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I am trying to solve a PDE similar to the advection-diffusion equation:

$$ \frac{\partial T}{\partial t} + (\vec{u} \cdot \nabla)T = D \Delta T $$

(where $\vec{u}$ is a known advecting vector field) with some boundary conditions on a domain bounded by two intersecting curves $h(x)$ and $b(x)$ (for example parabolas, but could be something else):

enter image description here

I also know that $h(x) = a b(x) + c$, with $a$ and $c$ real constants.

Boundary conditions can be written as follows:

  • $f(T, \vec{\nabla}T) = 0$ on $h$
  • $g(T, \vec{\nabla}T) = 0$ on $b$

As I am fine solving it on a square/rectangular domain, I first tried to find a mapping from this domain to a rectangular one, without success.

Hence, what approaches would you recommend? I'm fairly new to this, and the number of existing approaches is confusing.

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    $\begingroup$ Finite volume or finite element method can be used. $\endgroup$
    – cfdlab
    Commented Sep 11 at 16:07
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    $\begingroup$ Meshing your region then using an existing library is probably your best bet. CLAWPACK (Finite volumes) and deal.ii or FENICS (finite elements) are good bets that probably have example code for edvection diffusion equations. $\endgroup$
    – whpowell96
    Commented Sep 11 at 18:00
  • $\begingroup$ Thanks! I expected to use finite elements, but I don't get why finite volumes are also appropriate, more than finite differences? $\endgroup$
    – Liris
    Commented Sep 12 at 8:56
  • $\begingroup$ Finite elements can become unstable for advection-dominated problems. There are ways to alleviate this, but finite volumes tend to handle advection better than FEM "out of the box," so to speak. $\endgroup$
    – whpowell96
    Commented Sep 12 at 13:03
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    $\begingroup$ Check out parabolic coordinates, en.wikipedia.org/wiki/Parabolic_coordinates, it sounds like that's what you need there $\endgroup$
    – Maxim Umansky
    Commented Sep 26 at 19:09

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Consider the domain in the figure, limited by a(x) and b(x) from bottom and top; and $x_{min}$, $x_{max}$ from left and right; assume $b(x)>a(x)$ for $x \in [x_{min},x_{max}]$.

enter image description here

Now introduce new coordinates $\xi,\eta$, according to

$ \xi=(x-x_{min})/(x_{max}-x_{min}) \\ \eta=(y-a(x))/(b(x)-a(x)) $

It is easy to see that the original domain maps to a unit square in the $(\xi,\eta)$ coordinates. Now, with some standard algebra one can work out the transformation of the involved differential operators to the $(\xi,\eta)$ coordinates and solve the equation in the new coordinates.

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    $\begingroup$ Would this sort of method break at the corners at the left and right of OP’s diagram? $\endgroup$
    – whpowell96
    Commented Sep 28 at 18:22
  • $\begingroup$ @whpowell96 Yes, the corner points correspond to singularity of the metric transformation. But for simple kinds of boundary conditions (e.g., Dirichlet) one does not need to evaluate the metric transformation at the corner points, so one still can carry out a numerical solution. $\endgroup$
    – Maxim Umansky
    Commented Sep 29 at 3:50
  • $\begingroup$ On the boundaries, I have a 'non-flux through boundary' kinda of boundary condition. In some specific cases, it comes down to a Dirichlet boundary condition, but not in the general case ... I could however slightly change the domain, and add a (small) vertical wall at both edges instead of corners, while still being physically relevant I think. Could you explain how this relates to the suggested parabolic coordinates that you mentioned above? $\endgroup$
    – Liris
    Commented Sep 29 at 11:26
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    $\begingroup$ @Liris Adding a small gap at the end points is a good thing, this will allow setting independent BC on the top and bottom boundaries. The method proposed here applies to general a(x) and b(x), but it results in non-orthogonal coordinates which is a complication but nothing major. The parabolic coordinates are orthogonal, but they could be used only for specific parabolic form of a(x) and b(x). $\endgroup$
    – Maxim Umansky
    Commented Sep 29 at 13:58
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    $\begingroup$ @Liris No, with parabolic coordinates there is no singularity in the metric tensor anywhere, see the Wikipedia article. $\endgroup$
    – Maxim Umansky
    Commented Sep 29 at 14:50

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