Transport: Non-diffusive, flux conservative initial value problems and ...

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n+1
n+1/2
n
∆t
true characteristic
c(n)
xX
x
u(n+1/2)
u(n+1,j)
Figure 5.9: Schematic description of the semi-lagrangian advection scheme. x marks
the point at time n that lies on the true characteristic that will pass through our grid-point
j in time ∆t. X is the best approximation for this point found by taking a backwards
mid-point step from the future time; To get the concentration at this point we must use
some interpolation scheme. If there are no source terms, however, concentration remains
constant along a characteristic and c n+1
j = ˜c n . x is the position at time ∆t/2 where we
calculate the mid-point velocity.
point c n+1
j in time ∆t. The problem is to find this point because the solution
to Eq. (5.2.2) is that the concentration remains constant along characteristics i.e.
c n+1
j = ˜c. The important feature of transport equations, however, is that they can be
run backwards in time stably, thus rather than guessing at an old point and moving
it forward, we will start at our new point (j,n + 1) and use our ODE techniques to
shoot backwards in time. Fig. 5.9 (and 5.8d) uses a simple mid-point scheme. The
basic algorithm can be summarized as follows.
For each point j
1. given the velocity at our future point u n+1
j , find the fractional grid-point x
with coordinate ˜j = j − (∆tu n+1
j )/(2∆x) (i.e. take an Euler step back for
half a time step).
2. Using some interpolation scheme find the velocity at the half-time step u n+1/2
˜j
.
3. repeat the step nvit times using the new velocity to iterate and find a better
.
approximation to u n+1/2
˜j
4. When you’re happy use the centered velocity to find the fractional grid-point
X with coordinate ˜j ′ = j − (∆tu n+1/2
˜j
)/∆x)
5. Using another interpolation scheme, find the concentration at point ˜j ′ and
copy into c n+1
j
This formulation should be treated as a recipe that is easily varied. For example,
here we have used an iterative mid-point scheme, to find the point at time t − ∆t,
however, with only bit more work, a 4th order fixed step Runge-Kutta scheme
could also be easily employed.
j
∆x