These test cases solve the 2D advection equation. Three test cases are proposed:

- linear advection of a sinusoidal hills;
- block advection of a cosinusoidal hill;
- sheared advection of a cosinusoidal hill;
- steady linear advection

The Energy equation is used but results are valid for the species equations as well. The aims of this test cases are to:

- validate the discretization of the advection term of a cell advection/diffusion equation;
- compare available schemes.

#
Configurations

##
2D linear advection

Domain is a unit square \( \Omega = [0, 1]^2 \). The initial temperature field \( T^0(x,y) = \sin(2 \pi x) \sin(2 \pi y) \) is advected in a periodic domain with a constant velocity field \((u,v) = (1,1) \). Total time is equal to \( 10s \) (ten periods).

##
2D block advection

Domain is a unit square \( \Omega = [0, 1]^2 \). The initial temperature field \( T^0(x,y)=0.5(1 + \cos(\pi r)) \), with \( r=min \left( 1,4 \sqrt{(x - 0.25)^2 + (y - 0.25)^2} \right) \), is advected in a rotating velocity field \((u,v)\):

- \( u = -2 \pi \left(y - 0.5H \right) \)
- \( v = 2 \pi \left(x - 0.5H \right) \)

where \( H \) is domain height.

Total time, equal to \( 5s \), corresponds to five complete rotations of the initial temperature profile.

##
2D sheared advection

Domain is a unit square \( \Omega = [0, 1]^2 \). The initial temperature field \( T^0(x,y)=0.5(1 + \cos(\pi r)) \), with \( r=min\left(1,4\sqrt{(x - 0.25)^2 + (y - 0.25)^2}\right) \), is advected in a sheared velocity field \((u,v)\) with:

- \( u = sin^2(\pi x)sin(2\pi y)cos(\pi t/T) \)
- \( v = -sin^2(\pi y)sin(2\pi x)cos(\pi t/T) \)

Total time is equal to \( T=5s \). We expect a maximum flow deformation at \( T/2 \), while final temperature should be equal to the initial one.

##
2D steady linear advection

The flow is parallel to two solid lines that cross the domain. Temperature is equal to 50 between the two lines, equal to 100 elsewhere. Velocity field \((u,v)\) is equal to \((0.4, 0.6)\). The first line crosses the point (0, 0.4) and (0.4, 0.4), the second one the points (0.4, 0) and (1, 0.6). Boundary conditions are set accordingly.

#
Runtime parameters

#
Comments

#
Results