Proper advection equation for non-conservative values in polar cylindrical coordinates.

Question

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:

1. 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.
2. 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).

Answer 1

Yes, the above observations and derivations seem correct.

1. 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.

2. 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.

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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).