Download Donna Calhoun Boise State University The Riemann problem

The Riemann problem shallow water wave
systems
Donna Calhoun
Boise State University
PASI 2013
Valparaiso, Chile
January 2-13
Thursday, January 10, 13
Plan
Part I : Discussion of the Riemann solution for onedimensional linear and nonlinear shallow water wave
equations
Part II : Approximate Riemann solvers in GeoClaw;
discussion of accuracy, and extensions to higher
dimensions; f-wave approach to well-balancing.
Part III : Adaptive mesh refinement (AMR).
PASI : Tsunami Workshop Jan 2-14 2013
Thursday, January 10, 13
Shallow water wave equations
The shallow water wave equations, given by
is an example of a system of equations written in conservative
form. More generally, we can write PDEs in conservative form
as
qt + f (q)x = 0
These are typically derived from conservation laws for mass,
momentum, energy, species, and so on.
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Thursday, January 10, 13
GeoClaw
•
Based on solving the conservative form of the shallow
water wave equations using a finite volume method.
•
At the heart of many finite volume methods is a Riemann
solver which is used to compute numerical fluxes.
•
In GeoClaw, these are stored in files like rpn2_geo.f
and rpt2_geo.f
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Finite volume method
Assume a conservation law of the form
qt + f (q)x = 0
Define cell averages over the interval Ci = [xi
Qni
1
=
x
1/2 , xi+1/2 ]
q(x, tn ) dx
Ci
How does the average evolve?
d
dt
d
q(x, t) dx =
f (q(x, t)) dx
Ci
Ci dx
= f (q(xi 1/2 , t)) f (q(xi+1/2 , t))
PASI : Tsunami Workshop Jan 2-14 2013
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Finite volume method
Evolution of the cell average value :
d
dt
Z
q(x, t) dx = f (q(xi
1/2 , t))
f (q(xi+1/2 , t))
Ci
Integrate in time
⇤
q(x, tn+1 ) dx
=
Ci
+
⇤
q(x, tn ) dx
Ci
⇤ tn+1
tn
f (q(xi
1/2 , t))
⇥
f (q(xi+1/2 , t)) dt
PASI : Tsunami Workshop Jan 2-14 2013
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Finite volume method
Using numerical fluxes, we use the update formula :
n+1
Qi
=
n
Qi
t
n
Fi+1/2
x
n
Fi 1/2
⇥
Written as
this form resembles the conservation law :
PASI : Tsunami Workshop Jan 2-14 2013
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Numerical fluxes
We want to approximate the numerical flux.
Z tn+1
1
n
Fi 1/2 ⇡
f (q(xi 1/2 , t)) dt
t tn
For an explicit time stepping scheme, we try to find formulas
for the flux of the form
n
Fi 1/2
Qi
2
Qi
n
n
F(Qi , Qi 1 )
=
1
Qi
xi
Qi+1
1/2
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1d Riemann problem
At each cell interface, solve the hyperbolic problem with
special initial data, i.e.
qt + f (q)x = 0
subject to
q(x, 0) =
⇢
Qi
Qi
Qi
1
x < xi
x > xx
1/2
1/2
1
t=0
Qi
xi
1/2
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1d Riemann problem
Qi
t=0
1
Qi
xi
t>0
Qi
1/2
q⇤
Solve the conservation
law with piecewise
constant initial data
1
Qi
xi
1/2
PASI : Tsunami Workshop Jan 2-14 2013
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1d Riemann problem
q
t>0
Qi
⇤
1
Qi
xi
1/2
Numerical flux at cell interface is then approximated by
Fi
1/2
= f (q )
⇤
This is the classical Godunov approach for solving hyperbolic
conservation laws.
•
Resolves shocks and rarefactions
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Conservation?
Integrating over entire domain, we have
d
dt
Z
Z
xb
q(x, t) dx =
xa
xb
(f (q))x dx = f (q(xa , t))
f (q(xb , t))
xa
Discrete case
M
X
Qn+1
i
=
i=1
M
X
Qni
i=1
=
M
X
i=1
Qni
M
X
t
Fi+1/2
x i=1
t
FM +1/2
x
Fi
1/2
F1/2
Quantities are conserved up to fluxes at domain boundaries.
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Scalar advection
Consider the constant initial value problem
qt + ūqx = 0
It is easy to verify that
solves the initial value problem.
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Scalar advection
We can describe the problem in terms of how the solution
behaves along curves in the x-t plane.
We might look for curves
solution is constant or
along which the
Then we would get
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Characteristic curves
But this is true only if
or
Solution is constant along characteristic curves. For
,
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Characteristic curves
The solution can be traced back along characteristics. That is,
can be found by determining the
from which the
solution propagated. Solve
or
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Scalar advection
Consider the scalar advection equation :
qt + ūqx = 0
The solution travels along characteristic rays in the (x,t) plane
given by
. For u < 0 :
t=1
t=0
PASI : Tsunami Workshop Jan 2-14 2013
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Riemann problem for scalar advection
qt + ūqx = 0
subject to particular initial conditions
q(x, 0) =
q
qr
x<0
x>0
t
q
qr
Solution :
PASI : Tsunami Workshop Jan 2-14 2013
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Scalar Riemann Problem
qt + ūqx = 0
subject to initial conditions
q
qr
q(x, 0) =
x<0
x>0
x/t = ū
t
Solution to the
Riemann problem
q
qr
Solution :
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Scalar constant coefficient advection
Discontinuity propagates at speed
and has strength
t=1
t=0
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Solving constant coefficient linear systems
qt + Aqx = 0,
A 2 Rm⇥m
We assume that A has a complete set of eigenvectors and real
eigenvalues and so can be written as
A = R⇤R
R = [r , r , . . . r ]
1
2
m
1
⇤ = diag(
1
,
2
,...
m
)
Examples : Linearized shallow water wave equations,
constant coefficient acoustics, ...
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Solving a constant coefficient system
!
q t + A qx = 0
qt + R⇤R
1
Define characteristic variables
(x, t) = R
1
q(x, t),
A 2 Rm⇥m
qx = 0,
R
m
Assume that A is
diagonalizable
as
(x, 0) = R
1
q(x, 0)
Characteristic equations decouple into m scalar equations :
p
⇥t
+
p
⇥xp = 0,
p = 1, 2, . . . , m
Solution to characteristic equations are given by
⇥ (x, t) = ⇥ (x
p
p
p
t, 0)
PASI : Tsunami Workshop Jan 2-14 2013
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Solving a constant coefficient systems
p
⇥t
q t + A qx = 0
+
p
p
⇥x
=0
Solution for general initial conditions q(x, 0) :
m
q(x, t) = R ⇥(x, t) =
⇥ (x, t) r
p
p
⇥ (x
p
p=1
m
=
q(x, t)
p
t, 0) r
p
p=1
t
`p q(x
x
⇥ 3 (x
3
t, 0)
⇥ 2 (x
2
t, 0) ⇥ 1 (x
1
p
t, 0) = ! p (x
p
t, 0)
t, 0)
PASI : Tsunami Workshop Jan 2-14 2013
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Riemann problem for systems
Assume a constant coefficient system :
with piecewise constant initial data :
q(x, 0) =
q
qr
x<0
x>0
which can be decomposed as :
q` =
3
X
p=1
p p
!` r
qr =
3
X
p p
!r r
p=1
PASI : Tsunami Workshop Jan 2-14 2013
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Riemann problem for systems
t>0
Piecewise constant initial data
t=0
x1
q
q(x1 , t) =
qr
x=0
r +
1 1
r +
2 2
m
r
p=1
r
= q`
m
p p
q =
3 3
qr =
p p
r r
p=1
PASI : Tsunami Workshop Jan 2-14 2013
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Riemann problem for systems
q t + A qx = 0
t>0
t=0
x2
q
q(x2 , t) =
1 1
r r
qr
+
r +
2 2
m
r
p=1
r
m
p p
q =
3 3
qr =
p p
r r
p=1
PASI : Tsunami Workshop Jan 2-14 2013
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Riemann problem for systems
q t + A qx = 0
t>0
t=0
x3
q
q(x3 , t) =
1 1
r r
+
2 2
r r
m
r
p=1
+
3 3
r
m
p p
q =
qr
qr =
p p
r r
p=1
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Riemann problem for systems
q t + A qx = 0
t>0
t=0
q
x4
qr
q(x4 , t) =
1 1
r r
+
2 2
r r
m
r
p=1
= qr
m
p p
q =
+
3 3
r r
qr =
p p
r r
p=1
PASI : Tsunami Workshop Jan 2-14 2013
Thursday, January 10, 13
Riemann problem for systems
q t + A qx = 0
t>0
t=0
q
qr
q(x1 , t) =
1 1
r +
2 2
r +
3 3
q(x2 , t) =
1 1
r r
+
2 2
r +
3 3
q(x3 , t) =
1 1
r r
+
2 2
r r
+
3 3
q(x4 , t) =
1 1
r r
+
2 2
r r
+
3 3
r r
r
r
r
PASI : Tsunami Workshop Jan 2-14 2013
Thursday, January 10, 13
Riemann problem for systems
q t + A qx = 0
t>0
t=0
q (x, t)
p
q
qr
}
???
PASI : Tsunami Workshop Jan 2-14 2013
Thursday, January 10, 13
Riemann problem for systems
q t + A qx = 0
}
q (x, t)
p
q(x1 , t) =
1 1
q(x2 , t) =
1 1
r r
q(x2 , t)
r +
2 2
+
2 2
q(x1 , t) = (
r +
3 3
r +
3 3
1
r
r
r
1
)r
1
PASI : Tsunami Workshop Jan 2-14 2013
Thursday, January 10, 13
Riemann problem for systems
q t + A qx = 0
}
q (x, t)
p
q(x2 , t) =
1 1
r r
+
2 2
q(x3 , t) =
1 1
r r
+
2 2
r r
q(x3 , t)
q(x2 , t) = (
r +
3 3
+
3 3
2
r
r
r
2
)r
2
PASI : Tsunami Workshop Jan 2-14 2013
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Riemann problem for systems
q t + A qx = 0
q (x, t)
p
q(x3 , t) =
1 1
r r
q(x4 , t) =
1 1
r r
q(x4 , t)
+
2 2
r r
+
3 3
+
2 2
r r
+
3 3
r r
q(x3 , t) = (
3
r
r
3
)r
}
3
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Riemann problem for systems
q t + A qx = 0
q p (x, t)
3
⇥ ) rp
{
q(x1 , t) =
p
{
q(x4 , t)
(⇥rp
p=1
3
⇥
R
= qr
p p
r
p=1
q
PASI : Tsunami Workshop Jan 2-14 2013
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Riemann problem for systems
Solving the Riemann problem for linear problem
q t + A qx = 0
(1) Compute eigenvalues and eigenvectors of matrix A
(2) Compute characteristic variables by solving
R
= qr
q
(3) Use eigenvalues or “speeds” to determine piecewise
constant solution
q(x, t) = q⇧ +
p p
r
p:
p <x/t
= qr
p p
r
p:
p >x/t
PASI : Tsunami Workshop Jan 2-14 2013
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Numerical solution
qt + f (q)x = 0,
Qn+1
i
=
Qi
f (q) = Aq
t
n
Fi+1/2
x
Qni
Fin 1/2
⇥
2 2
wave ↵ r travels at
speed 2 > 0
q⇤
1
Qi
xi
1/2
Decompose jump in Q at the interface into waves :
q ⇤ = Qi
Fi
1/2
1 1
3 3
2 2
+
↵
r
=
Q
↵
r
↵
r
1
i
Z t+ t
1
⇡
f (q(xi 1/2 , t))dt = Aq ⇤
t t
PASI : Tsunami Workshop Jan 2-14 2013
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Example : Linearized shallow water
qt + A qx = 0,
A=
✓
U
g
◆
H
U
,
q=
✓
h
u
◆
Characteristic information :
Eigenvalues :
Eigenvectors :
1
=U
1
r =
✓
p
2
gH,
p
gH
g
◆
,
=U+
2
r =
p
gH
✓ p
gH
g
◆
PASI : Tsunami Workshop Jan 2-14 2013
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Example : Linearized shallow water
Characteristic variables :
1
Define :
✓
R = qr
= qr
↵1
↵2
◆
q`
1
=
2gH
✓
✓
◆
8
h`
>
>
q` =
>
>
u`
>
>
<
1 1
q
+
↵
r
`
q(x, t) =
>
✓
◆
>
>
>
hr
>
>
: qr =
u
`
!
p
gH
p
gH
2
q
=
=
H
H
hr
ur
◆✓
1
2
h`
u`
◆
p
gH
x/t < U
p
p
U
gH < x/t < U + gH
x/t > U +
p
gH
PASI : Tsunami Workshop Jan 2-14 2013
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Linear shallow water wave equations
Initial height and velocity
W1
W2 !
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Extending to nonlinear systems
q
⇤
t
q`
qr
q
⇤
1 1
q` = ↵ r
A(q
⇤
A(q
⇤
q` ) =
1
A(q ⇤
q` ) =
1
f (q) = Aq
!
1
1
1 )
(↵ r
(q ⇤
q` ) = ↵ Ar
⇤
f (q )
Rankine-Hugoniot
condition for the constant
coefficent linear system
q` )
`
f (q ) =
1
(q
⇤
q` )
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Rankine-Hugoniot conditions
For a 2x2 linear system, we have
A(q ⇤
A(qr
q` ) =
⇤
q )=
1
2
(q ⇤
q` )
(qr
⇤
q )
For f (q) = Aq , we can write this as :
The left and right states q` and qr as “connected” by an
⇤
q
intermediate state .
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Constant coefficient Riemann problem
q
q`
t
?
⇤
qr
⇤
We could have asked “Find an intermediate state q such
that
”
For f (q) = Aq , this leads to the eigenvalue problem that
we solved.
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Extending to nonlinear systems
t
q`
qr
Reminder : Solutions to the constant coefficient linear
system travel along characteristic curves
:
must be an eigenvalue of
, i.e.
Solution remains constant along straight lines
PASI : Tsunami Workshop Jan 2-14 2013
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Nonlinear shallow water wave equations
qt + f (q)x = 0
where
q=
✓
h
hu
◆
,
f (q) =
✓
hu
1
2
2
hu + 2 gh
◆
for smooth solutions, this can also be written as
where
0
f (q) =
✓
is the flux Jacobian matrix.
0
2
u + gh
1
2u
◆
Show!
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What changes in the nonlinear case?
qt + f (q)x = 0
We can still ask “Are there characteristic curves on which
the solution remains constant?
For smooth solutions, we have
where
is the flux Jacobian matrix.
Characteristics are governed by eigenvalues of the flux Jacobian
PASI : Tsunami Workshop Jan 2-14 2013
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Shallow water wave equations
q=
✓
◆
✓
hu
,
f (q) =
1
2
hu + 2 gh2
✓
◆
0
1
0
f (q) =
u2 + gh 2u
h
hu
◆
Eigenvalues and eigenvectors of the flux Jacobian
:
p
p
1
2
=u
gh,
= u + gh
✓
◆
✓
◆
1p
1p
1
2
r =
,
r =
u
gh
u + gh
Eigenvalues and eigenvectors depend on q!
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What can happen?
?
q`
t
qr
1-characteristics
?
q`
qr
2-characteristics
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Riemann solution for the SWE
Consider the case where
:
Assume that left and right
states are constant in this
infinitesimal box
Using the conservation law, we can write
Show!
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Riemann problem for SWE
This leads to
This is the required jump condition across shocks.
generally we can write this condition as
More
Rankine Hugoniot
jump condition
where
is the shock speed.
q`
qr
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Rarefaction waves
Consider the case where
Let
for
:
be the slope of the characteristic. We need to find
. Recall that
. Then
slope =
The denominator is
never 0!
q`
qr
Solve resulting ODE to get
in the centered rarefaction.
PASI : Tsunami Workshop Jan 2-14 2013
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Centered rarefaction
Solve the system of two ODEs (for SWE) :
subject to
Use Riemann invariants to solve for unknown constants.
q`
qr
PASI : Tsunami Workshop Jan 2-14 2013
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Solving the Riemann problem
q
t
q`
?
⇤
qr
Find a state such that
is connected to by a physically
correct 1-shock wave or 1-rarefaction, and
is connected to
by a physically correct 2-shock or 2-rarefaction.
The need to find a state that simultaneously satisfies both
conditions above means we have to solve something...
PASI : Tsunami Workshop Jan 2-14 2013
Thursday, January 10, 13
Solving the Riemann problem
ck
ho
1
2-s
n
o
i
t
c
a
-raref
Formulas for these
curves are relatively
simple.
sh
1oc
k
2-rarefaction
Curves represent states that can be connected to or by
a shock or a rarefaction. Use a nonlinear root-finder to find
the middle state .
•
Determine the structure of the rarefaction (if there is one).
PASI : Tsunami Workshop Jan 2-14 2013
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Riemann solution
1-rarefaction
t
q`
q
⇤
2-shock
qr
The structure of the Riemann solution depends on the initial conditions.
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Next :
Can we avoid the nonlinear solve?
How does this extend to the two dimensional shallow
water equations?
How does GeoClaw use Riemann solvers?
For details, see Finite Volume Methods for Hyperbolic Problems,
R. J. LeVeque (Cambridge University Press, 2002).
PASI : Tsunami Workshop Jan 2-14 2013
Thursday, January 10, 13
Lab sessions :
•
Experiment with linear SWE using ClawPack.
•
Experiment with nonlinear SWE using ClawPack.
Goals :
•
•
•
•
Learn about how Riemann solvers are used in Clawpack and
GeoClaw
Include bathymetry to see the effects of well-balancing.
Learn about various plotting parameters in Clawpack
Leave Chile with a simple 1d solver for the linear and
nonlinear shallow water wave equations.
See PASI website on Piazza for website describing project.
PASI : Tsunami Workshop Jan 2-14 2013
Thursday, January 10, 13