow converges (as would gas density accumulate, for example).
Now I try to solve the same system in axysimmetric cylindrical coordinates $(x, r)$. I found this form of 1D scalar advection equation in the literature:
$$
\frac{\partial \phi}{\partial t} + \frac{v}{r} \frac{\partial r\phi}{\partial r} = 0
$$
Or, in quasi-conservative form with source term (also, multiplied by $r$):
$$
\frac{\partial r\phi}{\partial t} + \frac{\partial r \phi v}{\partial r} = r \phi\frac{\partial v}{\partial r}
$$
(I derived this question by hand, so cannot 100% guarantee that is correct, though)
One exact solution is given in the paper:
$$
\phi(r,0) = \left\{\begin{array}{ll}
\frac{sin^4(\pi r)}{r}, & 0\leq r\leq 1\\
0, & r > 1
\end{array}\right\},
$$
$$
\phi(r,t) = \left\{\begin{array}{ll}
\frac{sin^4(\pi (r-vt))}{r}, & vt \leq r\leq vt + 1\\
0, & \text{otherwise.}
\end{array}\right\}
$$
Here is graphic representation of this solution for $v = 1$ at $t = 1$ from that paper:

It seems that original function shape is moved by 1 unit of distance away from symmetry axis and diminished in height by a factor of about ~3.
My questions are:

*

*Why does the height of the function shape diminish? In Cartesian coords, the original shape would just move and preserve it's shape. To my intuition, such phenomenon looks more like a conservation law behaviour in polar coords: function height decreases due to its base expansion so that total integral of the function remains constant.

*How should one properly describe an advection (or transport) of some non-additive value in cylindrical coords? I.e. of such a value that is simply "glued" to fluid particles and would not decrease in such divergent flow as shown above (or increase in convergent flow).

 A: Yes, the above observations and derivations seem correct.

*

*In cylindrical coordinates, the advection equation $\partial_t \phi + \frac{v}{r} \partial_r (r\phi) = 0$ for a given quantity $\phi(r,t)$ transported with uniform velocity $v\equiv 1$ m/s along the radial direction is a conservation law. In fact, it might be rewritten as
$$
\partial_t \phi + \text{div}(\phi \boldsymbol v) = 0 \, , \qquad \boldsymbol v= v \hat{\boldsymbol r}
$$
where $\hat{\boldsymbol r}$ is the radial unit vector.
Now, integration yields
$$
0 = \int_{\Omega} \left[\partial_t \phi + \text{div}(\phi \boldsymbol v)\right] \text{d}V = \frac{\text d}{\text d t} \int_{\Omega} \phi\, \text{d}V + \int_{\partial\Omega} \phi \boldsymbol v \cdot \boldsymbol{n}\, \text{d}S \, ,
$$
where we have used the divergence theorem. Here, $\boldsymbol n$ is the outgoing normal unit vector to the boundary $\partial\Omega$ of the spatial domain $\Omega$. Therefore, the variation in time of the total quantity $\int \phi \, \text{d}V$ can be expressed in terms of the flux $\phi \boldsymbol v$ passing through the boundary.


*If in-plane transport along a given direction is desired, a straightforward option would be to transform the initial Cartesian transport equation (a.k.a. the "color equation")
$$
\partial_t \phi + \text{div}(\phi \boldsymbol{v}) = 0, \qquad \boldsymbol{v} = u\hat{\bf x}+v\hat{\bf y}
$$
with $u$, $v$ constant, into cylindrical coordinates. This is done by using the relationships
\begin{aligned}
\hat{\bf x} &= \cos\theta\, \hat{\boldsymbol r} - \sin\theta\, \hat{\boldsymbol\theta} \\
\hat{\bf y} &= \sin\theta\, \hat{\boldsymbol r} + \cos\theta\, \hat{\boldsymbol\theta}
\end{aligned}
between unit vectors. Therefore, the velocity field becomes velocity $\boldsymbol{v} = v_r \hat{\boldsymbol r} + v_\theta \hat{\boldsymbol \theta}$ , and the transport equation $$
\partial_t \phi + v_r \partial_r \phi + \frac{v_\theta}{r} \partial_\theta \phi = 0
$$
$$
\text{with}\qquad v_r = u\cos\theta + v\sin\theta ,\qquad v_\theta = -u\sin\theta + v\cos\theta
$$
follows from the expression of the divergence operator in cylindrical coordinates.

Note: The product rule for the divergence operator reads (cf. vector calculus identities)
$$
 \text{div}(\phi \boldsymbol{v}) = \phi  \,\text{div}\, \boldsymbol{v} + (\text{grad}\, \phi)\cdot \boldsymbol{v} \, .
$$
By definition, the material derivative $\dot\phi$ in Eulerian coordinates reads
$$
\dot\phi = \partial_t \phi + (\text{grad}\, \phi)\cdot \boldsymbol{v} 
$$
so that the conservation law $\partial_t \phi =- \text{div}(\phi \boldsymbol{v})$ amounts to the evolution equation
$$
\dot \phi/\phi = -\text{div}\, \boldsymbol{v} \, .
$$
Therefore, the quantity $\dot \phi$ vanishes in incompressible or in uniform flow, in which cases the advection equation $\dot\phi = 0$ is equivalent to the conservation law. In particular, note that the conservation law is satisfied by the mass density $\phi = \rho$, owing to the balance principles of continuum physics (cf. integral form above).
