Model Task 0A:
Programming the 1D upstream
scheme
ATM 562 Fall 2015
Fovell
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Overview
• Program the 1D linear wave equation using the
upstream approximation in a periodic domain
• Task illustrates the basics of
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Model initialization
Time stepping
Application of boundary conditions
Amplitude and phase errors
Dependence on wavelength and time step
Avoidance of roundoff errors
• Use programming language (e.g., Fortran, C/C++,
Python) and visualization package (e.g., NCL, GrADS,
GNUplot, Excel*) of your choice
*Excel won’t be useful for model tasks ≥ 1
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Problem
• 1D linear wave equation
– When c > 0, wave translates to the right
• Upstream approximation is forward in time
and upwind in space
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Explicit form
• This approximation can be written explicitly as
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The initial time = 0 seconds
The time index n starts at 0
The space index runs from i = 1 to NX
The exact solution preserves its original shape
With periodic boundary conditions, the exact
solution will return to its original position after a
certain number of time steps
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Program structure
• Declare needed arrays
– I will call present value “u”, forecast value “up”
• Initialize parameters and constants (NX, n, dt, dx,
trigonometric p, etc.)
• Provide initial condition for u at n = 0 for all i
• Step ahead by time step dt, using upstream scheme to
create up
– i.e., time = time + dt … not the best choice (see next slide)
• Apply boundary conditions (BCs) on up
• Set for new time step (i.e., u = up)
• If time < timend, loop
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Tips
• Trigonometric p to machine precision can be
obtained with 4*arctan(1.0)
– To avoid confusion with nondimensional pressure
(also called “pi”), I call this “trigpi”
• To avoid roundoff error, advance time as time
= (n-1)*dt instead of time = time + dt (see
Appendix)
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Periodic boundary conditions
• For NX points, NX-2 will be real points, and
the other two will facilitate application of the
boundary conditions (BCs)
• For periodic BCs
u(1) = u(nx-1) and u(nx) = u(2)
– So, we loop from i = 2, nx-1 (the real points) and
then update the fake points
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Test problem
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Let c’ = c∆t/∆x (by definition)
Test: c = 1.0, dt = 1.0, dx = 1.0, so c’ = 1.0
NX = 52 (50 real points)
NT = 50 (timend = 50 since dt = 1.0)
Wavelength L = 50∆x (one sine wave in domain), with
amplitude 1.0
Execute for one revolution
Plot on next slide shows initial condition (= exact solution)
and upstream approximation after one revolution
For c’ = 1.0, there should be no significant error (so this is
your model test)
We will manipulate c’ by changing ∆t
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After 1 revolution
Upstream approximation to u_t = -cu_x
1.5
1.0
0.5
0.0
0
5
10
15
20
25
30
35
40
45
50
-0.5
-1.0
-1.5
exact
upstream (cfl=1, 1 rev)
(No significant error)
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Initial condition for u
wavelength = 50
! i.e., 50∆x
amp = 1.0
! amplitude
do i=2,nx-1
! Loop over real points
xi=float(i-2)
u(i) = amp*sin(2*xi*trigpi/wavelength)
print *,' i ',i,' u ',u(i)
enddo
C enforce periodic boundary conditions
u(nx)=u(2)
u(1)=u(nx-1)
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Exact solution for u at time step n
do i=2,nx-1
c exact solution at any time n
xi = float(i-2)-c*n*dt
uexact(i) = amp*sin(2*xi*trigpi/wavelength)
enddo
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Errors
• Amplitude error occurs if the original wave amplitude
is not preserved
• Phase error occurs if the simulated wave translates too
quickly or slowly
• Plots on next slide show amplitude error (left) and
phase error (right) as a function of wavelength
(horizontal axis, longer waves at left end)
• Plots on next slide show:
– Amplitude decays if c’ < 1, grows if c’ > 1, and shorter
waves handled worse than longer waves
– Waves propagate too slowly for all wavelengths; errors
reasonable for long waves but large for short waves
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Amplitude and phase errors
(Fig. 4.3 from notes)
0.9
0.96
1.0
> 1.0
0.9
0.6
Values < 1: damping
Values < 1: too slow
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Amplitude error
• For c’ = 1, upstream scheme has essentially zero
amplitude error
• For c’ ≠ 1, performance worst for shortest waves
– c’ > 1 unstable – solution grows with time
– Sampling the wave propagation better (i.e., c’ < 1)
makes the solution worse (counterintuitive!)
– For c’ < 1, performance worst at c’ = 0.5 for all waves
(even more counterintuitive!)
– At wavelength of 4∆x and c’ = 0.5, amplitude loss is
30% (0.7). That is per time step.
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Phase error
• For the exact solution, phase speed is independent of
wavelength. All waves move at the same speed, c. The
exact solution is nondispersive.
• Upstream scheme phase propagation is not absolutely
perfect even for c’ = 1
– For wavelengths > 4∆x, propagation too slow by about 4% (0.96)
– For c’ < 1, some wavelengths move too quickly, others too
slowly
• Again, the shorter the wavelength, the poorer the
performance
– 2∆x waves are stationary. (Really?)
• Since phase error depends on wavelength, the numerical
scheme is dispersive.
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Experiments
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Experiment #1
• Try values of ∆t < 1.0 (so c’ < 1) and compare
to exact solution for the 50∆x wave
• Observe how amplitude and phase error vary
with c’ after 1 revolution
• Compare your results to Fig. 4.3 (left)
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Experiment #2
• Try values of c’ > 1 (by increasing ∆t). What
happens? How does amplitude error after 1
revolution change with c’? This is linear
instability.
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Experiment #3
• Change the wavelength of the initial condition
to, say, 5∆x (wavelength = 5) and c’ = 0.5
(∆t = 0.5). From Fig. 4.3 from the notes,
expect to lose 20% amplitude per time step.
Plot maximum wave value vs. time. Do you?
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Warning: because of the periodic BCs, your initial condition has to have an integer
number of waves in the domain, or the BC will mishandle the wave. To explore
some wavelengths, it may be necessary to change NX.
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Experiment #4
• Compare solutions for the 50∆x wave with c’ =
0.5 and 0.25. Run each simulation for 50 time
units. According to Fig. 4.3 (left), the
amplitude error is supposed to be worse for c’
= 0.5. Is it? If not, why not?
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Experiment #5
• According to Fig. 4.3 (right), the 2∆x wave isn’t
supposed to move with time for any value of
c’. Does it move? If it does, why does it?
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Experiment #6
• An initial condition can consist of several waves, which
combine to make a certain shape. The exact solution, being
nondispersive, preserves that shape as the wave moves.
The upstream scheme, however, is dispersive, as it has a
different amplitude and phase error for different waves, so
the combined shape will change.
• Demonstrate this with an initial condition with two or more
waves, of possibly different amplitude. (Keep in mind you
need to have an integral number of waves in the initial
condition.) For example, combine a 50 and 10∆x wave,
each with initial amplitude of 1.0, and set c’=0.5.
Anticipate what the result would look like after 1 revolution
before running the model and checking your guess.
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Appendix: roundoff error
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Test #1
• For Fortran, a real-valued (single precision)
constant is 4 bytes long. A double precision
value is 8 bytes.
• Write a program that assigns 10 entries of an
single precision array to values of 0.1 to 0.9.
Print these values to at least 10 decimal
places. What do you observe?
• Now make the array elements double
precision. What happens?
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Test #2
• Let dt, time1 and time2 be single precision,
real variables
• Set dt to a value like 0.6, and time1 and
time2 to 0.
• Compare these after 20000 iterations (iter)
time1 = time1 + dt
time2 = float(iter)*dt
(For a model with a 4-5 sec time step, 20000 iterations represents about 1 day of integration.)
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Question for thought
• Most NWP models are single precision. Why?
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