theoretical handle on how to choose in an optimal way. These are:
<ul>
<li>The time step must satisfy a CFL condition
- $k\le \min_K \frac{c_kh_K}{\|\mathbf{u}\|_{L^\infty(K)}}$. Here, $c_k$ is
+ $k\le \min_K \frac{c_kh_K}{\|\mathbf{u}\|_{L^\infty(K)}}$. Here, $c_k$ is
dimensionless, but what is the right value?
<li>In the computation of the artificial viscosity,
@f{eqnarray*}
\frac{\|R_\alpha(T)\|_{L^\infty(K)}}{c(\mathbf{u},T)}
\right\},
@f}
- with $c(\mathbf{u},T) =
+ with $c(\mathbf{u},T) =
c_R\ \|\mathbf{u}\|_{L^\infty(\Omega)} \ \mathrm{var}(T)
\ |\mathrm{diam}(\Omega)|^{\alpha-2}$.
Here, the choice of the dimensionless numbers $\beta,c_R$ is of
value depends on that of the others, and most likely also on the space
dimension and polynomial degree of the finite element used for the
temperature. Below we'll discuss a few numerical experiments to choose
-constants.
+constants.
-<h4> Choosing <i>c<sub>k</sub></i> and β </h4>
+<h4> Choosing <i>c<sub>k</sub></i> and beta </h4>
These two constants are definitely linked in some way. The reason is easy to
-see: In the case of a pure advection problem,
+see: In the case of a pure advection problem,
$\frac{\partial T}{\partial t} + \mathbf{u}\cdot\nabla T = \gamma$, any
-explicit scheme has to satisfy a CFL condition of the form
+explicit scheme has to satisfy a CFL condition of the form
$k\le \min_K \frac{c_k^a h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$. On the other hand,
-for a pure diffusion problem,
+for a pure diffusion problem,
$\frac{\partial T}{\partial t} + \nu \Delta T = \gamma$,
explicit schemes need to satisfy a condition
$k\le \min_K \frac{c_k^d h_K^2}{\nu}$. So given the form of $\nu$ above, an
The find the right balance, the only way is to do a few computational
experiments. Here's what we did: We modified the program slightly to allow
-less mesh refinement (so we don't always have to wait that long) and to choose
+less mesh refinement (so we don't always have to wait that long) and to choose
$
\nu(T)|_K
=
big (i.e. choose a $c_k$ bigger than theoretically allowed) then we will get
exponential growth of the temperature. If we choose $\beta$ too small, then
the transport stabilization becomes insufficient and the solution will show
-significant oscillations but not exponential growth.
+significant oscillations but not exponential growth.
<h5>Results for Q<sub>1</sub> elements</h5>
The way to interpret these graphs goes like this: for $\beta=0.01$ and
$c_k=\frac 12,\frac 14$, we see exponential growth or at least large
-variations, but if we choose
-$k=\frac 18\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$
-or smaller, then the scheme is
-stable though a bit wobbly. For more artificial diffusion, we can choose
-$k=\frac 14\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$
+variations, but if we choose
+$k=\frac 18\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$
+or smaller, then the scheme is
+stable though a bit wobbly. For more artificial diffusion, we can choose
+$k=\frac 14\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$
or smaller for $\beta=0.03$,
-$k=\frac 13\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$
+$k=\frac 13\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$
or smaller for $\beta=0.1$, and again need
-$k=\frac 1{15}\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$
+$k=\frac 1{15}\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$
for $\beta=0.5$ (this time because much diffusion requires a small time
-step).
+step).
So how to choose? If we were simply interested in a large time step, then we
would go with $\beta=0.1$ and
choosing $\beta$ too small is bad: too little artificial diffusion leads to
over- and undershoots that aren't diffused away. On the other hand, for large
$\beta$, the minimum temperature drops below zero at the beginning but then
-quickly diffuses back to zero.
+quickly diffuses back to zero.
On the other hand, let's also look at the maximum temperature. Watching the
movie of the solution, we see that initially the fluid is at rest. The source
choose the time step very small to keep things under control. Too
large values of $\beta$ make for more diffusion, but again require
small time steps. The best value would appear to be $\beta=0.03$, as
-for the $Q_1$ element, and the we have to choose
+for the $Q_1$ element, and the we have to choose
$k=\frac 18\frac{h_K}{\|\mathbf{u}\|_{L^\infty(K)}}$ — exactly
half the size for the $Q_1$ element, a fact that may not be surprising
if we state the CFL condition as the requirement that the time step be
for each value of $\beta$ and find the best value of $\beta$. What one
finds is that for the same $\beta$ already used in 2d, the time steps
needs to be a bit small, by around a factor of 1.2 or so. This is
-easily explained: the time step restriction is
+easily explained: the time step restriction is
$k=\min_K \frac{ch_K}{\|\mathbf{u}\|_{L^\infty(K)}}$ where $h_K$ is
the <i>diameter</i> of the cell. However, what is really needed is the
distance between mesh points, which is $\frac{h_K}{\sqrt{d}}$. So a