Tsunami-Lab - TUMDepartment of Informatics V Alex Breuer Organization • Lab course • Teams: 3 students each • 4 assignments + project phase • Presentation of your work during

Department of Informatics V

Alex Breuer

Session 1: SWEs, f-Waves, SE-ToolsM. Bader, A. Breuer

Tsunami-Lab

Department of Informatics V

Alex Breuer

Organization• Lab course

• Teams: 3 students each

• 4 assignments + project phase

• Presentation of your work during big meetings (no exams)

• Tuesdays, 11 AM - 1 PM

• Grading

• Assignment: 10 points each, project phase: 20 points

• Presentations during big meetings

2

1. 1D-SWEs, f-Waves

2. Bathymetry, Dimensional-Splitting

3. netCDF, Tsunamis

4. MAC-Cluster, OpenMP

5. Project

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Schedule of Big Meetings

• Attendance is obligatory

• Each group has to prepare a presentation

3

Date Schedule10/14/14 Kickoff10/28/14 Presentation 111/11/14 Presentation 212/02/14 Presentation 312/16/14 Presentation 401/13/15 Report: Project phase01/20/15 Presentation Project

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SWEs in Differential Form

4

Prof. Dr. M. BaderDipl.-Math. A. BreuerS. Rettenberger, M. Sc.

Bachelor-Praktikum:Tsunami-Simulation April 2, 2013

In this assignment we are going to implement and test the most basic functionality of ourlab-course: The f-wave solver for the one-dimensional shallow water equations. The shallowwater equations are a system of nonlinear hyperbolic conservations laws with an optionalsource term:

h

hu

�

t

+

hu

hu2 + 1

2

gh2

�

x

= S(x, t). (1)

The quantities q = [h,hu]T are defined by h(x, t), the space-time dependent height of thewater column and hu(x, t), the space-time dependent momentum in spatial x-direction (uis the particle velocity). g the gravity constant (usually g := 9.81m/s2) and f := [hu,hu2 +1

2

gh2]T the flux function. As source term S(x, t) we will consider the e↵ect of space-dependentbathymetry only S(x) = [0,-ghB

x

]T , embedding of additional forces, such as friction or thecoriolis e↵ect is possible. Figure TODO illustrates the involved quantities.

To verify, that the most basic functionality of our program works as expected a proper(unit-) testing is required. We will do testing by a selection of standardized tests, for whicha solution is available.

Remark As units we use meters (m) and seconds (s) for all computations.

Literature

We discuss the basic ideas of numerics, software and strategies in our meetings, neverthelessmany important details can’t be covered in such a short time. We recommend a basic set ofliterature in each assignment as hint for your personal studies. In terms of this assignmentwe recommend the following list of books, papers and guides:

• Finite volume methods for hyperbolic problems, R. J. LeVeque, 2002

• Riemann solvers and numerical methods for fluid dynamics, E. F. Toro, 2009

• A wave propagation method for conservation laws and balance laws with spatially vary-

ing flux functions, D. S. Bale, 2003

• Thinking in C++: http://mindview.net/Books/TICPP/ThinkingInCPP2e.html

• git Documentation: http://git-scm.com/documentation

• Doxygen Manual : http://www.stack.nl/~dimitri/doxygen/manual

• CxxTest User Guide: http://cxxtest.com/guide.html

• SCons user Guide: http://www.scons.org/doc/production/HTML/scons-user.html

• Paraview Documentation: http://www.itk.org/Wiki/ParaView/Users_Guide/Table_Of_Contents

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SWEs as Finite Volumes

5

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Exact All Shock Solution

6

2.4 lineare riemann-probleme 15

Abbildung 3: Anfangswerte des skalaren Riemann-Problems. Zum Zeit-punkt t = 0 sind die beiden konstanten Zustände q

l

undq

r

durch eine Unstetigkeit im Punkt x = 0 getrennt.

den Überlegungen in Kapitel 2.2 wissen wir bereits, dass sichjeder der Anfangswerte x

0

entlang einer Charakteristik X

0

(t) =

x

0

+ ut bewegt. Da die Charakteristiken im Fall der konstantenTransportgleichung parallele Geraden waren, folgt direkt, dass Die Charakteristik

der Unstetigkeit legtdie Lösung desskalarenRiemann-Problemsfest.

sich die Unstetigkeit entlang der Geraden X

dis

(t) = ut fortsetzt,wir erhalten als Lösung des skalaren Riemann-Problems

q(x, t) = q(x - ut) =

8<

:q

l

x - ut < 0

q

r

x - ut > 0.(2.24)

Graphisch nimmt q folglich in allen Punkten, die in Abbildung4 links bzw. rechts von X

dis

(t) liegen, den Wert q

l

bzw. q

r

an.Als Verallgemeinerung untersuchen wir im Folgenden noch das

Abbildung 4: Lösung des Riemann-Problems für die Transportglei-chung in der x-t-Ebene. Die Charakteristik der Unste-tigkeit X

dis

(t) bestimmt die Lösung.

262 13 Nonlinear Systems of Conservation Laws

13.4 A Two-Shock Riemann Solution

The shallow water equations are genuinely nonlinear, and so the Riemann problem alwaysconsists of two waves, each of which is a shock or rarefaction. In Example 13.4 the solutionconsists of one of each. The following example shows that other combinations are possible.

Example 13.5. Consider the Riemann data

h(x, 0) ≡ h0, u(x, 0) =!

ul if x < 0,

−ul if x > 0.(13.13)

If ul > 0, then this corresponds to two streams of water slamming into each other, withthe resulting solution shown in Figure 13.7 for the case h0 = 1 and ul = 1. The solutionis symmetric in x with h(−x, t) = h(x, t) and u(−x, t) = −u(x, t) at all times. A shockwave moves in each direction, bringing the fluid to rest, since the middle state must haveum = 0 by symmetry. The solution to this problem is computed in Section 13.7.1.

The characteristic structure of this solution is shown in Figure 13.8. Note again that 1-characteristics impinge on the 1-shock while crossing the 2-shock, whereas 2-characteristicsimpinge on the 2-shock.

Note that if we look at only half of the domain, say x < 0, then we obtain the solutionto the problem of shallow water flowing into a wall located at x = 0 with velocity ul . Ashock wave moves out from the wall, behind which the fluid is at rest. This is now exactlyanalogous to traffic approaching a red light, as shown in Figure 11.2.

−1 −0.5 0 0.5 1

1

1.1

1.2

1.3

1.4

1.5

1.6Height

−1 −0.5 0 0.5 1

−0.4

−0.2

0

0.2

0.4

0.6Velocity

−1 −0.5 0 0.5 1

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

Vertically integrated pressure

−1 −0.5 0 0.5 10

0.2

0.4

0.6

0.8

1Particle paths in x–t plane

Fig. 13.7. Structure of the similarity solution of the two-shock Riemann problem for the shallowwater equations with ul = − ur . The depth h, velocity u, and vertically integrated pressure aredisplayed as functions of x/t . The structure in the x–t plane is also shown with particle pathsindicated for a set of particles with the spacing between particles inversely proportional to the depth.[claw/book/chap13/twoshock]

Riemann problem Shock-Shock Riemann solution

Source: R. LeVeque, Finite Volume Methods for Hyperbolic Problems, 2002

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Exact Solution with Rarefactions

7

2.4 lineare riemann-probleme 15

Abbildung 3: Anfangswerte des skalaren Riemann-Problems. Zum Zeit-punkt t = 0 sind die beiden konstanten Zustände q

l

undq

r

durch eine Unstetigkeit im Punkt x = 0 getrennt.

den Überlegungen in Kapitel 2.2 wissen wir bereits, dass sichjeder der Anfangswerte x

0

entlang einer Charakteristik X

0

(t) =

x

0

+ ut bewegt. Da die Charakteristiken im Fall der konstantenTransportgleichung parallele Geraden waren, folgt direkt, dass Die Charakteristik

der Unstetigkeit legtdie Lösung desskalarenRiemann-Problemsfest.

sich die Unstetigkeit entlang der Geraden X

dis

(t) = ut fortsetzt,wir erhalten als Lösung des skalaren Riemann-Problems

q(x, t) = q(x - ut) =

8<

:q

l

x - ut < 0

q

r

x - ut > 0.(2.24)

Graphisch nimmt q folglich in allen Punkten, die in Abbildung4 links bzw. rechts von X

dis

(t) liegen, den Wert q

l

bzw. q

r

an.Als Verallgemeinerung untersuchen wir im Folgenden noch das

Abbildung 4: Lösung des Riemann-Problems für die Transportglei-chung in der x-t-Ebene. Die Charakteristik der Unste-tigkeit X

dis

(t) bestimmt die Lösung.

13.2 Dam-Break and Riemann Problems 259

region is bounded by a hydraulic jump, where the depth of the water suddenly increasesand its speed abruptly decreases. By adjusting the flow rate or angle of the plate you shouldbe able to make the location of this shock wave move around. When the conditions arefixed, the shock is stationary and has zero propagation speed. This can be modeled by thetwo-dimensional shallow water equations, or in the radially symmetric case by the one-dimensional equations with additional source terms incorporated to model the geometriceffects as described in Section 18.9.

13.2 Dam-Break and Riemann Problems

Consider the shallow water equations (13.5) with the piecewise-constant initial data

h(x, 0) =!

hl if x < 0,

hr if x > 0,u(x, 0) = 0, (13.12)

where hl > hr ≥ 0. This is a special case of the Riemann problem in which ul = ur = 0,and is called the dam-break problem because it models what happens if a dam separatingtwo levels of water bursts at time t = 0. This is the shallow water equivalent of the shock-tube problem of gas dynamics (Section 14.13). We assume hr > 0.

Example 13.4. Figure 13.4 shows the evolution of the depth and fluid velocity for the dam-break problem with data hl = 3 and hr = 1. Figure 13.5 shows the structure of this solutionin the x–t plane. Water flows from left to right in a wedge that expands from the dam locationx = 0. At the right edge of this wedge, moving water with some intermediate depth hm andvelocity um > 0 slams into the stationary water with h = hr , accelerating it instantaneously

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5h at t = 0

−5 −4 −3 −2 −1 0 1 2 3 4 5−0.5

0

0.5

1

1.5

2hu at t = 0

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5h at t = 0.5

−5 −4 −3 −2 −1 0 1 2 3 4 5−0.5

0

0.5

1

1.5

2hu at t = 0.5

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5h at t = 2

−5 −4 −3 −2 −1 0 1 2 3 4 5−0.5

0

0.5

1

1.5

2hu at t = 2

Fig. 13.4. Solution of the dam-break Riemann problem for the shallow water equations withul = ur = 0. On the left is the depth h and on the right is the momentum hu. [claw/book/chap13/dambreak]

278 13 Nonlinear Systems of Conservation Laws

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.5

1

1.5

2h at t = 0

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

−0.5

0

0.5

1hu at t = 0

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.5

1

1.5

2h at t = 0.75

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

−0.5

0

0.5

1hu at t = 0.75

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.5

1

1.5

2h at t = 3

−5 −4 −3 −2 −1 0 1 2 3 4 5−1

−0.5

0

0.5

1hu at t = 3

Fig. 13.14. Solution of the Riemann problem for the shallow water equations with ul = −ur < 0.[claw/book/chap13/tworaref]

two rarefaction waves as shown in Figure 13.14. (Looking at only half the domain gives thesolution to the boundary-value problem with water flowing away from a wall.)

To solve this Riemann problem, we can proceed in a manner similar to what we didin Section 13.7.1 for the all-shock solution. There is an integral curve of r1 through ql

consisting of all states that can be connected to ql by a 1-rarefaction, and an integral curveof r2 through qr consisting of all states that can be connected to qr by a 2-rarefaction. Theseare illustrated in Figure 13.15(a) for the Riemann data

ul = −0.5, ur = 0.5, and hl = hr = 1. (13.54)

(a)

−0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

h = depth

hu =

mom

entu

m

ql

qrqm

(b)

−0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

h = depth

hu =

mom

entu

m

qlqr

qm

Fig. 13.15. (a) Construction of the all-rarefaction Riemann solution for the problem of Example 13.5.(b) The physically incorrect all-rarefaction Riemann solution for the dam-break problem of Exam-ple 13.4.

Rare-Rare Riemann solution Rare-Shock Riemann Solution

Source: R. LeVeque, Finite Volume Methods for Hyperbolic Problems, 2002

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f-waves: Splitting the flux jump

8

2.5 nichtlineare riemann-probleme 17

Abbildung 5: Lösung des Riemann-Problems für lineare Systeme inder x-t-Ebene (�1

< . . . < �

j

< 0 < . . . < �

n). An einemPunkt P(x, t) ergibt sich die Lösung durch seine Lage zuden Charakteristiken der Unstetigkeiten.

wieder aufgreifen werden. Dabei wird der Sprung q

r

- q

l

inEigenvektoren der Matrix A dargestellt. Durch Einsetzen derSummen aus den Gleichungen in (2.27) erhalten wir

q

r

- q

l

=nX

p=1

(wp

r

- w

p

l

)rp ⌘nX

p=1

↵

p

r

p =nX

p=1

r

p

↵

p. (2.31)

Der Sprung q

r

- q

l

wird also in Sprünge über die Unstetig-keiten zerlegt. Mit der Lösung ↵ = (↵1, . . . , ↵n)T des linearenGleichungssystems

R↵ = q

r

- q

l

(2.32)

und W

p ⌘ ↵

p

r

p hat die Zerlegung (2.30) in dieser Darstellungdie Form

q(x, t) = q

l

+X

p:�p

<x/t

W

p

= q

r

-X

p:�p

>x/t

W

p.(2.33)

2.5 nichtlineare riemann-probleme

Wir werden uns in diesem Kapitel mit den grundsätzlichen Ideenfür die exakte Lösung nichtlinearer Riemann-Probleme beschäfti-gen. Die hier präsentierte Theorie hat eher einen motivierendenCharakter (vgl. [37]), für eine ausführlichere Darstellung sei dahernoch einmal auf [30] verwiesen.

Aus Kapitel 2.2 wissen wir bereits, dass q in der sklaren Er-haltungsgleichung (2.7) eindeutig durch die Charakteristikengegeben ist, zumindest solange die Lösung glatt bleibt. Da die

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FV: Edge-local f-Waves1. Finite volume discretization at initial

time

2. F-Wave solver: Edge-local Riemann solutions

3. Cells update to the next time step

9

Q

n+1i = Q

ni -

�t

�x

⇣A

+�Q

ni-1/2 +A

-�Q

ni+1/2

⌘

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Choose Your Weapons• Code and libraries only tested with

Linux

• We won’t / can’t help you with other OS

• MAC-Cluster: Starting with assignment 4

• Keep limited toolchains of supercomputers in mind

10

Source: Top500.org, http://www.top500.org/statistics/list/

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SWE1D

• Git-repositoryhttps://github.com/TUM-I5/SWE1D

• 1D equivalent to SWE (2D code)https://github.com/TUM-I5/SWE

11

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Basics in SE• Git: http://git-scm.com

• Doxygen: http://www.stack.nl/~dimitri/doxygen/

• SCons: http://www.scons.org

• CxxTest: http://cxxtest.com

• ParaView: http://paraview.org, http://www.paraview.org/Wiki/ParaView/Git

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