An Example of Stabilization
This example uses the Heat Transfer interface. To illustrate the concepts, consider the problem
(3-5)
solved on the unit square. Equation 3-5 is discretized using 10 times 10 quadratic Lagrange elements. The boundary conditions are:
•
u = 1 for x = 0
•
u = 1 for y = 0
•
u = 0 for x = 1
•
u = 0 for y = 1
Figure 3-26 shows the mesh and boundary conditions. In general, using uniform meshes for transport problems is not recommended. Nevertheless, this example uses a uniform mesh to demonstrate the different stabilization techniques.
The expected solution rises slowly and smoothly from the left and lower boundaries and has sharp boundary layers along the upper and right boundaries. Figure 3-27 shows a reference solution obtained using 100-by-100 quadratic Lagrange elements with streamline diffusion and crosswind diffusion (see the next section). The arrows indicate the direction of β.
Figure 3-26: The computation domain, mesh, and boundary condition for Equation 3-5.
Figure 3-27: Reference solution of Equation 3-5. Solved using 100 times 100 quadratic elements with streamline diffusion and crosswind diffusion.
The cell Péclet number for this example is
Figure 3-28 displays the solution obtained using the mesh in Figure 3-26 and (unstabilized) Galerkin discretization. As can be expected with such a high Péclet number, the unstabilized solution shows little, if any, resemblance to the reference solution in Figure 3-27. The right plot in Figure 3-28 shows a cross-sectional plot along the dashed line, y = 0.8 and the corresponding reference solution. Notice that the unstabilized solution is destroyed by oscillations.
Figure 3-28: Equation 3-5 solved using unstabilized Galerkin formulation. The right plot compares the unstabilized solution (dashed line) along the dashed line in the left plot (y = 0.8) with the reference solution (solid line).
The Stabilization Techniques section explores how different stabilization techniques affect the solution of this example.